How to Find Power Series Solutions of Chebyshev's Equation?

[Ted123]

Homework Statement

Chebyshev's Equation is $$(1-x^2) y^{\prime\prime} - xy^{\prime} + c^2 y =0$$
where c is a real constant.

(a) Find 2 linearly independent power series solutions of Chebyshev's Equation at x=0: an even one and an odd one.

(b) Hence, using the ratio test, find the radius of convergence for both of these series.

The Attempt at a Solution

I've manipulated the power series down to:

$$\sum_{n=0}^{\infty} [(n+2)(n+1)a_{n+2} - n(n-1)a_n - na_n + c^2 a_n ] x^n = 0$$

which gives the recurrence relation:

$$a_{n+2} = \frac{n^2 - c^2}{(n+2)(n+1)}a_n$$

For the even solution:

We have arbitrary $$a_0$$
$$a_2 = -\frac{1}{2} c^2 a_0$$
$$a_4 = \frac{4-c}{12} a_2 = - \frac{c^2(4-c^2)}{24} a_0$$
$$a_6 = \frac{24-c^2}{30} a_4 = - \frac{c^2(4-c^2)(24-c^2)}{720} a_0$$

I'm struggling to solve this for a_n ...

For the odd solution:

We have arbitrary $$a_1$$
$$a_3 = \frac{1-c^2}{6} a_1$$
$$a_5 = \frac{9-c^2}{20} a_3 = \frac{(9-c^2)(1-c^2)}{120}$$

I'm also struggling to solve this for a_n ...

 

[Ted123]

I've just re-read the question and it says find the 2 linearly independent power series solutions at x=0.

So does this mean start from $$y'' + c^2y = 0\;?$$

In which case:

$$a_{n+2} = -\frac{c^2}{(n+2)(n+1)}a_n$$

EVEN:

$$a_0$$ arbitrary

$$a_2 = -\frac{c^2a_0}{2}$$

$$a_4 = -\frac{c^2a_2}{(4)(3)} = \frac{c^4a_0}{(4)(3)(2)}$$

$$a_6 = -\frac{c^2a_4}{(6)(5)} = -\frac{c^6a_0}{(6)(5)(4)(3)(2)}$$

ODD:

$$a_1$$ arbitrary

$$a_3 = -\frac{c^2a_1}{(3)(2)}$$

$$a_5 = -\frac{c^2a_3}{(5)(4)} = \frac{c^4a_1}{(5)(4)(3)(2)}$$

$$a_7 = -\frac{c^2a_5}{(7)(6)} = -\frac{c^6a_1}{(7)(6)(5)(4)(3)(2)}$$


So... it seems:

$$\displaystyle a_{2n} = (-1)^{n} \frac{c^{2n} a_0}{(2n)!}$$

$$\displaystyle a_{2n+1} = (-1)^{n} \frac{c^{2n}a_1}{(2n+1)!}$$

So it looks like the 2 linearly independent power series solutions (even and odd respectively) are?:

$$\displaystyle a_0 \sum_{n=0}^{\infty} (-1)^{n} \frac{c^{2n}}{(2n)!} x^{2n}$$

$$\displaystyle a_1 \sum_{n=0}^{\infty} (-1)^{n} \frac{c^{2n}}{(2n+1)!} x^{2n+1}$$

 

[vela]

Your first interpretation of the problem is correct. The series will have terms of the form a_n(x-x₀)ⁿ. The problem is telling you x₀=0.

$$a_{n+2} = \frac{n^2 - c^2}{(n+2)(n+1)}a_n$$

For the even solution:

We have arbitrary $$a_0$$
$$a_2 = -\frac{1}{2} c^2 a_0$$
$$a_4 = \frac{4-c}{12} a_2 = - \frac{c^2(4-c^2)}{24} a_0$$
$$a_6 = \frac{24-c^2}{30} a_4 = - \frac{c^2(4-c^2)(24-c^2)}{720} a_0$$

I'm struggling to solve this for a_n ...

You made one minor error: for a₆, you should have (16-c²) instead of (24-c²) in the numerator.

I think the best you'll be able to do is say

$$a_{2n} = -\frac{c^2(4-c^2)\cdots(4(n-1)^2-c^2)}{(2n)!}$$

For the odd solution:

We have arbitrary $$a_1$$
$$a_3 = \frac{1-c^2}{6} a_1$$
$$a_5 = \frac{9-c^2}{20} a_3 = \frac{(9-c^2)(1-c^2)}{120}$$

I'm also struggling to solve this for a_n ...