Just couldnt figure this out = (2)

[toni]

I just don't understand why it goes like this...why 'therefore'?

i know (x^2)f''(x)+xf'(x)+((x^2)-1)f(x)) = x(xf''(x)+f'(x)+xf(x)) - f(x), but why it goes to zero when taking (d^n/dx^n)?

 

[HallsofIvy]

"It is given that xf"+ f'+ xf= 0"

and then it concluded
\frac{d^n}{dx^n}(x^2f"+ xf '+ (x^2-1)f)= 0
for any positive integer n and for x in (-r, r).

There is clearly something missing. Where did that "r" come from? Was there something more before this?

 

[toni]

Sorry, i shouldve made things clearer.
Here is everything before this. r is the radius of convergence.

 

[toni]

Hmmm..it's actually on a solution sheet. I stuck at that point when i tried to go through it again before midterm...

the question is here, just in case. Thank you soooo much!

 

[toni]

can anyone tell me...

If xf"+f'+xf = 0

Is d^n/dx^n (xf"+f'+xf) = 0 ??

cos the solution says d^n/dx^n [x(xf"+f'+f)-f] = 0...that puzzles me. --'

 

[gabbagabbahey]

can anyone tell me...

If xf"+f'+xf = 0

Is d^n/dx^n (xf"+f'+xf) = 0 ??

Yes, since

\frac{d^n}{dx^n} (xf''+f'+xf)=\frac{d^n}{dx^n}(0)=0

; that is, since the quantity your taking the nth derivative of is by supposition always zero...

cos the solution says d^n/dx^n [x(xf"+f'+f)-f] = 0...that puzzles me. --'

That puzzles me too!...

\frac{d^n}{dx^n} [x(xf''+f'+xf)-f]=\frac{d^n}{dx^n}[(0)-f]=\frac{-d^nf}{dx^n} \neq 0

... perhaps that line in your solutions manual was a typo, and it was supposed to be anx^2f(x) instead of an (x^2-1)f(x)...are you sure you are looking at the solution to the given problem and nota different problem by accident?