Taylor Expansion: Computing x^2 + x^4/12

[robinegberts]

Hello friends,

I need to compute the taylor expansion of

$$\frac{x^4 e^x}{(e^x-1)^2}, $$
for $x<<1$, to find

$$ x^2 + \frac{x^4}{12}.$$

Can someone explain this to me?

Thanks!

 

[mfb]

What does your textbook say about Taylor expansions? Did you follow its steps, where did you run into problems?

I don't know if it is actual homework, but I think it fits better to our homework section.

 

[robinegberts]

Yes, thank you mfb.

I tried:

$$x^4 \frac{1+x+O(x^2)}{(x+O(x^2))^2} = x^2 + x^3,$$
which is clearly not the way to go. My main question is how to deal with that the expansion of $e^x-1$ will be squared and how that influences my choice of developing up till certain order terms. Can this even be done using standard series?

 

[mfb]

You'll need more terms in the expansion of the denominator.

Or calculate the derivatives, that takes more time but is less error-prone.

 

[pixel]

One thing I noticed is that if we write out the denominator we get

(e^x-1)² = e^2x - 2e^x + 1.

If we expand each exponential term to just first order in x, we get: 1 + 2x - 2 - 2x + 1 which = 0, whereas doing it as above you got x².

 

[Ray Vickson]

Yes, thank you mfb.

I tried:

$$x^4 \frac{1+x+O(x^2)}{(x+O(x^2))^2} = x^2 + x^3,$$
which is clearly not the way to go. My main question is how to deal with that the expansion of $e^x-1$ will be squared and how that influences my choice of developing up till certain order terms. Can this even be done using standard series?

That IS the way to go, but first you need to use the correct expansions for $e^x$, and you should keep 3 or 4 terms in the numerator and the enominator, just to be absolutely sure you have not neglected something important. Then, of course, you expand the denominator as
$$(1+Y)^{-2} = 1 - 2Y + [(-2)(-3)/2!]Y^2 +[(-2)(-3)(-4)/3!] Y^3 + \cdots,$$
where $1+Y$ is what you have left after expanding $e^x - 1$ and taking out the factor $x$.

 

[mfb]

One thing I noticed is that if we write out the denominator we get

(e^x-1)² = e^2x - 2e^x + 1.

If we expand each exponential term to just first order in x, we get: 1 + 2x - 2 - 2x + 1 which = 0, whereas doing it as above you got x².

That is fine, as the expansion does indeed start with 0 + 0x, you just missed the first nonzero term in your expansion.

To get three terms in the final expansion, we need three nonzero terms in the denominator, which means everything up to x⁴.

 

[robinegberts]

Clever Ray! Let's see whether I am getting it.

$$x^4 \frac{1+x+x^2/2}{(x+x^2/2+x^3/6)^2} = x^2 (1+x+x^2/2) (1+x/2+x^2/6)^{-2} = x^2 (1+x+x^2/2) (1-2(x/2+x^2/6)) = x^2 (1+x+x^2/2) (1-x-x^2/3) = (x^2+x^3+x^4/2)(1-x-x^2/3)=x^2-x^3-x^4/3+x^3-x^4+x^4/2=x^2-10/12 x^4. $$

Nearly there, did I miss something?

 

[mfb]

You'll need one more order for the (1+c)^-2 expansion: it adds x² terms, which then lead to x⁴ terms in the final result.

WolframAlpha gets -1/12 as prefactor for x⁴.

 

[ehild]

$$f(x) = \frac{x^4 e^x}{(e^x-1)^2}$$ can be written in a form easier to expand.
Divide numerator and denominator with e^x.
$$f(x) = \frac{x^4 }{(e^{x/2}-e^{-x/2})^2}=\frac{ x^4 }{4 \sinh^2(x/2)}$$
Using that $\sinh^2(x/2)=\frac{\cosh(x)-1}{2}$, $$f(x)=\frac{x^4}{2(\cosh(x)-1)}$$.
It is very easy to expand the cosh(x)-1 function. cosh(x)'=sinh(x), and sinh(x)' = cosh(x). Expand to the fourth power in x. It is two terms only as cosh is an even function.

 

[robinegberts]

Thanks for the help mfb!I give up... I am studying for a thermal physics test, and this really does not contribute to my understanding anymore. ;-)

 

[robinegberts]

Ok, so that yields me

$$f(x) = \frac{x^4}{x^2+x^4/12},$$ now what? ;-)

 

[mfb]

Just one step away:$$f(x) = \frac{x^4}{x^2+x^4/12} = x^2(1+x^2/12)^{-1} \approx x^2(1-x^2/12)$$

 

[robinegberts]

Ah, I understand. Wish I had more practise with these expansion. Thanks for the help all!

 

[SammyS]

One thing I noticed is that if we write out the denominator we get

(e^x-1)² = e^2x - 2e^x + 1.

.

Start with $\displaystyle \ \frac{x^4 e^x}{(e^x-1)^2}=\frac{x^4 }{e^{-x}(e^x-1)^2} \ .$

Then using pixel's expansion we have $\displaystyle\frac{x^4 }{e^{-x}(e^{2x}-2e^x+1)} \ .$

.And multiplying through by e^−x we have $\displaystyle\frac{x^4 }{(e^{x}-2+e^{-x})} \ ,$

which easily leads to ehild's result if that's where you would like to go.

However, using the well known series for e^x gives the following quite directly.

$\displaystyle\frac{x^4 }{\displaystyle 2\left(\frac{x^2}{2} + \frac{x^4}{4!} + \frac{x^6}{6!}+\dots \right)} \ ,$​

.
Use as many terms as necessary.