Math Is Fun Forum

yon

Guest

Hermite interpolation starting from Lagrange form

Hi everyone,

I'm having trouble deriving a Hermite formula starting from the Langrange form of an interpolation.
e.g. Suppose we know

and

.  What I actually want to achieve is a formula of the form:

I thought that if we started with a simple Lagrange formula for 4 points:

And we let x2 -> x0 and x3 -> x1, we'd get the result.  However, this does not seem to work.  Any advice?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Hermite interpolation starting from Lagrange form

Hi yon;

Welcome to the forum. The Lagrange form fits an interpolating polynomial while hermite is an osculating polynomial fit.

My question is did you try to use the Lagrange on the points you had even though some of them were for f'(x) ? Is that what you did?

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

yon

Guest

Re: Hermite interpolation starting from Lagrange form

Well, actually I just made up the Lagrange polynomial for 4 points (even though we only have 2), and the I took the limit.
e.g. for the first value:

However, this does not seem to work, as we get a 0 in the denominator by doing this...

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Hermite interpolation starting from Lagrange form

But you used the points that were from the derivatives and tried to fit an interpolating polynomial through them with the points from the function. This is where I think you went wrong.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

yon

Guest

Re: Hermite interpolation starting from Lagrange form

Do you have any suggestions on how it should be done?
What I'm actually trying to achieve is to find an quadrature formula for the region between x0 and x1 if we only know f0, f1, f'0 and f'1.
As the standard Newton-Cotes formula's were found starting from the Langrange interpolation notation, I thought this was the way to go.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Hermite interpolation starting from Lagrange form

Hi yon;

You are looking to just integrate a polynomial. I would do it by solving a set of simultaneous equations to find the coefficients of the polynomial you want. Do you have some actual points?

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

yon

Guest

Re: Hermite interpolation starting from Lagrange form

I don't have actual data.  I was just wondering if there was a simple way to derive a formula for it using the given parameters.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Hermite interpolation starting from Lagrange form

Can you solve a 4x4 set of simultaneous equations? Either by hand or by computer? All these problems reduce down to that. Honestly, differences, divided differences, lagrangian interpolation, all the things those guys in the past loved are nothing but curiosities really. They are like Cramer's rule, gone the way of the DoDo.

So, I ask again can you solve a 4x4 for any data set? I will show you how to set it up.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

yon

Guest

Re: Hermite interpolation starting from Lagrange form

I don't really get what you mean by '4x4 dataset solving'.
Could you give an example, please?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Hermite interpolation starting from Lagrange form

Hi yon;

Supposing you have 4 data points:

f(1) = 2
f(3) = 1
f '(1) = 1
f '(3) = 2

You try to fit a polynomial to them in this manner:

Now you substitute your points for x. This yields a 4x4 set of simultaneous equations.

Now you solve for a,b,c,d which are the coefficients of the wished for polynomial.

The polynomial that fits all four points is:

All collocation or interpolation problems can be done in this manner. Modern mathematics, experimental mathematics, encourages general methods.This approach has been called the Teakettle approach by the Soviet combinatoricist Villenkin.

In the past every mathematician sought a new approach for each problem, so we have differences, divided differences, Newton's form, Bessel's form, Gauss' form, Stirling's form, Lagrange's form, Euler's form etc.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

yon

Guest

Re: Hermite interpolation starting from Lagrange form

Ah, now I understand.  Thank you very much for the explanation!

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Hermite interpolation starting from Lagrange form

Hi yon;

Your welcome. Let me know how it works out.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.