# Differentiate n times

• Nov 6th 2011, 04:23 PM
alexmahone
Differentiate n times
\$\displaystyle (1-x^2)v''+2(n-1)xv'+2nv=0\$

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

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

My attempt:

Differentiating each side of the equation once,

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

I don't seem to be getting anywhere.
• Nov 6th 2011, 06:11 PM
MonroeYoder
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
• Nov 6th 2011, 06:16 PM
alexmahone
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.