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Science:Math Exam Resources/Courses/MATH152/April 2016/Question B 05 (a)/Solution 1

2018-03-26T08:15:27Z

ZIMINGYIN:

<math>z=-1/2+i\sqrt{3}/2</math>:

<math>a= -1/2, b=\sqrt{3}/2</math>, it has negative x-value and positive y-value so it should be in the '''second''' '''quadrant'''. Suppose the angle between <math>z</math> and <math>x</math>-axis is <math>\theta</math>, we have <math>\tan \theta = \frac{b}{a}= -\sqrt{3}</math>.

It's known <math> \tan x = - \tan ( \pi-x)</math> and <math> \tan ( \pi/3) = \sqrt{3}</math>, <math>\theta \in [90^{\circ}, 180^{\circ}]</math>. Thus the angle between <math>z</math> and <math>x</math>-axis is <math>\frac{2\pi}{3}</math>

[[File:Math Exam Resources Courses MATH152 April 2016 Question B 05 (a) 1.png|solution]]

Science:Math Exam Resources/Courses/MATH152/April 2016/Question B 05 (a)/Solution 1

2018-03-26T08:14:51Z

ZIMINGYIN:

<math>z=-1/2+i\sqrt{3}/2</math>:

<math>a= -1/2, b=\sqrt{3}/2</math>, it has negative x-value and positive y-value so it should be in the '''second''' '''quadrant'''. Suppose the angle between <math>z</math> and <math>x</math>-axis is <math>\theta</math>, we have <math>\tan \theta = \frac{b}{a}= -\sqrt{3}</math>.

It's known <math> \tan x = - \tan ( \pi-x)</math> and <math> \tan ( \pi/3) = \sqrt{3}</math>, <math>\theta \in [90^{\circ}, 180^{\circ}]</math>. Thus the angle between <math>z</math> and <math>x</math>-axis is <math>\\frac{2\pi}{3}</math>

[[File:Math Exam Resources Courses MATH152 April 2016 Question B 05 (a) 1.png|solution]]

File:Math Exam Resources Courses MATH152 April 2016 Question B 05 (a) 1.png

2018-03-26T08:14:28Z

ZIMINGYIN: User created page with UploadWizard

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Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 24/Hint 1

2018-03-26T07:39:49Z

ZIMINGYIN:

Definition of eigenvectors and eigenvalues.

In Matlab, command <math>[V,D] = eig(A)</math> returns diagonal matrix <math>D</math> of eigenvalues and matrix <math>V</math> whose columns are the corresponding right eigenvectors, so that <math>A*V = V*D</math>.

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 24/Solution 1

2018-03-26T07:39:22Z

ZIMINGYIN:

Following the hint, the diagonal matrix <math>D</math> is the eigenvalues and the column of <math>V</math> are the are the corresponding right eigenvectors.

Therefore, A has 3 eigenvalues: <math>1+2i</math>, <math>1-2i</math>, <math>-1</math>, and their corresponding eigenvectors are <math> [0.8165,0,0.4082-0.4082i]^T</math> , <math> [0.8165,0,0.4082+0.4082i]^T</math> , <math> [0.5774,0.5774,0.5774]^T</math>. The eigenvectors in <math>V</math> are normalized so that the 2-norm of each one is <math>1</math>.

So <math> (a)</math> is wrong. We have <math>-1</math>

<math> (b)</math> is wrong. We have <math>1+2i</math>

<math> (c) , (d)</math> is correct. A has three different eigenvalues and hence the three eigenvectors are independent, and they can form a set of basis.

<math> (e)</math> is correct. <math> [1,1,1]^{T} </math> is a scalar multiples of the last column vectors of <math>A</math>. The scalar multiplication of any eigenvector is a eigenvector as well, so <math>(e)</math> is correct

Thus, the answer is <math>\color{blue}(c),(d),(e)</math>.

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 24/Solution 1

2018-03-26T07:38:48Z

ZIMINGYIN:

Following the hint, the diagonal matrix <math>D</math> is the eigenvalues and the column of <math>V</math> are the are the corresponding right eigenvectors.

Therefore, A has 3 eigenvalues: <math>1+2i</math>, <math>1-2i</math>, <math>-1</math>, and their corresponding eigenvectors are <math> [0.8165,0,0.4082-0.4082i]</math> , <math> [0.8165,0,0.4082+0.4082i]</math> , <math> [0.5774,0.5774,0.5774]</math>. The eigenvectors in <math>V</math> are normalized so that the 2-norm of each one is <math>1</math>.

So <math> (a)</math> is wrong. We have <math>-1</math>

<math> (b)</math> is wrong. We have <math>1+2i</math>

<math> (c) , (d)</math> is correct. A has three different eigenvalues and hence the three eigenvectors are independent, and they can form a set of basis.

<math> (e)</math> is correct. <math> [1,1,1]^{T} </math> is a scalar multiples of the last column vectors of <math>A</math>. The scalar multiplication of any eigenvector is a eigenvector as well, so <math>(e)</math> is correct

Thus, the answer is <math>\color{blue}(c),(d),(e)</math>.

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 23/Hint 2

2018-03-25T21:12:21Z

ZIMINGYIN: Created page with "Instead of doing this for each column, we can row reduce all these systems simultaneously, by attaching all columns of <math>I</math> (i.e. the whole matrix <math> I</math>) o..."

Instead of doing this for each column, we can row reduce all these systems simultaneously, by attaching all columns of <math>I</math> (i.e. the whole matrix <math> I</math>) on the right of <math>A</math> in the augmented matrix and obtaining all columns of <math>X</math> (i.e. the whole inverse matrix) on the right of the identity matrix in the row-equivalent matrix:

<math>[ A | I ] \rightarrow [ I | X ]</math>.

If this procedure works out, i.e. if we are able to convert <math>A </math> to identity using row operations, then <math>A </math>is invertible and <math>A^{-1} = X</math>.

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 23/Solution 2

2018-03-25T21:08:49Z

ZIMINGYIN: Created page with "Alternatively, we can also use row reduction. Form the augmented matrix <math>[A|I]= </math> <math>\left[\begin{array}{ccc|ccc} 2 & 1 & 3 &1 &0 &0\\ 1 & 0 & 1 & 0 & 1&..."

Alternatively, we can also use row reduction.

Form the augmented matrix <math>[A|I]= </math> <math>\left[\begin{array}{ccc|ccc}
2 & 1 & 3 &1 &0 &0\\
1 & 0 & 1 & 0 & 1& 0\\
0 &1 & 2 & 0 & 0 &1
\end{array}\right]</math>

Interchange <math>R_2</math> and <math>R_1</math> :
<math>\left[\begin{array}{ccc|ccc}
1 & 0 & 1 & 0 & 1& 0\\
2 & 1 & 3 &1 &0 &0\\
0 &1 & 2 & 0 & 0 &1
\end{array}\right]</math>

<math> R_2 - 2R_1 \rightarrow R_2 </math> :
<math>\left[\begin{array}{ccc|ccc}
1 & 0 & 1 & 0 & 1& 0\\
0 & 1 & 1 &1 &-2 &0\\
0 &1 & 2 & 0 & 0 &1
\end{array}\right]</math>

<math> R_3 - R_2 \rightarrow R_3 </math> :
<math>\left[\begin{array}{ccc|ccc}
1 & 0 & 1 & 0 & 1& 0\\
0 & 1 & 1 &1 &-2 &0\\
0 &0 & 1 & -1 & 2 &1
\end{array}\right]</math>

<math> R_1 - R_3 \rightarrow R_1 </math> :
<math>\left[\begin{array}{ccc|ccc}
1 & 0 & 0 & 1 & -1& -1\\
0 & 1 & 1 &1 &-2 &0\\
0 &0 & 1 & -1 & 2 &1
\end{array}\right]</math>

<math> R_2 - R_3 \rightarrow R_2 </math> :
<math>\left[\begin{array}{ccc|ccc}
1 & 0 & 0 & 1 & -1& -1\\
0 & 1 & 0 &2 &-4 &-1\\
0 &0 & 1 & -1 & 2 &1
\end{array}\right]</math>

Therefore,

<math>A^{-1}= \begin{bmatrix}
1&-1&-1\\2&-4&-1\\-1&2&1
\end{bmatrix}.</math>

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 06 (b)

2018-03-25T20:24:27Z

ZIMINGYIN:

[[Category:MER QGQ flag]][[Category:MER RH flag]][[Category:MER RS flag]][[Category:MER RT flag]]







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 06 (b)/Solution 2

2018-03-25T20:24:05Z

ZIMINGYIN:

Alt Sol: using demand elasticity formula <math> E = \frac{p}{q}\frac{dq}{dp}</math>

<math> p +10q=168 </math>

Consider <math>q</math> is a function of <math>p</math>, and we take the derivative of <math>p</math>:

<math> 1+10 \frac{dq}{dp} =0</math>

<math>\frac{dq}{dp} = -\frac{1}{10}</math>

Therefore <math> E=-\frac{1}{10}\cdot\frac{p}{q}</math>

Recall when <math>E=-1</math>, we say the good is price unit elastic. This is the optimal price which means it maximizes revenue.

So setting <math>E=-1</math> gives us :

<math> -1 = -\frac{1}{10} \cdot \frac{p}{q} \rightarrow p=10q \rightarrow p = 168- p \rightarrow p = 84</math>

'''answer:''' <math>\color{blue}84</math>''

Thread:User talk:ZIMINGYIN/Q2b Math104 (2016) and others on this exam/reply

2018-03-18T08:33:18Z

ZIMINGYIN: Reply to Q2b Math104 (2016) and others on this exam

Two figures reupload now, added the source and author.

Ziming

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 02 (b)/Solution 1

2018-03-18T08:31:22Z

ZIMINGYIN:

[[File:Math Exam Resources Courses MATH104 December 2016 Question 2 (b) picture 3.jpg|solution]]

The first derivative of m is straightforward, if the derivative is positive, then <math>f'>0</math>.

For <math>f''</math>, we use the points of inflection (3 and 5 from question). 3 is one inflection point. From 0 to 3, <math>f'</math> goes from positive to negative, thus during that area, <math>f'' <0 </math>. For <math> t \in [3,5], f'' >0</math>. For <math> t >5, f'' <0</math>.

The part where <math>f', f''</math> are both positive: <math> \left( 4,5 \right)</math>

The part where <math>f', f''</math> are both negative: <math> \left( 2,3 \right)</math>

'''answer:''' <math>\color{blue}\left( 2,3 \right), \left( 4,5 \right)</math>''

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 02 (b)/Statement

2018-03-18T08:31:02Z

ZIMINGYIN:

The graph of the position of a particle is shown below, where <math> t </math> is measured in second and the dots are local extrema or points of inflection. Determine when the particle is
speeding up. Hint: a particle is "speeding up" when its velocity and acceleration have the same sign.

[[File:Math Exam Resources Courses MATH104 December 2016 Question 2 (b) picture 4.jpg|solution]]

File:Math Exam Resources Courses MATH104 December 2016 Question 2 (b) picture 4.jpg

2018-03-18T08:30:42Z

ZIMINGYIN: User created page with UploadWizard

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{{Information
|description={{en|1=solution}}
|date=2018-03-18 01:29:49
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|author=Shawn Desaulniers
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Science:Math Exam Resources/Courses/MATH104/December 2016/Question 02 (b)/Solution 1

2018-03-18T08:28:06Z

ZIMINGYIN:

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 02 (b)/Solution 1

2018-03-18T08:27:50Z

ZIMINGYIN:

[[File:Math Exam Resources Courses MATH104 December 2016 Question 2 (b) picture 3.png|solution]]

The first derivative of m is straightforward, if the derivative is positive, then <math>f'>0</math>.

For <math>f''</math>, we use the points of inflection (3 and 5 from question). 3 is one inflection point. From 0 to 3, <math>f'</math> goes from positive to negative, thus during that area, <math>f'' <0 </math>. For <math> t \in [3,5], f'' >0</math>. For <math> t >5, f'' <0</math>.

The part where <math>f', f''</math> are both positive: <math> \left( 4,5 \right)</math>

The part where <math>f', f''</math> are both negative: <math> \left( 2,3 \right)</math>

'''answer:''' <math>\color{blue}\left( 2,3 \right), \left( 4,5 \right)</math>''

File:Math Exam Resources Courses MATH104 December 2016 Question 2 (b) picture 3.jpg

2018-03-18T08:26:24Z

ZIMINGYIN: User created page with UploadWizard

=={{int:filedesc}}==
{{Information
|description={{en|1=solution}}
|date=2018-03-18 01:24:09
|source=final answers
|author=Shawn Desaulniers
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File:Math Exam Resources Courses MATH104 December 2016 Question 2 (b) picture 1.png

2018-03-18T08:19:43Z

ZIMINGYIN:

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|date=2018-03-17 22:50:51
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File:Math Exam Resources Courses MATH104 December 2016 Question 2 (b) picture 1.png

2018-03-18T08:19:08Z

ZIMINGYIN: ZIMINGYIN uploaded a new version of "File:Math Exam Resources Courses MATH104 December 2016 Question 2 (b) picture 1.png"

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|author=[[User:ZIMINGYIN|ZIMINGYIN]]
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Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 6 (a)/Solution 2

2018-03-18T08:11:51Z

ZIMINGYIN:

Following the second hint, Since matrix <math>A</math> is a orthogonal rotation matrix, therefore it has determinant <math>\pm 1</math>. For this question, the determinant is <math>1</math>.

Out of three eigenvalues, one of them must be 1 from the property of an orthogonal rotation matrix. Now suppose the left two eigenvalues are <math>z_1</math> and <math>z_2</math>. Since the sum of the eigenvalues is the trace of the matrix, and since their product is the determinant, we have <math> 1+z_1 +z_2 =1</math> and <math> z_1 \cdot z_2 = 1</math>.

<math>z_1 = i, z_2 = -i</math>

'''answer:''' <math>\color{blue}1,i,-i</math>''

Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 6 (a)/Solution 2

2018-03-18T08:11:32Z

ZIMINGYIN: Created page with "Following the second hint, Since matrix <math>A</math> is a orthogonal rotation matrix, therefore it has determinant <math>\pm 1</math>. For this question, the determinant is..."

Following the second hint, Since matrix <math>A</math> is a orthogonal rotation matrix, therefore it has determinant <math>\pm 1</math>. For this question, the determinant is <math>1</math>.

Out of three eigenvalues, one of them must be 1 from the property of an orthogonal rotation matrix. Now suppose the left two eigenvalues are <math>z_1</math> and <math>z_2</math>. Since the sum of the eigenvalues is the trace of the matrix, and since their product is the determinant, we have <math> 1+z_1 +z_2 =1</math> and <math> z_1 \cdot z_2 = 1</math>.

<math>z_1 = I, z_2 = -I</math>

'''answer:''' <math>\color{blue}1,i,-i</math>''

Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 6 (a)/Hint 2

2018-03-18T08:01:11Z

ZIMINGYIN: Created page with "We may also use the property of orthogonal rotation matrix is we notice the geometry of the question."

We may also use the property of orthogonal rotation matrix is we notice the geometry of the question.

Thread:Science talk:Math Exam Resources/Courses/MATH104/December 2016/Question 04 (a)/Add more explanation/reply

2018-03-18T07:43:02Z

ZIMINGYIN: Reply to Add more explanation

Hi Hyunju,

Sorry I finished them at this very late day... I added some explanation and let me know if it's still unclear.

Goodnight!
Ziming

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 04 (a)/Solution 1

2018-03-18T07:34:37Z

ZIMINGYIN:

The linear approximation of <math>f(x)</math> given a point <math>x =a</math> should be <math> T(x) = f(a) +f'(a) \cdot (x-a)</math>

Let <math> f(x) =\sqrt{x} </math>. The first thing we need to do is find a value of <math>x </math> "close to" 69 which will be 64 from the hint.

Next we'll need <math> f'(x) = \frac{1}{2\sqrt{x}}</math>, thus <math>f'(64) = \frac{1}{2 \cdot 8} = \frac{1}{16}</math>.

Therefore the linear approximation line is:

<math>
\begin{align}
T(x) = & f(a) +f'(a) \cdot (x-a)\\
=& f(64) +f'(64) \cdot (x-64)\\
=& 8 + \frac{1}{16} \cdot (x-64)
\end{align}
</math>

<math> T(69) = 8 + \frac{1}{16} \cdot (69-64)=8 + \frac{5}{16} = \frac{133}{16}</math>.

'''answer:''' <math>\color{blue}\frac{133}{16}</math>''

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 04 (a)/Hint 1

2018-03-18T07:31:41Z

ZIMINGYIN:

A simple way of approximating <math>\sqrt{69}</math> is to construct the linear approximation of (i.e., ''tangent line'' to) <math>f(x) = \sqrt{x}</math> at some point <math>(a, f(a))</math> for which <math>f(a)</math> is known exactly. Therefore, the point needs to satisfy two things: have a pretty straightforward square root and also close to <math>\sqrt{69}</math>. Thus we choose <math>a \approx 69.</math>

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 06 (b)/Solution 1

2018-03-18T07:27:58Z

ZIMINGYIN:

From demand curve, we have the relationship between <math>p</math> and <math>q</math> as <math>q= \frac{168-p}{10}</math>.

Following the hint, the revenue is

<math>
\begin{align}
R &= p \cdot q\\
R &= p \cdot \frac{168-p}{10}\\
\end{align}
</math>

If we wanna find the price that makes most revenue, then that point should be at the maximum of function <math>R</math>, where the derivative of function is zero.

Therefore setting <math>\frac{dR}{dp}= \frac{1}{10} \cdot (168 -p-p)=0</math> gives us <math>2p = 168 \rightarrow p=84</math>

'''answer:''' <math>\color{blue}84</math>''

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 06 (b)/Solution 1

2018-03-18T07:27:28Z

ZIMINGYIN:

From demand curve, we have the relationship between <math>p</math> and <math>q</math> as <math>q= \frac{168-p}{10}</math>.

Following the hint, the revenue is

<math>
\begin{align}
R &= p \cdot q\\
R &= p \cdot \frac{168-p}{10}\\
\frac{dR}{dp} &= \frac{1}{10} \cdot (168 -p-p)
\end{align}
</math>

If we wanna find the price that makes most revenue, then that point should be at the maximum of function <math>R</math>, where the derivative of function is zero.

Therefore setting <math>\frac{dR}{dp} =0</math> gives us <math>2p = 168 \rightarrow p=84</math>

'''answer:''' <math>\color{blue}84</math>''

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 06 (b)/Solution 1

2018-03-18T07:26:05Z

ZIMINGYIN:

The revenue is

<math>
\begin{align}
R &= p \cdot q\\
R &= p \cdot \frac{168-p}{10}\\
\frac{dR}{dp} &= \frac{1}{10} \cdot (168 -p-p)
\end{align}
</math>

If we wanna find the price that makes most revenue, then that point should be at the maximum of function <math>R</math>, where the derivative of function is zero.

Therefore setting <math>\frac{dR}{dp} =0</math> gives us <math>2p = 168 \rightarrow p=84</math>

'''answer:''' <math>\color{blue}84</math>''

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 06 (b)/Solution 2

2018-03-18T07:22:40Z

ZIMINGYIN: Created page with "Alt Sol: <math> p +10q=168 \rightarrow 1+10 \frac{dq}{dp} =0</math> <math>\frac{dq}{dp} = -\frac{1}{10} \rightarrow E=-\frac{1}{10}\cdot{p}{q}</math> setting <math>E=-1</mat..."

Alt Sol:
<math> p +10q=168 \rightarrow 1+10 \frac{dq}{dp} =0</math>

<math>\frac{dq}{dp} = -\frac{1}{10} \rightarrow E=-\frac{1}{10}\cdot{p}{q}</math>

setting <math>E=-1</math> gives us :

<math> -1 = -\frac{1}{10} \cdot \frac{p}{q} \rightarrow p=10q \rightarrow p = 168- p \rightarrow p = 84</math>

'''answer:''' <math>\color{blue}84</math>''

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 06 (b)/Solution 1

2018-03-18T07:22:08Z

ZIMINGYIN:

The revenue is

<math>
\begin{align}
R &= p \cdot q\\
R &= p \cdot \frac{168-p}{10}\\
\frac{dR}{dp} &= \frac{1}{10} \cdot (168 -p-p)
\end{align}
</math>

Setting <math>\frac{dR}{dp} =0</math> gives us <math>2p = 168 \rightarrow p=84</math>

'''answer:''' <math>\color{blue}84</math>''

Thread:Science talk:Math Exam Resources/Courses/MATH152/April 2017/Question B 02/Multiplication wrong/reply

2018-03-18T07:19:21Z

ZIMINGYIN: Reply to Multiplication wrong

Hi Alistair,

Thanks for pointing out the mistake. Hope now all are fixed[[http://wiki.ubc.ca/Science:Math_Exam_Resources/Courses/MATH152/April_2017/Question_B_02]]

Best,
Ziming

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

2018-03-18T07:15:54Z

ZIMINGYIN:

First, we need to know that equilibrium probability matrix is defined as a probability distribution, thus the sum of its component is 1. Exclude <math>(II)</math> from answer.

Second, since the total of transition probability from a state <math>I</math> to all other states must be 1, so that <math>\sum_{j=1}^S P_{i,j}=1.\,</math>. So <math>(C)</math> is not a transition probability matrix.

Then from the hint above,

(A): <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>(A)</math> is <math>(III)</math>

(B): <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>(B)</math> is <math>(III)</math>

(C): (VI)

(D): <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>(D)</math> is <math>(I)</math>

(E) <math>\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 0 \\ 1 \end{bmatrix}= \begin{bmatrix} 0 \\ 1 \end{bmatrix}</math>.
Thus corresponding equilibrium probability of <math>(E)</math> is <math>(I)</math>

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

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

2018-03-18T07:14:06Z

ZIMINGYIN:

First, we need to know that equilibrium probability matrix is defined as a probability distribution, thus the sum of its component is 1. Exclude <math>(ii)</math> from answer.

Second, since the total of transition probability from a state <math>I</math> to all other states must be 1, so that <math>\sum_{j=1}^S P_{i,j}=1.\,</math>. So <math>(C)</math> is not a transition probability matrix.

Then from the hint above,

(A): <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>(A)</math> is <math>(iii)</math>

(B): <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>(B)</math> is <math>(iii)</math>

(C): (VI)

(D): <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>(D)</math> is <math>(i)</math>

(E) <math>\begin{bmatrix} 0 \\ 1 \end{bmatrix}^T \cdot \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}^T</math>.
Thus corresponding equilibrium probability of (E) is(i)

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

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

2018-03-18T07:13:53Z

ZIMINGYIN:

First, we need to know that equilibrium probability matrix is defined as a probability distribution, thus the sum of its component is 1. Exclude <math>(ii)</math> from answer.

Second, since the total of transition probability from a state <math>I</math> to all other states must be 1, so that <math>\sum_{j=1}^S P_{i,j}=1.\,</math>. So <math>(C)</math> is not a transition probability matrix.

Then from the hint above,

(A): <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>(A)</math> is <math>(iii)</math>

(B): <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>(B)</math> is <math>(iii)</math>

(C): (VI)

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

Thus corresponding equilibrium probability of <math>(D)</math> is <math>(i)</math>

(E) <math>\begin{bmatrix} 0 \\ 1 \end{bmatrix}^T \cdot \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}^T</math>.
Thus corresponding equilibrium probability of (E) is(i)

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

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

2018-03-18T07:09:54Z

ZIMINGYIN:

First, we need to know that equilibrium probability matrix is defined as a probability distribution, thus the sum of its component is 1. Exclude <math>(ii)</math> from answer.

Second, since the total of transition probability from a state <math>I</math> to all other states must be 1, so that <math>\sum_{j=1}^S P_{i,j}=1.\,</math>. So <math>(C)</math> is not a transition probability matrix.

Then from the hint above,

(A): <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>(A)</math> is <math>(iii)</math>

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

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

Thus corresponding equilibrium probability of (B) is(iii)

(C): (VI)

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

<math>\begin{bmatrix} 1/2 \\ 1/2 \end{bmatrix}^T \cdot \begin{bmatrix} 9/10 & 0 \\ 1/10 & 1 \end{bmatrix} \neq \begin{bmatrix} 1/2 \\ 1/2 \end{bmatrix}^T</math>.

<math>\begin{bmatrix} 1/3 \\ 2/3 \end{bmatrix}^T \cdot \begin{bmatrix} 9/10 & 0 \\ 1/10 & 1\end{bmatrix} \neq \begin{bmatrix} 1/3 \\ 2/3 \end{bmatrix}^T</math>.

<math>\begin{bmatrix} 9/10 \\ 1/10 \end{bmatrix}^T \cdot \begin{bmatrix}9/10 & 0 \\ 1/10 & 1 \end{bmatrix} \neq \begin{bmatrix} 9/10 \\ 1/10 \end{bmatrix}^T</math>.

Thus (D) - (VII)

(E) <math>\begin{bmatrix} 0 \\ 1 \end{bmatrix}^T \cdot \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}^T</math>.
Thus corresponding equilibrium probability of (E) is(i)

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

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

2018-03-18T07:04:40Z

ZIMINGYIN:

If the probability of moving from <math>i</math> to <math>j</math> in one time step is <math>Pr(i|j)=P_{i,j}</math>, the stochastic matrix <math>P</math> is given by using <math>P_{i,j}</math> as the <math>i^{th}</math> row and <math>j^{th}</math> column element. And the equilibrium probability distribution <math>M</math> has the property that <math> PM = M</math>.

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

2018-03-18T07:00:18Z

ZIMINGYIN:

Thread:Science talk:Math Exam Resources/Courses/MATH103/April 2017/Question 07 (b)/Solution 1/solution style error/reply

2018-03-18T06:19:32Z

ZIMINGYIN: Reply to solution style error

Hi Rob,

For this one I changed the answer to a table now [[http://wiki.ubc.ca/Science:Math_Exam_Resources/Courses/MATH103/April_2017/Question_07_(b)]]

Best,
Ziming

Science:Math Exam Resources/Courses/MATH103/April 2017/Question 07 (b)/Solution 1

2018-03-18T06:18:07Z

ZIMINGYIN:

When <math>|f'(x_0)|=1</math>, the slope of that point should be parallel to <math> y=x</math> or <math>y=-x</math>. This means that a line that is flatter than <math> y = x</math> satisfies <math>|f'(x_0)| < 1 </math>, and therefore that the the fixed point is stable; a line that is steeper than <math> y = x </math> satisfies <math> |f'(x_0)| > 1 </math>
The coordinates for 4 fixed points is <math> (-5,-5), (0,0), (5,5), (10,10)</math>.

'''Answer:'''

{| class="wikitable" style="text-align:left; "
|-
! point
!stability
!reason
|-
|(i) <math>(-5,5)</math>|| unstable || <math> f'(-5)>1</math> ||
|-
|(ii) <math>(0,0) </math>||stable || <math> f'(0) = 0 <1 </math>||
|-
|(iii) <math>-(5,5)</math>|| unstable ||<math>f'(5) >1</math>||
|-
|(iv) <math>(10,10)</math> || stable || <math>f'(10) = 0 < 1</math>||
|-
|}

Science:Math Exam Resources/Courses/MATH103/April 2017/Question 07 (b)/Solution 1

2018-03-18T06:17:40Z

ZIMINGYIN:

When <math>|f'(x_0)|=1</math>, the slope of that point should be parallel to <math> y=x</math> or <math>y=-x</math>. This means that a line that is flatter than <math> y = x</math> satisfies <math>|f'(x_0)| < 1 </math>, and therefore that the the fixed point is stable; a line that is steeper than <math> y = x </math> satisfies <math> |f'(x_0)| > 1 </math>
The coordinates for 4 fixed points is <math> (-5,-5), (0,0), (5,5), (10,10)</math>.

'''Answer:'''

{| class="wikitable" style="text-align:left; "
|-
! point
!stability
!reason
|-
|(i) <math>(-5,5)</math>|| unstable || <math> f'(-5)>1</math> ||
|-
|(ii) <math>(0,0) </math>||stable || <math> f'(0) = 0 <1 </math>||
|-
|(iii) <math>-1 + 2^{-n}</math>|| unstable ||<math>f'(5) >1</math>||
|-
|(iv) <math>\ln (n)</math> || stable || <math>f'(10) = 0 < 1</math>||
|-
|}

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 02 (b)/Statement

2018-03-18T06:07:43Z

ZIMINGYIN:

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 02 (b)/Statement

2018-03-18T06:07:29Z

ZIMINGYIN:

[[File:Math Exam Resources Courses MATH104 December 2016 Question 2 (b) picture 1.png|question]]

The graph of the position of a particle is shown below, where <math> t </math> is measured in second and the dots are local extrema or points of inflection. Determine when the particle is
speeding up. Hint: a particle is "speeding up" when its velocity and acceleration have the same sign.

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 02 (b)/Statement

2018-03-18T06:07:18Z

ZIMINGYIN:

[[File:Math Exam Resources Courses MATH104 December 2016 Question 2 (b) picture 1.png|question]]
The graph of the position of a particle is shown below, where <math> t </math> is measured in second and the dots are local extrema or points of inflection. Determine when the particle is
speeding up. Hint: a particle is "speeding up" when its velocity and acceleration have the same sign.

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 02 (b)/Solution 1

2018-03-18T06:06:59Z

ZIMINGYIN:

[[File:Math Exam Resources Courses MATH104 December 2016 Question 2 (b) picture 2.png|solution]]

The first derivative of m is straightforward, if the derivative is positive, then <math>f'>0</math>.

For <math>f''</math>, we use the points of inflection (3 and 5 from question). 3 is one inflection point. From 0 to 3, <math>f'</math> goes from positive to negative, thus during that area, <math>f'' <0 </math>. For <math> t \in [3,5], f'' >0</math>. For <math> t >5, f'' <0</math>.

The part where <math>f', f''</math> are both positive: <math> \left( 4,5 \right)</math>

The part where <math>f', f''</math> are both negative: <math> \left( 2,3 \right)</math>

'''answer:''' <math>\color{blue}\left( 2,3 \right), \left( 4,5 \right)</math>''

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 02 (b)/Solution 1

2018-03-18T06:06:15Z

ZIMINGYIN:

[[File:File:Math Exam Resources Courses MATH104 December 2016 Question 2 (b) picture 2.png|thumb|solution]]

The first derivative of m is straightforward, if the derivative is positive, then <math>f'>0</math>.

For <math>f''</math>, we use the points of inflection (3 and 5 from question). 3 is one inflection point. From 0 to 3, <math>f'</math> goes from positive to negative, thus during that area, <math>f'' <0 </math>. For <math> t \in [3,5], f'' >0</math>. For <math> t >5, f'' <0</math>.

The part where <math>f', f''</math> are both positive: <math> \left( 4,5 \right)</math>

The part where <math>f', f''</math> are both negative: <math> \left( 2,3 \right)</math>

'''answer:''' <math>\color{blue}\left( 2,3 \right), \left( 4,5 \right)</math>''

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 02 (b)/Solution 1

2018-03-18T06:04:07Z

ZIMINGYIN:

[[File:Math Exam Resources Courses MATH104 December 2016 Question 2 (b) picture 2.png|thumb|solution]]

The first derivative of m is straightforward, if the derivative is positive, then <math>f'>0</math>.

For <math>f''</math>, we use the points of inflection (3 and 5 from question). 3 is one inflection point. From 0 to 3, <math>f'</math> goes from positive to negative, thus during that area, <math>f'' <0 </math>. For <math> t \in [3,5], f'' >0</math>. For <math> t >5, f'' <0</math>.

The part where <math>f', f''</math> are both positive: <math> \left( 4,5 \right)</math>

The part where <math>f', f''</math> are both negative: <math> \left( 2,3 \right)</math>

'''answer:''' <math>\color{blue}\left( 2,3 \right), \left( 4,5 \right)</math>''

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 02 (b)/Statement

2018-03-18T06:03:33Z

ZIMINGYIN:

[[File:Math Exam Resources Courses MATH104 December 2016 Question 2 (b) picture 1.png|thumb|question]]
The graph of the position of a particle is shown below, where <math> t </math> is measured in second and the dots are local extrema or points of inflection. Determine when the particle is
speeding up. Hint: a particle is "speeding up" when its velocity and acceleration have the same sign.

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 02 (b)/Statement

2018-03-18T06:03:23Z

ZIMINGYIN:

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 02 (b)/Statement

2018-03-18T06:02:24Z

ZIMINGYIN:

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 02 (b)/Statement

2018-03-18T06:02:17Z

ZIMINGYIN: