1. Convolutions

f(x) = e^-(x^2) and g(x) = e^-2(x^2).
Compute f*g (their convolution). Use it to show that if
h_t(x) = 1/(4 \pi t) e^{(-x^2)/4t}
we have h_t * h_s = h_(t+s)

My first thought was to take the Fourier transform of f and g - these are standard for example f^(k) = sqrt(\pi) e^{-(\pi ^2) k^2} - then use the property for the convolution of Fourier transform, namely (f*g)^(k) = f^(k) g^(k), but to no avail. Alternatively, I can't evaluate the integral when using the definition of convolution.
Sorry, LaTex doesn't seem to be working....

2. Originally Posted by davidmccormick f(x) = e^-(x^2) and g(x) = e^-2(x^2).
Compute f*g (their convolution). Use it to show that if
h_t(x) = 1/(4 \pi t) e^{(-x^2)/4t}
we have h_t * h_s = h_(t+s)

My first thought was to take the Fourier transform of f and g - these are standard for example f^(k) = sqrt(\pi) e^{-(\pi ^2) k^2} - then use the property for the convolution of Fourier transform, namely (f*g)^(k) = f^(k) g^(k), but to no avail. Alternatively, I can't evaluate the integral when using the definition of convolution.
The key to calculating the convolution is to complete the square. Take f(x) = e^{ax^2} and g(x) = e^{bx^2}.To find f*g, you want to integrate the exponential of at^2  b(xt)^2 (with respect to t). Complete the square to get at^2  b(xt)^2 = (a+b)[t  (bx)/(a+b)]^2  (abx^2)/(a+b). Now make the substitution u = t  (bx)/(a+b) in the integral. The answer should be \sqrt{\pi/(a+b)} times the exponential of (abx^2)/(a+b).

For the second part, there is a mistake in the question. The function h_t(x) should be given by h_t(x) = 1/(\sqrt{4 \pi t}) e^{(-x^2)/4t}. If you substitute a = 1/(4t) and b = 1/(4s) in the previous calculation, then everything works out nicely. Originally Posted by davidmccormick Sorry, LaTex doesn't seem to be working....
I hope it gets restored soon. It's a nightmare trying to read or write complicated maths without it.

3. Thanks for your great post. It solved the problem completely.

4. Originally Posted by Opalg I hope it gets restored soon. It's a nightmare trying to read or write complicated maths without it.
See post #8 on this thread. It helps.

-Dan

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