Unit  2
Numerical differentiation, integration and solution of differential equations
Formulae for derivatives:
Newton’s forward Difference formula:
This method is useful for interpolation near the beginning of a set of tabular values.
Where
Differentiating both side with respect to p, we get
h
This formula is applicable to compute the value of for non tabular values of x.
For tabular values of x , we can get formula by putting
Therefore
In similar manner we can get the formula for higher order by differentiating the previous order formulas
Again, differentiating with respect to p, we get
Hence
Also
And so on.
Example1: Given that
X  1.0  1.1  1.2  1.3 
Y  0.841  0.891  0.932  0.963 
Find at .
Here the first derivative is to be calculated at the beginning of the table, therefore forward difference formula will be used
Forward difference table is given below:
X  Y  
1.0
1.1
1.2
1.3  0.841
0.891
0.932
0.962 
0.050
0.041
0.031 
0.009
0.010 
0.001 
By Newton’s forward differentiation formula for differentiation
Here
Example2: Find the first and second derivatives of the function given below at the point :
X  1  2  3  4  5 
Y  0  1  5  6  8 
Here the point of the calculation is at the beginning of the table,
Forward difference table is given by:
X  Y  
1
2
3
4
5  0
1
5
6
8 
1
4
1
2 
3
3
1 
6
4

10

By Newton’s forward differentiation formula for differentiation
Here , 0.
Again
At
Example3: From the following table of values of x and y find for
X  1.00  1.05  1.10  1.15  1.20  1.25  1.30 
Y  1.0000  1.02470  1.04881  1.07238  1.09544  1.11803  1.14017 
Here the value of the derivative is to be calculated at the beginning of the table.
Forward difference table is given by
X  Y  
1.00
1.05
1.10
1.15
1.20
1.25
1.30  1.0000
1.02470
1.04881
1.07238
1.09544
1.11803
1.14017 
0.02470
0.02411
0.02357
0.02306
0.02259
0.02214 
0.00059
0.00054
0.00051
0.00047
0.00045 
0.00005
0.00003
0.00004
0.00002 
0.00002
0.00001
0.00002 
0.00003
0.00003 
0.00006 
From Newton’s forward difference formula for differentiation, we get
Here
=0.48763
Newton Backward Difference Method:
This method is useful for interpolation near the ending of a set of tabular values.
Where
Differentiating both side with respect to p, we get
This formula is applicable to compute the value of for non tabular values of x.
For tabular values of x, we can get formula by putting
Therefore
In similar manner we can get the formula for higher order by differentiating the previous order formulas
Differentiating both side with respect to p, we get
Also
Example1: Given that
X  0.1  0.2  0.3  0..4 
Y  1.10517  1.22140  1.34986  1.49182 
Find ?
Backward difference table:
X  Y  
0.1
0.2
0.3
0.4  1.10517
1.22140
1.34986
1.49182 
0.11623
0.12846
0.14196 
0.01223
0.01350 
0.00127 
Newton’s Backward formula for differentiation
Here
Example2: Given that
X  1.0  1.2  1.4  1.6  1.8  2.0 
Y  0  0.128  0.544  1.296  2.432  4.0 
Find the derivative of y at ?
The difference table is given below:
X  Y  
1.0
1.2
1.4
1.6
1.8
2.0  0
0.128
0.544
1.296
2.432
4.0 
0.128
0.416
0.752
0.136
1.568

0.288
0.336
0.384
0.432 
0.048
0.048
0.048 
0
0 
Since the point is at the beginning of the table therefore
From Newton’s forward difference formula for differentiation, we get
Here
Since the point is at the end of the table therefore
Backward difference table is:
X  Y  
1.0
1.2
1.4
1.6
1.8
2.0  0
0.128
0.544
1.296
2.432
4.000 
0.128
0.416
0.752
0.136
1.568 
0.288
0.336
0.384
0.432 
0.048
0.048
0.048 
0
0 
Newton’s Backward formula for differentiation
Key takeaways
Newton’s forward Difference formula:
Numerical Integration
Numerical integration is a process of evaluating or obtaining a definite integral from a set of numerical values of the integrand f(x). In case of function of single variable, the process is called quadrature.
Trapezoidal Method:
Let the interval [a, b] be divided into n equal intervals such that <<…. <=b.
Here .
To find the value of .
Setting n=1, we get
Or I =
The above is known as Trapezoidal method.
Note: In this method second and higher difference are neglected and so f(x) is a polynomial of degree 1.
Geometrical Significance: The curve y=f(x), is replaced by n straight lines with the points ();() and ();…….;() and ().
The area bounded by the curve y=f(x), the ordinates ,and the x axis is approximately equivalent to the sum of the area of the n trapeziums obtained.
Example1: State the trapezoidal rule for finding an approximate area under the given curve. A curve is given by the points (x, y) given below:
(0, 23), (0.5, 19), (1.0, 14), (1.5, 11), (2.0, 12.5), (2.5, 16), (3.0, 19), (3.5, 20), (4.0, 20).
Estimate the area bounded by the curve, the x axis and the extreme ordinates.
We construct the data table:
X  0  0.5  1.0  1.5  2.0  2.5  3.0  3.5  4.0 
Y  23  19  14  11  12.5  16  19  20  20 
Here length of interval h =0.5, initial value a = 0 and final value b = 4
By Trapezoidal method
Area of curve bounded on x axis =
Example2: Compute the value of
Using the trapezoidal rule with h=0.5, 0.25 and 0.125.
Here
For h=0.5, we construct the data table:
X  0  0.5  1 
Y  1  0.8  0.5 
By Trapezoidal rule
For h=0.25, we construct the data table:
X  0  0.25  0.5  0.75  1 
Y  1  0.94117  0.8  0.64  0.5 
By Trapezoidal rule
For h = 0.125, we construct the data table:
X  0  0.125  0.25  0.375  0.5  0.625  0.75  0.875  1 
Y  1  0.98461  0.94117  0.87671  0.8  0.71910  0.64  0.56637  0.5 
By Trapezoidal rule
[(1+0.5) +2(0.98461+0.94117+0.87671+0.8+0.71910+0.64+0.56637)]
Example3: Evaluate using trapezoidal rule with five ordinates
Here
We construct the data table:
X  0  
Y  0  0.3693161  1.195328  1.7926992  1.477265  0 
Key takeaways
 Numerical integration is a process of evaluating or obtaining a definite integral from a set of numerical values of the integrand f(x).
Overview
Generally fundamental theorem of calculus is used find the solution for definite integrals, but sometime integration becomes too hard to evaluate, numerical methods are used to find the approximated value of the integral.
Simpson’s rules are very useful in numerical integration to evaluate such integrals.
Here we will understand the concept of Simpson’s rule and evaluate integrals by using numerical technique of integration.
We find more accurate value of the integration by using Simpson’s rule than other methods
Simpson’s rule
We will study about Simpson’s onethird rule and Simpson’s threeeight rules.
But in order to get these two formulas, we should have to know about the general quadrature formula
General quadrature formula
The general quadrature formula is gives as
Simpson’s onethird and threeeighth formulas are derived by putting n = 2 and n = 3 respectively in general quadrature formula.
Simpson’s onethird and threeeighth formulas are derived by putting n = 2 and n = 3 respectively in general quadrature formula.
Simpson’s onethird rule
Put n = 2 in general quadrature formula
We get
Note the given interval of integration has to be divided into an even number of subintervals.
Simpson’s threeeighth rule
Put n = 3 in general quadrature formula
We get
Note the given interval of integration has to be divided into subintervals whose number n is a multiple of 3.
Example: Evaluate the following integral by using Simpson’s 1/3rd and 3/8th rule.
Solution
First, we will divide the interval into six parts, where width (h) = 1, the value of f(x) is given in the table below
x  0  1  2  3  4  5  6 
f(x)  1  0.5  0.2  0.1  1/17 = 0.05884  1/26 = 0.0385  1/37 = 0.027 
Now using Simpson’s 1/3rd rule
We get
And now
Now using Simpson’s 3/8th rule
Example: Find the approximated value of the following integral by using Simpson’1/3rd rule.
Solution
The table of the values
x  1  1.25  1.5  1.75  2 
f(x)  0.60653  0.53526  0.47237  0.41686  0.36788 
Now using Simpson’s 1/3rd rule
We get
Example1: Estimate the value of the integral
By Simpson’s rule with 4 strips and 8 strips respectively.
For n=4, we have
E construct the data table:
X  1.0  1.5  2.0  2.5  3.0 
Y=1/x  1  0.66666  0.5  0.4  0.33333 
By Simpson’s Rule
For n = 8, we have
X  1  1.25  1.50  1.75  2.0  2.25  2.50  2.75  3.0 
Y=1/x  1  0.8  0.66666  0.571428  0.5  0.444444  0.4  0.3636363  0.333333 
By Simpson’s Rule
Example2: Evaluate Using Simpson’s 1/3 rule with .
For , we construct the data table:
X  0  
0  0.50874  0.707106  0.840896  0.930604  0.98281  1 
By Simpson’s Rule
Example3: Using Simpson’s 1/3 rule with h = 1, evaluate
For h = 1, we construct the data table:
X  3  4  5  6  7 
9.88751  22.108709  40.23594  64.503340  95.34959 
By Simpson’s Rule
= 177.3853
Example1: Evaluate
By Simpson’s 3/8 rule.
Let us divide the range of the interval [4, 5.2] into six equal parts.
For h=0.2, we construct the data table:
X  4.0  4.2  4. 4  4.6  4.8  5.0  5.2 
Y=logx  1.3863  1.4351  1.4816  1.5261  1.5686  1.6094  1.6487 
By Simpson’s 3/8 rule
= 1.8278475
Example2: Evaluate
Let us divide the range of the interval [0,6] into six equal parts.
For h=1, we construct the data table:
X  0  1  2  3  4  5  6 
1  0.5  0.2  0.1  0.0588  0.0385  0.027 
By Simpson’s 3/8 rule
+3(0.0385) +0.027]
=1.3571
Error in Integration
Error in Trapezoidal method
The total error in trapezoidal method is given by
Let is the largest value of the n quantities on the righthand side of the above equation then
Error in Simpson’s Rule
The error in the Simpson’s rule is given by
Where is the largest value of the fourth derivative of y(x).
Error in Simpson’s 3/8 Rule
The error in this rule is given by
Where is the largest value of the derivative of y(x).
Key takeaways
 The general quadrature formula is gives as
2. Simpson’s onethird rule
3. Simpson’s threeeighth rule
Quadrature: It is the process to evaluate the value of the functions at the chosen point, to its exact value for polynomial up to higher degree as possible.
The general form Gaussian quadrature is given by
Where depends on the choice of n (number of points).
Also note that the possible polynomial of degree up to .
. …... (1)
For point
Which gives exact value of the polynomial up to degree degree
i.e.,
For point and using (1)
Which gives exact value of the polynomial up to degree degree
i.e.,
On solving we get
.
Gauss Quadrature 3point method:
The general form Gaussian quadrature is given by
Where depends on the choice of n (number of points).
Also note that the possible polynomial of degree up to .
. …... (1)
For point
Which gives exact value of the polynomial up to degree degree
i.e.,
On solving we get
.
For n=3,
Note: To evaluate
The above integral can be converted into Gauss quadrature by substituting
Hence .
Example: Evaluate
Here
Using =
Also
For
For
Hence
Here
By Gauss quadrature 3point rule
Example: Evaluate by 2point Gaussian rule.
Here
Using =
Also
For
For
Hence
Here
By Gauss quadrature 2point rule
=0.99847
Example: Solve by Gauss quadrature 3point method
Given
Here
Using =
Also
For
For
Hence
Here
By Gauss quadrature 3point rule
Hence
The general first order differential equation
…. (1)
With the initial condition … (2)
In general, the solution of first order differential equation in one of the two forms:
a) A series for y in terms of power of x, from which the value of y can be obtained by direct solution.
b) A set of tabulated values of x and y.
The case (a) is solved by Taylor’s Series or Picard method whereas case (b) is solved by Euler’s, Runge Kutta Methods etc.
Euler’s method:
In this method the solution is in the form of a tabulated values
Integrating both side of the equation (i) we get
Assuming that in this gives Euler’s formula
In general formula
, n=0,1, 2...
Error estimate for the Euler’s method
Example1: Use Euler’s method to find y (0.4) from the differential equation
with h=0.1
Given equation
Here
We break the interval in four steps.
So that
By Euler’s formula
, n=0,1,2,3 ……(i)
For n=0 in equation (i) we get
For n=1 in equation (i) we get
.01
For n=2 in equation (i) we get
For n=3 in equation (i) we get
Hence y (0.4) =1.061106.
Example2: Using Euler’s method solve the differential equation for y at x=1 in five steps
Given equation
Here
No. Of steps n=5 and so that
So that
Also
By Euler’s formula
, n=0,1,2,3,4 ……(i)
For n=0 in equation (i) we get
For n=1 in equation (i) we get
For n=2 in equation (i) we get
For n=3 in equation (i) we get
For n=4 in equation (i) we get
Hence
Example3: Given with the initial condition y=1 at x=0. Find y for x=0.1 by Euler’s method (five steps).
Given equation is
Here
No. Of steps n=5 and so that
So that
Also
By Euler’s formula
, n=0,1,2,3,4 ……(i)
For n=0 in equation (i) we get
For n=1 in equation (i) we get
For n=2 in equation (i) we get
For n=3 in equation (i) we get
For n=4 in equation (i) we get
Hence
Key takeaways
Euler’s method: , n=0,1, 2…...
Modified Euler’s Method:
Instead of approximating as in Euler’s method. In the modified Euler’s method, we have the iteration formula
Where is the nth approximation to .The iteration started with the Euler’s formula
Example1: Use modified Euler’s method to compute y for x=0.05. Given that
Result correct to three decimal places.
Given equation
Here
Take h = = 0.05
By modified Euler’s formula the initial iteration is
)
The iteration formula by modified Euler’s method is
(i)
For n=0 in equation (i) we get
Where and as above
For n=1 in equation (i) we get
For n=3 in equation (i) we get
Since third and fourth approximation are equal.
Hence y=1.0526 at x = 0.05 correct to three decimal places.
Example2: Using modified Euler’s method, obtain a solution of the equation
Given equation
Here
By modified Euler’s formula the initial iteration is
The iteration formula by modified Euler’s method is
(i)
For n=0 in equation (i) we get
Where and as above
For n=1 in equation (i) we get
For n=2 in equation (i) we get
For n=3 in equation (i) we get
Since third and fourth approximation are equal.
Hence y=0.0952 at x=0.1
To calculate the value of at x=0.2
By modified Euler’s formula the initial iteration is
The iteration formula by modified Euler’s method is
(ii)
For n=0 in equation (ii) we get
1814
For n=1 in equation (ii) we get
1814
Since first and second approximation are equal.
Hence y = 0.1814 at x=0.2
To calculate the value of at x=0.3
By modified Euler’s formula the initial iteration is
The iteration formula by modified Euler’s method is
(iii)
For n=0 in equation (iii) we get
For n=1 in equation (iii) we get
For n=2 in equation (iii) we get
For n=3 in equation (iii) we get
Since third and fourth approximation are same.
Hence y = 0.25936 at x = 0.3
Key takeaways
Modified Euler’s Method:
Solution of Second order ODE using 4th order RungeKutta method (Numerical), RungeKutta fourth order method (Numerical)
This method is more accurate than Euler’s method.
Consider the differential equation of first order
Let be the first interval.
A second order Runge Kutta formula
Where
Rewrite as
A fourth order Runge Kutta formula:
Where
Example1: Use Runge Kutta method to find y when x=1.2 in step of h=0.1 given that
Given equation
Here
Also
By Runge Kutta formula for first interval
Again
A fourth order Runge Kutta formula:
To find y at
A fourth order Runge Kutta formula:
Example2: Apply Runge Kutta fourth order method to find an approximate value of y for x=0.2 in step of 0.1, if
Given equation
Here
Also
By Runge Kutta formula for first interval
A fourth order Runge Kutta formula:
Again
A fourth order Runge Kutta formula:
Example3: Using Runge Kutta method of fourth order, solve
Given equation
Here
Also
By Runge Kutta formula for first interval
)
A fourth order Runge Kutta formula:
Hence at x = 0.2 then y = 1.196
To find the value of y at x=0.4. In this case
A fourth order Runge Kutta formula:
Hence at x = 0.4 then y=1.37527
Solution of Second order ODE using 4th order RungeKutta method:
The second order differential equation
Let then the above equation reduces to first order simultaneous differential equation
Then
This can be solved as we discuss above by Runge Kutta Method. Here for and for .
A fourth order Runge Kutta formula:
Where
Example1: Using Runge Kutta method of order four, solve to find
Given second order differential equation is
Let then above equation reduces to
Or
(say)
Or .
By Runge Kutta Method we have
A fourth order Runge Kutta formula:
Example2: Using Runge Kutta method, solve
for correct to four decimal places with initial condition .
Given second order differential equation is
Let then above equation reduces to
Or
(say)
Or .
By RungeKutta Method we have
A fourth order Runge Kutta formula:
And
.
Example3: Solve the differential equations
for
Using four order Runge Kutta method with initial conditions
Given differential equation are
Let
And
Also
By Runge Kutta Method we have
A fourth order Runge Kutta formula:
And
.
Key takeaways
A fourth order Runge Kutta formula:
Where
Problems involving second and higher order differential equations, in which the conditions at two or more points are specified, are called boundaryvalue problems.
Finite difference method
This method is used to solve bounded value problem.
The general twopoint linear boundary value problem
With boundary condition
To solve this problem by using finite difference, we divide the interval [a, b] into n sub interval so that .
The approximated function f(x) at the points
………
We use the following central difference formula:
We substitute these value in the given equation and apply the boundary condition to calculate the value of the unknown.
Example1: Solve the boundary value problem defined by
By finite difference method. Compare the solution at y (0.5) by taking h=0.5 and h=0.25.
Given equation
With boundary condition
By finite difference method
…. (iii)
Putting(iii) in (i) we get
…. (iv)
For h=0.5, here for which corresponds to
For i=1 in equation (iv) we get
For h=0.25, here
Which corresponds to
For i=1 in equation (iv) we get
For i=2 in equation (iv) we get
For i=3 in equation (iv) we get
From equation (v), (vi) and (vii) we get
On solving above triangular equation we get
Hence for h=0.5 we get y (0.5) =0.44444
And for h=0.25 we get y (0.5) =0.443674
Example2: Solve the bounded value problem
With boundary condition
Given equation
With
By finite difference method
…. (3)
Putting (3) in equation (1) we have
By finite difference method
…… (4)
Let h=1, we have
Corresponds to
For i=1 in equation (4) we get
For i=2 in equation (4) we get
For i=3 in equation (4) we get
From equation (5), (6) and (7) we get
On solving we get
Example3: Solve the boundary value problem
With y (0) =0 and y (2) =3.62686
Given equation …... (1)
With boundary condition y (0) =0 and y (2) =3.62686…. (2)
By finite difference method
…. (3)
Substituting (3) in equation (1) we get
…. (3)
Let h=0.5 then for
Which corresponds to
For i=1 in equation (3) we get
For i=2 in equation (3) we get
For i=3 in equation (3) we get
From equation (4), (5) and (6) we get
On solving we get
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 P. G. Hoel, S. C. Port and C. J. Stone, “Introduction to Probability Theory”, Universal Book Stall, 2003.
 S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
 W. Feller, “An Introduction to Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
 N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
 B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.
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