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Science:Math Exam Resources/Courses/MATH200/December 2012/Question 04/Hint 2

2014-09-17T19:59:39Z

VincentChan: Created page with "To handle the boundary case, break up the boundary into 4 cases: <math>\displaystyle x = -1</math>, <math>\displaystyle x = 3</math>, <math>\displaystyle y = -1</math>, and <m..."

To handle the boundary case, break up the boundary into 4 cases: <math>\displaystyle x = -1</math>, <math>\displaystyle x = 3</math>, <math>\displaystyle y = -1</math>, and <math>\displaystyle y = 1</math>. For the first line, you get a new function <math>\displaystyle g(y) = f(-1,y)</math>, which you can then optimize subject to the constraint <math>\displaystyle -1 \leq y \leq 1</math>. The other lines are similar.

Science:Math Exam Resources/Courses/MATH200/December 2012/Question 04/Hint 1

2014-09-17T19:20:29Z

VincentChan:

First, find the critical points of <math>\displaystyle f</math> (where <math>\displaystyle f_x = f_y = 0</math>) in the interior of the domain, then check the boundary. The extrema must be at one of the critical points, or on the boundary.

Science:Math Exam Resources/Courses/MATH200/December 2012/Question 03 (c)/Solution 1

2014-09-17T18:56:26Z

VincentChan:

We just plug in the values <math>\displaystyle (x,y) = (-1.02,0.97)</math> into
the linear approximation found in part b)
<math>\displaystyle f(x,y) \approx -1-\frac32 (y-1)</math>

<math>\displaystyle f(-1.02,0.97)\approx -1 - \frac32 (0.97-1) = -0.955</math>

[As a quick check, recall that <math>\displaystyle f(-1,1) = -1 </math>, so our answer should be close to this (assuming our approximation is good).]

Science:Math Exam Resources/Courses/MATH200/December 2012/Question 03 (c)/Hint 1

2014-09-17T18:54:23Z

VincentChan:

Just plug in the values <math>\displaystyle (x,y) = (-1.02,0.97)</math> into
<math>\displaystyle f(x,y)\approx -1-\frac32 (y-1)</math>

Science:Math Exam Resources/Courses/MATH200/December 2012/Question 03 (b)/Solution 1

2014-09-17T17:28:43Z

VincentChan:

What to do 

The linear approximation of a function <math>\displaystyle z=f(x,y)</math> in a point <math>\displaystyle (p_x,p_y)</math> is

<math>\displaystyle z = f(x,y) \approx f(p_x,p_y) + (x-p_x)\frac{\partial z}{\partial x}(p_x,p_y) + (y-p_y)\frac{\partial z}{\partial y}(p_x,p_y)</math>

 Find <math>\displaystyle f(-1,1)</math>

We have that <math>\displaystyle (p_x,p_y) = (-1,1)</math>. To find <math>\displaystyle f(-1,1)</math>, we must solve the equation <math>\displaystyle xyz+x+y^2+z^3 = 0</math>, where <math>\displaystyle (x,y) = (-1,1)</math> for the variable <math>\displaystyle z</math>.

<math>\displaystyle\begin{align}
(-1)1z + (-1) + 1^2 + z^3 &= 0\\
-z - 1 + 1 + z^3 &= 0\\
z(z^2-1) &= 0\\
z(z-1)(z+1) &= 0
\end{align}</math>

Hence, <math>\displaystyle z_1 = -1,\ z_2 = 0,\ z_3 = 1</math> are solution to this equations. With the knowledge,that <math>\displaystyle f(-1,1) < 0</math>, the value must be <math>\displaystyle z = f(-1,1) = -1</math>

 Find <math>\displaystyle \frac{\partial z}{\partial x}(-1,1)</math> 

We already know from part a), that <math>\displaystyle \frac{\partial z}{\partial x} = -\frac{yz+1}{xy+3z^2}</math>. Pluging in the point <math>\displaystyle (x,y,z) = (-1,1,-1)</math> gives
<math>\displaystyle \frac{\partial z}{\partial x}(-1,1) = 0 </math>

 Find <math>\displaystyle \frac{\partial z}{\partial y}(-1,1)</math> 

We know that <math>\displaystyle \frac{\partial z}{\partial y} = -\frac{F_y}{F_z} = -\frac{xz + 2y}{xy + 3z^2}</math>
Pluging in the point <math>\displaystyle (x,y,z) = (-1,1,-1)</math> gives
<math>\displaystyle \frac{\partial z}{\partial y}(-1,1) = -\frac32</math>

 The linear approximation 

Hence the linear approximation is
<math>\displaystyle f(x,y)\approx -1 -\frac32 (y-1)</math>

Science:Math Exam Resources/Courses/MATH200/December 2012/Question 03 (b)/Statement

2014-09-17T17:27:44Z

VincentChan:

Suppose that a function <math>\displaystyle z = f(x, y)</math> is implicitly deﬁned by an equation:
<math>\displaystyle xyz + x + y^2 + z^3 = 0</math>

(b) If <math>\displaystyle f(-1, 1) < 0</math>, find the linear approximation of the function <math>\displaystyle z = f(x, y)</math> at <math>\displaystyle (-1, 1)</math>.

Science:Math Exam Resources/Courses/MATH200/December 2012/Question 03 (b)/Hint 1

2014-09-17T17:27:16Z

VincentChan:

The linear approximation of a function <math>\displaystyle z=f(x,y)</math> at a point <math>\displaystyle (p_x,p_y)</math> is

<math>\displaystyle z = f(x,y) \approx f(p_x,p_y) + (x-p_x)\frac{\partial z}{\partial x}(p_x,p_y) + (y-p_y)\frac{\partial z}{\partial y}(p_x,p_y)</math>

Science:Math Exam Resources/Courses/MATH200/December 2012/Question 03 (b)/Hint 1

2014-09-17T17:26:54Z

VincentChan:

The linear approximation of a function <math>\displaystyle z=f(x,y)</math> in a point <math>\displaystyle (p_x,p_y)</math> is

<math>\displaystyle z = f(x,y) \approx f(p_x,p_y) + (x-p_x)\frac{\partial z}{\partial x}(p_x,p_y) + (y-p_y)\frac{\partial z}{\partial y}(p_x,p_y)</math>

Science:Math Exam Resources/Courses/MATH200/December 2012/Question 03 (a)/Solution 2

2014-09-17T17:12:27Z

VincentChan:

Implicitly differentiating term by term, we find

<math>
\begin{align}
\frac{\partial x}{\partial x}(yz) + \frac{\partial z}{\partial x}(xy) + \frac{\partial x}{\partial x} + 3z^2 \frac{\partial z}{\partial x} & = 0 \\
\frac{\partial z}{\partial x} (xy+3z^2) & = -(yz+1) \\
\frac{\partial z}{\partial x} & = -\frac{yz + 1}{xy + 3z^2}
\end{align}
</math>

Science:Math Exam Resources/Courses/MATH200/December 2012/Question 03 (a)/Solution 2

2014-09-17T17:12:17Z

VincentChan: Created page with "Implicitly differentiating term by term, we find <math> \begin{align} \frac{\partial x}{\partial x}(yz) + \frac{\partial z}{\partial x}(xy) + \frac{\partial x}{\partial x} ..."

Science:Math Exam Resources/Courses/MATH200/December 2012/Question 03 (a)/Hint 2

2014-09-17T17:08:13Z

VincentChan: Created page with "Alternatively, implicitly differentiate each term, keeping in mind that <math>\displaystyle x</math> and <math>\displaystyle z</math> are functions of <math>\displaystyle x</m..."

Alternatively, implicitly differentiate each term, keeping in mind that <math>\displaystyle x</math> and <math>\displaystyle z</math> are functions of <math>\displaystyle x</math>.

Science:Math Exam Resources/Courses/MATH200/December 2012/Question 01 (b)

2014-09-17T16:48:47Z

VincentChan:

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Science:Math Exam Resources/Courses/MATH200/December 2012/Question 02

2014-09-17T16:35:54Z

VincentChan:

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Science:Math Exam Resources/Courses/MATH200/December 2012/Question 02/Solution 1

2014-09-17T16:35:25Z

VincentChan:

We apply the Chain rule to calculate

<math>\displaystyle\begin{align}
\frac{\partial G}{\partial t} &= \frac{\partial F}{\partial z}\frac{\partial (At)}{\partial t} = \frac{\partial F}{\partial z} A\\
\frac{\partial G}{\partial \gamma} &= \frac{\partial F}{\partial x}\frac{\partial (\gamma + s)}{\partial \gamma} + \frac{\partial F}{\partial y}\frac{\partial (\gamma-s)}{\partial\gamma}
= \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y}\\
\frac{\partial G}{\partial s} &= \frac{\partial F}{\partial x}\frac{\partial (\gamma + s)}{\partial s} + \frac{\partial F}{\partial y}\frac{\partial (\gamma-s)}{\partial s}
=\frac{\partial F}{\partial x} - \frac{\partial F}{\partial y}
\end{align}</math>

For the second derivatives, we apply the Chain rule again to the already calculated derivatives

<math>\displaystyle\begin{align}
\frac{\partial^2G}{\partial\gamma^2} &= \frac{\partial}{\partial x}\frac{\partial F}{\partial x}\frac{\partial(\gamma + s)}{\partial \gamma} + \frac{\partial}{\partial y}\frac{\partial F}{\partial x}\frac{\partial(\gamma - s)}{\partial \gamma} + \frac{\partial}{\partial x}\frac{\partial F}{\partial y}\frac{\partial(\gamma + s)}{\partial \gamma} + \frac{\partial}{\partial y}\frac{\partial F}{\partial x}\frac{\partial(\gamma - s)}{\partial \gamma}\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \qquad = \frac{\partial^2F}{\partial x^2} + \frac{\partial^2F}{\partial y\partial x} + \frac{\partial^2F}{\partial x\partial y} + \frac{\partial F^2}{\partial y^2}\\

\frac{\partial^2G}{\partial s^2} &= \frac{\partial}{\partial x}\frac{\partial F}{\partial x}\frac{\partial(\gamma + s)}{\partial s} + \frac{\partial}{\partial y}\frac{\partial F}{\partial x}\frac{\partial(\gamma - s)}{\partial s} - \frac{\partial}{\partial x}\frac{\partial F}{\partial y}\frac{\partial(\gamma + s)}{\partial s} - \frac{\partial}{\partial y}\frac{\partial F}{\partial x}\frac{\partial(\gamma - s)}{\partial s}\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \qquad = \frac{\partial^2F}{\partial x^2} - \frac{\partial^2F}{\partial y\partial x} - \frac{\partial^2F}{\partial x\partial y} + \frac{\partial F^2}{\partial y^2}
\end{align}</math>

Now we set together

<math>\displaystyle \frac{\partial^2G}{\partial \gamma^2} + \frac{\partial G^2}{\partial s^2} =
\frac{\partial^2F}{\partial x^2} + \frac{\partial^2F}{\partial y\partial x} + \frac{\partial^2F}{\partial x\partial y} + \frac{\partial F^2}{\partial y^2} +
\frac{\partial^2F}{\partial x^2} - \frac{\partial^2F}{\partial y\partial x} - \frac{\partial^2F}{\partial x\partial y} + \frac{\partial F^2}{\partial y^2}
= 2\frac{\partial^2F}{\partial x^2} + 2\frac{\partial^2F}{\partial y^2}
= 2\frac{\partial F}{\partial z}
</math>

where the last step is already given by the problem.

With <math>\displaystyle \frac{\partial^2G}{\partial \gamma^2} + \frac{\partial G^2}{\partial s^2} = 2\frac{\partial F}{\partial z}</math> and <math>\displaystyle
\frac{\partial G}{\partial t} = A\frac{\partial F}{\partial z}</math>

we conclude that <math>\displaystyle A=2</math> is the value that <math>\displaystyle
\frac{\partial^2G}{\partial \gamma^2} + \frac{\partial G^2}{\partial s^2} = \frac{\partial G}{\partial t} </math>

Science:Math Exam Resources/Courses/MATH200/December 2012/Question 01 (a)

2014-09-09T17:07:30Z

VincentChan:

Science:Math Exam Resources/Courses/MATH200/December 2012/Question 01 (a)/Solution 1

2014-09-09T17:02:48Z

VincentChan:

The line <math>\displaystyle L</math> satisfies both plane equations. We solve the second one for <math>\displaystyle x</math>:

<math>\displaystyle\begin{align}
x &= y-2z\\
\end{align}</math>

and plug the result into the first equation:

<math>\displaystyle\begin{align}
(y - 2z)+y+z & = 6\\
z &= 2y-6
\end{align}</math>

Hence

<math>\displaystyle \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}y-2z\\y\\2y-6\end{pmatrix} = \begin{pmatrix}y-2(2y-6)\\y\\2y-6\end{pmatrix} = \begin{pmatrix}12-3y\\y\\2y-6\end{pmatrix} = \begin{pmatrix} 12\\0\\-6\end{pmatrix} + y\begin{pmatrix}-3\\1\\2\end{pmatrix}</math>

Then the line is

<math>\displaystyle L = \begin{pmatrix} 12\\0\\-6\end{pmatrix} + \lambda \begin{pmatrix}-3\\1\\2\end{pmatrix}</math>

In the <math>\displaystyle yz</math>-plane, <math>\displaystyle x=0</math>. Hence, for the first entry of the line <math>\displaystyle L</math> holds when
<math>\displaystyle 12-3\lambda = 0\quad\Rightarrow\quad \lambda = 4</math>

So, <math>\displaystyle L</math> intersects with the <math>\displaystyle yz</math>-plane in the point

<math>\displaystyle p_{yz} = \begin{pmatrix}12\\0\\-6\end{pmatrix} + 4\begin{pmatrix}-3\\1\\2\end{pmatrix} = \begin{pmatrix}0\\4\\2\end{pmatrix}</math>

In the <math>\displaystyle xz</math>-plane, <math>\displaystyle y=0</math>. Hence, for the second entry of the line <math>\displaystyle L</math> holds when
<math>\displaystyle \lambda = 0</math>

So, <math>\displaystyle L</math> intersects with the <math>\displaystyle xz</math>-plane in the point

<math>\displaystyle p_{xz} = \begin{pmatrix}12\\0\\-6\end{pmatrix} + 0\begin{pmatrix}-3\\1\\2\end{pmatrix} = \begin{pmatrix}12\\0\\-6\end{pmatrix}</math>

In the <math>\displaystyle xy</math>-plane, <math>\displaystyle z=0</math>. Hence, for the third entry of the line <math>\displaystyle L</math> holds when
<math>\displaystyle -6 + 2\lambda = 0\quad\Rightarrow\quad \lambda = 3</math>

So, <math>\displaystyle L</math> intersects with the <math>\displaystyle xy</math>-plane in the point

<math>\displaystyle p_{xy} = \begin{pmatrix}12\\0\\-6\end{pmatrix} + 3\begin{pmatrix}-3\\1\\2\end{pmatrix} = \begin{pmatrix}3\\3\\0\end{pmatrix}</math>

Science:Math Exam Resources/Courses/MATH200/December 2012/Question 01 (a)/Solution 1

2014-09-09T16:56:27Z

VincentChan:

The line <math>\displaystyle L</math> satisfies both plane equations. We solve the second one for <math>\displaystyle x</math>:

<math>\displaystyle\begin{align}
x &= y-2z\\
\end{align}</math>

and plug the result into the first equation:

<math>\displaystyle\begin{align}
(y - 2z)+y+z & = 6\\
z &= 2y-6
\end{align}</math>

Hence

<math>\displaystyle \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}y-2z\\y\\2y-6\end{pmatrix} = \begin{pmatrix}y-2(2y-6)\\y\\2y-6\end{pmatrix} = \begin{pmatrix}12-3y\\y\\2y-6\end{pmatrix} = \begin{pmatrix} 12\\0\\-6\end{pmatrix} + y\begin{pmatrix}-3\\1\\2\end{pmatrix}</math>

Then the line is

<math>\displaystyle L = \begin{pmatrix} 12\\0\\-6\end{pmatrix} + \lambda \begin{pmatrix}-3\\1\\2\end{pmatrix}</math>

In the <math>\displaystyle yz</math>-plane, <math>\displaystyle x=0</math>. Hence, for the first entry of the line <math>\displaystyle L</math> holds
<math>\displaystyle 12-3\lambda = 0\quad\Rightarrow\quad \lambda = 4</math>

So, <math>\displaystyle L</math> intersects with the <math>\displaystyle yz</math>-plane in the point

<math>\displaystyle p_{yz} = \begin{pmatrix}12\\0\\-6\end{pmatrix} + 4\begin{pmatrix}-3\\1\\2\end{pmatrix} = \begin{pmatrix}0\\4\\2\end{pmatrix}</math>

In the <math>\displaystyle xz</math>-plane, <math>\displaystyle y=0</math>. Hence, for the second entry of the line <math>\displaystyle L</math> holds
<math>\displaystyle \lambda = 0</math>

So, <math>\displaystyle L</math> intersects with the <math>\displaystyle xz</math>-plane in the point

<math>\displaystyle p_{xz} = \begin{pmatrix}12\\0\\-6\end{pmatrix} + 0\begin{pmatrix}-3\\1\\2\end{pmatrix} = \begin{pmatrix}12\\0\\-6\end{pmatrix}</math>

In the <math>\displaystyle xy</math>-plane, <math>\displaystyle z=0</math>. Hence, for the third entry of the line <math>\displaystyle L</math> holds
<math>\displaystyle -6 + 2\lambda = 0\quad\Rightarrow\quad \lambda = 3</math>

So, <math>\displaystyle L</math> intersects with the <math>\displaystyle xy</math>-plane in the point

<math>\displaystyle p_{xy} = \begin{pmatrix}12\\0\\-6\end{pmatrix} + 3\begin{pmatrix}-3\\1\\2\end{pmatrix} = \begin{pmatrix}3\\3\\0\end{pmatrix}</math>

Science:Math Exam Resources/Courses/MATH200/December 2012/Question 01 (a)/Solution 1

2014-09-09T16:54:50Z

VincentChan:

The line <math>\displaystyle L</math> satisfies both plane equations. We solve the second one for <math>\displaystyle x</math>:

<math>\displaystyle\begin{align}
x &= y-2z\\
\end{align}</math>

and plug the result into the first equation:

<math>\displaystyle\begin{align}
(y - 2z)+y+z & = 6\\
z &= 2y-6
\end{align}</math>

Hence

<math>\displaystyle \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}y-2z\\y\\2y-6\end{pmatrix} = \begin{pmatrix} 12\\0\\-6\end{pmatrix} + y\begin{pmatrix}-3\\1\\2\end{pmatrix}</math>

Then the line is

<math>\displaystyle L = \begin{pmatrix} 12\\0\\-6\end{pmatrix} + \lambda \begin{pmatrix}-3\\1\\2\end{pmatrix}</math>

In the <math>\displaystyle yz</math>-plane, <math>\displaystyle x=0</math>. Hence, for the first entry of the line <math>\displaystyle L</math> holds
<math>\displaystyle 12-3\lambda = 0\quad\Rightarrow\quad \lambda = 4</math>

So, <math>\displaystyle L</math> intersects with the <math>\displaystyle yz</math>-plane in the point

<math>\displaystyle p_{yz} = \begin{pmatrix}12\\0\\-6\end{pmatrix} + 4\begin{pmatrix}-3\\1\\2\end{pmatrix} = \begin{pmatrix}0\\4\\2\end{pmatrix}</math>

In the <math>\displaystyle xz</math>-plane, <math>\displaystyle y=0</math>. Hence, for the second entry of the line <math>\displaystyle L</math> holds
<math>\displaystyle \lambda = 0</math>

So, <math>\displaystyle L</math> intersects with the <math>\displaystyle xz</math>-plane in the point

<math>\displaystyle p_{xz} = \begin{pmatrix}12\\0\\-6\end{pmatrix} + 0\begin{pmatrix}-3\\1\\2\end{pmatrix} = \begin{pmatrix}12\\0\\-6\end{pmatrix}</math>

In the <math>\displaystyle xy</math>-plane, <math>\displaystyle z=0</math>. Hence, for the third entry of the line <math>\displaystyle L</math> holds
<math>\displaystyle -6 + 2\lambda = 0\quad\Rightarrow\quad \lambda = 3</math>

So, <math>\displaystyle L</math> intersects with the <math>\displaystyle xy</math>-plane in the point

<math>\displaystyle p_{xy} = \begin{pmatrix}12\\0\\-6\end{pmatrix} + 3\begin{pmatrix}-3\\1\\2\end{pmatrix} = \begin{pmatrix}3\\3\\0\end{pmatrix}</math>

Science:Math Exam Resources/Courses/MATH200/December 2012/Question 01 (a)/Solution 1

2014-09-09T16:53:55Z

VincentChan:

The line <math>\displaystyle L</math> satisfies both plane equations. We solve the second one for <math>\displaystyle x</math> and plug the result into the first one.

<math>\displaystyle\begin{align}
x &= y-2z\\

(y - 2z)+y+z & = 6\\
z &= 2y-6
\end{align}</math>

Hence

<math>\displaystyle \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}y-2z\\y\\2y-6\end{pmatrix} = \begin{pmatrix} 12\\0\\-6\end{pmatrix} + y\begin{pmatrix}-3\\1\\2\end{pmatrix}</math>

Then the line is

<math>\displaystyle L = \begin{pmatrix} 12\\0\\-6\end{pmatrix} + \lambda \begin{pmatrix}-3\\1\\2\end{pmatrix}</math>

In the <math>\displaystyle yz</math>-plane, <math>\displaystyle x=0</math>. Hence, for the first entry of the line <math>\displaystyle L</math> holds
<math>\displaystyle 12-3\lambda = 0\quad\Rightarrow\quad \lambda = 4</math>

So, <math>\displaystyle L</math> intersects with the <math>\displaystyle yz</math>-plane in the point

<math>\displaystyle p_{yz} = \begin{pmatrix}12\\0\\-6\end{pmatrix} + 4\begin{pmatrix}-3\\1\\2\end{pmatrix} = \begin{pmatrix}0\\4\\2\end{pmatrix}</math>

In the <math>\displaystyle xz</math>-plane, <math>\displaystyle y=0</math>. Hence, for the second entry of the line <math>\displaystyle L</math> holds
<math>\displaystyle \lambda = 0</math>

So, <math>\displaystyle L</math> intersects with the <math>\displaystyle xz</math>-plane in the point

<math>\displaystyle p_{xz} = \begin{pmatrix}12\\0\\-6\end{pmatrix} + 0\begin{pmatrix}-3\\1\\2\end{pmatrix} = \begin{pmatrix}12\\0\\-6\end{pmatrix}</math>

In the <math>\displaystyle xy</math>-plane, <math>\displaystyle z=0</math>. Hence, for the third entry of the line <math>\displaystyle L</math> holds
<math>\displaystyle -6 + 2\lambda = 0\quad\Rightarrow\quad \lambda = 3</math>

So, <math>\displaystyle L</math> intersects with the <math>\displaystyle xy</math>-plane in the point

<math>\displaystyle p_{xy} = \begin{pmatrix}12\\0\\-6\end{pmatrix} + 3\begin{pmatrix}-3\\1\\2\end{pmatrix} = \begin{pmatrix}3\\3\\0\end{pmatrix}</math>

Science:Math Exam Resources/Courses/MATH312/December 2008/Question 02 (a) iii

2014-07-11T07:50:37Z

VincentChan:

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Science:Math Exam Resources/Courses/MATH312/December 2008/Question 02 (a) ii

2014-07-11T07:49:43Z

VincentChan:

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Science:Math Exam Resources/Courses/MATH312/December 2008/Question 02 (a) i

2014-07-11T07:48:33Z

VincentChan:

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Science:Math Exam Resources/Courses/MATH312/December 2008/Question 01 (f)

2014-07-11T07:45:53Z

VincentChan:

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Science:Math Exam Resources/Courses/MATH312/December 2008/Question 01 (e)

2014-07-11T07:45:16Z

VincentChan:

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Science:Math Exam Resources/Courses/MATH312/December 2008/Question 01 (d)

2014-07-11T07:43:57Z

VincentChan:

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Science:Math Exam Resources/Courses/MATH312/December 2008/Question 01 (c)

2014-07-11T07:42:06Z

VincentChan:

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Science:Math Exam Resources/Courses/MATH312/December 2008/Question 01 (b)

2014-07-11T07:40:40Z

VincentChan:

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Science:Math Exam Resources/Courses/MATH312/December 2008/Question 01 (a)

2014-07-11T07:39:55Z

VincentChan:

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Science:Math Exam Resources/Courses/MATH312/December 2009/Question 01 (b)/Solution 1

2014-07-11T07:39:03Z

VincentChan:

Using long division, we see that

<math> \displaystyle a^{2} + b^{2} = (a+b)(a-b) + 2b^{2}</math>

and thus, the Euclidean algorithm states that

<math> \displaystyle \gcd(a^{2} + b^{2},a+b) = (a+b, 2b^{2})</math>.

Now, as ''a'' and ''b'' are coprime, we see that

<math> \displaystyle \gcd(a^{2} + b^{2},a+b) = (a+b, 2b^{2}) = (a+b,2)</math>. If this were not true then some factor of ''b'' would divide <math> \displaystyle a+b</math> and that means that this factor would also divide ''a'' which means that the factor must have been 1 since ''a'' and ''b'' were coprime. Finally, it's clear that <math> (a+b,2)</math> is 1 or 2. This completes the proof.

Science:Math Exam Resources/Courses/MATH312/December 2009/Question 01 (b)

2014-07-11T07:37:56Z

VincentChan:

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Science:Math Exam Resources/Courses/MATH312/December 2009/Question 06 (b)

2014-07-09T05:15:11Z

VincentChan:

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Science:Math Exam Resources/Courses/MATH312/December 2009/Question 06 (b)/Solution 1

2014-07-09T05:14:49Z

VincentChan:

The stated problem is equivalent to showing that <math> \displaystyle 0 \equiv a^{4n+1} - a \mod{10}</math> which is equivalent to showing that 10 divides <math> \displaystyle a(a^{4n}-1)</math>.

To do this, we break this into cases. Since we are looking modulo 10, we only have to discuss the cases when <math> \displaystyle a \in \{0,1,2,3,4,5,6,7,8,9\}</math> since all numbers reduce to one of these modulo 10. We proceed as suggested by the hints.

'''Case 1: <math> \displaystyle a=0</math>'''. This case clearly satisfies <math> \displaystyle a \equiv a^{4n+1} \mod{10}</math> by a simple substitution. We suppose that ''a'' is positive.

'''Case 2: <math> \displaystyle 2 \mid a</math>'''. In this case, notice that clearly 2 divides <math> \displaystyle a(a^{4n}-1)</math> so it suffices to show that 5 divides <math> \displaystyle a^{4n}-1</math>. Notice that 5 does not divide ''a'' and so <math> \displaystyle \gcd(a,5) = 1</math>. Thus Euler's Theorem applies giving us that

<math> \displaystyle 1 \equiv a^{\phi(5)} \equiv a^{4} \mod{5}</math>.

Hence

<math> \displaystyle a^{4n}-1 \equiv (a^{4})^{n} - 1 \equiv 1^{n}-1 \equiv 0 \mod{5}</math>

Thus, 5 divides <math> \displaystyle a^{4n}-1</math> and so 10 divides <math> \displaystyle a(a^{4n}-1)</math>.

'''Case 3: <math> \displaystyle a=5</math>'''. In this case, notice that clearly 5 divides <math> \displaystyle a(a^{4n}-1)</math> so it suffices to show that 2 divides <math> \displaystyle a^{4n}-1</math>. This last number is even since <math> \displaystyle a^{4n} = 5^{4n}</math> is odd. Thus 10 divides <math> \displaystyle a(a^{4n}-1)</math>.

'''Case 4: <math> \displaystyle 2 \nmid a</math> and <math> \displaystyle a \neq 5</math>'''. In this case, <math> \displaystyle \gcd(a,10) = 1</math>. Thus Euler's Theorem applies giving us that

<math> \displaystyle 1 \equiv a^{\phi(10)} \equiv a^{\phi(2)\phi(5)} \equiv a^{(1)(4)} \equiv a^{4} \mod{10}</math>.

Hence

<math> \displaystyle a(a^{4n}-1) \equiv a((a^{4})^{n} - 1) \equiv a(1^{n}-1)\equiv 0 \mod{10}</math>

Thus 10 divides <math> \displaystyle a(a^{4n}-1)</math> completing the proof.

Science:Math Exam Resources/Courses/MATH312/December 2009/Question 06 (b)/Solution 1

2014-07-09T05:13:44Z

VincentChan:

The stated problem is equivalent to showing that <math> \displaystyle 0 \equiv a^{4n+1} - a \mod{10}</math> which is equivalent to showing that 10 divides <math> \displaystyle a(a^{4n}-1)</math>.

To do this, we break this into cases. Since we are looking modulo 10, we only have to discuss the cases when <math> \displaystyle a \in \{0,1,2,3,4,5,6,7,8,9\}</math> since all numbers reduce to one of these modulo 10. We proceed as suggested by the hints.

'''Case 1: <math> \displaystyle a=0</math>'''. This case clearly satisfies <math> \displaystyle a \equiv a^{4n+1} \mod{10}</math> by a simple substitution. We suppose that ''a'' is positive.

'''Case 2: <math> \displaystyle 2 \mid a</math>'''. In this case, notice that clearly 2 divides <math> \displaystyle a(a^{4n}-1)</math> so it suffices to show that 5 divides <math> \displaystyle a^{4n}-1</math>. Notice that 5 does not divide ''a'' and so <math> \displaystyle \gcd(a,5) = 1</math>. Thus Euler's Theorem applies giving us that

<math> \displaystyle 1 \equiv a^{\phi(5)} \equiv a^{4} \mod{5}</math>.

Hence

<math> \displaystyle a^{4n}-1 \equiv (a^{4})^{n} - 1 \equiv 1^{n}-1 \equiv 0 \mod{5}</math>

Thus, 5 divides <math> \displaystyle a^{4n}-1</math> and so 10 divides <math> \displaystyle a(a^{4n}-1)</math>.

'''Case 3: <math> \displaystyle a=5</math>'''. In this case, notice that clearly 5 divides <math> \displaystyle a(a^{4n}-1)</math> so it suffices to show that 2 divides <math> \displaystyle a^{4n}-1</math>. This last number is even since <math> \displaystyle a^{4n} = 5^{4n}</math> is odd and the difference of two odd numbers is even. Thus 10 divides <math> \displaystyle a(a^{4n}-1)</math>.

'''Case 4: <math> \displaystyle 2 \nmid a</math> and <math> \displaystyle a \neq 5</math>'''. In this case, <math> \displaystyle \gcd(a,10) = 1</math>. Thus Euler's Theorem applies giving us that

<math> \displaystyle 1 \equiv a^{\phi(10)} \equiv a^{\phi(2)\phi(5)} \equiv a^{(1)(4)} \equiv a^{4} \mod{10}</math>.

Hence

<math> \displaystyle a(a^{4n}-1) \equiv a((a^{4})^{n} - 1) \equiv a(1^{n}-1)\equiv 0 \mod{10}</math>

Thus 10 divides <math> \displaystyle a(a^{4n}-1)</math> completing the proof.

Science:Math Exam Resources/Courses/MATH312/December 2009/Question 06 (b)/Hint 1

2014-07-09T05:04:09Z

VincentChan:

The stated problem is equivalent to showing that 10 divides <math> \displaystyle a(a^{4n}-1)</math>.

Science:Math Exam Resources/Courses/MATH312/December 2009/Question 06 (a)

2014-07-09T05:03:20Z

VincentChan:

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[[Category:MER Tag Irrationality]]

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Science:Math Exam Resources/Courses/MATH312/December 2009/Question 05 (b)

2014-07-09T04:52:24Z

VincentChan:

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[[Category:MER Tag Continued fraction]]

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Science:Math Exam Resources/Courses/MATH312/December 2009/Question 05 (a)

2014-07-09T04:36:51Z

VincentChan:

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[[Category:MER Tag Multiplicative functions]]
[[Category:MER Tag Fundamental theorem of arithmetic]]

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Science:Math Exam Resources/Courses/MATH312/December 2009/Question 05 (a)/Hint 1

2014-07-09T04:34:46Z

VincentChan:

Recall that the Mobius function is the function that evaluates to 1 if ''n'' is square-free and has an even number of prime factors, -1 if ''n'' is square-free and has an odd number of factors, and 0 if ''n'' has a square factor.

First show/recall that the Mobius function is multiplicative.

Science:Math Exam Resources/Courses/MATH312/December 2009/Question 04 (b)

2014-07-07T18:17:46Z

VincentChan:

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[[Category:MER Tag Wilson's theorem]]

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Science:Math Exam Resources/Courses/MATH312/December 2009/Question 04 (b)/Solution 1

2014-07-07T18:17:10Z

VincentChan:

We begin by multiplying the product by <math> \displaystyle (-1)^{p-1} \equiv 1 \mod{p}</math>. This gives

<math> \displaystyle
1^{2} \cdot 3^{2} \cdot 5^{2} \cdot ... \cdot (p-2)^{2} \equiv
1^{2} \cdot 3^{2} \cdot 5^{2} \cdot ... \cdot (p-2)^{2} \cdot (-1)^{p-1} \mod{p}</math>

Now, rewrite each square as <math> \displaystyle a^{2} = a\cdot a</math> and for each value of ''a'', distribute a power of -1. Notice that we have <math> \displaystyle (p-1)/2</math> terms in our product. So we need to take a factor of <math> \displaystyle (-1)^{(p-1)/2}</math> to accomplish this. This gives

<math>
\begin{align}
1^{2} \cdot &3^{2} \cdot 5^{2} \cdot ... \cdot (p-2)^{2} \cdot (-1)^{p-1} \\
&\equiv (1)(-1)(3)(-3)(5)(-5) ... (p-2)(-(p-2))\cdot (-1)^{(p-1)/2} \mod{p}
\end{align}
</math>

Now for each negative term, we add ''p'' to get.

<math>
\begin{align}
(1)(-1)&(3)(-3)(5)(-5) ... (p-2)(-(p-2))\cdot (-1)^{(p-1)/2} \\
&\equiv (1)(p-1)(3)(p-3)(5)(p-5)...(p-2)(2) \cdot (-1)^{(p-1)/2} \mod{p}
\end{align}
</math>

After rearranging the right hand side above, we see that

<math>
\begin{align}
(1)(p-1)&(3)(p-3)(5)(p-5)...(p-2)(2) \cdot (-1)^{(p-1)/2}\\
&\equiv (p-1)!\cdot (-1)^{(p-1)/2} \mod{p}
\end{align}
</math>

By Wilson's Theorem and combining the above, we have

<math>
\begin{align}
1^{2} \cdot &3^{2} \cdot 5^{2} \cdot ... \cdot (p-2)^{2} \\
&\equiv (p-1)!\cdot (-1)^{(p-1)/2} \mod{p}\\
&\equiv (-1)\cdot (-1)^{(p-1)/2} \mod{p} \\
&\equiv (-1)^{(p+1)/2} \mod{p}
\end{align}
</math>

and this was what we wanted to show.

Science:Math Exam Resources/Courses/MATH312/December 2009/Question 04 (a)

2014-07-07T18:13:16Z

VincentChan:

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[[Category:MER Tag Affine cryptosystem]]
[[Category:MER Tag Euclidean algorithm]]

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Science:Math Exam Resources/Courses/MATH312/December 2009/Question 03 (b)

2014-07-07T18:10:28Z

VincentChan:

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[[Category:MER Tag Modular arithmetic]]

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Science:Math Exam Resources/Courses/MATH312/December 2009/Question 03 (a)

2014-07-07T18:06:35Z

VincentChan:

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[[Category:MER Tag Euler's theorem]]
[[Category:MER Tag Chinese remainder theorem]]
[[Category:MER Tag Euclidean algorithm]]

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Science:Math Exam Resources/Courses/MATH312/December 2009/Question 02 (b)

2014-07-07T18:02:00Z

VincentChan:

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[[Category:MER Tag Carmichael number]]

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Science:Math Exam Resources/Courses/MATH312/December 2009/Question 02 (a)

2014-07-07T17:21:15Z

VincentChan:

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[[Category:MER Tag Miller's test]]

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Science:Math Exam Resources/Courses/MATH312/December 2009/Question 02 (a)/Solution 1

2014-07-07T17:20:26Z

VincentChan:

Recall that 2 passes Miller's test if

<math> \displaystyle 2^{d} \equiv 1 \mod{209}</math>

or for some value ''r''

<math> \displaystyle 2^{2^{r}d} \equiv -1 \mod{209}</math>

where <math> \displaystyle 209-1 = 208 = 2^{s}d</math> and <math> \displaystyle 0 \leq r \leq s-1</math>. If 2 fails the test, then 209 is not prime. For us, we see that <math> \displaystyle 208 = 2^{3}(26) = 2^{4}\cdot13</math> and so ''d'' is 13 and ''r'' is 4. We compute manually.

<math> \displaystyle
2^{13} \equiv 2^{8}\cdot 2^{5} \equiv (256)(32) \equiv (47)(32) \equiv 1504 \equiv 41 \not\equiv \pm 1 \mod {209}
</math>

and for the powers of 2,

<math> \displaystyle
2^{2\cdot 13} \equiv 41^{2} \equiv 1681 \equiv 9 \not\equiv \pm 1 \mod {209}
</math>

<math> \displaystyle
2^{4\cdot 13} \equiv 9^{2} \equiv 81 \not\equiv \pm 1 \mod {209}
</math>

<math> \displaystyle
2^{8\cdot 13} \equiv 81^{2} \equiv (2 \cdot 41-1)^2 \equiv 4 \cdot 41^2-4\cdot 41 + 1 \equiv 4 \cdot 9 - 164 + 1 \equiv 36+45+1 \equiv 82 \not\equiv \pm 1 \mod {209}
</math>

<math> \displaystyle
2^{16\cdot 13} \equiv 82^{2} \equiv (81+1)^{2} \equiv 81^{2} + 162 + 1 \equiv 82 + 162 + 1 \equiv 36 \not\equiv \pm 1 \mod {209}
</math>

and so the number 209 is not prime and 2 fails Miller's test. Thus 209 is composite. In fact <math> \displaystyle 209 = 11\cdot 19</math>.

Science:Math Exam Resources/Courses/MATH312/December 2009/Question 01 (a)

2014-07-07T16:42:24Z

VincentChan:

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[[Category:MER Tag Divisibility]]

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Science:Math Exam Resources/Courses/MATH312/December 2010/Question 05

2014-07-07T16:40:03Z

VincentChan:

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[[Category:MER Tag Divisibility]]

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Science:Math Exam Resources/Courses/MATH312/December 2010/Question 05

2014-07-07T16:38:04Z

VincentChan:

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[[Category:MER Tag Divisibility]]

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Science:Math Exam Resources/Courses/MATH312/December 2010/Question 05/Solution 1

2014-07-07T16:36:38Z

VincentChan:

We begin by noting that

<math> \displaystyle 2\cdot 11 = 22 = \phi(n) = n\prod_{p \mid n}\left(1 - \frac{1}{p}\right) = n\prod_{p \mid n}\frac{p-1}{p} = \prod_{p_{i}^{\alpha_{i}}\parallel n}p_{i}^{\alpha_{i}-1}(p-1)</math>

where above we used the Fundamental Theorem of Arithmetic to write

<math>
\displaystyle n = \prod_{p_{i}^{\alpha_{i}}\parallel n} p_{i}^{\alpha_{i}}.
</math>

Our ''main equation'' of discussion will be

<math> \displaystyle 2\cdot 11 = \prod_{p_{i}^{\alpha_{i}}\parallel n}p_{i}^{\alpha_{i}-1}(p_{i}-1)</math>.

We proceed in cases. First we discuss the powers of 2 in ''n''.

'''Case 1: <math> \displaystyle 8 \mid n </math>'''. In this case, notice that 4 divides the right hand side of the main equation but only 2 divides the left hand side. This is a contradiction.

'''Case 2: <math> \displaystyle 4 \parallel n </math>'''. In this case, notice that 2 divides the right hand side of the main equation and 2 divides the left hand side. However, if another odd prime divides ''n'', then the right hand side is divisible by an extra power of 2 which is a contradiction. The remaining subcase in this case is if no odd primes divde ''n'', that is, <math> \displaystyle n = 4 </math>. But since also <math> \displaystyle \phi(4) = 2 \neq 22 </math> this case cannot occur.

'''Case 3: <math> \displaystyle 2 \parallel n </math> or ''n'' is odd '''. In either of these cases, since <math>\displaystyle \phi</math> is multiplicative and as <math> \displaystyle \phi(2) = 1</math>, we have that either of these cases can occur, that is, if <math>\displaystyle \phi(n) = 22</math>, then <math>\displaystyle \phi(2n) = 22.</math> Therefore, without loss of generality, we suppose 2 does not divide <math>\displaystyle n</math>.

Now we deal with the odd primes. Notice that for each odd prime, we get a factor of 2 on the right hand side of the main equation. Thus only exactly one odd prime can occur. Thus <math> \displaystyle n = p^{\alpha} </math>. Since only 2 and 11 occur as prime factors on the left hand side of the equation, we know that we must have that <math> \displaystyle \alpha \leq 2 </math> (the right hand side of the main equation has an <math> \displaystyle p^{\alpha-1} </math> term so the factor could be 2). If <math> \displaystyle \alpha =2 </math>, then the prime ''p'' in question must be 11 since ''p'' divides the right hand side. So <math> \displaystyle n = 11^{2} </math> but this is inadmissable since <math> \displaystyle \phi(11^{2}) \neq 22</math>. Thus ''n'' is a prime. For primes, we have that <math> \displaystyle \phi(p) = p-1 </math>. This value we know is 22 and so <math> \displaystyle n = 23 </math>.

Thus combining the two discussions shows us that either <math> \displaystyle n = 23 </math> or <math> \displaystyle n = 46 </math> completing the question.