1
Question:Find Fourier transformation of the function $$e^{-\alpha x^2/2}sin(\beta x)$$
with $$\alpha>0,\beta>0$$
with $$\alpha>0,\beta>0$$
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.
Pages: [1]
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!
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
$\int_\gamma(z^2+3z +4) dz$,
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]