### Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

### Messages - Pengyun Li

Pages: [1] 2
1
##### Quiz 5 / Quiz 5 - 5101 A
« on: November 05, 2020, 07:16:45 PM »
Question: Find the Laurent series for the given function about the indicated point. Also, give the residue of the function at the point. $z\cos{\frac{1}{z}}$ at $z = \infty$.

Let $w = \frac{1}{z}$, then we are finding Laurent series for $\frac{\cos{w}}{w}$ at $w = 0$.

$\cos{w} = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}w^{2n}$,

so $z\cos{\frac{1}{z}} = \frac{\cos{w}}{w} = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}w^{2n-1} = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}(\frac{1}{z})^{2n-1}$.

Note that $\frac{\cos{w}}{w} = \frac{1}{w}(1-\frac{w^2}{2!}+\frac{w^4}{4!}-...) = \frac{1}{w}- \frac{w}{2!}+\frac{w^3}{4!}-...$,
so the residue of the function is 1 as the coefficient for the term $\frac{1}{w}$ is 1.

$z\cos{\frac{1}{z}} = \frac{1}{\frac{1}{z}}- \frac{\frac{1}{z}}{2!}+\frac{(\frac{1}{z})^3}{4!}-... = z-\frac{1}{z(2!)}+\frac{1}{z^3(4!)}-...$

2
##### Quiz 4 / Quiz-5101-D
« on: October 22, 2020, 07:12:31 PM »
Question: Evaluate the given integral using Cauchy’s Formula or Theorem: $\int_{|z|=2}\frac{e^z}{z(z-3)}dz$.

For $z(z-3)=0$, $z=0$ or $z=3$, where only $z=0$ is bounded by $|z|=2$, thus $z_0=0$.

$\int_{|z|=2}\frac{e^z}{z(z-3)}dz = \int_{|z|=2}\frac{\frac{e^z}{z-3}}{z-0}dz$

(By Cauchy's formula)

$= 2\pi i f(z_0)$, where $f(z) = \frac{e^z}{z-3}$ is analytic on $\mathbb{C}$,

Thus, $2\pi i f(z_0) = 2\pi i \cdot \frac{e^0}{0-3} = -\frac{2\pi i}{3}$.

Therefore, $\int_{|z|=2}\frac{e^z}{z(z-3)}dz = -\frac{2\pi i}{3}$.

3
##### Quiz 2 / Quiz 2- Lec 5101-A
« on: October 01, 2020, 07:08:28 PM »
$\textbf{Question}$: Find the limit of each function at the given point, or explain why it does not exist: $f(z) = (z-2)log|z-2|$ at $z_0 = 2$.

$\textbf{Answer}$: Since $z_0= 2$, let $z' = z-2$.
$$\lim_{z\to z_0} f(z) = \lim_{z' \to 0} f(z) =z'\log |z'|$$
$$=z' (\ln |z'| +i\cdot 0) = \frac{ln |z'|}{\frac{1}{z'}}$$
By L'Hôpital's Rule, $$= \frac{\frac{1}{z'}}{-\frac{1}{(z')^2}}=\lim_{z'\to 0} (-z') = 0$$.

Therefore, the limit of the function is 0 at $z_0=2$.

4
##### Quiz 2 / Re: Quiz 2 Session 6101
« on: October 01, 2020, 05:56:14 PM »
I think the answer should be convergent instead? But correct me if I'm wrong

$(\frac{1+i}{\sqrt{2}})^n = e^{i\frac{n\pi}{4}} = \cos (\frac{n\pi}{4})+i\sin (\frac{n\pi}{4})$.

When $n = 1$, $\frac{1+i}{\sqrt{2}} = \frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}$.

When $n = 2$, $(\frac{1+i}{\sqrt{2}})^2 = 0+i$.

When $n = 3$, $(\frac{1+i}{\sqrt{2}})^3 = -\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}$.
......

Then you will find that the result cycles for every 8 members.

Denote the result for $n=1$ as $Z_1$, $n=2$ as $Z_2$ , etc.,

$Z_1 + Z_5 + Z_9 +.... = (\frac{1}{\sqrt{2}}-\frac{1}{5\sqrt{2}}+\frac{1}{9\sqrt{2}}-....)+i(\frac{1}{\sqrt{2}}-\frac{1}{5\sqrt{2}}+\frac{1}{9\sqrt{2}}-.....)$. It is very obvious to see that this series is convergent.

Similarly, we can figure out that $(Z_2 + Z_6 + Z_{10} +.... )$, $(Z_3 + Z_7 + Z_{11}+.... )$, $(Z_4 + Z_8 + Z_{12} +.... )$ all convergent.

Thus, $\sum_{n=1}^{\infty}(\frac{1+i}{\sqrt{2}})^n$ is convergent.

Also, $\lim_{n\to\infty} \frac{1}{n}\rightarrow 0$, is convergent.

Therefore, $\sum_{n=1}^\infty \frac{1}{n}(\frac{1+i}{\sqrt{2}})^n$  is convergent.

5
##### Chapter 1 / Re: how to solve question 33 in textbook section 12
« on: September 29, 2020, 09:40:44 PM »
According to what has been provided in Textbook: The translation of $\beta$ of $C$ is the circle centered at $z_0+\beta$ with the same radius. If the line L is $\{z:Re(Az+B) = 0\}$, then the translation of L by $\beta$ is the line $\{w: Re(Aw+B-A\beta) = 0\}$

6
##### Quiz 1 / Quiz 1- Lec 5101-A
« on: September 24, 2020, 07:06:24 PM »
$\textbf{Question:}$ Describe the locus of points z satisfying the given equation: $Re(z^2) = 4$.

$\textbf{Answer:}$

Let $z=x+iy$,

$z^2 = (x+iy) (x+iy) = x^2-y^2+(2xy)i$.

Thus, $Re(z^2) = x^2-y^2=4$.

Therefore, the locus of points z is a hyperbola with equation $x^2-y^2=4$.

7
##### Quiz-6 / Re: Q6 TUT 0801
« on: November 18, 2018, 02:43:08 PM »
Firstly, we need to find the eigenvalues and eigenvectors.

$det(A-\lambda I) = det\left|\begin{matrix}1-\lambda & 1 & 1 \\ 2 & 1-\lambda & -1 \\ -8 & -5 & -3-\lambda\end{matrix}\right| = (\lambda + 2)(\lambda - 2)(\lambda + 1)$

Thus, the eigenvalues are $\lambda_1 = -2, \lambda_2 = -1, \lambda_3 = 2$

When $\lambda_1 = -2, (A - \lambda I) = (A + 2I) = \left(\begin{matrix}3 & 1 & 1 \\ 2 & 3 & -1 \\ -8 & -5 & -1\end{matrix}\right)$

\begin{equation*}
\begin{pmatrix}
3 & 1 & 1\\
2 & 3 & -1\\
-8 & -5 & -1
\end{pmatrix}
\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=0
\end{equation*}
Let x_1 = t
\begin{equation*}
\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=t
\begin{pmatrix}
-4\\
5\\
7
\end{pmatrix}
\end{equation*}

Therefore, the corresponding eigenvector $\vec{v_1} = \left(\begin{matrix}-4 \\ 5 \\ 7\end{matrix}\right)$

Similarly, when $\lambda_2 = -1, \lambda_3 = 2$, we can get $\vec{v_2} = \left(\begin{matrix}-3 \\ 4 \\ 2\end{matrix}\right) and \ \vec{v_3} = \left(\begin{matrix}0 \\ -1 \\ 1\end{matrix}\right)$ respectively.

Therefore, the general solution $X(t) = c_1 e^{-t} \left(\begin{matrix}-3 \\ 4 \\ 2\end{matrix}\right) + c_2 e^{2t} \left(\begin{matrix}0 \\ -1 \\ 1\end{matrix}\right) + c_3 e^{-2t} \left(\begin{matrix}-4 \\ 5 \\ 7\end{matrix}\right)$

8
##### Quiz-5 / Re: Q5 TUT 0201
« on: November 04, 2018, 01:23:04 AM »
Hi Guanyao Liang,

Your answer is very close, but I think you messed up a calculation. The integral of $\tan(t)$ is $-\ln|\cos(t)|$, not $\ln|\sec(t)|$.

Therefore the solution should be:

$y = c_1 + c_2\cos(t) + c_3\sin(t) - \ln|\cos(t)| - \sin(t)\ln|\sec(t) + \tan(t)|$

Hi Michael, $\ln{|sec(t)|} and -\ln{|cos(t)|}$  are the same thing...

9
##### Quiz-5 / Re: Q5 TUT 0301
« on: November 02, 2018, 03:51:40 PM »
Isolate $x_2$ in first equation we get $x_2 = \frac{4}{3}x'_1 - \frac{5}{3}x_1$

Differentiate both sides with respect to t we get $x'_2 = \frac{4}{3}x''_1 - \frac{5}{3}x'_1$

Sub into second equation , we get $x''_1 - \frac{5}{2}x'_1 + x_1 = 0$

Characteristic equation is $r^2 - \frac{5}{2} r + 1 = 0$,

hence $r_1 = \frac{1}{2}, r_2 = 2$

General solution for $x_1$ is $x_1 = c_1e^\frac{t}{2}+ c_2 e^{2t}$

Plug into $x_2 = \frac{4}{3}x'_1 - \frac{5}{3}x_1$, we get $x_2 = -c_1 e^\frac{t}{2}+ c_2 e^{2t}$,

So, $x_1 = c_1e^\frac{t}{2}+ c_2e^{2t}$, $x_2 = -c_1 e^\frac{t}{2} + c_2 e^{2t}$

Plug in $x_1(0) = -2, x_2(0) = 1$,

$c_1 + c_2 = -2, -c_1 + c_2 = 1$

hence $c_1 = -\frac{3}{2}, c_2 = -\frac{1}{2}$

Therefore, $x_1 = -\frac{3}{2} e^\frac{t}{2}-\frac{1}{2} e^{2t}$,
$x_2 = \frac{3}{2} e^\frac{t}{2} -\frac{1}{2} e^{2t}$

10
##### Quiz-4 / Re: Q4 TUT 0201
« on: October 26, 2018, 05:41:52 PM »
Let $L(y) = x^2 y'' - 3xy' + 4y$

$y_1(x) = x^2, y_1'(x) = 2x, y_1''(x)= 2$

Sub it into $L(y)$, then $L(y) = 2x^2 - 3x(2x) + 4x^2 = 0$, hence $y_1(x) = x^2$ satisfies the homo equation.

$y_2(x) = x^2 ln(x), y_2'(x) = x + 2xln(x), y_2''(x) = 3 + 2ln(x)$

Sub it into $L(y)$, then $L(y) = x^2(3 + 2ln(x)) - 3x(x + 2xln(x) + 4x^2 ln(x) = 0$, hence $y_2(x) = x^2 ln(x)$ satisfies the homo equation.

$W[y_1, y_2](x) = \left|\begin{matrix}y_1(x) & y_2(x) \\ y_1'(x) & y_2'(x)\end{matrix}\right|= \left|\begin{matrix}x^2 & x^2 ln(x) \\ 2x & x + 2xln(x)\end{matrix}\right| = x^3$

Let the particular solution $y_p(x) = u_1 y_1(x) + u_2 y_2(x)$,

Since $x^2 y'' - 3xy' + 4y = x^2 ln(x)$, thus $y'' - \frac{3}{x}y' + \frac{4}{x^2}y = ln(x)$

$p(x) = - \frac{3}{x}, q(x) = \frac{4}{x^2}, g(x) = ln(x)$

$u_1 = -\int{\frac{y_2(x)g(x)}{W[y_1, y_2](x)}}dx = - \int{\frac{x^2 ln(x)ln(x)}{x^3}}dx = - \frac{1}{3}(ln(x))^3 +c_1$

$u_2 = \int{\frac{y_1(x)g(x)}{W[y_1, y_2](x)}}dx = \int{\frac{x^2 ln(x)}{x^3}}dx = \frac{(ln(x)^2)}{2} +c_2$

$y(x) = (- \frac{1}{3}(ln(x))^3 +c_1) x^2 + (\frac{(ln(x)^2)}{2}+c_2)x^2 ln(x) = \frac{1}{6}x^2(ln(x))^3 + c_1 x^2 + c_2 x^2 ln(x)$

The particular solution is thus

$y_p(x) = \frac{1}{6}x^2(ln(x))^3$

11
##### Quiz-3 / Re: Q3 TUT 0201
« on: October 12, 2018, 06:07:28 PM »
$W(x, xe^x) = \left|\begin{matrix}x & xe^x \\ x' & (xe^x)'\end{matrix}\right|= \left|\begin{matrix}x & xe^x \\ 1 & x^2e^x+e^x\end{matrix}\right| = x(x^2e^x+e^x) - xe^x = x^3e^x$

12
##### Thanksgiving Bonus / Re: Thanksgiving bonus 7
« on: October 07, 2018, 01:21:47 PM »
$y =xy'+\sqrt{(y')^2+1}$

Since we already know that $\psi(y)=\sqrt{(y')^2+1}, and \ \psi(y) = y'$

Plug $p = y'$,

we get $y = xp+\sqrt{p^2+1}$

So $\psi(p) = p,\psi'(p) = 1$

And $\psi(p) = \sqrt{p^2+1},\psi'(p) = \frac{p}{\sqrt{p^2+1}}$

Differentiate the equation w.r.t. $x$,

$pdx=pdx+(x\psi'(p)+\psi'(p))dp$

$0=(x+\frac{p}{\sqrt{p^2+1}})dp$

$dp = 0$

$\int1 dp=p=C$

Thus, $y = cx+\sqrt{c^2+1}$, is the general solution.

To get the singular solution in the parametric form,

we know that $x = -\psi'(p) = -\frac{p}{\sqrt{p^2+1}}$, hence, $\sqrt{p^2+1} =-\frac{p}{x}$,

since $y = xp+\psi(p)=xp+\sqrt{p^2+1}$, we can derive that $y = \frac{1}{\sqrt{p^2+1}} = -\frac{x}{p}$

Therefore, the singular solution in the parametric form s:
$\begin{cases} x = - \frac{p}{\sqrt{p^2+1}} \\ y = \frac{1}{\sqrt{p^2+1}} \end{cases}$

Since $x = -\frac{p}{\sqrt{p^2+1}}$, we can derive that $p =\pm\frac{x}{\sqrt{1-x^2}}$,

Sub into $y = -\frac{x}{p}$, we can get that:

$y = \pm\sqrt{1-x^2}$

13
##### Thanksgiving Bonus / Re: Thanksgiving bonus 7
« on: October 06, 2018, 08:18:45 PM »
$y =xy'+\sqrt{(y')^2+1}$

Since we already know that $\psi(y)=\sqrt{(y')^2+1}, and \ \psi(y) = y'$

Plug $p = y'$,

we get $y = xp+\sqrt{p^2+1}$

So $\psi(p) = p,\psi'(p) = 1$

And $\psi(p) = \sqrt{p^2+1},\psi'(p) = \frac{p}{\sqrt{p^2+1}}$

Differentiate the equation w.r.t. $x$,

$pdx=pdx+(x\psi'(p)+\psi'(p))dp$

$0=(x+\frac{p}{\sqrt{p^2+1}})dp$

$dp = 0$

$\int1 dp=p=C$

Thus, $y = cx+\sqrt{c^2+1}$, is the general solution.

To get the singular solution in the parametric form,

we know that $x = -\psi'(p) = -\frac{p}{\sqrt{p^2+1}}$, hence, $\sqrt{p^2+1} =-\frac{p}{x}$,

since $y = xp+\psi(p)=xp+\sqrt{p^2+1}$, we can derive that $y = \frac{1}{\sqrt{p^2+1}} = -\frac{x}{p}$

Therefore, the singular solution in the parametric form s:
$\begin{cases} x = - \frac{p}{\sqrt{p^2+1}} \\ y = - \frac{1}{p}{x} \end{cases}$
No it is not!!! I wrote solution in the parametric form already. Now the question is to exclude $p$ , expressing $y$ via $x$

14
##### Thanksgiving Bonus / Re: Thanksgiving bonus 1
« on: October 05, 2018, 11:15:44 PM »
Corrected:

We want to find a second order equation with the fundamental system of solutions $\{y_1(x),y_2(x)\} = \{\frac{1}{x+1}, \frac{1}{x-1}\}$.

Then $y_1=\frac{1}{x+1}, y_2=\frac{1}{x-1}$.

$\implies$ $W(y,y_1,y_2) = W(y, \frac{1}{x+1},\frac{1}{x-1})$ = $\left|\begin{matrix}y & \frac{1}{x+1} & \frac{1}{x-1}\\ y' &-\frac{1}{(x+1)^2} &-\frac{1}{(x-1)^2}\\ y'' &\frac{2}{(x+1)^3}& \frac{2}{(x-1)^3} \end{matrix}\right|= 0.$

$\implies$ Solve the determinant : $y\ \left|\begin{matrix} -\frac{1}{(x+1)^2} & -\frac{1}{(x-1)^2}\\ \frac{2}{(x+1)^3} &\frac{2}{(x-1)^3}\end{matrix}\right| - y^{'}\ \left|\begin{matrix} \frac{1}{x+1} & \frac{1}{x-1}\\ \frac{2}{(x+1)^3} &\frac{2}{(x-1)^3}\end{matrix}\right| + y^{''}\ \left|\begin{matrix} \frac{1}{x+1} & \frac{1}{x-1}\\ -\frac{1}{(x+1)^2} &-\frac{1}{(x-1)^2}\end{matrix}\right|$

= $y\ (-\frac{4}{(x+1)^3(x-1)^3}) - y^{'}\frac{8x}{(x+1)^3(x-1)^3} + y^{''}\frac{-2}{(x+1)^2(x-1)^2} = 0$

Then we multiply both sides with $(x-1)^3(x+1)^3$ to get:

$y(-4) - y^{'}(8x) + y^{''}(2-2x^2) = 0$

Our final equation is: $2y+4xy'+(x^2-1)y'' = 0$.

15
##### Thanksgiving Bonus / Re: Thanksgiving bonus 1
« on: October 05, 2018, 10:37:55 PM »
Hi Sir, I figured out the problem and corrected the answer below! Thank you!

Pages: [1] 2