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


by Gregg
Tags: 1 or xn
Gregg
Gregg is offline
#19
Jun21-10, 11:06 AM
P: 463
I understand that

where do you get [tex] (-a)^{{N-1}\over{N}} [/tex] from?
ross_tang
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#20
Jun22-10, 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{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)}
[/tex]
You don't understand why there is a factor of [tex](-a)^{\frac{N-1}{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}}^{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)
[/tex]
[tex]
\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)
[/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 N-1 factors, so you are take N-1 factors out instead. Hope you can understand.
Gregg
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#21
Jun22-10, 05:40 AM
P: 463
Quote Quote by ross_tang View Post
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{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)}
[/tex]
You don't understand why there is a factor of [tex](-a)^{\frac{N-1}{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}}^{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)
[/tex]
[tex]
\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)
[/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 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
JJacquelin
JJacquelin is offline
#22
Jun23-10, 06:28 AM
P: 746
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.
arildno
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#23
Jun23-10, 12:18 PM
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Quote Quote by JJacquelin View Post
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.
Gregg
Gregg is offline
#24
Jun23-10, 12:19 PM
P: 463
Quote Quote by JJacquelin View Post
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.
jackmell
jackmell is offline
#25
Jun23-10, 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^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}
[/tex]

and then ask how must the multi-valued antiderivative be interpreted so that only the end-points 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 multi-valued logarithms or hypergeometric expressions? For me, that comparison is a crucial part of doing mathematics.
Gregg
Gregg is offline
#26
Jun23-10, 04:46 PM
P: 463
Is [tex] a_j = \frac{z_j}{N} [/tex] ?

Is it supposed to be obvious?
jackmell
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#27
Jun23-10, 06:40 PM
P: 1,666
Quote Quote by Gregg View Post
Is [tex] a_j = \frac{z_j}{N} [/tex] ?

Is it supposed to be obvious?
No. It's not that. My [itex]a_j[/itex] is just the coefficients of the partial fraction decomposition. Ross in post 10 showed a compact way of computing these coefficients which he called [itex]b_n[/itex] which looks like he's using the Residue Theorem to compute: the coefficients are just the residues for each zero (pole). And my [itex]z_j[/itex] are the roots of [itex]x^N+1[/itex]. Sorry I didn't make that more clear above.
Gregg
Gregg is offline
#28
Jun24-10, 03:08 AM
P: 463
Quote Quote by jackmell View Post
No. It's not that. My [itex]a_j[/itex] is just the coefficients of the partial fraction decomposition. Ross in post 10 showed a compact way of computing these coefficients which he called [itex]b_n[/itex] which looks like he's using the Residue Theorem to compute: the coefficients are just the residues for each zero (pole). And my [itex]z_j[/itex] are the roots of [itex]x^N+1[/itex]. Sorry I didn't make that more clear above.
Oh right I just thought that it was since

[tex] \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}[/tex]

and

[tex]-\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}[/tex]
JJacquelin
JJacquelin is offline
#29
Jun24-10, 04:51 AM
P: 746
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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jackmell
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#30
Jun24-10, 02:08 PM
P: 1,666
Quote Quote by Gregg View Post
Oh right I just thought that it was since

[tex] \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}[/tex]

and

[tex]-\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}[/tex]
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.
JJacquelin
JJacquelin is offline
#31
Jun25-10, 01:24 PM
P: 746
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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