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Nicholi.

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- Thread starter nshiell
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Nicholi.

- #2

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Does anyone know a book or website or wants to type out the answer for me?

Nicholi.

Could you not write out the convolution explicitly (even for two standard gaussians) ?

- #3

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int(exp(-x^2)*exp(-(r-x)^2))dx

there is the limits are +/- infinity

there is the limits are +/- infinity

- #4

CompuChip

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[tex]\exp\left[ -x^2 \right] \exp\left[ -(r-x)^2 \right] =

\exp\left[ -2 x^2 + 2 r x - r^2 \right] =

\exp\left[ -2 (x + r/2)^2 \right] \exp\left[ - r^2 / 2 \right]

[/tex]

where the last term does not depend on

- #5

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Thanks!

- #6

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[tex]\exp\left[ -x^2 \right] \exp\left[ -(r-x)^2 \right] =

\exp\left[ -2 x^2 + 2 r x - r^2 \right] =

\exp\left[ -2 (x + r/2)^2 \right] \exp\left[ - r^2 / 2 \right]

[/tex]

where the last term does not depend onxand the first one is just another Gaussian, but centered around a different point.

This only shows c) here below.

Another argument, besides direct computation, is by standard properties of Fourier transform:

a) fourier transform of a gaussian is a gaussian

b) fourier transform of convolution product = pointwise product of fourier transforms (up to choice of normalisation)

c) pointwise product of gaussians is gaussian.

d) inverse fourier transform of a gaussian is gaussian.

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