Numerical Methods

1. Numerical Methods Presentation on • Numerical Differentiation and Integration: Gauss Quadrature

2. The need for Gauss Quadrature Evaluating integrals are common in engineering problems. Sometimes, it is difficult to evaluate complex integrals. Therefore, techniques are devised in order to simplify these.

3. From Trapezoidal rule, Trapezoidal rule evaluates an integral by approximating the function of the curve to a straight line. And then finding the area under the straight line.

4. Larger error (Using Trapezoidal rule) Reduced error (Taking an improved estimate)

5. Method of Undetermined Coefficients Constant function Linear function Quadratic function Cubic function

6. Assuming that a new variable xd is related to the original x in a linear fashion For x­d=-1, For x­d=1, Then, Therefore, When differentiated, Then, the above two equations can be substituted in to the equation to be integrated.

7. Table for finding Weighting factors and Function Arguments

8. Example 1: Use three-point formula to evaluate it. Solution: Given: a=-1 and b=1; f(x) = x+3x2+x5 From table for three points, C0=0.5555556 C1=0.8888889 C2=0.5555556 X0=-0.774596669 X1=0 X2=0.774596669 I=C0f(X0) +C1f(X1) +C2f(X2) f (X0)=.7465 f(X1)=0 f(X2)=2.8537762 I=[0.5555556*.7465+.8888889*0+0.5555556*2.8537762]=2.001

9. Example 2: Given: a=-1 and b=1; f(x) = x+3x2+x5 From table for three points, C0=0.5555556 C1=0.8888889 C2=0.5555556 X0=-0.774596669 X1=0 X2=0.774596669 I=C0f(X0) +C1f(X1) +C2f(X2) X0`= ((b-a)/2+ X0 (b+a)/2 ) =4.6762 X1`= ((b-a)/2+ X1 (b+a)/2) =7 X2`= ((b-a)/2+ X2 (b+a)/2)=9.3238 f (X0`)=2306.24373 f (X1`)=16961 f (X2`)=70733.71 I=3[0.5555556*2306.24373+.8888889*16961+0.5555556*70733.71] I= 166,962.5999

10. Error Analysis for Gauss Quadrature

11. See this