network profile
| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 9 months |
| seen | Jan 10 at 9:25 | |
| stats | profile views | 49 |
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Aug 28 |
asked | To show Taylor series of a Fourier transform $\hat{f }$ converges to $\hat{f}$ |
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Aug 28 |
awarded | Editor |
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Aug 28 |
comment |
find a certain $L^1(R)$ function I forgot to mention it's compactly supported |
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Aug 28 |
revised |
find a certain $L^1(R)$ function added 45 characters in body |
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Aug 28 |
asked | find a certain $L^1(R)$ function |
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Aug 26 |
comment |
convolution of a function with itself equals itself sorry again ...that's just riemannn-lebesgue lemma .. |
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Aug 26 |
comment |
convolution of a function with itself equals itself sorry , what I want to ask is : how do you get $\hat{f}$ goes to zero at infinity ? and why fourier transform maps a L1 funciton to a contiuous one ? is that some propositon ? Is it possible for you to show the solution ? |
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Aug 26 |
comment |
convolution of a function with itself equals itself Hi , I've added some working above , still need more help though.. |
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Aug 26 |
comment |
convolution of a function with itself equals itself I cant see the relation between $\hat{f}$ is continuous and f is zero function .. |
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Aug 26 |
comment |
convolution of a function with itself equals itself Now If I can put the limit under the integral and show $lim_{B\to \infty}\frac{sin(B2\pi(x-t))}{\pi(x-t)}$ is in both L1(R) and L2(R), then by replacing $f(x)$ by $lim_{B\to \infty}\frac{sin(B2\pi(x-t))}{\pi(x-t)}$, can I get the conclusion ? The reason that I need to show the function is in L1(R) is because to get the first equality , I assumed $f(x)$ is absolutely integrable. |
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Aug 26 |
comment |
convolution of a function with itself equals itself For part (2) , now my thinking is : $f(x) =\int\hat{f(w)}e^{2\pi iwx} = lim_{B\to \infty}\int_{-B}^B\hat{f(w)}e^{2\pi iwx} $ . After a few lines of change order of integral , this is equal to $lim_{B\to \infty}\int_Rf(t)\frac{sin(B2\pi(x-t))}{\pi(x-t)}dt $ . |
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Aug 26 |
awarded | Student |
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Aug 26 |
asked | convolution of a function with itself equals itself |
