W10 Numerical Calculus

8/2/2019 W10 Numerical Calculus

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Centre for Computer Technology

ICT114Mathematics for

Computing

Week 10

Numerical Differentiation andIntegration


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March 20, 2012 Copyright Box Hill Institute

Objectives

Review week 9

Numerical Differentiation

Newtons Forward Difference formula Newtons Backward Difference formula

Numerical Integration

Trapezoid rule

Simpsons one third rule


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Newton's Method

Using an initial guess at the root and the

slope of f(x), Newton's method usesextrapolation to estimate where f(x)crosses the x axis. This method converges

very quickly, but it can diverge if f(x) = 0 isencountered during iterations.

(f(x) is the differential of f(x))


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Algorithm

initialize: x1 = . . .

for k= 2, 3, . . .xk= xk-1- f(xk-1)/f(xk-1)

if converged, stop

end


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Secant Method

The secant method approximates f(x)

from the value of f(x) at two previous

guesses at the root. It is as fast as theNewton's method but can also fail atf(x)=0.


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Algorithm

initialize: x1 = . . ., x2 = . . .

for k= 2, 3 .. .

xk+1 = xk- f(xk)(xk- xk-1)/(f(xk) - f(xk-1))If f(xk+1)


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Newtons ForwardDifference Formula


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Numerical Differentiation

We discuss Newtons forward

difference formula in detail. This is suitable for differentiation for

the the values towards the beginning

of the table


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Forward Difference formula (1)

For functions tabulated with constant interval h,

E f(x) = f (x+h)

E2 f(x)= E (E f(x)) = E f(x+h)=f(x+2h) Like this, Epf(x) = f (x + ph) Again, f(x) = f(x+h) - f(x) Hence f(x+h) = f (x) + f(x)

= ( 1 + ) f(x) That is, E f(x) = ( 1 + ) f(x) or simply, E ( 1 + )


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Forward Difference formula (2)

Hence,

f(x0 +p.h) = Epf (x0)

= (1 +)p f(x0) = ( 1 + p +pC2

2 +pC33+ .... ) f(x0)

= ( 1 + p + p(p-1)/2! 2

+ p(p-1)(p-2)/3! 3+ ) f(x0)


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Forward difference formula (3)

Putting, x = x0 + ph,

df df dp 1 df

---- = ---- . ---- = ---- . -----

dx dp dx h dp


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Forward difference formula (4)

So, f/(x)= (1/h) [ + (2p-1)/2. 2

+ (3p2 - 6p +2)/6 3

+ (4p3

-18p2

+22p-6)/24 4

+] Putting p = 0,

f/(x)= (1/h) [ 1/22 ++1/3

31/44+.]

This is Newtons forward differenceformula for differentiation suitable forvalues given in the table


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An example : f(x) = ex (1)

x f(x)

1.0 2.718282

1.2 3.320117

1.4 4.055200

1.6 4.953032

1.8 6.049646


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x f(x)

1.0 2.718282

0.601835

1.2 3.320117

0.735083

1.4 4.055200

0.897832

1.6 4.953032

1.096615

1.8 6.049646



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March 20, 2012Copyright Box Hill Institute

x f(x) 2

1.0 2.718282

0.601835

1.2 3.320117 0.133248

0.735083

1.4 4.055200 0.162749

0.897832

1.6 4.953032 0.198783

1.096615

1.8 6.049646



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x f(x) 2 3

1.0 2.718282

0.601835

1.2 3.320117 0.133248

0.735083 0.029501

1.4 4.055200 0.162749

0.897832 0.036034

1.6 4.953032 0.198783

1.096615

1.8 6.049646



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x f(x) 2 3 4

1.0 2.718282

0.601835

1.2 3.320117 0.133248

0.735083 0.029501

1.4 4.055200 0.162749 0.006532

0.897832 0.036034

1.6 4.953032 0.198783

1.096615

1.8 6.049646



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Example (continued)

To find the differential coefficient for x=1.0

From the table, = 0.601835, 2 = 0.133248

3

= 0.029501, and 4

= 0.006532 Here h = 0.2

The approximate value of the diff coeff =

(1/0.2) [ 1/2

2 +1/3

31/4

4 ]

= 2.717060

The true value is 2.718282


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Formula for value not in table

Suppose we want to find derivative at apoint not given in the table,say at x=1.1

So, ph = 0.1. Then, as h=0.2, p=0.5

Putting p=0.5,we get

f/(x)= (1/h) [ + 0 * 20.25/6 3+ 1/24 4+]


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Forward difference formula

So,

f/(1.1)= (1/0.2) [ + - 0.25/6 3

+ 1/12 4+]= 5*[0.601835 - 0.029501* 0.25/6 +

0.006532/12]

= 3.005750The true value is = 3.004166


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Newtons BackwardDifference Formula


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Backward difference formula

Backward difference formula is given by

f/(x)= (1/h) [ +1/22+1/3

3

+ 1/44+..] Where, f(x) = f(x) f(x-h),

2 f(x) = ( f(x)), and so on

This is suitable for finding derivativetowards the end of the table


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An example : f(x) = ex

x f(x) 2 3 4

1.0 2.718282

0.601835

1.2 3.320117 0.133248

0.735083 0.029501

1.4 4.055200 0.162749 0.006532

0.897832 0.036034

1.6 4.953032 0.198783

1.096615

1.8 6.049646


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Backward difference formula

To find the differential coefficient for x=1.8

From the table, = 1.096615, 2 = 0.198783

3

= 0.036033, and

4

= 0.006532 Here h = 0.2

The approximate value of the diff coeff =

(1/0.2) [ +1/2

2 +1/3

3 +1/4

4 ]

= 6.048252

The true value is 6.049647


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Some remarks

There is a central difference formula forfinding differential coefficient when the

values are around middle of the table. Wedid not discuss that.

It may be remembered that at timesnumerical differentiation might be veryinaccurate when there are largefluctuations.


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Trapezoid Rule


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March 20, 2012

Copyright Box Hill Institute

Numerical Integration

Value of a definite integral within its limitsis the area under the curve in the limits

In numerical integration, the function isapproximated by a polynomial, and thearea under the polynomial is taken as thevalue of the integral

We study two simple rules (1) Trapezoidalrule and (2) Simpson's one-third rule


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Trapezoid rule (1)

Area below the curve is approximated by aTrapezium

f(x)

a b

Value of integral of f (x) between a and b is the area

under the curve between a and b


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Trapezoid rule (2)

Area below the curve is approximated by aTrapezium

f(x)

a b

Value of the integral is approximated as

= area of the trapezium

= [f(a) +f(b)] . (ba)


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Trapezoid rule (3)

Now two intervals

Trapezoid Rule for two intervals

f(x)

a0=a a1 a2 = b

Value of the integral is better approximated by

= area of trapezium 1 + area of trapezium 2

= [f(a0) +f(a1)].(a1a0)+ [f(a1) +f(a2)](a2a1)


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Trapezoid rule (4)

If a1=a0 + h , and, a2 =a1+ h= a0 +2h,

The approximate area under the curve(when there are two intervals)

= h/2 [ f(a0) + f(a1) ] +h/2 [ f(a1) + f(a2) ]

= h/2 [ f(a0) + 2 f(a1) + f(a2) ]


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Trapezoid rule (5)

When there are n equidistant intervals, theapproximate value of the integral is equal

to =(h/2) [ f(a0) + 2 f(a1) + 2f(a2) +.

+ 2 f(an-1) + f(an)]


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x f(x)

1.0 2.718282

1.2 3.320117

1.4 4.055200

1.6 4.953032

1.8 6.049646


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Example (continued)

To evaluate the value of integral

f(x) = exp(x) between x= 1.0 and 1.8

Suppose we take interval h = 0.4 Then a0 = 1.0, a1 =1.4, a2= 1.8

The value is given by

(0.4/2)[2.718282 + (2) 4.055200 + 6.049646]

= 3.375666


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Example (low interval length)

Suppose we take interval h = 0.2

Then a0 = 1.0, a1 =1.2, a2= 1.4 , a3 =1.6, a4= 1.8


(0.2/2) [2.718282 + (2) 3.320117

+ (2) 4.055200 + (2) 4.953032

+ 6.049646 ]

= 3.342463

Actual value is 3.331366


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Simpsons One Third Rule


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Simpsons one third rule (1)

In trapeziod rule,the curve is approximatedby a line

In Simpsons rule , the curve isapproximated by a second degreepolynomial

It requires even number of intervals


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Consider three equidistant points, a ,(a+b)/2 and b, and a second degreepolynomial f(x)= c

0+c

1x +c

2x2

So, we have, f(a) = c0 +c1 a +c2 a2,

f((a+b)/2) = c0 +c1 ((a+b)/2) +c2 ((a+b)/2)2

And, f(b) = c0 +c1 b +c2 b2


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On integration of the second degreepolynomial , the value of the integralequals

c0 (b a) +c1 (b2 a2)/2 +c2 (b

3 - a3)/3

This can be shown equal to

((b a )/ 6)[f(a) + 4 f((a+b)/2) +f(b)]= ( h /3 ) [f(a) + 4 f((a+b)/2) +f(b)]


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If we denote the points as a0, a1, a2 and his the interval length, the value of theintegral is (h/3)[ f(a

0) + 4f (a

1) + f(a

2)]

In general case, the value equals

(h/3)[ (f(a0) + 4f (a1) + f(a2))

+ (f(a2) + 4f (a3) + f(a4))+ (f(a4) + 4f (a5) + f(a6)) + .]


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This may be compactly written as

(h/3)[ ( f(a0) + f (an) )

+ 4 (f(a1) + f (a3) + f(a5)+ f(an-1) )+ 2 (f(a2) + f (a4) + f(a6) + .. f(an-2) ) ]

Remember that the number of intervalshave to be even


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To evaluate the value of integral

f(x) = exp (x) between x= 1.0 and 1.8

Suppose we take interval h = 0.4 Then a0 = 1.0, a1 =1.4, a2= 1.8


(0.4/3)[2.718282 + (4) 4.055200 + 6.049646]

= 3.331831


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Example (less interval length)

Suppose we take interval h = 0.2

Then a0 = 1.0, a1 =1.2, a2= 1.4 , a3 =1.6, a4= 1.8


(0.2/3) [ (2.718282 + 6.049646)

+ 4 ( 3.320117 + 4.953032)

+ 2 ( 4.055200) ]

= 3.331395

As noted earlier, the actual value is 3.331366


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Summary

Newtons Forward Difference Formula

f/

(x)= (1/h) [ 1

/22 +

+1

/33

1/44+.]

Newtons Backward Difference Formulaf/(x)= (1/h) [ +1/22+1/3

3

+ 1/44+..]


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Summary

Trapezoid Rule

= (h/2) [ f(a0) + 2 f(a1) + 2f(a2) +.

+ 2 f(an-1) + f(an)]

Simpsons One Third Rule (the number ofintervals have to be even)

= (h/3)[ ( f(a0) + f (an) )+ 4 (f(a1) + f (a3) + f(a5)+ f(an-1) )

+ 2 (f(a2) + f (a4) + f(a6) + .. f(an-2) ) ]


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References

H L Verma and C W Gross : Introduction toQuantitative Methods,John Wiley

JB Scarborough : Numerical MathematicalAnalysis, Jon Hopkins Hall, New Jersey

Gerald W. Recktenwald, Numerical Methodswith MATLAB, Implementation and Application,Prentice Hall

Murray Spiegel, John Schiller, Alu Srinivasan,

Probability and Statistics, Schaums easyOutlines

http://mathworld.wolfram.com

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