Frobenius method aplied to simple harmonic oscillator problem

[noblegas]

Homework Statement

P4-1. The Method of Frobenius: Sines and Cosines. The solutions to the differential
equation
y"+ y = 0
can be expressed in terms of our familiar sine and cosine: y(x) = Acos(x) + Bsin(x) .
Use the Method of Frobenius to solve the above differential equation for the even
solution (cosine) and odd solution (sine). By doing so, express the cosine and sine each
as power series.

Homework Equations

The Attempt at a Solution

I think the first thing I would have to do is convert my trigonometric function to a power series function. sin(x)=x-x^3/3!+x^5/5!+... , cos(x)=1-x^2/2!+x^4/4! +... and y(x) now equals : y(x)=x-x^3/3!+x^5/5!+1-x^2/2!+x^4/4! +... and now I can apply the Method of frobenius? y''+y=A(-1)^n/(2n!)(x^2n+(2n(2n-1)*x^2n-1)+B(-1)^n/(2n+1!)*(x^2n+1 +(2n)(2n+1)x^2n-1) where sin(x)=(-1)^n*x^2n/(2n)! , and cos(x)=(-1)^n*x^2n/(2n+1)!. My expression does not look like it would be equal to zero.

 

[Yitzach]

You are working the problem backwards. You want to assume a power series expansion is a solution to the problem.
$$y=\Sigma_{i=0}^\infty a_nx^n$$
You can then take the derivative of that twice so you can find y". You can then plug the series into the equation and solve for $$a_n$$ and you will end up with something of the form $$a_{n+2}=C_na_n$$
Each possible set of coefficients will be either the sine series or the cosine series coefficients.
Wikipedia has an article on the method.

 

[noblegas]

You are working the problem backwards. You want to assume a power series expansion is a solution to the problem.
$$y=\Sigma_{i=0}^\infty a_nx^n$$
You can then take the derivative of that twice so you can find y". You can then plug the series into the equation and solve for $$a_n$$ and you will end up with something of the form $$a_{n+2}=C_na_n$$
Each possible set of coefficients will be either the sine series or the cosine series coefficients.
Wikipedia has an article on the method.

I thought I did that! I converted my trigonomentric functions to functions of the Maclarin series, which are power series functions

 

[Yitzach]

I thought I did that! I converted my trigonomentric functions to functions of the Maclarin series, which are power series functions

You want to derive the MacLaurin series, not use them to reach the answer.
a_(2n) will be a_n of the cosine MacLaurin series. a_(2n+1) will be a_n of the sine Maclarin series.

By working it backward you would end up proving that the results of the method work. http://en.wikipedia.org/wiki/Frobenius_method has an example that if you use the steps in the example you will come to the answer.

Need to do:
1. Assume $$y=\Sigma^\infty_{i=0}a_nx^n$$ is the solution.
2. Differentiate twice with respect to x the series assumed in 1.
3. PnP (Plug and Play/chug) the answers into the equation.
4. Make indices line up, collect like terms behind the Sigma and assemble in ascending order the terms in front of the Sigma there.
$$a_0+a_1x+\Sigma^\infty_{i=2}a_nx^n$$
a_0 and a_1 are the starting values of the odd and even elements given initial conditions.
5. Separate a_n and a_(n+1) from a_(n+2) to yield a recursive formula for the coefficients. I think there should not be an a_(n+1) in this case. I'm a little fog on this step but a good ordinary differential equations book (which I have) should be able to fill in the gaps.
6. Go through a few iterations and find a pattern. The pattern should be the MacLaurin series for sine with even n and cosine for odd n.

Are trying to do:
1. Take MacLaurin series for sin and cosine.
2. Differentiate twice with respect to x.
3. PnP into y"+y=0.
4. Rearrange terms so it all lines up.
5. Discover that all terms come to naught. QED

But it does look like you are doing 1-3 correctly. I can't tell for certain because it is written on one line.
When you get done with four of what you are trying to do, you should have this or something similar:
$$\Sigma^\infty_{i=0}\frac{(-1)^{n}x^{2n}}{(2n)!}-\frac{(-1)^{n}x^{2n}}{(2n)!}+\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}-\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$$

Can you identify that what you are doing is different from what you need to do? Do you need help manipulating the series to make the indices line up or differentiating the power series? Do you need a better explanation of step 5 of what needs to be done?

 

[paranormal]

Your attempt at a solution is on the right track, but there are a few errors. Here is a corrected version:

To apply the Method of Frobenius, we first need to convert the trigonometric functions to power series. We can do this by using the Maclaurin series expansions for sine and cosine:

sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...

cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + ...

Next, we can substitute these power series into the given differential equation:

y'' + y = 0

(x - x^3/3! + x^5/5! - x^7/7! + ...)'' + (x - x^3/3! + x^5/5! - x^7/7! + ...) = 0

Expanding the derivatives and simplifying, we get:

(1 + x^2/2! + x^4/4! + ...) + (x - x^3/3! + x^5/5! - x^7/7! + ...) = 0

We can now collect like terms and group them by the power of x:

1 + (x - x)/1! + (x^2 - x^2)/2! + (x^3 - x^3)/3! + ... = 0

Since all the terms cancel out, we are left with 1 = 0, which is not a valid solution. This means that the even solution (cosine) does not exist for this differential equation.

For the odd solution (sine), we can follow the same steps:

y'' + y = 0

(x - x^3/3! + x^5/5! - x^7/7! + ...)'' + (x - x^3/3! + x^5/5! - x^7/7! + ...) = 0

Expanding the derivatives and simplifying, we get:

(1 + x^2/2! + x^4/4! + ...) + (x - x^3/3! + x^5/5! - x^7/7! + ...) = 0

Collecting like terms and grouping by the