Interpolation and You: A Brief Overview of Some Interpolation Tools

1 / 22

# Interpolation and You: A Brief Overview of Some Interpolation Tools - PowerPoint PPT Presentation

Interpolation and You: A Brief Overview of Some Interpolation Tools. By: Mark Coose Joetta Swift Micah Weiss. What Problems Can Interpolation Solve?. Given a table of values, find a simple function that passes through the given points exactly

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Interpolation and You: A Brief Overview of Some Interpolation Tools' - Gabriel

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Interpolation and You: A Brief Overview of Some Interpolation Tools

By:

Mark Coose

Joetta Swift

Micah Weiss

What Problems Can Interpolation Solve?
• Given a table of values, find a simple function that passes through the given points exactly
• Given a table of experimental data, find a formula that approximates the data and possibly filters out errors
• Given an arbitrary function f, find an approximation in the form of a simpler function g
How Do We Find An Interpolating Polynomial?
• Lagrange form
• Newton’s form
• Cubic Splines
Lagrange Form of the Interpolating Polynomial
• A linear combination of n+1 polynomials
• Each term is a polynomial of degree n
Consider the Table

Begin by finding n+1=3 cardinal functions li(x)

Or,

• Each cardinal polynomial term is degree n
• To add one point we must recalculate each cardinal function from scratch
Newton Form of Interpolating Polynomial
• Data points can be added without recalculating existing sequence
• Pn(xi)=f(xi) for i=0,1,…,n
• The polynomial pn is of deg ≤ n
• Pn(x) = a0 + a1(x-x0) + a2(x-x0)(x-x1) + … + an(x – xn-1)

The beauty of Newton’s form of the interpolating polynomial is that it allows us to add points and extend the polynomial without redoing previous calculations. The additional point only requires us to calculate the three values shown in red.

• The function can oscillate wildly
• Each polynomial can be of degree n (computationally expensive for large n = number of points).
Splines

Definition: A function is called a spline of degree k if

• The domain of S is an interval[a,b]
• There are points ti such that a=t0<t1<…<tn=b and S is a polynomial of degree at most k on each subinterval [ti, ti+1]
• S, S’, S’’,…,S(k-1) are all continuous functions on [a,b]
Splines
• Cubic Splines
• Cubic Splines are almost always used over polynomial interpolation because of its increased accuracy (smaller error) and the final curve is smoother than polynomial interpolation
Cubic Spline Smoothness Property

If S is the natural cubic spline function that interpolates a twice-continuous differentiable function f at knots a=t0<t1<…<tn=b, then

What Does This Mean?
• A function with large second derivatives is subject to wild oscillations. The Smoothness Property ensures that the spline S oscillates less than the function that it interpolates.
• An example is shown, here.
Proof of Smoothness Property
• Let g(x)=f(x)-s(x)

Then, f”(x)=g”(x)+s”(x)

And,

Apply integration by parts
• Let u=s’’(x) and dv=g’’(x)
• Then,
• Since, this is a natural cubic spline, s’’=0, so the first term is equal to zero.
• Which leaves
Integration by Parts cont.
• [break this up into sub integrals]

=

• (s’’’ is a constant within this integral)

=

• [g(t(i) + 1) – g(t(i))] = 0
• because g(t(i) + 1) = g(t(i)) = 0

So,

• But,
• So,