Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 4

f, g

s = x - t

to obtain

h (k) = \int_{R} \int_{R} f (t) g (x - t) e^{- ik x} d t d x = \int_{R} \int_{R} f (t) g (s) e^{- ik (s + t)} d s d t = [\int_{R} f (t) e^{- ik t} d t] [\int_{R} g (s) e^{- ik s} d s] .

This proves the convolution theorem with the stated normalization:

h (k) = f (k) g (k) .

b

f (k) = \int_{- 1/2}^{1/2} e^{- ik x} d x = \frac{2 sin ( k /2 )}{k}, f (0) = 1 .

For the even triangular function, the odd sine part integrates to zero. Two elementary integrations by parts or one evaluation of the cosine integral give

g (k) = 2 \int_{0}^{1} (1 - x) cos (k x) d x = \frac{2 ( 1 - c o s k )}{k ^{2}} = \frac{4 sin ^{2} ( k /2 )}{k ^{2}}, g (0) = 1 .

The values at zero are the respective areas, agreeing with the removable limits. Therefore

g = f^{2}

(f * f) (x) = g (x) = {1 - ∣ x ∣, 0, ∣ x ∣ < 1, ∣ x ∣ \geq 1.

There is also a direct interpretation: the convolution is the length of overlap of the intervals

(- 1/2, 1/2)

and

(x - 1/2, x + 1/2)

, giving exactly that triangular profile.

\int_{R} ∣ h (x) ∣^{2} d x = \frac{1}{2 π} \int_{R} ∣ h (k) ∣^{2} d k .

Apply it to

h = g

. Its squared norm is

2 \int_{0}^{1} (1 - x)^{2} d x = 2/3

, so

\frac{2}{3} = \frac{1}{2 π} \int_{R} \frac{16 s i n ^{4} ( k /2 )}{k ^{4}} d k .

Under

k = 2 x

d k

becomes

2 sin^{4} x d x / x^{4}

. Consequently

\int_{- \infty}^{\infty} \frac{sin ^{4} x}{x ^{4}} d x = \frac{2 π}{3} .

The integrand has continuous value

1

at zero and decays integrably at infinity, so this is an ordinary convergent improper integral.

There are no discussions about this article yet.

There are currently no matching articles.