EXAMPLES
1. Let a function f(x) is known at the points x = 0, 2, 3, 5 & 6 as follows
x
0
2
3
5
6
f(x)
1
3
7
21
31
    the find the value of f'(4.1)

    Sol : The numbers of points n = 5 i.e., 4th degree polynomial which passes through all the five points can be obtained by langrages interpolation

    now 

    4
    f4'(x) = P4'(x) =
    S
    		 Li'(x)fi
    i=0

     
    where Li'(x) = 
    (x - x0)(x - x1). . . (x - x4)
    (x - xi)[(x - x0). . . (x - x4)]x = xk

    f'(4.1) = 
      

    ((4.1-3.0)(4.1-5.0)(4.1-6.0)+(4.1-2.0)(4.1-5.0)(4.1-6.0)+(4.1-2.0)(4.1-3.0)(4.1-6.0)+(4.1-2.0)(4.1-3.0)(4.1-5.0))  * 1.0 +
    ((0.0-2.0)(0.0-3.0)(0.0-5.0)(0.0-6.0))
     
    ((4.1-3.0)(4.1-5.0)(4.1-6.0)+(4.1-0.0)(4.1-5.0)(4.1-6.0)+(4.1-0.0)(4.1-3.0)(4.1-6.0)+(4.1-0.0)(4.1-3.0)(4.1-5.0))  * 3.0 +
    ((2.0-0.0)(2.0-3.0)(2.0-5.0)(2.0-6.0))
     
    ((4.1-2.0)(4.1-5.0)(4.1-6.0)+(4.1-0.0)(4.1-5.0)(4.1-6.0)+(4.1-0.0)(4.1-2.0)(4.1-6.0)+(4.1-0.0)(4.1-2.0)(4.1-5.0))   * 7.0 +
    ((3.0-0.0)(3.0-2.0)(3.0-5.0)(3.0-6.0))
     
    ((4.1-2.0)(4.1-3.0)(4.1-6.0)+(4.1-0.0)(4.1-3.0)(4.1-6.0)+(4.1-0.0)(4.1-2.0)(4.1-6.0)+(4.1-0.0)(4.1-2.0)(4.1-3.0))   * 21.0 +
    ((5.0-0.0)(5.0-2.0)(5.0-3.0)(5.0-6.0))
     
    ((4.1-3.0)(4.1-5.0)(4.1-6.0)+(4.1-2.0)(4.1-5.0)(4.1-6.0)+(4.1-2.0)(4.1-3.0)(4.1-6.0)+(4.1-2.0)(4.1-3.0)(4.1-5.0))  * 31.0 
    ((6.0-0.0)(6.0-2.0)(6.0-3.0)(6.0-5.0)) 
           =    -0.005533327  +    0.46699977  +    -5.2523327  +    13.8921995  +    -1.9013343

           =  7.1999984

    Error E(x) is given by
     

      p'(x)
     + 
    		 p(x)
     
     En(x)  = 
    		 f(n+1)(x)
    		   f(n+2)(x)
     
    (n + 1)!
      (n + 1)!
    		  

    =   0.13496177 T   - 0.0011124568 U

    where T = f(n+1)(x) and U = f(n+2)(x)
    x0 < x <xn
2. Evenly spaced data points
x
1.3
1.9
2.5
3.10
3.7
4.3
4.9
f(x)
3.669
6.686
12.182
22.198
40.447
73.7
134.29
find the value of f'(3.3)

Sol : The numbers of points n = 7 i.e., 6th degree polynomial which passes through all the seven points can be obtained by langrages interpolation

now 

6
f6'(x) = P6'(x) =
S
 Li'(x)fi
i=0

 
where Li'(x) = 
(x - x0)(x - x1). . . (x - x6)
(x - xi)[(x - x0). . . (x - x6)]x = xk

f'(3.3) =  -0.044457488  +    0.64200306  +    -4.073202  +    -31.464916  +    75.59536  +

   -16.377771  +    2.8399248
           =  27.116947
 
 

Error E(x) is given by
 

  p'(x)
 + 
 p(x)
 
 En(x)  = 
 f(n+1)(x)
   f(n+2)(x)
 
(n + 1)!
  (n + 1)!
  

=   -5.6888934E-5 T   -  1.8996838E-4 U 

     
    where T = f(n+1)(x) and U = f(n+2)(x)
    x0 < x <xn
3. The data points are
x
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
f(x)
.123060
.105706
.089584
.074764
.61277
.049126
.038288
.028722
.020371
.013164
.007026
find the value of f'(2.4)

Sol : The numbers of points n = 11 i.e., 10th degree polynomial which passes through all the eleven points can be obtained by langrages interpolation

now 

10
f10'(x) = P10'(x) =
S
 Li'(x)fi
i=0

 
where Li'(x) = 
(x - x0)(x - x1). . . (x - x10)
(x - xi)[(x - x0). . . (x - x10)]x = xk
f'(2.4) =  0.0014650004  +    -0.016778681  +    0.09598279  +    -0.42722163  - .22468448  +    0.58951324  +    -0.19144018  +    0.054708514  +    -0.010913037  + 
  0.0012537113  +    -5.576195E-5
           =  -0.12817052
 
 

Error E(x) is given by
 

  p'(x)
 + 
 p(x)
 
 En(x)  = 
 f(n+1)(x)
   f(n+2)(x)
 
(n + 1)!
  (n + 1)!
  

=   1.8150002E-10 T  +  0.0 U

     
    where T = f(n+1)(x) and U = f(n+2)(x)
    x0 < x <xn
4. The data points are
x
0.15
0.21
0.23
0.27
0.32
0.35
f(x)
0.1761
0.3222
0.3617
0.4317
0.5031
0.5441
find the value of f'(0.242)

Sol : The numbers of points n = 6 i.e., 5th degree polynomial which passes through all the six points can be obtained by langrages interpolation

now 

5
f5'(x) = P5'(x) =
S
 Li'(x)fi
i=0

 
where Li'(x) = 
(x - x0)(x - x1). . . (x - x5)
(x - xi)[(x - x0). . . (x - x5)]x = xk
f'(0.242) =   0.046252206  +    -2.7550204  +    -5.6957664  +    11.520535  + 
-1.7202352  +    0.40043205
           =  1.7961981
      Error E(x) is given by
 
  p'(x)
 + 
 p(x)
 
 En(x)  = 
 f(n+1)(x)
   f(n+2)(x)
 
(n + 1)!
  (n + 1)!
  

=   -1.5240344E-8 T   - 1.1573452E-11 U 

     
    where T = f(n+1)(x) and U = f(n+2)(x)
    x0 < x <xn
5. Find f'(0.72), f'(1.33) & f'(0.5) for the data
x
0.3
0.5
0.7
0.9
1.1
1.3
1.5
f(x)
0.3985
0.6598
0.9147
1.1611
1.3971
1.6212
1.8325
 
Sol : The numbers of points n = 7 i.e., 6th degree polynomial which passes through all the seven points can be obtained by langrages interpolation

now 

6
f6'(x) = P6'(x) =
S
 Li'(x)fi
i=0

 
where Li'(x) = 
(x - x0)(x - x1). . . (x - x6)
(x - xi)[(x - x0). . . (x - x6)]x = xk
f'(0.72) =   0.051045507  +    -0.91976225  +    -3.6326413  +    8.211387  + 
-3.3100252  +    0.98771214  +    -0.13707396
           =  1.2506417

f'(1.33) = -0.08420189  +    1.0314666  +    -4.660655  +    11.322042  +    -18.015987  +

   8.705107  +    2.7815008
           =  1.0792731
f'(0.5) = -0.33208334  +    -4.233717  +    11.433749  +    -9.675832  +    5.821248  + 
-2.0265005  +    0.30541652
           = 1.2922815

     Error E(x) is given by
 

  p'(x)
 + 
 p(x)
 
 En(x)  = 
 f(n+1)(x)
   f(n+2)(x)
 
(n + 1)!
  (n + 1)!
  

E(0.72) =  5.39515E-5 T  +  1.1346211E-8 U
E(1.33) =  3.15478E-4 T  +  -5.390054E-8 U
E(0.5) =  2.605206E-4 T  +  -0.000000000 U

     
    where T = f(n+1)(x) and U = f(n+2)(x)
    x0 < x <xn
Solution of Transcendental Equations | Solution of Linear System of Algebric Equations | Interpolation & Curve Fitting
Numerical Differentiation & Integration | Numerical Solution of Ordinary Differetial Equations
Numerical Solution of Partial Differential Equations