Fast Fourier Transform with Negative Integer Exponent The Next CEO of Stack OverflowObtain...
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Fast Fourier Transform with Negative Integer Exponent
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Given $f(x)=ax+b+frac{c}{x}$ and $N$, I'd like to ask how to calculate $sum_{i=1}^{N}f(x)^i$ efficiently using fast Fourier transform?
combinatorics computational-complexity convolution fast-fourier-transform
edited Mar 17 at 18:00
Max
9711319
asked Mar 17 at 17:30
Hang WuHang Wu
501311
$endgroup$
add a comment |
$begingroup$
Given $f(x)=ax+b+frac{c}{x}$ and $N$, I'd like to ask how to calculate $sum_{i=1}^{N}f(x)^i$ efficiently using fast Fourier transform?
combinatorics computational-complexity convolution fast-fourier-transform
edited Mar 17 at 18:00
Max
9711319
asked Mar 17 at 17:30
Hang WuHang Wu
501311
$endgroup$
$begingroup$
It works same way as with polynomials. Just place constant term in middle.
$endgroup$
– mathreadler
Mar 17 at 18:34
add a comment |
$begingroup$
Given $f(x)=ax+b+frac{c}{x}$ and $N$, I'd like to ask how to calculate $sum_{i=1}^{N}f(x)^i$ efficiently using fast Fourier transform?
combinatorics computational-complexity convolution fast-fourier-transform
edited Mar 17 at 18:00
Max
9711319
asked Mar 17 at 17:30
Hang WuHang Wu
501311
$endgroup$
Given $f(x)=ax+b+frac{c}{x}$ and $N$, I'd like to ask how to calculate $sum_{i=1}^{N}f(x)^i$ efficiently using fast Fourier transform?
combinatorics computational-complexity convolution fast-fourier-transform
combinatorics computational-complexity convolution fast-fourier-transform
edited Mar 17 at 18:00
Max
9711319
asked Mar 17 at 17:30
Hang WuHang Wu
501311
edited Mar 17 at 18:00
Max
9711319
asked Mar 17 at 17:30
Hang WuHang Wu
501311
edited Mar 17 at 18:00
Max
9711319
edited Mar 17 at 18:00
Max
9711319
edited Mar 17 at 18:00
Max
9711319
9711319
asked Mar 17 at 17:30
Hang WuHang Wu
501311
asked Mar 17 at 17:30
Hang WuHang Wu
501311
asked Mar 17 at 17:30
Hang WuHang Wu
501311
501311
$begingroup$
It works same way as with polynomials. Just place constant term in middle.
$endgroup$
– mathreadler
Mar 17 at 18:34
add a comment |
$begingroup$
It works same way as with polynomials. Just place constant term in middle.
$endgroup$
– mathreadler
Mar 17 at 18:34
$begingroup$
It works same way as with polynomials. Just place constant term in middle.
$endgroup$
– mathreadler
Mar 17 at 18:34
$begingroup$
It works same way as with polynomials. Just place constant term in middle.
$endgroup$
– mathreadler
Mar 17 at 18:34
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Power series multiplication is convolution in terms of coefficients, but multiplication in Fourier transform domain of coefficients.
This is thanks to the Convolution Theorem, which can be written for example
$$(f * g)(t) = mathcal F^{-1}(mathcal F(f)cdot mathcal F(g))(t)$$
answered Mar 17 at 18:12
mathreadlermathreadler
15.3k72263
$endgroup$
add a comment |
$begingroup$
No need to FFT! Just write $$sum_{i=1}^{N}f(x)^i=f(x){1-f^{N+1}(x)over 1-f(x)}=left(ax+b+frac{c}{x}right){1-left(ax+b+frac{c}{x}right)^{N+1}over 1-left(ax+b+frac{c}{x}right)}$$
answered Mar 17 at 21:03
Mostafa AyazMostafa Ayaz
18.3k31040
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Power series multiplication is convolution in terms of coefficients, but multiplication in Fourier transform domain of coefficients.
This is thanks to the Convolution Theorem, which can be written for example
$$(f * g)(t) = mathcal F^{-1}(mathcal F(f)cdot mathcal F(g))(t)$$
answered Mar 17 at 18:12
mathreadlermathreadler
15.3k72263
$endgroup$
add a comment |
$begingroup$
Power series multiplication is convolution in terms of coefficients, but multiplication in Fourier transform domain of coefficients.
This is thanks to the Convolution Theorem, which can be written for example
$$(f * g)(t) = mathcal F^{-1}(mathcal F(f)cdot mathcal F(g))(t)$$
answered Mar 17 at 18:12
mathreadlermathreadler
15.3k72263
$endgroup$
add a comment |
$begingroup$
Power series multiplication is convolution in terms of coefficients, but multiplication in Fourier transform domain of coefficients.
This is thanks to the Convolution Theorem, which can be written for example
$$(f * g)(t) = mathcal F^{-1}(mathcal F(f)cdot mathcal F(g))(t)$$
answered Mar 17 at 18:12
mathreadlermathreadler
15.3k72263
$endgroup$
Power series multiplication is convolution in terms of coefficients, but multiplication in Fourier transform domain of coefficients.
This is thanks to the Convolution Theorem, which can be written for example
$$(f * g)(t) = mathcal F^{-1}(mathcal F(f)cdot mathcal F(g))(t)$$
answered Mar 17 at 18:12
mathreadlermathreadler
15.3k72263
answered Mar 17 at 18:12
mathreadlermathreadler
15.3k72263
answered Mar 17 at 18:12
mathreadlermathreadler
15.3k72263
answered Mar 17 at 18:12
mathreadlermathreadler
15.3k72263
15.3k72263
add a comment |
add a comment |
$begingroup$
No need to FFT! Just write $$sum_{i=1}^{N}f(x)^i=f(x){1-f^{N+1}(x)over 1-f(x)}=left(ax+b+frac{c}{x}right){1-left(ax+b+frac{c}{x}right)^{N+1}over 1-left(ax+b+frac{c}{x}right)}$$
answered Mar 17 at 21:03
Mostafa AyazMostafa Ayaz
18.3k31040
$endgroup$
add a comment |
$begingroup$
No need to FFT! Just write $$sum_{i=1}^{N}f(x)^i=f(x){1-f^{N+1}(x)over 1-f(x)}=left(ax+b+frac{c}{x}right){1-left(ax+b+frac{c}{x}right)^{N+1}over 1-left(ax+b+frac{c}{x}right)}$$
answered Mar 17 at 21:03
Mostafa AyazMostafa Ayaz
18.3k31040
$endgroup$
add a comment |
$begingroup$
No need to FFT! Just write $$sum_{i=1}^{N}f(x)^i=f(x){1-f^{N+1}(x)over 1-f(x)}=left(ax+b+frac{c}{x}right){1-left(ax+b+frac{c}{x}right)^{N+1}over 1-left(ax+b+frac{c}{x}right)}$$
answered Mar 17 at 21:03
Mostafa AyazMostafa Ayaz
18.3k31040
$endgroup$
No need to FFT! Just write $$sum_{i=1}^{N}f(x)^i=f(x){1-f^{N+1}(x)over 1-f(x)}=left(ax+b+frac{c}{x}right){1-left(ax+b+frac{c}{x}right)^{N+1}over 1-left(ax+b+frac{c}{x}right)}$$
answered Mar 17 at 21:03
Mostafa AyazMostafa Ayaz
18.3k31040
answered Mar 17 at 21:03
Mostafa AyazMostafa Ayaz
18.3k31040
answered Mar 17 at 21:03
Mostafa AyazMostafa Ayaz
18.3k31040
answered Mar 17 at 21:03
Mostafa AyazMostafa Ayaz
18.3k31040
18.3k31040
add a comment |
add a comment |
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$begingroup$
It works same way as with polynomials. Just place constant term in middle.
$endgroup$
– mathreadler
Mar 17 at 18:34