Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 215 / 1 / b

Table of contents
- Solution b

Solution

On the diagonal,

P^{k} (x, x) - π (x) = π (x) \sum_{i \geq 2} λ_{i}^{k} f_{i} (x)^{2},

so every summand is nonnegative. Let

m = ⌈ t_{rel} ⌉

. For every

i \geq 2

,

λ_{i}^{m + 1} \leq λ_{2}^{t_{rel}} = (1 - \frac{1}{t _{rel}})^{t_{rel}} \leq e^{- 1} .

Hence

\frac{1}{1 - λ _{i}} \leq \frac{e}{e - 1} \frac{1 - λ _{i}^{m + 1}}{1 - λ _{i}} = \frac{e}{e - 1} \sum_{k = 0}^{m} λ_{i}^{k} .

Multiply by

π (x) f_{i} (x)^{2}

and sum over

i \geq 2

to obtain the required inequality.

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