In this activity we will explore polynomial interpolation of data.
You should know that two points determine a straight line. We will generalize this to show that points determine a parabola, points determine a 3rd degree polynomial, and in general points determine a unique degree polynomial which pass through (or interpolate) the data points.
Let’s start with a detailed example.
whose graph passes through the points
If you solved the last problem by solving a system of three linear equations in three unknowns by hand, you probably had a slightly painful time. You could imagine that interpolating more data points using higher degree polynomials could be even more painful.
Thankfully, Linear Algebra will come to the rescue!
Sometimes generalizing and abstracting a problem can actually help you solve it.
Before learning how to solve the totally general problem of interpolating points with a degree polynomial, we will resolve our original example problem using some new ideas.
Let be the vector space of real polynomials of degree two. Consider the function defined by
So far we have only rephrased the problem: we haven’t really done anything to help us solve it.
Here is the key idea:
Choosing the right basis makes life awesome
So we will search for a basis of which makes the job of finding interpolating polynomials easier. If we could find polynomials , , and which satisfy , and , then the problem would be resolved by the linearity of . To see this, note that (if we have already found such polynomials) then we can find a polynomial passing through and simply by using .
Now use this new basis to confirm your answer to the original problem.
We have done a lot of work, and it has a payoff. Now if I had some different data points, I could use this new basis to ease computation of an interpolating polynomial. For instance:
Now we will replay the whole story to solve a more general question:
Which quadratic interpolates the points ?
As before, we will
- Define a linear transformation by
- Find a basis of with , and
- Then will pass through the points and .
Now we can generalize to any number of points!
- We want to find a polynomial of degree which interpolates the points .
- Define a linear map by
- Find a basis of with the property that .
- In particular, since has as roots, then we must have
- Thus solves the interpolation problem.
- Putting it all together,
is the unique polynomial of degree which passes through the points
This final formula is called the Lagrange Interpolation Formula