MATH 417 Lecture 8

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Quiz

Find $p_{4}(x)$ interpolating polynomial of $2\cos {(\pi \,x)}+6x+14$ for $x\in \left\{0,1,2,3\right\}$

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 $f(x)=\cos {x}$.

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

We could use an interpolating polynomial $p_{n}(x)$ such that

$f(x)=p_{n}(x)+{\frac {f^{(n+1)}(\xi (x))}{(n+1)!}}\,\prod _{i=0}^{n}\left(x-x_{i}\right)$

In which case we can approximate $f'(x)$.

$f'(x_{j})=p_{n}'(x_{j})+{\frac {f^{(n+1)}(\xi )}{(n+1)!}}\prod _{i\neq j}\left(x_{j}-x_{i}\right)$

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

$f'(x_{0})-{\frac {f(x_{0}+h)-f(x_{0})}{h}}={\frac {h}{2}}\,f''(\xi )\in O(h)$

Example

Let's differentiate $p_{2}(x)=f(x_{0})\,{\frac {(x-x_{1})(x-x_{2})}{(x_{0}-x_{1})(x_{0}-x_{2})}}+f(x_{1})\,{\frac {(x-x_{0})(x-x_{2})}{(x_{1}-x_{0})(x-x_{2})}}+f(x_{2})\,{\frac {(x-x_{0})(x-x_{1})}{(x_{2}-x_{0})(x_{2}-x_{1})}}$:

$p_{2}'(x)={\frac {f(x_{0})}{(x_{0}-x_{1})(x_{0}-x_{2})}}\,\left(x-x_{2}+x-x_{1}\right)+{\frac {f(x_{1})}{(x_{1}-x_{0})(x_{1}-x_{2})}}\,\left(x-x_{2}+x-x_{0}\right)+{\frac {f(x_{2})}{(x_{2}-x_{0})(x_{2}-x_{1})}}\,\left(x-x_{1}+x-x_{0}\right)$

If the points are equidistant at distance $h$,

$p_{2}'(x)={\frac {-3f(x_{0})+4f(x_{0}+h)-f(x_{0}+2h)}{2h}}$

To estimate the error, we use

$f'(x_{0})-{\frac {-3f_{0}+4f_{1}-f_{2}}{2h}}={\frac {f'''(\xi )}{6}}h\cdot 2h={\frac {h^{2}}{3}}\,f'''(\xi )=O(h^{2})$

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.