### Author Topic: Problem 1--simplified  (Read 20912 times)

#### Bowei Xiao

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##### Problem 1--simplified
« on: November 01, 2012, 02:11:55 PM »
I guess in problem 1 part c and d...you actually want to type X^n exp() and so does d? Guess there's typo there..

Yes, thanks, it was typo, but I decided for a sake of simplicity to take $n=1$. V.I.
« Last Edit: November 04, 2012, 05:03:20 PM by Victor Ivrii »

#### Calvin Arnott

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##### Re: Problem 1
« Reply #1 on: November 03, 2012, 08:37:38 PM »
In part d. of this question, the Fourier transform of $x^ne^{-\alpha |x|}\cos(\beta x)$, I can find quite a few representations for $F(k)$ but none of them have a really tractable closed form. Is it OK to write an answer for the transform in the form of operators acting on a function?- this form of the solution is easily the most elegant in appearance.

#### Victor Ivrii

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##### Re: Problem 1- simplified
« Reply #2 on: November 04, 2012, 05:01:40 PM »
In part d. of this question, the Fourier transform of $x^ne^{-\alpha |x|}\cos(\beta x)$, I can find quite a few representations for $F(k)$ but none of them have a really tractable closed form. Is it OK to write an answer for the transform in the form of operators acting on a function?- this form of the solution is easily the most elegant in appearance.

If you can (c) then you can (d); however to make things simpler I changed $n$ to $1$ in both (c) and (d).

#### Hanqing Liu

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##### Re: Problem 1--simplified
« Reply #3 on: November 04, 2012, 05:34:18 PM »
In part b and d, Does it matter if we change the trig into its complex form?

#### Victor Ivrii

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##### Re: Problem 1--simplified
« Reply #4 on: November 04, 2012, 05:51:43 PM »
In part b and d, Does it matter if we change the trig into its complex form?

#### Hanqing Liu

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##### Re: Problem 1--simplified
« Reply #5 on: November 04, 2012, 05:59:43 PM »
Sorry

Also, just to clarify, the 1/2pi term in front of FT could be a part of either Fourier integral or Fourier transform, where should we usually put it?

#### Victor Ivrii

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##### Re: Problem 1--simplified
« Reply #6 on: November 04, 2012, 06:08:12 PM »
Sorry

Also, just to clarify, the 1/2pi term in front of FT could be a part of either Fourier integral or Fourier transform, where should we usually put it?

Usually in FT, not in FI=IFT.

As I mentioned sometimes both carry $\frac{1}{\sqrt{2\pi}}$. AFAIAC you need to be just consistent.

#### Hanqing Liu

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##### Re: Problem 1--simplified
« Reply #7 on: November 04, 2012, 06:40:54 PM »
Does the condition "beta>0" matter in this question? Since it doesn't really enter the integral evaluation.

#### Victor Ivrii

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##### Re: Problem 1--simplified
« Reply #8 on: November 04, 2012, 06:51:38 PM »
Does the condition "beta>0" matter in this question? Since it doesn't really enter the integral evaluation.

No, it does not (you may guess why).

#### Hanqing Liu

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##### Re: Problem 1--simplified
« Reply #9 on: November 04, 2012, 07:04:04 PM »
Does the condition "beta>0" matter in this question? Since it doesn't really enter the integral evaluation.

No, it does not (you may guess why).

We are worried about the possibility of the definite integrals values go to infinity.

When we evaluate the definite integrals, only the sign of the real part (alpha) matters, since beta is timed by i, its sign doesn't matter.

But then again, I think it's necessary to impose condition alpha and beta belongs to R

#### Victor Ivrii

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##### Re: Problem 1--simplified
« Reply #10 on: November 04, 2012, 07:09:59 PM »
But then again, I think it's necessary to impose condition alpha and beta belongs to R

$\alpha >0$ implies $\alpha \in \mathbb{R}$ by definition.

#### Aida Razi

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##### Re: Problem 1--simplified
« Reply #11 on: November 07, 2012, 09:30:00 PM »
Solution is attached!

#### Fanxun Zeng

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##### Re: Problem 1--simplified
« Reply #12 on: November 07, 2012, 09:30:57 PM »
Question 1 a solution attached

#### Calvin Arnott

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##### Re: Problem 1--simplified
« Reply #13 on: November 07, 2012, 09:33:54 PM »
I ended up solving with $x^n$ in this problem. Notice that the messy integration in part b. could have been avoided by using properties of the Fourier transform as is done later in the problem.

I take the definition of the Fourier transform $\hat{f}$ for a function $f$ to be:  $F\left(k\right) = \hat{f}\left(k\right) = \int_{-\infty}^{\infty}f\left(x\right) e^{-i k x}dx,$ with inverse Fourier transform $f\left(x\right) = \check{F}\left(x\right) = \int_{-\infty}^{\infty}F\left(k\right) e^{i k x}\frac{dk}{2 \pi}$.

Problem

Let: $\alpha > 0, \beta > 0, n\in\mathbb{N}$. Compute the Fourier transform for:

a. $e^{-\alpha |x|}$

$$F\left(k\right) = \int_{-\infty}^{\infty}e^{-\alpha |x|} e^{-i k x}dx = \int_{0}^{\infty}e^{-\alpha \left(x\right)} e^{-i k x}dx + \int_{-\infty}^{0}e^{-\alpha \left(-x\right)} e^{-i k x}dx$$

$$= \int_{0}^{\infty}e^{-\left(\alpha + i k\right)x}dx + \int_{-\infty}^{0}e^{\left(\alpha - i k \right)x}dx = \frac{1}{-\left(\alpha + i k \right)}e^{-\left(\alpha + i k\right)x}\Bigr|_{x=0}^{\infty} + \frac{1}{\left(\alpha- i k\right)}e^{\left(\alpha - i k\right)x}\Bigr|_{x=-\infty}^{0}$$

$$= \lim_{x \to +\infty} \frac{1}{-\left(\alpha + i k \right)}e^{-\left(\alpha + i k\right)x} - \frac{1}{-\left(\alpha + i k \right)}e^{-\left(\alpha + i k\right)\left(0\right)} + \frac{1}{\left(\alpha- i k\right)}e^{\left(\alpha - i k\right)\left(0\right)} - \lim_{x \to -\infty} \frac{1}{\left(\alpha- i k\right)}e^{\left(\alpha - i k\right)x}$$

$$= 0 + \frac{1}{\left(\alpha + i k \right)} + \frac{1}{\left(\alpha- i k\right)} - 0 = \frac{\left(\alpha- i k\right)+\left(\alpha + i k \right)}{\left(\alpha + i k \right)\left(\alpha- i k\right)} = \frac{2\alpha}{\alpha ^2 - k^2}$$

$$\implies \frac{2\alpha}{\alpha ^2 - k^2} = F\left(k\right) \text{ is our Fourier transform }\blacksquare$$

b. i) $e^{-\alpha |x|}\cos\left(\beta x\right)$

$$F\left(k\right) = \int_{-\infty}^{\infty}e^{-\alpha |x|} \cos\left(\beta x\right) e^{-i k x}dx = \int_{0}^{\infty}e^{-\alpha \left(x\right)} e^{-i k x} \cos\left(\beta x\right)dx + \int_{-\infty}^{0}e^{-\alpha \left(-x\right)} e^{-i k x} \cos\left(\beta x\right)dx$$

$$\text{Now, } \forall\{a,b\}\in\mathbb{C}: \int e^{a x} \cos\left(b x\right) dx = \frac{e^{a x}}{a^2 + b^2}\left(b \sin\left(b x\right) + a \cos\left(b x\right)\right) \text{ so out integral is:}$$

$$\int_{0}^{\infty}e^{-\left(\alpha + i k\right) x} \cos\left(\beta x\right)dx + \int_{-\infty}^{0}e^{\left(\alpha -i k\right) x} \cos\left(\beta x\right)dx$$

$$= \frac{e^{\left(-\left(\alpha + i k\right)\right) x}}{\left(-\left(\alpha + i k\right)\right)^2 + \beta^2}\left(\beta \sin\left(\beta x\right) + \left(-\left(\alpha + i k\right)\right) \cos\left(\beta x\right)\right)\Bigr|_{x=0}^{\infty} + \frac{e^{\left(\alpha -i k\right) x}}{\left(\alpha -i k\right)^2 + \beta^2}\left(\beta \sin\left(\beta x\right) + \left(\alpha -i k\right) \cos\left(\beta x\right)\right)\Bigr|_{x=-\infty}^{0}$$

$$= \lim_{x \to +\infty} \frac{e^{-\left(\alpha + i k\right) x}}{\left(\alpha + i k\right)^2 + \beta^2}\left(\beta \sin\left(\beta x\right) - \left(\alpha + i k\right) \cos\left(\beta x\right)\right) - \frac{e^{-\left(\alpha + i k\right) \left(0\right)}}{\left(\alpha + i k\right)^2 + \beta^2}\left(\beta \sin\left(\beta \left(0\right)\right) -\left(\alpha + i k\right) \cos\left(\beta \left(0\right)\right)\right)$$

$$+ \frac{e^{\left(\alpha -i k\right) \left(0\right)}}{\left(\alpha -i k\right)^2 + \beta^2}\left(\beta \sin\left(\beta \left(0\right)\right) + \left(\alpha -i k\right) \cos\left(\beta \left(0\right)\right)\right) - \lim_{x \to -\infty} \frac{e^{\left(\alpha -i k\right) x}}{\left(\alpha -i k\right)^2 + \beta^2}\left(\beta \sin\left(\beta x\right) + \left(\alpha -i k\right) \cos\left(\beta x\right)\right)$$

$$= 0 - \frac{-\left(\alpha + i k\right)}{\left(\alpha + i k\right)^2 + \beta^2} + \frac{\left(\alpha -i k\right)}{\left(\alpha -i k\right)^2 + \beta^2} - 0$$

$$\implies \frac{\left(\alpha + i k\right)}{\left(\alpha + i k\right)^2 + \beta^2} + \frac{\left(\alpha -i k\right)}{\left(\alpha -i k\right)^2 + \beta^2} = F\left(k\right) \text{ is our Fourier transform } \blacksquare$$

b. ii) $e^{-\alpha |x|}\sin\left(\beta x\right)$

$$F\left(k\right) = \int_{-\infty}^{\infty}e^{-\alpha |x|} \sin\left(\beta x\right) e^{-i k x}dx = \int_{0}^{\infty}e^{-\alpha \left(x\right)} e^{-i k x} \sin\left(\beta x\right)dx + \int_{-\infty}^{0}e^{-\alpha \left(-x\right)} e^{-i k x} \sin\left(\beta x\right)dx$$

$$\text{Now, } \forall\{a,b\}\in\mathbb{C}: \int e^{a x} \sin\left(b x\right) dx = \frac{e^{a x}}{a^2 + b^2}\left(a \sin\left(b x\right) - b \cos\left(b x\right)\right) \text{ so out integral is:}$$

$$\int_{0}^{\infty}e^{-\left(\alpha + i k\right) x} \sin\left(\beta x\right)dx + \int_{-\infty}^{0}e^{\left(\alpha -i k\right) x} \sin\left(\beta x\right)dx$$

$$= \frac{e^{\left(-\left(\alpha + i k\right)\right) x}}{\left(-\left(\alpha + i k\right)\right)^2 + \beta^2}\left(\left(-\left(\alpha + i k\right)\right) \sin\left(\beta x\right) - \beta \cos\left(\beta x\right)\right)\Bigr|_{x=0}^{\infty}$$
$$+ \frac{e^{\left(\alpha -i k\right) x}}{\left(\alpha -i k\right)^2 + \beta^2}\left(\left(\alpha -i k\right) \sin\left(\beta x\right) - \beta \cos\left(\beta x\right)\right)\Bigr|_{x=-\infty}^{0}$$

$$= \lim_{x \to +\infty} \frac{e^{-\left(\alpha + i k\right) x}}{\left(\alpha + i k\right)^2 + \beta^2}\left(-\left(\alpha + i k\right) \sin\left(\beta x\right) - \beta \cos\left(\beta x\right)\right) - \frac{e^{-\left(\alpha + i k\right) \left(0\right)}}{\left(\alpha + i k\right)^2 + \beta^2}\left(-\left(\alpha + i k\right) \sin\left(\beta \left(0\right)\right) - \beta \cos\left(\beta \left(0\right)\right)\right)$$

$$+ \frac{e^{\left(\alpha -i k\right) \left(0\right)}}{\left(\alpha -i k\right)^2 + \beta^2}\left(\left(\alpha -i k\right) \sin\left(\beta \left(0\right)\right) - \beta \cos\left(\beta \left(0\right)\right)\right) - \lim_{x \to -\infty} \frac{e^{\left(\alpha -i k\right) x}}{\left(\alpha -i k\right)^2 + \beta^2}\left(\left(\alpha -i k\right) \sin\left(\beta x\right) - \beta \cos\left(\beta x\right)\right)$$

$$= 0 - \frac{-\beta}{\left(\alpha + i k\right)^2 + \beta^2} + \frac{- \beta}{\left(\alpha -i k\right)^2 + \beta^2} - 0$$

$$\implies \frac{\beta}{\left(\alpha + i k\right)^2 + \beta^2} - \frac{\beta}{\left(\alpha -i k\right)^2 + \beta^2} = F\left(k\right) \text{ is our Fourier transform } \blacksquare$$

c. $x^n e^{-\alpha |x|}$

We have for for any function $f\left(x\right)$ with Fourier transform $F\left(k\right),$ the transform of $g\left(x\right) = x f\left(x\right)$  is given by: $G\left(k\right) = i\frac{dF}{dk}$.  Moreover, it's clear that for: $g\left(x\right) = x^n f\left(x\right)$, $G\left(k\right) = i^n\frac{d^n F}{d k^n}$.

Proof:} We proceed by induction. Our base case is obvious: for $n = 1$, $g\left(x\right) = x^n f\left(x\right) = x f\left(x\right) \implies G\left(k\right) = i\frac{dF}{dk} = i^n\frac{d^n F}{d k^n}$. Suppose we have for some $n \in \mathbb{N}$, $h\left(x\right) = x^n f\left(x\right) \implies H\left(k\right) = i^n\frac{d^n F}{d k^n}.$ Then, $g\left(x\right) = x^{n+1} f\left(x\right) = x h\left(x\right) \implies G\left(k\right) = i \cdot i^n\partial_k H\left(k\right) = i^{n+1} \partial_k \frac{d^n F}{d k^n} = i^{n+1}\frac{d^{n+1} F}{d k^{n+1}}$ and $\forall n \in \mathbb{N}: g\left(x\right) = x^n f\left(x\right) \implies i^n\frac{d^n F}{d k^n}$, as needed $\square$

Now, we found in part a. that the Fourier transform for $f\left(x\right) = e^{-\alpha |x|}$ is given by: $F\left(k\right) = \frac{2\alpha}{\alpha ^2 - k^2}$. Examining the derivatives of $F\left(k\right)$:

$$F\left(k\right) = \frac {2\alpha} {\alpha^2-k^2}, F'\left(k\right) = \frac{4 k \alpha }{\left(\alpha ^2-k^2\right)^2}, F''\left(k\right) = 2 \left(\frac{-1}{\left(\alpha-k \right)^3}+\frac{1}{\left(\alpha+k \right)^3}\right)$$

$$F'''\left(k\right) = 6 \left( \frac{1}{\left(\alpha-k\right)^4}-\frac{1}{\left(\alpha+k \right)^4}\right), F^{\left(4\right)}\left(k\right) = 24 \left(\frac{-1}{\left(\alpha-k \right)^5}+\frac{1}{ \left(\alpha+k \right)^5}\right)$$

$$\text{So for: } n \in \mathbb{N}: F^n \left(k\right) = n! \left(\frac{\left(-1\right)^{n+1}}{\left(\alpha-k\right)^{n+1}}+\frac{\left(-1\right)^n}{\left(\alpha+k\right)^{n+1}} \right) \text{, and:}$$

$$g\left(x\right) = x^n f\left(x\right) = x^n e^{-\alpha |x|} \implies G\left(k\right) = i^n\frac{d^n F}{d k^n} = i^n n! \left(\frac{\left(-1\right)^{n}}{\left(\alpha-k\right)^{n+1}}+\frac{\left(-1\right)^{n+1}}{\left(\alpha+k\right)^{n+1}} \right) \text{ is our transform } \blacksquare$$

d. i) $x^n e^{-\alpha |x|}\cos\left(\beta x\right)$

Two properties of the Fourier transform are that for: $g\left(x\right) = e^{i a x} f\left(x\right)$ the Fourier transform of $g\left(x\right)$ is given by $G\left(k\right) = F\left(k-a\right)$, and that the transform is linear: $h\left(x\right) = a f\left(x\right) + b g\left(x\right) \implies H\left(k\right) = a F\left(k\right) + b G\left(k\right)$

$$\text{Now, } \cos{\beta x} = \frac{1}{2}\left(e^{i \beta x} + e^{- i \beta x}\right) \text{ so we write:}$$

$$x^n e^{-\alpha |x|}\cos\left(\beta x\right) = \frac{1}{2}\left(e^{i \beta x} x^n e^{-\alpha |x|} + e^{- i \beta x} x^n e^{-\alpha |x|}\right)$$

$$\text{Then our transform is given by: } G\left(k\right) = \frac{1}{2}\left(F\left(k-\beta\right) + F\left(k+\beta\right)\right) \text{ where: }$$

$$F\left(k\right) = i^n n! \left(\frac{\left(-1\right)^{n}}{\left(\alpha-k\right)^{n+1}}+\frac{\left(-1\right)^{n+1}}{\left(\alpha+k\right)^{n+1}} \right) \text{ is our transform from part c. for } x^n e^{-\alpha |x|}$$

$$\implies G\left(k\right) = \frac{1}{2} i^n n! \left( \frac{\left(-1\right)^{n}}{\left(\alpha+ \beta - k\right)^{n+1}}+\frac{\left(-1\right)^{n+1}}{\left(\alpha -\beta + k\right)^{n+1}} + \frac{\left(-1\right)^{n}}{\left(\alpha -\beta -k\right)^{n+1}}+\frac{\left(-1\right)^{n+1}}{\left(\alpha+\beta +k\right)^{n+1}} \right) \blacksquare$$

ii) $x^n e^{-\alpha |x|}\sin\left(\beta x\right)$

Proceeding as in part i) we use that: $\sin{\beta x} = \frac{1}{2 i}\left(e^{i \beta x} - e^{- i \beta x}\right)$ to write:

$$g\left(x\right) = x^n e^{-\alpha |x|}\sin\left(\beta x\right) = \frac{1}{2i}\left(e^{i \beta x} x^n e^{-\alpha |x|} - e^{- i \beta x} x^n e^{-\alpha |x|}\right)$$

$$\text{Which yields the transform: } G\left(k\right) = \frac{1}{2i}\left(F\left(k-\beta\right) - F\left(k+\beta\right)\right) \text{ with: } F\left(k\right) = i^n n! \left(\frac{\left(-1\right)^{n}}{\left(\alpha-k\right)^{n+1}}+\frac{\left(-1\right)^{n+1}}{\left(\alpha+k\right)^{n+1}} \right) \text{ and so:}$$

$$G\left(k\right) = \frac{1}{2 i} i^n n! \left( \frac{\left(-1\right)^{n}}{\left(\alpha+ \beta - k\right)^{n+1}}+\frac{\left(-1\right)^{n+1}}{\left(\alpha -\beta + k\right)^{n+1}} - \frac{\left(-1\right)^{n}}{\left(\alpha -\beta -k\right)^{n+1}}+\frac{\left(-1\right)^{n+1}}{\left(\alpha+\beta +k\right)^{n+1}} \right) \blacksquare$$
« Last Edit: November 18, 2012, 04:50:53 PM by Calvin Arnott »

#### Calvin Arnott

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##### Re: Problem 1--simplified
« Reply #14 on: November 07, 2012, 09:34:13 PM »
Part 2 of solution.

*edited to png
« Last Edit: November 07, 2012, 11:42:27 PM by Calvin Arnott »