Differential equation with frobenius method

In summary, the first solution is just the power series given by your expression. The second solution is the power series for the function sin(x).
  • #1
JohanL
158
0
Im trying to solve

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

For the first solution i get with frobenius method

[tex]y_1(x) = sum_{p=0}^\infty \frac{a_0(-1)^px^(2p+1)}{(2p+1)!}
[/tex]

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?
 
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  • #2
why did the latex become so strange? I couldn't preview it.
 
  • #3
Assuming your calculations are correct, the first solution is just the power series given by your expression.
 
  • #4
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
[tex]\Sigma \frac{a_0(-1)^p x^{2p+1}}{(2p+1)!}[/tex]
Exactly what do you mean by (2p+1)! ?
 
  • #5
HallsofIvy said:
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
[tex]\Sigma \frac{a_0(-1)^p x^{2p+1}}{(2p+1)!}[/tex]
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
 
  • #6
arildno said:
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.
 
  • #7
HallsofIvy said:
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
[tex]\Sigma \frac{a_0(-1)^p x^{2p+1}}{(2p+1)!}[/tex]
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.
 
  • #8
JohanL said:
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).
 
  • #9
topsquark said:
(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)= 2pp! as you said and (2n+1)!= 1*3*5*...*(2n+1)= 1*2*3*4*...*2n*(2n+1)/(2*4*...*(2n))= (2n+1)!/(2nn!)
 
  • #10
arildno said:
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.
 
  • #11
Generally speaking, even linear equations with variable coefficients have solutions that cannot be written in terms of elementary functions.
 
  • #12
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.
 
  • #13
I'm pretty sure it can't!

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

That's close to being sin(x) but that p! in the numerator makes it quite different.
 
  • #14
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?
 
  • #15
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.
 
  • #16
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)
 
  • #17
arildno said:
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.
 

1. What is the Frobenius method?

The Frobenius method is a technique for solving ordinary differential equations that cannot be solved by standard methods. It involves assuming a solution in the form of a power series and then using a recursion relation to determine the coefficients of the series.

2. When is the Frobenius method used?

The Frobenius method is typically used when the differential equation has a singular point, such as at the origin, infinity, or a branch point. It is also useful for equations with variable coefficients or higher-order equations.

3. What are the steps involved in solving a differential equation using the Frobenius method?

The steps for solving a differential equation with the Frobenius method are: 1) Assume a solution in the form of a power series, 2) Substitute the series into the equation and equate coefficients of like powers, 3) Use a recursion relation to determine the coefficients, 4) Determine the radius of convergence of the series, and 5) Use the known coefficients to construct the general solution.

4. What is a recursion relation?

A recursion relation is an equation that relates the coefficients of a power series solution. It is used in the Frobenius method to determine the coefficients of the series by expressing each coefficient in terms of the previous ones.

5. What are some advantages and disadvantages of using the Frobenius method?

Some advantages of using the Frobenius method include its ability to solve a wide range of differential equations, its simplicity compared to other methods, and its use of power series solutions which can be easily manipulated. However, some disadvantages include the need for a singular point in the equation, the potential for the series to converge slowly or not at all, and the possibility of obtaining complex solutions.

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