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-   -   y' = 1/(x^N+1) (http://www.physicsforums.com/showthread.php?t=411218)

 Gregg Jun19-10 10:47 PM

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

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

 arildno Jun20-10 05:10 AM

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

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 Jun20-10 07:16 AM

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

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 Jun20-10 12:29 PM

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

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 Jun20-10 12:33 PM

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

Quote:
 Quote by jostpuur (Post 2768821) 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 Jun20-10 06:08 PM

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

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 Jun20-10 06:17 PM

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

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

 arildno Jun20-10 06:20 PM

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

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

 Gregg Jun20-10 07:06 PM

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

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

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 Jun21-10 12:58 AM

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

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 question in Voofie for step by step details.

 arildno Jun21-10 02:24 AM

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

Quote:
 Quote by ross_tang (Post 2769508) 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 question in Voofie 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 Jun21-10 07:12 AM

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

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 Jun21-10 07:34 AM

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

Quote:
 Quote by Gregg (Post 2769685) 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 Jun21-10 08:07 AM

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

Quote:
 Quote by ross_tang (Post 2769508) 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 question in Voofie 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 Jun21-10 08:14 AM

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

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 bn

 ross_tang Jun21-10 08:47 AM

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

@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 Jun21-10 09:22 AM

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

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 Jun21-10 10:09 AM

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

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:

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

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