Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 11G / d

Table of contents
- Solution d

Solution

Every prime in

(X, X]

is at least

X

. Consequently the primorial satisfies

P (X) \geq (X)^{π (X) - π (X)} .

Combining this with part (c) and taking logarithms gives

(π (X) - π (X)) \frac{l o g X}{2} \leq X lo g 4,

so

π (X) \leq π (X) + \frac{2 X l o g 4}{l o g X} \leq X + \frac{2 X l o g 4}{l o g X} .

For

X \geq 2

, the inequality

lo g X \leq X

implies

X \leq \frac{X}{l o g X} .

Thus the prime-counting upper bound from a primorial estimate yields

π (X) \leq (1 + 2 lo g 4) \frac{X}{l o g X} .

We may take

c = 1 + 2 lo g 4

.

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version