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evinda
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Hello! (Wave)
$$(1-x^2)y''-2xy'+p(p+1)y=0, p \in \mathbb{R} \text{ constant } \\ -1 < x<1$$
At the interval $(-1,1)$ the above differential equation can be written equivalently
$$y''+p(x)y'+q(x)y=0, -1<x<1 \text{ where } \\p(x)=\frac{-2x}{1-x^2} \\ q(x)= \frac{p(p+1)}{1-x^2}$$
$p,q$ can be written as power series $\sum_{n=0}^{\infty} p_n x^n, \sum_{n=0}^{\infty} q_n x^n$ respectively with center $0$ and $\sum_{n=0}^{\infty} p_n x^n=p(x)$ and $\sum_{n=0}^{\infty} q_nx^n=q(x), \ \forall -1<x<1$
$$p(x)= \sum_{n=0}^{\infty} (-2) x^{2n+1}, -1<x<1$$
$$q(x)= \sum_{n=0}^{\infty} p(p+1) x^{2n}, \forall -1<x<1$$Since $p,q$ can be written as power series with center $0$ and radius of convergence $1$, it's logical to look for a solution of the differential equation of the form$$y(x)=\sum_{n=0}^{\infty} a_n x^n \text{ with radius of convergence } R>0$$
$$-2xy'(x)= \sum_{n=1}^{\infty} -2n a_n x^n$$
$$y''(x)= \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n \\ -x^2y''(x)=\sum_{n=2}^{\infty} -n(n-1)a_nx^n$$We have:
$$\sum_{n=0}^{\infty} \left[ (n+2)(n+1) a_{n+2}-n(n-1)a_n-2na_n+p(p+1)a_n\right]x^n=0, \forall x \in (-R,R)$$It has to hold: $(n+2)(n+1)a_{n+2}-n(n-1)a_n-2na_n+p(p+1)a_n=0, \forall n=0,1,2, \dots$
Thus: $$a_{n+2}=-\frac{(p-n)(p+n+1)}{(n+1)(n+2)}a_n, \forall n=0,1,2, \dots$$So the solution can be written in the following form, right?
$$y(x)= \sum_{k=0}^{\infty} a_{2k} x^{2k}+ \sum_{k=0}^{\infty} a_{2k+1} x^{2k+1}$$
where :
$$a_{2k}= \frac{\prod_{j=0}^{2k-1} (j+(-1)^{j+1} p)}{(2k)!}a_0$$
and
$$a_{2k+1}= \frac{\prod_{j=1}^{2k} (j+(-1)^j p)}{(2k+1)!}a_1$$
If it is like that, applying the ratio test we would get:For $n=2k$:
$$\left| \frac{\frac{\prod_{j=0}^{2k+1} (j+(-1)^{j+1} p) a_0 x^{2k+2}}{(2k+2)!}}{\frac{\prod_{j=0}^{2k-1} (j+(-1)^{j+1} p) a_0 x^{2k}}{(2k)!}}\right| = \left| \frac{(2k-p)(2k+1+p) x^2}{(2k+1)(2k+2)} \right| \to |x^2|<1$$So the series $\sum_{k=0}^{\infty} a_{2k} x^{2k}$ converges for all $x$ such that $-1<x<1$.
Similarly, we show that the series $\sum_{k=0}^{\infty} a_{2k+1} x^{2k+1}$ converges for $-1<x<1$.
Is it right so far? (Thinking)
According to my lecture notes, if $p \in \mathbb{R} \setminus{\mathbb{Z}}$ then the power series at the right of $a_0,a_1$ have radius of convergence $1$ and so they define functions $y_1(x), y_2(x)$, that are infinitely many times differentiable at $(-1,1)$ .
If $p$ is a positive integer, then one of the series terminates and becomes polynomial.
For example if $p=5$, then $a_7, a_9, a_{11}, \dots, a_{2n+1}=0$ and so $\sum_{k=0}^{\infty} a_{2k+1} x^{2k+1}= a_1 x+ a_3 x^3+ a_5 x^5$, right?So does this mean that in such a case the following does not hold?
The power series at the right of $a_0,a_1$ have radius of convergence $1$ and so they define functions $y_1(x), y_2(x)$, that are infinitely many times differentiable at $(-1,1)$?
$$(1-x^2)y''-2xy'+p(p+1)y=0, p \in \mathbb{R} \text{ constant } \\ -1 < x<1$$
At the interval $(-1,1)$ the above differential equation can be written equivalently
$$y''+p(x)y'+q(x)y=0, -1<x<1 \text{ where } \\p(x)=\frac{-2x}{1-x^2} \\ q(x)= \frac{p(p+1)}{1-x^2}$$
$p,q$ can be written as power series $\sum_{n=0}^{\infty} p_n x^n, \sum_{n=0}^{\infty} q_n x^n$ respectively with center $0$ and $\sum_{n=0}^{\infty} p_n x^n=p(x)$ and $\sum_{n=0}^{\infty} q_nx^n=q(x), \ \forall -1<x<1$
$$p(x)= \sum_{n=0}^{\infty} (-2) x^{2n+1}, -1<x<1$$
$$q(x)= \sum_{n=0}^{\infty} p(p+1) x^{2n}, \forall -1<x<1$$Since $p,q$ can be written as power series with center $0$ and radius of convergence $1$, it's logical to look for a solution of the differential equation of the form$$y(x)=\sum_{n=0}^{\infty} a_n x^n \text{ with radius of convergence } R>0$$
$$-2xy'(x)= \sum_{n=1}^{\infty} -2n a_n x^n$$
$$y''(x)= \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n \\ -x^2y''(x)=\sum_{n=2}^{\infty} -n(n-1)a_nx^n$$We have:
$$\sum_{n=0}^{\infty} \left[ (n+2)(n+1) a_{n+2}-n(n-1)a_n-2na_n+p(p+1)a_n\right]x^n=0, \forall x \in (-R,R)$$It has to hold: $(n+2)(n+1)a_{n+2}-n(n-1)a_n-2na_n+p(p+1)a_n=0, \forall n=0,1,2, \dots$
Thus: $$a_{n+2}=-\frac{(p-n)(p+n+1)}{(n+1)(n+2)}a_n, \forall n=0,1,2, \dots$$So the solution can be written in the following form, right?
$$y(x)= \sum_{k=0}^{\infty} a_{2k} x^{2k}+ \sum_{k=0}^{\infty} a_{2k+1} x^{2k+1}$$
where :
$$a_{2k}= \frac{\prod_{j=0}^{2k-1} (j+(-1)^{j+1} p)}{(2k)!}a_0$$
and
$$a_{2k+1}= \frac{\prod_{j=1}^{2k} (j+(-1)^j p)}{(2k+1)!}a_1$$
If it is like that, applying the ratio test we would get:For $n=2k$:
$$\left| \frac{\frac{\prod_{j=0}^{2k+1} (j+(-1)^{j+1} p) a_0 x^{2k+2}}{(2k+2)!}}{\frac{\prod_{j=0}^{2k-1} (j+(-1)^{j+1} p) a_0 x^{2k}}{(2k)!}}\right| = \left| \frac{(2k-p)(2k+1+p) x^2}{(2k+1)(2k+2)} \right| \to |x^2|<1$$So the series $\sum_{k=0}^{\infty} a_{2k} x^{2k}$ converges for all $x$ such that $-1<x<1$.
Similarly, we show that the series $\sum_{k=0}^{\infty} a_{2k+1} x^{2k+1}$ converges for $-1<x<1$.
Is it right so far? (Thinking)
According to my lecture notes, if $p \in \mathbb{R} \setminus{\mathbb{Z}}$ then the power series at the right of $a_0,a_1$ have radius of convergence $1$ and so they define functions $y_1(x), y_2(x)$, that are infinitely many times differentiable at $(-1,1)$ .
If $p$ is a positive integer, then one of the series terminates and becomes polynomial.
For example if $p=5$, then $a_7, a_9, a_{11}, \dots, a_{2n+1}=0$ and so $\sum_{k=0}^{\infty} a_{2k+1} x^{2k+1}= a_1 x+ a_3 x^3+ a_5 x^5$, right?So does this mean that in such a case the following does not hold?
The power series at the right of $a_0,a_1$ have radius of convergence $1$ and so they define functions $y_1(x), y_2(x)$, that are infinitely many times differentiable at $(-1,1)$?