Maclaurin Series When 0 is in the Denominator?

[Ascendant0]

Homework Statement

Find the Maclaurin series for $2x/(e^{2x}-1)$

Relevant Equations

Maclaurin: $f(x) = f(0) + f'(0) + f''(0) ...$

I have tried a few things and can't figure this out. If I separate the top and bottom, the top obviously quickly goes to 0, so there's no series after 1 derivative (f'(2x) = 2, then f''(2x) = 0), so I can't separate them and do anything with it

And if I do it with $2x(e^{2x}-1)^{-1}$, every single derivative of that comes out to a 0 in the denominator

This would lead me to think there's no Maclaurin for this set. But in the back of the solutions manual, sure enough they have one. And we can't do it any other way - this problem specifically asks for the Maclaurin series.

I don't know what else I can do, as no matter how I take the derivatives, they end up having a 0 in the denominator???

 

[Ascendant0]

I think I know what to do now. The only thing I could possibly come up with is long division of the numerator and denominator series. I don't see any other possible way, not with what we've been taught. It's just odd that it asked for the Maclaurin specifically, but it's not even possible to do it that way

 

[mjc123]

Try expanding e^2x - 1. Let x = δ, where δ << 1. Evaluate the expression as δ becomes very small.

 

[PeroK]

You could expand $e^{2x}$ as a Maclaurin series and produce an equivalent function for $x\ne 0$.

Fun fact: a few years ago I met Maclaurin's great, great, great ... grandson.

 

[PeroK]

Do you only need the first three terms?

The function appears to be infinitely differentiable at $x =0$, in the sense that we can define suitable derivatives at $x =0$ to be the limit of the derivative as $x \to 0$.

 

[fresh_42]

We must show that $f(0)$ exists anyway. With $\dfrac{2x}{e^{2x}-1}=\dfrac{2}{e^x+1}\cdot \dfrac{x}{e^x-1}$ we can concentrate on the second term, i.e. $x+y=e^xy.$ Then $(1-x-y)+y'=e^xy'$ and $-2+x+y-2y'+y''=e^xy'',$ etc. All are expressions $p\left(x,y,\ldots ,y^{(n-1)}\right)+y^{(n)}=e^xy^{(n)}$ with a linear polynomial $p$ in all its variables. Maybe we can conclude by induction that if $y^{(k)}(0)$ exists for all $k<n$ then $y^{(n)}(0)$ exists, too, by using the same argument as with $y(0).$

Edit: $y(0)=1\, , \,y'(0)=-\dfrac{1}{2}\, , \,\ldots$ deduced from the requirement that $p\left(0,y(0),\ldots,y^{(n-1)}(0)\right)=0$ in order to get a continuous solution for $p\left(x,y,\ldots ,y^{(n-1)}\right)+y^{(n)}=e^xy^{(n)}$.

 

[FactChecker]

Homework Statement: Find the Maclaurin series for $2x/(e^{2x}-1)$
Relevant Equations: Maclaurin: $f(x) = f(0) + f'(0) + f''(0) ...$

This is not the correct formula for the Maclaurin series.

I have tried a few things and can't figure this out. If I separate the top and bottom, the top obviously quickly goes to 0, so there's no series after 1 derivative (f'(2x) = 2,

That is wrong. The derivative at x=0 is negative.

then f''(2x) = 0), so I can't separate them and do anything with it

And if I do it with $2x(e^{2x}-1)^{-1}$, every single derivative of that comes out to a 0 in the denominator

Although I would not recommend it for this problem, you should be aware of the L'Hospital's Rule.

For this problem, I would try long division of the Maclaurin series of the numerator by the Maclaurin series of the denominator. Even that seems to get messy after a couple of terms, so there might be a better way.

 

[PeroK]

One approach is to let $g(x) = \frac 1 {f(x)}$, cancel the $x$ using the series for $e^x$, then express the derivatives of $f(x)$ at $0$ in terms of the derivatives of $g(x)$ at zero.

It only gets complicated from the third derivative upwards.

PS another trick is to calculate the series for $\frac x {e^x -1}$, then replace $x \to 2x$ as a final step.

 

[fresh_42]

One approach is to let $g(x) = \frac 1 {f(x)}$, cancel the $x$ using the series for $e^x$, then express the derivatives of $f(x)$ at $0$ in terms of the derivatives of $g(x)$ at zero.

It only gets complicated from the third derivative upwards.

PS another trick is to calculate the series for $\frac x {e^x -1}$, then replace $x \to 2x$ as a final step.

I still think one can derive a recursion for $p\left(x,y,\ldots ,y^{(n-1)}\right)+y^{(n)}=e^xy^{(n)}$ and $y^{(n)}(0)$ such that $p\left(0,y(0),\ldots ,y^{(n-1)}(0)\right)=0$ which is necessary and sufficient for smoothness (see my post #6). The substitution may not be necessary since the other factor of $f(x)$ is $\dfrac{2}{e^x+1}$ which is conveniently one at zero and smooth. $y=\dfrac{x}{e^x-1}$ and therefore $x+y=e^xy$ is all that counts. One has to show the existence of the derivatives including $y(0)=1$ anyway.

 

[PeroK]

I still think one can derive a recursion for $p\left(x,y,\ldots ,y^{(n-1)}\right)+y^{(n)}=e^xy^{(n)}$ and $y^{(n)}(0)$ such that $p\left(0,y(0),\ldots ,y^{(n-1)}(0)\right)=0$ which is necessary and sufficient for smoothness (see my post #6). The substitution may not be necessary since the other factor of $f(x)$ is $\dfrac{2}{e^x+1}$ which is conveniently one at zero and smooth. $y=\dfrac{x}{e^x-1}$ and therefore $x+y=e^xy$ is all that counts. One has to show the existence of the derivatives including $y(0)=1$ anyway.

For $x \ne 0$:
$$g(x) = \frac{e^x -1}x = 1 + \frac x {2!} + \frac {x^2}{3!} + \dots$$So, if we define $f(x)=\frac 1{g(x)}$ we are fine.

 

[Ascendant0]

You could expand $e^{2x}$ as a Maclaurin series and produce an equivalent function for $x\ne 0$.

Fun fact: a few years ago I met Maclaurin's great, great, great ... grandson.

I did try that, but I believe I'm using it wrong somehow. It would be easy for me if it weren't for the $- 1$ combined with the $e^{2x}$ in the denominator. That is what is throwing me off and keeping me from seeing how to do this

So, I made $u = 2x$ to use the expansion for $e^u$, then put x back into the equation after expansion, so:

$1 +2x + 2x^2 +4x^3/3 + 2x^4/3$

But now matter what I do with that, the answer is coming out wrong. I thought maybe since the "-1" in the denominator becomes irrelevant as $x \rightarrow \inf$, I could just keep this in the denominator of the original problem under the 2x and try to simplify that way. That would give me:

$2x/1 + 2x/{2x} + 2x/{2x^2} + 2x(3)/{4x^3} ...$ and simplifies to:

$2x + 1 +1/x + 3/{2x^2} ...$ but that is not the right answer, so I know I'm doing it wrong, I'm just not sure how or why?

I also considered because of the $-1$ in the denominator to do a double substitution for series expansion, so to place the expansion I got for $e^{2x}$ and put that in a $(1+x)^p$, where it would be another $u$ substitution of $u=e^{2x}$ for $(u -1)^{-1}$, but I don't feel that is right at all, and not even sure how I would combine both of them without it being an algebra nightmare

I've tried a few other things as well, but nothing I'm doing is getting me even close to the correct answer. What am I doing wrong here?

 

[Ascendant0]

This is not the correct formula for the Maclaurin series.

That is wrong. The derivative at x=0 is negative.

Although I would not recommend it for this problem, you should be aware of the L'Hospital's Rule. View attachment 351369
For this problem, I would try long division of the Maclaurin series of the numerator by the Maclaurin series of the denominator. Even that seems to get messy after a couple of terms, so there might be a better way.

Ok, I am even more confused now as far as the statement "that is not the correct formula for the Maclaurin series", as it is exactly what it states the Maclaurin as in our book (attached image) literally right before the problems that tell you to specifically use the Maclaurin series. So you're saying the book is wrong??? So what is wrong about it?

As far as the $lim f'(x)/g'(x)$, I don't get that either. If I were to take the derivative of the top and bottom of this equation, I would get $2/(2e^{2x})$. So with that one, it would simplify to $(e^{2x})^{-1}$, and would be extremely simple from there. But, I tried that as well, and that gives me the wrong answer too. So, why is that approach not working for this equation?

And as far as the derivative being a negative number at x = 0, again this is confusing me. I took the derivative by hand yesterday, and even doubled checked it today, and the derivative of the equation is:

$2/(e^{2x}-1) - 4e^{2x}x/((e^{2x}-1)^2)$

Plugging $x=0$ into that gives you $2/0 - 0/1$, so it's undefined due to the $2/0$. How are you getting a different, negative answer?

Lastly, when you state to try division of the expansions of both the numerator and the denominator, what kind of infinite series expansion would there be for $2x$ of the numerator? I thought that with a problem such as this, you'd find the expansion of the denominator, and the whole thing would be multiplied by $2x$, so basically multiplying each fraction by $2x$.

If you could help make sense of those things for me, it would be greatly appreciated.

 

[PeroK]

I did try that, but I believe I'm using it wrong somehow. It would be easy for me if it weren't for the $- 1$ combined with the $e^{2x}$ in the denominator. That is what is throwing me off and keeping me from seeing how to do this

So, I made $u = 2x$ to use the expansion for $e^u$, then put x back into the equation after expansion, so:

$1 +2x + 2x^2 +4x^3/3 + 2x^4/3$

But now matter what I do with that, the answer is coming out wrong. I thought maybe since the "-1" in the denominator becomes irrelevant as $x \rightarrow \inf$, I could just keep this in the denominator of the original problem under the 2x and try to simplify that way. That would give me:

$2x/1 + 2x/{2x} + 2x/{2x^2} + 2x(3)/{4x^3} ...$ and simplifies to:

$2x + 1 +1/x + 3/{2x^2} ...$ but that is not the right answer, so I know I'm doing it wrong, I'm just not sure how or why?

I also considered because of the $-1$ in the denominator to do a double substitution for series expansion, so to place the expansion I got for $e^{2x}$ and put that in a $(1+x)^p$, where it would be another $u$ substitution of $u=e^{2x}$ for $(u -1)^{-1}$, but I don't feel that is right at all, and not even sure how I would combine both of them without it being an algebra nightmare

I've tried a few other things as well, but nothing I'm doing is getting me even close to the correct answer. What am I doing wrong here?

As was pointed out on another thread, unless you post your calculations no one can see what's wrong.

Try using the product/quotient rule to get the derivatives of $f(x)$ in terms of the reciprocal function $g(x)$, as in post 10.

Note that higher derivatives of a reciprocal get quite complicated. So, it gets hard to get more than the first three terms, up to $x^2$.

Show your working!

 

[Ascendant0]

As was pointed out on another thread, unless you post your calculations no one can see what's wrong.

Try using the product/quotient rule to get the derivatives of $f(x)$ in terms of the reciprocal function $g(x)$, as in post 10.

Note that higher derivatives of a reciprocal get quite complicated. So, it gets hard to get more than the first three terms, up to $x^2$.

Show your working!

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?

What I just tried before I read this post of yours was a substitution within a substitution.

Step 1: Based on the expansion $e^x = \sum x^n/n!$, I took $e^{2x} = e^u$ (substituted $u = 2x$) and got $1+u+u^2/2 + u^3/6 + u^4/24 ...$. If you're confused on how I got that, it is in Mathematical Methods by Boas, page 26. Hopefully, you know the expansion I'm applying, because there is no other work I can show for that part other than literally writing out the algebra term by term, which should be self-evident. I'm using the terms in the book, as our TA is very critical and expects things done exactly how they are in that section of the textbook

Step 2: Then, I took the the expansion I got from that, left it as the value $u$ for now, and additionally substituted $e^u = v$ so I could take $(v - 1 )^{-1}$ (the denominator of the original equation I'm supposed to expand) and expanded that series by a similar series expansion for an equation like that in the book, which gave me $-1-v-v^2-v^3-v^4-v^5...$

Step 3: In attempting to revert it back to $x$, I substituted back in $v = e^u$, and $u = 2x$. This is where everything got extremely convoluted, as I then have infinite series within an infinite series, so things like ""$(infinite series)^2$ " and so on, as should hopefully be evident by the equations I got from step 1 and step 2

Step 4: I tried to cancel out each value as I was able to, combine values as able, and it was completely wrong, but also extremely sloppy due to the infinite series within increasing exponentials, so showing all that work would be extremely sloppy and most likely confusing, but if you feel it was heading in the right direction and should get me the right answer, I'll take the time to type all that out. But it's all basic algebra from there, and it's just not getting me the right answer

Some of the things you suggested are over my head and beyond the scope of what we've covered so far, so I'm trying to use the things you've suggested that I understand and know how to do. And also like I said, the TA is very critical when it comes to the homework, and wants us using the methods and equations in each sub-section. So, I can't do something entirely different, or else he's just going to mark points off for it. There isn't a single person in class who got higher than a 75 on the first 3 homeworks we've had so far. Literally.

As far as the suggestions in posts 8 to 10, I'm a bit confused as to what is being used there and how? When you say to use the reciprocal and cancel x, are you suggesting to take the reciprocal of the portion I plan to expand, the denominator $e^{2x} - 1$ and expand that, then afterwards, set that over the $2x$ and cancel the values and reduce as I'm able to? Then I suppose after I do that is when I would take the reciprocal again to bring it back to the correct values? That is how I'm interpreting your suggestion.

I've expanded both $e^{2x}$ and $e^{2x}-1$ and tried using them both multiple ways at least a dozen times now. It keeps getting me nowhere.

Taking the series of $e^{2x}-1$ gives me $2x + 2x^2 + 4x^3/3 + 2x^4/3 ...$. If I set that over the $2x$ and simplify, I get $2x/2x + 2x^2/2x + 4x^3/(3*2x) ...$ and it simplifies down to $1 + x + 2x^2/3 ...$. Taking the reciprocal of that doesn't even come close to giving me the right answer, so I guess I'm just not understanding exactly what you're suggesting I do?

Like I said, some of what you guys state goes over my head, so I try to take the parts I do understand and run with those. And I am most certainly doing a ton of work. I've put at least five hours into this already between yesterday and today, and no matter what I do, I still can't get the right answer. I emailed our TA, but he can take days sometimes to get back, and I won't see the professor until Tues, and we have an exam on Thurs, so I need to get this stuff down. It's on expansions, including complex numbers. The complex numbers part has been easy, I grasp all that. It's this one particular problem that is for some reason stumping me, but it is a homework problem, so I need to make sure I know it, as the prof said the review materials for our exam is the homework, so to review that. And if this isn't making sense to me, it means I'm not fully grasping it, but I am most certainly trying

 

[PeroK]

There are a lot of words in that post. Forget the $2x$ for now. I don't see any derivatives being calculated. The Maclaurin series is all derivatives. As a first step, post the first and second derivatives of $f(x)$ in terms of $g(x)$, as in post 10.

 

[PeroK]

The strategy is based on the derivatives of $g(x)$ being easy to calculate at $x=0$.

The alternative is to use a binomial expansion to get the reciprocal Maclaurin series. PS that may not work.

 

[FactChecker]

View attachment 351381

Ok, I am even more confused now as far as the statement "that is not the correct formula for the Maclaurin series", as it is exactly what it states the Maclaurin as in our book (attached image) literally right before the problems that tell you to specifically use the Maclaurin series. So you're saying the book is wrong??? So what is wrong about it?

This is what I see in your original post:

That is not the Maclaurin series.

As far as the $lim f'(x)/g'(x)$, I don't get that either. If I were to take the derivative of the top and bottom of this equation, I would get $2/(2e^{2x})$.

Right. (But in this case you should not use 'f' to represent the numerator. You are already using 'f' to represent the original function. Use something else.)
So this gives $f(0) = [2x/(e^{2x}-1)]|_{x=0} = [2/(2e^{2x}]|_{x=0} = 1$, which is the correct limit at the undefined point, x=0.

So with that one, it would simplify to $(e^{2x})^{-1}$, and would be extremely simple from there. But, I tried that as well, and that gives me the wrong answer too. So, why is that approach not working for this equation?

L'Hospital's Rule does work. You should check your work again.

And as far as the derivative being a negative number at x = 0, again this is confusing me. I took the derivative by hand yesterday, and even doubled checked it today, and the derivative of the equation is:

$2/(e^{2x}-1) - 4e^{2x}x/((e^{2x}-1)^2)$

Plugging $x=0$ into that gives you $2/0 - 0/1$,

How do you get 1 in the denominator of the second term? Check your work. You will have to apply L'Hospital's Rule more than once. That happens a lot.

so it's undefined due to the $2/0$. How are you getting a different, negative answer?

When you split up the numerator and simplified the first term, you created the 2/0 problem. If you left the numerator together, you would get 0/0 and know to apply L'Hospital's Rule.

Lastly, when you state to try division of the expansions of both the numerator and the denominator, what kind of infinite series expansion would there be for $2x$ of the numerator? I thought that with a problem such as this, you'd find the expansion of the denominator, and the whole thing would be multiplied by $2x$,

That is correct if you are talking about the expansion of 1/(expansion of denominator), which requires long division of 1 by the expansion of the denominator. To do that, I would truncate the denominator expansion to a reasonably short polynomial, do the long division, and get a truncated answer of a few terms.

 

[fresh_42]

One can prove that
$$
y^{(n)}\left(e^x-1\right)=(-1)^{n+1}\cdot (n-x)+\sum_{k=0}^{n-1} \binom{n}{k}(-1)^{n-k}y^{(k)}\\
$$
with $y(x)=\dfrac{x}{e^x-1}.$ Note that $f(x)=\dfrac{2}{e^x+1}\cdot \dfrac{x}{e^x-1}=\dfrac{2}{e^x+1}\cdot y(x).$ This proves smoothness of $f(x)$ and provides all values of $y^{(n)}(0)$ recursively. The first ones are $y^{(0)}(0)=1,y^{(1)}(0)=-1/2 , y^{(2)}(0)=1/6, y^{(3)}(0)=0,y^{(4)}(0)=-1/30.$

I suppose that similar can be done with the first factor so that the solution is just a Cauchy product of the two series. Or we can use
$$
f^{(n)}(x)=2^ny^{(n)}(2x)
$$
so $f^{(n)}(0)=2^ny^{(n)}(0)$ and the series stands.

 

[FactChecker]

Another approach would be to use the geometric series, $1/(1-r)=1+r+r^2+r^3+....$
Set $r=e^{2x}$ and $-2x/(1-e^{2x})=-2x/(1-r)=-2x(1+r+r^2+r^3...)=-2x(1-e^{2x}+e^{4x}+...$.
Now truncate the Maclaurin series for the exponential for the number of terms needed: $e^{kx}=1+(kx)+(kx)^2/2!+(kx)^3/3!+...$, substitute it in with the appropriate choices of k and combine the terms with equal powers of x.
EDIT: But this is a less basic approach than some others and uses some of the basic Taylor series expansions.

 

[PeroK]

I got a valid series up to $x^4$ using the approach I recommended. I checked it using couple of values for $x$.

I suggest we don't get bogged down with the technicalities of proving a valid series exists.

 

[fresh_42]

I got a valid series up to $x^4$ using the approach I recommended. I checked it using couple of values for $x$.

I suggest we don't get bogged down with the technicalities of proving a valid series exists.

I like my recursive solution better. It provides arbitrary exactness without dealing with complicated derivatives or inversions. It would be a pity not to use the recursions that $ye^x$ generates. But, yes, that's an opinion.

 

[PeroK]

I like my recursive solution better. It provides arbitrary exactness without dealing with complicated derivatives or inversions. It would be a pity not to use the recursions that $ye^x$ generates. But, yes, that's an opinion.

It's the OP's decision!

 

[FactChecker]

Since this is a homework problem, we should probably concentrate on techniques that are being practiced in the homework. That might eliminate many of the approaches that are being discussed here. IMO, we need more feedback to go further.

 

[Ascendant0]

For $x \ne 0$:
$$g(x) = \frac{e^x -1}x = 1 + \frac x {2!} + \frac {x^2}{3!} + \dots$$So, if we define $f(x)=\frac 1{g(x)}$ we are fine.

Ok, so since this is the post you mentioned, where $g(x) = (e^x-1)/x$, the first derivative would be:

$g'(x) = -(e^x-1)/x^2 + e^x/x$

The second derivative would then be:

$g''(x) = (2(e^x-1)/x^3 -2e^x/x^2+ e^x/x$

Before I go any further, is this what you were asking, because this is how I interpreted it?

If this is right, please help clarify to me... I get that with $f(x) = 1/g(x)$, you're just flipping it. I mean that is clear. But, flipping it would give $g(x) = (e^{2x}-1)/2x$. So, are you basically taking the original equation, and substituting as well, so essentially in $g(x)$, it's a new $x_1$ where $x_1 = 2x$? I want to make sure I'm following you here, and that is how I'm viewing everything you suggested so far. Am I on the right track so far?

 

[PeroK]

The idea is to get rid of the indeterminate forms at $x=0$ by using:
$$g(x) = 1 + \frac x {2!} + \frac {x^2}{3!} + \dots$$

 

[PeroK]

The next two steps are:

Calculate the derivatives of $g$ at $x =0$.

Express the derivatives of $f(x) = \frac 1 {g(x)}$ in terms of the derivatives of $g(x)$.

 

[Ascendant0]

The idea is to get rid of the indeterminate forms at $x=0$ by using:
$$g(x) = 1 + \frac x {2!} + \frac {x^2}{3!} + \dots$$

Ok, so I'm a bit lost on this here. All these values have some value of $x^a$ in the denominator, so I know the goal is to correct that, but I'm not sure how you're suggesting to do so from that? If I plug them into the MacLaurin series, they will still have values with zeros in the denominator, so I'm unsure about the next step you're suggesting?

 

[PeroK]

First, we have:
$$f'(x) =-\frac{g'(x)}{[g(x)]^2}$$That gives us the first terms in the series for $f(x)$:
$$f(x) = f(0)+f'(0)x \dots = \frac 1 {g(0)} -\frac{g'(0)}{g(0)^2}x \dots$$You can get the term in $x^2$ by the same approach. And so on. Although the higher derivatives get more complicated.

 

[PeroK]

You could do the first two terms to start with and check it's the same as the book answer.

 

[Ascendant0]

First, we have:
$$f'(x) =-\frac{g'(x)}{[g(x)]^2}$$That gives us the first terms in the series for $f(x)$:
$$f(x) = f(0)+f'(0)x \dots = \frac 1 {g(0)} -\frac{g'(0)}{g(0)^2}x \dots$$You can get the term in $x^2$ by the same approach. And so on. Although the higher derivatives get more complicated.

Oh, ok, I'm starting to see what you're saying now. Yea, unfortunately I never would've gotten to that myself. We've never done math like that yet, but from what I'm getting from what you have there, you're basically using the chain rule for the $g(0)$, so you take the derivative of $g(0)^{-1}$ first, the outside, giving the $-g(0)^{-2}$, and then take the derivative of the inside, giving you $g'(0)$, and then the $x$ in the term simply comes from the series term $x$, then $x^2$ for the next term, etc.

Ok, I'm going to go through that. Seems like a lot of tedious algebra involved in that, but that'll be good practice for me anyway. Since that's outside of the scope of what we're covering now (the course is "Mathematical Methods in Physics", 3rd year undergrad), I already know the TA won't accept that on the homework, but I like seeing different ways of handling it like this. Plus, our professor just emailed the class yesterday letting us know he's talking to the TA, so hopefully he'll ease off on being so harsh on homeworks after.

Our methods currently in this chapter up to what we've done so far has been expanding various equations as infinite series, specific series for specific equations, like $e^x$ and $cosx$, basics like that, and then how to combine those expansions via multiplication, addition, division (including synthetic), etc. It all seemed pretty straightforward, until this specific problem, and I've talked to eight different students in group chat. Not a single one knows how to do it. The ones that thought they did got the wrong answers. Most just skipped it and gave up.

 

[fresh_42]

Not a single one knows how to do it. The ones that thought they did got the wrong answers. Most just skipped it and gave up.

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

 

[FactChecker]

To get the Maclaurin series for $\frac {e^x-1}{x}$, you can start by getting the Maclaurin series for $e^x$, which is very easy. Then simple subtraction and division gives the desired result.

 

[fresh_42]

To get the Maclaurin series for $\frac {e^x-1}{x}$, you can start by getting the Maclaurin series for $e^x$, which is very easy. Then simple subtraction and division gives the desired result.

Yes, but I want to know what comes up without division, and $f^{(n)}(0)$ is still a problem for say, $n=6.$ I have an error of $6.5 \cdot 10^{-5}$ at $x=0.5$ I'm just curious.

 

[PeroK]

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

The series I got was:
$$\frac{2x}{e^{2x}-1} = 1-x + \frac{x^2}3 - \frac{8x^4}{15} \dots$$PS I forgot to divide the last term by $4!$. That should be, as below:
$$\frac{2x}{e^{2x}-1} = 1-x + \frac{x^2}3 - \frac{x^4}{45} \dots$$

 

[fresh_42]

The series I got was:
$$\frac{2x}{e^{2x}-1} = 1-x + \frac{x^2}3 - \frac{8x^4}{15} \dots$$

Thanks. That's three percent at $x=0.5.$ Mine is
$$
f(x)=1-x+\dfrac{1}{3}x^2-\dfrac{1}{45}x^4+ \dfrac{2}{945}x^6 +O(x^7)
$$
I guess I was too long into complexity theory to like divisions. Whenever I can I use multiplications instead. The two solutions are generally interesting. Yours is more analytical, mine more algebraically.

 

[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:
$$a_n = -a_0b_n - a_1b_{n-1} \dots - a_{n-1}b_1$$

 

[Dan_Krz]

$$\lim_{x\to 0} \frac{2x}{e^{2x}-1} = \lim_{x\to 0} \frac{2}{2e^{2x}} = 1$$
$$P_0\left(x\right) = f\left(0\right) = 1$$
$$f'\left(0\right) = -1$$
$$\frac{x}{e^{x}-1} = \sum_{n=0}^{\infty} \frac{B_n}{n!} x^n$$
$$P_\infty\left(x\right) = \sum_{n=0}^{\infty} \frac{2^n B_n}{n!} x^n$$