Solving the Equation y'(x) = 1/(x^N+1)

[Gregg]

How would I solve y'(x) = \frac{1}{x^N+1} ?

 

[arildno]

Note that the polynomials x^{n}+1 can be factorized in terms of the nth-roots of (-1), i.e, as:
x_{j,n}=e^{i\pi\frac{1+2{j}}{n}}, i=\sqrt{(-1)}, j=0,...n-1

You can then factorize your polynomial denominator, use fractional decomposition, and perform termwise integration.
Be particularly aware of the pitfalls involved in complex logarithms.

 

[arildno]

Let us take the case with n=4.
Then, the roots are:
x_{0,4}=e^{i\frac{\pi}{4}}=\frac{1}{\sqrt{2}}(1+i)
x_{1,4}=e^{i\frac{3\pi}{4}}=\frac{1}{\sqrt{2}}(-1+i)
x_{2,4}=e^{i\frac{5\pi}{4}}=\frac{1}{\sqrt{2}}(-1-i)
x_{3,4}=e^{i\frac{7\pi}{4}}=\frac{1}{\sqrt{2}}(1-i)

We may factorize x^4+1 into 4 complex-valued first-order polynomials:
\frac{1}{x^{4}+1}=\frac{1}{(x-\frac{1}{\sqrt{2}}(1+i))(x-\frac{1}{\sqrt{2}}(1-i))(x-\frac{1}{\sqrt{2}}(-1+i))(x-\frac{1}{\sqrt{2}}(-1-i))}

If you wish to work with real valued polynomials, multiply together the complex conjugates, and get:
\frac{1}{x^{4}+1}=\frac{1}{(x^{2}-\sqrt{2}x+1)(x^{2}+\sqrt{2}x+1)}

A partial fractions decompositions would then proceed as follows:
\frac{1}{x^{4}+1}=\frac{Ax+B}{(x^{2}-\sqrt{2}x+1)}+\frac{Cx+D}{(x^{2}+\sqrt{2}x+1)}

You may then determine what A,B, C and D must be by multplying the whole equation with x^4+1, getting, by rearrangement:
0=(A+C)x^{3}+(\sqrt{2}{A}-\sqrt{2}C+B+D)x^{2}+(A+C+B-D)x+(B+D-1)
The coefficients to each power of x must be 0, yielding:

B=D=1/2, C=-A=\frac{1}{2\sqrt{2}}

In general, your anti-derivatives will be sums of logaritms and arctan-functions.

 

[jostpuur]

I see that when N is chosen and fixed, the task can be completed with finite amount of effort. But is there more general results? Is there theory about these coefficients that you encounter in fractional decomposition?

 

[arildno]

I see that when N is chosen and fixed, the task can be completed with finite amount of effort. But is there more general results? Is there theory about these coefficients that you encounter in fractional decomposition?

Probably.

I don't know of it, though I'm sure a generalized result, in terms, perhaps, of a complex finite sum as a function of N has been made by somebody

 

[Gregg]

I could see the cases N=2,3 etc. but Mathematica gives


\int \frac{dx}{x^N+1} = x\left({}_2 F_1 \left(\frac{1}{N},1,1+\frac{1}{N},-x^N\right)\right)

\text{Hypergeometric2F1}(a,b,c,z) = \, _2F_1(a,b;c;z)

 

[arildno]

And that is the closed-form solution, with a truly nasty function called Hypergeometric2F1 in its expression..

 

[arildno]

Definitions of your 2F1 is found as expressions (16) and (17) at wolfram mathworld:
http://mathworld.wolfram.com/HypergeometricFunction.html

 

[Gregg]

\, _2F_1(a,b;c;z) = 1+\frac{a b}{1!c}z + \frac{a(a+1)b(b+1)}{2!c(c+1)}z^2 +\cdots

So I think it looks something like this

\, _2F_1(a,b;c;z) =\displaystyle 1+\sum _{k=0}^{\infty } \frac{\prod _{j=0}^{(k)} \frac{(a+j)(b+j)}{(c+j)}}{(k+1)!}z^k


\int \frac{1}{x^N+1} \, dx = x\left(\displaystyle 1+\sum _{k=0}^{\infty } \frac{\prod _{j=0}^k \frac{(\frac{1}{N}+j)(1+j)}{(1+\frac{1}{N}+j)}}{(k+1)!}(-x^N)^{(k+1)}\right)

 

[ross_tang]

Actually, this integral can be computer if N is integer, without using hypergeometric function.

\int \frac{dx}{x^N+ a}=\sum _{k=0}^{N-1} b_k\ln \left(x-(-a)^{\frac{1}{N}}e^{\frac{2k \pi }{N}i}\right)

and

b_n=\frac{1}{(-a)^{\frac{N-1}{N}}\prod _{\underset{j\neq k}{j=0}}^{N-1} \left(e^{\frac{2n \pi }{N}i}-e^{\frac{2j \pi }{N}i}\right)}

Please refer to this http://www.voofie.com/content/82/how-to-find-the-integral-of-1xna/" for step by step details.

 

[arildno]

Actually, this integral can be computer if N is integer, without using hypergeometric function.

\int \frac{dx}{x^N+ a}=\sum _{k=0}^{N-1} b_k\ln \left(x-(-a)^{\frac{1}{N}}e^{\frac{2k \pi }{N}i}\right)

and

b_n=\frac{1}{(-a)^{\frac{N-1}{N}}\prod _{\underset{j\neq k}{j=0}}^{N-1} \left(e^{\frac{2n \pi }{N}i}-e^{\frac{2j \pi }{N}i}\right)}

Please refer to this http://www.voofie.com/content/82/how-to-find-the-integral-of-1xna/" for step by step details.

Thank you, ross tang.
I mentioned, in post 2, that by using factorization by means of first order complex polynomials, you'd get a sum of complex logarithms as your answer.

It is nice to see the closed form solution in this case as well.

 

[Gregg]

If N is an integer isn't this even simpler?

y=\int \frac{1}{x^N+1} dx


y=\int (1-x^N+x^{2N}-x^{3N}+O(x^{4N})) dx


y=x-\frac{x^{N+1}}{N+1}+\frac{x^{2N+1}}{2N+1}-\frac{x^{3N+1}}{3N+1}+O(x^{4N+1})

y=\displaystyle \sum_{k=0}^{\infty} \frac{x^{kN+1}}{kN+1}

 

[arildno]

If N is an integer isn't this even simpler?

y=\int \frac{1}{x^N+1} dx


y=\int (1-x^N+x^{2N}-x^{3N}+O(x^{4N})) dx


y=x-\frac{x^{N+1}}{N+1}+\frac{x^{2N+1}}{2N+1}-\frac{x^{3N+1}}{3N+1}+O(x^{4N+1})

y=\displaystyle \sum_{k=0}^{\infty} \frac{x^{kN+1}}{kN+1}

Is an infinite series simpler??

Furthermore, the anti-derivative you make there implies that |x|<1

 

[Gregg]

Actually, this integral can be computer if N is integer, without using hypergeometric function.

\int \frac{dx}{x^N+ a}=\sum _{k=0}^{N-1} b_k\ln \left(x-(-a)^{\frac{1}{N}}e^{\frac{2k \pi }{N}i}\right)

and

b_n=\frac{1}{(-a)^{\frac{N-1}{N}}\prod _{\underset{j\neq k}{j=0}}^{N-1} \left(e^{\frac{2n \pi }{N}i}-e^{\frac{2j \pi }{N}i}\right)}

Please refer to this http://www.voofie.com/content/82/how-to-find-the-integral-of-1xna/" for step by step details.

How do you go from

\sum _{k=0}^{N-1} b_k\prod _{j=0j\neq k}^{N-1} \left(A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2n \pi )}-A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2j \pi )}\right)=1

to

b_n=\frac{1}{(-a)^{\frac{N-1}{N}}\prod _{j=0j\neq k}^{N-1} \left(e^{\frac{i}{N}(\theta +2n \pi )}-e^{\frac{i}{N}(\theta +2j \pi )}\right)}

 

[Gregg]

Ah you actually go from

\sum _{k=0}^{N-1} b_k\prod _{j=0j\neq k}^{N-1} \left(A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2n \pi )}-A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2j \pi )}\right)=1


\Rightarrow b_n=\frac{1}{(-a)^{\frac{N-1}{N}}\prod _{j=0j\neq k}^{N-1} \left(e^{\frac{i}{N}(2n \pi )}-e^{\frac{i}{N}(2j \pi )}\right)}


I can see

\sum _{k=0}^{N-1} b_k=\frac{1}{(-a)^{\frac{1}{N}}\prod _{j=0j\neq k}^{N-1} \left(e^{\frac{i}{N}(2n \pi )}-e^{\frac{i}{N}(2j \pi )}\right)}


But I can't see how you get b_n

 

[ross_tang]

@Gregg,

Have you try putting x = \sqrt[N]{A}e^{\frac{\theta +2n \pi }{N}i}?

If you do that, you can see for every other term in the sum, i.e. when k \neq n

b_k\prod _{\underset{j\neq k}{j=0}}^{N-1} \left(\sqrt[N]{A}e^{\frac{\theta +2n \pi }{N}i}-\sqrt[N]{A}e^{\frac{\theta +2j \pi }{N}i}\right) = 0

Since there is a factor of

\sqrt[N]{A}e^{\frac{\theta +2n \pi }{N}i}-\sqrt[N]{A}e^{\frac{\theta +2n\pi }{N}i}

when j = n.

Only when k = n, the term j = n is gone, since the sum excluded the factor.

 

[Gregg]

Can you write it as this?

\sum _{k=0}^{N-1} b_k=\frac{A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2n \pi )}-A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2k \pi )}}{\prod _{j=0}^{N-1} \left(A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2n \pi )}-A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2j \pi )}\right)}

So every time that k \ne n it is possible for j=n and therefore a factor of zero in the product. With k=n since j \ne k = n there is no zero factor.

b_n=\frac{1}{\prod _{j=0k\neq j}^{N-1} \left(A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2n \pi )}-A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2j \pi )}\right)}


Then i get

b_n=\frac{1}{ A^{\frac{1}{N}}e^{i \frac{\theta }{N}}\prod _{j=0k\neq j}^{N-1} \left(e^{\frac{(2n i \pi )}{N}}-e^{\frac{(2j i \pi )}{N}}\right)}

b_n=\frac{1}{(-a)^{\frac{1}{N}}\prod _{j=0k\neq j}^{N-1} \left(e^{\frac{(2n i \pi )}{N}}-e^{\frac{(2j i \pi )}{N}}\right)}

which is different so what have I missed?

 

[ross_tang]

I am sorry. I think my notation confused you.

I wrote this as the equation:

\sum _{k=0}^{N-1} b_k\prod _{j=0j\neq k}^{N-1} \left(A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2n \pi )}-A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2j \pi )}\right)=1

But in fact, i really means this:

\sum _{k=0}^{N-1} \left( b_k\prod _{j=0j\neq k}^{N-1} \left(A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2n \pi )}-A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2j \pi )}\right)\right)=1

 

[Gregg]

I understand that

where do you get (-a)^{{N-1}\over{N}} from?

 

[ross_tang]

I am sorry. I have little bit of typo in the answer.
It should be like this:
<br /> \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)}<br />
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:
<br /> \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)<br />
<br /> \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)<br />
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]

I am sorry. I have little bit of typo in the answer.
It should be like this:
<br /> \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)}<br />
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:
<br /> \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)<br />
<br /> \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)<br />
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]

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]

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]

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]

I would approach it entirely from the perspective of complex analysis:

<br /> \begin{aligned}<br /> \int_C \frac{dz}{z^N-1}&amp;=\sum_{j=0}^{N-1}\int_C \frac{a_j}{z-z_j}dz\\<br /> &amp;=\sum_{j=0}^{N-1} a_j\log(z-z_j)\biggr|_{c_a}^{c_b} \\<br /> &amp;=\sum_{j=0}^{N-1} a_j\big(\log(c_b-z_j)-\log(c_a-z_j)\big)\\<br /> \end{aligned}<br />

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&#039;=\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.

 

[Gregg]

Is a_j = \frac{z_j}{N} ?

Is it supposed to be obvious?

 

[jackmell]

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.

 

[Gregg]

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}

 

[JJacquelin]

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 development 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.

 

[jackmell]

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.

 

[JJacquelin]

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 doesn't 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)