# Just couldnt figure this out = (2)

1. Oct 31, 2008

### 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)?

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2. Nov 1, 2008

### HallsofIvy

Staff Emeritus
"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?

3. Nov 2, 2008

### toni

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

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4. Nov 2, 2008

### 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!

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5. Nov 2, 2008

### 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. --'

6. Nov 3, 2008

### gabbagabbahey

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...

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 an$x^2f(x)$ 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?