Maclaurin Series When 0 is in the Denominator?

[Ascendant0]

May I ask whether you know the expected result? I'm curious to compare it with my solution.

I'm really pressed for time, but yes, I got it. The TA also got back to me earlier today and showed me another method, one that was very fast and simple, but I never would've figured it out on my own. I'll try to post it after the exam later this week

The series I got for this one is:

$1 - x + x^2/3 - x^4/45 ...$

Same as yours, and it's consistent with the one in the book, plus I know for sure the method he wants to see on the exam now as well

I greatly appreciate the help on this one, all of you.

 

[FactChecker]

The TA also got back to me earlier today and showed me another method, one that was very fast and simple

I'm curious about the method.

 

[fresh_42]

I'm curious about the method.

I think I found it.
$$
f(x)=-x+x\operatorname{coth}(x)
$$
The series for the cotangent hyperbolic is a bit tricky but can be looked up. E.g., Wikipedia gives me:
$$
\operatorname{coth}(x)-\dfrac{1}{x}=\dfrac{1}{3}x-\dfrac{1}{45}x^3+\dfrac{2}{945}x^5-\dfrac{1}{4725}x^7 +\ldots =\sum_{k=1}^\infty (-1)^{k+1}\cdot\dfrac{2}{\pi^{2k}}\cdot \zeta(2k)\cdot x^{2k-1}
$$
and therefore
$$
f(x)=1-x+\dfrac{1}{3}x^2-\dfrac{1}{45}x^4+\dfrac{2}{945}x^6-\dfrac{1}{4725}x^8 +\ldots
$$
The radius of convergence is $\pi.$

 

[PeroK]

I think I found it.
$$
f(x)=-x+x\operatorname{coth}(x)
$$
The series for the cotangent hyperbolic is a bit tricky but can be looked up. E.g., Wikipedia gives me:
$$
\operatorname{coth}(x)-\dfrac{1}{x}=\dfrac{1}{3}x-\dfrac{1}{45}x^3+\dfrac{2}{945}x^5-\dfrac{1}{4725}x^7 +\ldots =\sum_{k=1}^\infty (-1)^{k+1}\cdot\dfrac{2}{\pi^{2k}}\cdot \zeta(2k)\cdot x^{2k-1}
$$
and therefore
$$
f(x)=1-x+\dfrac{1}{3}x^2-\dfrac{1}{45}x^4+\dfrac{2}{945}x^6-\dfrac{1}{4725}x^8 +\ldots
$$
The radius of convergence is $\pi.$

That's neat, but I'm sceptical that's what would be expected.

 

[vela]

I explained exactly what I tried, exactly what steps I took. I'm lost on what it is some of you want on here. I showed you the equations I got and what steps I took. What else am I supposed to show?

Your work showing intermediate results. It's great that you described what you think you did correctly, but that can often be different than what you actually did.

I'm curious about the method.

I'd guess it's something like this. Let $u=2x$. Then we have
$$\frac{2x}{e^{2x}-1} = \frac{u}{u+u^2/2! + u^3/3!+\dots} = \frac{1}{1+\underbrace{u/2 + u^2/6 + \dots}_r}.$$ Then using the Maclaurin series for $1/(1+r)$, just start collecting powers of $u$ to figure out the first few terms of the series.

 

[PeroK]

Your work showing intermediate results. It's great that you described what you think you did correctly, but that can often be different than what you actually did.


I'd guess it's something like this. Let $u=2x$. Then we have
$$\frac{2x}{e^{2x}-1} = \frac{u}{u+u^2/2! + u^3/3!+\dots} = \frac{1}{1+\underbrace{u/2 + u^2/6 + \dots}_r}.$$ Then using the Maclaurin series for $1/(1+r)$, just start collecting powers of $u$ to figure out the first few terms of the series.

The funny thing is that the function in this question is effectively the generating function for the Bernoulli numbers:

https://www.efunda.com/math/special_numbers/special_numbers.cfm#bernoulli

Quite a homework assignment!

 

[fresh_42]

The funny thing is that the function in this question is effectively the generating function for the Bernoulli numbers:

https://www.efunda.com/math/special_numbers/special_numbers.cfm#bernoulli

Quite a homework assignment!

I was surprised, too, that I basically found a recursion for $\zeta(2n)$ after I saw the series expansion of $\coth(x).$ The functions here even have names and real physical meanings!

https://en.wikipedia.org/wiki/Brillouin_and_Langevin_functions

 

[fresh_42]

Fun fact: I did the math to figure out what my recursively defined sequence $(b_0,b_1,b_2,\ldots)$ would be in terms of the zeta-function. It turned out that $y^{(2n)}(0)=b_{2n}=B_{2n},$ i.e. exactly the Bernoulli numbers.

 

[FactChecker]

Your work showing intermediate results. It's great that you described what you think you did correctly, but that can often be different than what you actually did.


I'd guess it's something like this. Let $u=2x$. Then we have
$$\frac{2x}{e^{2x}-1} = \frac{u}{u+u^2/2! + u^3/3!+\dots} = \frac{1}{1+\underbrace{u/2 + u^2/6 + \dots}_r}.$$ Then using the Maclaurin series for $1/(1+r)$, just start collecting powers of $u$ to figure out the first few terms of the series.

That's a very nice observation. It might work for any $1/(a_0+a_1x^+a_2x^2+...)$ problem. Set $(a_0+a_1x^+a_2x^2+...)=1+(-1+a_0+a_1x^+a_2x^2+...)=1+r \text{ with } r=-1+a_0+a_1x^+a_2x^2+...$.

 

[fresh_42]

That's a very nice observation.

This entire problem is ...

Quite a homework assignment!

... but is very interesting. We already have found three different solutions and hope ...

I'm curious about the method.

... to see a fourth. I never had a connection to Bernoulli numbers and looked them up every time I met them. But
$$
B_{n}=\left(\dfrac{x}{e^x-1}\right)^{(n)}(0)
$$
might be a way to memorize them after I did all these calculations on the right. And the most interesting fact was the formula
$$
M=N\cdot m\cdot L\left(\dfrac{mB}{k_B T}\right) \text{ with } L(x)=\dfrac{f(x)+x-1}{x}
$$
with $m$ the magnetic moment, $B$ the magnetic field, $k_B$ Boltzmann, $T$ temperature, $N$ amount of substance and $M$ the paramagnetism.

So, although undoubtedly ...

Quite a homework assignment!

... it bears many interesting aspects.

 

[vela]

That's a very nice observation. It might work for any $1/(a_0+a_1x^+a_2x^2+...)$ problem. Set $(a_0+a_1x+a_2x^2+...)=1+(-1+a_0+a_1x^+a_2x^2+...)=1+r \text{ with } r=-1+a_0+a_1x^+a_2x^2+...$.

You don't want a constant term in $r$ otherwise all the terms in the expansion will contribute to all orders of $x$. Instead, factor out the lowest-order term present. For example, if $a_0 \ne 0$, you have
$$\frac{1}{a_0+a_1x+a_2x^2+\dots} = \frac{1}{a_0}\frac{1}{1 + b_1 x+ b_2 x^2 + \dots} $$ where $b_i = a_i/a_0$. If $r$ is of order $x$, then the linear term of the series only gets a contribution from $r$, the quadratic term gets contributions from $r$ and $r^2$, and so on.

 

[FactChecker]

You don't want a constant term in $r$ otherwise all the terms in the expansion will contribute to all orders of $x$. Instead, factor out the lowest-order term present.

Good point.

For example, if $a_0 \ne 0$, you have
$$\frac{1}{a_0+a_1x+a_2x^2+\dots} = \frac{1}{a_0}\frac{1}{1 + b_1 x+ b_2 x^2 + \dots} $$ where $b_i = a_i/a_0$.

I thought about that and wanted something that might work if there was no constant term. Now that you have pointed out the problem if there is a constant term in r, I see that it would be better to divide the denominator by the leading term, including any power of x.

 

[PeroK]

Good point.

I thought about that and wanted something that might work if there was no constant term. Now that you have pointed out the problem if there is a constant term in r, I see that it would be better to divide the denominator by the leading term, including any power of x.

To flesh this out a bit. Suppose we have $f(x) = \frac 1 {g(x)}$, where:
$$f(x) = a_0 + a_1x + a_2x^2 + \dots$$$$g(x) = 1 + b_1x + b_2x^2 + \dots$$Note that:
$$f(x) = 1 - f(x)(g(x) - 1)$$Hence:
$$a_0 + a_1x + a_2x^2 + \dots = 1 - (a_0 + a_1x + a_2x^2 + \dots)(b_1x + b_2x^2 + \dots)$$$$= 1 - (a_0b_1)x - (a_0b_2 + a_1b_1)x^2 - (a_0b_3 + a_1b_2 + a_2b_1)x^3 - \dots$$And we have a recursive relationship between the $a_n$ and $b_n$:
$$a_0 = 1, a_1 = -b_1, a_2 = -(b_2 - b_1^2) \dots$$In general:
$$a_n = -a_0b_n - a_1b_{n-1} \dots - a_{n-1}b_1$$