- #1
Gregg
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How would I solve [tex] y'(x) = \frac{1}{x^N+1} [/tex] ?
Probably.jostpuur said:I see that when [itex]N[/itex] 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?
ross_tang said:Actually, this integral can be computer if N is integer, without using hypergeometric function.
[tex]\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)[/tex]
and
[tex]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)}[/tex]
Please refer to this http://www.voofie.com/content/82/how-to-find-the-integral-of-1xna/" for step by step details.
Gregg said:If N is an integer isn't this even simpler?
[tex] y=\int \frac{1}{x^N+1} dx [/tex]
[tex] y=\int (1-x^N+x^{2N}-x^{3N}+O(x^{4N})) dx [/tex]
[tex] y=x-\frac{x^{N+1}}{N+1}+\frac{x^{2N+1}}{2N+1}-\frac{x^{3N+1}}{3N+1}+O(x^{4N+1}) [/tex]
[tex] y=\displaystyle \sum_{k=0}^{\infty} \frac{x^{kN+1}}{kN+1} [/tex]
ross_tang said:Actually, this integral can be computer if N is integer, without using hypergeometric function.
[tex]\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)[/tex]
and
[tex]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)}[/tex]
Please refer to this http://www.voofie.com/content/82/how-to-find-the-integral-of-1xna/" for step by step details.
ross_tang said: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.
JJacquelin said: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.
JJacquelin said: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.
Gregg said:Is [tex] a_j = \frac{z_j}{N} [/tex] ?
Is it supposed to be obvious?
jackmell said: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.
I suggest you read post 9 by Gregg.
May be apparently !A point was made about a finite series being more desireable though.
Gregg said: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]
Is that true for the general case? If so, sorry if I said it wasn't.
The equation represents a first-order ordinary differential equation, where the derivative of the unknown function y with respect to the independent variable x is equal to the inverse of x raised to the N+1 power.
The purpose of solving this equation is to find the general solution for the unknown function y, which can then be used to model and understand various physical phenomena.
The value of N can be any real number, as long as it is not equal to -1. This is because the denominator of the equation cannot be equal to zero.
This equation can be solved using various methods, such as separation of variables, integrating factors, or substitution. The specific method used will depend on the form of the equation and the skills of the solver.
This equation can be used to model various physical phenomena, such as population growth, radioactive decay, and chemical reactions. It can also be applied in engineering and economics to analyze systems and make predictions.