3 ms·
Is the proof even valid? The claim is that the integral (1) is bounded between 0 and a positive number decaying to 0 as n tends to infinity, which sounds like a
by rich_sasha 2y ago
Is the proof even valid? The claim is that the integral (1) is bounded between 0 and a positive number decaying to 0 as n tends to infinity, which sounds like a contradiction.
But the integral (1) is itself a function of n, since both f and F are. Not a fixed number. So the inequality is eminently satisfiable.
In other words, if we call (1) I, they show
0 < I < π^n a^n / n!
which is bad. But I depends on n so this really should read
0 < I_n < π^n a^n / n!
which is fine... as far as the proof shows.
But perhaps I'm missing something...
- ducttapecrown 2y agoThink of a collection of integrals, one for each n, rather than a function depending on multiple variables. The contradiction is that the number F(pi) + F(0) is both a positive integer and arbitrarily small. There is no contradiction until n is large enough that F(pi) + F(0) is bounded below 1, hence the statement at the start: we'll pick a large enough value of n later.
- rich_sasha 2y agoBut F(pi) + F(0) depends on n, rather than being a number, just this isn't explicitly spelled out. So all the proof shows is that the integral (1) converges to 0 rather quickly, as a function of n. The claim of the proof is that I=F(0)+F(π) is simultaneously positive and arbitrarily small, which is not possible for a fixed number, but totally possible if you see I as a sequence indexed by n.
- alimw 2y ago> The claim of the proof is that I=F(0)+F(π) is simultaneously positive and arbitrarily small, which is not possible for a fixed number, but totally possible if you see I as a sequence indexed by n. This is not possible because I_n must be an integer for every n.
- rich_sasha 2y agoAh yes, that's the key bit. Thanks!
- deleted 2y ago[deleted]