find which function corresponds to $\displaystyle{\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} x^{2k}}$

Question

I want to find which function corresponds to $\displaystyle{\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} x^{2k}}$.

That's what I have tried:

$$a_{2k-1}=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_{2k}=\frac{(-1)^k}{(2k)!}, \ \ \ \ \ \ k\in \mathbb{N}^{*}$$

$$\sqrt[2k-1]{|a_{2k-1}|}=0, \sqrt[2k]{|a_{2k}|}=\sqrt[2k]{|\frac{(-1)^k}{(2k)!}|}=\frac{1}{\sqrt[2k]{(2k)!}} \to 0$$

Therefore, $\sqrt[k]{|a_k|} \to 0$,so $\lim \sup \sqrt[k]{|a_k|}=0 \Rightarrow R=+\infty$,so the power series converges at $(-\infty,+\infty)$.

Let $\displaystyle{s(x)=\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} x^{2k}}, x \in \mathbb{R}$

$$s'(x)=\sum_{k=1}^{\infty} \frac{(-1)^k}{(2k)!}2k x^{2k-1}= \dots=- \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2k-1)!} x^{2k-1}$$

We consider the power series: $$\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2k-1)!} x^{2k-1}$$

$$a_{2k}=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_{2k-1}=\frac{(-1)^{k-1}}{(2k-1)!}, \ \ \ \ \ \ k\in \mathbb{N}^{*}$$

$$\sqrt[2k]{|a_{2k}|}=0, \sqrt[2k-1]{|a_{2k-1}|}=\sqrt[2k-1]{|\frac{(-1)^{k-1}}{(2k-1)!}|}=\frac{1}{\sqrt[2k-1]{(2k-1)!}} \to 0$$

So,$R=+\infty $,therefore the powerseries converges at $(-\infty,+\infty)$.

$\exists c(x):\mathbb{R} \to \mathbb{R} \text{ such that } c(x)=\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2k-1)!} x^{2k-1}$

We see that $ \displaystyle{ s'(x)=-c(x), x \in \mathbb{R}}$

$$c'(x)= \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2k-1)!} (2k-1) x^{2k-2}= \dots= \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{(2k)!}=s(x)$$

We see that $\displaystyle{ c'(x)=s(x) \Rightarrow -s''(x)=s(x) \Rightarrow s''(x)+s(x)=0}$

$$\text{ The characteristic equation is } m^2+1=0 \Rightarrow m=\pm i$$

So,we have $\displaystyle{ s(x)=c_1 \cos{x}+ c_2 \sin{x}}$

$$s(0)=1 \Rightarrow c_1=1 , s'(0)=0 \Rightarrow c_2=0 $$

So,we have $s(x)=\cos{x} \text{ and } c(x)= \sin{x}$.

Could you tell me if it is right???

user71352 · Answer 1 · 2014-05-18T19:26:07.983

3

Hint: Note that $e^{z}=\sum_{k=0}^{\infty}\frac{z^{k}}{k!}$ so

$$e^{z}+e^{-z}=2\sum_{k=0}^{\infty}\frac{z^{2k}}{(2k)!}$$

edited May 18 '14 at 19:26

answered May 18 '14 at 19:20

user71352

13,038

So,is the way I found it wrong? – evinda May 18 '14 at 19:21
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@evinda, what you have done looks fine to me. – achille hui May 18 '14 at 19:22
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Sorry for the wait I'm currently verifying it. – user71352 May 18 '14 at 19:22
A ok,no problem!!!! – evinda May 18 '14 at 19:23
But,can I just derivate the relation $$s'(x)=-c(x)$$ ?? – evinda May 18 '14 at 19:23
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As noted earlier there appears to be nothing wrong. – user71352 May 18 '14 at 19:27
But why do we know that $s'$ is differentiable,so that we can take the derivative of the relation above? – evinda May 18 '14 at 19:28
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You can demonstrate that differentiating a series is okay be appealing to uniform convergence results. One of which is here: http://en.wikipedia.org/wiki/Uniform_convergence#To_differentiability – user71352 May 18 '14 at 19:31
But...if we have a power series,isn't the convergence only uniform,at the case when the power series is defined at a closed interval?? Or am I wrong?? – evinda May 18 '14 at 19:34
Uniform convergence doesn't have to be strictly on a closed interval. It can be the entire real line. For example the power series expansion for $e^{x}$ is valid on the entire real line. – user71352 May 18 '14 at 19:36
Ok,but how do we know that $s'$ is differentiable, although we don't know which function it is?? We can't check the convergence..or am I wrong?? – evinda May 18 '14 at 19:38
By looking at partial sums of $s'$ we can find a candidate for the derivative of $s'$. We then check for uniform convergence. – user71352 May 18 '14 at 19:46

find which function corresponds to $\displaystyle{\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} x^{2k}}$

1 Answers1