Chapters 5 and 6: Numerical Integration

Definition of Integral by Upper and Lower Sums n=number of partitions

Application to monotone decreasing function Divide [a,b] into n subintervals using n+1 equally spaced points h=(b-a)/n is subinterval width

“composite trapezoid rule”

How would this change for monotone increasing function? Divide [a,b] into n subintervals using n+1 equally spaced points h=(b-a)/n is subinterval width How would this change for monotone increasing function?

How would this change for monotone increasing function? “composite trapezoid rule” How would this change for monotone increasing function?

Derive the trapezoid rule for any f(x)

Replace f(x) on [a,b] by line p(x) Area under p(x) on [a,b] is a trapezoid = (b-a)(f(a)+f(b))/2 Pseudo code for trapezoid rule with arbitrarily spaced points Trap_Arbitrary_Points(x,f) n=length(x); sum=0; for i=2 to n sum=sum+(xk – xk-1)(fk + fk-1)/2; Usually x1 = a and xn = b

Pseudo code for trapezoid rule with arbitrarily spaced points Trap_Arbitrary_Points(x,f) n=length(x); sum=0; for i=2 to n sum=sum+(xk – xk-1)(fk + fk-1)/2; Write MatLab code for Trap_Arbitrairy_Points(x,f) Test on with 10 logarithmically spaced points between 1 and 5

Replace f(x) on [a,b] by line p(x) “composite” trapezoid rule is a special case of Trap-Arbitrary-Points Samples of the integrand are equally spaced in the range of integration Pseudo code for composite trapezoid rule Composite_Trap_Rule(fh,a,b,npts) x=linspace(a,b,npts); call Trap_Arbitrary_Points(x,fh(x));

Pseudo code for composite trapezoid rule Composite_Trap_Rule(fh,a,b,npts) x=linspace(a,b,npts); call Trap_Arbitrary_Points(x,fh(x)) Write MatLab code for Composite_Trap_Rule(fh,a,b,npts) Test on with 10 equally spaced points between 1 and 5

For given number of points, accuracy of numerical integration depends on where the integrand is evaluated Accuracy of numerical integration also depends on the variable of integration Consider approximated by the trapezoid rule with (a) equally spaced points in the interval 1<x<5 (b) logarithmically spaced points in the interval 1<x<5 (c) equally spaced points with integration variable y = ln(x)

Changing variable of integration

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)

Compare 3 ways to evaluate

Replacing f(x) by polynomial p(x) is a common numerical integration strategy Area under p(x) on [a,b] is integral of the polynomial between a and b, which can be done analytically if we know the coefficients. Given values of the integrand f(x) at n points we can determine a polynomial of degree n-1 that passes through those points. If we are given only f(a) and f(b) we can determine the equation of a line (degree=1) that passes through them.

Which is the same result that we got using the area of trapezoid

Replace f(x) on [a,b] by line p(x) Pseudo code for composite trapezoid rule without call to Trap_Arbitrary_Points CTR(fh,a,b,npts) h=(b-a)/(npts-1) CTR=(fh(a)+fh(b))h/2 if npts > 2 CTR=CTR+h(sum of “internal” samples of integrand) Every numerical integration formula has the form: (range of integration) (average of integrand on that range)

Use Taylor formula to derive an expression for the expected error in the trapezoid rule for any function with a continuous second derivative

Error in Trap Rule: single subinterval of size h Taylor-series expression for exact result

Error in Trap Rule: single subinterval of size h From previous slide Difference between exact and trapezoid rule

Error in Trap Rule: n subinterval of size h 1 n is number of subinterval Equals number of points -1

For the trapezoid rule Error estimate e = (b – a)3|f ‘’(x)|/12n2 applies to any continuous integrand f(x) on [a,b] Error estimate e = (b – a)[f(b) - f(a)]/2n applies to any monotone increasing integrand f(x) on [a,b] Error estimate e = (b – a)[f(a) - f(b)]/2n applies to any monotone decreasing integrand f(x) on [a,b] Exact error can evaluated when the anti-derivative of the integrand f(x) is known

Assignment 2, Due 2/5/15 Use the error formulas derived in class for the composite trapezoid rule applied to the integral to obtain 2 estimates of the number of sub-intervals needed to have an error less than 10-4 Write a MabLab function for the composite trapezoid rule with function handle and number of subintervals as parameters. Use this code to discover which of the error estimates is more realistic.

Log10(|traprule-exact|)

Problems from text on application of error formulas

Problem 5.1-6 text p 188 Integrand is monotone decreasing. Estimate the number of points needed for accuracy 0.5x10-4 using the error formula derived from upper and lower sums.

Problem 5.1-6 text p 188 n > (b-a) [f(a)-f(b)]/2e n > (5-2) [1/log10(2)-1/log10(5)]/(1x10-4 ) = 56,738

Problem 5.2-2 text p 200 Estimate the error if the integral is evaluated by the composite trapezoid rule with 3 points |error| < (b-a)3|f’’(x)|max /12n2 Plot |f’’(x)| to estimate maximum value on [0, 1]

f ``(x) = (6x2 -2)/(x2 +1)3 |f’’(x)| X

Problem 5.2-2 text p 200 Estimate the error if the integral is evaluated by the composite trapezoid rule with 3 points |error| < (b-a)3|f’’(x)|max /12n2 = (1)(2)/((12)(9)) = 1/54 Correct answer is 1/24. What is wrong?

Most numerical integration formulas have the form Example: Trapezoid rule w0 = wn = 1/2n wk = 1/n if 1< k<n-1 n = number of intervals One approach to deriving such formulas: Replace the integrand by the Lagrange interpolation formula

p(x) “Lagrange interpolation formula” is a polynomial of degree n that passes through a set of n+1 user-supplied point {xk,f(xk)}. p(x) Regardless of where {xk} are in [a,b], this integration formula will be exact for polynomials of degree up to n. Usually {xk} are chosen to make the formula exact for higher-degree polynomials.

Simpson’s rule for one pair of subintervals

Same result by simpler approach

n pairs of subintervals

Problem 6.1-4 text p228 Evaluated by composite Simpson’s rule with 5 points Bound the error |error| = (b-a)h4|f(4)(x)|/180 a < x < b (text p221) h = width subintervals Since |error| proportional to |f(4)(x)|, Simpson’s rule is exact for a cubic, one degree higher than we should expect from number of samples of the integrand

Problem 6.1-4 text p228, 6th ed Evaluated by composite Simpson’s rule with 5 points Bound the error |error| = (b-a)h4|f(4)(x)|/180 a < x < b (text p221) h = width subintervals In this case, h = 0.25 f(4)(x) = 24/x5 |f(4)(x)| < 24 1<x<2 |error| = (1)(0.25)4 24/180 = 0.00052

Another example of deriving integration formulas Why is this different from Simpson’s rule?

Same example by simpler approach

Which is likely to be more accurate?

Problems from text p240 on deriving integration formulas 6.2-5 6.2-6 6.2-9

Example 6.2-6 text p240 Find A and B by requiring formula to be exact for f(x) = 1 and f(x) = x

Example 6.2-6 text p240 f(x) = 1, f `(x) = 0, f(x) = x, f `(x) = 1, B = (b2 - a2)/2 - Aa = (b2-a2)/2 - (b-a)a

Assignment 3, Due 2/12/15: 1. How many samples of the integrand are required to estimate with error less than 0.001 by the composite trapezoid rule and the composite Simpson’s rule. Show all derivation steps 2. Problem 6.2.5 p240, 6th edition; exercise 5.4.5 p248, 7th edition. Show all derivation steps.

Let {x0, x1, …, xn} be n+1 points on [a, b] where integrand evaluated. Will be exact for polynomials of degree n Trapezoid rule: (n = 1) exact for a line In general, error proportional to |f’’(z)| Simpson’s rule: (n = 2) exact for a parabola In general, error proportional to |f(4)(z)| How could this happen?

0 < k < n number of nodes = degree of polynomial Gauss quadrature extents Simpson’s rule idea 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. 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 0 < k < n number of nodes = degree of polynomial

Proof of Gauss quadrature theorem

For any set of n+1 nodes on [a,b] zeros

For any set of n+1 nodes on [a,b], Summary of Gauss Quadrature Theorem For any set of n+1 nodes on [a,b], is exact for polynomials up to degree n For the special set of node in Gauss quadrature, the formula is correct of polynomials of degree at most 2n+1 Nodes and weights, Ai, depend on both [a,b] and n Scaling property of integrals lessens the impact of these dependences

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

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

Using plan (b)

Using plan (b)

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

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

From text p236, 6th edition Note: n = number of intervals Example: note symmetry of points and weights

(no anti-derivative) with 2 nodes

Pseudo-code GQ2345(a,b,f) % Gauss quadrature for integral of f between a and b with 2-5 points calculate scale factor and coefficients in x(y) for each number of points calculate points and weights by formulas in text calculate where in [a,b] integrand will be sampled calculate samples of integrand calculate weighted average of integrand samples multiply by scale factor return values Including symmetry of points and weights is optional

Assignment 4, Due 2/19/15: Write MATLAB codes for composite Simpson’s rule and Gauss-quadrature Make a table that compares the relative error (percent difference from exact value) in approximated by the composite trapezoid rule, the composite Simpson’s rule and Gauss quadrature when the integrand is evaluated at 2, 3, 4, and 5 points, if possible.

Quiz #4 2-19-15 Deriving integration formulas

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

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?

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

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

Note less symmetry than with Gauss quadrature Use first and last columns For n > 6 see me

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

Assignment 5, Due 2/24/15: Write MATLAB codes to approximate the integral after the transformation y = ln(1/x) changes it into a form where Laguerre integration can be applied. Make a graph that compares, as a function of the number of points where the integrand is evaluated, the relative error of the Laguerre approximation with results obtained when the integral is evaluated by Gauss quadrature without a change of variable.

Suggested problems from the text on numerical integration   Chapter 5.1 Problems 1, 6 Computer problem 2 Chapter 5.2 Problems 2, 4, 7, 12, 13, 15, 16 Computer problems 2 and 5(by Laguerre and Gauss quadrature) Chapter 6.1 Problems 2, 4, Chapter 6.2 Problems 1, 5, 6, 9, 11, 12 Computer problems 2, 3, 6, 8, 10

Computer problems 6.2-2 and 6.2-3 p241 Formula (5) p234 is 3-point Gauss quadrature

exact=2 quad=1.418302 Trap rule 800 =0.946803

CP 5.1-2 If the integrand cannot be defined in the command window with an inline statement, use an m-file. Use @ to inform MatLab.

Computer problem 5.2-2 p203 Approximate by Gauss quadrature

= p/2 Computer problem 5.2-5 p204 = sqrt(p/2) = sqrt(p/2)/2 Approximate by Laguerre quadrature

= p/2 Computer problem 5.2-5 p204 = sqrt(p/2) = sqrt(p/2)/2 Approximate by Laguerre quadrature

Computer problem 6.2-6 p242 By 32-point trapezoid rule and 2-point Gauss quadrature

Computer problem 6.2-8 p242

Computer problem 6.2-8 p242 quad=2.0348159 quad=0.4268024 quad with upper limit of 10 =0.89298

Computer problem 6.2-8 p242

Computer problem 6.2-10 p243