Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 27K / b

Table of contents
- Solution b

Solution

Put

H (u) = \frac{4 s i n ^{2} ( u /2 )}{u ^{2}}, H (0) = 1.

This function is bounded near zero and is

O (u^{- 2})

at infinity, so

H \in L^{1} (R)

and the defining integral for

g

converges absolutely.

For the triangular function

h (x) = (1 - ∣ x ∣) 1_{[- 1, 1]} (x),

direct integration gives

h (u) = \int_{- 1}^{1} (1 - ∣ x ∣) e^{i ux} d x = 2 \int_{0}^{1} (1 - x) cos (ux) d x = \frac{2 ( 1 - cos u )}{u ^{2}} = \frac{4 sin ^{2} ( u /2 )}{u ^{2}} = H (u) .

The Fourier inversion theorem, or equivalently the Fourier transform of a triangular function, therefore yields

g (x) = (1 - ∣ x ∣) 1_{[- 1, 1]} (x) .

In particular

g (x) = 0

whenever

∣ x ∣ > 1

. Finally,

∥ g ∥_{2}^{2} = 2 \int_{0}^{1} (1 - x)^{2} d x = \frac{2}{3} .

This also agrees with the Plancherel theorem applied to

h

and

H

.

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version