Differential equation with frobenius method

[JohanL]

Im trying to solve

y''(x) + xy'(x) + y = 0

For the first solution i get with frobenius method

y_1(x) = sum_{p=0}^\infty \frac{a_0(-1)^px^(2p+1)}{(2p+1)!} <br />

Im not sure if its correct but i think so.

This must be the taylor series for some function.
Can someone help me with which function or where i can find it?

 

[JohanL]

why did the latex become so strange? I couldn't preview it.

 

[arildno]

Assuming your calculations are correct, the first solution is just the power series given by your expression.

 

[HallsofIvy]

Strictly speaking, this is not "Frobenious' method", just a power series solutions. Frobenious' method applies when the point about which you are expanding is a regular singular point.
I take it what you have is
\Sigma \frac{a_0(-1)^p x^{2p+1}}{(2p+1)!}
Exactly what do you mean by (2p+1)! ?

 

[topsquark]

Strictly speaking, this is not "Frobenious' method", just a power series solutions. Frobenious' method applies when the point about which you are expanding is a regular singular point.
I take it what you have is
\Sigma \frac{a_0(-1)^p x^{2p+1}}{(2p+1)!}
Exactly what do you mean by (2p+1)! ?

(2p+1)! is the "odd factorial function" defined as 1*3*...*(2p+1).

Similarly, (2p)! is 2*4*...*(2p)=(2^p)p!

I have occasionally found it convenient to extend this concept to such products as "(3p+2)!" = 2*5*8*...*(3p+2) though I don't know if this is standard.

-Dan

 

[JohanL]

Assuming your calculations are correct, the first solution is just the power series given by your expression.

yes, but now I am looking for what function that power series is the taylor series for.

 

[JohanL]

Strictly speaking, this is not "Frobenious' method", just a power series solutions. Frobenious' method applies when the point about which you are expanding is a regular singular point.
I take it what you have is
\Sigma \frac{a_0(-1)^p x^{2p+1}}{(2p+1)!}
Exactly what do you mean by (2p+1)! ?

1*3*...*(2p-1)*(2p+1) = (2p+1)!

Ok i thought Frobenius method was when exanding about any ordinary or regular singular point.
So strictly speaking you can't solve that equation with Frobenius method because it don't have any regular singular points?
But the more important thing for me is for what function that is a taylor series.

 

[arildno]

yes, but now I am looking for what function that power series is the taylor series for.

A function can perfectly well be uniquely described as the solution of some differential equation, and that there doesn't exist any other representations of them.
Some functions, as yours, also allow a (deducible) power series representation , but it doesn't follow from this that there exist any other name for the function. If you like, you could call it Johann(x).

 

[HallsofIvy]

(2p+1)! is the "odd factorial function" defined as 1*3*...*(2p+1).

Similarly, (2p)! is 2*4*...*(2p)=(2^p)p!

I have occasionally found it convenient to extend this concept to such products as "(3p+2)!" = 2*5*8*...*(3p+2) though I don't know if this is standard.

-Dan

I have never seen any good reason to do that. (2n)!= 2*4*...(2p)= 2^pp! as you said and (2n+1)!= 1*3*5*...*(2n+1)= 1*2*3*4*...*2n*(2n+1)/(2*4*...*(2n))= (2n+1)!/(2ⁿn!)

 

[JohanL]

A function can perfectly well be uniquely described as the solution of some differential equation, and that there doesn't exist any other representations of them.
Some functions, as yours, also allow a (deducible) power series representation , but it doesn't follow from this that there exist any other name for the function. If you like, you could call it Johann(x).

Yes that's true.
I was wondering if it was possible to write that power series in closed form (not sure if that's the correct english word for it). ln [(1+x)/(1-x)] or something. If that is possible it simplifies other calculations for me.
There must be some program where its possible to do things like that?
This series looked so simple so i thought someone could give me a hint or tell me the answer right away.

 

[HallsofIvy]

Generally speaking, even linear equations with variable coefficients have solutions that cannot be written in terms of elementary functions.

 

[JohanL]

ok, but if the solution can be written in terms of elementary functions how do you get it in this form.
Im pretty sure this one can be written in terms of elementary functions.

 

[HallsofIvy]

I'm pretty sure it can't!

As I said before, I would have written this as
\Sigma \frac{a_0(-1)^p 2^p p! x^{2p+1}}{(2p+1)!}

That's close to being sin(x) but that p! in the numerator makes it quite different.

 

[JohanL]

ok. thank you.
it looked like it could be written with elementary functions :) to me

You can't use mathematica or something to check that in an easy way?

 

[JohanL]

You could write the series in closed form.

y1 = exp(-x^2/2)*sqrt(pi/2)*Erfi[x/sqrt(2)]

where Erfi is the imaginary error function, Erf[iz]/i

and then the second solution becomes

y2 = -exp(-x^2/2)

You only had to write the series in mathematica and you got it in closed form.

 

[arildno]

Well, but to most, the imaginary error function doesn't count as an ELEMENTARY function.
You would also get your solution in "closed" form if you defined your power series as Johann(x) or arildno(2x+47)

 

[JohanL]

Well, but to most, the imaginary error function doesn't count as an ELEMENTARY function.
You would also get your solution in "closed" form if you defined your power series as Johann(x) or arildno(2x+47)

It was Halsofivy who introduced elementary functions in this thread.
I wrote
"I was wondering if it was possible to write that power series in closed form. If that is possible it simplifies other calculations for me.
There must be some program where its possible to do things like that?
"
And the closed form i got from mathematica simplified the calculations a lot. I am not sure how much Johann(x) would simplify the calculations. For example, the derivative of the error function can be written in terms of elementary functions, 2exp[-x^2]/sqrt(pi). Of course the derivative of Johann(x) then also can be written in terms of elementary functions but to get this you still have to work with the series solution so Johann(x) doesn't simplify anything in that case.
But all this doesn't matter anymore :)
Thanks for your posts.