The maximum value (peak) of multiple self-convolution of rectangular function The Next CEO of...
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The maximum value (peak) of multiple self-convolution of rectangular function
The Next CEO of Stack OverflowMultiple self-convolution of rectangular function - integral evaluationConvolution with multiple step functionsInfinite self-convolution for a functionNeed help with the convolution of two complex functionsMultiple self-convolution of rectangular function - integral evaluationConvolution of a function with itselfHolder continuity of the convolution of a Holder continuous functionConvolution of an integrable function an $L^infty$ functionCalculating the convolution of a piecewise functionConvolution of two rectangular pulsesMaximal value of integral over a fixed-length interval of the convolution
$begingroup$
In Multiple self-convolution of rectangular function - integral evaluation, formula for self-rectangular function of rectangular function seems to have been derived. How do we prove that this formula is accurate?
Also, what is the formula for the maximum value of multiple self-convolution of rectangular function?
integration fourier-analysis improper-integrals convolution
edited Apr 13 '17 at 12:21
Community♦
1
asked Apr 27 '15 at 14:51
LamasLamas
1
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add a comment |
$begingroup$
In Multiple self-convolution of rectangular function - integral evaluation, formula for self-rectangular function of rectangular function seems to have been derived. How do we prove that this formula is accurate?
Also, what is the formula for the maximum value of multiple self-convolution of rectangular function?
integration fourier-analysis improper-integrals convolution
edited Apr 13 '17 at 12:21
Community♦
1
asked Apr 27 '15 at 14:51
LamasLamas
1
$endgroup$
add a comment |
$begingroup$
In Multiple self-convolution of rectangular function - integral evaluation, formula for self-rectangular function of rectangular function seems to have been derived. How do we prove that this formula is accurate?
Also, what is the formula for the maximum value of multiple self-convolution of rectangular function?
integration fourier-analysis improper-integrals convolution
edited Apr 13 '17 at 12:21
Community♦
1
asked Apr 27 '15 at 14:51
LamasLamas
1
$endgroup$
In Multiple self-convolution of rectangular function - integral evaluation, formula for self-rectangular function of rectangular function seems to have been derived. How do we prove that this formula is accurate?
Also, what is the formula for the maximum value of multiple self-convolution of rectangular function?
integration fourier-analysis improper-integrals convolution
integration fourier-analysis improper-integrals convolution
edited Apr 13 '17 at 12:21
Community♦
1
asked Apr 27 '15 at 14:51
LamasLamas
1
edited Apr 13 '17 at 12:21
Community♦
1
asked Apr 27 '15 at 14:51
LamasLamas
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
asked Apr 27 '15 at 14:51
LamasLamas
1
asked Apr 27 '15 at 14:51
LamasLamas
1
asked Apr 27 '15 at 14:51
LamasLamas
1
1
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
I'll just address your first question.
Let
$$
h(t) = begin{matrix} 1 &; text{if} ; t in [0,1] \ 0 &; text{else}end{matrix}
$$
and let
$$
g_n(t) = (h*h*cdots*h)(t)
$$
be the n-fold convolution of $h$. Observe that
$$
widehat{g_n}(w) = left(frac{1}{iw}right)^nleft(1-e^{-iw}right)^n =: hat{b}(w)hat{c}(w)
$$
where $hat{f}(w) = int_{mathbb{R}} f(t) e^{-iwt} dt$. So then by the convolution theorem, we have
$$
g_n(t) = (b*c)(t)
$$
To find $b(t)$ and $c(t)$, you will have to find these distributional inverse transforms, and use the binomial theorem for $hat{c}(w)$. I'll leave this to you. You should find that:
$$
g_n(t) = frac{1}{2(n-1)!}sum_{j=0}^n begin{pmatrix} n \ j end{pmatrix} (-1)^j (t-j)^{n-1} text{sign}(t-j)
$$
You of course would have to show this is equivalent to your linked question.
answered Apr 28 '15 at 2:36
ChesterChester
1,004913
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add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
I'll just address your first question.
Let
$$
h(t) = begin{matrix} 1 &; text{if} ; t in [0,1] \ 0 &; text{else}end{matrix}
$$
and let
$$
g_n(t) = (h*h*cdots*h)(t)
$$
be the n-fold convolution of $h$. Observe that
$$
widehat{g_n}(w) = left(frac{1}{iw}right)^nleft(1-e^{-iw}right)^n =: hat{b}(w)hat{c}(w)
$$
where $hat{f}(w) = int_{mathbb{R}} f(t) e^{-iwt} dt$. So then by the convolution theorem, we have
$$
g_n(t) = (b*c)(t)
$$
To find $b(t)$ and $c(t)$, you will have to find these distributional inverse transforms, and use the binomial theorem for $hat{c}(w)$. I'll leave this to you. You should find that:
$$
g_n(t) = frac{1}{2(n-1)!}sum_{j=0}^n begin{pmatrix} n \ j end{pmatrix} (-1)^j (t-j)^{n-1} text{sign}(t-j)
$$
You of course would have to show this is equivalent to your linked question.
answered Apr 28 '15 at 2:36
ChesterChester
1,004913
$endgroup$
add a comment |
$begingroup$
I'll just address your first question.
Let
$$
h(t) = begin{matrix} 1 &; text{if} ; t in [0,1] \ 0 &; text{else}end{matrix}
$$
and let
$$
g_n(t) = (h*h*cdots*h)(t)
$$
be the n-fold convolution of $h$. Observe that
$$
widehat{g_n}(w) = left(frac{1}{iw}right)^nleft(1-e^{-iw}right)^n =: hat{b}(w)hat{c}(w)
$$
where $hat{f}(w) = int_{mathbb{R}} f(t) e^{-iwt} dt$. So then by the convolution theorem, we have
$$
g_n(t) = (b*c)(t)
$$
To find $b(t)$ and $c(t)$, you will have to find these distributional inverse transforms, and use the binomial theorem for $hat{c}(w)$. I'll leave this to you. You should find that:
$$
g_n(t) = frac{1}{2(n-1)!}sum_{j=0}^n begin{pmatrix} n \ j end{pmatrix} (-1)^j (t-j)^{n-1} text{sign}(t-j)
$$
You of course would have to show this is equivalent to your linked question.
answered Apr 28 '15 at 2:36
ChesterChester
1,004913
$endgroup$
add a comment |
$begingroup$
I'll just address your first question.
Let
$$
h(t) = begin{matrix} 1 &; text{if} ; t in [0,1] \ 0 &; text{else}end{matrix}
$$
and let
$$
g_n(t) = (h*h*cdots*h)(t)
$$
be the n-fold convolution of $h$. Observe that
$$
widehat{g_n}(w) = left(frac{1}{iw}right)^nleft(1-e^{-iw}right)^n =: hat{b}(w)hat{c}(w)
$$
where $hat{f}(w) = int_{mathbb{R}} f(t) e^{-iwt} dt$. So then by the convolution theorem, we have
$$
g_n(t) = (b*c)(t)
$$
To find $b(t)$ and $c(t)$, you will have to find these distributional inverse transforms, and use the binomial theorem for $hat{c}(w)$. I'll leave this to you. You should find that:
$$
g_n(t) = frac{1}{2(n-1)!}sum_{j=0}^n begin{pmatrix} n \ j end{pmatrix} (-1)^j (t-j)^{n-1} text{sign}(t-j)
$$
You of course would have to show this is equivalent to your linked question.
answered Apr 28 '15 at 2:36
ChesterChester
1,004913
$endgroup$
I'll just address your first question.
Let
$$
h(t) = begin{matrix} 1 &; text{if} ; t in [0,1] \ 0 &; text{else}end{matrix}
$$
and let
$$
g_n(t) = (h*h*cdots*h)(t)
$$
be the n-fold convolution of $h$. Observe that
$$
widehat{g_n}(w) = left(frac{1}{iw}right)^nleft(1-e^{-iw}right)^n =: hat{b}(w)hat{c}(w)
$$
where $hat{f}(w) = int_{mathbb{R}} f(t) e^{-iwt} dt$. So then by the convolution theorem, we have
$$
g_n(t) = (b*c)(t)
$$
To find $b(t)$ and $c(t)$, you will have to find these distributional inverse transforms, and use the binomial theorem for $hat{c}(w)$. I'll leave this to you. You should find that:
$$
g_n(t) = frac{1}{2(n-1)!}sum_{j=0}^n begin{pmatrix} n \ j end{pmatrix} (-1)^j (t-j)^{n-1} text{sign}(t-j)
$$
You of course would have to show this is equivalent to your linked question.
answered Apr 28 '15 at 2:36
ChesterChester
1,004913
answered Apr 28 '15 at 2:36
ChesterChester
1,004913
answered Apr 28 '15 at 2:36
ChesterChester
1,004913
answered Apr 28 '15 at 2:36
ChesterChester
1,004913
1,004913
add a comment |
add a comment |
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