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### Messages - RunboZhang

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16
##### Test 2 / Re: possible typo on 2020 Question 3 Morning sitting
« on: October 27, 2020, 09:58:45 AM »
I agree with you, I think the solution has some small typos.

17
##### Test 1 / LEC0201-Retest-ALT-Y-Question2
« on: October 23, 2020, 04:52:15 PM »
$\textbf {Problem:} \\\\$
$\text{(a) Find Wronskian } W(y_1, y_2)(x) \text{ of a fundamental set of solutions} y_1(x), y_2(x) \text{ for ODE} \\$
$\begin{gather} y'' + 3tan(x)y' + (1 + 3tan^{2}{x})y = 0 \end{gather}$
$\text{(b) Check that } y_1(x) = cos(x) \text{ is a solution and find another linearly independent solution.}\\\\$
$\text{(c) Write the general solution, and find solution such that} y(0) = 0 \text{ , } y'(0) = 12\\\\$

$\textbf{Solution: } \\\\$
$\text{(a):}\\\\$

\begin{gather} \begin{aligned} y'' + 3 tan(x) \cdot y' + (1 + 3tan^{2}(x))y = 0 \\\\ \implies p(x) = 3tan(x) \\\\ \end{aligned} \end{gather}

\begin{gather} \begin{aligned} w &= c \cdot e^{- \int_ (p(x)) \, dx} \\\\ &= c \cdot e^{- \int_ (3tan(x)) \, dx} \\\\ &= c \cdot e^{3ln|cos(x)| + c'} \\\\ &= c \cdot cos^{3}(x) + c \cdot e^{c'} \end{aligned} \end{gather}

$\text{Let c = 1, then we have: }$

\begin{gather} \begin{aligned} w = cos^{3}(x) \end{aligned} \end{gather}

$\text{(b):}\\\\$

\begin{gather} \begin{aligned} y_1 = cos(x) \ y_1' = - sin(x) \ y_1'' = - cos(x) \end{aligned} \end{gather}

$\text{Plug in the equation, we have: }$

\begin{gather} \begin{aligned} y'' + 3 tan(x) \cdot y' + (1 + 3tan^{2}(x))y &= -cos(x) + 3 \frac{sin(x)}{cos(x)} \cdot (- sin(x)) + (1 + 3 \frac{sin^{2}(x)}{cos^{2}(x)}) cos(x) \\\\ &= - cos(x) - 3\frac{sin^{2}(x)}{cos(x)} + cos(x) + 3 \frac{sin^{2}(x)}{cos(x)} \\\\ &= 0 \\\\ \end{aligned} \end{gather}

$\text{Thus we have proved} y_1(x) = cos(x) \text{ is a solution.} \\\\$

$\text{Now we solve for } y_2 \text{: } \\\\$

\begin{gather} \begin{aligned} W(x,y) &= cos^{3}(x) \\\\ &= cos(x) \cdot y_2' - (- sin(x)) \cdot y_2 \\\\ &= cos(x) \cdot y_2' + sin(x) \cdot y_2 \\\\ \end{aligned} \end{gather}

$\text{Now we solve for } y_2$  \text{ : }

\begin{gather} \begin{aligned} cos^{3}(x) &= cos(x) \cdot y_2' - (- sin(x)) \cdot y_2 \\\\ cos(x) &= \frac{y_2'}{cos(x)} + \frac{sin(x)}{cos^{2}(x)} y_2 \\\\ &= (\frac{y_2}{cos(x)})' \\\\ y_2 &= cos(x) \cdot \int(cos(x)) ,\ dx \\\\ &= cos(x) \cdot (sin(x) + c)\\\\ &= cos(x) \cdot sin(x) \\\\ &= c \cdot cos(x) \end{aligned} \end{gather}

$\text{(c):}\\\\$

$\text{We have: } y(0) = 0 \ \text{ and plug in our general form, we have: }$

\begin{gather} \begin{aligned} 0 &= c_1 + c_2 \cdot 0 \\\\ c_1 &= 0 \end{aligned} \end{gather}

$\text{Then our general form becomes: } y = c_2(cos(x) \cdot sin(x))$
$\text{Also, we have initial value: } y'(0) = 12 \ \text{ By plugging it in, we have: }$

\begin{gather} \begin{aligned} 12 &= c_2 \cdot (-sin0 \cdot sin0 + cos0 \cdot cos0) &= c_2 \end{aligned} \end{gather}

$\text{Therefore we got our final answer: } y = 12 \cdot sin(x) \cdot cos(x)$

18
##### Test 1 / LEC0201-Retest-ALT-Y-Question1
« on: October 23, 2020, 04:18:51 PM »
$\textbf {Problem:} \\\\$
$\text{(a) Find integrating factor and then a general solution of ODE} \\$
$\begin{gather} (1+xy+y^{2}) + (1+xy+x^{2})y' = 0 \end{gather}$
$\text{(b) Also, find a solution satisfying y(1) = 1}\\\\$

$\textbf{Solution: } \\\\$
$\text{(a):}\\\\$
$\text{We have:}$

\begin{gather} \begin{aligned} M_y = x+2y \\\\ N_x = y+2x \end{aligned} \end{gather}

$\text{Since} M_y \ne N_x \ \text{, thus equation is not exact.}$

\begin{gather} \begin{aligned} \frac{N_x - M_y}{M \cdot x - N \cdot y} &{}= \frac{y+2x-x-2y}{x+x^{2}y+xy^{2}-y-xy^{2}-x^{2}y} \\\\ &{} = \frac{x - y}{x - y} \\\\ &{} = 1 \end{aligned} \end{gather}

$\text{Now, we have our integrating factor } \mu \ \text{computed as follow: }$

\begin{gather} \begin{aligned} \mu &{} = e^{\int {1} \, dxy} \\\\ &{} = e^{xy} \end{aligned} \end{gather}

$\text{Now multiplying the original equation with our integrating factor, we have: }$

\begin{gather} \begin{aligned} e^{xy} + e^{xy} \cdot xy + e^{xy} \cdot y^{2} + (e^{xy} + e^{xy} \cdot xy + e^{xy} \cdot x^{2})y' = 0 \end{aligned} \end{gather}

$\text{Lastly, solve for general solution F(x,y): }$

\begin{gather} \begin{aligned} F(x,y) &{} = \int_{0}^{x} (e^{xy} + e^{xy} \cdot xy + e^{xy} \cdot y^{2}) \, dx + h'(y) \\\\ &{} = x \cdot e^{xy} + y \cdot e^{xy} = c \end{aligned} \end{gather}

$\text{(b):}\\\\$
$\text{By plugging in initial value condition, we have: }$
\begin{gather} \begin{aligned} y(1) = 1 \\\\ c = e + e = 2e\\\\ \implies x \cdot e^{xy} + y \cdot e^{xy} = 2e \end{aligned} \end{gather}

19
##### Quiz 4 / Quiz-5101-C
« on: October 23, 2020, 03:48:21 PM »
$\textbf {Problem:} \\\\$
$\text{Evaluate the given integral using the technique of Example 10 of Section 2.3:} \\$
$\begin{gather} \int_{\gamma} e^{z}\, dz \end{gather}$
$\text{where}\ \gamma \ \text{is the semicircle from -1 to 1 passing through i.}$

$\textbf{Solution: } \\\\$
$\text{We have integrand} f(z) = e^{z} \text{, and it is the derivative of }F(z)=e^{z} . \\\\$
$\text{This is valid when } F(z) \text{is analytic on domain D.}\\\\$
$\text{Indeed, both} F(z) \text{ and } f(z) \text{ is analytic on the semicircle.}\\\\$
$\text{Therefore, we have}\\\\$

\begin{gather} \begin{aligned} \int_{\gamma} e^{z}\, dz &{} = \int_{\gamma} f(z)\, dz \\\\ &{} = \int_{\gamma} F'(z)\, dz \\\\ &{} = \text{F(endpoint) \m F(initialpoint)} \\\\ &{} = F(1) - F(-1) \\\\ &{} = e - e^{-1} \end{aligned} \end{gather}

$\text{Therefore} \ e - e^{-1} \ \text{is our final answer.}$

20
##### Test 1 / Re: past test question
« on: October 16, 2020, 09:07:29 AM »
I used ratio test to solve this problem. Since this one has double factorial in its series, so check the limit of $|\frac {a_{n+2}}{a_n}|$ would be helpful. After expanding the limit, you will see most of the terms will be canceled out. What's left is $\frac{1}{n+2}$, and it equals to 0 as n approaches infinity. It also equals to $\frac {1}{R}$. Thus our radius of convergence is infinity.

Another method is to solve this series by splitting it into its even part and odd part. Then use ratio test to get the radius of convergence of both parts. The smaller radius will be the final answer.

21
##### Test 1 / Re: TT1 2020S Main Sitting
« on: October 15, 2020, 06:19:19 PM »
Actually it is $log(2^{\frac {1}{2}})$. But by one property of log function, $log(2^{\frac {1}{2}}) = \frac{1}{2} \cdot log(2)$

22
##### Test 1 / Re: 2020TT1 Deferred Sitting #1
« on: October 14, 2020, 09:04:13 PM »
Since $\pm (2n+1) \pi i$ can be on both positive y-axis and negative y-axis. Thats why we have m instead of 2m. If we have 2m (period of $2\pi$), we would have either positive or negative y-axis only, which does not include every case.

23
##### Test 1 / Re: 2020 TT1 Main Setting - Problem 1b
« on: October 14, 2020, 07:37:18 PM »
Firstly, for the first quadrant, we have $Re(z) > 0$ and $Im(z) > 0$.
Secondly, we have $z = \frac{1}{12}log(2) + (\frac{\pi}{8} + \frac{2\pi }{6} n)i$, $n \in \mathbb{Z}$ by part (a).
By combining the previous two conclusions we have, $Re(z) = \frac{1}{12}log(2) > 0$ since $log(2) > 0$. Also $Im(z) = (\frac{\pi}{8} + \frac{2\pi }{6} n) > 0$ when $n \ge 0$.
Therefore, as long as we have a non-negative $n$, our $z$ is in the first quadrant of complex plane.

24
##### Test 1 / Re: 2020TT1 Deferred Sitting #1
« on: October 14, 2020, 07:20:41 PM »
Firstly, we have the formula $log(z) = ln|z| + i \cdot arg(z)$ and $z:=log(\pm(2n+1)\pi i)$.
Then by plugging $z$ into that formula we have $ln((2n+1)\pi) + i\cdot arg(\pm(2n+1)\pi i)$.
Note that $\pm(2n+1)\pi i$ is always on the y-axis/imaginary-axis.
Thus $i\cdot arg(\pm(2n+1)\pi i) = i\cdot (\frac{1}{2} + m) \cdot \pi$. Plus we need $m\pi$ here since it is periodic.
Also we will not have $2m\pi$ since it does not include all the case we have in this question. We need to remember we have to include the all y-axis.

About part b of this problem, I did by setting n=0,1,2,... and try m correspondingly. You will see the restriction we have: $D \in {z:|z|<3}$ does not include many cases. Only when $n=0$ and $m=\pm1$ work here (n goes beyond 1 will break our restriction).

25
##### Test 1 / Re: 2018 test 1 variant C 2b
« on: October 13, 2020, 01:51:14 PM »
Sorry I didn't not make myself clear on this, I think the answer has a typo. The denominator in the original screenshot should be $(2n+2)\cdot(2n+1)$

26
##### Test 1 / Re: 2018 test 1 variant C 2b
« on: October 13, 2020, 01:00:11 PM »
No. I think it should be $(2n+2) \cdot (2n+1)$in the denominator. The last line of my equation in previous comment is the same stuff. Since (2n+2)=2(n+1), then cancel out the same term on denominator and numerator. If I didn't make myself clear pls comment below

27
##### Test 1 / Re: 2018 test 1 variant C 2b
« on: October 13, 2020, 12:03:03 PM »
I believe the equation can be written as:
$= |z| \cdot \frac{(n+1)^{3}}{(2n+2)(2n+1)}$
$= |z| \cdot \frac{(n+1)^{2}}{2 \cdot (2n+1)}$

28
##### Test 1 / Re: 2018 test 1 variant C 2b
« on: October 13, 2020, 09:45:35 AM »
It will be clearer if you expand the factorial:
$2n! = 2n \cdot (2n-1) \cdot ... \cdot 1$,
$(2n+2)! = (2n+2) \cdot (2n+1) \cdot (2n) \cdot ... \cdot 1 = (2n+2) \cdot (2n+1) \cdot (2n)!$
Thus $\frac {(2n)!}{(2n+2)!} = \frac {1}{(2n+2) \cdot (2n+1)}$

29
##### Quiz 3 / LEC 5101 Quiz3 Question 5A
« on: October 09, 2020, 10:06:44 PM »
Problem:
Show that $F(z) = e^{z}$ maps the strip $S = \{x+iy: -\infty < x < \infty, -\pi/2 \leq y \leq \pi/2\}$ onto the region $D = \{w = s + it: s \geq 0, w \ne 0\}$ and that $F$ is one-to-one on $S$. Furthermore, show that $F$ maps the boundary of $S$ onto all the boundary of $Q$ except $\{w = 0\}$. Explain what happens to each of the horizontal lines $\{Im z = \pi/2\}$ and $\{Im z = -\pi/2\}$.

Proof:
Firstly, we are going to expand the equation and prove that $F$ maps the strip $S$ onto the region $D$.
Let $F(z) = z_0 = u_0 + iv_0$.
$F(z) = e^{z} = e^{x+iy} = e^{x} \cdot e^{iy} = e^{x} \cdot (cos(y) + isin(y)) = e^{x}cos(y) + ie^{x}sin(y) = u_0 + iv_0$.
Note $x \in \mathbb{R}, y \in [-\pi/2, \pi/2]$, we may conclude that $e^{x} \ge 0, cos(y) \in [0,1], and \sin(y) \in [-1,1]$.
Then we have $Re(z_0) = u_0 = e^{x}cos(y) \ge 0$.
Although we get $Re(z_0) = u_0 = 0$ when $x = \pi/2$ or $-\pi/2, Im(z_0) \ne 0$. Thus after mapping, $z_0$ takes all positive x-axis except origin.
Moreover, we have $Im(z_0) = v_0 = e^{x}sin(y) \in \mathbb{R}$.
Since $z_0$ is an arbitrary element we chose from S, we have proved S gets mapped onto D.

Secondly, we are proving $F(z)$ is one-to-one mapping on S.
Equivalently, WTS $F(z_1) = F(z_2) \Longrightarrow z_1 = z_2$.
$\begin{gather} F(z_1) = F(z_2) \\ e^{x_1}\cdot e^{iy_1} = e^{x_2} \cdot e^{iy_2} \\ {\frac{e^{x_1} \cdot e^{iy_1}}{e^{x_2}\cdot e^{iy_2}}} = 1 \\ e^{x_1-x_2} \cdot e^{i(y_1-y_2)} = 1 \end{gather}$
That implies $Re(z_1) = Re(z_2)$ and $Im(z_1) = Im(z_2)$, alternatively, $x_1 = x_2$ and $y_1 = y_2$
Therefore we have proved F is one-to-one on S.

Lastly, we will discuss what happens to the boundary $\{Im z = \pi/2\}$ and $\{Im z = -\pi/2\}$.
When $y = \frac {\pi}{2}$:
$\begin{gather} \begin{split} F(x + i\pi/2) & = e^{x} \cdot e^{\frac {i\pi}{2}} \\ & = e^{x} \cdot (cos(\pi/2) + isin(\pi/2)) \\ & = ie^{x} \end{split} \end{gather}$

When $y = \frac {-i\pi}{2}$:
$\begin{gather} \begin{split} F(x - i\pi/2) & = e^{x} \cdot e^{\frac {-i\pi}{2}} \\ & = e^{x} \cdot (cos(-\pi/2) + isin(-\pi/2)) \\ & = -ie^{x} \end{split} \end{gather}$

Since $e^{x}$ > 0, boundaries $\{Im z = \pi/2\}$ and $\{Im z = -\pi/2\}$ is mapped to positive y-axis and negative y-axis except the origin 0.

The pic below is the domain before and after mapping.

30
##### Chapter 2 / Question on 2.1 Example 10
« on: October 02, 2020, 08:08:28 PM »
Hi guys, I just went through chapter 2.1 and found that the example 10 is quite confusing. I am particularly wondering, firstly, why the derivative is du/dx+i dv/dx? Why can't we take the derivative with respect to y? It says Theorem 3 implies the rationale of taking the derivative. Indeed it does, but I could not find how does Theorem 3 indicate the computation of the derivative. Furthermore, by Cauchy-Riemann equation we have du/dx=dv/dy, and du/dy=-dv/dx, if we do take the derivative with respect to y, then the derivative of log(z) would have real part of z and imaginary part of z switched in numerator. Does this cause a problem?

And secondly, it has drawn the conclusion that function log(z) is not analytic on any domain D that contains a simple closed curve that surrounds the origin. So I think log(z) on complex plane is similar as log(x) on real plane, and they both does not have derivative at 0. I have a rough idea about its reasoning but I dont know if I am on the correct track, correct me if I am wrong pls.

Btw I have highlighted the two parts that I have mentioned in my question in the picture below.

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