Derivatives and Intergration, Lecture notes of Numerical Methods in Engineering

Download Derivatives and Intergration and more Lecture notes Numerical Methods in Engineering in PDF only on Docsity!

NUMERICAL DIFFERENTIATION AND INTEGRATION

Hoang Hai Ha

CONTENT

1 NUMERICAL DIFFERENTIATION

FINDING DERIVATIVE IN GENERAL

Given a function y = f ( x ) by the table:

xk x 0 x 1 · · · xn yk = f ( xk ) y 0 y 1 · · · yn

To find f ′( ¯ x ), we interpolate it by a function G ( x ) and apply

f ′( ¯ x ) ≈ G ′( ¯ x )

DIFFERENCE FORMULA FOR 2 POINTS

Consider the table of data points for y = f ( x ):

x x 0 x 1 y y 0 y 1 where y 0 = f ( x 0 ) and y 1 = f ( x 1 ) = f ( x 0 + h ). The polynomial interpolation Lagrange is

L ( x ) =

x − x 0 h

y 1 −

x − x 1 h

y 0 ,

where h = x 1 − x 0. Then, for all ∀ x ∈ [ x 0 , x 1 ] we have

f ′( x ) ≈

y 1 − y 0 h

f ( x 0 + h ) − f ( x 0 ) h

Forward difference formula:

f ′( x 0 ) ≈

y 1 − y 0 h

f ( x 0 + h ) − f ( x 0 ) h

Backward difference formula:

f ′( x 0 ) ≈

f ( x 0 ) − f ( x 0 − h ) h

EXAMPLE 1.

Use forward and backward difference formulas to approximate derivative of f ( x ) = ln x at x 0 = 1.8 with h = 0..

THE CASE OF 3 NODES

Consider the table where nodes are equally spaced x x 0 x 1 x 2 y y 0 y 1 y 2 where y 0 = f ( x 0 ), y 1 = f ( x 1 ) = f ( x 0 + h ), y 2 = f ( x 2 ) = f ( x 0 + 2 h ) Polynomial interpolation Lagrange for this table is

L ( x ) ==

( x − x 0 )( x − x 1 ) 2 h^2

y 2 −

( x − x 0 )( x − x 2 ) h^2

y 1 +

( x − x 1 )( x − x 2 ) 2 h^2

y 0.

L ′( x ) =

x − x 0 2 h^2

( y 2 − 2 y 1 ) +

x − x 1 h^2

( y 2 + y 0 ) +

x − x 2 2 h^2

( y 0 − 2 y 1 )

L ′′( x ) =

y 2 − 2 y 1 + y 0 h^2

The formula

f ′( x 0 ) ≈ L ′( x 0 ) =

− 3 y 0 + 4 y 1 − y 2 2 h

is called forward difference formul, and rewritten by

f ′( x 0 ) =

− 3 f ( x 0 ) + 4 f ( x 0 + h ) − f ( x 0 + 2 h ) 2 h

At x 1 , f ′( x 1 ) ≈ L ′( x 1 ) =

y 2 − y 0 2 h

is called centered difference formula

and written in the form

f ′( x 0 ) ≈

f ( x 0 + h ) − f ( x 0 − h ) 2 h

EXAMPLE 1.

Compute y ′(1) with h = 0.5 for function y = e − x^ , using center difference formula, compare to the exact value(take the value found by the pocket calculator as the exact value.)

EXAMPLE 1.

Compute y ′(1) with h = 0.5 for function y = e − x^ , using center difference formula, compare to the exact value(take the value found by the pocket calculator as the exact value.)

SOLUTION

f ′(1) =

f (1 + 0.5) − f (1 − 0.5) 2 × 0.

Exact value: −0.367879441171442.

NUMERICAL INTEGRATION

We usually find a definite integral by Newton-Leibniz formula Z (^) b

a

f ( x ) d x = F ( x )

b a

= F ( b ) − F ( a ), F ′( x ) = f ( x ).

But this method does not work in two cases: it is impossible to find F or we don’t know f ( x ) explicitly. ALTERNATIVE METHOD We replace f ( x ) by its interpolation polynomial Pn ( x ) on [ a , b ] and approximate (^) Z b

a

f ( x ) d x ≈

Z (^) b

a

Pn ( x ) d x

TRAPEZOIDAL FORMULA

To derive the approximation for

Z (^) b

a

f ( x ) d x we replace f ( x ) by its polynomial interpolation passing through 2 points ( a , f ( a )) và ( b , f ( b )) Then

P 1 ( x ) = f ( a ) +

f ( b ) − f ( a ) b − a

( x − a )

TRAPEZOIDAL FORMULA

To derive the approximation for

Z (^) b

a

f ( x ) d x we replace f ( x ) by its polynomial interpolation passing through 2 points ( a , f ( a )) và ( b , f ( b )) Then

P 1 ( x ) = f ( a ) +

f ( b ) − f ( a ) b − a

( x − a )

Z (^) b

a

P 1 ( x ) d x = f ( a ) x +

f ( b ) − f ( a ) b − a

μ x^2 2

− ax

b

a

We get the following which is called the Trapezoidal formula

Z (^) b

a

f ( x ) d x ≈

b − a 2

f ( a ) + f ( b )

COMPOSITE TRAPEZOIDAL FORMULA

Dividing [ a , b ] into n sub-intervals with step h =

b − a n

. Then a = x 0 , x 1 = x 0 + h ,... , xk = x 0 + kh ,... , xn = x 0 + nh và yk = f ( xk ), k = 0, 1,... , n Applying the trapezoidal rule for each interval [ xk xk + 1 ]

Z^ b

a

f ( x ) d x =

Z^ x^1

x 0

f ( x ) d x +

Z^ x^2

x 1

f ( x ) d x +... +

Z^ xn

xn − 1

f ( x ) d x

≈ h ·

y 0 + y 1 2

• h ·

y 1 + y 2 2

+... + h ·

yn − 1 + yn 2

h 2

y 0 + 2 y 1 + 2 y 2 + .... + 2 yn − 1 + yn