uniform convergence on (1,a)

**hmmmm** · October 15th 2011, 09:47 AM

The function $f_n=\frac{x^n}{x^n+1}$ converges uniformly to $f(x)=1$ on the interval $(1,a)$ .

Proof.

taking x=a

as $|\frac{a^n}{a^n+1}-1|<\epsilon\rightarrow \frac{1}{1+a^n}<\epsilon \rightarrow ln(\frac{1-\epsilon}{\epsilon})<ln(a) n$

So can I take $N={ln(\frac{1-\epsilon}{\epsilon})}$ and then I have that:

$\forall\epsilon>0\exists N(={ln(\frac{1-\epsilon}{\epsilon})}) s.t \forall n\geq N \ \forall x\in(1,a)\ \mbox{we have}\ |f_n(x)-f(x)|<\epsilon$

is this correct or is it wildly off- uniform convergence is kind of killing me

thanks for any help

**Drexel28** · October 15th 2011, 07:00 PM

Originally Posted by hmmmm

The function $f_n=\frac{x^n}{x^n+1}$ converges uniformly to $f(x)=1$ on the interval $(1,a)$ .

Proof.

taking x=a

as $|\frac{a^n}{a^n+1}-1|<\epsilon\rightarrow \frac{1}{1+a^n}<\epsilon \rightarrow ln(\frac{1-\epsilon}{\epsilon})<ln(a) n$

So can I take $N={ln(\frac{1-\epsilon}{\epsilon})}$ and then I have that:

$\forall\epsilon>0\exists N(={ln(\frac{1-\epsilon}{\epsilon})}) s.t \forall n\geq N \ \forall x\in(1,a)\ \mbox{we have}\ |f_n(x)-f(x)|<\epsilon$

is this correct or is it wildly off- uniform convergence is kind of killing me

thanks for any help

I don't quite understand what you're doing here, why are you choosing this $a$ ? Are you secretly plugging this in because it's the max, and you just omitted this? Give us more words, explaining your choices etc. and then we can help you decide of what you did is correct. A quick check: you should be proving that for every $\varepsilon>$ there exists $N\in\mathbb{N}$ such that $\|f_n(x)-f(x)\|_\infty<\varepsilon$ where $\|\cdot\|_\infty$ is the infinity norm. This 'looks' like what you're doing, but please fill in more holes.

**hmmmm** · October 16th 2011, 02:45 AM

The function $f_n=\frac{x^n}{x^n+1}$ converges uniformly to $f(x)=1$ on the interval $(1,a)$ .

Proof.

taking x=a as this will be the max of the $f_n(x)$ so if the max is less then epsilon then $f_n(x)$ will be for all other values of x.

as $|\frac{a^n}{a^n+1}-1|<|\frac{a^n}{a^n+1}-1|<\epsilon$

and as $\frac{a^n}{a^n+1}-1=\frac{-1}{1+a^n}$

and so:

$|\frac{1}{1+a^n}|<\epsilon \rightarrow ln(\frac{1-\epsilon}{\epsilon})<ln(a) n$

So can I take $N={ln(\frac{1-\epsilon}{\epsilon})}$ and then I have that:

$\forall\epsilon>0\exists N(={ln(\frac{1-\epsilon}{\epsilon})}) s.t \forall n\geq N \ \forall x\in(1,a)\ \mbox{we have}\ |f_n(x)-f(x)|<\epsilon$

I think this is a bit better? Sorry about it being a bit of a mess I am a little confused by this. Thanks very much for the help

thanks for any help

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