Register to reply 
Y' = 1/(x^N+1)by Gregg
Tags: 1 or xn 
Share this thread: 
#19
Jun2110, 11:06 AM

P: 463

I understand that
where do you get [tex] (a)^{{N1}\over{N}} [/tex] from? 


#20
Jun2210, 02:42 AM

P: 65

I am sorry. I have little bit of typo in the answer.
It should be like this: [tex] \Rightarrow b_k=\frac{1}{(a)^{\frac{N1}{N}}\prod _{\underset{j\neq k}{j=0}}^{N1} \left(e^{\frac{2k \pi }{N}i}e^{\frac{2j \pi }{N}i}\right)} [/tex] You don't understand why there is a factor of [tex](a)^{\frac{N1}{N}}[/tex]. Actually it is just property of the product notation. The step you missed is this one: [tex] \prod _{\underset{j\neq k}{j=0}}^{N1} \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) [/tex] [tex] \Rightarrow \left(\sqrt[N]{A}e^{\frac{\theta }{N}i}\right)^{N1}\prod _{\underset{j\neq k}{j=0}}^{N1} \left(e^{\frac{2 n \pi }{N}i}e^{\frac{2j \pi }{N}i}\right) [/tex] When you take out a factor in the product sign, you are not just taking 1 factor out. Since in the product, there are N1 factors, so you are take N1 factors out instead. Hope you can understand. 


#21
Jun2210, 05:40 AM

P: 463




#22
Jun2310, 06:28 AM

P: 756

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. 


#23
Jun2310, 12:18 PM

Sci Advisor
HW Helper
PF Gold
P: 12,016




#24
Jun2310, 12:19 PM

P: 463




#25
Jun2310, 12:26 PM

P: 1,666

I would approach it entirely from the perspective of complex analysis:
[tex] \begin{aligned} \int_C \frac{dz}{z^N1}&=\sum_{j=0}^{N1}\int_C \frac{a_j}{zz_j}dz\\ &=\sum_{j=0}^{N1} a_j\log(zz_j)\biggr_{c_a}^{c_b} \\ &=\sum_{j=0}^{N1} a_j\big(\log(c_bz_j)\log(c_az_j)\big)\\ \end{aligned} [/tex] and then ask how must the multivalued antiderivative be interpreted so that only the endpoints of the contour [itex] (c_a, c_b)[/itex], 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: [tex]y'=\frac{1}{x^4+1},\quad y(0)=y_0[/tex] Now, how does the numeric solution to that compare with all those multivalued logarithms or hypergeometric expressions? For me, that comparison is a crucial part of doing mathematics. 


#26
Jun2310, 04:46 PM

P: 463

Is [tex] a_j = \frac{z_j}{N} [/tex] ?
Is it supposed to be obvious? 


#27
Jun2310, 06:40 PM

P: 1,666




#28
Jun2410, 03:08 AM

P: 463

[tex] \frac{1}{3 (z1)}\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^31}[/tex] and [tex]\frac{1}{4(z+1)}+\frac{1}{4(z1)}\frac{i}{4(z+i)}+\frac{i}{4(zi)}=\frac{1}{z^41}[/tex] 


#29
Jun2410, 04:51 AM

P: 756

Hello !
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. 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. 


#30
Jun2410, 02:08 PM

P: 1,666




#31
Jun2510, 01:24 PM

P: 756

Hello Gregg and Jackmell
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) 


Register to reply 