https://wiki.ubc.ca/api.php?action=feedcontributions&feedformat=atom&user=GregMayerUBC Wiki - User contributions [en]2026-06-03T00:48:51ZUser contributionsMediaWiki 1.43.8https://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_3/3.2_Taylor_Series/3.2.04_The_Maclaurin_Series_Expansion_for_sin&diff=330672Science:Infinite Series Module/Units/Unit 3/3.2 Taylor Series/3.2.04 The Maclaurin Series Expansion for sin2014-11-02T14:48:46Z<p>GregMayer: /* An Alternate Explanation */</p>
<hr />
<div>==Example==<br />
<br />
Find the Taylor series expansion for sin(''x'') at ''x'' = 0, and determine its radius of convergence.<br />
<br />
==Complete Solution==<br />
<br />
Again, before starting this problem, we note that the Taylor series expansion at ''x'' = 0 is '''equal''' to the Maclaurin series expansion.<br />
<br />
===Step 1: Find Coefficients===<br />
<br />
Let ''f''(''x'') = sin(''x''). To find the Maclaurin series coefficients, we must evaluate<br />
<br />
<math>\Bigg(\frac{d^{k}}{dx^{k}}\sin(x)\Bigg)\Bigg|_{x=0}</math><br />
<br />
for ''k'' = 0, 1, 2, 3, 4, ...<br />
<br />
Calculating the first few coefficients, a pattern emerges:<br />
<br />
<math>\begin{align}<br />
f(0) &= \sin(0) = 0 \\<br />
f'(0) &= \cos(0) = 1 \\<br />
f''(0) &= -\sin(0) = 0 \\<br />
f'''(0) &= -\cos(0) = -1 \\<br />
f^{(4)}(0) &= \sin(0) = 0 \\<br />
f^{(5)}(0) &= \cos(0) = 1 <br />
\end{align}</math><br />
<br />
The coefficients alternate between 0, 1, and -1. You should be able to, for the n<sup>th</sup> derivative, determine whether the n<sup>th</sup> coefficient is 0, 1, or -1.<br />
<br />
===Step 2: Substitute Coefficients into Expansion===<br />
<br />
Thus, the Maclaurin series for sin(''x'') is <br />
<br />
<math>\begin{align}<br />
\sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!} (x-0)^k <br />
&= f(0) + \frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots \\<br />
&= 0+ \Big( \frac{1}{1!}x\Big) + 0 + \Big( \frac{-1}{3!}x^3\Big)+0+\Big( \frac{1}{5!}x^5 \Big)+ \ldots \\<br />
&= x -\frac{x^3}{3!}+\frac{x^5}{5!} - \ldots<br />
\end{align}</math><br />
<br />
===Step 3: Write the Expansion in Sigma Notation===<br />
<br />
From the first few terms that we have calculated, we can see a pattern that allows us to derive an expansion for the ''n''<sup>th</sup> term in the series, which is<br />
<br />
<math>\frac{(-1)^n}{(2n+1)!}x^{2n+1},\quad n=0,1,2,3,\ldots</math><br />
<br />
Substituting this into the formula for the Taylor series expansion, we obtain<br />
<br />
<math>\sin(x) =\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!}x^{2k+1} = x -\frac{x^3}{3!}+\frac{x^5}{5!} - \ldots</math><br />
<br />
===Radius of Convergence===<br />
<br />
The ratio test gives us:<br />
<br />
<math>\begin{align}<br />
\lim_{k \rightarrow \infty} \Big| \frac{(-1)^{k+1}}{(2(k+1)+1)!}x^{2(k+1)+1} \Bigg/\frac{(-1)^k}{(2k+1)!}x^{2k+1} \Big | <br />
&= \lim_{k \rightarrow \infty} \frac{(2k+1)!}{(2k+3)!} |x|^2 \\<br />
&= \lim_{k \rightarrow \infty} \frac{1}{(2k+3)(2k+2)} |x|^2 \\<br />
&= 0<br />
\end{align} </math><br />
<br />
Because this limit is zero for all real values of ''x'', the radius of convergence of the expansion is the set of all real numbers. <br />
<br />
==Explanation of Each Step==<br />
<br />
===Step 1===<br />
<br />
Maclaurin series coefficients, ''a<sub>k</sub>'' can be calculated using the formula (that comes from the definition of a Taylor series)<br />
<br />
<math>a_k = \frac{f^{(k)}(0)}{k!}</math><br />
<br />
where ''f'' is the given function, and in this case is sin(''x''). In step 1, we are only using this formula to calculate the first few coefficients. We can calculate as many as we need, and in this case were able to stop calculating coefficients when we found a pattern to write a general formula for the expansion. <br />
<br />
===Step 2===<br />
<br />
Step 2 was a simple substitution of our coefficients into the expression of the Taylor series. <br />
<br />
===Step 3===<br />
<br />
A helpful step to find a compact expression for the ''n''<sup>th</sup> term in the series, is to write out more explicitly the terms in the series that we have found:<br />
<br />
<math><br />
\sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!} (x-0)^k = (+1)\cdot \frac{x^1}{1!} + (-1)\cdot \frac{x^3}{3!}+(+1)\cdot \frac{x^5}{5!} + \ldots <br />
</math><br />
<br />
We have discovered the sequence 1, 3, 5, ... in the exponents and in the denominator of each term. We may then find a way to convert this sequence that we have discovered, into the sequence ''k''=0, 1, 2, ... that appears in the final summation. The simple transform 2''k''+1 performs this transformation for us. <br />
<br />
===Step 4===<br />
<br />
This step was nothing more than substitution of our formula into the formula for the ratio test. Because we found that the series converges for all ''x'', we did not need to test the endpoints of our interval. If however we did find that the series only converged on an interval with a finite width, then we may need to take extra steps to determine the convergence at the boundary points of the interval. <br />
<br />
===An Alternate Explanation===<br />
<br />
The following Khan Acadmey video provides a similar derivation of the Maclaurin expansion for sin(x) that you may find helpful. <br />
<br />
{| border="1" cellspacing="0" cellpadding="4" align="center" width="570px"<br />
|- style="background-color:#f0f0f0;"<br />
! Sine Taylor Series at 0<br />
|-<br />
| {{#widget:YouTube|id=LlKvubIqHc8|height=315|width=560}}<br />
|-<br />
! Derivation of the Maclaurin series expansion for sin(''x'').<br />
|-<br />
! This video can be found on the [http://www.khanacademy.org/ Kahn Academy website], and carries a Creative Commons copyright (CC BY-NC-SA 3.0).<br />
|}<br />
<br />
==Possible Challenges==<br />
<br />
===What if we Need the Taylor Series of sin(x) at Some Other Point?===<br />
<br />
The Maclaurin series of sin(''x'') is '''only''' the Taylor series of sin(''x'') at ''x'' = 0. If we wish to calculate the Taylor series at any '''other''' value of ''x'', we can consider a variety of approaches. <br />
<br />
Suppose we wish to find the Taylor series of sin(''x'') at ''x'' = ''c'', where ''c'' is any real number that is not zero. We could find the associated Taylor series by applying the same steps we took here to find the Macluarin series. That is, calculate the series coefficients, substitute the coefficients into the formula for a Taylor series, and if needed, derive a general representation for the infinite sum. <br />
<br />
Another approach could be to use a trigonometric identity. Consider this approach<br />
<br />
<math>\begin{align}<br />
\sin(x) &= \sin(x + c - c) \\<br />
&= \sin(x+c)\cos(c) - \cos(x+c)\sin(c) \\<br />
&= \sin(u)\cos(c) - \cos(u)\sin(c), \quad u = x+c<br />
\end{align} </math><br />
<br />
The functions cos(''u'') and sin(''u'') can be expanded in with a Maclaurin series, and cos(''c'') and sin(''c'') are constants. We will see the Maclaurin expansion for cosine on the next page.<br />
<br />
===How Many Terms do I Need to Calculate?===<br />
<br />
It can be difficult to find an expression for the ''n''<sup>th</sup> term in the series that allows us to write out a compact expression for an infinite sum. In our example here, we only calculated three terms. It may be helpful in other problems to write out a few more terms to find a useful pattern. <br />
<br />
==Summary==<br />
<br />
To summarize, we found the Macluarin expansion of the sine function. <br />
<br />
{| border="1" cellspacing="0" cellpadding="4" align="center"<br />
|- style="background-color:#f0f0f0;"<br />
! The Maclaurin Expansion of sin(x)<br />
|-<br />
| The Maclaurin series expansion for sin(''x'') is given by<br />
<br />
<math>\sin(x) =\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!}x^{2k+1} = x -\frac{x^3}{3!}+\frac{x^5}{5!} - \ldots</math><br />
<br />
This formula is valid for all real values of ''x''.<br />
|}<br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 3]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330671Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T14:41:24Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
<b>Blue and Green Triangles</b><br />
<br />
Assume</b> that the one blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <br />
<br />
<math>1+3(1/9).</math><br />
<br />
<b>Blue, Green, and Yellow Triangles</b><br />
<br />
Each of the 12 = 3•4 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + (3\cdot4)(1/9^2).</math><br />
<br />
<b>Blue, Green, Yellow, and Red Triangles</b><br />
<br />
Each of the 48 = 3•4•4 red triangles has 1/9 the area of a yellow triangles, or <math>1/(9^3)</math> the area of a blue triangle. The area of the blue, green, yellow, and red triangles is<br />
<br />
<math>1 + 3(1/9) + (3\cdot4)(1/9^2) + (3\cdot4\cdot4)(1/9^3).</math><br />
<br />
<b>Total Area</b><br />
<br />
The total area of the snowflake uses the infinite sequence<br />
<br />
<math>\Big\{3\cdot4^{n-1}\Big(\frac{1}{9}\Big)^{n}\Big\}_{n=1}^{\infty}</math>.<br />
<br />
We will add all the terms of the series together, and add 1, to produce the following sum<br />
<br />
<math>1 + \sum_{n=1}^{\infty}3 \cdot 4^{n-1} \Big(\frac{1}{9}\Big)^{n}.</math><br />
<br />
Seeing that this is a geometric series with a = 1/3 and r = 4/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>1 + \frac{a}{1-r} = 1 + \frac{3}{5} = 8/5.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330670Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T14:40:00Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
<b>Blue and Green Triangles</b><br />
<br />
Assume</b> that the one blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <br />
<br />
<math>1+3(1/9).</math><br />
<br />
<b>Blue, Green, and Yellow Triangles</b><br />
<br />
Each of the 12 = 3•4 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + (3\cdot4)(1/9^2).</math><br />
<br />
<b>Blue, Green, Yellow, and Red Triangles</b><br />
<br />
Each of the 48 = 3•4•4 red triangles has 1/9 the area of a yellow triangles, or <math>1/(9^3)</math> the area of a blue triangle. The area of the blue, green, yellow, and red triangles is<br />
<br />
<math>1 + 3(1/9) + (3\cdot4)(1/9^2) + (3\cdot4\cdot4)(1/9^3).</math><br />
<br />
<b>Total Area</b><br />
<br />
The total area of the snowflake uses the infinite sequence<br />
<br />
<math>\Big\{3\cdot4^{n-1}\Big(\frac{1}{9}\Big)^{n}\Big\}_{n=1}^{\infty}</math>.<br />
<br />
We will add all the terms of the series together, and add 1, to produce the following sum<br />
<br />
<math>1 + \sum_{n=1}^{\infty}3 \cdot 4^{n-1} \Big(\frac{1}{9}\Big)^{n}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 4/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{9}{5}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330669Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T14:31:59Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
<b>Blue and Green Triangles</b><br />
<br />
Assume</b> that the one blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <br />
<br />
<math>1+3(1/9).</math><br />
<br />
<b>Blue, Green, and Yellow Triangles</b><br />
<br />
Each of the 12 = 3•4 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + (3\cdot4)(1/9^2).</math><br />
<br />
<b>Blue, Green, Yellow, and Red Triangles</b><br />
<br />
Each of the 48 = 3•4•4 red triangles has 1/9 the area of a yellow triangles, or <math>1/(9^3)</math> the area of a blue triangle. The area of the blue, green, yellow, and red triangles is<br />
<br />
<math>1 + 3(1/9) + (3\cdot4)(1/9^2) + (3\cdot4\cdot4)(1/9^3).</math><br />
<br />
<b>Total Area</b><br />
<br />
The total area of the snowflake uses the infinite sequence<br />
<br />
<math>\Big\{3\cdot4^{n-1}\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math>.<br />
<br />
We will add all the terms of the series together, and add 1, to produce the following sum<br />
<br />
<math>1 + \sum_{n=1}^{\infty}3 \cdot 4^{n-1} \Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 4/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{9}{5}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330668Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T14:30:13Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
<b>Blue and Green Triangles</b><br />
<br />
Assume</b> that the one blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <br />
<br />
<math>1+3(1/9).</math><br />
<br />
<b>Blue, Green, and Yellow Triangles</b><br />
<br />
Each of the 12 = 3•4 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + (3\cdot4)(1/(9^2).</math><br />
<br />
<b>Blue, Green, Yellow, and Red Triangles</b><br />
<br />
Each of the 48 = 3•4•4 red triangles has 1/9 the area of a yellow triangles, or <math>1/(9^3)</math> the area of a blue triangle. The area of the blue, green, yellow, and red triangles is<br />
<br />
<math>1 + 3(1/9) + (3\cdot4)(1/(9^2)) + (3\cdot4\cdot4)(1/9^3).</math><br />
<br />
<b>Total Area</b><br />
<br />
The total area of the snowflake uses the infinite sequence<br />
<br />
<math>\Big\{3\cdot4^{n-1}\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math>.<br />
<br />
We will add all the terms of the series together, and add 1, to produce the following sum<br />
<br />
<math>1 + \sum_{n=1}^{\infty}3 \cdot 4^{n-1} \Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 4/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{9}{5}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330667Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T14:29:43Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
<b>Blue and Green Triangles</b><br />
<br />
Assume</b> that the one blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <br />
<br />
<math>1+3(1/9).</math><br />
<br />
<b>Blue, Green, and Yellow Triangles</b><br />
<br />
Each of the 12 = 3•4 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + (3\cdot4)(1/(9^2).</math><br />
<br />
<b>Blue, Green, Yellow, and Red Triangles</b><br />
<br />
Each of the 48 = 3•4•4 red triangles has 1/9 the area of a yellow triangles, or <math>1/(9^3)</math> the area of a blue triangle. The area of the blue, green, yellow, and red triangles is<br />
<br />
<math>1 + 3(1/9) + (3\cdot4)(1/(9^2)) + (3\cdot4\cdot4)(1/9^3).</math><br />
<br />
<b>Total Area</b><br />
<br />
The total area of the snowflake uses the infinite sequence<br />
<br />
<math>\Big\{3\cdot4^{n-1}\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math>.<br />
<br />
We will add all the terms of the series together, and add 1, to produce the following sum<br />
<br />
<math>1 + \sum_{n=1}^{\infty}3 4^{n-1} \Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 4/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{9}{5}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330666Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T14:27:04Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
<b>Blue and Green Triangles</b><br />
<br />
Assume</b> that the one blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <br />
<br />
<math>1+3(1/9).</math><br />
<br />
<b>Blue, Green, and Yellow Triangles</b><br />
<br />
Each of the 12 = 3•4 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + (3\cdot4)(1/(9^2).</math><br />
<br />
<b>Blue, Green, Yellow, and Red Triangles</b><br />
<br />
Each of the 48 = 3•4•4 red triangles has 1/9 the area of a yellow triangles, or <math>1/(9^3)</math> the area of a blue triangle. The area of the blue, green, yellow, and red triangles is<br />
<br />
<math>1 + 3(1/9) + (3\cdot4)(1/(9^2)) + (3\cdot4\cdot4)(1/9^3).</math><br />
<br />
<b>Total Area</b><br />
<br />
The total area of the snowflake uses the infinite sequence<br />
<br />
<math>\Big\{3\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math>.<br />
<br />
We will add all the terms of the series together, and add 1, to produce the following sum<br />
<br />
<math>1 + 3(1/9) + 9(1/27) + \ldots = \sum_{n=1}^{\infty}3\Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 1/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{27}{8}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330665Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T14:18:43Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
<b>Blue and Green Triangles</b><br />
<br />
Assume</b> that the one blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <br />
<br />
<math>1+3(1/9).</math><br />
<br />
<b>Blue, Green, and Yellow Triangles</b><br />
<br />
Each of the 12 = 3•4 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + (3\cdot4)(1/(9^2)).</math><br />
<br />
<b>Blue, Green, Yellow, and Red Triangles</b><br />
<br />
Each of the 48 = 3•4•4 red triangles has 1/9 the area of a yellow triangles, or <math>1/(9^3)</math> the area of a blue triangle. The area of the blue, green, yellow, and red triangles is<br />
<br />
<math>1 + 3(1/9) + (3\cdot4)(1/(9^2)) + (3\cdot4\cdot4)(1/9^3)).</math><br />
<br />
<b>Total Area</b><br />
<br />
The total area of the snowflake uses the infinite sequence<br />
<br />
<math>\Big\{3\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math>.<br />
<br />
We will add all the terms of the series together, and add 1, to produce the following sum<br />
<br />
<math>1 + 3(1/9) + 9(1/27) + \ldots = \sum_{n=1}^{\infty}3\Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 1/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{27}{8}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330664Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T14:15:49Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
<b>Blue and Green Triangles</b><br />
<br />
Assume</b> that the one blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <br />
<br />
<math>1+3(1/9).</math><br />
<br />
<b>Blue, Green, and Yellow Triangles</b><br />
<br />
Each of the 12 = 3•4 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + (3\cdot4)(1/(9^2)).</math><br />
<br />
<b>Blue, Green, Yellow, and Red Triangles</b><br />
<br />
Each of the 48 = 3•4•4 red triangles has 1/9 the area of a yellow triangles, or <math>1/(9^3)</math> the area of a blue triangle. The area of the blue, green, yellow, and red triangles is<br />
<br />
<math>1 + 3(1/9) + (3\cdot4)(1/(9^2)) + (3\cdot4\cdot4)(1/9^3)).</math><br />
<br />
<b>Total Area</b><br />
<br />
The total area of the snowflake is given by taking the infinite sequence<br />
<br />
<math>\Big\{3\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math><br />
<br />
and adding its terms together to produce the following sum<br />
<br />
<math>1 + 3(1/9) + 9(1/27) + \ldots = \sum_{n=1}^{\infty}3\Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 1/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{27}{8}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330663Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T14:14:45Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
<b>Blue and Green Triangles</b><br />
<br />
Assume</b> that the one blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <br />
<br />
<math>1+3(1/9).</math><br />
<br />
<b>Blue, Green, and Yellow Triangles</b><br />
<br />
Each of the 12 = 3•4 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + (3\cdot4)(1/(9^2)).</math><br />
<br />
<b>Blue, Green, Yellow, and Red Triangles</b><br />
<br />
Each of the 48 = 3•4•4 red triangles has 1/9 the area of a yellow triangles, or <math>1/(9^3)</math> the area of a blue triangle. The area of the blue, green, yellow, and red triangles is<br />
<br />
<math>1 + 3(1/9) + (3\cdot4)(1/(9^2)) + (3\cdot4\cdot4)(9^3)).</math><br />
<br />
<b>Total Area</b><br />
<br />
The total area of the snowflake is given by taking the infinite sequence<br />
<br />
<math>\Big\{3\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math><br />
<br />
and adding its terms together to produce the following sum<br />
<br />
<math>1 + 3(1/9) + 9(1/27) + \ldots = \sum_{n=1}^{\infty}3\Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 1/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{27}{8}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330662Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T14:13:07Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
<b>Blue and Green Triangles</b><br />
<br />
Assume</b> that the one blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <br />
<br />
<math>1+3(1/9).</math><br />
<br />
<b>Blue, Green, and Yellow Triangles</b><br />
<br />
Each of the 12 = 3•4 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + 3\cdot4(1/(9^2)).</math><br />
<br />
<b>Blue, Green, Yellow, and Red Triangles</b><br />
<br />
Each of the 48 = 3•4•4 red triangles has 1/9 the area of a yellow triangles, or <math>1/(9^3)</math> the area of a blue triangle. The area of the blue, green, yellow, and red triangles is<br />
<br />
<math>1 + 3(1/9) + (3\cdot4)(1/81) + (3\cdot4\cdot4)(1/81).</math><br />
<br />
<b>Total Area</b><br />
<br />
The total area of the snowflake is given by taking the infinite sequence<br />
<br />
<math>\Big\{3\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math><br />
<br />
and adding its terms together to produce the following sum<br />
<br />
<math>1 + 3(1/9) + 9(1/27) + \ldots = \sum_{n=1}^{\infty}3\Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 1/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{27}{8}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330661Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T14:08:12Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
<b>Blue and Green Triangles</b><br />
<br />
Assume</b> that the one blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <br />
<br />
<math>1+3(1/9).</math><br />
<br />
<b>Blue, Green, and Yellow Triangles</b><br />
<br />
Each of the 12 = 3•4 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + 3\cdot4(1/81).</math><br />
<br />
<b>Blue, Green, Yellow, and Red Triangles</b><br />
<br />
Each of the 48 = 3•4•4 red triangles has 1/9 the area of a yellow triangles, or <math>1/(9^3)</math> the area of a blue triangle. The area of the blue, green, yellow, and red triangles is<br />
<br />
<math>1 + 3(1/9) + 3\cdot4(1/81) + 3\cdot4\cdot4(1/81).</math><br />
<br />
<b>Total Area</b><br />
<br />
The total area of the snowflake is given by taking the infinite sequence<br />
<br />
<math>\Big\{3\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math><br />
<br />
and adding its terms together to produce the following sum<br />
<br />
<math>1 + 3(1/9) + 9(1/27) + \ldots = \sum_{n=1}^{\infty}3\Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 1/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{27}{8}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330660Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T14:07:21Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
<b>Blue and Green Triangles</b><br />
<br />
Let's assume</b> that the blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <br />
<br />
<math>1+3(1/9).</math><br />
<br />
<b>Blue, Green, and Yellow Triangles</b><br />
<br />
Each of the 12 = 3•4 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + 3\cdot4(1/81).</math><br />
<br />
<b>Blue, Green, Yellow, and Red Triangles</b><br />
<br />
Each of the 48 = 3•4•4 red triangles has 1/9 the area of a yellow triangles, or <math>1/(9^3)</math> the area of a blue triangle. The area of the blue, green, yellow, and red triangles is<br />
<br />
<math>1 + 3(1/9) + 3\cdot4(1/81) + 3\cdot4\cdot4(1/81).</math><br />
<br />
<b>Total Area</b><br />
<br />
The total area of the snowflake is given by taking the infinite sequence<br />
<br />
<math>\Big\{3\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math><br />
<br />
and adding its terms together to produce the following sum<br />
<br />
<math>1 + 3(1/9) + 9(1/27) + \ldots = \sum_{n=1}^{\infty}3\Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 1/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{27}{8}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330659Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T14:06:16Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
<b>Blue and Green Triangles</b><br />
<br />
Let's assume</b> that the blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <br />
<br />
<math>1+3(1/9).</math><br />
<br />
<b>Blue, Green, and Yellow Triangles</b><br />
<br />
Each of the 12 = 3•4 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + 3\cdot4(1/81).</math><br />
<br />
<b>Blue, Green, Yellow, and Red Triangles</b><br />
<br />
Each of the 48 = 3•4•4 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, yellow, and red triangles is<br />
<br />
<math>1 + 3(1/9) + 3\cdot4(1/81) + 3\cdot4\cdot4(1/81).</math><br />
<br />
<b>Total Area</b><br />
<br />
The total area of the snowflake is given by taking the infinite sequence<br />
<br />
<math>\Big\{3\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math><br />
<br />
and adding its terms together to produce the following sum<br />
<br />
<math>1 + 3(1/9) + 9(1/27) + \ldots = \sum_{n=1}^{\infty}3\Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 1/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{27}{8}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330658Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T14:04:06Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
<b>Blue and Green Triangles</b><br />
<br />
Let's assume</b> that the blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <br />
<br />
<math>1+3(1/9).</math><br />
<br />
<b>Blue, Green, and Yellow Triangles</b><br />
<br />
Each of the 12 = 3•4 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + 3\cdot4(1/81).</math><br />
<br />
<b>Blue, Green, Yellow, and Red Triangles</b><br />
<br />
Each of the 48 = 3•4•4 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, yellow, and red triangles is<br />
<br />
<math>1 + 3(1/9) + 3\cdot4\cdot4(1/81).</math><br />
<br />
<b>Total Area</b><br />
<br />
The total area of the snowflake is given by taking the infinite sequence<br />
<br />
<math>\Big\{3\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math><br />
<br />
and adding its terms together to produce the following sum<br />
<br />
<math>1 + 3(1/9) + 9(1/27) + \ldots = \sum_{n=1}^{\infty}3\Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 1/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{27}{8}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330657Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T14:02:02Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
<b>Blue and Green Triangles</b><br />
<br />
Let's assume</b> that the blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <math>1+3(1/9)=4/3.</math><br />
<br />
<b>Blue, Green, and Yellow Triangles</b><br />
<br />
Each of the 12 = 3•4 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + 3\cdot4(1/81) = 9/9 + 3/9 + 1/9 = 13/9.</math><br />
<br />
<b>Blue, Green, Yellow, and Red Triangles</b><br />
<br />
Each of the 48 = 3•4•4 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, yellow, and red triangles is<br />
<br />
<math>1 + 3(1/9) + 3\cdot4\cdot4(1/81) = 9/9 + 3/9 + 1/9 = 13/9.</math><br />
<br />
<b>Total Area</b><br />
<br />
The total area of the snowflake is given by taking the infinite sequence<br />
<br />
<math>\Big\{3\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math><br />
<br />
and adding its terms together to produce the following sum<br />
<br />
<math>1 + 3(1/9) + 9(1/27) + \ldots = \sum_{n=1}^{\infty}3\Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 1/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{27}{8}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330656Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T14:00:26Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
<b>Blue and Green Triangles</b><br />
<br />
Let's assume</b> that the blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <math>1+3(1/9)=4/3.</math><br />
<br />
<b>Blue, Green, and Yellow Triangles</b><br />
<br />
Each of the 12 = 3•4 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + 3\cdot4(1/81) = 9/9 + 3/9 + 1/9 = 13/9.</math><br />
<br />
<b>Blue, Green, Yellow, and Red Triangles</b><br />
<br />
Each of the <math>48 = 4\cdot3</math> yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, yellow, and red triangles is<br />
<br />
<b>Total Area</b><br />
<br />
The total area of the snowflake is given by taking the infinite sequence<br />
<br />
<math>\Big\{3\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math><br />
<br />
and adding its terms together to produce the following sum<br />
<br />
<math>1 + 3(1/9) + 9(1/27) + \ldots = \sum_{n=1}^{\infty}3\Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 1/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{27}{8}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330655Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T13:59:02Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
<b>Blue and Green Triangles</b><br />
<br />
Let's assume</b> that the blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <math>1+3(1/9)=4/3.</math><br />
<br />
<b>Blue, Green, and Yellow Triangles</b><br />
<br />
Each of the <math>12 = 4\cdot3</math> yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + 3\cdot4(1/81) = 9/9 + 3/9 + 1/9 = 13/9.</math><br />
<br />
<b>Blue, Green, Yellow, and Red Triangles</b><br />
<br />
Each of the <math>48 = 4\cdot3</math> yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, yellow, and red triangles is<br />
<br />
<b>Total Area</b><br />
<br />
The total area of the snowflake is given by taking the infinite sequence<br />
<br />
<math>\Big\{3\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math><br />
<br />
and adding its terms together to produce the following sum<br />
<br />
<math>1 + 3(1/9) + 9(1/27) + \ldots = \sum_{n=1}^{\infty}3\Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 1/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{27}{8}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330654Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T13:58:38Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
<b>Blue and Green Triangles</b><br />
Let's assume</b> that the blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <math>1+3(1/9)=4/3.</math><br />
<br />
<b>Blue, Green, and Yellow Triangles</b><br />
<br />
Each of the <math>12 = 4\cdot3</math> yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + 3\cdot4(1/81) = 9/9 + 3/9 + 1/9 = 13/9.</math><br />
<br />
<b>Blue, Green, Yellow, and Red Triangles</b><br />
<br />
Each of the <math>48 = 4\cdot3</math> yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, yellow, and red triangles is<br />
<br />
<b>Total Area</b><br />
<br />
The total area of the snowflake is given by taking the infinite sequence<br />
<br />
<math>\Big\{3\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math><br />
<br />
and adding its terms together to produce the following sum<br />
<br />
<math>1 + 3(1/9) + 9(1/27) + \ldots = \sum_{n=1}^{\infty}3\Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 1/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{27}{8}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330653Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T13:57:38Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
===Blue and Green Triangles===<br />
Let's <b>assume</b> that the blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <math>1+3(1/9)=4/3.</math><br />
<br />
===Blue, Green, and Yellow Triangles===<br />
<br />
Each of the <math>12 = 4\cdot3</math> yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + 3\cdot4(1/81) = 9/9 + 3/9 + 1/9 = 13/9.</math><br />
<br />
===Blue, Green, Yellow, and Red Triangles===<br />
<br />
Each of the <math>48 = 4\cdot3</math> yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, yellow, and red triangles is<br />
<br />
===Total Area===<br />
<br />
The total area of the snowflake is given by taking the infinite sequence<br />
<br />
<math>\Big\{3\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math><br />
<br />
and adding its terms together to produce the following sum<br />
<br />
<math>1 + 3(1/9) + 9(1/27) + \ldots = \sum_{n=1}^{\infty}3\Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 1/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{27}{8}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330652Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T13:41:07Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
===Blue and Green Triangles===<br />
Let's assume that the blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <math>1+3(1/9)=4/3.</math><br />
<br />
===Blue, Green, and Yellow Triangles===<br />
<br />
Similarly, each of the 12 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + 3\cdot4(1/81) = 9/9 + 3/9 + 1/9 = 13/9.</math><br />
<br />
===Blue, Green, Yellow, and Red Triangles===<br />
<br />
<br />
===Total Area===<br />
<br />
The total area of the snowflake is given by taking the infinite sequence<br />
<br />
<math>\Big\{3\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math><br />
<br />
and adding its terms together to produce the following sum<br />
<br />
<math>1 + 3(1/9) + 9(1/27) + \ldots = \sum_{n=1}^{\infty}3\Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 1/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{27}{8}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330651Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T13:39:11Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Let's assume that the blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <math>1+3(1/9)=4/3.</math><br />
<br />
Similarly, each of the 12 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + 3\cdot4(1/81) = 9/9 + 3/9 + 1/9 = 13/9.</math><br />
<br />
The total area of the snowflake is given by taking the infinite sequence<br />
<br />
<math>\Big\{3\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math><br />
<br />
and adding its terms together to produce the following sum<br />
<br />
<math>1 + 3(1/9) + 9(1/27) + \ldots = \sum_{n=1}^{\infty}3\Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 1/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{27}{8}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330650Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T13:38:47Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Let's assume that the blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <math>1+3(1/9)=4/3.</math><br />
<br />
Similarly, each of the 12 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + 4\cdot4(1/81) = 9/9 + 3/9 + 1/9 = 13/9.</math><br />
<br />
The total area of the snowflake is given by taking the infinite sequence<br />
<br />
<math>\Big\{3\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math><br />
<br />
and adding its terms together to produce the following sum<br />
<br />
<math>1 + 3(1/9) + 9(1/27) + \ldots = \sum_{n=1}^{\infty}3\Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 1/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{27}{8}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330649Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T13:37:13Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Let's assume that the blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <math>1+3(1/9)=4/3.</math><br />
<br />
Similarly, each of the 12 yellow triangles has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + 12(1/81) = 9/9 + 3/9 + 1/9 = 13/9.</math><br />
<br />
The total area of the snowflake is given by taking the infinite sequence<br />
<br />
<math>\Big\{3\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math><br />
<br />
and adding its terms together to produce the following sum<br />
<br />
<math>1 + 3(1/9) + 9(1/27) + \ldots = \sum_{n=1}^{\infty}3\Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 1/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{27}{8}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330648Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T13:35:12Z<p>GregMayer: </p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Let's assume that the blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <math>1+3(1/9)=4/3.</math><br />
<br />
Similarly, each yellow triangle has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + 9(1/81) = 9/9 + 3/9 + 1/9 = 13/9.</math><br />
<br />
The total area of the snowflake is given by taking the infinite sequence<br />
<br />
<math>\Big\{3\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math><br />
<br />
and adding its terms together to produce the following sum<br />
<br />
<math>1 + 3(1/9) + 9(1/27) + \ldots = \sum_{n=1}^{\infty}3\Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 1/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{27}{8}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330647Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T13:29:43Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Let's assume that the blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <math>1+3(1/9)=4/3.</math><br />
<br />
Similarly, each yellow triangle has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + 9(1/81) = 1 + 1/3 + 1/9 = 9/9 + 3/9 + 1/9 = 13/9.</math><br />
<br />
The total area of the snowflake is given by taking the infinite sequence<br />
<br />
<math>\Big\{3\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math><br />
<br />
and adding its terms together to produce the following sum<br />
<br />
<math>1 + 3(1/9) + 9(1/27) + \ldots = \sum_{n=1}^{\infty}3\Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 1/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{27}{8}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Units/Unit_1/1.3_Infinite_Series/1.3.08_Koch_Snowflake_Example&diff=330646Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.08 Koch Snowflake Example2014-11-02T13:24:44Z<p>GregMayer: /* Solution */</p>
<hr />
<div>==Problem==<br />
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).<br />
<br />
[[File:KochIterations.png|thumb|400px|Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).]]<br />
<br />
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:<br />
<br />
* divide the line segment into three segments of equal length<br />
* draw an equilateral triangle that has the middle segment from step 1 as its base and points outward<br />
* remove the line segment that is the base of the triangle from step 2<br />
<br />
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.<br />
<br />
==Solution==<br />
<br />
Now, to derive an expression for the area of our construction at the <math>n^{th}</math> iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.<br />
<br />
[[File:Koch_Snowflake_Triangles.png|The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).|thumb]]<br />
<br />
Let's assume that the blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of <math>1+3(1/9)=4/3.</math><br />
<br />
Similarly, each yellow triangle has 1/9 the area of a green triangle, or <math>1/(9^2)</math> the area of a blue triangle. The area of the blue, green, and yellow triangles is<br />
<br />
<math>1 + 3(1/9) + 9(1/81) = 5/3.</math><br />
<br />
The total area of the snowflake is given by taking the infinite sequence<br />
<br />
<math>\Big\{3\Big(\frac{1}{9}\Big)^{n-1}\Big\}_{n=1}^{\infty}</math><br />
<br />
and adding its terms together to produce the following sum<br />
<br />
<math>1 + 3(1/9) + 9(1/27) + \ldots = \sum_{n=1}^{\infty}3\Big(\frac{1}{9}\Big)^{n-1}.</math><br />
<br />
Seeing that this is a geometric series with a = 1 and r = 1/9, we immediately conclude that this series converges and is equal to<br />
<br />
<math>\frac{a}{1-r} = \frac{27}{8}.</math><br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Unit 1]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=User:GregMayer&diff=209757User:GregMayer2012-12-06T03:09:10Z<p>GregMayer: </p>
<hr />
<div>I was a postdoctoral fellow in the Department of Mathematics at UBC Vancouver, from May 2011 to May 2012. My work was focused on developing curriculum for undergraduate level courses, much of which ends up being in the UBC Wiki. These days I am a postdoc at Georgia Tech, developing curriculum for online math courses for high school teachers and their students, and TAing online math courses. <br />
<br />
For more about me, check out [http://gsmayer.ca my website], and you can contact me via email at gsmayer@gmail.com.</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Thread:Science_talk:Math_Exam_Resources/Courses/MATH152/April_2011/Question_A15/_1,2_and_2,1&diff=209755Thread:Science talk:Math Exam Resources/Courses/MATH152/April 2011/Question A15/ 1,2 and 2,12012-12-06T03:08:03Z<p>GregMayer: New thread: [1,2] and [2,1]</p>
<hr />
<div>the question asks us to project onto the vector with direction [1,2], not [2,1]. is there a mistake in the solution?</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Syllabus/Development_of_The_ISM&diff=168736Science:Infinite Series Module/Syllabus/Development of The ISM2012-05-14T21:21:32Z<p>GregMayer: </p>
<hr />
<div>For the benefit of those who would like to more about how the ISM was created and is delivered, the ISM was initially developed over a period from roughly May 2011 to December 2011 by members of the [http://www.math.ubc.ca/ Department of Mathematics] at the [http://ubc.ca University of British Columbia] with the support of the [http://ctlt.ubc.ca Centre for Teaching and Learning Technology]. <br />
<br />
===Delivery Systems===<br />
<br />
The content management system [http://wordpress.org WordPress] was used to present material in the ISM. A set of WordPress [http://wordpress.org/extend/plugins/ plugins] were added to extend the functionality of the site:<br />
<br />
* Simple CSS: to allow customizability of aesthetic elements (lists, tables, fonts, etc)<br />
* FlexiPages: to offer the "Site Map" on the right sidebar<br />
* NextPage: to include the "previous" and "next" buttons on most pages<br />
<br />
The [http://wiki.ubc.ca/Science:Infinite_Series_Module content of the ISM] lives on the [http://wiki.ubc.ca UBC Wiki] (which is an installation of a [http://mediawiki.org MediaWiki]) on the Faculty of Science "namespace". One advantage of having content on the UBC Wiki is that it can be more easily imported into as many websites as desired. <br />
<br />
All equations are rendered as images in the UBC Wiki. This approach has these advantages: <br />
<br />
* every image has an automatically generated [http://en.wikipedia.org/wiki/Alt_attribute alt tag] in LaTeX, that allows users, who often have a disability, to be able use a [http://en.wikipedia.org/wiki/Screen_reader screen reader] to access equation content, <br />
* pages on the ISM site can be converted to PDF format for printing, and <br />
* all modern browsers can view these images <br />
<br />
=== Graphics ===<br />
<br />
[http://www.inkscape.org Inkscape], Adobe Photoshop, [http://en.wikipedia.org/wiki/Grapher Grapher], and even Microsoft PowerPoint were used to create the various graphical elements you will find throughout the ISM. Most mathematical symbols were generated in MediaWiki. When symbols were needed in images, [http://www.codecogs.com/latex/eqneditor.php CodeCogs] was used to create mathematical symbols in [http://en.wikipedia.org/wiki/Scalable_Vector_Graphics SVG] files that could be imported into graphics that appear throughout the ISM. <br />
<br />
=== Video ===<br />
<br />
For infinite series, there is an abundance of free educational videos freely available on the internet, and so development time was not focused on the development of video. Most of the video in the ISM was sourced from the [http://www.khanacademy.org/ Khan Academy], [http://patrickjmt.com/ Patrick JMT], and the [http://mathispower4u.yolasite.com/ Mathispower4u] sites. <br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Syllabus]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Syllabus/For_Instructors&diff=168317Science:Infinite Series Module/Syllabus/For Instructors2012-05-10T21:21:51Z<p>GregMayer: </p>
<hr />
<div>For the purposes of the long term maintenance of the infinite series module, the following information provides UBC instructors with an idea of how to use and maintain its content. <br />
<br />
==Functionality==<br />
<br />
The content management system [http://wordpress.org WordPress] was used to present material in the module, but the [http://wiki.ubc.ca/Science:Infinite_Series_Module content of the infinite series module] is pulled automatically from the [http://wiki.ubc.ca UBC Wiki] (which is an installation of a [http://mediawiki.org MediaWiki]) on the Faculty of Science "namespace", which is a restricted access area on the wiki.<br />
<br />
Changes made on the wiki are automatically reflected on the UBC Blog. The blog imports its content from the wiki.<br />
<br />
==Editor Access==<br />
<br />
Editing the infinite series module content is ''currently'' intended for instructors only. You may want to edit content on the blog and/or the wiki.<br />
<br />
* If you would like to edit content on the UBC Wiki, contact the [http://wiki.ubc.ca/Help:Contents#Support CTLT Wiki Support] to request access to the Faculty of Science namespace. CTLT will direct you to the appropriate person to allow you access to the faculty of science namespace, but circa May 2012, Andrea Han (han@science.ubc.ca) was the contact in CTLT to contact to provide access to content on the Science Namespace.<br />
* If you would like to edit content on the UBC Blog, please contact Rajiv Gupta, [http://gsmayer.ca Greg Mayer], or [http://blogs.ubc.ca/support/ CTLT blog support] to request access to the blog. <br />
<br />
By no means does the infinite series module have to remain a resource that can only be edited by instructors. <br />
<br />
==Editing Guidelines==<br />
<br />
When editing content in the wiki, you may want to take the following into account: <br />
<br />
* changes made on the wiki are automatically made on the UBC Blog, but not in the PDF version of the module<br />
* removing a page on the Blog does not remove content on the wiki<br />
<br />
Therefore,<br />
<br />
* if there is content you want to remove for a particular offer of your course, remove it from the Blog, not the Wiki, so that it is available for future offers<br />
* if there is content that contains errors, you need only make changes on the UBC Wiki and a PDF version of the module content (if one is available for download)<br />
<br />
PDF versions of the module content can be generated with the [http://wiki.ubc.ca/Help:Books UBC Wiki book creator]. <br />
<br />
==Google Analytics==<br />
<br />
Google Analytics was installed in January 2012 to record usage statistics. If you would like to gain access to the Analytics account, [http://gsmayer.ca Greg Mayer] or Rajiv Gupta may have the Google Analytics account credentials.<br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Syllabus]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Syllabus/For_Instructors&diff=168314Science:Infinite Series Module/Syllabus/For Instructors2012-05-10T21:18:32Z<p>GregMayer: </p>
<hr />
<div>For the purposes of the long term maintenance of the infinite series module, the following information provides UBC instructors with an idea of how to use and maintain its content. <br />
<br />
==Functionality==<br />
<br />
The content management system [http://wordpress.org WordPress] was used to present material in the module, but the [http://wiki.ubc.ca/Science:Infinite_Series_Module content of the infinite series module] is pulled automatically from the [http://wiki.ubc.ca UBC Wiki] (which is an installation of a [http://mediawiki.org MediaWiki]) on the Faculty of Science "namespace", which is a restricted access area on the wiki.<br />
<br />
Changes made on the wiki are automatically reflected on the UBC Blog. The blog imports its content from the wiki.<br />
<br />
==Editor Access==<br />
<br />
Editing the infinite series module content is ''currently'' intended for instructors only. <br />
<br />
* If you are an instructor and would like to edit content on the UBC Wiki, contact CTLT and ask for permission to edit content on the UBC please contact the [http://wiki.ubc.ca/Help:Contents#Support CTLT Wiki Support] to request access to the Faculty of Science namespace. They will direct you to the appropriate person to allow you access to the wiki, but circa May 2012, Andrea Han (han@science.ubc.ca) was the contact in CTLT to contact to provide access to content on the Science Namespace.<br />
* If you would like to edit content on the UBC Blog, please contact CTLT blog support ( to request access to the blog. <br />
<br />
By no means does the infinite series module have to remain a resource that can only be edited by instructors. <br />
<br />
==Editing Guidelines==<br />
<br />
When editing content in the wiki, you may want to take the following into account: <br />
<br />
* changes made on the wiki are automatically made on the UBC Blog, but not in the PDF version of the module<br />
* removing a page on the Blog does not remove content on the wiki<br />
<br />
Therefore,<br />
<br />
* if there is content you want to remove for a particular offer of your course, remove it from the Blog, not the Wiki, so that it is available for future offers<br />
* if there is content that contains errors, you need only make changes on the UBC Wiki and a PDF version of the module content (if one is available for download)<br />
<br />
PDF versions of the module content can be generated with the [http://wiki.ubc.ca/Help:Books UBC Wiki book creator]. <br />
<br />
==Google Analytics==<br />
<br />
Google Analytics was installed in January 2012 to record usage statistics. If you would like to gain access to the Analytics account, [http://gsmayer.ca Greg Mayer] or Rajiv Gupta may have the Google Analytics account credentials.<br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Syllabus]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Syllabus/For_Instructors&diff=168311Science:Infinite Series Module/Syllabus/For Instructors2012-05-10T21:13:20Z<p>GregMayer: </p>
<hr />
<div>For the purposes of the long term maintenance of the UBC Wiki, the following information provides UBC instructors with an idea of how to use and maintain content. <br />
<br />
==Functionality==<br />
<br />
The content management system [http://wordpress.org WordPress] was used to present material in the module, but the [http://wiki.ubc.ca/Science:Infinite_Series_Module content of the infinite series module] lives on the [http://wiki.ubc.ca UBC Wiki] (which is an installation of a [http://mediawiki.org MediaWiki]) on the Faculty of Science "namespace", which is a restricted access area on the wiki.<br />
<br />
==Editor Access==<br />
<br />
Editing the infinite series module content is ''currently'' intended for instructors only. <br />
<br />
* If you are an instructor and would like to edit content on the UBC Wiki, contact CTLT and ask for permission to edit content on the UBC please contact the [http://wiki.ubc.ca/Help:Contents#Support CTLT Wiki Support] to request access to the Faculty of Science namespace. They will direct you to the appropriate person to allow you access to the wiki, but circa May 2012, Andrea Han (han@science.ubc.ca) was the contact in CTLT to contact to provide access to content on the Science Namespace.<br />
* If you would like to edit content on the UBC Blog, please contact CTLT blog support ( to request access to the blog. <br />
<br />
By no means does the infinite series module have to remain a resource that can only be edited by instructors. <br />
<br />
==Editing Guidelines==<br />
<br />
When editing content in the wiki, you may want to take the following into account: <br />
<br />
* changes made on the wiki are automatically made on the UBC Blog, but not in the PDF version of the module<br />
* removing a page on the Blog does not remove content on the wiki<br />
<br />
Therefore,<br />
<br />
* if there is content you want to remove for a particular offer of your course, remove it from the Blog, not the Wiki, so that it is available for future offers<br />
* if there is content that contains errors, you need only make changes on the UBC Wiki and a PDF version of the module content (if one is available for download)<br />
<br />
PDF versions of the module content can be generated with the [http://wiki.ubc.ca/Help:Books UBC Wiki book creator]. <br />
<br />
==Google Analytics==<br />
<br />
Google Analytics was installed in January 2012 to record usage statistics. If you would like to gain access to the Analytics account, [http://gsmayer.ca Greg Mayer] or Rajiv Gupta may have the Google Analytics account credentials.<br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Syllabus]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Syllabus/ForInstructors&diff=168308Science:Infinite Series Module/Syllabus/ForInstructors2012-05-10T21:08:27Z<p>GregMayer: moved Science:Infinite Series Module/Syllabus/ForInstructors to Science:Infinite Series Module/Syllabus/For Instructors</p>
<hr />
<div>#REDIRECT [[Science:Infinite Series Module/Syllabus/For Instructors]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Syllabus/For_Instructors&diff=168307Science:Infinite Series Module/Syllabus/For Instructors2012-05-10T21:08:27Z<p>GregMayer: moved Science:Infinite Series Module/Syllabus/ForInstructors to Science:Infinite Series Module/Syllabus/For Instructors</p>
<hr />
<div>For the purposes of the long term maintenance of the UBC Wiki, the following information provides UBC instructors with an idea of how to use and maintain content. <br />
<br />
==Functionality==<br />
<br />
The content management system [http://wordpress.org WordPress] was used to present material in the module, but the [http://wiki.ubc.ca/Science:MATH105_Probability content of the probability module] lives on the [http://wiki.ubc.ca UBC Wiki] (which is an installation of a [http://mediawiki.org MediaWiki]) on the Faculty of Science "namespace". <br />
<br />
==Editor Access==<br />
<br />
Editing the infinite series module content is ''currently'' intended for instructors only. <br />
<br />
* If you are an instructor and would like to edit content on the UBC Wiki, contact CTLT and ask for permission to edit content on the UBC please contact the [http://wiki.ubc.ca/Help:Contents#Support CTLT Wiki Support] to request access to the Faculty of Science namespace. They will direct you to the appropriate person to allow you access to the wiki, but circa May 2012, Andrea Han (han@science.ubc.ca) was the contact in CTLT to contact to provide access to content on the Science Namespace.<br />
* If you would like to edit content on the UBC Blog, please contact CTLT blog support ( to request access to the blog. <br />
<br />
==Editing Guidelines==<br />
<br />
When editing content in the probability module, you may want to take the following into account: <br />
<br />
* changes made on the Wiki are automatically made on the UBC Blog, but not in the PDF version of the module<br />
* removing a page on the Blog does not remove it on the wiki<br />
<br />
Therefore,<br />
<br />
* if there is content you want to remove for a particular offer of MATH 105, remove it from the Blog, not the Wiki, so that it is available for future offers<br />
* if there is content that contains errors, you need only make changes on the UBC Wiki and a PDF version of the module content (if one is available for download)<br />
<br />
PDF versions of the module content can be generated with the [http://wiki.ubc.ca/Help:Books UBC Wiki book creator]. <br />
<br />
==Google Analytics==<br />
<br />
Google Analytics was installed in January 2012 to record usage statistics. If you would like to gain access to the Analytics account, [http://gsmayer.ca Greg Mayer], [mailto:malabika@math.ubc.ca Malabika Pramanik], and other past MATH 105 instructors may have the Google Analytics account credentials.<br />
<br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Syllabus]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Syllabus/For_Instructors&diff=168306Science:Infinite Series Module/Syllabus/For Instructors2012-05-10T21:08:16Z<p>GregMayer: </p>
<hr />
<div>For the purposes of the long term maintenance of the UBC Wiki, the following information provides UBC instructors with an idea of how to use and maintain content. <br />
<br />
==Functionality==<br />
<br />
The content management system [http://wordpress.org WordPress] was used to present material in the module, but the [http://wiki.ubc.ca/Science:MATH105_Probability content of the probability module] lives on the [http://wiki.ubc.ca UBC Wiki] (which is an installation of a [http://mediawiki.org MediaWiki]) on the Faculty of Science "namespace". <br />
<br />
==Editor Access==<br />
<br />
Editing the infinite series module content is ''currently'' intended for instructors only. <br />
<br />
* If you are an instructor and would like to edit content on the UBC Wiki, contact CTLT and ask for permission to edit content on the UBC please contact the [http://wiki.ubc.ca/Help:Contents#Support CTLT Wiki Support] to request access to the Faculty of Science namespace. They will direct you to the appropriate person to allow you access to the wiki, but circa May 2012, Andrea Han (han@science.ubc.ca) was the contact in CTLT to contact to provide access to content on the Science Namespace.<br />
* If you would like to edit content on the UBC Blog, please contact CTLT blog support ( to request access to the blog. <br />
<br />
==Editing Guidelines==<br />
<br />
When editing content in the probability module, you may want to take the following into account: <br />
<br />
* changes made on the Wiki are automatically made on the UBC Blog, but not in the PDF version of the module<br />
* removing a page on the Blog does not remove it on the wiki<br />
<br />
Therefore,<br />
<br />
* if there is content you want to remove for a particular offer of MATH 105, remove it from the Blog, not the Wiki, so that it is available for future offers<br />
* if there is content that contains errors, you need only make changes on the UBC Wiki and a PDF version of the module content (if one is available for download)<br />
<br />
PDF versions of the module content can be generated with the [http://wiki.ubc.ca/Help:Books UBC Wiki book creator]. <br />
<br />
==Google Analytics==<br />
<br />
Google Analytics was installed in January 2012 to record usage statistics. If you would like to gain access to the Analytics account, [http://gsmayer.ca Greg Mayer], [mailto:malabika@math.ubc.ca Malabika Pramanik], and other past MATH 105 instructors may have the Google Analytics account credentials.<br />
<br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Syllabus]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Syllabus/For_Instructors&diff=168305Science:Infinite Series Module/Syllabus/For Instructors2012-05-10T21:04:05Z<p>GregMayer: </p>
<hr />
<div>For the purposes of the long term maintenance of the UBC Wiki, the following information provides UBC instructors with an idea of how to use and maintain content. <br />
<br />
==Functionality==<br />
<br />
The content management system [http://wordpress.org WordPress] was used to present material in the module, but the [http://wiki.ubc.ca/Science:MATH105_Probability content of the probability module] lives on the [http://wiki.ubc.ca UBC Wiki] (which is an installation of a [http://mediawiki.org MediaWiki]) on the Faculty of Science "namespace". <br />
<br />
==Editor Access==<br />
<br />
''Editing'' the infinite series module content is currently intended for instructors only. If you are an instructor and would like to edit content on the UBC Wiki, contact CTLT and ask for permission to edit content on the UBC please contact the [http://wiki.ubc.ca/Help:Contents#Support CTLT Wiki Support] to request access to the Faculty of Science namespace. They will direct you to the appropriate person to allow you access to the wiki. If you would like to edit content on the UBC Blog, please contact [mailto:malabika@math.ubc.ca Malabika Pramanik] to request access to the blog. <br />
<br />
==Editing Guidelines==<br />
<br />
When editing content in the probability module, you may want to take the following into account: <br />
<br />
* changes made on the Wiki are automatically made on the UBC Blog, but not in the PDF version of the module<br />
* removing a page on the Blog does not remove it on the wiki<br />
<br />
Therefore,<br />
<br />
* if there is content you want to remove for a particular offer of MATH 105, remove it from the Blog, not the Wiki, so that it is available for future offers<br />
* if there is content that contains errors, you need only make changes on the UBC Wiki and a PDF version of the module content (if one is available for download)<br />
<br />
PDF versions of the module content can be generated with the [http://wiki.ubc.ca/Help:Books UBC Wiki book creator]. <br />
<br />
==Google Analytics==<br />
<br />
Google Analytics was installed in January 2012 to record usage statistics. If you would like to gain access to the Analytics account, [http://gsmayer.ca Greg Mayer], [mailto:malabika@math.ubc.ca Malabika Pramanik], and other past MATH 105 instructors may have the Google Analytics account credentials.<br />
<br />
<br />
[[Category:ISM]]<br />
[[Category:ISM Syllabus]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module/Syllabus/For_Instructors&diff=168304Science:Infinite Series Module/Syllabus/For Instructors2012-05-10T21:03:23Z<p>GregMayer: Created page with "For the purposes of the long term maintenance of the UBC Wiki, the following information provides UBC instructors with an idea of how to use and maintain content. ==Function..."</p>
<hr />
<div>For the purposes of the long term maintenance of the UBC Wiki, the following information provides UBC instructors with an idea of how to use and maintain content. <br />
<br />
==Functionality==<br />
<br />
The content management system [http://wordpress.org WordPress] was used to present material in the module, but the [http://wiki.ubc.ca/Science:MATH105_Probability content of the probability module] lives on the [http://wiki.ubc.ca UBC Wiki] (which is an installation of a [http://mediawiki.org MediaWiki]) on the Faculty of Science "namespace". <br />
<br />
==Editor Access==<br />
<br />
''Editing'' the infinite series module content is currently intended for instructors only. If you are an instructor and would like to edit content on the UBC Wiki, contact CTLT and ask for permission to edit content on the UBC please contact the [http://wiki.ubc.ca/Help:Contents#Support CTLT Wiki Support] to request access to the Faculty of Science namespace. They will direct you to the appropriate person to allow you access to the wiki. If you would like to edit content on the UBC Blog, please contact [mailto:malabika@math.ubc.ca Malabika Pramanik] to request access to the blog. <br />
<br />
==Editing Guidelines==<br />
<br />
When editing content in the probability module, you may want to take the following into account: <br />
<br />
* changes made on the Wiki are automatically made on the UBC Blog, but not in the PDF version of the module<br />
* removing a page on the Blog does not remove it on the wiki<br />
<br />
Therefore,<br />
<br />
* if there is content you want to remove for a particular offer of MATH 105, remove it from the Blog, not the Wiki, so that it is available for future offers<br />
* if there is content that contains errors, you need only make changes on the UBC Wiki and a PDF version of the module content (if one is available for download)<br />
<br />
PDF versions of the module content can be generated with the [http://wiki.ubc.ca/Help:Books UBC Wiki book creator]. <br />
<br />
==Google Analytics==<br />
<br />
Google Analytics was installed in January 2012 to record usage statistics. If you would like to gain access to the Analytics account, [http://gsmayer.ca Greg Mayer], [mailto:malabika@math.ubc.ca Malabika Pramanik], and other past MATH 105 instructors may have the Google Analytics account credentials.<br />
<br />
[[Category:MATH105]]<br />
[[Category:MATH105 Probability]]<br />
[[Category:MATH105 About]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Infinite_Series_Module&diff=168301Science:Infinite Series Module2012-05-10T20:58:40Z<p>GregMayer: </p>
<hr />
<div>The infinite Series Module (ISM) is a set of assessments and learning resources that support students in acquiring knowledge and skills working with infinite sequences and series. The ISM is being developed and is maintained by the Department of Mathematics at UBC.<br />
<br />
The pages in this wiki are a repository of resources that are used by a set of websites used by various courses in the Mathematics Department at UBC: <br />
<br />
* MATH 101: [http://blogs.ubc.ca/infiniteseriesmodule blogs.ubc.ca/infiniteseriesmodule]<br />
* MATH 105: [http://blogs.ubc.ca/ism105/ blogs.ubc.ca/ism105]<br />
* MATH 300: [http://blogs.ubc.ca/ismcomplexvariables/ blogs.ubc.ca/ismcomplexvariables]<br />
<br />
== Index of Pages in the ISM ==<br />
Items in the following index are ordered alphabetically. <br />
<dpl> <br />
titlematch={{PAGENAME}}/% <br />
namespace={{NAMESPACE}} <br />
shownamespace=false <br />
</dpl><br />
<br />
[[Category:ISM]]<br />
[[Category:Mathematics]]</div>GregMayerhttps://wiki.ubc.ca/index.php?title=User:GregMayer&diff=168278User:GregMayer2012-05-10T19:02:48Z<p>GregMayer: </p>
<hr />
<div>I was a postdoctoral fellow in the Department of Mathematics at UBC Vancouver, from May 2011 to May 2012. My work was focused on developing curriculum for undergraduate level courses, much of which ends up being in the UBC Wiki. <br />
<br />
For more about me, check out [http://gsmayer.ca my website], and you can contact me via email at gsmayer@gmail.com.</div>GregMayerhttps://wiki.ubc.ca/index.php?title=User:GregMayer&diff=168277User:GregMayer2012-05-10T19:02:34Z<p>GregMayer: </p>
<hr />
<div>I was a postdoctoral fellow in the Department of Mathematics at UBC Vancouver, from May 2011-2012. My work was focused on developing curriculum for undergraduate level courses, much of which ends up being in the UBC Wiki. <br />
<br />
For more about me, check out [http://gsmayer.ca my website], and you can contact me via email at gsmayer@gmail.com.</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH104/December_2011/Question_01_(g)/Solution_1&diff=167499Science:Math Exam Resources/Courses/MATH104/December 2011/Question 01 (g)/Solution 12012-05-01T23:50:03Z<p>GregMayer: </p>
<hr />
<div>The given function is increasing where its first derivative is positive. Taking the derivative yields<br />
<br />
:<math>\displaystyle f'(x) = 2x - 1/x</math><br />
<br />
We must now solve the inequality for ''x'' yields the interval where ''f'' is increasing. We will only be interested in positive values of ''x'', as negative values of ''x'' are not in the domain of the given function (because of the natural logarithm). Therefore,<br />
<br />
:<math>\begin{align}<br />
0 &< 2x - 1/x , \quad x > 0\\<br />
1/x &< 2x \\<br />
1 & < 2x^2 \\<br />
\frac{1}{2} &< x^2 \\<br />
\end{align}<br />
</math><br />
Therefore, we have <br />
<br />
:<math><br />
x > + \frac{1}{\sqrt{2}}<br />
</math><br />
<br />
The negative solution was ruled out because of the requirement that x be a positive number. <br />
<br />
The interval on which f is increasing is<br />
<br />
:<math><br />
(\frac{1}{\sqrt{2}}, \infty)<br />
</math></div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH104/December_2011/Question_01_(i)/Solution_1&diff=167497Science:Math Exam Resources/Courses/MATH104/December 2011/Question 01 (i)/Solution 12012-05-01T23:45:09Z<p>GregMayer: </p>
<hr />
<div>To determine the equation of the tangent line we will need the slope of the line at the given point. To find the slope at the given point, we may use implicit differentiation and find an expression for y'. <br />
<br />
:<math>\begin{align} <br />
\frac{d}{dx} \Big( xy^2+2xy \Big) &= \frac{d}{dx} (8) \\<br />
y^2 + 2xyy'+2y +2xy' &= 0 \\<br />
\end{align}</math><br />
<br />
At the point (1,2), this equation becomes<br />
<br />
:<math>\begin{align} <br />
(2)^2 + 2(1)(2)y'+2(2)+2(1)y' &= 0 \\<br />
4+4y'+4+2y'&=0\\<br />
6y'&=-8\\<br />
y'&=-4/3<br />
\end{align}</math><br />
<br />
Using the "point-slope" equation of a straight line with slope ''m'' that passes through the point (''x<sub>0</sub>'',''y<sub>0</sub>'')<br />
<br />
:<math> \displaystyle y-y_0 = m(x-x_0)</math><br />
<br />
we have <br />
<br />
:<math> \begin{align} <br />
y-(2) &= (-\frac{4}{3})\big(x-(1)\big) \\<br />
y &= -\frac{4}{3}x + \frac{10}{3} \\<br />
\end{align}</math><br />
<br />
which is the equation of the tangent line at (1,2).</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH104/December_2011/Question_01_(i)/Solution_1&diff=167496Science:Math Exam Resources/Courses/MATH104/December 2011/Question 01 (i)/Solution 12012-05-01T23:44:33Z<p>GregMayer: </p>
<hr />
<div>To determine the equation of the tangent line we will need the slope of the line at the given point. To find the slope at the given point, we may use implicit differentiation and find an expression for y'. <br />
<br />
:<math>\begin{align} <br />
\frac{d}{dx} \Big( xy^2+2xy \Big) &= \frac{d}{dx} (8) \\<br />
y^2 + 2xyy'+2y +2xy' &= 0 \\<br />
\end{align}</math><br />
<br />
At the point (1,2), this equation becomes<br />
<br />
:<math>\begin{align} <br />
(2)^2 + 2(1)(2)y'+2(2)+2(1)y' &= 0 \\<br />
4+4y'+4+2y'&=0\\<br />
6y'&=-8\\<br />
y'&=-4/3<br />
\end{align}</math><br />
<br />
Using the "point-slope" equation of a straight line with slope ''m'' that passes through the point (''x<sub>0</sub>'',''y<sub>0</sub>'')<br />
<br />
:<math> \displaystyle y-y_0 = m(x-x_0)</math><br />
<br />
we have <br />
<br />
:<math> \begin{align} <br />
y-(2) &= (-\frac{4}{3})\big(x-(1)\big) \\<br />
y &= -\frac{4}{3}x + 4 \\<br />
\end{align}</math><br />
<br />
which is the equation of the tangent line at (1,2).</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH104/December_2011/Question_01_(i)/Solution_1&diff=167495Science:Math Exam Resources/Courses/MATH104/December 2011/Question 01 (i)/Solution 12012-05-01T23:43:54Z<p>GregMayer: </p>
<hr />
<div>To determine the equation of the tangent line we will need the slope of the line at the given point. To find the slope at the given point, we may use implicit differentiation and find an expression for y'. <br />
<br />
:<math>\begin{align} <br />
\frac{d}{dx} \Big( xy^2+2xy \Big) &= \frac{d}{dx} (8) \\<br />
y^2 + 2xyy'+2y +2xy' &= 0 \\<br />
\end{align}</math><br />
<br />
At the point (1,2), this equation becomes<br />
<br />
:<math>\begin{align} <br />
(2)^2 + 2(1)(2)y'+2(2)+2(1)y' &= 0 \\<br />
4+4y'+4+2y'&=0\\<br />
6y'&=-8\\<br />
y'&=-4/3<br />
\end{align}</math><br />
<br />
Using the "point-slope" equation of a straight line with slope ''m'' that passes through the point (''x<sub>0</sub>'',''y<sub>0</sub>'')<br />
<br />
:<math> \displaystyle y-y_0 = m(x-x_0)</math><br />
<br />
we have <br />
<br />
:<math> \begin{align} <br />
y-(2) &= (-2)\big(x-(1)\big) \\<br />
y &= -2x + 4 \\<br />
\end{align}</math><br />
<br />
which is the equation of the tangent line at (1,2).</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH104/December_2011/Question_01_(i)/Solution_1&diff=167494Science:Math Exam Resources/Courses/MATH104/December 2011/Question 01 (i)/Solution 12012-05-01T23:43:21Z<p>GregMayer: </p>
<hr />
<div>To determine the equation of the tangent line we will need the slope of the line at the given point. To find the slope at the given point, we may use implicit differentiation and find an expression for y'. <br />
<br />
:<math>\begin{align} <br />
\frac{d}{dx} \Big( xy^2+2xy \Big) &= \frac{d}{dx} (8) \\<br />
y^2 + 2xyy'+2y +2xy' &= 0 \\<br />
\end{align}</math><br />
<br />
At the point (1,2), this equation becomes<br />
<br />
:<math>\begin{align} <br />
(2)^2 + 2(1)(2)y'+2(2)+2(1)y' &= 0 \\<br />
4+4y'+4+2y'&=0\\<br />
6y'&=-2<br />
\end{align}</math><br />
<br />
Using the "point-slope" equation of a straight line with slope ''m'' that passes through the point (''x<sub>0</sub>'',''y<sub>0</sub>'')<br />
<br />
:<math> \displaystyle y-y_0 = m(x-x_0)</math><br />
<br />
we have <br />
<br />
:<math> \begin{align} <br />
y-(2) &= (-2)\big(x-(1)\big) \\<br />
y &= -2x + 4 \\<br />
\end{align}</math><br />
<br />
which is the equation of the tangent line at (1,2).</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH104/December_2011/Question_01_(i)/Solution_1&diff=167493Science:Math Exam Resources/Courses/MATH104/December 2011/Question 01 (i)/Solution 12012-05-01T23:41:41Z<p>GregMayer: </p>
<hr />
<div>To determine the equation of the tangent line we will need the slope of the line at the given point. To find the slope at the given point, we may use implicit differentiation and find an expression for y'. <br />
<br />
:<math>\begin{align} <br />
\frac{d}{dx} \Big( xy^2+2xy \Big) &= \frac{d}{dx} (8) \\<br />
y^2 + 2xyy'+2y +2xy' &= 0 \\<br />
\end{align}</math><br />
<br />
At the point (1,2), this equation becomes<br />
<br />
:<math>\begin{align} <br />
(2)^2 + (1)(2)(2)y'+2(2)+2(1)y' &= 0 \\<br />
4+4y'+4+2y'&=0\\<br />
6y'&=-2<br />
\end{align}</math><br />
<br />
Using the "point-slope" equation of a straight line with slope ''m'' that passes through the point (''x<sub>0</sub>'',''y<sub>0</sub>'')<br />
<br />
:<math> \displaystyle y-y_0 = m(x-x_0)</math><br />
<br />
we have <br />
<br />
:<math> \begin{align} <br />
y-(2) &= (-2)\big(x-(1)\big) \\<br />
y &= -2x + 4 \\<br />
\end{align}</math><br />
<br />
which is the equation of the tangent line at (1,2).</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH104/December_2011/Question_02_(a)&diff=164577Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (a)2012-04-18T18:50:08Z<p>GregMayer: </p>
<hr />
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{{MER Question page}}</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH104/December_2011/Question_02_(a)/Statement&diff=164576Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (a)/Statement2012-04-18T18:49:33Z<p>GregMayer: Created page with "Consider the function :<math>f(x) = \frac{x^3+x^2-2x-3}{x^2-3}.</math> Its first and second derivatives are given by :<math>f'(x) = \frac{(x^2-1)(x^2-6)}{(x^2-3)^2}, \qquad..."</p>
<hr />
<div>Consider the function<br />
<br />
:<math>f(x) = \frac{x^3+x^2-2x-3}{x^2-3}.</math><br />
<br />
Its first and second derivatives are given by<br />
<br />
:<math>f'(x) = \frac{(x^2-1)(x^2-6)}{(x^2-3)^2}, \qquad f''(x) = \frac{2x(x^2+9)}{(x^2-3)^3}.</math><br />
<br />
Find all ''x'' such that f'(x) = 0 or f'(x) does not exist.</div>GregMayerhttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH104/December_2011/Question_01_(l)&diff=164575Science:Math Exam Resources/Courses/MATH104/December 2011/Question 01 (l)2012-04-18T18:45:39Z<p>GregMayer: </p>
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