# Derivative Problem

• Nov 4th 2010, 09:07 AM
TheCoffeeMachine
Derivative Problem
Challenge problem: Suppose [LaTeX ERROR: Compile failed] , and denote [LaTeX ERROR: Compile failed] by [LaTeX ERROR: Compile failed] .

For what [LaTeX ERROR: Compile failed] is it true that [LaTeX ERROR: Compile failed] ?

Moderator approved CB
• Nov 4th 2010, 02:08 PM
Opalg
Spoiler:
By laboriously differentiating y twice, you can check that $(1-x^2)y^{(2)} - xy^{(1)} + y^{(0)} + 2x = 0$. Differentiate that to get $(1-x^2)y^{(3)} - 3xy^{(2)} + 0y^{(1)} + 2 = 0$. One more differentiation shows that $(1-x^2)y^{(4)} - 5xy^{(3)} - 3y^{(0)} = 0$. 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 $n\geqslant2$.
• Nov 5th 2010, 05:52 AM
TheCoffeeMachine
Quote:

Originally Posted by Opalg
Spoiler:
By laboriously differentiating y twice, you can check that $(1-x^2)y^{(2)} - xy^{(1)} + y^{(0)} + 2x = 0$. Differentiate that to get $(1-x^2)y^{(3)} - 3xy^{(2)} + 0y^{(1)} + 2 = 0$. One more differentiation shows that $(1-x^2)y^{(4)} - 5xy^{(3)} - 3y^{(0)} = 0$. 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 $n\geqslant2$.

Very nice - I didn't think of induction.
Spoiler:
After we have the relation $(1-x^2)y^{(2)} - xy^{(1)} + y^{(0)} + 2x = 0$

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

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

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

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

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

Thus it holds for $n\ge 2.$