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Real roots of Fourier transform... |
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| May29-06, 02:55 PM | #1 |
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Real roots of Fourier transform...
If we define the function:
[tex] F[w]=\int_{-\infty}^{\infty}dxe^{-iwx}g(x) [/tex] my question is..what would be the criterion to decide if F[w] has all the roots real (w=w*) and how is derived?..thanks. |
| Jun7-06, 11:17 PM | #2 |
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If anybody knows the answer to this question, I would be very interested in learning the answer also!
At least in the case of a similar transformation, a Laplace like transform, Int[e^(-s*sqrt(t))*sin(pi*sqrt(t)),0, infinity] = T{sin(pi*sqrt(t))} = 4*pi*s/(s^2 + pi^2)^2 Int[e^(-s*sqrt(t+c))*sin(pi*sqrt(t+c)),0, infinity] = T{sin(pi*sqrt(t+c))} = q(s)*e^[-sqrt(c)*s]/(s^2 + pi^2)^2, where q(s) is a cubic polynomial in s, c is an arbitrary positive constant, and t is greater than or equal to 0. It seems that the information about the zeros of the sine functions above is contained in the residue at the poles of there s-like-transforms, T{f(t)} = F(s), while the kind of function f(t) is, is indicated by the position and order of the poles of F(s). This is only a guess, based on the fact that the transformations above yield functions of s with a common denominator of (s^2 + pi^2)^2, but a different expression in the variable s in the numerator. The two original sine functions in the real variable t have zeros in different locations, but the position and order of the singularities of their s-like-transforms above are identical. As a result, I would presume that all the information about the zeros of the original sine functions in t would have to be tied up in the residues of the singularities of the s-like-transforms What are your thoughts? Inquisitively, Edwin G. Schasteen |
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