1. Numerical Integration (Chapters 5 & 6, C&K 6th edition)
Code development
trapezoid rule
Simpson’s rule
Gauss quadrature
Laguerre quadrature
Analysis
changing the variable of integration
estimating error in numerical integration
uniform method to derive numerical integration

2. Derive the trapezoid rule

3. Replace integrand f(x) on [a,b] by line p(x)
Area of a trapezoid
Multiple intervals with f(x) on [xk-1,xk] replaced by a line
Usually x1 = a and xn = b

4. Trapezoid rule with integrand evaluated at n arbitrary points
in the range of integration:
Basic idea behind MatLab code:
Input 2 arrays of length n:
x = values where the integrand is evaluated.
f = values of the integrand at those points
In a for-loop, accumulate the area of n-1 trapezoids
Output sum as trapezoid rule approximation to integral

5. MatLab code for trapezoid rule integration with integrand evaluated at n
arbitrary points in the range of integration
function A=trap_arbitraty_points(x,f)
n=length(x);
sum=0;
for k=2:n
sum=sum+(x(k)-x(k-1))*(f(k)+f(k-1))/2;
end
A=sum;

6. We will approximate
by the trapezoid rule in several forms
and compare their accuracy as percent
difference from the exact value.
How do we find the exact value of an integral?
What is the anti-derivative of e-x ?

7. Write a script to apply trapezoid rule function to estimate
with 10 equally spaced points and compare to exact value

8. Pseudo-code for script with equally spaced values of x.
Definition a function handle for the integrand
Exact = result using anti-derivative
Use linspace(1,5,10) to create array of points where integrand will be evaluated
A = trap_arbitrary_points(x, integrand(x))
Percent difference=100abs((A-exact)/exact)
Note that integrand is evaluated inside the call to trapezoid rule.
When an array is passed to a function, MatLab returns an arrary

9. MatLab script to apply trapezoid rule to estimate
with 10 equally spaced points.
exp(-x);
exact=exp(1)-exp(5);
x=linspace(1,5,10);
A1=trap_arbitrary_points(x,integrand(x));
PD1=100*abs((A1-exact)/exact);
disp([A1,exact, PD1])

10. Expand previous script to estimate
with 10 “logarithmically” spaced points between 1 and 5.
Logarithmic spacing puts more points at smaller values of x
in the range of integration ([1,5] in this case)

11. Extended MatLab script to apply trapezoid rule to
exp(-x);
exact=exp(1)-exp(5);
x=linspace(1,5,10);
A1=trap_arbitrary_points(x,integrand(x));
PD1=100*abs((A1-exact)/exact);
disp([A1,exact,PD1])
y=linspace(0,log(5),10)
newx=exp(y)
A2 = trap_arbitrary_points(newx,integrand(newx))
PD2 = 100*abs((A2-exact)/exact)
disp(PD2)

12. What is the difference between the two choices of
where the integrand is evaluated?
Plot the 2 choices
What is your conclusion?
Can you rationalize the difference in accuracy?

13. equal spacing
x-value
log spacing
index

14. Changing variable of integration

15. Use the transformation y = ln(x) to change the variable of integration
Use the transformation y = ln(x) to change the variable of integration. dy = dx/x
x(y)= exp(y)
Integrand is a function of y through x(y)

16. x(y)= exp(y) Extended MatLab script to apply trapezoid rule to
Use the integration variable y with 10 equally spaced points
between 0 and ln(5). Recall log(x) is the function name for
natural logirithms.

17. Extended MatLab script to apply trapezoid rule to
exp(-x);
exact=exp(1)-exp(5);
x=linspace(1,5,10);
A1=trap_arbitrary_points(x,integrand(x));
PD1=100*abs((A1-exact)/exact);
disp([A1,exact,PD1])
y=linspace(0,log(5),10)
newx=exp(y)
A2 = trap_arbitrary_points(newx,integrand(newx))
PD2 = 100*abs((A2-exact)/exact)
disp(PD2)
xofy=newx
A3 = trap_arbitrary_points(y, xofy.*exp(-xofy))
PD3=100*abs((A3-exact)/exact)
disp(PD3)

18. Summary of results to estimate
by the trapezoid rule with 10 points when the points are chosen in the following ways:
1. Equally spaced on [1, 5]
2. Logarithmically spaced (xk = exp(yk) where yk= linspace(0,ln(5),10))
3. Equally spaced on [0, ln(5)] in the new integration variable y = ln(x).
Calculate the percent difference from the exact value in each case
Results for percent difference: (1) , (2) , (3)

19. What we learned from this exercise
For a given numerical-integration formula and the number of points where the integrand will be evaluated, two additional factors affect the accuracy of approximation.
1. Where the integrand is evaluated.
2. Which integration variable is used.

20. Composite trapezoid rule

21. Composite trapezoid rule
Trapezoid rule for n arbitrarily spaced points
Can be simplifies for equally spaced points due to the common width of trapezoid base, (b-a)/(npt-1). In the sum over averages of trapezoid heights, f(x1)=f(a) and f(xn)=f(b) occur once. All other values of the integrand occur twice.
w1 = wnpt = ½
wk = 1 2<k<npt-1
Write a MatLab code for composite trapezoid rule

22. MatLab code for composite trapezoid rule
function A=ctraprule(fh,a,b,npts)
h=(b-a)/(npts-1);
x=a;
sum=0;
for i=2:npts-1
x=x+h;
sum=sum+fh(x);
end %contribution from internal points
sum=sum+(fh(a)+fh(b))/2;
A=h*sum;
end

23. Assignment 5, Due 2/27/19
Assume is the “exact” value of the integral.
Estimate this integral by the composite trapezoid rule with 3, 5, and 7 points. Calculate the percent difference from the exact value in each case using PD=100*abs((exact-trap)/exact).

24. Estimating absolute error in composite trapezoid rule
More points in the range of integration were the integrand is evaluated
makes the trapezoid rule more accurate. To find a relation between error
and number of points:
Use Taylor formula to derive an analytic expression for
and the trapezoid rule approximation h[f(a+h) + f(a)]/2.
Difference will be error in the composite trapezoid rule for one interval
of width h.
Generalize to n intervals between a and b.
Based on math of next 3 slides, |e| = (b-a)3|f “(x)|/12n2 where a<x<b
To get an upper bound on |e|, use maximum value of |f “(x)| in the range of
integrations
Note: n = # of intervals = # of points - 1.

25. Estimate error in Trap Rule: single subinterval of size h
a<x1<a+h
By fundamental theorem of calculous
exact result but we don’t know the value of x1

26. Estimate error in Trap Rule: single subinterval of size h
a<x2<a+h
Trap rule approximation
Exact from previous slide
Difference between
exact and trap rule

27. Estimate error in Trap Rule: n subinterval of size h1
n is number of subinterval

28. Example: upper bound on the error in approximating
by the composite trapezoid rule with 10 points
|e| < (b-a)3|f “(x)|max /12n2
Plot |f “(x)| between 2 and 5 to find its maximum value

29. Note: maximum does no occur at either end point of integration.
fplot(‘abs(sin(x))’,[2,5])
|f “(x)|max = 1
|sin(x)|
radians (by default)
Note: maximum does no occur at either end point of integration.
Note: max(|sin(x)|) is not equal to max(sin(x)) for 2<x<5

30. (b-a)=3, |f “(x)|max = 1, n = 9
|e| < (b-a)3|f “(x)|max/12n2 ~
What is the actual absolute error in approximating
by the composite trapezoid rule with 10 points
Get the exact value by anti derivative
Calculate trapezoid-rule estimate
Actual absolute error = abs(traprule-exact)

31. Exact value =
Trap-rule 10 pts =
Actual absolute error = ~ 1% of exact value
Upper bound on absolute error = ~ 4% of exact value

32. Deriving numerical integration methods

33. Most numerical integration formulas have the form
which is a weighted average of values of the integrand.
Example: composite Trapezoid rule
w0 = wn = (b-a)/(2n)
wk = (b-a)/n if 1< k<n-1
n = number of subintervals
One approach to deriving such formulas is to replace the
integrand by a polynomial.

34. replace integrand with line (polynomial of degree = 1)2
This approach to deriving numerical integration approximations is hard to generalize to polynomials of degree higher than 1.

35. Derive Simpson’s rule with 1 pair of subintervals for range of integration [-a,a]
Find weights that give exact results for polynomials up to degree 2

36. Derive Simpson’s rule with 1 pair of subintervals for range of integration [-a,a]
Find weights that give exact results for polynomials up to degree 2

37. Composite Simpson’s rule:
Generalized result for 1 pair of subintervals on [-a,a] to
n pairs of subintervals on [a,b]

38. n pairs of subintervals

39. MatLab code for composite Simpson’s rule:
Layout of points when npairs =4 i=
|-----|------| | |------| |------| | n=2npairs=8
a a+h b

40. Review: MatLab code for composite Simpson’s rule: npairs =4
|-----|------| | |------| |------| | n=2npairs=8
a a+h b
function A = simprule(fh,a,b,npairs)
n=2*npairs; %number of sub intervals
h=(b-a)/n;
sum_even=0;
for i=2:2:n-2
sum_even=sum_even+fh(a+i*h);
end
sum_odd=0;
for i=1:2:n-1
sum_odd=sum_odd+fh(a+i*h);
A=h*(fh(a)+4*sum_odd+2*sum_even+fh(b))/3;

41. Assignment 6, Due 3/6/19
Approximate the integral
by Simpson rules with 3, 5, and 7 equally spaced points on [1,3]. Calculate the percent difference from the “exact” value, , in each case.
Compare with results using the trapezoid rule (assignment #5).

42. Let {x0, x1, …, xn} be n+1 points on [a, b] where integrand evaluated, then the numerical integration formula
Will be exact for polynomials of degree at least n.
Derive weights by a system of n+1 linear equations obtained by requiring the formula to be exact for
f(x) = 1, x, x2, …xn
Trapezoid rule: (n = 1) exact for a line
In general, error proportional to |f’’(z)|
Simpson’s rule: (n = 2) exact for a cubic
In general, error proportional to |f(4)(z)|

43. Gauss quadrature
Gauss quadrature extents the idea of Simpson’s rule to its logical extreme: Give up all freedom about where the integrand will be evaluated to maximize the degree of polynomial for which formula is exact.
Theorem: nodes {x0, x1, …, xn} exist on [a, b] such that
is exact for polynomials of degree at most 2n+1
Nodes are zeros of polynomial g(x) of degree n+1 defined by
for 0 < k < n
How many coefficients are in a polynomial of degree n+1?

44. Proof of Gauss quadrature theorem

45. Exact for any set of n+1 nodes on [a,b]
Exact for polynomial degree 2n+1

46. Example of Gauss quadrature
3 nodes on [-1, 1] such that approximation
Step in deriving this quadrature result
1. Find cubic g(x) such that
for 0 < k < 2
2. Find roots of g(x)
3. Find the weights A0, A1 and A2

47. Degree of g(x) = number of nodes
conditions on g(x) =
number of nodes
Number of conditions not
sufficient to determine all the
coefficients
Terms with odd powers of x
in g(x) do not contribute
due to symmetry

48. Determines ratio of C1 and C3 only.
As noted before, number of
conditions on g(x) insufficient
to determine all 4 coefficients
of a cubic but are sufficient to
determine the zeros

49. Using plan (b)
Compare with table 6.1, text p236

50. Table from C&K 6th ed. p236
Note symmetry of points and weights
1.odd number always contains 0
2.nonzero points symmetric about origin always have same weight

51. Points and weights of Gauss quadrature are determined for
Every integral from a to b can transformed into integral from -1 to 1

52. Nodes and weights determined
for [-1,1] can be used for [a,b]
Given values of y from table, this x(y)
determines values on [a,b] where
integrand will be evaluated

53. Example: Review: Gauss quadrature with n+1 points
Points and weights from 2, 3, 4, and 5 points:table 6.1, text p236
Example:

54. (no anti-derivative)
with 2 nodes

55. Assignment 7, Due 3/20/19:
Write MatLab functions for Gauss quadrature with 2, 3, 4 and
5 points. Apply these function to the integral
Take as the exact value. Hand in a copy of the command window where you called the Gauss functions and calculated the percent differences from the exact value (100|(exact-Gauss)/exact|). Compare percent differences using Gauss quadrature with 3 and 5 points to results using trapezoid and Simpson’s rules (HW 5 and 6).

56. Table from C&K 6th ed. p236
Note symmetry of points and weights
1.odd number always contains 0
2.nonzero points symmetric about origin always have same weight

57. fh(x);
fh(x);

59. MatLab code for Gauss Quadrature with 2, 3, 4 and 5 point
function [g2,g3,g4,g5]=GQ2345(fh,a,b);
c1=(b-a)2; c2=(a+b)2;
y2=sqrt(1/3);
g2=fh(c2+c1*y2)+fh(c2-c1*y2);
y3=sqrt(3/5); w3=5/9; g3=fh(c2)*8/9;
g3=g3+w3*(fh(c2+c1*y3)+fh(c2-c3*y3));
y4(1)=sqrt((3-4*sqrt(0.3))/7); y4(2)=sqrt((3+4*sqrt(0.3))/7);
w4(1)=0.5+sqrt(10/3)/12; w4(2)=0.5-sqrt(10/3)/12; g4=0;
for i=1:2
g4=g4+w4(i)*(fh(c2+c1*y4(i))+fh(c2-c1*y4(i));
end
y5(1)=sqrt((5-2*sqrt(10/7))/9); y5(2)=sqrt((5+2*sqrt(10/7))/9);
w5(1)=0.3*(5*sqrt(0.7)-0.7)/(5*sqrt(0.7)-2));
w5(2)=0.3*(5*sqrt(0.7)+0.7)/(5*sqrt(0.7)+2)); g5=fh(c2)*128/225;
g5=g5+w5(i)*((fh(c2+c1*y5(i))+fh(c2-c1*y5(i));
g2=c1*g2; g3=c1*g3; g4=c1*g4; g5=c1*g5;

60. More extensive tables of nodes and weights than given in text.
Note that symmetry used to compress the table
Verify that values for n = 2 to 5 are the same as n = 1 to 4 in table on
P236 of text

61. Why is the integral
hard to evaluate by numerical integration?
Does the integral have a finite value?
If so, can it be estimated by trapezoid rule,
Simpson’s rule and Gauss quadrature?

62. Change the variable of integration in
by the transformation y = ln(1/x) = -ln(x)

63. Change the variable of integration in
y = ln(1/x) = -ln(x)
when x=0 y=infinite
when x=1 y=0
dy = -dx/x
x(y) = exp(-y)

64. By arguments similar to the Guass-quadrature theorem,
Laguerre derived an approximation to infinite integrals
in which the integrand is evaluated at the zeroes of the
Laguerre polynomials
This formula only can be applied when f(x) is rapidly
decreasing with increasing x, which is required for the integral to be finite

65. Note:
Only applies to 1 type of integral
Less symmetry than with Gauss quadrature
Use first and last columns

66. Pseudo-code LQ2345(f)
% Leguerre quadrature for integral between 0 and infinity with 2-5 points
enter points and weights from table
initialize sum
for each number of points
calculate integrand where it is to be sampled
calculate weighted average of integrand samples
return values

68. Gauss with x as integration variable
Laguerre with y = -ln(x) as integration variable