Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics II / 1 / a / iii

Table of contents
- Solution iii

Solution

It is useful to prove the contrapositive. Suppose

k

is a peak of a sequence

(a_{n})

and

m

is a peak of a sequence

(- a_{n})

. The second condition says

a_{j} \geq a_{m}

for every

j > m

. Hence

j > max {k, m} ⟹ a_{m} \leq a_{j} \leq a_{k} .

Only finitely many earlier terms remain, and their absolute values have a finite maximum. Together these give a bound for every term, so

(a_{n})

is a bounded sequence. Consequently,

(a_{n}) unbounded ⟹ (a_{n}) or (- a_{n}) has no peaks .

 Ancestors (12)

 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