# Why doesn't $\sinh(x)$'s Taylor Series divide by half?

Learning about Taylor Series, I have the problem sinh(x). Obviously, $\sinh(x) = \dfrac{e^x - e^{-x}}2$. I basically did it all correctly, since most of it cancels when compared to $e^x$'s taylor series. But that $\cfrac 12$ that's in the problem is throwing me off. Why isn't the summation series:

$$\sum 2\frac{x^{2x+1}}{(2n + 1)!}$$

That (* 2) on the bottom is what is confusing me. How is that getting cancelled out? All of the positive terms of n are getting cancelled out in e$^x$'s taylor series, but we still have these left. And they're still being divided by 2.

What I'm getting for the series itself written out, is:

$\cfrac 12\left[x + \cfrac {x^3}{3!} + \cfrac{x^5}{5!}\right]$

Why is that 2 just forgotten about?

• You are forgetting that you will obtain twice the odd coefficients, that is when you subtract them you will have say $2x$ for the first term not $x$. Thus this cancels with the $2$ in the denominator. Apr 18, 2018 at 1:55
• @TriatticusJust to make sure I'm understanding this correctly. It's because I'm re-defining n? Because I threw out half the results for summation of e$^x$ and are adding in extras? Apr 18, 2018 at 1:58
• It has nothing to do with redefining $n$. When you perform the subtraction the odd coefficients will be doubled because you are subtracting a negative which becomes addition so that $$\small{(1 + x + (1/2)x^2 + (1/3!)x^3+\dots) - (1 - x + (1/2)x^2-(1/3!)x^3 +\dots) = 2x + 2(1/3!)x^3 +\dots)}$$ Apr 18, 2018 at 2:01
• MathJax tutorial Apr 18, 2018 at 2:35
• Divide by two? Multiply by half? Divide in half? Apr 18, 2018 at 13:51

Write out the Taylor series for $e^x$ and $e^{-x}$:

\begin {align*} e^x &= \sum_{n=0}^\infty \frac{x^n}{n!} \\ e^{-x} &= \sum_{n=0}^\infty \frac{(-1)^n x^n}{n!} \end {align*}

Subtract them and you get $$e^x - e^{-x} = \sum_{n=0}^\infty \frac{1-(-1)^n}{n!} \, x^n$$

When $n$ is even, the coefficient is zero, and when $n$ is odd, it is $\frac{2}{n!}$. So the $\frac{1}{2}$ cancels this 2.

• Why are you getting 1 - (-1)$^n$. And where is the x$^n$ coming from in the bottom half? editing If I split it up I get x$^n$/n! - x$^n$/n! - (-1)$^n$. So wouldn't it be (-1)$^n$ / n! times 1/2? Am I going backwards here? Apr 18, 2018 at 2:03
• If you have two power series $f(x) = \sum a_n x^n$ and $g(x) = \sum b_n x^n$, then when you add them, you just add the coefficients: $f(x) + g(x) = \sum (a_n + b_n) x^n$. It's just like adding polynomials.
– Nick
Apr 18, 2018 at 2:06

Write out the first few terms of each series

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} \cdots$$

$$e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \frac{x^5}{5!} + \cdots$$

Subtract the two series and you get

$$e^x - e^{-x} = 2x + \frac{2x^3}{3!} + \frac{2x^5}{5!} + \cdots$$

Notice how each odd power coefficient doubles? Divide the above by $2$ and you obtain the $\sinh x$ series

The following manipulation can be proven valid as all of the series are absolutely convergent on all of $\mathbb{R}$ (actually on all of $\mathbb{C}$). \begin{align} \sinh x&=\frac{e^x - e^{-x}}{2} \\ &=\frac{1}{2}\left[\sum_{n\in\mathbb{N}}\frac{x^n}{n!} - \sum_{n\in\mathbb{N}} \frac{(-1)^nx^n}{n!}\right] \\ &=\frac{1}{2} \sum_{n\in\mathbb{N}} \left(1-(-1)^n\right) \frac{x^n}{n!} \end{align} Now, notice that when $n$ is even, then the summand will vanish. When $n$ is odd, there will be a factor of $2$. So, \begin{align} \sinh x &= \frac{1}{2}\sum_{n\in 2\mathbb{N}+1}2\cdot \frac{x^n}{n!} \\ &= \sum_{n\in 2\mathbb{N}+1} \frac{x^n}{n!} \\ &= \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} \end{align}