High-precision IFFT

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Hi! I need some high precision computations, so I use the function vpa() with about 32 digits. My final result is obtained by making ifft(x, 'symmetric'). The problem is that I need to transform x to double again, loosing my required precision. Is there any way to overcome this? I think the only way is to make my own ifft function, I am correct? Thanks!
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Demian Augusto Vera
Demian Augusto Vera on 25 Jan 2022
Thanks for your answer, Paul. The symbolic array x is just a high-precision complex array of numers; no symbolic expression involved.

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Accepted Answer

Star Strider
Star Strider on 24 Jan 2022
Since it appears that these calculations are all symbolic, one option would be the ifourier function, or if you are using your own Fourier transform integration code, simply reversing the sign of the argument of the exponent will produce the inverse transform with essentially the same code, although with a different variable of integration (ω rather than t).
That way, everything remains symbolic, with the desired precision.
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Star Strider
Star Strider on 25 Jan 2022
As always, my pleasure!

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More Answers (1)

Paul
Paul on 25 Jan 2022
Edited: Paul on 25 Jan 2022
Given that x is an array of high precision numbers, not symbolic expressions, it should be straightforward to implement the ifft sum. It might not be efficient and it might be slow, but it should work. The one thing I'm not sure about is how well this will work to ensure that the time domain sequence, X[n], is real, which I think is the expectation based on the symmetric flag in the call to numerical ifft. If the result does have a small imaginary part, it can always be removed I suppose, but I don't know what that indicates about precision of the solution you're trying to obtain. OTOH, I'm curious if you are ensuring that the sequence x is exactly conjugate symmetric in the first place, to whatever precision you're using. Or the ifft sum can be implemented assuming x is conjugate symmetric.
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Demian Augusto Vera
Demian Augusto Vera on 25 Jan 2022
Thanks, Paul. The theory indicates that it is conjugate symmetric. Using double precision that's ensured, as I can see. Don't know if using high-precision arithmetic it remains conjugate symmetric, but I think it has to.
Thanks again for your help answers.

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