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Maybe is a silly question but I am confused...so I hope someone can help me.

Is the convergence of the Taylor series uniform?

To be more specific. We know for example that

$\displaystyle{ e^x = \sum_{n=0}^{\infty} \frac{x^n}{ n!} \quad}$ , $\displaystyle{ \sin x = \sum_{n=0}^{\infty} \frac{ (-1)^{n+1} }{ (2n+1)!} x^{2n+1} }$

Now the question is: Does these series of functions converges uniformly to $e^x$ and $\sin x$ respectively?

And one more question: Every Taylor series converges (pointiwise) for every $x$ so has radius of convergence $\infty$ right?

These questions came up when I was studying and saw I my notes that the lecture interchange the summation and the integration of the Taylor series, for example

$\displaystyle{ \frac{1}{ 1+ x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2n}, |x|<1 \implies \arctan x = \int_{0}^{x} \sum_{n=0}^{\infty} (-1)^n t^{2n} dt= \sum_{n=0}^{\infty} (-1)^n \int_{0}^{x} t^{2n} dt }$

and I can understand how can we do this if the convergence in not uniform, which in this example is not since $\displaystyle{ \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} }$ pointwise but not uniform.

passenger
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1 Answers1

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Hints

For your first question: Weirestrass $\;M$-test;

For your second question: no. For example, the series of $\,\frac1{1-x}\;$ converges only for $\,|x|<1\,$ , and the convergence is uniform for $\,|x|\leq<1\;$ ...

DonAntonio
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  • For the first question I did that and I get that the convergence is uniform on compact intervals is that correct? For the example with the $\arctan x$ how can we interchange the sum and the integral? – passenger Jun 02 '13 at 11:20
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    The M-test talks of "sets" on which the conditions are fulfilled, so I think in both first cases you get uniform convergence everywhere where you can bound $,x,$ . Of course, compact sets are handy here...:) . For the arctangent case you can interchange as long as you can apply the monotone or the domianted convergence theorems...which you can within the series's convergence radius. – DonAntonio Jun 02 '13 at 11:37
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    Thank you very much for your answer! It was very helpful! – passenger Jun 02 '13 at 11:55