MATH 417 Lecture 12

« previous | Thursday, February 20, 2014 | next »

Quiz

Approximate $f^{\prime }(x+3h)$ given $f(x_0-h)$, $f(x_0)$, and $f(x_0+3h)$ to within $O(h^2)$ accuracy

Lagrange

Find $P_2(x)$ (which has error $O(h^3)$) and differentiate

Taylor Series

Recall that the taylor series for $f$ centered at $a$ is

$f(x)=f(a)+(x-a)f^{\prime }(a)+\dots$

In our case, we want to take the function at the point $x_0-h$ centered around $x_0+3h$:

$f(x_0-h)=f((x_0+3h)-4h)=f(x_0+3h)+\frac{(-4h{)}^1}{1!}{f}^{\prime }(x_0+3h)+\frac{(-4h{)}^2}{2!}{f}^{″}(x_0+3h)+O(h^3)$

Integration

Suppose we are given $x_0$ and $x_1$. Our goal for this example is to approximate $\underset{-1}{\overset{1}{\int }}f(x)\mathrm{d}x$ with a linear function $A_{0}f(x_0)+A_{1}f(x_1)$.

We can tell that DAC ≥ 1

• Let $f(x)=1$, then $\int f(x)\mathrm{d}x=\int 1\mathrm{d}x=A_0\cdot 1+A_1\cdot 1=2$
• Let $f(x)=x$, then $\int f(x)\mathrm{d}x=\int x\mathrm{d}x=A_0\cdot x_0+A_1\cdot x_1={\frac{x^2}{2}|}_{-1}^1$

we solve the equations above and come up with the trapezoidal rule.

The Best Rule

$\underset{-1}{\overset{1}{\int }}f(x)\mathrm{d}x\approx Q_n(f)=\sum _{i=1}^n{A}_{i}f(x_i)$

This requires solving for $2n+2$ unknowns: $A_0,A_1,\dots ,A_n,x_0,x_1,\dots ,x_n$.

Theorem. [Gaussian Rule]. There is only one rule with degree of accuracy $2n+1$ based on $n+1$ points.

Proof.

quod erat demonstrandum

For two points, the gaussian rule is:

$\underset{-1}{\overset{1}{\int }}f(x)\mathrm{d}x\approx f(-\frac{\sqrt{3}}{3})+f\left(\frac{\sqrt{3}}{3}\right)$

This has DAC = 3

Legendre Polynomials

$L_n(x)=\frac{{\mathrm{d}}^n}{\mathrm{d}{x}^n}((1-x^2{)}^n)=c(x^n+\dots )=0$

This polynomial has $n$ real-valued zeroes $x_0,x_1,\dots ,x_{n-1}$

1. $L_1(x)=-2x$ with root $x=0$, so the Gaussian rule is $\underset{-1}{\overset{1}{\int }}f(x)\mathrm{d}x=A_{0}f(x_0)$. $A_0$ can be found to be $2$
2. $L_2(x)=4(3x^2-1)$ with roots $x=\pm \frac{\sqrt{3}}{3}$, hence the Gaussian rule above.
3. $L_3(x)=24x(3-5x^2)$ with zeroes $x\in \{0,\pm \sqrt{\frac{3}{5}}\}$, so the Gaussian rule is $\underset{-1}{\overset{1}{\int }}f(x)\mathrm{d}x\approx A_{0}f(-\sqrt{\frac{3}{5}})+A_{1}f(0)+A_{2}f\left(\sqrt{\frac{3}{5}}\right)$. The coefficients can be found to be $\frac{5}{9}$, $\frac{8}{9}$, and $\frac{5}{9}$, respectively.

Note: the coefficients and points are always symmetric (about a symmetric interval)

Suppose we scale the gaussian rule for 3 points to the interval $(0,10)$:

$\underset{0}{\overset{10}{\int }}f(x)\mathrm{d}x\approx \frac{10-0}{2}(\frac{5}{9}f(5+\frac{\sqrt{3}}{3}\cdot 5)+\frac{8}{9}f(5)+\frac{5}{9}f(5-\frac{\sqrt{3}}{3}\cdot 5))$

We shift the points by 5 (to the midpoint of the