This document discusses numerical methods for interpolation and solving algebraic and transcendental equations. It provides examples of using Newton's forward and backward interpolation formulas to find function values between given data points. Lagrange's interpolation formula is also introduced as a way to find function values using all the given data points, whether they are equally spaced or not. Three examples are worked through applying these interpolation methods to find unknown function values from tables of x and y data pairs.

1. NUMERICAL METHODS

2. INTERPOLATION AND
SOLUTION OF ALGEBRAIC AND
TRANSCENDENTAL EQUATIONS.
INTRODUCTION

3. Interpolation denotes the process
of computing the value of a
function for any value of the
independent variable within the
interval for which some values are
given.

4. Interpolation is a
fundamental operation in
mathematics it has been
described as a art of
reading
between the lines of the
tables.

5. Suppose the following table represents
a set of corresponding values of two
equations X and y .
X : x₀ x₁ x₂………………..xn
Y : y₀ y₁ y ₂……………… yn.
We may require the value of y=yr
corresponding to a value x=xr between
x0 and xn .

6. Suppose the following table represents a set of
corresponding values of two equations
X and y .
X : x₀ x₁ x₂………………..xn
Y : y₀ y₁ y ₂……………… yn.
We may require the value of y=yr corresponding to a
value x=xr between x0 and xn . Thus interpolation
denotes the process of computing the value of a
function for any value of the independent variable
within the interval for which some values are given.

7. The term extrapolation is used to
denote the process of finding the
values outside this interval.
However the term interpolation
is supplied to both process.
Example : Y=X3+2X+5

8. Let yX=f(x) be a function which takes a value
y0,y1,y2………….yn
corresponding to the value x0,x1,x2…….xn of the
independent variable of x.
let the value of x be at the equi-distant
intervals.
yp=y0+p∆y0+p(p-1)∆2y0+ p(p-1)(p-2)∆3y0+
2ḷ 3ḷ
………………………… p(p-1(p-2)……(p-(n-1))∆ny0
nḷ
It is known as newtons forward interpolation
formula.

9. This formula is used to interpolate the value of y
near the beginning of a set of tabular value.
Problem
1.Following are the data from the steam table.
Temperature C° 140 150 160 170
180
Pressure 3.685 4.854 6.302 8.076
10.225
Using newtons formula find the pressure of a
steam for a temperature 142°?

10. Solution:
x y ∆y ∆2y ∆3y ∆4y
140 3.685 1.169 0.279 0.047 0.002
150 4.854 1.448 0.326 0.049
160 6.302 1.774 0.375
170 8.076 2.149
180 10.225
yp=y0+p∆y0+p(p-1)∆2y0+ p(p-1)(p-2)∆3y0+…………………………
2ḷ 3ḷ
p(p-1(p-2)……(p-(n-1))∆ny0
nḷ
y0=3.685 p=x-x0
h
∆y0=1.169 =(142-140)/10=0.2
∆2y0=0.279
∆3y0=0.047
∆4y0=0.002
Yp=3.685+0.2(1.169)+0.2(0.2-1)0.279+0.2(0.2-1)(0.2-2)0.047+0.2(0.2-
2 6 24
1)(0.2-2)(0.2-3)0.002
24
=3.685+0.2338+0.0223+0.0022+0.00007
=3.89863
=3.899

11. 2.Following data gives the value x and y. find the value of y when x is 21. From the table
X 20 23 26 29
Y 0.3420 0.3907 0.4384 0.4848
Solution:
x y ∆y ∆2y ∆3y
120 0.3420 0.0487 -0.001 0.0003
23 0.3907 0.0477 -0.0013
26 0.4384 0.0464
29 0.4848
Formula
yp=y0+p∆y0+p(p-1)∆2y0+ p(p-1)(p-2)∆3y0+………………………… p(p-1(p-2)……(p-(n-1))∆ny0
2ḷ 3ḷ nḷ
Y0=0.3420 p=x-x0
h
∆y0=0.0487 =(21-20)/3=0.3
∆2y0=-0.001
∆3y0=0.0003
yp= 0.3420+0.3(0.0487)+0.3(0.3-1)(-0.001)+0.3(0.3-1)(0.3-2)0.0003
2 6
=0.3420+0.01461+0.000105+0.00001785
=0.3567

12. 3.Using newtons forward interpolation formula. Calculate x is 1.02 upto 3 decimal places in the table given
below
X 1.0 1.1 1.2 1.3 1.4
F(x) 0.841 0.89 1.932 0.964 0.985
Solution
X f(x) ∆y ∆2y ∆3y ∆4y
1.0 0.841 0.05 -0.009 0 -0.002
1.1 0.891 0.041 -0.0009 -0.002
1.2 0.932 0.032 -0.011
1.3 0.964 0.021
1.4 0.985
yp=y0+p∆y0+p(p-1)∆2y0+ p(p-1)(p-2)∆3y0+………………………… p(p-1(p-2)……(p-(n-1))∆ny0
2ḷ 3ḷ nḷ
y0=0.841 p=x-x0
h
∆y0=0.05 =(1.02-1)/0.1=0.2
∆2y0=-0.009
∆3y0=0
∆4y0=-0.002
Yp=0.841+0.2(0.05)+0.2(0.2-1)(-0.009)+0.2(0.2-1)(0.2-2)0+0.2(0.2-1)(0.2-2)(0.2-3)(-0.009)
2 6 24
=0.841+0.01+0.00072+0+0.0000672
=0.851

13. NEWTONS GREGORY BACKWARD INTERPOLATION FORMULA
Let yx=f(x) be a function which takes the value y0,y1,y2………….yn corresponding to the value
x0,x1,x2…….xn of the independent variable x. let the value of x be at equi-distance
intervals.
Let x=xn + ph
Yp=yn+ pṼyn+p(p+1)Ṽ2yn+ p(p+1)(p+2) Ṽ3yn +………………………… p(p+1(p+2)……(p+(n-1)) Ṽnyn
2ḷ 3ḷ nḷ
Note:
This formula is used to interpolate the value of y near the end of a set of tabular values .
Problem:

14. From the following table of half yearly premium of policies maturing at qunnial ages estimate the
premium for polices maturing at age 63
Age(x) : 45 50 55 60 65
Premium(y): 114.84 96.16 83.32 74.48 64.48
Solution:
Age premium Ṽy Ṽ2y Ṽ3y Ṽ4y
45 114.84 -18.68 5.84 -1-84 0.68
50 96.16 -12.84 4 -1.16
55 83.32 -8.84 2.84
60 74.48 -6
65 68.4
Formula:
Yp=yn+ pṼyn+p(p+1)Ṽ2yn+ p(p+1)(p+2) Ṽ3yn +………………………… p(p+1(p+2)……(p+(n-1)) Ṽnyn
2ḷ 3ḷ nḷ
yn=68.48
Ṽyn=-6
Ṽ2yn=2.84
Ṽ3yn=-1.16
Yp=68.48+(-0.4)(-6)+(-0.4)(0.4+1)2.84+(-0.4)(-0.2+1)(-0.4+2)(-1.16)+(-0.4)(-0.2+1)(-0.4+2)(0.4+3)(0.68)
2 6 24
=68.48+2.4-0.3400+0.07424-0.02828
=70.58

15. 2. The amount A of a substance remaining reacting of interveral of time t in certain chemical
experiment is given below
T(min) 2 5 8 11
A(gm) 94.8 87.9 81.3 75.1
Obtain the value of A where t=9 mints using newton backward difference interpolation formula.
T A Ṽy Ṽ2y Ṽ3y
2 94.8 -6.9 0.3 0.1
5 87.9 -6.6 0.4
8 81.3 -6.2
11 75.1
Yp=yn+ pṼyn+p(p+1)Ṽ2yn+ p(p+1)(p+2) Ṽ3yn +………………………… p(p+1(p+2)……(p+(n-1)) Ṽnyn
2ḷ 3ḷ nḷ
yn=75.1
Ṽyn=-6.2
Ṽ2yn=0.4
Ṽ3yn=0.1
Yp=75.1+(-0.6)(-6.2)+(-0.6)(0.6+1)0.4+(-0.6)(-0.6+1)(-0.6+2)(0.1)
2 6
=75.1-3.72-0.048-0.0056+4.12
=79.20

16. 3.Using newtons formula the value of x=27
X 10 15 20 25 30
Y 35.4 32.2 29.1 26 23.1
Solution:
X y Ṽy Ṽ2y Ṽ3y Ṽ4y
10 35.4 -3.2 0.1 -0.1 0.3
15 32.2 -3.1 0 0.2
20 29.1 -3.1 0.2
25 26 -2.9
30 23.1
Yp=yn+ pṼyn+p(p+1)Ṽ2yn+ p(p+1)(p+2) Ṽ3yn +………………………… p(p+1(p+2)……(p+(n-1)) Ṽnyn
2ḷ 3ḷ nḷ
yn=23.1
Ṽyn=-2.9
Ṽ2yn=0.2
Ṽ3yn=0.2
Ṽ4yn=0.3
Yp=23.1+(-0.6)(-2.9)+(-0.6)(0.6+1)0.2+(-0.6)(-0.6+1)(-0.6+2)(0.2) +(-0.6)(-0.6+1)(-
2 6 24
0.6+2)(0.3)
=23.1+1.74+-0.024-0.01008-0.0112
=24.7947

17. LAGRANGE’S INTERPOLATION
FORMULA
Let f(x) = y be a function which takes the values y0,y1,y2………………yn
corresponding to the value x = x0,x1,x2…….xn .Since there are n+1 values of y
corresponding to n+1 value of x.
we can represent the function f(x) by a polynomial in x of degree n.
y=f(x)=(x-x1)(x-x2)……….(x-xn)y0 + (x-x0)(x-x2)……….(x-xn)y1 +……+(x-x0)(x-x1)……….(x-xn-1)yn
(x0-x1)(x0-x2)……….(x0-xn) (x1-x0)(x1-x2)…….(x1-xn) (xn-x0)(xn-x1)…….(xn-xn-1)
is known as lagrange’s interpolation formula.
It can be used whether the values x0,x1,x2…….xn are equally spaced or not and whether the
differences of y become smaller or not.

18. 1.Given the values
x :14 17 31 35
f(x): 64 44 39.1
find the value of f(x) corresponding of x=27
solution:
since there are only fours corresponding pairs of values given in the polynomial
representing the data is
x0=14 x1=17 x2=31 x3=35
y0=68.7 y1=64 y2=44 y3=39.1
by lagrange’s interpolation formula
y=f(x)=(x-x1)(x-x2)……….(x-xn)y0 + (x-x0)(x-x2)……….(x-xn)y1 +……+(x-x0)(x-
x1)……….(x-xn-1)yn
(x0-x1)(x0-x2)……….(x0-xn) (x1-x0)(x1-x2)…….(x1-xn) (xn-x0)(xn-
x1)…….(xn-xn-1)
= (27-17)(27-31)(27-35)68.7 +(27-17)(27-31)(27-35)64 + (27-17)(27-31)(27-35)44
(14-17)(14-31)(14-35) (17-14)(17-31)(17-35) (31-14)(31-17)(31-35)
+(27-14)(27-31)(27-17)39.1
(35.4-14)(35-17)(35-31)
=20.5266+35.2169+48.0672-13.4479
=49.3104