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### Messages - Tianfangtong Zhang

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1
##### Final Exam / Re: FE-P1
« on: December 14, 2018, 12:25:45 PM »
$M = 2x \sin(y) +1$

$N = 4 x^2 \cos(y) + 3x \cot(y) + 5 \sin(2y)$

$M_y = 2x\cos(y)$

$N_x = 8x\cos(y) + 3cos(y)$

$\frac{M_y - N_x} {M} = \frac{-6x\cos(y) - 3cos(y)}{2x\sin(y) + 1} = -3\cot(y)$

$\mu = e^{-\int -3\cot(y) dy } = \sin^3(y)$

Then multiply $sin^3(y)$ on M and N.

Then we get $M = sin^3(y)(2x \sin(y) +1)$, $N =sin^3(y)(4 x^2 \cos(y) + 3x \cot(y) + 5 \sin(2y))$

$M_y = \sin^2(y) \cos(y) (8x \sin(y) + 3)$, and $N_x = \sin^2(y) \cos(y) (8x \sin(y) + 3)$

Thus $M_y = N_x$, exact.

Then $\psi_x = M$

$\psi = \int M dx = \int \sin^3(y)(2x \sin(y) +1) dx = \sin^3(y)(x^2\sin(y) + x) + h(y)$

$N = \psi_y = 4x^2\sin^3(y)\cos(y) + x\sin^2(y)\cos(y) + h'(y)$

$h'(y) = 10\sin^4cos(y)$

$h(y) = \int 10\sin^4cos(y) dy = 2\sin^5(y) + c$

Then $\psi = \sin^3(y)(x^2\sin(y) + x) + 2\sin^5(y) = c$

2
##### Quiz-5 / Re: Q5 TUT 5102
« on: November 03, 2018, 01:33:22 AM »
thank you, fix it

3
##### Quiz-5 / Re: Q5 TUT 5101
« on: November 02, 2018, 04:15:18 PM »
$$\tan(z) = \frac{\sin(z)}{\cos (z)} = \frac{z - \frac{1}{3!}z^3 + \frac{z^5}{5!} - \frac{z^7}{7!}...}{1 - \frac{z^2}{2!}+\frac{z^4}{4!}...} = a_0 + a_1z + a_2z^2 +a_3z^3 + ...$$

Thus
$(a_0 + a_1z + a_2z^2 +a_3z^3 + ...)(1 - \frac{z^2}{2!}+\frac{z^4}{4!}...) = z - \frac{1}{3!}z^3 + \frac{z^5}{5!} - \frac{z^7}{7!}...$

Thus we get

$a_0 = 0$

$a_1 = 1$

$a_2 = 0$

.
.
.

Therefore $\tan(z) = z + \frac{1}{3}z^3 + \frac{2}{15}z^5 ....$

4
##### Quiz-5 / Re: Q5 TUT 5301
« on: November 02, 2018, 03:43:00 PM »
\begin{align*}
\frac{z+2}{z+3} &= \frac{z+3-1}{z+3} \\ &= 1-\frac{1}{z+3}\\
&= 1- \frac{1}{z+1+2} \\ &= 1 - \frac{1}{2} \frac{1}{1+\frac{z+1}{2}}\\
&= 1 - \frac{1}{2} \frac{1}{1-\frac{-(z+1)}{2}}\\
&= 1 - \sum_{n=0}^{\infty}(\frac{-(z+1)}{2})^n \\
&= 1 - \sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}(z-(-1))^n
\end{align*}

5
##### Quiz-5 / Re: Q5 TUT 5201
« on: November 02, 2018, 03:41:27 PM »
\begin{align*}
\frac{z+2}{z+3} &= \frac{z+3-1}{z+3} \\ &= 1-\frac{1}{z+3}\\
&= 1- \frac{1}{z+1+2}\\ &= 1 - \frac{1}{2} \frac{1}{1+\frac{z+1}{2}}\\
&= 1 - \frac{1}{2} \frac{1}{1-\frac{-(z+1)}{2}}\\
&= 1 - \sum_{n=0}^{\infty}(\frac{-(z+1)}{2})^n \\
&= 1 - \sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}(z-(-1))^n
\end{align*}

6
##### Quiz-5 / Re: Q5 TUT 0101
« on: November 02, 2018, 03:29:12 PM »
Let $z^2(1-\cos(z)) = 0$

Then $z = 0$ or $\cos(z) = 1$

thus $z = 2k\pi$

case $1$: $z = 0$

let $f(z) = z^2$ and $h(z) = 1-\cos(z)$

Then $f(0)=0$

$f^{'}(0) = 2z|_{z=0} = 0$

$f^{''}(0)=2\neq 0$

thus order = 2

Then $h(0) = 0$

$h^{'}(0) = 2\sin(z)|_{z=0} = 0$

$f^{''}(0)=\cos(z)\neq 0$

thus order = 2

Therefore order(0) = 4

case $2$: $z = 2k\pi$ ($k\neq 0)$

let $f(z) = z^2$ and $h(z) = 1-\cos(z)$

Then $f(z) = z^2 = (2k\pi)^2 \neq 0$

thus order = 0

$h(z) = 1- \cos(z) = 0$

$h^{'}(z) = \sin(z) = 0$

$h^{''} = \cos(z) = 0$

thus order = 2

Therefore order($2k\pi$) = 2, $k\neq0$

7
##### Quiz-5 / Re: Q5 TUT 5102
« on: November 02, 2018, 03:20:04 PM »
homogeneous equation is
$y^{'''} - y^{'} = 0$

then $r^3 - r = 0$

$r = 0, \pm 1$

thus, $y_{c}(t) = c_{1} + c_{2}e^t + c_{3}e^{-t}$

\begin{align*}
W(t) &=

\begin{bmatrix}
1 & e^t & e^{-t} \\
0 & e^t & -e^{-t}\\
0 & e^t & e^{-t}
\end{bmatrix} \\
\\

&= 1*(-1)^{1+1}
\begin{bmatrix}
e^t & -e^{-t}\\
e^t & e^{-t}
\end{bmatrix}
+ e^t(-1)^{2+1}
\begin{bmatrix}
0 & -e^{-t}\\
0 & e^{-t}
\end{bmatrix}
+ e^{-t}(-1)^{3+1}
\begin{bmatrix}
0 & e^t\\
0 & e^t
\end{bmatrix}\\
\\

&= e^t*e^t - (-e^{-t})e^t + 0 + 0 \\
\\

&= 1 + 1 \\
\\

&= 2
\end{align*}

\begin{align*}
W_{1}(t) &=
\begin{bmatrix}
0 & e^t & e^{-t}\\
0 & e^t & -e^{-t}\\
1 & e^t & e^{-t}
\end{bmatrix}\\
\\

&=1*(-1)^{1+3}
\begin{bmatrix}
e^t & e^{-t}\\
e^t & -e^{-t}
\end{bmatrix}\\
\\

&= -1-1 \\
\\

&= -2
\end{align*}

\begin{align*}
W_{2}(t) &=
\begin{bmatrix}
1 & 0 & e^{-t}\\
0 & 0 & -e^{-t}\\
0 & 1 & e^{-t}
\end{bmatrix}\\
\\

&=1*(-1)^{1+1}
\begin{bmatrix}
0 & -e^{-t}\\
1 & e^{-t}
\end{bmatrix}\\
\\

&= e^{-t}
\end{align*}
\\

\begin{align*}
W_{3}(t) &=
\begin{bmatrix}
1 & e^t & 0\\
0 & e^t & 0\\
0 & e^t & 1
\end{bmatrix}\\
\\

&=1*(-1)^{1+1}
\begin{bmatrix}
e^{t} & 0\\
e^{t} & 1
\end{bmatrix}\\
\\

&= e^{t}
\end{align*}

\begin{align*}
y_{p}(t) &= y_{1}\int \frac{g(t)W_{1}(t)}{W(t)} dt + y_{2}\int \frac{g(t)W_{2}(t)}{W(t)} dt + y_{3}\int \frac{g(t)W_{3}(t)}{W(t)} dt \\
\\

&= 1\int \frac{\cosh(t)(-2)}{2)} dt + e^t\int \frac{\cosh(t)e^{-t}}{2} dt + e^{-t}\int \frac{\cosh(t)e^t}{2} dt \\
\\

&= -1\sinh(t) + \frac{1}{2}e^t\int \cosh(t)e^{-t} dt + \frac{1}{2}e^{-t}\int \cosh(t) e^t dt \\
\\

&= -\sinh(t) + \frac{1}{2}e^t \int \cosh(t)e^{-t}dt + \frac{1}{2}e^{-t}\int \cosh(t)e^t dt \\
\\

&= -\sinh(t) + \frac{1}{2}e^t(\frac{t}{2} - \frac{e^{-2t}}{4}) + \frac{1}{2}e^{-t}(\frac{t}{2}+ \frac{e^{2t}}{4})
\end{align*}

Therefore $y(t) = c_{1} + c_{2}e^t + c_{3}e^{-t} -\sinh(t) + \frac{1}{2}e^t(\frac{t}{2} - \frac{e^{-2t}}{4}) + \frac{1}{2}e^{-t}(\frac{t}{2}+ \frac{e^{2t}}{4})$

8
##### Quiz-5 / Re: Q5 TUT 0401
« on: November 02, 2018, 03:19:43 PM »

The homogeneous equation is $y^{'''} + y^{'} = 0$

$r^3 + r = 0$

$r = 0, \pm i$

Thus the roots are $r = 0, \pm i$

Thus $y_{c} = c_1 + c_2\cos(t) + c_3\sin(t)$

\begin{equation*}
W(t) =
W(1, \cos (t), \sin (t))
= \begin{bmatrix}
1 & \cos(t) & \sin(t) \\
0 & -\sin(t) & \cos(t) \\
0 & -\cos(t) &-\sin(t)
\end{bmatrix}
= \sin^{2}(t) + \cos^{2} (t)
= 1
\end{equation*}

\begin{equation*}
W_{1}(t) =
\begin{bmatrix}
0 & \cos(t) & \sin(t) \\
0 & -\sin(t) & \cos(t) \\
1 & -\cos(t) &-\sin(t)
\end{bmatrix}
= \sin^{2}(t) + \cos^{2} (t)
= 1
\end{equation*}

\begin{equation*}
W_{2}(t) =
\begin{bmatrix}
1 & 0 & \sin(t) \\
0 & 0 & \cos(t) \\
1 & 1 &-\sin(t)
\end{bmatrix}
= -\cos (t)
\end{equation*}

\begin{equation*}
W_{3}(t) =
\begin{bmatrix}
1 & \cos (t) & 0 \\
0 & -\sin(t) & 0 \\
0 & -\cos(t) & 1
\end{bmatrix}
= -\sin (t)
\end{equation*}

\begin{align*}
y_{p}(t) &= y_{1}\int \frac{g(t)W_{1}(t)}{W(t)} dt + y_{2}\int \frac{g(t)W_{2}(t)}{W(t)} dt + y_{3}\int \frac{g(t)W_{3}(t)}{W(t)} dt \\
\\
&= 1\int \frac{\sec(t) 1}{1} dt + \cos (t)\int \frac{\sec (t) (-\cos(t))}{1} dt + \sin(t)\int \frac{\sec(t)(-\sin(t))}{1} dt\\
\\
&= \ln (\sec(t)+\tan(t))-t\cos(t) + \sin(t)\ln(\cos(t))
\end{align*}

Thus, the general solution is

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

9
##### Thanksgiving bonus / Re: Thanksgiving bonus 4
« on: October 06, 2018, 10:58:53 PM »
sorry, there is a typo. The solution should be the given function doesn't represent a locally sourceless and an irrotational flow

10
##### Thanksgiving bonus / Re: Thanksgiving bonus 4
« on: October 06, 2018, 10:32:20 PM »
$cosh(x)cos(y)+isinh(x)sin(y)$
Then
$u=cosh(x)cos(y)$
$v=sinh(x)sin(y)$
$\frac{du}{dx} = sinh(x)cos(y)$
$\frac{du}{dy}= -cosh(x)sin(y)$
$\frac{dv}{dx} = cosh(x)sin(y)$
$\frac{dv}{dy} = sinh(x)cos(y)$
since $\frac{du}{dx} \neq - \frac{dv}{dy}$
$\frac{du}{dy} \neq \frac{dv}{dx}$
Thus, the flow is not globally sourceless and is not irrotational.

11
##### Quiz-2 / Re: Q2 TUT 0101
« on: October 05, 2018, 07:22:31 PM »
$$g(z) = \frac{4z^6-7z^3}{(Z^2-4)^3} = \frac{4z^6-7z^3}{z^6-12z^4+48z^2-64}$$
we can divide $z^6$ on both numerator and denominator.
Then we can get
$$g(z) = \frac{4-7z^{-1}}{z-12z^{-2}+48z^{-4}-64z^{-6}}$$
Then as $z\to\infty$
$$\lim_{z\to\infty} f(z) = \frac{4 - 0}{1 - 0 + 0 - 0} = 4$$

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