Test007

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@Axel Vogt Thanks, that worked! However, is it possible to get 0 for F[f(x) g(x)] = 1/sqrt(2Pi) × (F[f(x)]∗F[g(x)])(k), where the asterisk (∗) stands for convolution? This is what I got:
 

restart

with(inttrans)

left := fourier(f(x)*g(x), x, k)/sqrt(2*Pi)

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

(1)

"f(k):=1/(sqrt(2 Pi))fourier(f(x),x,k)"

proc (k) options operator, arrow, function_assign; fourier(f(x), x, k)/sqrt(2*Pi) end proc

(2)

"(g)(k):=1/(sqrt(2 Pi))fourier(g(x),x,k)"

proc (k) options operator, arrow, function_assign; fourier(g(x), x, k)/sqrt(2*Pi) end proc

(3)

right := (int(`#mover(mi("f"),mo("ˆ"))`(y)*`#mover(mi("g"),mo("ˆ"))`(k-y), y = -infinity .. infinity))/sqrt(2*Pi)

(1/2)*2^(1/2)*(int((1/2)*fourier(f(x), x, y)*fourier(g(x), x, k-y)/Pi, y = -infinity .. infinity))/Pi^(1/2)

(4)

simplify(convert(left, int)-convert(right, int))

(1/2)*2^(1/2)*(int(f(y)*g(y)*exp(-I*y*k), y = -infinity .. infinity)-(int((1/2)*(int(f(x)*exp(-I*x*y), x = -infinity .. infinity))*(int(g(x)*exp(-I*x*(k-y)), x = -infinity .. infinity))/Pi, y = -infinity .. infinity)))/Pi^(1/2)

(5)

``


 

Download Fourier_of_Product.mw

@Rouben Rostamian I was hoping for some way to specify the basis and have Maple evaluate the formula in a single command (like FourierSeries).

@nm Thanks! The right BC will also be 0 because of the where condition of the eigenvalues.

@Rouben Rostamian NVM, I was testing it with 0..l instead of -Pi..Pi.

@Rouben Rostamian Thanks, the i replacement worked. Did you change some other settings though since I get a different answer for the sine series?

@Carl Love Thanks, but could you tell me how you came up with the constants? Is there a general procedure for this and does it apply to other ODEs that have piecewise solutions like this (e.g. ay'' + by' + cy = 0)?

@Carl Love Why? Shouldn't Maple fix this as it's a common scenario?

@tomleslie So? OP typed in x^5 + 2*x^3 + 3, which satisfies the requirement if he defined it the alternate way.

@tomleslie Hmm, isn't it also possible to define it as

    f(x) := x^5+2*x^3+3

So I think OP had a legitimate cause for confusion.

@Carl Love Thanks, it works now! I had an extra minus sign for G in my code.

@Carl Love pdetest returns [2 t − 2 x, 0] for me.

@Carl Love Is the result not 0? Is that a bug then?

What happens if there are conflicting assumptions?

@Carl Love The second solution works as intended. Thanks!

@acer Thanks. My original question was incomplete. How can I get Maple to give g(x) - g(-x) as 0?

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