# Find the Error

1. Oct 29, 2016

### Lilia

1. The problem statement, all variables and given/known data
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.

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

3. 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?

2. Oct 29, 2016

### Ray Vickson

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. Oct 29, 2016

### Staff: Mentor

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.
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. Oct 29, 2016

### Lilia

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: Oct 29, 2016
5. Oct 29, 2016

### lurflurf

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. Oct 30, 2016

### Lilia

Isn't that the function deviation in the range x=x1÷x5? I need to find that?

Last edited: Oct 30, 2016
7. Oct 30, 2016

### lurflurf

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.