MTH603_3

Document Sample
MTH603_3 Powered By Docstoc
					                       Assignment: # 3 (Spring 2010)
                             Mth603 (Numerical Analysis)
                                    Lecture: 23 – 28
Total Marks = 25




Q. No. 1. (Marks 10).
Find the first and second order derivatives of f(x) at x = 1.5 if:


             x        1.5            2.0      2.5        3.0           3.5            4.0


           f(x)       3.37          7.00    13.62       24.00         38.87          59.00


Since x=1.5 appear at beginning of the table, it is appropriat to use formulae based on


forward differences to find the derivatives. The difference table for the given data is


depicted


X            Y               f ( x)        2 f ( x)     3 f ( x)           4 f ( x)       5 f ( x)

1.5          3.37


2.0          7.00            3.63


2.5          13.62           6.62          2.99


3.0          24.00           10.38         3.76          0.77


3.5          38.87           14.87         4.49          0.73            -0.04
4.0             59.00            20.13            5.26             0.77     0.04    0.08




H=0.5


Using forward difference formua for


           1             2 f ( x)  3 f ( x)  4 f ( x)  5 f ( x)
Df ( x)  [f ( x)                                              ]
           h                2            3         4          5
             1          2.99 0.77 0.04 0.08
Df ( x)        [3.63                               ]
           0.5            2         3       4        5
Df ( x)  4.8354
              1                          11            5
D 2 f ( x)  2 [ 2 f ( x)   3 f ( x)   4 f ( x)   5 f ( x)]
             h                           12            6
                1                      11            5
D 2 f ( x)        2
                     [2.99  0.77  ( 0.04)  (0.08)]
             (0.5)                     12            6
               1
D 2 f ( x)        [2.99  0.77  0.0367  0.0667]
             0.25
               1
D 2 f ( x)        [2.22  0.0367  0.0667]  8.467
             0.25
Q. No. 2.        (Marks 8).

Let P3(x) be the interpolating polynomial for the data (0, 0), (0.5, y), (1, 3) and (2, 2).
Find y if the coefficient of x3 in P3(x) is 6.

Solution:-
Solving for P3(x) gives
d

P3  x  
            x  0  x  1 x  2  y   x  0  x  0.5 x  2  3
             0.5 0.5 1.5                1 0.5 1

   x  0  x  0.5 x  1 2
           2 1.51
P3  x  
             x  0  x  1 x  2  y   x  0  x  0.5 x  2  3
                      0.375                          0.5

   x  0  x  0.5 x  1 2
                 3
                       x  1 x  2      x  0.5 x  2  3
                                        y                      
P3  x    x  0    0.375                    0.5        
                       x  0.5  x  1                       
                                          2                    
                     
                                 3                            
                                                                 
                       x  1 x  2           x  0.5 x  2  30
                                        y1000                        
                            375                        5          
P3  x    x  0 
                       x  0.5  x  1                             
                                          2                          
                     
                                 3                                  
                                                                       
                       x 2  3x  2           x 2  2.5x  1 30
                                       y1000                      
                       375                            5        
P3  x    x  0                                                
                       x  1.5 x  0.5 
                             2
                                                                    
                                 3
                                            2                       
                                                                   
            1              30      2  3
P3  x           y1000             x ..........
             375          5   3 
6   2.67 y  6  0.67 
6  6  0.67   2.67 y 
11.33  2.67 y
11.33
      y
 2.67
y  4.24



Q. No. 3.       (Marks 7).


The following data are given for the polynomial P(x) of unknown degree.


                   x                   0                   1               2


                 P(x)                  2                  -1               4
Determine the coefficient of x2 in P(x) if all the third order differences are 1.


                         x                   P(x)          1st differences   2nd differences


                        0                      2


                        1                     -1               -3                    8/2=4


                        2                      4               5




P2  x  
              x  1 x  2  2   x  0  x  2  (1)   x  0  x  1 (4)
              0  1 0  2  1  0 1  2                2  0  2  1
P2  x    
             x   2
                       3x  2 
                                   2   x2  2x   2  x2  x 
                 2
P2  x   4 x  7 x  2
              2

				
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
views:38
posted:7/10/2010
language:English
pages:4