Question

Prove the Rodrigues formula following the steps below: Let v = x^(n)exp(−x). First show
that xdv/dx = (n − x)v. Then derive that
xd^(n+2)v/dx^(n+2)+ (1 + x)d^(n+1)v/dx(n+1)+ (n + 1)d^(n)v/dx^n= 0 . (4)
Next show that y(x) =exp(x)/n! d^(n)v/dx^(n) is a solution of Laguerre’s differential equation xy"(x)+(1-x)y'(x)+ny=0
using (4)

Answer 1

Let v = x^(n)exp(−x).

First show that x dv/dx = (n − x)v
\[v = x^n~e^{−x}\]
\[\frac{dv}{dx} = -x^{n}~e^{−x}\]
\[x\frac{dv}{dx} = -x^{n+1}~e^{−x}\]
\[x\frac{dv}{dx} = (-x)v\]

maybe i aint reading this right

Answer 2

forgot to drop the n

Answer 3

i have managed the first part i just not sure about the show y(x) is a solution using (4)

Answer 4

well, when i catch up to you, lets see what i can make of it :)

\[v = x^n~e^{−x}\]
\[v' = nx^{n-1}~e^{−x}-x^n~e^{−x}\]
\[v' = (nx^{n-1}-x^n)~e^{−x}\]
\[v' = (nx^{-1}-1)~x^ne^{−x}\]
\[xv' = (n-x)v\]

Answer 5

ok thanks for the help :)... its bugging me aha

Answer 6

Then derive that
\[x \frac{d^{n+2}v}{dx^{n+2}}+ (1 + x)\frac{d^{n+1}v}{dx^{n+1}}+ (n + 1)\frac{d^{n}v}{dx^n} = 0\]
\[x v^{(n+2)}+ (1 + x)v^{(n+1)} + (n + 1)v^{(n)} = 0\]
.................................

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

murmur ....

Answer 7

\[y =\frac{e^x}{n!} v^{(n)}\]
\[y^{(1)} =\frac{e^x}{n!} (v^{(n)}+v^{(n+1)})\]
\[y^{(2)} =\frac{e^x}{n!} (v^{(n)}+2v^{(n+1)}+v^{(n+2)})\]

replace the ys and thier derivatives with these and see if it goes to zero

Answer 8

assuming of course that i derived those correctly ;)

Answer 9

ok so the question is essentially saying plug the given y in to laguerres DE?

Answer 10

yes, and you should be able to do some simplifications with the v parts from the first portion of it

Answer 11

ok let me try this

Answer 12

\[xy"+(1-x)y'+ny=0\]
\[x(\frac{e^x}{n!} (v^{(n)}+2v^{(n+1)}+v^{(n+2)}))\]\[~~+(1-x)(\frac{e^x}{n!} (v^{(n)}+v^{(n+1)}))\]\[~~~~+n(\frac{e^x}{n!} v^{(n)})=0\]
\[\frac{e^x}{n!}\]that factors out and never equals zero so its redundant, leaving us with

\[x((v^{(n)}+2v^{(n+1)}+v^{(n+2)}))\]\[~~+(1-x)( (v^{(n)}+v^{(n+1)}))\]\[~~~~+n( v^{(n)})=0\]

\[xv^{(n)}+2xv^{(n+1)}+xv^{(n+2)}\]\[~~+(1-x)v^{(n)}+(1-x)v^{(n+1)}\]\[~~~~+n v^{(n)}=0\]

Answer 13

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

Answer 14

Which equals (4) :) you are a lifesaver amistre64. God it was such a basic thing haha. Thanks for the help though much appreciated!!