1
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$
    – Triatticus
    Apr 18, 2018 at 1:55
  • $\begingroup$ @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? $\endgroup$ Apr 18, 2018 at 1:58
  • $\begingroup$ 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 $$\small{(1 + x + (1/2)x^2 + (1/3!)x^3+\dots) - (1 - x + (1/2)x^2-(1/3!)x^3 +\dots) = 2x + 2(1/3!)x^3 +\dots)}$$ $\endgroup$
    – Triatticus
    Apr 18, 2018 at 2:01
  • $\begingroup$ MathJax tutorial $\endgroup$ Apr 18, 2018 at 2:35
  • $\begingroup$ Divide by two? Multiply by half? Divide in half? $\endgroup$ Apr 18, 2018 at 13:51

3 Answers 3

9
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$ Apr 18, 2018 at 2:03
  • 1
    $\begingroup$ 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. $\endgroup$
    – Nick
    Apr 18, 2018 at 2:06
8
$\begingroup$

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

$\endgroup$
7
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.