1. ## Derivative Problem

Challenge problem: Suppose $\displaystyle \displaystyle y = \sqrt{1-x^2}\arcsin{x}$, and denote $\displaystyle \displaystyle \frac{d^ky}{dx^k}$ by $\displaystyle \displaystyle y_{k}$.

For what $\displaystyle \displaystyle n$ is it true that $\displaystyle \displaystyle (1-x^2)y_{n+2}-(2n+1)xy_{n+1}-(n^2-1)y_{n} = 0$?

Moderator approved CB

2. Spoiler:
By laboriously differentiating y twice, you can check that $\displaystyle (1-x^2)y^{(2)} - xy^{(1)} + y^{(0)} + 2x = 0$. Differentiate that to get $\displaystyle (1-x^2)y^{(3)} - 3xy^{(2)} + 0y^{(1)} + 2 = 0$. One more differentiation shows that $\displaystyle (1-x^2)y^{(4)} - 5xy^{(3)} - 3y^{(0)} = 0$. But that is the given relation for n=2.

Next, if you differentiate the given relation, assuming it is true for n, you find that it also holds for n+1. Therefore by induction it holds for all $\displaystyle n\geqslant2$.

3. Originally Posted by Opalg
Spoiler:
By laboriously differentiating y twice, you can check that $\displaystyle (1-x^2)y^{(2)} - xy^{(1)} + y^{(0)} + 2x = 0$. Differentiate that to get $\displaystyle (1-x^2)y^{(3)} - 3xy^{(2)} + 0y^{(1)} + 2 = 0$. One more differentiation shows that $\displaystyle (1-x^2)y^{(4)} - 5xy^{(3)} - 3y^{(0)} = 0$. But that is the given relation for n=2.

Next, if you differentiate the given relation, assuming it is true for n, you find that it also holds for n+1. Therefore by induction it holds for all $\displaystyle n\geqslant2$.
Very nice - I didn't think of induction.
Spoiler:
After we have the relation $\displaystyle (1-x^2)y^{(2)} - xy^{(1)} + y^{(0)} + 2x = 0$

We can write $\displaystyle (1-x^2)y^{(2)} - xy^{(1)} + y^{(0)} = -2x$

Applying Leibniz's rule $\displaystyle (f \cdot g)^{(n)}=\sum_{k=0}^n {n \choose k} f^{(k)} g^{(n-k)}$ to the LHS gives:

$\displaystyle (1-x^2)y^{n+2}-2nxy^{n+1}-xy^{n+1}-n(n-1)y^{n}-ny^n+y^{n}$

$\displaystyle = (1-x^2)y^{n+2}-(2n+1)xy^{n+1}-(n^2-1)y^{n}$

For the RHS, if $\displaystyle u = -2x$, then $\displaystyle u_{1} = -2$ and $\displaystyle u_{n} = 0$, whenever $\displaystyle n\ge 2$.

Thus it holds for $\displaystyle n\ge 2.$