NUMERICAL DIFFERENTIATIONNUMERICAL DIFFERENTIATIONThe derivative of f (x) at x0 is:

h

xfhxfxf Limith

)( 0000

An approximation to this is:

h

xfhxfxf )( 000

for small

values of h.

Forward Difference Formula

810 . and ln)(Let xxxfFind an approximate value for

81.f

0.1 0.5877867

0.6418539

0.5406720

0.01 0.5877867

0.5933268

0.5540100

0.0010.5877867

0.5883421

0.5554000

).( 81f ).( hf 81 hfhf ).().( 8181

h

The exact value of

555081 .. f

Assume that a function goes through three points:

., and ,,, 221100 xfxxfxxfx

)()( xPxf

221100 xfxLxfxLxfxLxP

Lagrange Interpolating Polynomial

21202

10

12101

20

02010

21

))(())((

))(())((

))(())((

xfxxxx

xxxx

xfxxxx

xxxx

xfxxxx

xxxxxP

221100 xfxLxfxLxfxLxP

)()( xPxf

21202

10

12101

20

02010

21

))((2

))((2

))((2

xfxxxx

xxx

xfxxxx

xxx

xfxxxx

xxxxP

20000

000

10000

000

00000

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

2212020 22

23 xf

hhxf

hhxf

hhxP

2100 4321 xfxfxfh

xP

hxfhxfxfh

xf 24321

0000

Three-point formula:

hxxhxx 0201 and If the points are equally spaced with x0 in the middle:

20000

000

10000

000

00000

0000

)()()()(2

)()()()(2

)(()()()(2

xfhxhxxhx

hxxx

xfhxhxxhx

hxxx

xfhxxhxx

hxhxxxP

2212020 220 xf

hhxf

hhxf

hxP

hxfhxfh

xf 000 21

Another Three-point formula:

Alternate approach (Error estimate)Take Taylor series expansion of f(x+h) about x: xfhxfhxfhxfhxf 3

32

2

!32

xfhxfhxfhxfhxf 33

22

!32

xfhxfhxf

hxfhxf 3

22

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

hOxfh

xfhxf

hOh

xfhxfxf

h

xfhxfxf Forward

Difference Formula

xfhxfhhO 32

2

!32

xfhxfhxfhxfhxf 33

22

!38

2422

xfhxfhxfhxfhxf 33

22

!38

2422

xfhxfhxf

hxfhxf 3

22

!34

22

22

..........

....... (2)

xfhxfhxf

hxfhxf 3

22

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

xfhxfhxf

hxfhxf 3

22

!34

22

22

..........

....... (2)

2 X Eqn. (1) – Eqn. (2)

xfhxfhxf

hxfhxf

hxfhxf

43

32

!46

!32

222

24

33

2

!46

!32

2342

hOxf

xfhxfhxf

hxfhxfhxf

22342 hOxf

hxfhxfhxf

22342 hO

hxfhxfhxfxf

h

xfhxfhxfxf 2342

Three-point Formula xfhxfhhO 4

33

22

!46

!32

Take Taylor series expansion of f(x+h) about x: xfhxfhxfhxfhxf 3

32

2

!32Take Taylor series expansion of f(x-h) about x: xfhxfhxfhxfhxf 3

32

2

!32

The Second Three-point Formula

Subtract one expression from another

xfhxfhxfhhxfhxf 66

33

!62

!322

xfhxfhxfhhxfhxf 66

33

!62

!322

xfhxfhxf

hhxfhxf 6

53

2

!6!32

xfhxfhh

hxfhxfxf 65

32

!6!32

22 hOh

hxfhxfxf

xfhxfhhO 65

32

2

!6!3

h

hxfhxfxf 2

Second Three-point Formula

h

xfhxfxf Forward

Difference Formula

Summary of Errors

xfhxfhhO 32

2

!32Error term

h

xfhxfhxfxf 2342

First Three-point Formula

Summary of Errors continued

xfhxfhhO 43

32

2

!46

!32Error

term

h

hxfhxfxf 2

Second Three-point Formula

xfhxfhhO 65

32

2

!6!3Error term

Summary of Errors continued

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

Using the Forward Difference formula:

0 0 01f x f x h f xh

12 21 2011 17148957 1477811201

23708450

ff . f.

. ...

Using the 1st Three-point formula:

032310.22 855030.19

148957.174778112.1432.01

)2.2()1.2(4)2(31.0212

ffff

hxfhxfxfh

xf 24321

0000

Using the 2nd Three-point formula:

228790.22

703199.12148957.172.01

)9.1()1.2(1.0212

fff

The exact value of

22.167168 :is 2f

hxfhxfh

xf 000 21

Comparison of the results with h = 0.1

Formula ErrorForward 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 is 22.167168

Second-order Derivative

xfhxfhxfhxfhxf 33

22

!32

xfhxfhxfhxfhxf 33

22

!32Add these two equations.

xfhxfhxfhxfhxf 44

22

!42

222

2 4

2 42 22 2 4!h hf x h f x f x h f x f x

xfhxf

hhxfxfhxf 4

22

2 !422

xfhh

hxfxfhxfxf 42

22

!422

2

2 2h

hxfxfhxfxf

NUMERICAL INTEGRATIONNUMERICAL INTEGRATION

b

a

dxxf )( area under the curve f(x) between . to bxax

In many cases a mathematical expression for f(x) is unknown and in some cases even if f(x) is known its complex form makes it difficult to perform the integration.

y

xx 0 = a x 0 = b

f(a)

f(b)f(x)

y

xx 0 = a x 0 = b

f(a)

f(b)

f(x)

Area of the trapezoidThe length of the two parallel sides of the trapezoid are: f(a) and f(b)

The height is b-a

)()(

)()()(

bfafh

bfafabdxxfb

a

2

2

Simpson’s Rule:

2

0

2

0

x

x

x

xdxxPdxxf )()(

hxxhxx 20201 and

2 2

0 0

2

0

2

0

1 20

0 1 0 2

0 21

1 0 1 2

0 12

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 )

)()(

)()(

210 43

2

0

2

0

xfxfxfh

dxxPdxxfx

x

x

x

y

xx 0 = a x 0 = b

f(a)

f(b)

f(x)

y

xx 0 = a x 0 = b

f(a)

f(b)

f(x)

y

xx 0 = a x 0 = b

f(a)

f(b)

f(x)

y

xx 0 = a x 0 = b

f(a)

f(b)

f(x)

Composite Numerical Integration

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.

In Left Riemann sum, the left-side sample of the function is used as the height of the individual rectangle.

a b

Left Riem ann Sumy

xx

1

2

3 2

1i

x b a / nx ax a xx a x

x a i x

1

nbia i

f x dx f x x

In Right Riemann sum, the right-side sample of the function is used as the height of the individual rectangle.a b

Right Riem ann Sumy

xx

1

2

3

23

i

x b a / nx a xx a xx a x

x a i x

1

nbia i

f x dx f x x

In the Midpoint Rule, the sample at the middle of the subinterval is used as the height of the individual rectangle.

a b

M idpoint Ruley

xx

1

2

3

2 1 1 22 2 1 22 3 1 2

2 1 2i

x b a / nx a x /x a x /x a x /

x a i x /

1

nbia i

f x dx f x x

Composite Trapezoidal Rule:Divide the interval into n subintervals and apply Trapezoidal Rule in each subinterval.

)()()( bfxfafhdxxfn

kk

b

a

1

122

where

nkkhaxn

abh k , ... , , ,for and 210

by dividing the interval into 20 subintervals.

Find πsin

0(x)dx

2021020

20

20

....., , , , ,π

π

kkkhax

nabh

n

k

9958861202040

2219

1

1

10

.

πsinπsinsinπ

)()()sin(π

k

n

kk

k

bfxfafhdxx

Composite Simpson’s Rule:Divide the interval into n subintervals and apply Simpson’s Rule on each consecutive pair of subinterval. Note that n must be even.

)()(

)()(

/

)/(

bfxf

xfafhdxxf

n

kk

n

kk

b

a2

112

12

12

4

23

nkkhaxn

abh k , ... , , ,for and 210

where

by dividing the interval into 20 subintervals.

Find πsin

0(x)dx

2021020

2020

....., , , , ,π

π

kkkhax

nabhn

k

000006220124

2022060

10

1

9

10

.

)πsin(π)(sin

πsinsinπ)sin(π

k

k

k

kdxx