Finite Difference (Interpolating Polynomial)

In summary, you are trying to find the second derivative of a function at a particular point, but you don't know how to do it. You simplify the first differences of the function and then take the limit as the derivative becomes zero.
  • #1
planauts
86
0

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!
 
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  • #2
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.
 
  • #3
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...
 
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  • #4
planauts said:
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.
 
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  • #5
planauts said:
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.
 
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  • #6
LCKurtz said:
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.
 
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  • #7
planauts said:
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.
 
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  • #8
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.
 

FAQ: Finite Difference (Interpolating Polynomial)

What is a finite difference interpolating polynomial?

A finite difference interpolating polynomial is a mathematical function that is used to approximate a given set of data points by constructing a polynomial that passes through those points. It is based on the concept of finite differences, which is the difference between two consecutive data points. The polynomial is then used to estimate the value of the function at any point within the given range.

How is a finite difference interpolating polynomial calculated?

To calculate a finite difference interpolating polynomial, the data points are first arranged in a table and the finite differences are calculated. The coefficients of the polynomial are then determined using the finite differences and the divided difference method. The polynomial is then constructed using these coefficients and can be used to approximate the function at any point within the given range.

What are the advantages of using finite difference interpolating polynomials?

Finite difference interpolating polynomials have several advantages, including being relatively easy to calculate and implement. They also provide a good approximation for smooth functions and can be used to estimate the value of a function at any point within the given range. Additionally, they do not require any prior knowledge of the underlying function and can handle missing or noisy data points.

What are the limitations of finite difference interpolating polynomials?

While finite difference interpolating polynomials have many advantages, they also have some limitations. One major limitation is that they can only approximate smooth functions, so they may not provide accurate results for functions with sharp changes. They also require a large number of data points to accurately approximate the function, which can be time-consuming and computationally expensive.

How are finite difference interpolating polynomials used in practical applications?

Finite difference interpolating polynomials have various applications in fields such as physics, engineering, and finance. They are commonly used to approximate and analyze data in situations where the underlying function is unknown or difficult to obtain. They are also used in numerical methods, such as solving differential equations, and in creating mathematical models for real-world problems.

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