w10 numerical calculus
TRANSCRIPT
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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
An example : f(x) = ex (2)
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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
An example : f(x) = ex (3)
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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
An example : f(x) = ex (4)
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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
An example : f(x) = ex (5)
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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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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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An example : f(x) = ex
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
The value is given by
(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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Simpsons one third rule (2)
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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Simpsons one third rule (3)
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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Simpsons one third rule (4)
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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Simpsons one third rule (5)
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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An example : f(x) = ex
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
(0.4/3)[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
The value is given by
(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