What is the Lagrangian Interpolation Formula for Approximating Functions?

In summary, we are considering the Lagrange Polynomial approximation and we want to show that it can be written as p(x)=\psi(x) \sum_{k=0}^n\frac{f(x_k)}{(x-x_k)\psi^\prime(x)}, where \psi(x)=\prod_{i=0}^n x-x_i. We evaluate p(x) and \psi^\prime(x) and find that we need to multiply \psi(x) by (x-x_k) in order to get the correct denominator.
  • #1
stvoffutt
15
0

Homework Statement


Consider the Lagrange Polynomial approximation [tex]p(x) =\sum_{k=0}^n f(x_k)L_k(x)[/tex] where [tex]L_k(x)=\prod_{i=0,i\neq k}^n \frac{x-x_i}{x_k-x_i}[/tex]
Let [tex]\psi(x)=\prod_{i=0}^n x-x_i[/tex]. Show that [tex]p(x)=\psi(x) \sum_{k=0}^n\frac{f(x_k)}{(x-x_k)\psi^\prime(x)}[/tex]


Homework Equations


None. Just plug in and see if it pops out.


The Attempt at a Solution


I just evaluated what [itex]p(x) = \sum_{k=0}^n f(x_k)L_k(x)[/itex] and [itex]\psi^\prime(x)[/itex]. Writing out some terms of p(x):[tex]p(x)=f(x_0)L_0(x_0)+f(x_1)L_1(x_1)+\cdots+f(x_n)L_n(x_n)[/tex]
[tex]=f(x_0)\frac{\prod_{i=0,i\neq k}^n x-x_i}{\prod_{i=0,i\neq k}^n x_o-x_i}+\cdots+f(x_n)\frac{\prod_{i=0,i\neq k}^n x-x_i}{\prod_{i=0,i\neq k}^n x_n-x_i}[/tex]
I find that this is just
[tex]p(x)=\prod_{i=0,i\neq k}^nx-x_i \sum_{k=0}^n \frac{f(x_k)}{\prod_{i=0,i\neq k}^n x_k-x_i}[/tex]
The product outside of the sum is just [itex]\psi(x)[/itex]. I then evaluate [itex]\psi^\prime(x)[/itex]:
[tex]\psi^\prime(x)=\prod_{i=0,i\neq k=0}^nx-x_i +\prod_{i=0,i\neq k=1}^nx-x_i +\cdots +\prod_{i=0,i\neq k=n}^nx-x_i [/tex]
Evaluating this at [itex]x_k[/itex] we get exactly the denominator from above. I cannot figure out where the extra [itex]x-x_k[/itex] term comes from. I suspect my error is in taking the derivative but I have looked this over for many hours and cannot find my mistake.
 
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  • #2
Wow, never mind. I feel dumb. What I pulled out of the sum is not [itex]\psi(x)[/itex]. You need to multiply that product by [itex](x-x_k)[/itex] for it to be [itex]\psi(x)[/itex]. Fun with definite products - not my strong suit.
 

1. What is Lagrangian interpolation and how does it work?

Lagrangian interpolation is a method used to approximate a function using a polynomial that passes through a set of given data points. It works by constructing a polynomial of degree n-1, where n is the number of data points, and using it to approximate the function at any point within the given range.

2. When is Lagrangian interpolation used in scientific research?

Lagrangian interpolation is commonly used in scientific research when there is a need to approximate a continuous function using a limited set of data points. This can occur in various fields such as physics, engineering, and economics.

3. What are the advantages of using Lagrangian interpolation?

One of the main advantages of Lagrangian interpolation is its simplicity and ease of implementation. It also provides a good approximation of the function within the given range of data points. Additionally, it can be used for both interpolation (approximating values within the given data points) and extrapolation (predicting values outside the given data points).

4. Are there any limitations to using Lagrangian interpolation?

Yes, there are some limitations to using Lagrangian interpolation. One of the main limitations is that the accuracy of the approximation heavily depends on the distribution of the data points. If the data points are clustered in certain areas, the approximation may not be as accurate. Additionally, as the degree of the polynomial increases, the approximation may become more prone to error.

5. Are there any alternative methods to Lagrangian interpolation?

Yes, there are other interpolation methods that can be used, such as Newton's divided differences interpolation and spline interpolation. Each method has its own advantages and limitations, and the choice of method depends on the specific requirements of the problem at hand.

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