2 sol'ns to Airy Equation y" - xy = 0: one is an almost-factorial power series; the other is similar

[mario99]

[math]y'' - xy = 0[/math]
I found two solutions to Airy's equation using Power Series.

One solution

[math]y_1(x) = a_0\sum_{n=0}^{\infty}\frac{1}{...9 \cdot 8 \cdot 6 \cdot 5 \cdot 3 \cdot 2} \ x^{3n}[/math]
As you can see in the denominator it is almost a factorial. I can't find a suitable formula for that. I also don't know how to find the value of [math]a_0[/math] to use this sum to approximate the values of Airy's function to some decimal places.

The second solution is similar with a little modification.

 

[Steven G]

Can't you just factor out the denominator? Can you define the denominator a little better?

 

[khansaheb]

Did you derive the solution or did you find the solution somewhere else?

Please refer to

http://www.sosmath.com/diffeq/series/series04/series04.html

for complete solution.

 

[blamocur]

You have only one unknown constant [imath]a_0[/imath], but a typical second order equation should have two.

 

[blamocur]

It would be easier for us to help you if you posted how you got to your formula, which, I believe, is incorrect, or at least incomplete.

 

[mario99]

Thank you everyone for helping me.

Can't you just factor out the denominator? Can you define the denominator a little better?

I saw the link sent by khansaheb. It suggests the factor of [math](2\cdot 3)(5\cdot 6)......(3k - 1)(3k)[/math]

Did you derive the solution or did you find the solution somewhere else?

Please refer to

http://www.sosmath.com/diffeq/series/series04/series04.html

for complete solution.

I did derive it by myself. It looks exactly the same as the steps in the link, except the factoring part.

You have only one unknown constant [imath]a_0[/imath], but a typical second order equation should have two.

It has two as the second unknown constant (a_1) is belong to the second solution.

It would be easier for us to help you if you posted how you got to your formula, which, I believe, is incorrect, or at least incomplete.

It is the same derivation as the link sent by khansaheb, except for the factorization. You can say that it is incomplete as I should not write 2*3*5*6*8*9....but instead I should represent it with a factorization containing the index n.




I came up with a brilliant idea but it needs some modification to work correctly. I have already mentioned that the denominator is almost a factorial. What about making it factorial and writing the missing numbers in the numerator. Something like this

[math]\frac{...7 \cdot 4 \cdot 1}{(3n)!}[/math]