 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-1/4/d/solution

Evenness means the negative half of the function is determined as well: this is the triangular function

q (x) = 1 - ∣ x ∣

∣ x ∣ \leq 1

, zero outside. Its transform is real and even:

q (k) = 2 \int_{0}^{1} (1 - x) cos (k x) d x .

Integration by parts, or integrating the two terms separately, gives

q (k) = \frac{2 ( 1 - cos k )}{k ^{2}} = (\frac{sin ( k /2 )}{k /2})^{2}, q (0) = 1.

The apparent singularity at zero is removable, and its limiting value is the area under

q

 New to topics? Read the docs here!