MATH 417 Lecture 12

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Quiz

Approximate $f'(x+3h)$ 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

$f(x)=f(a)+(x-a)\,f'(a)+\dots$

In our case, we want to take the function at the point $x_{0}-h$ centered around $x_{0}+3h$:

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

Integration

Suppose we are given $x_{0}$ and $x_{1}$. Our goal for this example is to approximate $\int _{-1}^{1}f(x)\,\mathrm {d} x$ 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 $\int f(x)\,\mathrm {d} x=\int x\,\mathrm {d} x=A_{0}\cdot x_{0}+A_{1}\cdot x_{1}=\left.{\frac {x^{2}}{2}}\right|_{-1}^{1}$

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

The Best Rule

$\int _{-1}^{1}f(x)\,\mathrm {d} x\approx Q_{n}(f)=\sum _{i=1}^{n}A_{i}\,f(x_{i})$

This requires solving for $2n+2$ unknowns: $A_{0},A_{1},\ldots ,A_{n},x_{0},x_{1},\ldots ,x_{n}$.

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:

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

This has DAC = 3

Legendre Polynomials

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

This polynomial has $n$ real-valued zeroes $x_{0},x_{1},\ldots ,x_{n-1}$

1. $L_{1}(x)=-2x$ with root $x=0$, so the Gaussian rule is $\int _{-1}^{1}f(x)\,\mathrm {d} x=A_{0}f(x_{0})$. $A_{0}$ 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 $x\in \left\{0,\pm {\sqrt {\frac {3}{5}}}\right\}$, so the Gaussian rule is $\int _{-1}^{1}f(x)\,\mathrm {d} x\approx A_{0}\,f\left(-{\sqrt {\frac {3}{5}}}\right)+A_{1}\,f(0)+A_{2}\,f\left({\sqrt {\frac {3}{5}}}\right)$. 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 $\left(0,10\right)$:

$\int _{0}^{10}f(x)\,\mathrm {d} x\approx {\frac {10-0}{2}}\left({\frac {5}{9}}\,f\left(5+{\frac {\sqrt {3}}{3}}\cdot 5\right)+{\frac {8}{9}}\,f(5)+{\frac {5}{9}}\,f\left(5-{\frac {\sqrt {3}}{3}}\cdot 5\right)\right)$

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