Paper 3, Section II, D

Numerical Analysis | Part II, 2012

The inverse discrete Fourier transform Fn1:RnRn\mathcal{F}_{n}^{-1}: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n} is given by the formula

x=Fn1y, where xl=j=0n1ωnjlyj,l=0,,n1\mathbf{x}=\mathcal{F}_{n}^{-1} \mathbf{y}, \quad \text { where } \quad x_{l}=\sum_{j=0}^{n-1} \omega_{n}^{j l} y_{j}, \quad l=0, \ldots, n-1

Here, ωn=exp(2πi/n)\omega_{n}=\exp (2 \pi i / n) is the primitive root of unity of degree nn and n=2p,p=1,2,n=2^{p}, p=1,2, \ldots

(i) Show how to assemble x=F2m1y\mathbf{x}=\mathcal{F}_{2 m}^{-1} \mathbf{y} in a small number of operations if the Fourier transforms of the even and odd parts of y\mathbf{y},

x(E)=Fm1y(E),x(O)=Fm1y(O)\mathbf{x}^{(E)}=\mathcal{F}_{m}^{-1} \mathbf{y}^{(E)}, \quad \mathbf{x}^{(O)}=\mathcal{F}_{m}^{-1} \mathbf{y}^{(O)}

are already known.

(ii) Describe the Fast Fourier Transform (FFT) method for evaluating x\mathbf{x}, and draw a relevant diagram for n=8n=8.

(iii) Find the costs of the FFT method for n=2pn=2^{p} (only multiplications count).

(iv) For n=4n=4 use the FFT method to find x=F41y\mathbf{x}=\mathcal{F}_{4}^{-1} \mathbf{y} when:

(a) y=(1,1,1,1)\mathbf{y}=(1,1,-1,-1),

(b) y=(1,1,1,1)\mathbf{y}=(1,-1,1,-1).

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