Re: Differentiate n times

actually you're doing just fine =). After differentiating once we have n=1. So the $\displaystyle (1-x^2)v^{(3)}$ corresponds to the $\displaystyle (1-x^2)v^{n+2}$

Then $\displaystyle -2xv''$ corresponds to $\displaystyle -2xv^{n+1}$

Now the part you prolly didnt realize, since $\displaystyle n=1$, we have $\displaystyle 2(n-1)(xv''+v')=2(1-1)(xv''+v')=0$

And now $\displaystyle 2nv' = 2v'$ which corresponds to $\displaystyle n(n+1)v^{(n)} = 1(1+1)v^{(1)}=2v'$

So all of your terms follow the pattern perfectly... now just keep differentiating

Re: Differentiate n times

Quote:

Originally Posted by

**MonroeYoder** actually you're doing just fine =). After differentiating once we have n=1. So the $\displaystyle (1-x^2)v^{(3)}$ corresponds to the $\displaystyle (1-x^2)v^{n+2}$

Then $\displaystyle -2xv''$ corresponds to $\displaystyle -2xv^{n+1}$

Now the part you prolly didnt realize, since $\displaystyle n=1$, we have $\displaystyle 2(n-1)(xv''+v')=2(1-1)(xv''+v')=0$

And now $\displaystyle 2nv' = 2v'$ which corresponds to $\displaystyle n(n+1)v^{(n)} = 1(1+1)v^{(1)}=2v'$

So all of your terms follow the pattern perfectly... now just keep differentiating

Differentiating each side of the equation once,

$\displaystyle (1-x^2)v^{(3)}-2xv''+2(n-1)(xv''+v')+2nv'=0$

$\displaystyle (1-x^2)v^{(3)}+2(n-2)xv''+2(2n-1)v'=0$

Differentiating again,

$\displaystyle (1-x^2)v^{(4)}-2xv^{(3)}+2(n-2)(xv^{(3)}+v'')+2(2n-1)v''=0$

$\displaystyle (1-x^2)v^{(4)}+2(n-3)xv^{(3)}+6(n-1)v''=0$

$\displaystyle (1-x^2)v^{(4)}+2(n-3)xv^{(3)}+3(2n-2)v''=0$

Differentiating again,

$\displaystyle (1-x^2)v^{(5)}-2xv^{(4)}+2(n-3)(xv^{(4)}+v^{(3)})+6(n-1)v^{(3)}=0$

$\displaystyle (1-x^2)v^{(5)}+2(n-4)xv^{(4)}+(8n-12)v^{(3)}=0$

$\displaystyle (1-x^2)v^{(5)}+2(n-4)xv^{(4)}+4(2n-3)v^{(3)}=0$

**Claim**: After $\displaystyle m$ differentiations, the equation becomes

$\displaystyle (1-x^2)v^{(m+2)}+2(n-m-1)xv^{(m+1)}+(m+1)(2n-m)v^{(m)}=0$

Substituting $\displaystyle m=n$, we get the required result.