Solve Newton Interpolating Polynomial for Error

In summary: Can anyone explain why this is the case?In summary, the conversation is about finding the Newton interpolating polynomial and the error for the function y=1/(x^1/4) using 5 given x-values and their corresponding y-values. The formula Ln-1(x)=f(x1)+f(x1,x2)(x-x1)+... is used to write the Newton interpolating polynomial, but the speaker is unsure of how to find the error without the formula. They ask for help and someone suggests using the mean value argument to approximate the error. There is also a discussion about the difference between deviation and error. The conclusion is that the calculated error is quite pessimistic and does not accurately model the error for the given function
  • #1
Lilia
48
0

Homework Statement


y=1/(x^1/4). I'm given 5 x's and 5y's. I need to write Newton interpolating polynomial and find the error.

Homework Equations


Ln-1(x)=f(x1)+f(x1,x2)(x-x1)+...

The Attempt at a Solution


With the formula above I wrote the Newton interpolating polynomial but I can't find the error since I don't have the formula. Can anyone help?
 
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  • #2
Lilia said:

Homework Statement


y=1/(x^1/4). I'm given 5 x's and 5y's. I need to write Newton interpolating polynomial and find the error.

Homework Equations


Ln-1(x)=f(x1)+f(x1,x2)(x-x1)+...

The Attempt at a Solution


With the formula above I wrote the Newton interpolating polynomial but I can't find the error since I don't have the formula. Can anyone help?

First show your work.

When you say you "don't have the formula..", what do you mean? Are you unsure how to find the Newton polynomial, or unsure how to find the error? Have you consulted your textbook and/or course notes? Have you looked on-line for topics related to "Newton interpolation"?
 
  • #3
Lilia said:

Homework Statement


y=1/(x^1/4). I'm given 5 x's and 5y's. I need to write Newton interpolating polynomial and find the error.

Homework Equations


Ln-1(x)=f(x1)+f(x1,x2)(x-x1)+...
What does this mean?
Is "Ln-1(x)" supposed to be "ln-1(x)" (i.e., ex)? How is what you wrote relevant to your problem?
Also, you wrote f(x1) and f(x1, x2). Is f a function of one variable or is it a function of two variables? It can't be both.
Lilia said:

The Attempt at a Solution


With the formula above I wrote the Newton interpolating polynomial but I can't find the error since I don't have the formula. Can anyone help?
The formula for what? It looks like you're supposed to approximate ##f(x) = \frac 1 {x^{1/4}}## using a term or two from the Taylor series.
 
  • #4
x | 1 | 2 | 5 | 16 | 39
------------------------------
y | 1 | 0.8 | 0.6 | 0.5 | 0.4

I found the Newton interpolating polynomial:
P4(x)=1+(-0.2)*(x-1) + (0.04)*(x-1)(x-2) + (-0.003)*(x-1)(x-2)(x-5) + A(x-1)(x-2)(x-5)(x-16)

I wrote this looking at an example we wrote in class. But we haven't written the error in any example nor I have the formula. But in this exercise I need to find the error.

f(x1,x2) etc. these are the divided differences which are used to make up the Newton Interpolating polynomial, -0.2, 0.04 etc in this case, which I calculated too

Should I calculate the error with this formula? R(x) = f(x)-Ln-1(x)≤ Mn/n! * (x-x1)(x-x2)*...*(x-xn)?
 
Last edited:
  • #5
The error term is
f(x,x1,x2,x3,x4,,x5)(x-x1)(x-x2)(x-x3)(x-x4)(x-x5)
often we use a mean value argument to replace f(x,x1,x2,x3,x4,,x5) with
$$\frac{f^{(5)}(\xi)}{5!}$$
where the derivative is evaluated at an unknown point between the largest and smallest of
x,x1,x2,x3,x4,x5
 
  • #6
lurflurf said:
The error term is
f(x,x1,x2,x3,x4,,x5)(x-x1)(x-x2)(x-x3)(x-x4)(x-x5)
often we use a mean value argument to replace f(x,x1,x2,x3,x4,,x5) with
$$\frac{f^{(5)}(\xi)}{5!}$$
where the derivative is evaluated at an unknown point between the largest and smallest of
x,x1,x2,x3,x4,x5
Isn't that the function deviation in the range x=x1÷x5? I need to find that?
 
Last edited:
  • #7
The deviation and error are the same right? This is an interesting question. If I calculated correctly M is 9945/1024~10 but that is quite pessimistic as 0.0002 models the error well. Overall that is a quite bad approximation especially around 30.
 

Related to Solve Newton Interpolating Polynomial for Error

1. What is Newton interpolating polynomial?

Newton interpolating polynomial is a form of polynomial interpolation that is used to approximate a function using a set of data points. It involves finding a polynomial that passes through all the given data points and can be used to estimate the value of the function at any point within the given range.

2. How is Newton interpolating polynomial calculated?

The Newton interpolating polynomial is calculated using the Newton's divided difference formula, which is based on the finite differences between the data points. This formula involves recursively calculating the divided differences until the polynomial is found.

3. What is the purpose of solving Newton interpolating polynomial for error?

Solving Newton interpolating polynomial for error allows us to estimate the error in our approximation of the function. This is important because it helps us determine how accurate our polynomial interpolation is and whether we need to use more data points or a higher degree polynomial to improve our approximation.

4. How do you interpret the error in Newton interpolating polynomial?

The error in Newton interpolating polynomial is typically expressed as a function of the maximum value of the n+1th derivative of the function being approximated, where n is the degree of the polynomial. This means that the error decreases as the number of data points or the degree of the polynomial increases.

5. Can Newton interpolating polynomial be used for any function?

Yes, Newton interpolating polynomial can be used to approximate any function as long as the function is continuous within the given range and the data points are evenly spaced. However, for highly oscillatory or discontinuous functions, other interpolation methods may be more suitable.

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