MA6452 STATISTICS AND NUMERICAL METHODS UNIT IV NUMERICAL DIFFERENTIATION AND INTEGRATION By Ms. K. Vijayalakshmi Assistant Professor Department of Applied Mathematics SVCE
NUMERICAL DIFFERENCE:
1.NEWTON S FORWARD DIFFERENCE FORMULA TO COMPUTE THE DERIVATIONS: Newton s forward difference formula for equal intervals is where
Special Case:-
2. NEWTON S BACKWARD DIFFERENCES FORMULA TO COMPUTE THE DERIVATIVES
Special Case:-
3.MAXIMA AND MINIMA OF THE INTERPOLATING POLYNOMIAL
Newton s forward interpolation formula is where u=
For y to be a maximum or minimum
4. The following data gives the velocity of a particle for 2 seconds at an interval of 5 seconds. Find the initial acceleration using the entire date. time(sec) 0 5 10 15 20 Volocity(m/sec) 0 3 14 69 228
Solution: Let v denote velocity and a denote acceleration Then a= Initial acceleration Here x=t, y=v
x y y 0 0 5 3 3 8 10 14 11 44 36 24 15 69 55 104 60 20 228 159
Now h=5
5.From the following table, find the value of x for which y is minimum and find this value ofy. X -2-1 0 1 2 3 4 Y 2-0.25 0-0.25 2 15.75 56 Solution: Since we have equal intervals, we use Newton s forward difference formula.
x y Δy 2 2- -1-0.25-2.25 2.50 0 0 1-0.25 2 2 3 15.75 4 56 0.25-0.25 2.25 13.75 40.25-0.50 2.50 11.50 26.50-3 3 9 15 6 6 6 0 0
For minimum value of y
u=1.4
6. Using the following date, find, and the maximum value of X 0 2 3 4 7 9 f(x) 4 26 58 112 466 922 Solution: Since the x values are not equally spaced, we use Newton s divided difference formula.
x 0 4 2 26 3 58 4 112 7 466 9 922 32 54 118 228 7 11 16 22 1 1 1 0 0
By Newton s divided difference formula,
i.e., is maximum if But the roots of this equation are imaginary. Hence there is no extremum value.
7.NUMERICAL INTEGRATION The process of computing a definite integral from a set of tabulated values of the integrand is called numerical integration. This process when applied to a function of a single variable is known as quadrature.
8.Quadrature formula for equidistant ordinates (or Newton s Cote s formula)
9.TRAPEZOIDAL RULE
10.ERROR IN TRAPEZOIDAL RULE Hence the error in the Trapezoidalrule is of order
11.SIMSON S S ONE THIRD RULE This is known as Simpson s one-third rule.
12.ERROR IN SIMPSON S S ONE-THIRD RULE The total error is given by where is the largest of the error in Simpson s rule of order
Note: 1.While applying Simpson s rule, the number of sub-intervals should be even. 2.While applying Simpson s rule, the number of sub-intervals should be a multiple of 3
. 15.Evaluate using Trapezoidal rule with h=0.2. Hence obtain an approximate value of Solution: Let Here h=0.2 in the interval (0,1)
Now, the values of y are given below: x 0 0.2 0.4 0.6 0.8 1.0 Y 1 0.9615 0.8621 0.7353 0.6098 0.5 By Trapezoidal rule
=0.7837-------(1)
By actual integration
To find the value of from (1) and (2) we get
16.Evaluate by dividing the range into six equal parts using Solution: We form the table with
Note: X 0 1/6 2/6 3/6 4/6 5/6 1 (Sinx)/x 1 0.9954 0.9816 0.9589 0.9276 0.8882 0.8415
17.Romerg s Method To evaluate I systematically, we take
18.Use Romberg s method to compute correct to 4 decimal places. Solution: We take = 0.5, 0.25 and 0.125 successively and evaluate the given integral using Trapezoidal rule. i) When h= 0.5, the values of
x : 0 0.5 1.0 Y : 1 0.8 0.5
(ii) When h=0.25, the values of x : 0 0.25 0.5 0.75 1.0 y : 1 0.9412 0.8 0.64 0.5
(iii) When h=0.125, the values of x: 0 0.125 0.25 0.375 0.5 0.625 0.75 0.875 1.0 y: 1 0.9846 0.9412 0.8767 0.8000 0.7191 0.6400 0.5664 0.5
Thus we have Now using
and
The table of these values is 0.7750 0.7854 0.7828 0.7855 0.7855 0.7848 Hence the value of the integral=0.7855
19. Find the approximate value of using composite trapezoidal rule with and then Romberg s method. Solution: We take h=0.5, 0.25 and 0.125 successively and evaluate the given integral using Trapezoidal rule.
(i) When h=0.5, the value of x 0 1/2 1 y 1 2/3 1/2 By trapezoidal rule
(ii) When h=0.25, the value of x 0 1/4 1/2 3/4 1 y 1 4/5 2/3 4/7 1/2 =0.69702
(iii) When h=0.125, the value of x 0 1/8 2/8 3/8 4/8 5/8 6/8 7/8 1 y 1 8/9 4/5 8/11 2/3 8/13 4/7 8/15 1/2
Thus we have Now using
The table of these values is 0.70833 0.69325 0.6702 0.69312 0.69315 0.69412 Hence the value of the integral = 0.6931
20.Two-point Gaussian quadrature formula and this is exact for polynomials of degree upto3
21.Three- point Gaussian formula which is exact for polynomials of degree upto5
Note: by the linear transformation
by two-point and three-point Gaussian formula and compare with the exact value. Solution: By two-point Gaussian formula
By Three-point Gaussian formula
The exact value is
23.Using 2-point Gaussian quadrature, evaluate Solution: Here a=0, b=1
We get equivalent integral
By Gaussian two-point formula,
23.Double integration Consider the table x y
By trapezoidal rule,
by using Trapezoidal rule with h = k = 0.25 Solution: We first form the table for
y x 1 1.25 1.5 1.75 2 1 1 0.5 0.44444 0.4 0.36364 0.33333 0.40619 1.25 0.44444 0.4 0.3634 0.33333 0.30769 0.36826 1.5 0.4 0.36364 0.33333 0.30769 0.28571 0.33688 1.75 0.36364 0.33333 0.30769 0.28571 0.26667 0.31047 2 0.33333 0.30769 0.28571 0.26667 0.25 0.28783
By Trapezoidal rule,
Applying this rule to each row, we get
Now applying trapezoidal rule to =0.34067
25.Apply Simpson s s rule to evaluate the integral Take h = 0.2, k = 0.3 Solution: We form the table for
y x 4 4.2 4.4 I 2 0.125 0.1190 0.1136 0.0476 2.3 0.1087 0.1035 0.0988 0.0414 2.6 0.0962 0.0916 0.0874 0.0367
to each row of the table, with h=0.2
Now we apply Simpon srule to