Finite Difference (Interpolating Polynomial)

[planauts]

Homework Statement

http://puu.sh/1QFsA

Homework Equations

The Attempt at a Solution

I'm actually not sure how to do this question. How do i find Δx^2. I kind of understand the question but I don't know how to prove it. I know that Δy becomes dy when the width becomes infinitesimally small (Ʃ (infinity)). And y'' is basically the second derivative... but I have no clue how to go about verifying this...

This is my attempt:
http://puu.sh/1QFqE

I, myself, think it looks more like gibberish. Could someone nudge me to the right direction, please? Thanks!

 

[LCKurtz]

You have calculated $y_0,\,y_1,\,y_2$. The first differences are $y_1-y_0, y_2-y_1$, as you have calculated. Simplify them both by adding the fractions. Then take their difference and simplify it similarly. That will give you $\Delta^2y_0$. Then divide that by $\Delta^2x$ and take the limit as $\Delta x\rightarrow 0$. Just push it through and determine whether it gives the second derivative of $\frac 1 x$ at $x=x_0$ or not.

 

[planauts]

But the question asks for http://puu.sh/1QKd3 , and you are saying to divide http://puu.sh/1QKeC by http://puu.sh/1QKdV . The square is associated with x, rather than delta. Also, when I take the second difference of x, I get zero...

 

[Ray Vickson]

But the question asks for http://puu.sh/1QKd3 , and you are saying to divide http://puu.sh/1QKeC by http://puu.sh/1QKdV . The square is associated with x, rather than delta. Also, when I take the second difference of x, I get zero...

You don't need $\Delta x^2$ (not $\Delta^2 x$); that is just "notation", the same as $dx^2$ is merely a notation in differentiation. We have $\Delta y / \Delta x$ = first divided-difference, and so $\Delta^2 y / \Delta x^2$ = second divided difference = divided difference of the divided differences.

 

[LCKurtz]

But the question asks for http://puu.sh/1QKd3 , and you are saying to divide http://puu.sh/1QKeC by http://puu.sh/1QKdV . The square is associated with x, rather than delta. Also, when I take the second difference of x, I get zero...

Yes, I meant $\Delta x^2 = (\Delta x)^2$ as stated in the problem. That's why I almost (and should have) suggested you use $h$ instead of $\Delta x$ when working that problem. Using $h^2$ there would have been no typo and no misunderstanding.

 

[planauts]

You have calculated $y_0,\,y_1,\,y_2$. The first differences are $y_1-y_0, y_2-y_1$, as you have calculated. Simplify them both by adding the fractions. Then take their difference and simplify it similarly. That will give you $\Delta^2y_0$. Then divide that by $\Delta^2x$ and take the limit as $\Delta x\rightarrow 0$. Just push it through and determine whether it gives the second derivative of $\frac 1 x$ at $x=x_0$ or not.

http://puu.sh/1QMxk

I'm still very confused about the "Then divide that by Δ^2x and take the limit as Δx→0."

I got it simplified using CAS, and replaced Δx by h like you suggested. However, how do I convert that into an derivative. If I just take the limit as h->0, then it would be 0 and the derivative that I need is 2/x^2.

 

[LCKurtz]

http://puu.sh/1QMxk

I'm still very confused about the "Then divide that by Δ^2x and take the limit as Δx→0."

I got it simplified using CAS, and replaced Δx by h like you suggested. However, how do I convert that into an derivative. If I just take the limit as h->0, then it would be 0 and the derivative that I need is 2/x^2.

Your steps look OK. Weren't you asked to divide by $(\Delta x)^2=h^2$ before you let $h\rightarrow 0$? And hopefully you will get the second derivative.

 

[planauts]

Thanks a lot! It makes sense now. The question was so confusing but it is actually a very simple question!

Thanks again everyone for your help.