UBC Wiki - User contributions [en]

Science:Math Exam Resources/Courses/MATH 180/December 2017/Question 06/Solution 1

2024-12-18T08:13:34Z

NicholasHu:

Observe that if <math>V(r)= \dfrac{A}{e^{Br}}-\dfrac{C}{r^6}</math> approaches positive/negative infinity, then either <math>\dfrac{A}{e^{Br}}</math> or <math>-\dfrac{C}{r^6}</math> approaches positive/negative infinity, as <math>r\to a</math>, for some <math>a</math>. In other words, if <math>r=a</math> is a vertical asymptote of <math>V</math>, then <math>r=a</math> is a vertical asymptote of <math>\dfrac{A}{e^{Br}}</math> or of <math>-\dfrac{C}{r^6}</math>.

Recall that the exponential function can never be zero, so <math>\dfrac{A}{e^{Br}}</math> is finite for any <math>r</math>. On the other hand, the rational function <math>-\dfrac{C}{r^6}</math> is not defined at <math>r=0</math>, which makes <math>r=0</math> a candidate for a vertical asymptote of <math>-\dfrac{C}{r^6}</math> and hence of <math>V</math>.
In fact, <math display="block">\lim\limits_{r\to0^+}V(r)=\lim\limits_{r\to0^+}\left(\dfrac{A}{e^{Br}}-\dfrac{C}{r^6}\right)=A-\lim\limits_{r\to0^+}\dfrac{C}{r^6}=-\infty.</math> So <math display="inline">r=0</math> is the only vertical asymptote.

For horizontal asymptotes, since <math display="inline">V(r)</math> is only defined for <math display="inline">r>0</math>, we find that <math display="block">\lim\limits_{r\to+\infty}V(r)=\lim\limits_{r\to +\infty}\left(\dfrac{A}{e^{Br}}-\dfrac{C}{r^6}\right)=0-0=0.</math> So the only horizontal asymptote is <math display="inline">V=0</math>.

'''Answer:''' the vertical asymptote is <math display="inline">{\color{blue}r=0}</math> and the horizontal asymptote is <math display="inline">{\color{blue}V=0}</math>.

Science:Math Exam Resources/Courses/MATH100 A/December 2023/Question 25/Solution 1

2024-12-10T09:39:46Z

NicholasHu:

We compute the partial derivatives and set them equal to 0:

<math>
\begin{align}
\frac{\partial f}{\partial x}(x,y) &= 2 - 4x - y^2 = 0\\
\frac{\partial f}{\partial y}(x,y) &= 0 - 0 - 2xy = 0.
\end{align}
</math>

We now solve for <math>x</math> and <math>y</math>. Let's start with the second equation. If the point <math>(x,y)</math> satisfies <math>-2xy = 0</math>, then either <math>x=0</math> or <math>y=0</math>.

# Let us take the first option and suppose <math>x = 0</math>. Then the first equation then becomes <math>2 - y^2 = 0</math>, for which the solutions are <math> y = \pm \sqrt{2}</math>. Thus, the points <math>(0, -\sqrt{2})</math> and <math>(0, \sqrt{2})</math> are critical points of <math>f</math>.
# We shouldn't forget the second option, <math>y = 0</math>. In this case, the first equation becomes <math>2-4x = 0</math>, for which the solution is <math>x = \tfrac{1}{2}</math>. Thus, another critical point of <math>f</math> is <math>(\tfrac{1}{2},0)</math>.

The final list of critical points is <math>(0, -\sqrt{2}), (0, \sqrt{2}), (\tfrac{1}{2}, 0)</math>.

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 18/Solution 1

2024-04-24T23:02:18Z

NicholasHu:

We write these three vectors as the column vectors of a matrix. The determinant of a matrix is zero if its column vectors are dependent:

:<math>
\begin{vmatrix}
a & 1 & 4 \\
0 & 2 & 1 \\
1& 1 & 3
\end{vmatrix} = 0.
</math>

If we expand the first column to calculate the determinant, we have

:<math> a \cdot \begin{vmatrix}

2 & 1 \\1 & 3
\end{vmatrix} + 1 \cdot \begin{vmatrix}
1 & 4 \\
2 & 1 \\

\end{vmatrix} = 0.</math>

This implies <math> a(2\times 3-1\times 1) + 1(1\times 1- 2\times 4) = 5a - 7 = 0.</math> Thus, <math>\color{blue}a= \frac{7}{5}.</math>

Science talk:Math Exam Resources/Courses/MATH152/April 2022/Question B4 (c)

2024-04-24T22:58:25Z

NicholasHu: Talk page autocreated when first thread was posted

Thread:Science talk:Math Exam Resources/Courses/MATH152/April 2022/Question B4 (c)/QBS

2024-04-24T22:58:25Z

NicholasHu: New thread: QBS

Sign error. Please fix.

Science:Math Exam Resources/Courses/MATH152/April 2022/Question B4 (c)

2024-04-24T22:58:02Z

NicholasHu:

[[Category:MER QGQ flag]][[Category:MER QGH flag]][[Category:MER QBS flag]][[Category:MER CT flag]]







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH152/April 2015/Question A 09/Hint 1

2024-04-24T22:50:23Z

NicholasHu:

Recall that an eigenvector of a matrix <math> \mathbf{A} </math> with corresponsing eigenvalue <math> \mathbf{\lambda} </math> is a non-zero vector <math> \mathbf{v} </math> that satisfies <math> \mathbf{A}\mathbf{v}=\mathbf{\lambda}\mathbf{v}.</math>

Science:Math Exam Resources/Courses/MATH101/April 2011/Question 01 (b)/Solution 1

2024-04-22T23:47:53Z

NicholasHu:

Using Hint 2, we may immediately rewrite <math>f(t)</math> as

:<math>f(t)=-\int_2^{t^4} \sqrt{1+x^3} \, dx.</math>

We would like to use Hint 1, but it would only be applicable if the upper limit of integration were <math>t</math>, which in this case is not true. So we define

:<math> g(u) = -\int_2^u \sqrt{1 + x^3} \, dx </math>.

From Hint 1, we see immediately that <math> g'(u) = -\sqrt{1 + u^3}</math>. We also see that <math> f(t) = g(t^4)</math>. Using the chain rule, we can then compute that

:<math>f'(t) = g'(t^4) \cdot (t^4)' = -\sqrt{1 + (t^{4})^3} \cdot 4t^3 = -4t^3\sqrt{1 + t^{12}}.</math>

Science:Math Exam Resources/Courses/MATH101/April 2011/Question 01 (b)/Hint 2

2024-04-22T23:42:30Z

NicholasHu:

Recall that

:<math> \int_b^a f(t) \, dt = -\int_a^b f(t) \, dt.</math>

Science:Math Exam Resources/Courses/MATH101/April 2011/Question 01 (b)/Hint 2

2024-04-22T23:42:02Z

NicholasHu:

Recall that

:<math> \int_a^b f(t) \, dt = -\int_b^a f(t) \, dt</math>

Science:Math Exam Resources/Courses/MATH101/April 2011/Question 01 (b)/Hint 1

2024-04-22T23:41:35Z

NicholasHu:

Remember that the Fundamental Theorem of Calculus says that

:<math> \frac{d}{dx}\int_a^x f(t) \, dt = f(x) </math>.

Science:Math Exam Resources/Courses/MATH101/April 2017/Question 11 (a)/Solution 1

2024-04-21T05:37:44Z

NicholasHu:

We can easily see that when <math>x=2</math>,

:<math>\sum_{n=0}^\infty A_n(x - 1)^n=\sum_{n=0}^\infty A_n</math>.

Since the radius of convergence is given as <math>R = \frac 32</math> for the power series <math>\sum_{n=0}^\infty A_n(x - 1)^n</math>, the series converges when <math>|x-1|<\frac 32</math>.

Therefore, <math>|2-1|=1<\frac 32</math> implies the convergence of the series <math>\sum_{n=0}^\infty A_n</math>.

'''Answer:''' <math>\color{blue}\sum_{n=0}^\infty A_n \ \text{converges}.</math>

Science talk:Math Exam Resources/Courses/MATH100 A/December 2023/Question 25

2024-04-21T05:33:07Z

NicholasHu: Talk page autocreated when first thread was posted

Thread:Science talk:Math Exam Resources/Courses/MATH100 A/December 2023/Question 25/QBS

2024-04-21T05:33:07Z

NicholasHu: New thread: QBS

Solution is incorrect; please revise.

Science:Math Exam Resources/Courses/MATH100 A/December 2023/Question 25

2024-04-21T05:30:59Z

NicholasHu:

Science:Math Exam Resources/Courses/MATH101/April 2017/Question 04 (b)/Solution 1

2024-04-18T21:25:05Z

NicholasHu:

Using the information given above all we need to find is <math>K</math> and replace the values of <math>a = 0,\, b= \pi,\, n = 6</math> in the formula for the error <math>\frac{K(b - a)^3}{24n^2}</math>, as the difference between the 2 numbers will, at most, be the error.

Thus we must first find the 2nd derivative of <math> f(x) = x\sin x</math>:

<math>f ^{\prime}(x) = \sin x + x\cos x</math>

<math> f ^ {\prime \prime}(x) = \cos x + \cos x -x\sin x</math>.

We must now bound the derivative. Using the inequatility <math>|x+y| \leq |x|+|y|</math>, we have:

<math>|2\cos x -x\sin x| \leq 2|\cos x| +|x||\sin x|</math>. Furthermore we know that <math> |\cos x|\leq 1</math>, <math>|\sin x| \leq 1</math> and, in our domain, <math> |x| \leq \pi</math>, which makes us conclude that <math>|f ^ {\prime \prime}(x)| \leq \pi +2</math> on <math> [0,\pi]</math>. (Note that this does not necessarily mean the maximum of <math>|f''(x)|</math> is <math>\pi+2</math>. The value <math>\pi + 2</math> is only an ''upper bound''; other bounds are possible.)

Thus the difference is given by <math>\frac{(\pi+2)(\pi-0)^3}{24 (6)^ 2}</math>

'''Answer:''' <math>\color{blue}\frac{(\pi+2)\pi^ 3}{24 (6)^ 2}</math> (note that this is only one possible answer; see the note above)

Science:Math Exam Resources/Courses/MATH101/April 2016/Question 02 (c)/Solution 1

2024-04-15T23:17:26Z

NicholasHu:

First consider when <math>q\geq 0</math>.

For <math>n\geq 3</math>, we have <math>\log n\geq 1</math>, which implies <math>(\log n)^q \geq 1</math> and <math>\frac{(\log n)^q}{n}> \frac{1}n</math>.

By the ''comparison test'', we have

<math>\sum_{n=1}^\infty \frac{(\log n)^q}{n}
\geq\sum_{n=3}^\infty \frac{(\log n)^q}{n}
\geq\sum_{n=3}^\infty \frac{1}{n} = +\infty.
</math>

Therefore, for <math>q\geq 0</math>, the given series diverges.

On the other hand, for <math>q<0</math>, note that <math>\frac{(\log x)^q}{x}</math> is continuous, positive, and decreasing on <math>[2,\infty)</math>.

In this case, we'll use the integral test.

For <math>q<0</math> but <math>q\neq -1</math>, we have

<math>\begin{align}
\int_2^{\infty} \frac{(\log x)^q}{x} dx
&= \int_{\log 2}^{\infty} y^q dy
= \lim_{R\to \infty} \int_{\log 2}^{R} y^q dy\\
&= \lim_{R\to \infty} \frac {1}{q+1} y^{q+1}\bigg|^{R}_{\log 2}
=\lim_{R\to \infty} \frac {1}{q+1}\left( R^{q+1} - (\log 2)^{q+1}\right) <+\infty,
\end{align}</math>

provided that <math>q< -1</math>,

Here, the second equality is obtained from the substitution <math> y= \log x </math>.

Using <math> \log 1 = 0 </math>,

<math>
\sum_{n=1}^{\infty} \frac{(\log n)^q}{n} = \sum_{n=2}^{\infty} \frac{(\log n)^q}{n}.
</math>

Then, by the ''integral test'', the given series is convergent for <math>q< -1</math>; ,

For <math>q=-1</math>, the integral can be computed by

<math>\begin{align}
\int_2^{\infty} \frac{1}{x \log x } dx
&= \lim_{R\to \infty} \int_{\log 2}^{R} \frac 1 y dy\\
&= \lim_{R\to \infty} \ln y \bigg|^{R}_{\log 2}
=\lim_{R\to \infty} \left( \ln R - \log 2 \right) =+\infty,
\end{align}</math>.

Therefore, again by the ''integral test'', the given series diverges when <math>q= -1</math>.

To sum, the answer is L.

Science:Math Exam Resources/Courses/MATH101/April 2015/Question 04 (b)/Solution 1

2024-04-15T23:11:32Z

NicholasHu:

Using partial fraction decomposition, we can write the integrand as

:<math>\frac{1}{x^4 + x^2} = \frac{1}{x^2(x^2 + 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2 + 1}</math>

for some constants <math>A, B, C, D</math>. We find that <math>A = 0, B = 1, C = 0, D = -1</math>, which implies that

:<math> \int \frac{1}{x^4 + x^2} \, dx = \int \frac{1}{x^2} \, dx -\int \frac 1{1+x^2} \, dx.</math>

The first integral on the right-hand side of the equation can be easily evaluated:

:<math>\int \frac{1}{x^2} \, dx = -\frac 1x + C.</math>

On the other hand, to compute the second integral, we can use the substitution <math> x=\tan u</math>.
Then <math>1+x^2 = 1+ \tan^2 u = \sec^2 u </math> and <math> dx = \sec^2 u \, du </math>, which gives

:<math>
\int \frac 1{1+x^2} \, dx = \int \frac 1{\sec^2 u} \sec^2 u \, du = \int 1 \, du = u + C = \arctan x + C.</math>

Combining these, we obtain

:<math> \int \frac{1}{x^4 + x^2} dx = \color{blue} -\frac 1x -\arctan x + C. </math>

Science:Math Exam Resources/Courses/MATH101/April 2015/Question 04 (b)/Solution 1

2024-04-15T23:10:40Z

NicholasHu:

Using partial fraction decomposition, we can write the integrand as

:<math>\frac{1}{x^4 + x^2} = \frac{1}{x^2(x^2 + 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2 + 1}</math>

for some constants <math>A, B, C, D</math>. We find that <math>A = 0, B = 1, C = 0, D = -1</math>, which implies that

:<math> \int \frac{1}{x^4 + x^2} \, dx = \int \frac{1}{x^2} \, dx -\int \frac 1{1+x^2} \, dx.</math>

The first integral on the right-hand side of the equation can be easily evaluated:

:<math>\int \frac{1}{x^2} \, dx = -\frac 1x + C.</math>

On the other hand, to compute the second integral, we can use the substitution <math> x=\tan u</math>.
Then <math>1+x^2 = 1+ \tan^2 u = \sec^2 u </math> and <math> dx = \sec^2 u \, du </math>, which gives

:<math>
\int \frac 1{1+x^2} \, dx = \int \frac 1{\sec^2 u} \sec^2 u \, du = \int 1 \, du = u + C = \arctan x + C</math>.

Combining these, we obtain

:<math> \int \frac{1}{x^4 + x^2} dx = \color{blue} -\frac 1x -\arctan x + C. </math>

Science:Math Exam Resources/Courses/MATH100/December 2016/Question 13 (a)/Solution 2

2023-12-17T06:13:48Z

NicholasHu: Created page with "We know (or can calculate) that the Taylor expansion of <math>\sin(x)</math> centred at <math>x = 0</math> is :<math> \sin(x) = x - \frac{x^3}{3!} + \cdots. </math> Hence the Taylor expansion of <math>f(x) = (x+1) \sin(x)</math> centred at <math>x = 0</math> is :<math> (x+1)\sin(x) = (x+1)\left(x - \frac{x^3}{3!} + \cdots\right) = x^2 - \frac{x^4}{3!} + x - \frac{x^3}{3!} + \cdots. </math> Discarding the terms of degree higher than 3, we obtain the third degree Taylor po..."

We know (or can calculate) that the Taylor expansion of <math>\sin(x)</math> centred at <math>x = 0</math> is
:<math>
\sin(x) = x - \frac{x^3}{3!} + \cdots.
</math>
Hence the Taylor expansion of <math>f(x) = (x+1) \sin(x)</math> centred at <math>x = 0</math> is
:<math>
(x+1)\sin(x) = (x+1)\left(x - \frac{x^3}{3!} + \cdots\right) = x^2 - \frac{x^4}{3!} + x - \frac{x^3}{3!} + \cdots.
</math>
Discarding the terms of degree higher than 3, we obtain the third degree Taylor polynomial
:<math>
\color{blue}
T_3(x) = x + x^2 - \frac{x^3}{6}.
</math>

Science:Math Exam Resources/Courses/MATH100/December 2015/Question 08 (b)/Solution 1

2023-12-15T07:03:33Z

NicholasHu:

Let <math>r</math> be the radius of the circle at the top of the water and <math>h</math> be the height of the water.

Then, using the picture in part (a) with a property of similar triangles gives the ratio between <math>r</math> and <math>h</math> as

<math> \frac rh = \frac {10}{16} \implies r = \frac 58 h.</math>

On the other hand, by the volume formula for a cone, the volume <math>V</math> of water can be written as

<math>
V = \frac 13 \pi r^2 h = \frac 13 \pi \cdot \left(\frac 58 h\right)^2\cdot h =\frac 13 \pi\cdot \left(\frac 58 \right)^2 \cdot h^3.
</math>

Therefore, by differentiating with respect to <math>t</math> on both sides, we obtain

<math>
\frac {dV}{dt} = \frac 13 \pi\cdot \left(\frac 58 \right)^2 \cdot 3h^2 \cdot \frac {dh}{dt} = \pi \cdot \left(\frac 58 \right)^2\cdot h^2 \cdot \frac {dh}{dt} .
</math>

Since the volume changes at a rate of <math>2\,\mathrm{m}^3/\mathrm{min}</math>, the rate of change of the height of the water when the height is <math>10\,\mathrm{m}</math> is

<math>
2\,\mathrm{m}^3/\mathrm{min} = \pi \left(\frac 58 \right)^2\cdot (10\,\mathrm{m})^2 \cdot \frac {dh}{dt}\bigg|_{h=10} \implies \frac {dh}{dt}\bigg|_{h=10} = \frac {2 \cdot 8^2}{5^2 \cdot 10^2 \pi }\,\mathrm{m}/\mathrm{min}=\color{blue}{\frac {2^5}{5^4 \pi } \,\mathrm{m}/\mathrm{min}}.
</math>

Science:Math Exam Resources/Courses/MATH220/April 2005/Question 05/Solution 1

2023-12-14T04:18:41Z

NicholasHu:

In the base case when <math> n=1 </math>, we have <math> |a_1|\leq |a_1| </math>, so the statement is true.

For the inductive step, we assume that <math> \left|\sum_{i=1}^n a_i \right|\leq \sum_{i=1}^n|a_i| </math> holds for some (fixed) <math> n \in \mathbb{N}</math>. Then we show that it also holds with <math>n</math> replaced by <math>n+1</math>:

:<math>
\begin{align}
\left|\sum_{i=1}^{n+1}a_i\right|&=\left|\sum_{i=1}^{n}a_i+a_{n+1}\right|\\
&\leq \left|\sum_{i=1}^{n}a_i\right|+|a_{n+1}|\\
& \leq \sum_{i=1}^n|a_i|+|a_{n+1}|\\
&= \sum_{i=1}^{n+1}|a_i|.
\end{align}</math>

The first inequality is true by the triangle inequality and the second by the induction hypothesis. This finishes the inductive step and so the result is true for all <math> n\in \mathbb{N} </math> by induction.

Science:Math Exam Resources/Courses/MATH220/April 2005/Question 05/Hint 1

2023-12-14T04:14:27Z

NicholasHu:

Use the fact that <math> \sum_{i=1}^na_i=\sum_{i=1}^{n-1}a_i+a_n </math>.

Science:Math Exam Resources/Courses/MATH307/December 2008/Question 04/Solution 1

2023-12-12T03:35:21Z

NicholasHu:

To begin with, we note that the columns of ''A'' are already orthogonal,

:<math> \begin{bmatrix}
0 \\ 1 \\ 0 \\ -1 \\ 0
\end{bmatrix} \cdot \begin{bmatrix}
1 \\ 0 \\ -1 \\ 0 \\ 1
\end{bmatrix} = 0. </math>

So we can simply put the normalized columns of ''A'' into the matrix ''Q'':

:<math> Q = \begin{bmatrix}
0 & \frac1{\sqrt 3} \\
\frac1{\sqrt 2} & 0 \\
0 & -\frac1{\sqrt 3} \\
-\frac1{\sqrt 2} & 0 \\
0 & \frac1{\sqrt 3}
\end{bmatrix}. </math>

As a last step, we calculate ''R = Q<sup>T</sup>A'' to find

:<math> R = \begin{bmatrix}
\sqrt 2 & 0 \\ 0 & \sqrt 3
\end{bmatrix}. </math>

(Note that the question asks for ''the'' QR decomposition, not ''a'' QR decomposition. This is the ''unique'' QR decomposition of ''A'' such that ''Q'' has orthonormal columns and ''R'' is square upper triangular with positive diagonal entries.)

Science:Math Exam Resources/Courses/MATH307/December 2008/Question 04/Hint 3

2023-12-12T03:22:34Z

NicholasHu:

As a last step, since ''A = QR'' and ''Q'' has orthonormal columns, we can find ''R'' by computing ''R = Q<sup>T</sup>A''.

Science:Math Exam Resources/Courses/MATH307/December 2008/Question 07/Solution 1

2023-12-12T03:19:56Z

NicholasHu:

The infinity norm of a matrix is the infinity norm of the vector whose entries are the 1-norm of each row, so it will be

:<math>\displaystyle \left \| A \right \|_{\infty} = \max\{2+1+0, 1+2+1, 0+1+2\} = 4. </math>

The 2-norm of a matrix is the largest of the absolute values of the eigenvalues. So we need to calculate the eigenvalues of ''A'' as the roots of the characteristic polynomial

:<math>\begin{align}
\det(A-\lambda I) &= \det\begin{vmatrix}
-2-\lambda & 1 & 0 \\
1 & -2-\lambda & 1 \\
0 & 1 & -2-\lambda \end{vmatrix} \\
&= (-2-\lambda) \det\begin{vmatrix}
-2-\lambda & 1 \\
1 & -2-\lambda \end{vmatrix} - (1)\det\begin{vmatrix}
1 & 0 \\
1 & -2-\lambda \end{vmatrix} \\
&= (-2-\lambda)((-2-\lambda)(-2-\lambda)-1)-(-2-\lambda) \\
&= (-2-\lambda)(\lambda^2+4\lambda+2)
\end{align}</math>

Hence, the eigenvalues of ''A'' are <math>\displaystyle \lambda_1 = -2</math> and <math>\lambda_{2,3} = \frac{-4\pm\sqrt{16-8}}2 = -2\pm\sqrt 2</math>. The 2-norm of ''A'' is the largest of the absolute values of these and therefore <math>\left \| A \right \|_{2 } = |{-2}-{\sqrt 2}| = 2+\sqrt{2}</math>.

Science:Math Exam Resources/Courses/MATH220/December 2009/Question 05 (b)/Solution 1

2023-08-15T09:48:28Z

NicholasHu:

Let <math> \displaystyle S(n) </math> be the statement that

:<math>
5^{n} + 2(11^{n})
</math>

is a multiple of 3. We prove this statement is true for all nonnegative integers <math>n</math> using mathematical induction. Notice that <math> \displaystyle S(0) </math> is true since

:<math>
5^{0} + 2(11^{0}) = 1 + 2(1) = 3 = 3(1).
</math>

Now, we assume that <math> \displaystyle S(k) </math> is true for some integer <math>k \geq 0</math> and show that <math> \displaystyle S(k+1) </math> is true. When <math> \displaystyle n = k+1 </math>, we can start with the induction hypothesis

:<math>
5^{k} + 2(11^{k}) = 3m.
</math>

for some integer <math>m</math>. The induction hypothesis is equivalent to <math>5^k = 3m-2(11^{k})</math>. To see that <math> \displaystyle S(k+1) </math> is true, we then notice that

:<math>
\begin{align}
5^{k+1} + 2(11^{k+1}) &= 5(5^{k}) + 2(11^{k+1}) \\
&= 5(3m - 2(11^{k})) + 2(11^{k+1}) \\
&= 15m - 10(11^{k}) + 22(11^{k}) \\
&= 15m + 12(11^{k}) \\
&= 3(5m + 4(11^{k})),
\end{align}
</math>

which shows that <math> \displaystyle S(k+1) </math> is true. Hence <math> \displaystyle S(n) </math> is true for all nonnegative integers <math>n</math> by the principle of mathematical induction.

Science:Math Exam Resources/Courses/MATH220/December 2009/Question 05 (b)/Hint 1

2023-08-15T09:23:20Z

NicholasHu:

The proof writes itself after you set it up correctly. Remember that to prove a statement <math>\displaystyle S(n) </math> is true for all nonnegative integers <math>n</math>, you must do three things:

# Prove that <math>\displaystyle S(0) </math> is true.
# Assume that <math>\displaystyle S(k) </math> is true for some integer <math>k \geq 0</math>.
# Show that <math>\displaystyle S(k+1) </math> is true based on the assumption above.

Science:Math Exam Resources/Courses/MATH221/December 2009/Question 09/Solution 1

2023-06-28T06:38:06Z

NicholasHu:

The coefficients <math>a, b, c</math> of the least squares fit are given by
:<math>
X^T X \begin{bmatrix} a \\ b \\ c \end{bmatrix} = X^T y,
</math>
where
:<math>
X =
\begin{bmatrix}
x_1^0 & x_1^1 & x_1^2 \\
x_2^0 & x_2^1 & x_2^2 \\
x_3^0 & x_3^1 & x_3^2 \\
x_4^0 & x_4^1 & x_4^2 \\
\end{bmatrix}
=
\begin{bmatrix}
0^0 & 0^1 & 0^2 \\
1^0 & 1^1 & 1^2 \\
2^0 & 2^1 & 2^2 \\
3^0 & 3^1 & 3^2 \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 0 \\
1 & 1 & 1 \\
1 & 2 & 4 \\
1 & 3 & 9 \\
\end{bmatrix}
</math>
and
:<math>
y =
\begin{bmatrix}
y_1 \\ y_2 \\ y_3 \\ y_4
\end{bmatrix} =
\begin{bmatrix}
1 \\ 0 \\ 5 \\ 6
\end{bmatrix}.
</math>

Now
:<math>
X^T X =
\begin{bmatrix}
98 & 36 & 14 \\
36 & 14 & 6 \\
14 & 6 & 4
\end{bmatrix}
</math>
and
:<math>
X^T y =
\begin{bmatrix}
74 \\ 28 \\ 12
\end{bmatrix}
</math>.

Solving the resulting linear system, we find that <math> a = \frac{1}{2}, b = \frac{1}{2}, c = \frac{1}{2}</math>, so the least squares fit is
<math>\color{blue} y = \frac{1}{2} + \frac{1}{2} x + \frac{1}{2}x^2</math>.

Science:Math Exam Resources/Courses/MATH101/April 2018/Question 11 (ii)/Hint 1

2023-04-25T23:39:07Z

NicholasHu:

For the region below a curve <math>y=f(x)</math> and above a curve <math>y=g(x)</math> on the interval <math>[a,b]</math>, the <math>y</math>-coordinate of the centroid is given by <math>\bar{y}=\frac{1}{A}\int_a^b \left[\frac{f(x) + g(x)}{2}\right][f(x) - g(x)]\;dx,</math> where <math>A</math> is the area of the region (this formula can be found in the [https://personal.math.ubc.ca/~CLP/CLP2/clp_2_ic/subsection-32.html CLP-2 Integral Calculus textbook]).

Science:Math Exam Resources/Courses/MATH101/April 2018/Question 11 (ii)/Solution 1

2023-04-25T23:30:57Z

NicholasHu:

We will use the formula from the hint with <math display="inline">f(x)=\frac{4}{5}(\sqrt{|x|}+\sqrt{1-x^2})</math> and <math display="inline">g(x)=\frac{4}{5}(\sqrt{|x|}-\sqrt{1-x^2})</math> and <math display="inline">a=-1,b=1</math>; these were obtained already in part (i).

Then <math display="inline">f(x)+g(x) =\frac{8}{5}\sqrt{|x|}</math> and <math display="inline">f(x)-g(x) =\frac{8}{5}\sqrt{1-x^2}</math>. Moreover, we know from part (i) that <math display="inline">A=4\pi/5</math>. Thus, the <math display="inline">y</math>-coordinate of the centroid of <math display="inline">\mathcal{H}</math> is precisely
<math display="block">\overline{y} = \frac{1}{\frac{4}{5} \pi} \int_{-1}^1 \frac{\frac{8}{5}\sqrt{|x|}}{2} \cdot \frac{8}{5} \sqrt{1-x^2}\;dx=\frac{8}{5\pi} \int_{-1}^1\sqrt{|x|(1-x^2)}\;dx.</math>
Since the integrand is an even function, we have
<math display="block">\frac{8}{5\pi} \int_{-1}^1\sqrt{|x|(1-x^2)}= 2 \cdot \frac{8}{5\pi}\int_{0}^1\sqrt{|x|(1-x^2)}\;dx = \frac{16}{5\pi} \int_{0}^1\sqrt{x-x^3}\;dx,</math>
using the fact that <math display="inline">|x|=x</math> for <math display="inline">x\geq0</math>.

'''Answer:''' The correct constant is <math display="inline">{\color{blue}b=\frac{16}{5\pi}}</math>.

Science:Math Exam Resources/Courses/MATH101/April 2018/Question 11 (ii)/Hint 1

2023-04-25T23:25:59Z

NicholasHu:

Science:Math Exam Resources/Courses/MATH152/April 2017/Question B 02/Solution 1

2023-04-25T23:19:30Z

NicholasHu:

First, we need to know that equilibrium probability is defined as a probability vector, i.e., the sum of its components is 1. Exclude <math>\text{(II)}</math> from answer.

Second, from the definition of the probability transition matrix, we have <math>\sum_{i=1}^2 P_{i,j}=1.\,</math>. (Total probability should be 1) So <math>\text{(C)}</math> is not a transition probability matrix. (i.e., <math>\text{(VI)}</math>.)

Recall that when <math>P</math> has a single eigenvector for the eigenvalue <math>1</math>, we call the probability eigenvector as the ''equilibrium probability''.
Since we only have <math>2\times 2</math> transition matrices, the maximum number of eigenvectors corresponding to the eigenvalue <math>1</math> is <math>2</math> and in that case actually <math>P\bold{x} = \bold{x}</math> for any <math>\bold{x}</math>. i.e., <math>P = I</math>. Therefore <math>\text{(E)}</math> has no equilibrium probability.

Since the remained probability transition matrices are not identity matrix and we only have limited number of options for equilibrium probabilities, instead of finding eigen-vectors corresponding to the eigenvalue <math>1</math>, we find equilibrium probabilities by trials and errors,

<math>\text{(A)}</math>: <math> \begin{bmatrix} 1/3 & 2/3 \\ 2/3 & 1/3 \end{bmatrix} \cdot \begin{bmatrix} 0 \\ 1 \end{bmatrix} \neq \begin{bmatrix} 0 \\ 1 \end{bmatrix}</math>.

<math>\begin{bmatrix} 1/3 & 2/3 \\ 2/3 & 1/3 \end{bmatrix} \cdot \begin{bmatrix} 1/2 \\ 1/2 \end{bmatrix}= \begin{bmatrix} 1/2 \\ 1/2 \end{bmatrix}</math>.

Thus corresponding equilibrium probability of <math>\text{(A)}</math> is <math>\text{(III)}</math>

<math>\text{(B)}</math>: <math> \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{bmatrix} \cdot \begin{bmatrix} 0 \\ 1 \end{bmatrix} \neq \begin{bmatrix} 0 \\ 1 \end{bmatrix}</math>.

<math>\begin{bmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{bmatrix} \cdot \begin{bmatrix} 1/2 \\ 1/2 \end{bmatrix} = \begin{bmatrix} 1/2 \\ 1/2 \end{bmatrix}</math>.

Thus corresponding equilibrium probability of <math>\text{(B)}</math> is <math>\text{(III)}</math>

<math>\text{(D)}</math>: <math> \begin{bmatrix} 9/10 & 0 \\ 1/10 & 1 \end{bmatrix} \cdot \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}</math>.

Thus corresponding equilibrium probability of <math>\text{(D)}</math> is <math>\text{(I)}</math>

'''Answer:''' <math>\color{blue}\text{A-(III), B-(III), C-(VI), D-(I), E-(VIII)}</math>''

Science:Math Exam Resources/Courses/MATH221/December 2011/Question 06 (a)/Solution 1

2023-04-25T22:41:56Z

NicholasHu:

First we must determine whether the vectors are orthogonal. We know that two vectors are orthogonal if their dot product is zero.

Consider <math>v_1 \cdot v_2</math>. This is equal to

: <math>(1/2)(1/2) + (1/2)(1/2) + (1/2)(-1/2) + (1/2)(-1/2) = 0.</math>

All other dot products are calculated in the same way and all vanish. Hence <math>\{v_1, v_2, v_3, v_4\}</math> is an orthogonal set.

Since an orthogonal set of nonzero vectors must be linearly independent, and a set of 4 linearly independent vectors in <math>\mathbb{R}^4</math> must be a basis of <math>\mathbb{R}^4</math>, we conclude that <math>\{v_1, v_2, v_3, v_4\}</math> is an orthogonal basis of <math>\mathbb{R}^4</math>.

Science:Math Exam Resources/Courses/MATH101/April 2008/Question 03 (b)/Solution 1

2023-04-25T22:34:37Z

NicholasHu:

As in the hint, let <math> u = \sqrt{x} </math> so that <math> du = \frac{dx}{2\sqrt{x}} = \frac{dx}{2u} </math>. This gives

<math>
\int \cos\sqrt{x} \, dx = \int 2u\cos(u)\, du
</math>

To solve this last integral, we use integration by parts. Let

<math>
\begin{align}
w = u \quad&\quad v = \sin(u) \\
dw = du \quad & \quad dv = \cos(u)du
\end{align}
</math>

Then, we have

<math>
\begin{align}
\int 2u\cos(u)\, du &= 2\int u\cos(u)\, du\\
&= 2u\sin(u) - 2\int\sin(u)\, du\\
&= 2u\sin(u) + 2\cos(u) + C
\end{align}
</math>

and hence

<math>
\begin{align}
\int\cos(\sqrt{x})\, dx &= 2\sqrt{x}\sin(\sqrt{x}) + 2\cos(\sqrt{x}) + C.
\end{align}
</math>

Science:Math Exam Resources/Courses/MATH110/April 2012/Question 05 (e)/Solution 1

2023-04-25T22:32:34Z

NicholasHu:

We first find the derivative of ''f(x)''. Using the quotient rule,

: <math>f'(x) = \frac{4x(x^2 - 1) - (2x)(2x^2)}{(x^2 - 1)^2} = \frac{4x^3 - 4x - 4x^3}{(x^2 - 1)^2} = \frac{-4x}{(x^2 - 1)^2}</math>.

Since the denominator is always positive on the domain (squaring <math>x^2 - 1</math> always yields a positive number), we only need check whether the numerator is positive or negative. When <math>x < 0</math>, <math>-4x</math> is positive and when <math>x > 0</math>, <math>-4x</math> is negative. Thus ''f'' is increasing on the interval <math>(-\infty, -1)\cup (-1, 0)</math> and decreasing on the interval <math>(0, 1)\cup (1, \infty)</math>.

Science:Math Exam Resources/Courses/MATH312/December 2012/Question 02 (c)/Solution 1

2023-04-20T07:17:28Z

NicholasHu:

The answer is '''true'''.

Recall that 7 passes Miller's test if

<math> \displaystyle 7^{d} \equiv 1 \mod{25}</math>

or the following holds for some ''r''

<math> \displaystyle 7^{2^{r}d} \equiv -1 \mod{25}</math>

where <math> \displaystyle 25-1 = 24 = 2^{s}d = 2^{3}\cdot 3</math> and <math> \displaystyle 0 \leq r \leq 3-1</math>.

If 7 passes the test, then 25 is a probable prime. If it fails the test, then the number is not prime.

So we check manually:

<math> \displaystyle
\begin{align}
7^{3} &\equiv 49 \cdot 7 \equiv -1 \cdot 7 \not\equiv 1 \mod{25}\\
7^{2^1 \cdot3} &\equiv 49^{3} \equiv (-1)^3 \equiv -1 \mod{25}
\end{align}
</math>

Thus, 7 passes Miller's test. (Note that 25 is not prime but still 7 passes Miller's test).

Science:Math Exam Resources/Courses/MATH312/December 2012/Question 02 (c)/Hint 1

2023-04-20T07:15:32Z

NicholasHu:

Recall that 7 passes Miller's test if

<math> \displaystyle 7^{d} \equiv 1 \mod{25}</math>

or the following holds for some ''r''

<math> \displaystyle 7^{2^{r}d} \equiv -1 \mod{25}</math>

where <math> \displaystyle 25-1 = 24 = 2^{s}d</math> for <math>s = 3</math> and <math>d = 3</math>, and <math> \displaystyle 0 \leq r \leq s-1</math>.

If 7 passes the test, then 25 is a probable prime. If it fails the test, then the number is not prime.

Science:Math Exam Resources/Courses/MATH100/December 2016/Question 12/Solution 1

2022-12-14T22:49:55Z

NicholasHu:

The diagram of the cylinder inside the cone is as the following
[[File:MATH10016Q12.png|thumb|cylinder in a cone|left]]

where <math>R=1</math> m and <math>H=2</math> m are the base radius and height of the cone.

The two right riangles in the diagram are similar, so we have the following ratios:

<math>\begin{align}\frac{r}{R}=\frac{H-h}{H} &\Rightarrow \frac{r}{1}=\frac{2-h}{2}\\
& \Rightarrow r=\frac 12(2-h), \qquad 0<r<1 , 0<h<2
\end{align}</math>

Now we know that the volume of the cylinder is the area of the base times height i.e.

<math>V=\pi r^2 h =\frac \pi 4 h (2-h)^2 = \frac{\pi}{4}(4h-4h^2+h^3)</math>

To maximize the volume we need to find the critical points of <math>V</math>. We use the power rule to get

<math>V'(h)=\frac \pi 4 \left( 3h^2-8h+4 \right)=0</math>

Using the quadratic formula (or factoring) gives <math>h=2, h=\frac 23</math>. For <math>h=2</math> the volume becomes 0, so the maximum must be attained at <math>h=\frac 23</math>, and hence <math> r=\frac 12(2-\frac 23)=\frac 23</math>, therefore,

<math> V=\pi \times \frac 49 \times \frac 23=\frac 8{27} \pi</math>

Note that in general, we need to check the endpoints for <math>h</math> or <math>r</math>, where in this example make the volume equal to 0.

The dimension of the cylinder is therefore <math>\color{blue}h=\frac 23, r=\frac 23</math>.

Science:Math Exam Resources/Courses/MATH100/December 2016

2022-12-14T22:41:55Z

NicholasHu:

{{MER Exam page|creation_box=true}}

[[Category:MER Exam QG flag]]
[[Category:MER Exam Page]]

Science:Math Exam Resources/Courses/MATH100/December 2016/Question 2 (b)

2022-12-14T22:41:05Z

NicholasHu: Replaced content with "{{ delete }}"

Science:Math Exam Resources/Courses/MATH100/December 2016/Question 02 (b)/Solution 1

2022-12-14T22:40:17Z

NicholasHu:

By the quotient rule,
:<math>\begin{align}
g'(x)
&= \frac{(2x)(2x-1) - (x^2+3)(2)}{(2x-1)^2}\\
&= \frac{4x^2-2x -2x^2-6}{(2x-1)^2}\\
&= {\color{blue}\frac{2(x^2-x-3)}{(2x-1)^2}} .\\\end{align}</math>

Science:Math Exam Resources/Courses/MATH100/December 2016/Question 02 (a)/Solution 1

2022-12-14T22:36:12Z

NicholasHu:

By the product rule,
:<math> \begin{align}
f'(x) &= 2x e^x + x^2 e^x = {\color{blue}xe^x(2+x)}.
\end{align}</math>

Science:Math Exam Resources/Courses/MATH100/December 2016/Question 02 (b)/Solution 1

2022-12-14T22:35:23Z

NicholasHu: Created page with "By the quotient rule, :<math>\begin{align} g'(x) &= \frac{(2x-1)(2x) - 2(x^2+3)}{(2x-1)^2}\\ &= \frac{4x^2-2x -2x^2-6}{(2x-1)^2}\\ &= {\color{blue}\frac{2(x^2-x-3)}{(2..."

By the quotient rule,
:<math>\begin{align}
g'(x)
&= \frac{(2x-1)(2x) - 2(x^2+3)}{(2x-1)^2}\\
&= \frac{4x^2-2x -2x^2-6}{(2x-1)^2}\\
&= {\color{blue}\frac{2(x^2-x-3)}{(2x-1)^2}} \,.\\\end{align}</math>

Science:Math Exam Resources/Courses/MATH100/December 2016/Question 02 (a)/Hint 1

2022-12-14T22:34:14Z

NicholasHu:

Use the product rule.

Science:Math Exam Resources/Courses/MATH100/December 2016/Question 02 (b)/Hint 1

2022-12-14T22:33:53Z

NicholasHu: Created page with "Use the quotient rule."

Use the quotient rule.

Science:Math Exam Resources/Courses/MATH100/December 2016/Question 02 (a)/Statement

2022-12-14T22:33:31Z

NicholasHu: Removed redirect to Science:Math Exam Resources/Courses/MATH100/December 2016/Question 02 (b)/Statement

Find the derivative of <math>f(x) = x^2 e^{x}</math>.

Science:Math Exam Resources/Courses/MATH100/December 2016/Question 02 (a)/Statement

2022-12-14T22:31:50Z

NicholasHu: NicholasHu moved page Science:Math Exam Resources/Courses/MATH100/December 2016/Question 02 (a)/Statement to Science:Math Exam Resources/Courses/MATH100/December 2016/Question 02 (b)/Statement

#REDIRECT [[Science:Math Exam Resources/Courses/MATH100/December 2016/Question 02 (b)/Statement]]

Science:Math Exam Resources/Courses/MATH100/December 2016/Question 02 (b)/Statement

2022-12-14T22:31:50Z

Find the derivative of <math>g(x) = \dfrac{x^2+3}{2x-1}</math>.

Science:Math Exam Resources/Courses/MATH100/December 2016/Question 02 (b)

2022-12-14T22:31:10Z

NicholasHu: Removed redirect to Science:Math Exam Resources/Courses/MATH100/December 2016/Question 02 (c)

[[Category:MER QGQ flag]][[Category:MER QGH flag]][[Category:MER QGS flag]][[Category:MER CT flag]]







{{MER Question page}}