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.


Topics - duoyizhang

Pages: [1]
1
Quiz-6 / Qz6-TWO-C
« on: March 25, 2021, 01:45:34 PM »
Question:Find Fourier transformation of the function $$e^{-\alpha x^2/2}sin(\beta x)$$
with $$\alpha>0,\beta>0$$

2
Quiz-5 / Qz5-THREE-E
« on: March 16, 2021, 01:27:02 AM »
The question is:Decompose f(x) = xcos(x)  into full Fourier series on interval [0, pi].
My confusion is how to decompose it on an interval like  [0,𝑙] rather than[-l,l]
Firstly,I compute f(x) into  full Fourier series on interval [-pi, pi] by the formula,Which is $$f1=f(x)=-\frac{sinx}{2}+\sum_{n=2}^\infty\frac{2n(-1)^{n}}{n^{2}-1}$$
Then what should we do to compute f(x) on[0,pi],I tried to use the property that f(x)is an odd function but it seems to be wrong.
Any help will be appreciated!

3
Quiz 4 / Quiz4 Lec0101_C
« on: October 27, 2020, 02:45:59 PM »
Evaluate the definite trigonometric integral making use of the technique of Examples 6 and 7 in this Section 2.3.
$\int_{0}^{2\pi}{\frac{d(\theta)}{5+3cos(\theta)}}$

4
Quiz 3 / Quiz3 Lec0101_B
« on: October 09, 2020, 12:01:57 PM »
The question is :Compute the following line integral:

$\int_\gamma(z^2+3z +4) dz$,

where $\gamma$ is the circle ${|z|=2}$,oriented counterclockwise.

Here is my solution:

f(z)=$(z^2+3z +4)$,

$\gamma(t)=2e^{it}$ ,$0 \leq t \leq 2\pi$

${\gamma(t)}'$=$2ie^{it}$

$f(\gamma(t))=4e^{2it}+6e^{it}+4$

$\int_\gamma f(z) dz$=$\int_{0}^{2\pi} f(\gamma(t))\gamma(t)'dt$

                     =$\int_{0}^{2\pi} (4e^{2it}+6e^{it}+4)2ie^{it}dt$
                     
                     =2i$\int_{0}^{2\pi} (4e^{3it}+6e^{2it}+4e^{it})dt$
                     
                     =$2i(\frac{4}{3i}e^{3it}+\frac{6}{2i}e^{2it}+4e^{it})|_{0}^{2\pi}$
                     
                     =0
                     
Since $e^{it}|_{0}^{2\pi}$=$cos(2\pi)+isin(2\pi)-cos(0)-isin(0)$
                         
                          =0
                         
Similarly, $e^{3it}|_{0}^{2\pi}$= $e^{2it}|_{0}^{2\pi}$=0

Pages: [1]