06 Numerical Differentiation Integration

8/11/2019 06 Numerical Differentiation Integration

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NUMERICAL DIFFERENTIATION

The derivative of f(x)at x0is:

h

xfhxfxf

Limit

h

)(00

00

An approximation to this is:

h

xfhxfxf

)(00

0

for small values of h.

Forward Difference

Formula


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810

.andln)(Let xxxf

Find an approximate value for 81.f

0.1 0.5877867 0.6418539 0.5406720

0.01 0.5877867 0.5933268 0.5540100

0.001 0.5877867 0.5883421 0.5554000

).( 81f ).( hf 81h

fhf ).().( 8181 h

The exact value of 555081 .. f


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Assume that a function goes through three points:

.,and,,, 221100 xfxxfxxfx

)()( xPxf

221100 xfxLxfxLxfxLxP

Lagrange Interpolating Polynomial


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2

1202

10

12101

20

0

2010

21

))((

))((

))((

))((

))(())((

xfxxxx

xxxx

xfxxxx

xxxx

xfxxxxxxxxxP

221100 xfxLxfxLxfxLxP


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)()( xPxf

2

1202

10

1

2101

20

0

2010

21

))((

2

))((

2

))((2

xf

xxxx

xxx

xfxxxx

xxx

xfxxxx

xxxxP


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2

0000

000

1

0000

000

0

0000

0000

)()2()2(

)(2

)2()()(

)2(2

)2()(

)2()(2

xfhxhxxhx

hxxx

xfhxhxxhx

hxxx

xfhxxhxx

hxhxxxP

If the points are equally spaced, i.e.,

hxxhxx 20201

and


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2212020

2

2

2

3xf

h

hxf

h

hxf

h

hxP

2100 432

1xfxfxf

hxP

hxfhxfxfhxf 2432

1

0000

Three-point formula:


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hxxhxx

0201and

If the points are equally spaced with x0in the middle:

2

0000

000

1

0000

000

0

0000

0000

)()()(

)(2

)()()(

)(2

)(()(

)()(2

xf

hxhxxhx

hxxx

xfhxhxxhx

hxxx

xfhxxhxx

hxhxxxP


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2212020

22

0xf

h

hxf

h

hxf

h

xP

hxfhxfh

xf 000

2

1

Another Three-point formula:


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Alternate approach (Error estimate)

Take Taylor series expansion of f(x+h) about x:

xfh

xfh

xfhxfhxf3

32

2

!32

xf

h

xf

h

xfhxfhxf

33

22

!32

xf

hxf

hxf

h

xfhxf 32

2

!32 .............. (1)


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hOxf

h

xfhxf

hOh

xfhxfxf

h

xfhxfxf

Forward DifferenceFormula

xfh

xf

h

hO

32

2

!32


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xfh

xfh

xfhxfhxf3

32

2

!3

8

2

422

xfh

xfh

xfhxfhxf3

32

2

!3

8

2

422

xfhxfhxfh

xfhxf 322

!3

4

2

2

2

2

................. (2)


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xf

hxf

hxf

h

xfhxf 32

2

!32.............. (1)

xf

hxf

hxf

h

xfhxf 32

2

!3

4

2

2

2

2

................. (2)

2 X Eqn. (1)Eqn. (2)


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xfhxfhxf

h

xfhxf

h

xfhxf

4

3

3

2

!46

!32

2

22

2

43

32

!4

6

!3

22

342

hOxf

xfh

xfh

xf

h

xfhxfhxf


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2

2

342hOxf

h

xfhxfhxf

22

342 hOh

xfhxfhxfxf

hxfhxfhxfxf

2342

Three-point Formula

xfh

xfh

hO4

33

22

!4

6

!3

2


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Take Taylor series expansion of f(x+h) about x:

xfh

xf

h

xfhxfhxf3

32

2

!32

Take Taylor series expansion of f(x-h) about x:

xfh

xfh

xfhxfhxf3

32

2

!32

The Second Three-point Formula

Subtract one expression from another

xfh

xfh

xfhhxfhxf6

63

3

!6

2

!3

22


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xfh

xfh

xfhhxfhxf6

63

3

!6

2

!3

22

xfh

xfh

xfh

hxfhxf 65

32

!6!32

xfhxfh

h

hxfhxfxf6532

!6!32


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2

2hO

h

hxfhxfxf

xfh

xfh

hO6

53

22

!6!3

h

hxfhxfxf

2

Second Three-point Formula


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h

xfhxfxf

Forward Difference

Formula

Summary of Errors

xfh

xfh

hO3

22

!32Error term


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h

xfhxfhxfxf

2

342

First Three-point Formula

Summary of Errors continued

xfh

xfh

hO4

33

22

!4

6

!3

2Error term


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h

hxfhxfxf

2

Second Three-point Formula

xfh

xfh

hO6

53

22

!6!3Error term

Summary of Errors continued


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Example:

xxexf

Find the approximate value of 2f

1.9 12.703199

2.0 14.778112

2.1 17.148957

2.2 19.855030

x xf

with 10.h


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Using the Forward Difference formula:

0 0 0

1f x f x h f x

h

12 2 1 2

0 11

17 148957 14 7781120 1

23 708450

f f . f

.

. ..

.


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Using the 1st Three-point formula:

032310.22

855030.19

148957.174778112.1432.0

1

)2.2()1.2(4)2(31.02

12

ffff

hxfhxfxf

h

xf 243

2

10000


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Using the 2nd Three-point formula:

228790.22

703199.12148957.172.0

1

)9.1()1.2(1.02

12

fff

The exact value of 22.167168:is2f

hxfhxf

h

xf 000

2

1


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Comparison of the results with h= 0.1

Formula Error

Forward Difference 23.708450 1.541282

1st Three-point 22.032310 0.134858

2nd Three-point 22.228790 0.061622

2f

The exact value of 2f is22.167168


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Second-order Derivative

xfh

xfh

xfhxfhxf3

32

2

!32

xfh

xfh

xfhxfhxf3

32

2

!32

Add these two equations.

xfh

xfh

xfhxfhxf4

42

2

!4

2

2

22

2 42 42 222 4!

h hf x h f x f x h f x f x


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NUMERICAL INTEGRATION

b

a

dxxf )( area under the curve f(x)between

.to bxax

In many cases a mathematical expression for f(x)isunknown and in some cases even if f(x)is known its

complex form makes it difficult to perform the integration.


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y

xx0= a x0= b

f(a)

f(b) f(x)


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y

xx0= a x0= b

f(a)

f(b)

f(x)


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Area of the trapezoid

The length of the two parallel sides of the trapezoid

are: f(a)and f(b)The height isb-a

)()(

)()()(

bfaf

h

bfafabdxxf

b

a

2

2


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Simpsons Rule:

2

0

2

0

x

x

x

x

dxxPdxxf )()(

hxxhxx 20201

and


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2 2

0 0

2

0

2

0

1 2

0

0 1 0 2

0 2

1

1 0 1 2

0 1

2

2 0 2 1

x x

x x

x

x

x

x

( x x )( x x )P x dx f x dx

( x x )( x x )( x x )( x x )

f x dx( x x )( x x )

( x x )( x x )f x dx

( x x )( x x )


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)()(

)()(

2104

3

2

0

2

0

xfxfxfh

dxxPdxxf

x

x

x

x


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y

xx0= a x0= b

f(a)

f(b)

f(x)


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y

xx0= a x0= b

f(a)

f(b)

f(x)


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y

xx0= a x0= b

f(a)

f(b)

f(x)


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y

xx0= a x0= b

f(a)

f(b)

f(x)


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Composite Numerical Integration


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Riemann Sum

The area under the curve is subdivided into n

subintervals. Each subinterval is treated as a

rectangle. The area of all subintervals are added

to determine the area under the curve.

There are several variations of Riemann sum as

applied to composite integration.

L ft Ri S


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In Left Riemann

sum, the left-

side sample of

the function is

used as the

height of the

individual

rectangle.

a b

Left Riemann Sumy

x

x

1

2

32

1i

x b a / n

x a

x a x

x a x

x a i x

1

nb

ia

i

f x dx f x x

Ri ht Ri S


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In Right Riemann

sum, the right-side

sample of the

function is used as

the height of the

individual rectangle.

a b

Right Riemann Sumy

x

x

1

2

3

2

3

i

x b a / n

x a x

x a x

x a x

x a i x

1

nb

ia

i

f x dx f x x

Mid i t R l


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In the Midpoint

Rule, the sample at

the middle of the

subinterval is usedas the height of the

individual

rectangle.

a b

Midpoint Ruley

x

x

1

2

3

2 1 1 2

2 2 1 2

2 3 1 2

2 1 2i

x b a / n

x a x /

x a x /

x a x /

x a i x /

1

nb

ia

if x dx f x x


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Composite Trapezoidal Rule:

Divide the interval into nsubintervals and apply

Trapezoidal Rule in each subinterval.

)()()( bfxfafh

dxxfn

kk

b

a

1

1

2

2

where

nkkhaxnabh

k ,...,,,forand 210


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by dividing the interval into 20 subintervals.

Find

sin0

(x)dx

20210

20

20

20

.....,,,,,

kk

khax

nabh

n

k


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9958861

20

20

40

2

2

19

1

1

10

.

sin

sinsin

)()()sin(

k

n

k k

k

bfxfafh

dxx

C it Si R l


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Composite Simpsons Rule:

Divide the interval into nsubintervals and apply

Simpsons Rule on each consecutive pair of subinterval.

Note that nmust be even.

)()(

)()(

/

)/(

bfxf

xfaf

h

dxxf

n

k

k

n

kk

b

a

2

1

12

12

12

4

23

where


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nkkhaxn

abh

k,...,,,forand 210

where

by dividing the interval into 20 subintervals.

Find

sin0

(x)dx

2021020

2020

.....,,,,,

kkkhax

n

abhn

k


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0000062

20

124

20

220

60

10

1

9

10

.

)sin()(

sin

sinsin

)sin(

k

k

k

kdxx

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06 Numerical Differentiation Integration