Uniform convergence

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June 2nd 2008, 05:09 AM
asi123

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Uniform convergence

Hello, I need to investigate Uniform convergence for these two using Weierstrass M-test, 10x in advance.
Assaf.
June 2nd 2008, 04:39 PM
ThePerfectHacker

For the first problem note $x^n + x^{-n} \leq 2$ .
Thus, $|f_n(x)| \leq \frac{2n^2}{\sqrt{n!}}$ and $\sum_{n=0}^{\infty} \frac{2n^2}{\sqrt{n!}} < \infty$ .

I did not do the second problem but I would imagine you need the fact $\log \left( 1 + \frac{x^2}{n\log^2 n} \right) \leq \log \left( 1+\frac{a^2}{n\log^2 n} \right)$ .
Now let us hope that $\sum_{n=2}^{\infty} \log \left( 1+\frac{a^2}{n\log^2 n} \right) < \infty$ .
June 2nd 2008, 08:50 PM
asi123

it is

is is < Inf, I used my calculator to see, the question is how I show it...
10x.
June 2nd 2008, 08:57 PM
ThePerfectHacker

Quote:

Originally Posted by asi123

is is < Inf, I used my calculator to see, the question is how I show it...
10x.

It is! And there is an easy way to show it. (Happy)

You need the fact that $\log (1+x) \leq x$ for $x\geq 0$ .
Thus,
$0\leq \log \left( 1 + \frac{a^2}{n^2\log n} \right) \leq \frac{a^2}{n^2\log n}$
And of course, by applying Betrand's test
$\sum_{n=2}^{\infty}\frac{a^2}{n^2\log n} < \infty$
June 3rd 2008, 07:50 AM
Isomorphism

Quote:

Originally Posted by ThePerfectHacker

For the first problem note $x^n + x^{-n} \leq 2$ .

$x^n + x^{-n} \leq 2$
How is it true generally? (Worried)
June 3rd 2008, 07:55 AM
ThePerfectHacker

Quote:

Originally Posted by Isomorphism

$x^n + x^{-n} \leq 2$
How is it true generally? (Worried)

It should be $x^n + x^{-n}\geq 2$ . But that does not solve the problem then.
In that case we should maybe consider $x^n+x^{-n} \leq 2^n+2^{-n}$ .

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