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##### Quiz-6 / Qz6-TWO-C

« Last post by**duoyizhang**on

*March 25, 2021, 01:45:34 PM*»

Question:Find Fourier transformation of the function $$e^{-\alpha x^2/2}sin(\beta x)$$

with $$\alpha>0,\beta>0$$

with $$\alpha>0,\beta>0$$

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Question:Find Fourier transformation of the function $$e^{-\alpha x^2/2}sin(\beta x)$$

with $$\alpha>0,\beta>0$$

with $$\alpha>0,\beta>0$$

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f(x) = e^−α|x|sin(βx)

Answer: f1(x) = e^−α|x|

f1^(k) = (2/pi)^1/2(α/α^2 + k^2)

f^(k) = F(f(x)) = -i/2(F(e^−α|x|e^iβx)-F(e^−α|x|e^-iβx))

F(e^−α|x|e^iβx) = (2/pi)^1/2(α/α^2 +(k^2- β^2)^2)

F(e^−α|x|e^-iβx))= (2/pi)^1/2(α/α^2 +(k^2+β^2)^2)

f^(k) = -i/(2pi)^1/2(α/α^2 +(k^2- β^2)^2 - α/α^2 +(k^2+β^2)^2 )

Answer: f1(x) = e^−α|x|

f1^(k) = (2/pi)^1/2(α/α^2 + k^2)

f^(k) = F(f(x)) = -i/2(F(e^−α|x|e^iβx)-F(e^−α|x|e^-iβx))

F(e^−α|x|e^iβx) = (2/pi)^1/2(α/α^2 +(k^2- β^2)^2)

F(e^−α|x|e^-iβx))= (2/pi)^1/2(α/α^2 +(k^2+β^2)^2)

f^(k) = -i/(2pi)^1/2(α/α^2 +(k^2- β^2)^2 - α/α^2 +(k^2+β^2)^2 )

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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!

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Solution is allowed to be discontinuous.

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I had a question about problem $1$ of my Test$2$. (My version was called alternative-F, night section).There was something confusing about this problem that I realized during the test.

Here is the problem:

\begin{equation}

\nonumber

\left\{ \begin{aligned}

& u_{tt}-u_{xx}=0, &&0< t < \pi, 0 < x < \pi, &(1.1) \\\

&u|_{t=0}= 2\cos (x), && 0< x < \pi, &(1.2)\\

&u_t|_{t=0}= 0, && 0< x < \pi, &(1.3)\\

&u|_{x=0}= u|_{x= \pi}=0, && 0< t < \pi. &(1.4)

\end{aligned}

\right.

\end{equation}

So, on the $t-x$ diagram, the lines $t = 0$ and $x = \pi$ intersect at ($t=0,x=\pi$), which will be a boundary point of the region where $0<t<x<\pi$.

If this point (i.e. ($t=0,x=\pi$)) on the diagram is approached by the line $t = 0$, equation $(1.2)$ is used to conclude that the value of $u(x,t)$ approaches $-2$.

On the other hand, if the point is approached from the line $x = \pi$, $u(x,t)$ should become zero, which is in contradiction with the other boundary condition. It seems this would make it impossible to incorporate conditions $(1.2)$ and $(1.4$) at the same time. I was hoping someone could clear my confusion regarding this problem.

Here is the problem:

\begin{equation}

\nonumber

\left\{ \begin{aligned}

& u_{tt}-u_{xx}=0, &&0< t < \pi, 0 < x < \pi, &(1.1) \\\

&u|_{t=0}= 2\cos (x), && 0< x < \pi, &(1.2)\\

&u_t|_{t=0}= 0, && 0< x < \pi, &(1.3)\\

&u|_{x=0}= u|_{x= \pi}=0, && 0< t < \pi. &(1.4)

\end{aligned}

\right.

\end{equation}

So, on the $t-x$ diagram, the lines $t = 0$ and $x = \pi$ intersect at ($t=0,x=\pi$), which will be a boundary point of the region where $0<t<x<\pi$.

If this point (i.e. ($t=0,x=\pi$)) on the diagram is approached by the line $t = 0$, equation $(1.2)$ is used to conclude that the value of $u(x,t)$ approaches $-2$.

On the other hand, if the point is approached from the line $x = \pi$, $u(x,t)$ should become zero, which is in contradiction with the other boundary condition. It seems this would make it impossible to incorporate conditions $(1.2)$ and $(1.4$) at the same time. I was hoping someone could clear my confusion regarding this problem.

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Hi, this is my answer for QUIZ3 TWO-C in section 5301. Hope this can help you out!

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Everything is correct. You need to look carefully at limits in the integrals

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$\int_0^\infty$. I fixed it

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I am not sure what should this integral integrate over. Is it -∞ to +∞ or 0 to +∞?

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I believe as t->0^{+} and x>0, the integral I marked in red should be sqrt(pi), then U(x, t) should be 1/2. I do not know how we get 1.