1.   NUMERICAL METHODS

2. INTERPOLATION AND SOLUTION OF ALGEBRAIC AND TRANSCENDENTAL EQUATIONS
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. NEWTON GREGORY FORWARD INTERPOLATION
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°
Pressure
Using newtons formula find the pressure of a steam for a temperature 142°?

10. Solution:
x y ∆y ∆2y ∆3y ∆4y
yp=y0+p∆y0+p(p-1)∆2y0+ p(p-1)(p-2)∆3y0+­…………………………
2ḷ ḷ
p(p-1(p-2)……(p-(n-1))∆ny0­ nḷ
y0= p=x-x0 h
∆y0= =( )/10=0.2
∆2y0=0.279
∆3y0=0.047
∆4y0=0.002
Yp= (1.169)+0.2(0.2-1) (0.2-1)(0.2-2) (0.2-
1)(0.2-2)(0.2-3)0.002 24 = =
=3.899

11. 2. Following data gives the value x and y
2.Following data gives the value x and y. find the value of y when x is 21. From the table X Y
Solution:
x y ∆y ∆2y ∆3y
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 ḷ nḷ
Y0= p=x-x0 h
∆y0= =(21-20)/3=0.3
∆2y0=-0.001
∆3y0=0.0003
yp= (0.0487)+0.3(0.3-1)(-0.001)+0.3(0.3-1)(0.3-2)0.0003 =
=0.3567

12. 3. Using newtons forward interpolation formula. Calculate x is 1
3.Using newtons forward interpolation formula. Calculate x is 1.02 upto 3 decimal places in the table given
below X
F(x)
Solution
X f(x) ∆y ∆2y ∆3y ∆4y
yp=y0+p∆y0+p(p-1)∆2y0+ p(p-1)(p-2)∆3y0+­………………………… p(p-1(p-2)……(p-(n-1))∆ny0­
2ḷ ḷ nḷ
y0= p=x-x0 h
∆y0= =(1.02-1)/0.1=0.2
∆2y0=-0.009
∆3y0=0
∆4y0=-0.002
Yp= (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) =
=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 ḷ 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) :
Premium(y):
Solution:
Age premium Ṽy Ṽ2y Ṽ3y Ṽ4y
Formula:
Yp=yn+ pṼyn+p(p+1)Ṽ2yn+ p(p+1)(p+2) Ṽ3yn +­………………………… p(p+1(p+2)……(p+(n-1)) Ṽnyn ­
2ḷ ḷ 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) =
=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)
A(gm)
Obtain the value of A where t=9 mints using newton backward difference interpolation formula.
T A Ṽy Ṽ2y Ṽ3y
Yp=yn+ pṼyn+p(p+1)Ṽ2yn+ p(p+1)(p+2) Ṽ3yn +­………………………… p(p+1(p+2)……(p+(n-1)) Ṽnyn ­
2ḷ ḷ 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 =
=79.20

16. 3.Using newtons formula the value of x=27Y
Solution:
X y Ṽy Ṽ2y Ṽ3y Ṽ4y
Yp=yn+ pṼyn+p(p+1)Ṽ2yn+ p(p+1)(p+2) Ṽ3yn +­………………………… p(p+1(p+2)……(p+(n-1)) Ṽnyn ­
2ḷ ḷ 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)(-
0.6+2)(0.3) = =

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.
LAGRANGE’S INTERPOLATION FORMULA

18. 1. Given the values x. :14. 17. 31. 35. f(x):. 64. 44. 39
1.Given the values x : f(x): 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= 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) (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 ( )(35-17)(35-31) = =