Series Uniform Continuity proof

**gutnedawg** · November 20th 2010, 06:17 PM

Let f(x)= sum [(3^(n) + cos(n))/n!]X^n

1. Prove that for every x in [-10,10] the sum converges
2. Show that for every E >0 there's an N independant of x in [-10,10] such that
|f(x) - sum [(3^(n) + cos(n))/n!]X^n | < E
3. Use 2 together with the fact that polynomials are contiuous everywhere to show that f is continuous in [-10,10]

ratio test for part 1, not sure how to fully show it
2 I'm not sure what to do here

the sum in part two is from 0 to N

**FernandoRevilla** · November 21st 2010, 12:51 AM

Originally Posted by gutnedawg

Let f(x)= sum [(3^(n) + cos(n))/n!]X^n
ratio test for part 1, not sure how to fully show it

If the series is $\sum_{n=0}^{+\infty}u_n(x)$ then, express:

$\left |{\dfrac{u_{n+1}(x)}{u_n(x)}}\right |=\left | \dfrac{3+\cos (n+1)/3^{n+1}}{1+\cos n/3^{n+1}}\cdot{\dfrac{1}{n+1}} \right|\left |{x}\right |\rightarrow{0}\;(n\rightarrow{+\infty})$

This series converges on $\mathbb{R}$ , as a consequence on $[-10,10]$

2 I'm not sure what to do here
the sum in part two is from 0 to N

Find a bound $|u_n(x)|\leq a_n$ on $[-10,10]$ , prove that $\sum_{n=0}^{+\infty}a_n$ converges and use Cauchy's criterion of convergence.

Regards.

Fernando Revilla

**gutnedawg** · November 21st 2010, 01:51 PM

how would this work in part 2? I was under the impression that it would be just like a limit proof of sorts with the difference in the abs being sum N to infinity of * and then solve for N then using n>N show the proof.

**FernandoRevilla** · November 21st 2010, 08:51 PM

Originally Posted by gutnedawg

how would this work in part 2? I was under the impression that it would be just like a limit proof of sorts with the difference in the abs being sum N to infinity of * and then solve for N then using n>N show the proof.

On $[-10,10]$ we have:

$\left |{u_n(x)}\right |\leq \ldots=\dfrac{3^n+1}{n!}10^n=\alpha_n$

and the series $\sum_{n=0}^{+\infty}\alpha_n <br />$ is convergent. So, $\sum_{n=0}^{+\infty}u_n(x) <br />$ uniformly converges to $f(x)$ on $[-10,10]$ as a consequence of the Weierstrass criterion (which is deduced from Cauchy criterion).

Regards.

Fernando Revilla

**gutnedawg** · November 30th 2010, 05:11 PM

I am not allowed to use anything to do with uniform convergence, this would not be difficult if I was allowed to use these theorems =[

**FernandoRevilla** · December 1st 2010, 08:39 AM

No problem, as $\sum_{n=o}^{\infty}\alpha_n$ is convergent,

$\forall \epsilon >0\;\exists n_0\in \mathbb{N}: n\geq m\geq n_o\Rightarrow\alpha_m+\alpha_{m+1}+\ldots+\alpha_ n<\epsilon$

Then

$n\geq m\geq n_0\Rightarrow|u_m(x)|+|u_{m+1}(x)|+\ldots+|u_n(x) |<\epsilon$ for all $x\in [-10,10]$

A fortiori,

$n\geq m\geq n_0\Rightarrow|u_m(x)+u_{m+1}(x)+\ldots+u_n(x)|<\e psilon$ for all $x\in [-10,10]$

Could you continue?.

Regards.

Fernando Revilla

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