# University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg35

## EGM6321 - Principles of Engineering Analysis 1, Fall 2009

Mtg 35: Thurs, 12Nov09

### Page 35-1

Area $=\sum$  Quadrilaterals

Cubature; CUBE; Volume $=\sum$  cubes

$I(f):=\int _{-1}^{1}f(x)\,dx\$

$I_{n}(f):=\sum _{j=1}^{n}w_{j}f(x_{j})\,dx\$  with $\left\{x_{j}\right\}\$  the roots for $P_{n}(x)=0\$  , where n is the degree of $P_{n}(x)\$  and $w_{j}\$  being the weight

$-1

### Page 35-2

 \displaystyle {\begin{aligned}I(f)=I_{n}(f)+E_{n}(f)\end{aligned}} (1)

 \displaystyle {\begin{aligned}w_{j}={\frac {-2}{(n+1)P_{n}'(x_{j})P_{n+1}(x_{j})}}\end{aligned}} (2)

where j=1,2,...,n

 \displaystyle {\begin{aligned}E_{n}(f)={\frac {2^{2n+1}(n!)^{4}}{(2n+1)\left[(2n)!\right]^{2}}}{\frac {f^{(2n)}(\eta \ )}{(2n)!}}\end{aligned}} (3)

for $\eta \ \in \left[-1,+1\right]\$

Ex: $n=2\$  (2 point interpolation)

Eq.(3) P.31-3 $P_{2}(x)={\frac {1}{2}}(3x^{2}-1)\$

$\Rightarrow \ x_{1,2}=\pm \ {\frac {1}{\sqrt {3}}}\$

Eq.(4) P.31-3 $P_{2}'(x)=3x,P_{3}(x)={\frac {1}{2}}(5x^{3}-3x)\$

$W_{1}={\frac {2}{(2+1)(3)({\frac {-1}{\sqrt {3}}}){\frac {1}{2}}\left[5({\frac {-1}{\sqrt {3}}})^{3}-3({\frac {-1}{\sqrt {3}}})\right]}}=1\$

### Page 35-3

$W_{2}=1\$

HW: verify table for Gauss Legendre quadrature in wikipedia, analytical expression of $\left\{x_{j}\right\}\$  and $\left\{w_{j}\right\},j=1,...,n\$  and $n=1,...,5\$  (n=integration points) after verifying the expression for $P_{n}(x)\$  with $n=1,...,6\$ ; (see HW p31-3 )

Evaluate numerically $\left\{x_{j}\right\}\$  and $\left\{w_{j}\right\}\$  and compute results with Abram & Stegum (see lecture plan)

Question: How does Gauss Legendre quadrature compare to other quadrature methods, e.g. trapezoidal rule?

Answer: Look at $E_{n}(f)\$ , Eq.(3) P.35-2. Consider $f\in \mathrm {P} \ _{2n-1}\$ ...