Unit6
Numerical Methods
Interpolation with an unequal interval
Divided Difference:
In the case of interpolation, when the value of the arguments are not equispaced (unequal intervals) we use the class of differences called divided differences.
Definition: The difference which is defined by taking into consideration the change in the value of the argument are known asdivided differences.
Let be a function defined as
…….  
………… 
Where are unequal i.e. it is a case of unequal interval.
The Firstorder divided differences are:
And so on.
The secondorder divided difference is:
And so on.
Similarly, the nth order divided difference is:
With the help of these we construct the divided difference table:
X  f(x)  




Newton’s Divided difference Formula:
Let be a function defined as
…….  
………… 
Where are unequal i.e. it is a case of unequal interval.
.
Example1:Using Newton’s divided difference formula, find the values of from the following table:
x  4  5  7  10  11  13 
f(x)  48  100  294  900  1210  2028 
We construct the divided difference table is given by:
x  f(x)  Firstorder divide difference  Secondorder divide difference  Thirdorder divide difference  Fourthorder divide difference 
4
5
7
10
11
13  48
100
294
900
1210
2028 



0
0 
By Newton’s Divided difference formula
.
Now, putting in above we get
.
Example2: The following values of the function f(x) for values of x are given:
Find the value of and also the value of x for which f(x) is maximum or minimum.
We construct the divide difference table:
x  f(x)  Firstorder divide difference  Secondorder divide difference  Thirdorder divide difference 
1
2
7
8  4
5
5
4 


0 
By Newton’s divided difference formula
.
Putting in above we get
For maximum and minimum of , we have
Or
Example3: Find a polynomial satisfied by , by the use of Newton’s interpolation formula with a divided difference.
x  4  1  0  2  4 
F(x)  1245  33  5  9  1335 
Here
We will construct the divided differencetable:
x  F(x)  Firstorder divided difference  Secondorder divided difference  Thirdorder divided difference  Fourthorder divided difference 
4
1
0
2
4  1245
33
5
9
1335 




By Newton’s divided difference formula
.
This is the required polynomial.
Newton Forward Difference formula:
This method is useful for interpolation near the beginning of a set of tabularvalues.
Where
Example1: Using Newton’s forward difference formula, find the sum
Putting
It follows that
Since isafourthdegree polynomial inn.
Further,
By Newton Forward Difference Method
Example2:Givenfind, by usingNewton forwardinterpolationmethod.
Let, then
0.7071  0.7660    0.8192  0.8660 
The table of forward finite difference is given below:
45
50
55
60  0.7071
0.7660
0.8192
0.8660 
0.0589
0.0532
0.0468 
0.0057
0.0064 
0.0007 
By Newton forward difference method
Here initial value = 45, difference ofintervalh = 5 andthe value to be calculated at x=52.
By Formula
Example3: Find the missing term in the following:
0  1  2  3  4  
1  3  9  ?  81 
Let
First, we construct the forward difference table:
0
1
2
3
4  1
3
9
81 
2
6

4


Now,
Newton Backward Difference Method:
This method is useful for interpolation near the ending of a set of tabular values.
Where
Example1:Find from the following table:
0.20  0.22  0.24  0.26  0.28  0.30  
1.6596  1.6698  1.6804  1.6912  1.7024  1.7139 
Consider the backward difference method
0.20
0.22
0.24
0.26
0.28
0.30  1.6596
1.6698
1.6804
1.6912
1.7024
1.7139 
0.0102
0.0106
0.0108
0.0112
0.0115 
0.0004
0.0002
0.0004
0.0003 
0.0002
0.0002
0.0001 
0.0004
0.0003 
0.0007 
Here
By Newton backward difference formula
Example2:The following table gives the amount of a chemicaldissolved in water:
Temp.  
Solubility  19.97  21.51  22.47  23.52  24.65  25.89 
Compute the amount dissolve at
Consider the following backward difference table:
Temp.x  Solubility y  
10
15
20
25
30
35  19.97
21.51
22.47
23.52
24.65
25.89 
1.54
0.96
1.05
1.13
1.24 
0.58
0.09
0.08
0.11 
0.67
0.01
0.03 
0.68
0.04 
0.72 
Here
ByNewtonBackwarddifference formula
Example3:The following are the marks obtained by492candidatein a certain examination
Marks  040  4045  4550  5055  5560  6065 
No.ofcandidates  210  43  54  74  32  79 
Findout the number of candidates:
a) Who secured more than 48butnot more than50 marks?
b) Who secured less than 48butnotless than 45 marks?
Consider theforward difference table given below:
Marksupto x  No. of candidates y  
40
45
50
55
60
65  210
210+43=253
253+54=307
307+74=381
381+32=413
413+79= 492 
43
54
74
32
79 
11
20
42
47 
9
62
89 
71
151 
222 
Here
By Newton Forward Difference formula
f
a) No. ofcandidate secured more than 48butnot more than50 marks
b) No. ofcandidate secured less than 48 butnotlessthan 45 marks
Lagrange interpolation
Given a set of values of x and y, the process of finding the value of xfor a certain value of y is called inverse interpolation.
Lagrange’s Inverse interpolation:
Let , bedefinedfunction we get
x  …..  
f(Y)  …… 
Where the interval is not necessarily equal. We assume f(x) is a polynomial of degree n.Then Lagrange’s inverse interpolation formula is given by
Example1: Use the inverse interpolation to find the value of x at for the following data:
X  1  3  4 
Y  4  12  19 
Here , we have thedata
Lagrange’s inverse interpolation formula is given by
.
Example2:Use the inverse Lagrange’s method to find the root of the equation , give data
X  30  34  38  42 
F(x)  30  13  3  18 
Here , we have thedata
Also.
Lagrange’s inverse interpolation formula is given by
Thus the approximate root of the given equation is .
Example3: Find the value of x at for the following data:
X  1  2  4  5  8 
Y  1.000  0.500  0.250  0.200  0.125 
Here , we have thedata
Also.
Lagrange’s inverse interpolation formula is given by
Thus the value.
Key takeaways
2. Newton Backward Difference Method:
This method is useful for interpolation near the ending of a set of tabular values.
Where
3. Then Lagrange’s inverse interpolation formula is given by
Newton’s forward Difference formula:
This method is useful for interpolation near the beginning of a set of tabularvalues.
Where
Differentiating both sides with respect to p, we get
h
This formula is applicable to compute the value of for nontabular values of x.
For tabular values of x, we can get formula by putting
Therefore
Similarly, we can get the formula for higherorder 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, thereforeforward differenceformula will be used
Forward differencetable 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: Findthe first and secondderivatives of the function given below at thepoint :
X  1  2  3  4  5 
Y  0  1  5  6  8 
Here the point of the calculation isat the beginning of thetable,
Forward difference table is givenby:
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 followingtable 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 thevalue of the derivative is to becalculated atthe 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 sides with respect to p, we get
This formula is applicable to compute the value of for nontabular values of x.
For tabular values of x, we can get formula by putting
Therefore
Similarly, we can get the formula for higherorder by differentiating the previous order formulas
Differentiating both sides 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 ?
Thedifference table isgiven 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 pointisat the beginning of the table therefore
From Newton’s forward difference formula for differentiation we get
Here
Since thepoint is at theend of thetabletherefore
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
Numerical integration is a process of evaluating or obtaining a definite integral from a set of numerical values of the integrand f(x).In the case of the function of a single variable, the process is called quadrature.
Trapezoidal Method:
Let the interval 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 the Trapezoidal method.
Note: In this method second and higher differences 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 xaxis 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:
Estimate the area bounded by the curve, the xaxis, 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 xaxis =
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 
Simpson’s Rule:
Let the interval be divided into n equal intervals such that <<….<=b.
Here.
To find the value of.
Setting n = 2,
Which is known as Simpson’s 1/3 rule or Simpson’s rule.
Note: In this rule third and higher differences are neglected a so f(x) is a polynomial of degree 2.
Example1: Estimate the value of the integral
by Simpson’s rule with 4 strips and 8 strips respectively.
For n=4, we have
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
Simpson’s 3/8 rule
Let the interval be divided into n equal intervals such that <<….<=b.
Here.
To find the value of .
Setting n=3, we get
Is known as Simpson’s 3/8 rule which is not as accurate as Simpson’s rule.
Note: In this rule, the fourth and higher differences are neglected and so f(x) is a polynomial of degree 3.
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 
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 Simpson’s 3/8 Rule
The error in this rule is given by
Where is the largest value of the derivative of y(x).
Error in Trapezoidal method
The total error in the 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 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).
Euler’s Method:
The general Firstorder differential equation
With the initial condition
Inthismethod,the solution is in the formof tabulatedvalues.
Integrating both side of the equation (i) we get
Assuming thatinthisgives Euler’s formula
In generalformula
, n=0,1,2,…..
Error estimate for the Euler’s method
Example1:Use Euler’s method tofind y(0.4) from the differential equation
with h=0.1
Givenequation
Here
Webreaktheinterval into four steps.
So that
By Euler’sformula
, n=0,1,2,3 ……(i)
For n=0 in equation (i) weget
For n=1 in equation (i) weget
.01
For n=2 in equation (i) weget
For n=3 in equation (i) weget
Hence y(0.4)=1.061106.
Example2:Using Euler’s method solves the differential equation for y at x=1 in five steps
Given equation
Here
No. of stepsn=5 andso that
So that
Also
By Euler’sformula
, n=0,1,2,3,4 ……(i)
For n=0 in equation (i) weget
For n=1in equation (i) weget
For n=2in equation (i) weget
For n=3in equation (i) weget
For n=4in equation (i) weget
Hence
Example3:Given withthe initial conditiony=1 at x=0.Findy for x=0.1 byEuler’s method(fivesteps).
Given equation is
Here
No. of stepsn=5 and so that
So that
Also
By Euler’sformula
, n=0,1,2,3,4 ……(i)
For n=0 in equation (i) weget
For n=1 in equation (i) weget
For n=2 in equation (i) weget
For n=3 in equation (i) weget
For n=4 in equation (i) weget
Hence
Modified Euler’sMethod:
Instead ofapproximating as in Euler’s method.Inthemodified Euler’smethod, wehave the iteration formula
Where is the nth approximation to .The iteration started with Euler’sformula
Example1:Use modifiedEuler’s methodto compute yforx=0.05.Given that
The result correctto three decimal places.
Givenequation
Here
Takeh= = 0.05
By modified Euler’sformula, theinitial iteration is
)
The iterationformula bymodified Euler’s methodis
(i)
For n=0in equation(i)we get
Where andasabove
For n=1in equation(i)we get
For n=3in equation(i)we get
Sincethe third andfourth approximation is equal.
Hence y=1.0526at x =0.05correcttothree decimal places.
Example2: Using modified Euler’s method,obtain a solution ofthe equation
Given equation
Here
By modified Euler’sformula, theinitial iteration is
The iterationformula bymodified Euler’s methodis
(i)
For n=0in equation(i)we get
Where andasabove
For n=1in equation(i)we get
For n=2 in equation(i)we get
For n=3in equation(i)we get
Since the third and fourth approximation is equal.
Hence y=0.0952 atx=0.1
To calculatethe valueof at x=0.2
By modified Euler’sformula, theinitial iteration is
The iterationformula bymodified Euler’s methodis
(ii)
For n=0in equation(ii)we get
1814
For n=1in equation(ii)we get
1814
Since the first and second approximations areequal.
Hencey= 0.1814atx=0.2
Tocalculate the valueof at x=0.3
By modified Euler’sformula, theinitial iteration is
The iterationformula bymodified Euler’s methodis
(iii)
For n=0in equation(iii)we get
For n=1in equation(iii)we get
For n=2 in equation(iii)we get
For n=3 in equation(iii)we get
Since the third and fourth approximations are the same.
Hence y= 0.25936 at x = 0.3
This method is more accurate than Euler’s method.
Consider the differential equation of Firstorder
Let be the first interval.
A Secondorder RungeKutta formula
Where
Rewriteas
A fourthorderRungeKutta formula:
Where
Example1: Use RungeKutta method tofind y when x=1.2in the step of h=0.1 given
that
Given equation
Here
Also
ByRungeKutta formula for firstinterval
Again
A fourthorderRungeKutta formula:
Tofind y at
A fourthorderRungeKutta formula:
Example2: Apply RungeKutta fourthorder method to find an approximate value ofy forx=0.2 in the step of 0.1,if
Givenequation
Here
Also
ByRungeKutta formula for firstinterval
A fourthorderRungeKutta formula:
Again
A fourthorderRungeKutta formula:
Example3: Using the RungeKutta method of fourthorder, solve
Given equation
Here
Also
ByRungeKutta formula for firstinterval
)
A fourthorderRungeKutta formula:
Hence at x= 0.2 then y= 1.196
To find thevalue of y at x=0.4. In this case
A fourthorderRungeKutta formula:
Hence atx= 0.4 theny=1.37527
The solution of Secondorder ODE using the 4th order RungeKutta method:
The Secondorder differential equation
Let then the above equation reduces to a Firstorder simultaneous differential equation
Then
This can be solved as we discuss above by RungeKutta Method. Here for and for .
A fourthorderRungeKutta formula:
Where
Example1:UsingRungeKutta method of order four, solve to find
Given Secondorder differential equation is
Let then above equation reduces to
Or
(say)
Or .
By RungeKutta Method we have
A fourthorderRungeKutta formula:
Example2: Using the RungeKutta method, solve
for correct to four decimal places with initial condition .
Given Secondorder differential equation is
Let then above equation reduces to
Or
(say)
Or .
By RungeKutta Method we have
A fourthorderRungeKutta formula:
And
.
Example3: Solve the differential equations
for
Using the four order RungeKutta method with initial conditions
Given differential equation are
Let
And
Also
By RungeKutta Method we have
A fourthorderRungeKutta formula:
And
.
Predictor and corrector
Predicator and corrector method uses more than one previous step to calculate the next value of y.
A general form predictor and corrector method is
Where are the constants to be determined.
Note
References
1. Erwin Kreyszig, Advanced Engineering Mathematics, 9thEdition, John Wiley & Sons, 2006.
2. N.P. Bali and Manish Goyal, A textbook of Engineering Mathematics, Laxmi Publications.
3. Higher engineering mathematic, Dr. B.S. Grewal, Khanna publishers