Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2008 / Paper 1 / 4 / a

Table of contents
- Solution a
- i a
  - Solution i
- ii a
  - Solution ii

Solution

Use the angular-frequency Fourier transform convention

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

For the piecewise smooth functions here, the inverse gives the function at continuity points and the mean of one-sided limits at a jump. Values assigned at individual points do not affect the transform.

i

Solution

The even extension is the triangular function

f (x) = (1 - ∣ x ∣)

for

∣ x ∣ < 1

and zero outside. Its Fourier transform is real and even:

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

At zero the removable limiting value is one, also obtained directly by integrating the triangle. This is the Fourier transform of a triangular function.

ii

Solution

For the odd extension, the cosine integral vanishes and the sine integral doubles. Integration by parts gives

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

The value at

k = 0

is zero by the odd function symmetry, or by the Maclaurin series of the sine. The transform is purely imaginary and odd.

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