# y' = 1/(x^N+1)

by Gregg
Tags: 1 or xn
 P: 463 I understand that where do you get $$(-a)^{{N-1}\over{N}}$$ from?
 P: 65 I am sorry. I have little bit of typo in the answer. It should be like this: $$\Rightarrow b_k=\frac{1}{(-a)^{\frac{N-1}{N}}\prod _{\underset{j\neq k}{j=0}}^{N-1} \left(e^{\frac{2k \pi }{N}i}-e^{\frac{2j \pi }{N}i}\right)}$$ You don't understand why there is a factor of $$(-a)^{\frac{N-1}{N}}$$. Actually it is just property of the product notation. The step you missed is this one: $$\prod _{\underset{j\neq k}{j=0}}^{N-1} \left(\sqrt[N]{A}e^{\frac{\theta }{N}i}\left(e^{\frac{2 n \pi }{N}i}-e^{\frac{2j \pi }{N}i}\right)\right)$$ $$\Rightarrow \left(\sqrt[N]{A}e^{\frac{\theta }{N}i}\right)^{N-1}\prod _{\underset{j\neq k}{j=0}}^{N-1} \left(e^{\frac{2 n \pi }{N}i}-e^{\frac{2j \pi }{N}i}\right)$$ When you take out a factor in the product sign, you are not just taking 1 factor out. Since in the product, there are N-1 factors, so you are take N-1 factors out instead. Hope you can understand.
P: 463
 Quote by ross_tang I am sorry. I have little bit of typo in the answer. It should be like this: $$\Rightarrow b_k=\frac{1}{(-a)^{\frac{N-1}{N}}\prod _{\underset{j\neq k}{j=0}}^{N-1} \left(e^{\frac{2k \pi }{N}i}-e^{\frac{2j \pi }{N}i}\right)}$$ You don't understand why there is a factor of $$(-a)^{\frac{N-1}{N}}$$. Actually it is just property of the product notation. The step you missed is this one: $$\prod _{\underset{j\neq k}{j=0}}^{N-1} \left(\sqrt[N]{A}e^{\frac{\theta }{N}i}\left(e^{\frac{2 n \pi }{N}i}-e^{\frac{2j \pi }{N}i}\right)\right)$$ $$\Rightarrow \left(\sqrt[N]{A}e^{\frac{\theta }{N}i}\right)^{N-1}\prod _{\underset{j\neq k}{j=0}}^{N-1} \left(e^{\frac{2 n \pi }{N}i}-e^{\frac{2j \pi }{N}i}\right)$$ When you take out a factor in the product sign, you are not just taking 1 factor out. Since in the product, there are N-1 factors, so you are take N-1 factors out instead. Hope you can understand.
Ahh that is much clearer now! I didn't think about that product
 P: 714 Hello ! in order to find a primitive of 1/((x^N)+1), let t=x^N and z = y/x where z(x) is the new function to find. This leads to z as an Euler's hypergeometric integral with variable t. Finally, back to y and x, we obtain the solution : y = x*F(a,b;c;X) + constant where F is the Gauss hypergeometric function (usually noted 2F1 in the hypergeometric functions classification) a = 1 b = 1/N c =1+(1/N) X = -x^N Using the general series definition of the hypergeometric function, it is easy to express y(x) in terms of a rather simple infinite series.
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P: 11,935
 Quote by JJacquelin Hello ! in order to find a primitive of 1/((x^N)+1), let t=x^N and z = y/x where z(x) is the new function to find. This leads to z as an Euler's hypergeometric integral with variable t. Finally, back to y and x, we obtain the solution : y = x*F(a,b;c;X) + constant where F is the Gauss hypergeometric function (usually noted 2F1 in the hypergeometric functions classification) a = 1 b = 1/N c =1+(1/N) X = -x^N Using the general series definition of the hypergeometric function, it is easy to express y(x) in terms of a rather simple infinite series.
I suggest you read post 9 by Gregg.
P: 463
 Quote by JJacquelin Hello ! in order to find a primitive of 1/((x^N)+1), let t=x^N and z = y/x where z(x) is the new function to find. This leads to z as an Euler's hypergeometric integral with variable t. Finally, back to y and x, we obtain the solution : y = x*F(a,b;c;X) + constant where F is the Gauss hypergeometric function (usually noted 2F1 in the hypergeometric functions classification) a = 1 b = 1/N c =1+(1/N) X = -x^N Using the general series definition of the hypergeometric function, it is easy to express y(x) in terms of a rather simple infinite series.
Earlier on in the thread we found the same 2F1 solution as you did an infinite series representation aswell. A point was made about a finite series being more desireable though.
 P: 1,651 I would approach it entirely from the perspective of complex analysis: \begin{aligned} \int_C \frac{dz}{z^N-1}&=\sum_{j=0}^{N-1}\int_C \frac{a_j}{z-z_j}dz\\ &=\sum_{j=0}^{N-1} a_j\log(z-z_j)\biggr|_{c_a}^{c_b} \\ &=\sum_{j=0}^{N-1} a_j\big(\log(c_b-z_j)-\log(c_a-z_j)\big)\\ \end{aligned} and then ask how must the multi-valued antiderivative be interpreted so that only the end-points of the contour $(c_a, c_b)$, can be used in the expression above for all reasonable contours even ones which loop around multiple times. :) Also, if it were mine, I'd check it with something real: $$y'=\frac{1}{x^4+1},\quad y(0)=y_0$$ Now, how does the numeric solution to that compare with all those multi-valued logarithms or hypergeometric expressions? For me, that comparison is a crucial part of doing mathematics.
 P: 463 Is $$a_j = \frac{z_j}{N}$$ ? Is it supposed to be obvious?
P: 1,651
 Quote by Gregg Is $$a_j = \frac{z_j}{N}$$ ? Is it supposed to be obvious?
No. It's not that. My $a_j$ is just the coefficients of the partial fraction decomposition. Ross in post 10 showed a compact way of computing these coefficients which he called $b_n$ which looks like he's using the Residue Theorem to compute: the coefficients are just the residues for each zero (pole). And my $z_j$ are the roots of $x^N+1$. Sorry I didn't make that more clear above.
P: 463
 Quote by jackmell No. It's not that. My $a_j$ is just the coefficients of the partial fraction decomposition. Ross in post 10 showed a compact way of computing these coefficients which he called $b_n$ which looks like he's using the Residue Theorem to compute: the coefficients are just the residues for each zero (pole). And my $z_j$ are the roots of $x^N+1$. Sorry I didn't make that more clear above.
Oh right I just thought that it was since

$$\frac{1}{3 (z-1)}-\frac{(-1)^{1/3}}{3 \left(z+(-1)^{1/3}\right)}+\frac{(-1)^{2/3}}{3 \left(z-(-1)^{2/3}\right)} = \frac{1}{z^3-1}$$

and

$$-\frac{1}{4(z+1)}+\frac{1}{4(z-1)}-\frac{i}{4(z+i)}+\frac{i}{4(z-i)}=\frac{1}{z^4-1}$$
P: 714
Hello !

 I suggest you read post 9 by Gregg.
All right, but of course don't directly develop in series the general hypergeometric F(a,b;c,X) function.
First, simplify it, since parameters are particular (a=1, c=1+b). It reduces into particular hypergeometric functions of lower level : Beta incomplete (in the complex range) or Lerch function.
The series developement of the Lerch function directly leads to very simple series.
(i.e. : joint page)
Of course, everybody is aware that these series can be obtained on a very simple way by developing in series 1/(1+x^N) or (x^-N)/(1+x^-N) before integration.
 A point was made about a finite series being more desireable though.
May be apparently !
But if the finite series contains functions like log, or polylog, or etc., each of these functions is an infinite series, so, the computation would be a finite sum of infinite series. Why only one special function like hypergeometric (or better, the Lerch function in the present case) be less desirable ?
Well, this is my viewpoint. But I understand that many people prefer using more common functions than only one special function.
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P: 1,651
 Quote by Gregg Oh right I just thought that it was since $$\frac{1}{3 (z-1)}-\frac{(-1)^{1/3}}{3 \left(z+(-1)^{1/3}\right)}+\frac{(-1)^{2/3}}{3 \left(z-(-1)^{2/3}\right)} = \frac{1}{z^3-1}$$ and $$-\frac{1}{4(z+1)}+\frac{1}{4(z-1)}-\frac{i}{4(z+i)}+\frac{i}{4(z-i)}=\frac{1}{z^4-1}$$
Ok then. That's interesting. I didn't know that. Perhaps it is as you indicated. Is that true for the general case? If so, sorry if I said it wasn't.
P: 714
Hello Gregg and Jackmell

 Is that true for the general case? If so, sorry if I said it wasn't.
Yes that's true for the general case. But with + instead of - at denominator, the general term of the series will not be real. On theoretical viewpoint, it does'nt matter : after integration the general term will be a complex logarithm. On the other hand, further simplifications will be laborious.
In order to remain in the real range, it's better to start with a similar series, but with a quadric at denominator. The integation isn't too difficult and leads to a general term including real logarithm and arctangent. (i.e.: attachement)
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