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These are questions asked by Test007

Is there a way to verify the following Fourier transform property: F[f(x) exp(x)](k) = F[f(x)](k − b)? This is what I tried:



constants := constants, b

false, gamma, infinity, true, Catalan, FAIL, Pi, b


left := fourier(f(x)*exp(I*b*x), x, k)/sqrt(2*Pi)

(1/2)*2^(1/2)*fourier(f(x)*exp(I*b*x), x, k)/Pi^(1/2)


right := fourier(f(x), x, k-b)/sqrt(2*Pi)

(1/2)*2^(1/2)*fourier(f(x), x, k-b)/Pi^(1/2)



(1/2)*2^(1/2)*(fourier(f(x)*exp(I*b*x), x, k)-fourier(f(x), x, k-b))/Pi^(1/2)




Download First_Shifting_Theorem.mw

Is there a way to specify options or something to change the definition of fourier and invfourier to have a 1/sqrt(2Pi) factor instead? I don't want to manually multiply the result by 1/sqrt(2Pi) every time in case I forget to do it and it leads to a mistake in the future.

Why is pdsolve's 'generalsolution' option giving the particular solution u(x, y) = 0 instead of the general solution u(x, y) = A sin(x) sin(2 y) + sin(2 x) sin(y) for the attached problem?


Is there a way to convert a Fourier series (from the OrthogonalExpansions package) automatically into the sum of odd/even terms if the even/odd terms are 0 respectively?




FourierSeries(x-2*Pi, x = 0 .. 4*Pi, infinity, series = cosine); simplify(subs(i = n, %))

Sum(8*((-1)^n-1)*cos((1/4)*n*x)/(n^2*Pi), n = 1 .. infinity)


Sum((-16*cos((1/4)*(2*k-1)*x))*(1/((2*k-1)^2*Pi)), k = 1 .. infinity)

Sum(-16*cos((1/4)*(2*k-1)*x)/((2*k-1)^2*Pi), k = 1 .. infinity)



Download FourierSeries.mw

Is there a way to get the generalized Fourier series with respect to an arbitrary orthogonal system (instead of just sin, cos, exp, etc.) using the OrthogonalExpansions package or any other way in Maple?

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