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Maclaurin series for

$$\frac{x}{e^x-1}$$

The answer is

$$1-\frac x2 + \frac {x^2}{12} - \frac {x^4}{720} + \cdots$$

How can i get that answer?

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  • $\begingroup$ Do you mean the $5$ terms only? $\endgroup$ Feb 23, 2013 at 8:21
  • $\begingroup$ Yes, I need to find untill x^4's term $\endgroup$
    – Bek Abdik
    Feb 23, 2013 at 8:26
  • 1
    $\begingroup$ For the whole series, see en.wikipedia.org/wiki/Bernoulli_number#Generating_function $\endgroup$ Feb 23, 2013 at 8:33
  • $\begingroup$ I was typing it out, but Ittay Weiss is faster. It will work very nicely for the first few coefficients. $\endgroup$ Feb 23, 2013 at 8:35
  • $\begingroup$ The more often you compute derivatives, the faster you will become at computing derivatives. Conversely, if you avoid computing derivatives because it is slow, you will find that you find that computing derivatives is slow. $\endgroup$
    – user14972
    Feb 23, 2013 at 15:59

4 Answers 4

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Let $f(x)=\frac {x}{e^x-1}$ and consider the product $(e^x-1)\cdot f(x)=x$. Since $f$ is infinitely differentiable it follows that it has a Taylor series. Now, the Taylor series for $e^x-1$ is $$\sum_{k=1}^\infty \frac{x^k}{k!} $$(obtained immediately from the Taylor series for $e^x$). Thus, if $$\sum _{k=0}^\infty c_kx^k$$ is the Taylor series for $f(x)$ then $$(\sum _{k=1}^\infty \frac {x^k}{k!})\cdot (\sum _{k=0}^\infty c_kx^k)=x$$ and dividing by $x$ yields $$(\sum _{k=0}^\infty \frac {x^{k}}{(k+1)!})\cdot (\sum _{k=0}^\infty c_kx^k)=1,$$ from which, by expanding and equating coefficients, we obtain

$\frac{1}{1!}\cdot c_0 = 1$,

$\frac{1}{1!}\cdot c_1 + \frac{1}{2!}\cdot c_0 = 0$

$\vdots$

$\sum _{j=0}^m\frac{1}{(j+1)!}\cdot c_{m-j}=0$

$\vdots$

You solve these equations inductively to obtain the values for the $c_k$.

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One way is to write $e^x-1 $ as $1 + x + x^2/2 + ... - 1$ and then factor out $x$ and cancel up the top and expand it as geometric series and collect the coefficients of like powers.

$\displaystyle \begin{align*} e^x - 1 &= x + \frac{x^2}{2!} + \frac{x^3}{3!} + o(x^4)\\ \frac{x}{e^x - 1} &= \frac{1}{1 + \left( \frac x 2 + \frac{x^2}{6} + o(x^3) \right )} \\ &= 1 - \left( \frac x 2 + \frac{x^2}{6} + o(x^3) \right ) + \left( \frac x 2 + \frac{x^2}{6} + o(x^3) \right )^2 - \left( \frac x 2 - \frac{x^2}{6} + o(x^3) \right )^3 + \left( \frac x 2 + \frac{x^2}{6} + o(x^3) \right )^4... \\ &= 1 - \frac{x}{2} + x^2 \left( \frac 1 4 - \frac 1 6 \right ) + x^3 \left(-\frac{1}{4!} + 2 \cdot \frac 1 2 \cdot \frac 1 6 - \frac{1}{2^3} \right ) + x^4 \left(-\frac{1}{5!} + \frac{1}{6^2} + 2 \cdot \frac 12 \cdot \frac{1}{4!} + \frac{1}{2^4}\right )+o(x^5) \end{align*} $

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This is not a straightforward solution, but I added this to show that we have other ways if we know some properties of the function.

Method 2. Using the Taylor series of the logarithm, we have

\begin{align*} \frac{x}{e^x - 1} &= \frac{\log(1+(e^x - 1))}{e^x - 1} \\ &= \sum_{n=0}^{\infty} \frac{(-1)^n}{n+1} (e^x - 1)^n \\ &= \sum_{n=0}^{\infty} \frac{(-1)^n}{n+1} x^n \left( \frac{e^x - 1}{x} \right)^n . \end{align*}

Since we only want to extract terms up to degree 4, we can focus on the following expansion:

\begin{align*} \frac{x}{e^x - 1} &= 1 - \frac{x}{2} \left( 1 + \frac{x}{2} + \frac{x^2}{6} + \frac{x^3}{24} + \cdots \right) + \frac{x^2}{3} \left( 1 + \frac{x}{2} + \frac{x^2}{6} + \cdots \right)^2 \\ &\qquad - \frac{x^3}{4} \left(1 + \frac{x}{2} + \cdots \right)^3 + \frac{x^4}{5} \left(1 + \cdots \right)^4 \end{align*}

Method 3. We decompose the function into the odd part and the even part:

\begin{align*} \frac{x}{e^x - 1} &= \color{red}{\frac{1}{2}\left( \frac{x}{e^x - 1} - \frac{(-x)}{e^{-x} - 1} \right) } + \color{blue}{\frac{1}{2}\left( \frac{x}{e^x - 1} + \frac{(-x)}{e^{-x} - 1} \right)} \\ &= \color{red}{-\frac{x}{2}} + \color{blue}{\frac{x}{2} \cdot \frac{e^x + 1}{e^x - 1}} \\ &= \color{red}{-\frac{x}{2}} + \color{blue}{\frac{\cosh (x/2)}{\left(\frac{\sinh(x/2)}{x/2}\right)}} \end{align*}

Expanding both the numerator and the denominator of the blue-colored term,

$$ \cosh(x/2) = 1 + \frac{x^2}{8} + \frac{x^4}{384} + \cdots, \qquad \frac{\sinh (x/2)}{x/2} = 1 + \frac{x^2}{24} + \frac{x^4}{1920} + \cdots. $$

Thus using the same trick as other answers we find that

$$ \frac{x}{e^x - 1} = -\frac{x}{2} + \left( 1 + \frac{x^2}{8} + \frac{x^4}{384} + \cdots \right)\left[ 1 - \left( \frac{x^2}{24} + \frac{x^4}{1920} + \cdots \right) + \left( \frac{x^2}{24} + \cdots \right)^2 - \cdots \right]. $$

So the burden of calculation reduces greatly.

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An other approach would be to take the inverted first: $$ \frac{e^x-1}{x} = 1 + \frac{x}{2} + \frac{x^2}{6} +\frac{x^3}{24} + \frac{x^4}{120} \cdots $$ and then you go: $$\frac{x}{e^x-1} = (\frac{e^x-1}{x})^{-1} = [1 + (\frac{x}{2} + \frac{x^2}{6} +\frac{x^3}{24} +\frac{x^4}{120})]^{-1} $$ and then expand this as in the binomial expansion $(1+u)^{-1}$ and since here $ u <1$ the previous gets expanded to: $$ 1 - (\frac{x}{2} + \frac{x^2}{6} +\frac{x^3}{24} +\frac{x^4}{120}) +(\frac{x}{2} + \frac{x^2}{6} +\frac{x^3}{24})^2 - (\frac{x}{2} + \frac{x^2}{6})^3 + (\frac{x}{2})^4 \tag1 $$ Notice that i only take terms accordingly until the desired $x^4$ terms neglecting the other higher order terms. Now keep expanding $(1)$ while still not taking into account calculations that will result in terms higher than $x^4$ (to avoid unnecessary trouble) and it won't be long until you reach the correct result.

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