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Show that every function $e^{inx}$ can be uniformly approximated on $[-\pi,\pi]$ by polynomials in $x$.

Using the power series expansion, $$e^{inx}=1+inx+\frac{(inx)^2}{2!}+\frac{(inx)^3}{3!}+\ldots$$ Let $s_k(x)$ denote the sum up to the $k$th term.

Also, since $e^{inx}=\cos(nx)+i\sin(nx)$, the function is periodic of period $2\pi$.

I want to show that the sequence $s_k(x)$ uniformly approximate $e^{inx}$. To do that, I need to show that for all $\epsilon>0$, there exists $k$ such that $|e^{inx}-s_k(x)|=\left|\sum_{i=k}^\infty\dfrac{(inx)^k}{k!}\right|<\epsilon$ for all $x\in[-\pi,\pi]$.

How can I show that?

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It seems to me that $s_k(x)$ is a sequence of polynomials which approximate $e^{inx}$. Doesn't it settle your question? – a12345 Nov 2 '13 at 2:19
@a12345 Right, please see my edit. – Mika H. Nov 2 '13 at 4:01
up vote 1 down vote accepted

I think this is standard real analysis, we bound

$$|\sum_{j=k}^{\infty}\frac{{[inx]}^{j}}{j!}|\le \sum^{\infty}_{j=k}\frac{[n\pi]^{j}}{j!}\le \sum^{\infty}_{j=k}\frac{M^{j}}{j!},M\ge n\pi$$ Here I used a different symbol $j$ to avoid the same symbol $i$ being used twice.

We choose $k$ large enough that $2M<k$. Then we have $M^{j}\le k^{j}$. The above summation now is bounded by $$\sum^{\infty}_{j=k}\frac{k^{j}}{j!}$$ where we have $$a_{i+1}=a_{i}\frac{k}{i+1},i\ge k$$ So this is bounded by a geometric series. Now the to make it less than $\epsilon$ we just need to choose another partial sum with the beginning index $k'$ much, much greater than $k$.

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Why do you need $2M<k$? Also, since $j$ is the index of the sum, you cannot "choose" it... – Mika H. Nov 2 '13 at 4:34
Just let $k$ be large enough, though $M<k$ is suffice. We can choose another partial sum whose beginning index is large enough. Sorry if this is not clear. – Bombyx mori Nov 2 '13 at 4:47

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