While finding a second solution with wronskian for bessel

[mertymertcan]

Hello guys, I'm new here. i was working on a mathematical methods in physics book and there is a part that i don't understand. so i want to ask if anyone knows... while finding a second solution for bessel diff. eq.(for m=0) the book used wronskian method. in the method there is J^2 bin the denominator. so it becomes J^-2 and the series is made such that

{1 - (x^2)/4 + (x^4)/64 - (x^6)/2304 + ... }^-2 = {1 + (x^2)/2 + (x^6)/32 + ...}

the part that i didnt understand how can someone conver minus 2 power into right hand side:S. please help me, is there a general formula for this?

 

[jvicens]

Hello and welcome to the forum! It's great to have you here. I can understand your confusion about converting the -2 power into the right-hand side of the equation. This is a common technique used in solving differential equations, and it's called the Frobenius method.

The general formula for converting a negative power into the right-hand side is given by the Frobenius method:

y(x) = x^r ∑n=0 an(x-x0)^(n+r)

Where r is the negative power and x0 is the point where the series is expanded. In your case, r = -2 and x0 = 0.

To better understand this, let's take a closer look at your equation:

{1 - (x^2)/4 + (x^4)/64 - (x^6)/2304 + ... }^-2 = {1 + (x^2)/2 + (x^6)/32 + ...}

We can rewrite the left-hand side as:

{1 - (x^2)/4 + (x^4)/64 - (x^6)/2304 + ... }^-2 = (1 - (x^2)/4 + (x^4)/64 - (x^6)/2304 + ...)^-2

Now, using the Frobenius method, we can expand the right-hand side as:

(1 - (x^2)/4 + (x^4)/64 - (x^6)/2304 + ...)^-2 = (1 + (x^2)/2 + (x^6)/32 + ...)^-2

= (1 + (x^2)/2 + (x^6)/32 + ...)^2

= 1 + (2x^2)/2 + (2x^6)/32 + ...

= 1 + x^2 + (x^6)/16 + ...

= 1 + (x^2)/2 + (x^6)/32 + ...

As you can see, the negative power has been converted into the right-hand side using the Frobenius method. I hope this explanation helps you understand the process better. If you have any further questions, please don't hesitate to ask. Good luck with your book!