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

1. Jun 19, 2010

### Gregg

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

2. Jun 20, 2010

### arildno

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.

Last edited: Jun 20, 2010
3. Jun 20, 2010

### arildno

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.

Last edited: Jun 20, 2010
4. Jun 20, 2010

### jostpuur

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?

5. Jun 20, 2010

### arildno

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

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

6. Jun 20, 2010

### Gregg

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

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

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

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

7. Jun 20, 2010

### arildno

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

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

8. Jun 20, 2010

### arildno

9. Jun 20, 2010

### Gregg

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

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

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

Last edited: Jun 20, 2010
10. Jun 21, 2010

### ross_tang

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 http://www.voofie.com/content/82/how-to-find-the-integral-of-1xna/" [Broken] for step by step details.

Last edited by a moderator: May 4, 2017
11. Jun 21, 2010

### arildno

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

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.

Last edited by a moderator: May 4, 2017
12. Jun 21, 2010

### Gregg

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

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

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

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

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

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

13. Jun 21, 2010

### arildno

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

Is an infinite series simpler??

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

14. Jun 21, 2010

### Gregg

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

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

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

Last edited by a moderator: May 4, 2017
15. Jun 21, 2010

### Gregg

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

Last edited: Jun 21, 2010
16. Jun 21, 2010

### ross_tang

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.

17. Jun 21, 2010

### Gregg

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?

Last edited: Jun 21, 2010
18. Jun 21, 2010

### ross_tang

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:

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

19. Jun 21, 2010

### Gregg

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

I understand that

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

20. Jun 22, 2010

### ross_tang

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

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.

21. Jun 22, 2010

### Gregg

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

Ahh that is much clearer now! I didn't think about that product

22. Jun 23, 2010

### JJacquelin

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

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. Jun 23, 2010

### arildno

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

I suggest you read post 9 by Gregg.

24. Jun 23, 2010

### Gregg

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

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.

25. Jun 23, 2010

### jackmell

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

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.

Last edited: Jun 23, 2010