Question 705054

SCQMEDIUM

An integral transform f~(x)\tilde{f}(x) of a function f(x)f(x) can be regarded as a result of applying an operator FF to the function such that (Ff)(x)=f~(x)=∫−∞∞dye−ixyf(y)(Ff)(x) = \tilde{f}(x) = \int_{-\infty}^{\infty} dye^{-ixy} f(y). If II is the identity operator, then the operator F4F^4 is given by

(A)

(2Ï€)4I(2\pi)^4 I

(B)

(2Ï€)I(2\pi) I

(C)

II

(D)

(2Ï€)2I(2\pi)^2 I

Detailed Solution

The operator FF is the Fourier transform. F2(f(x))=2πf(−x)F^2(f(x)) = 2\pi f(-x) and F4(f(x))=F2(2πf(−x))=2π×2πf(x)=(2π)2If(x)F^4(f(x)) = F^2(2\pi f(-x)) = 2\pi \times 2\pi f(x) = (2\pi)^2 I f(x).

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