# Binomial expansion

1. Oct 4, 2014

### Appleton

1. The problem statement, all variables and given/known data
Prove that, if x is so small that
$x^6$ and higher powers of x may be neglected, then $\frac{e^{2x}-1}{e^{2x}+1}\approx x-\frac{x^3}{3}+\frac{2x^5}{15}$
2. Relevant equations

3. The attempt at a solution

$\\ \frac{e^{2x}-1}{e^{2x}+1}\approx\frac{2x+\frac{(2x)^2}{2}+\frac{(2x)^3}{3!}+ \frac{(2x)^4}{4!}+ \frac{(2x)^5}{5!}}{2+2x+\frac{(2x)^2}{2}+\frac{(2x)^3}{3!}+ \frac{(2x)^4}{4!}+ \frac{(2x)^5}{5!}} \\ \approx \frac{x+x^2+\frac{2x^3}{3}+ \frac{x^4}{3}+ \frac{2x^5}{15}}{1+x+x^2+\frac{2x^3}{3}+ \frac{x^4}{3}+ \frac{2x^5}{15}} \\ \approx \frac{x-\frac{x^3}{3}+\frac{2x^5}{15}+x^2+x^3+ \frac{x^4}{3}}{1+x+x^2+\frac{2x^3}{3}+ \frac{x^4}{3}+ \frac{2x^5}{15}} \\$
At this point I am unable to isolate
$x-\frac{x^3}{3}+\frac{2x^5}{15} \\$
The only other approaches that occur to me are:
$e^{2x}(e^{2x} +1)^{-1}-(e^{2x}+1)^{-1}$
From which I could yield a series of ascending powers of $e^{2x}$, however this doesn't seem to lead anywhere useful.
Alternatively:
$\frac{e^{2x}-1}{e^{2x}+1}\approx 1-\frac{1}{1+x+x^2+\frac{2x^3}{3}+ \frac{x^4}{3}+ \frac{2x^5}{15}} \\$
was another expression I looked at but didn't seem to go anywhere.
Part a) of this question asks me to express $ln{\frac{1+x}{1-x}}$ as a series of ascending powers of x up to and including $x^5$. I got $2(x+\frac{x^3}{3}+\frac{x^5}{5}).$ Given that the two problems form 2 parts of the same question and the form of the expressions look so similar and ln and e are closely related, I would expect there to be some strong relationship between the route to finding the answers to both these questions.

2. Oct 4, 2014

### Orodruin

Staff Emeritus
I would start from this insight. What happens if you take the logarithm of the entire expression?

3. Oct 4, 2014

### Ray Vickson

It is easier and neater to first expand
$$f(t)= \frac{t}{t+2}$$
in powers of $t$, then substitute $t = e^{2x}-1$ afterward. Since $t$ is of "order" $x$ in magnitude, to get results accurate up to order $x^6$ you need keep only powers up to $t^6$; do you see why? Then just substitute the series for $e^{2x}$ and grind and grind.

4. Oct 4, 2014

### Appleton

Thanks for your replies. I did look at taking the logarithm of
$\frac{e^{2x}-1}{e^{2x}+1}\approx x-\frac{x^3}{3}+\frac{2x^5}{15}$
Which gave me
$ln(e^{2x}-1) - ln(e^{2x}+1)\approx ln(x-\frac{x^3}{3}+\frac{2x^5}{15})$
But I couldn't progress with it because I couldn't see how to expand $ln(e^{2x}-1)$
So, deploying Ray's method, I guess I am meant to simplify the following and then substitute the expansion series for $e^{2x}$ right?
$\frac{e^{2x}-1}{2}-\frac{(e^{2x}-1)^2}{4}+\frac{(e^{2x}-1)^3}{8}-\frac{(e^{2x}-1)^4}{32}+\frac{(e^{2x}-1)^5}{192}-\frac{(e^{2x}-1)^6}{1536}$
Seems pretty laborious, I guess that's what you mean by 'grind and grind'.

5. Oct 5, 2014

### vela

Staff Emeritus
I'd start by restoring some symmetry to the function:
$$\frac{e^{2x}-1}{e^{2x}+1} = \frac{e^x(e^x-e^{-x})}{e^x(e^x+e^{-x})} = \frac{e^x-e^{-x}}{e^x+e^{-x}}.$$ If you now expand the numerator and denominator as series, the algebra's actually not too bad.

6. Oct 5, 2014

### Appleton

Thanks for explaining that. I am still quite interested to know what Orodruin had in mind when he suggested taking the logarithm of the whole expression.

7. Oct 5, 2014

### Orodruin

Staff Emeritus
To be quite honest, I think I was tired and not thinking very straight when I wrote that. Ray's or vela's proposals seem much more straightforward.

Edit: Looking at the computations, vela's method seems to be the one with less algebra, simply because you only have to consider three terms in the numerator and denominator, respectively, due to the symmetry.

Last edited: Oct 5, 2014
8. Oct 5, 2014

### PeroK

Alternatively, you could note that:

$\frac{e^{2x}-1}{e^{2x}+1} = tanh(x)$

Then derive the Taylor series about 0 for $tanh$. Setting $f(x) = tanh(x)$ and expressing higher derivates in terms of lower derivatives keeps the algebra under control.

$f'(x) = 1 - f(x)^2$

$f''(x) = -2f(x)f'(x)$

$f'''(x) = -2f'(x)^2 -2f(x)f''(x)$

Etc.

9. Oct 5, 2014

### Appleton

Sorry to string this out, unfortunately i'm unable to derive the answer from this method, after expanding your expression I get $\frac{x+\frac{x^3}{3!}+\frac{x^5}{5!}}{1+\frac{x^2}{2}+\frac{x^4}{4!}}$ and can't see how to progress from here.

10. Oct 5, 2014

### vela

Staff Emeritus
Use the fact that $\frac{1}{1+t} = 1-t+t^2-\cdots$.

11. Oct 6, 2014

### Appleton

Never would have thought of that. Good trick. I'll try and remember that. Thanks.