How evaluate this $f(x)={x}^{2}+\frac{1}{1+2{x}^{4}}$ with fourier tranform

ormaybesaladqh
2022-10-23
Answered

How evaluate this $f(x)={x}^{2}+\frac{1}{1+2{x}^{4}}$ with fourier tranform

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ehedem26

Answered 2022-10-24
Author has **13** answers

The Fourier transform of f(x) doesn't exist in the usual sense, but since f can be viewed as a tempered distribution, we can interpret the Fourier transform in that setting. (I'm using the normalization $\hat{f}(\omega )={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{e}^{-i\omega t}f(t)\phantom{\rule{thinmathspace}{0ex}}dt$. If you're using something else, the answer is a little different.)

First of all, Fourier transform of 1 is $2\pi \delta (\omega )$. Hence

$$\begin{array}{rl}t& \stackrel{\mathcal{F}}{\to}2\pi i{\delta}^{\prime}(\omega )\\ {t}^{2}& \stackrel{\mathcal{F}}{\to}2\pi {i}^{2}{\delta}^{\u2033}(\omega )=-2\pi {\delta}^{\u2033}(\omega ).\end{array}$$

The second term is less problematic, and exists in the usual sense. It is a standard exercise in residue calculus to compute the Fourier transform of $\frac{1}{1+2{x}^{4}}$. The result (and especially all the intermediate steps) are very messy though. I get:

$$\hat{f}(\omega )=-2\pi {\delta}^{\u2033}(\omega )+\{\begin{array}{ll}\pi q{e}^{q\omega}(\mathrm{cos}q\omega -\mathrm{sin}q\omega ),& \omega <0\\ \pi q{e}^{-q\omega}(\mathrm{sin}q\omega -\mathrm{cos}q\omega ),& \omega \ge 0\end{array}$$

where $q={2}^{1/4}/2$

First of all, Fourier transform of 1 is $2\pi \delta (\omega )$. Hence

$$\begin{array}{rl}t& \stackrel{\mathcal{F}}{\to}2\pi i{\delta}^{\prime}(\omega )\\ {t}^{2}& \stackrel{\mathcal{F}}{\to}2\pi {i}^{2}{\delta}^{\u2033}(\omega )=-2\pi {\delta}^{\u2033}(\omega ).\end{array}$$

The second term is less problematic, and exists in the usual sense. It is a standard exercise in residue calculus to compute the Fourier transform of $\frac{1}{1+2{x}^{4}}$. The result (and especially all the intermediate steps) are very messy though. I get:

$$\hat{f}(\omega )=-2\pi {\delta}^{\u2033}(\omega )+\{\begin{array}{ll}\pi q{e}^{q\omega}(\mathrm{cos}q\omega -\mathrm{sin}q\omega ),& \omega <0\\ \pi q{e}^{-q\omega}(\mathrm{sin}q\omega -\mathrm{cos}q\omega ),& \omega \ge 0\end{array}$$

where $q={2}^{1/4}/2$

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But I don't know how to solve such cases

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Where ${f}_{1}(s)$ is in the form of $\frac{{f}^{2}(s)}{{f}^{3}(s)}$, ${f}_{2}(s)={k}_{0}s$, and ${f}_{0}(s)$ is in the form of ${k}_{1}+\frac{{f}^{4}(s)}{s({s}^{2}+{\alpha}_{0}^{2}){f}^{2}(s)}$

$${f}_{1}(s)X(s)+{f}_{2}(s)({e}^{i\varphi}X(s-i{\alpha}_{1})+{e}^{-i\varphi}X(s+i{\alpha}_{1}))={f}_{0}(s)$$

Where ${f}_{1}(s)$ is in the form of $\frac{{f}^{2}(s)}{{f}^{3}(s)}$, ${f}_{2}(s)={k}_{0}s$, and ${f}_{0}(s)$ is in the form of ${k}_{1}+\frac{{f}^{4}(s)}{s({s}^{2}+{\alpha}_{0}^{2}){f}^{2}(s)}$

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b.$16\frac{{d}^{2}y}{{dx}^{2}}-8\frac{dy}{dx}+y=0$

c.$3\frac{{d}^{2}y}{{dx}^{2}}+2\frac{dy}{dx}=0$

d.$\frac{{d}^{2}y}{{dx}^{2}}=-4y$

a.

b.

c.

d.

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