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What is the cost of computation of FFT? (Note: ‘N’ is the number of grid points).(a) N(b) log ⁡N(c) N log ⁡N(d) \(\frac{N^2}{2} \)I have been asked this question during an internship interview.This interesting question is from Finite Difference Methods topic in chapter Finite Difference Methods of Computational Fluid Dynamics

Answer»

The CORRECT choice is (c) N LOG ⁡N

To explain I would say: The cost of COMPUTATION is reduced by FFT. FFT has a cost of computation of Nlog⁡ N orders. This is much lower than N^2, especially when N is large. This reduces the PROBLEM of computation cost in the Spectral method.



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