Differentiation-Discrete Functions - Numerical Methods - Lecture Slides, Slides of Mathematical Methods for Numerical Analysis and Optimization

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Differentiation-Discrete Functions

Forward Difference Approximation

x

f x x f x

x

f x

lim  

For a finite '^ Δ x '

x

f x x f x

f x

Example 1

The upward velocity of a rocket is given as a function of time in Table 1.

Using forward divided difference, find the acceleration of the rocket at.

t v(t) s m/s 0 0 10 227. 15 362. 20 517. 22.5 602. 30 901.

Table 1 Velocity as a function of time

t  16 s

Example 1 Cont.

 

   

t

t t

a ti i i

 1 

ti  15

1

 t  ti   ti

To find the acceleration at , we need to choose the two values closest to , that also bracket to evaluate it. The two

points are and.

t  16 s

t  16 s

t  15 s t  20 s

ti  1  20

t  16 s

Solution

Direct Fit Polynomials

' n  1 '  x (^) 0 , y 0  , x 1 , y 1  , x 2 , y 2 , , xn , yn 

nth Pn   x  a 0  a 1 x  an  1 xn ^1  anxn

n ^  ^  n (^ )  a 1 ^2 a 2 x   n^ ^1  a^ n  1 xn ^2  nanxn ^1 dx

dP x P x 

In this method, given (^) data points one can fit a (^) order polynomial given by

To find the first derivative,

Similarly other derivatives can be found.

Example 2-Direct Fit Polynomials

The upward velocity of a rocket is given as a function of time in Table 2.

Using the third order polynomial interpolant for velocity, find the acceleration of the rocket at.

t v(t) s m/s 0 0 10 227. 15 362. 20 517. 22.5 602. 30 901.

Table 2 Velocity as a function of time

t  16 s

Example 2-Direct Fit Polynomials cont.

such that

Writing the four equations in matrix form, we have

       

3 3

2

v 10  227. 04  a 0  a 110  a 2 10  a 10

       

3 3

2

v 15  362. 78  a 0  a 115  a 2 15  a 15

       

3 3

2

v 20  517. 35  a 0  a 1 20  a 2 20  a 20

       

3 3

2

v 22. 5  602. 97  a 0  a 1 22. 5  a 2 22. 5  a 22. 5

3

2

1

0

a

a

a

a

Example 2-Direct Fit Polynomials cont.

Solving the above four equations gives

a 0  4. 3810

a 1  21. 289

a 2  0. 13065

a 3  0. 0054606

Hence

 

4. 3810 21. 289 0. 13065 2 0. 0054606 3 , 10 22. 5

3 3

2 0 1 2      

    t t t t

v t a at a t a t

Example 2-Direct Fit Polynomials cont.

,

The acceleration at t=16 is given by

 16   v   t (^) t  16

dt

d

a

Given that

  t   4. 3810  21. 289 t  0. 13065 t^2  0. 0054606 t^3 , 10  t  22. 5

  v   t

dt

d

a t 

 4. 3810 21. 289 t 0. 13065 t^2 0. 0054606 t^3  dt

d     

 21. 289  0. 26130 t  0. 016382 t^2 , 10  t  22. 5

a  16   21. 289  0. 26130  16   0. 016382  16 ^2

 29. 664 m/s^2

Lagrange Polynomial

In this method, given ^ x 1 (^) , y 1 ,^ ,^ xn , yn , one can fit a  n  1  th order Lagrangian polynomial given by



n

i

f n x Li x f xi

0

where ‘ n ’ in fn ( x ) stands for the nth order polynomial that approximates the function

y  f ( x ) given at ( n  1 ) data points as ^ x 0 , y 0  ,^ x 1 , y 1 ,......,^ ^ xn  1 , yn  1  ,^ xn , yn  , and



 

 

n

j i

j (^) i j

j i x x

x x L x 0

( )

Li ( x )^ a weighting function that includes a product of^ (^ n ^1 ) terms with terms of

j  i^ omitted.

    

    

    

 2  2 0 2 1

1 1 0 1 2

0 0 1 0 2

2

2 2 2 f x x x x x

f x x x x x

f x x x x x

f x  

  

  

 

  ^ 

  ^ 

  ^ 

2 0 2 1

1 0 1 1 0 1 2

0 0 2 0 1 0 2

2 2 1 2 2 2 f x x x x x

f x x x x x x x x

f x x x x x x x x

f x x x x  

    

    

   

Differentiating again would give the second derivative as

Lagrange Polynomial Cont.

Example 3

Determine the value of the acceleration at using the second order Lagrangian polynomial interpolation for velocity.

t v(t) s m/s 0 0 10 227. 15 362. 20 517. 22.5 602. 30 901.

Table 3 Velocity as a function of time

t  16 s

The upward velocity of a rocket is given as a function of time in Table 3.