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$$\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}$$

I am having difficulty finding the function that represents this series.

I have only found radius of convergence which is $(-\infty,+\infty)$ from the ratio test.

apnorton
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Rudolf
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4 Answers4

6

Define $$f(x)=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!}$$ hence $$f'(x)=\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$$ and then $$f''(x)=\sum_{n=1}^{\infty}\frac{x^{2n-1}}{(2n-1)!}=f(x)$$

Now note that $f(0)=0$ and $f'(0)=1$. Therefore we need to solve $f''(x)-f(x)=0$. Using Laplace transforms we get $$s^2F(s)-sf(0)-f'(0)-F(s)=0$$ or that $$F(s)=\frac{1}{s^2-1}=\frac12\frac{1}{s-1}-\frac12\frac{1}{s+1}$$ and taking an inverse we obtain $$f(x)=\frac{e^x}{2}-\frac{e^{-x}}{2}=\sinh(x)$$

GFauxPas
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Math-fun
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5

Let $f(x)$ be the function represented by your series. Then

$$f(x) = \frac{1}{2}\sum_{n = 0}^\infty \frac{1 - (-1)^n}{n!}x^n$$

since $1 - (-1)^n$ is $0$ when $n$ is even and $2$ when $n$ is odd. Thus

$$f(x) = \frac{1}{2}\sum_{n = 0}^\infty \frac{x^n}{n!} - \frac{1}{2}\sum_{n = 0}^\infty \frac{(-x)^n}{n!}, $$

which simplifies to

$$f(x) = \frac{1}{2}e^x - \frac{1}{2}e^{-x}.$$

kobe
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3

That is the Taylor series of $\sinh$, hyperbolic sine.

The derivation of this comes from:

$$D_x \sinh x = \cosh x$$

$$D_x \cosh x = \sinh x$$

and:

$$\sinh 0 = 0$$

$$\cosh 0 = 1$$

GFauxPas
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This happens to be precisely the odd terms of: $$e^x=\sum_{n\ge0}\frac{x^n}{n!}$$

Let's be a bit more general. Let's say we have a function $A$, defined by: $$A(x)=\sum_{n\ge0}a_nx^n$$ where $a_n$ is some sequence. How do we get only the odd terms? Let's call the function generated by the odd terms $A_O$. ("O" for "odd.") So, we want this function: $$A_O(x)=\sum_{n\ge0}a_{2n+1}x^{2n+1}$$ Well, here's a little trick. Let's write out $A(x)$, like this:
$A(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+\dotsb$
And, now, write out $A(-x)$:
$A(-x)=a_0-a_1x+a_2x^2-a_3x^3+a_4x^4-a_5x^5+\dotsb$
And, finally, find $A(x)-A(-x)$.
$A(x)-A(-x)=2a_1x+2a_3x^3+2a_5x^5+\dotsb$
because everything else cancels. This is just twice what we want! So: $$A_O(x)=\frac{A(x)-A(-x)}2$$

And, to solve your own problem, we have:

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

Akiva Weinberger
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  • By the way, we also have $A_E(x)=\dfrac{A(x)+A(-x)}2$, where $A_E$ is the function taken from the even terms. (You can check that $A_O$ is an odd function and that $A_E$ is an even function, by the way. And, as you'd expect, $A_E(x)+A_O(x)=A(x)$ for all $x$.) – Akiva Weinberger Mar 18 '15 at 01:49
  • One last thing: $\dfrac{e^x-e^{-x}}2$ is also known as "the hyperbolic sine," written $\sinh(x)$. $\dfrac{e^x+e^{-x}}2$ is also known as "the hyperbolic cosine," written $\cosh(x)$. They have similar identities to the normal sine and cosine; for example, $\cosh^2(x)-\sinh^2(x)=1$, and the sum and difference formulae are similar, too, as are the the differentiation formulae. – Akiva Weinberger Mar 18 '15 at 01:53