Method for finding a complex series?

[Ascendant0]

We are working with series currently, and some of the problems ask you to create a general term and write it in series summation form. Some of it is a no-brainer, but other ones, I'm at a loss as to how they expect us to get the answer without a ton of trial and error.

For example, there is one where the series is:

$x - x^5/10 +x^9/216 - x^13/9360$

So, 2/3 of this I can do no problem. Since the sign alternates and starts with a +, it will have a $(-1)^n$ and start at 0. The exponents on the top are just $x^{4n+1}$. Again, no problems figuring that out, it seemed fairly obvious.

But then that denominator... Finding a series for 1, 10, 216, and 9360, that is where I got stuck. After trying for about 20mins, I looked at the solution, and for the denominator, it gave $(2n)!(4n+1)$. I also notice that in every general term, the denominator includes a multiple of the exponential value, like this one with the $4n+1$, but I'm not sure what property is causing that, and how to look at it in an way where I could see why.

One of them even has the exponential value two times in the denominator, and there's no way I ever would've figured that out. Even with this one, if I already knew of the $4n+1$, figuring out $2!$ with that by trial and error would've been a serious challenge. I'm assuming there must be some kind of method, as well as indicator when the exponent value will also be in the denominator?

 

[anuttarasammyak]

By factorization
$$1=1$$
$$10=2*5$$
$$216=2^3*3^3=2^3*3*9=2*3*4*9$$
$$9360=2^4*3^2*5*13=2*3*4*5*6*13$$
Finding factor sequence of 1,5,x,13, x would be 9. 5 in the 4th suggests factorization is inculded. But I am not sure I can find it without your answer.

[EDIT] In the 6th line
error : factorization correct: factorial

 

[Ascendant0]

By factorization
$$1=1$$
$$10=2*5$$
$$216=2^3*3^3=2^3*3*9=2*3*4*9$$
$$9360=2^4*3^2*5*13=2*3*4*5*6*13$$
Finding factor sequence of 1,5,x,13, x would be 9. 5 in the 4th suggests factorization is inculded. But I am not sure I can find it without your answer.

Ah, ok I see now. Thank you for the help. While it still seems to be a bit of a challenge, that at least helps significantly. I appreciate it

 

[fresh_42]

I don't think these are reasonable exercises. They remind me of IQ tests when they ask for the next number in a given sequence. Finding a series for a function makes sense since it proves that the function is analytical. The other way around is guesswork with in general no unique solution. I wouldn't spend too much time on it.

However, if you decided to do so, then it would be a good idea to work with arrays, in your case
$$
\begin{array}{|c|c|c|c|c|c|}
\hline k & 0 & 1 & 2 & 3 & 4\\
\hline n & 1 & 5 & 9 & 13 & 17\\
\hline q^{-1} &1& -10 & 216 & -9360& \\
\hline
\end{array}
$$
to keep track of position $(k),$ power $(n),$ and quotients $(q)$ as $k,n$ might play a role. Starting at zero makes sense since most series expansions start at zero. Here we have $n=4k+1$ and a factor $(-1)^k.$ Factoring the quotients is a must and terms like $k!\, , \,n!\, , \,\binom n k\, , \,(n-k)!$ might occur, in your case even $(2k)!$ Those things can be done quickly and you may succeed or not. But do not waste time if not.

Your other series $f(x)=\dfrac{2x}{e^{2x}-1}=a_0+a_1x+a_2x^2+a_3x^3+\ldots$ is an example where it can be complicated up to impossible to find the coefficients. I found the coefficients by the recursion
$$
b_n=(-1)^{n}\left(1-\dfrac{1}{n+1}\sum_{k=0}^{n-1} \binom{n+1}{k}(-1)^{k}b_k\right)\, , \,a_n=\dfrac{2^nb_n}{n!}
$$
I have no idea of how to reconstruct a general term from
$$
1\, , \,-1\, , \,\dfrac{1}{3}\, , \,0\, , \,-\dfrac{1}{45}\, , \,0\, , \,\dfrac{2}{945}\, , \,0\, , \,-\dfrac{1}{4725}\, , \,\ldots
$$

 

[FactChecker]

I don't think these are reasonable exercises. They remind me of IQ tests when they ask for the next number in a given sequence. Finding a series for a function makes sense since it proves that the function is analytical. The other way around is guesswork with in general no unique solution. I wouldn't spend too much time on it.

I agree completely. There are many profound things to learn in mathematics. Identifying patterns like this does not seem like one of those things.

 

[anuttarasammyak]

Your other series f(x)=2xe2x−1=a0+a1x+a2x2+a3x3+… is an example where it can be complicated up to impossible to find the coefficients. I found the coefficients by the recursion
bn=(−1)n(1−1n+1∑k=0n−1(n+1k)(−1)kbk),an=2nbnn!
I have no idea of how to reconstruct a general term from

BTW from the sequence you show I expect f(x)+x is even function.
$$f(x)+x=x \coth x$$
confirmed.

[EDIT]

ref. https://math.stackexchange.com/questions/1109021/approximate-cothx-around-x-0

So your way of expansion has relation with Rieman zeta function. Interesting.
Someone who has strong interest on zeta fuction might be able to find rule of sequence.

 

[fresh_42]

BTW from the sequence you show I expect f(x)+x is even function. Of cource it is not. Not all the odd powers more than x^3 are zero.

Sure? I was wondering, too. WA said that $f(x)-\sum_{k=0}^6a_kx^k$ is even. Solving the recursion looks like a nightmare. I'm currently thinking of whether I should write a program for it to check a few more numbers. I did the ones above manually.

 

[anuttarasammyak]

Sory I have edited my post #6 for correction before noticing #7.

 

[FactChecker]

For example, there is one where the series is:

$x - x^5/10 +x^9/216 - x^13/9360$

So, 2/3 of this I can do no problem. Since the sign alternates and starts with a +, it will have a $(-1)^n$ and start at 0. The exponents on the top are just $x^{4n+1}$. Again, no problems figuring that out, it seemed fairly obvious.

But then that denominator... Finding a series for 1, 10, 216, and 9360, that is where I got stuck. After trying for about 20mins, I looked at the solution, and for the denominator, it gave $(2n)!(4n+1)$. I also notice that in every general term, the denominator includes a multiple of the exponential value, like this one with the $4n+1$, but I'm not sure what property is causing that, and how to look at it in an way where I could see why.

You should be careful here. Obviously, just because you see a pattern in the first few coefficients does not mean that it will continue with that one pattern. There may be other patterns that match or it might not continue in any fixed pattern.

 

[Mark44]

For example, there is one where the series is:
$x - x^5/10 +x^9/216 - x^13/9360$

I'm pretty sure the exponent on the last term is meant to be 13.
Corrected, the above would be $x - x^5/10 +x^9/216 - x^{13}/9360$

In LaTeX, if the exponent, subscript, superscript, etc. is more than one character, put braces around the characters.

 

[Ascendant0]

Thanks for all the feedback, as well as additional strategies.

And I agree, with how much we have to learn, I don't see it productive to have us do something so tedious. That's why I figured maybe there was some method that made it much easier, but I just wasn't aware of it yet. It's hard for me to catch this professor during his office hours, so I greatly appreciate the help here