Day 19 in Mat360

Today:

• Announcements:
  • Your first project is due.

  • You have a new homework assigned, due next Tuesday (concerning interpolation).

  • Your homework is graded. Let's talk a little about these three problems and Taylor Polynomials.
    • p. 178, #2: what's the relative error? Now bound it. Michael W.
    • p. 178, #3 (ditto -- I'm going to put Brooke on the hot seat -- look out, Brooke!:)
    • Lauren's p. 181, #6 (and #3)

  • Would you all mind if we push the text back a week, to the 6th of April?

    I want to cover root-finding, and interpolation, but I feel like we won't have had enough time to cover all the interpolation issues.

  • One way or another, we're going to have to move along to differentiation next time!:0

• Today:
  • We wrap up an important example: order of convergence the Secant Method (Theorem 3.5, p. 99)

    Even though the order of convergence is less, the routine may run in about the same time as Newton's (depends on the problem, of course).

  • The "linear spline" error term (section 5.5): the linear spline interpolator is $C^0$: we have continuity only.

    The authors observe that, for a much-used and computationally intense function, it may be quite a bit faster to compute it up front for a boatload of points, and then use an interpolator to give a pretty good approximation for general use.

    That's the idea that's being explored here, for $e^x$. Suppose we want to compute it on the interval $[0,1]$ with an error of at most $10^{-6}$: how many points would a linear interpolator require?

  • The general polynomial interpolator error term is given in Theorem 5.3, p. 193. A picture of this error term (or rather a bound on the error term) is given in Figure 5.11, p. 195.

  • An example of cubic splines: how do we intepolate two points and their slopes using a cubic? We can derive the so-called "Hermite spline" as a variant of Lagrange interpolation, featuring "get out of the way" functions!

    Here's another picture of how we might build one of these.

    • One advantage of Hermite polynomials is that the derivative of the spline interpolates the first derivative -- the interpolator is $C^1$.

    • One disadvantage of Hermite polynomials is that we need to know the derivative (! we asked for it!) of the function. Question: if we don't have a function, e.g. just data, how might we proceed?
    Other methods are, of course, possible. Some popular splines include "natural" and "clamped" cubic splines. These achieve other objectives.

    I can think of several ways of approaching the fitting of this data with the Hermite interpolant. How might you go about it?

    You might not be surprised to learn that about the easiest formulation is Newton's, only we're going to repeat some abscissa:

    $f[x_1]+(x-x_1)(f[x_1,x_1]+ (x-x_1)(f[x_1,x_1,x_2]+(x-x_2)f[x_1,x_1,x_2,x_2]$

    Since we'll need this over and over, we should go ahead and simplify those divided difference coefficients, using the assumption that $f'$ is known (use the recurrence relation at the top of p. 215 to do so).

    This will have the effect of fitting the two endpoints, and the derivatives of $f$ at the two endpoints.

    I'd like to discuss my implementation of this in Mathematica.

    What error are we making when we use Hermite interpolation? Proposition 5.6 gives us the answer (p. 219):

    $(x-x_1)^2(x-x_2)^2f[x_1,x_1,x_2,x_2,x]$

    which is equal to a derivatives times some stuff:

    $\frac{f^{(4)}(\xi)}{{4!}}(x-x_1)^2(x-x_2)^2$

    where $\xi$ is some unknown number between the minimum and maximum of $\{x_1,x_2,x\}$.

    While my Hermite spline does the job (approximating Sine for all real numbers), one might do the same job with Taylor polynomials.

• Notes from Class:
  • Polynomial Interpolation schemes
  • Proof of Order of Convergence of the Secant Method