Differentiate n times

**alexmahone** · November 6th 2011, 04:23 PM

$(1-x^2)v''+2(n-1)xv'+2nv=0$

Differentiate each side of the equation $n$ times in succession to obtain

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

My attempt:

Differentiating each side of the equation once,

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

I don't seem to be getting anywhere.

**MonroeYoder** · November 6th 2011, 06:11 PM

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

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

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

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

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

**alexmahone** · November 6th 2011, 06:16 PM

Originally Posted by MonroeYoder

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

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

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

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

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

Differentiating each side of the equation once,

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

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

Differentiating again,

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

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

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

Differentiating again,

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

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

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

Claim: After $m$ differentiations, the equation becomes

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

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

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