Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 1 / 14A / b / iii

Table of contents
- Solution iii

Solution

For the weight

w = x

, substitution

x = e^{s}

gives

⟨ y_{m}, y_{n} ⟩_{w} = \int_{1}^{e} \frac{s i n ( mπ l o g x ) s i n ( nπ l o g x )}{x} d x = \int_{0}^{1} sin (mπ s) sin (nπ s) d s .

The product-to-sum formula makes this zero when

m \neq = n

and

1/2

when

m = n

. Therefore

⟨ y_{m}, y_{n} ⟩_{w} = \frac{1}{2} δ_{mn} .

In particular the eigenfunctions are orthogonal, and multiplying each by

2

gives an orthonormal basis for the weighted space.

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