MATH 417 Lecture 8

« previous | Thursday, February 6, 2014 | next »

Quiz

Find $p_4(x)$ interpolating polynomial of $2\cos (\pi x)+6x+14$ for $x\in \{0,1,2,3\}$

The $6x+14$ can be omitted for the computation of the and added at the end

Chapter 4: Numerical Integration and Differentiation

Suppose we wanted to find the length of the curve $f(x)=\cos x$.

$\underset{a}{\overset{b}{\int }}\sqrt{1+{\sin }^{2}x}\mathrm{d}x$ is an elliptic integral that is impossible to find.

We could use an interpolating polynomial $p_n(x)$ such that

$f(x)=p_n(x)+\frac{f^{(n+1)}(\xi (x))}{(n+1)!}\prod _{i=0}^n(x-x_i)$

In which case we can approximate $f^{\prime }(x)$.

$f^{\prime }(x_j)={p_n}^{\prime }(x_j)+\frac{f^{(n+1)}(\xi )}{(n+1)!}\prod _{i\ne j}(x_j-x_i)$

We find that the error between two points is approximately linear in regards to the distance between them:

$f^{\prime }(x_0)-\frac{f(x_0+h)-f(x_0)}{h}=\frac{h}{2}{f}^{″}(\xi )\in O(h)$

Example

Let's differentiate $p_2(x)=f(x_0)\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}+f(x_1)\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x-x_2)}+f(x_2)\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}$:

${p_2}^{\prime }(x)=\frac{f(x_0)}{(x_0-x_1)(x_0-x_2)}(x-x_2+x-x_1)+\frac{f(x_1)}{(x_1-x_0)(x_1-x_2)}(x-x_2+x-x_0)+\frac{f(x_2)}{(x_2-x_0)(x_2-x_1)}(x-x_1+x-x_0)$

If the points are equidistant at distance $h$,

${p_2}^{\prime }(x)=\frac{-3f(x_0)+4f(x_0+h)-f(x_0+2h)}{2h}$

To estimate the error, we use

$f^{\prime }(x_0)-\frac{-3f_0+4f_1-f_2}{2h}=\frac{f^{‴}(\xi )}{6}h\cdot 2h=\frac{h^2}{3}{f}^{‴}(\xi )=O(h^2)$

The moral of the story: For 3 equidistant points on a parabola, the slope of the tangent line at the middle point is equal to the slope of the secant line of the first and last points.