NUMERICAL METHODS
Systems Of Linear Algebraic Equations
Systems of equations arise in all branches of engineering and science. These equations
may be algebraic, transcendental (i.e., involving trigonometric, logarithmic, exponential,
etc., functions), ordinary differential equations, or partial differential equations. The
equations may be linear or nonlinear.
NUMERICAL METHODS
Systems of non-linear Equations
Nonlinear equations arise in many physical problems. Finding their roots, or zeros is a
common problem. The problem can be stated as follows:
Given the continuous nonlinear function f(x), find the value of x = α such that f(α) = 0,
where α is the root, or zero, of the nonlinear equation.
Nonlinear equations are solved by iterative methods. A trial solution is assumed, the trial
solution is substituted into the nonlinear equation to determine the error or mismatch, and the
mismatch is used in some systematic manner to generate an improved estimate of the
solution.
NUMERICAL METHODS
The workhorse methods of choice for solving nonlinear equations are Newton’s method, the
secant method, the Bisection method, and fixed point iteration. Nonlinear equations occur
throughout engineering and science. Nonlinear equations also arise in other areas of
numerical analysis.
Implicit methods for solving nonlinear differential equations yield
nonlinear difference equations
Note Bisection and Newton- Raphson are based on the intermediate value theorem (IVT), If
𝑓(𝑥) is continuous in the interval (a,b ) where 𝑓 𝑎 𝑓 𝑏 < 0, (𝑓 𝑎 𝑎𝑛𝑑 𝑓 𝑏 ) have
different sign, there exist at least a solution C 𝜖( a, b) where 𝑓 𝑐 = 0
NUMERICAL METHODS
• EXP 1: Find the possible sub-interval that contains the solution of 𝑓 𝑥 =
𝑥3
3
− 𝑥 2 + 1 in the
interval (-2, 3) with step size (h) = 1 using the intermediate value theorem. Use 4 decimal
points in your calculation
Bisection method
This method is a root-finding method that applies to any continuous functions with two known
values of opposite signs.
The interval defined by these two values is bisected and a sub-interval in which the function
changes sign is selected.
This sub-interval must contain the root. These two steps are repeatedly executed until the root
is in the form of the required precision level.
This method is also called as interval halving method, the binary method, or the dichotomy
method.
The Method: Explained
Let f be a continuous function defined on an interval [a,b] where f(a) and f(b) have opposite
signs.
The bisection method is applicable for solving the equation 𝑓(𝑥)=0 for a real variable x.
At each step, the interval is divided into two parts/halves by computing the midpoint, 𝑐𝑖 =
𝑎𝑖 +𝑏𝑖
2
and the value of f(c) at that point.
Unless the root is c, there are two possibilities: 𝑓(𝑎) and: 𝑓(𝑐) have opposite signs and
bracket a root,
OR: 𝑓(𝑐) and: 𝑓(𝑏) have opposite signs and bracket a root.
One of the sub-intervals is chosen as the new interval to be used in the next step.
This process is carried out again and again until the interval is sufficiently small.
Bisection METHODS
If 𝑓(𝑎) and 𝑓(𝑐) have opposite signs, then the value of b is replaced by c.
If 𝑓(𝑏) and 𝑓(𝑐) have opposite signs, then the value of a is replaced by c.
In the case, that: 𝑓(𝑐) = 0, c will be taken as the solution and the process stops
Steps
1. Using the two initial guesses, find the root in some interval
𝑐𝑖 =
𝑎𝑖 +𝑏𝑖
2
The size of the interval will decrease until the value of 𝑓 𝑐𝑖 = 0 0r 𝑏𝑖 − 𝑐𝑖 ≤ 𝜀 indicating
to stop the iteration where 𝜀 is the error of tolerance
This method always coverage
NUMERICAL METHODS
Figure
NUMERICAL METHODS
EXP II Find the solution of 𝑒 −𝑥 − 𝑥 = 0 in the interval (0,1) using the bisection method. Use 3 decimal places in
all calculations
Solution
The root of 𝑓 𝑥 is approximately 0.567
NUMERICAL METHODS
EXP III : Find the root of x4-x-10=0 accurate of ε=0.015 in the interval (1.5, 2) by using the bisection method,
use 3 decimal places in all calculation
𝒊
𝒂𝒊
𝒃𝒊
𝒄𝒊 =
𝒂𝒊 + 𝒃𝒊
𝟐
𝒇(𝒂𝒊 )
𝒇(𝒃𝒊 )
𝒇(𝒄𝒊 )
𝒂𝒊 − 𝒄𝒊 < 𝜺
𝒃𝒊 − 𝒄𝒊 < 𝜺
0
1.5
2
1.75
-6.438
4
-2.370
2-1.75 = 0.25
1
1.75
2
1.875
-2.371
4
0.485
1.75-1.875 = -0.125
2
3
NUMERICAL METHODS
EXERCISE
a) Find the approximate root of equation x3-x-4 = 0
b) Determine the root of the equation, f(x)=x3–x–2 for x ∈ [1,2].
c) You have a spherical storage tank containing oil. The tank has a diameter of 6 m. You are asked to calculate the
height, h, to which a dipstick 8m long would be wet with oil when immersed in the tank when it contains 4 m3
of oil.
Dipstick
Spherical Storage Tank
r
h
The equation that gives the height, h, of liquid in the spherical tank for the given volume and radius is given by
f h  h3  9h 2  3.8197  0
NUMERICAL METHODS
Use the bisection method of finding roots of equations to find the height, h, to which the
dipstick is wet with oil. Find the absolute relative approximate error at the end of each
iteration and the number of significant digits at least correct at the end of each iteration.
Newton-Raphson Method
gg
AB
tan(  
AC
f(x)
f(xi)
f ( xi )
f ' ( xi ) 
xi  xi 1
B
C

xi+1
A
xi
f ( xi )
xi 1  xi 
f ( xi )
X
Newton-Raphson Method
Only one initial guess 𝑥𝑖 is needed to start the iteration
Formula 𝑥𝑖+1 = 𝑥𝑖 −
𝑓(𝑥)
𝑓′(𝑥)
Stop the iteration when the value of 𝑓(𝑥𝑖) = 0 or 𝑥𝑖 − 𝑥𝑖−1 ≤ 𝜀 where 𝜀 is the error of
tolerance
If the method converges, it is much faster than the bisection method
Newton-Raphson Method
EXP IV: Use Newton Raphson method to estimate the root of 𝑓 𝑥 = 𝑥 2 + 2𝑥 − 5 = 0
accurate to 𝜀 < 0.001 and given an initial point x0 = 1. Use 4 decimal places in all calculation
𝒙𝒊
i
𝒇(𝒙𝒊)
𝒇′(𝒙𝒊)
0
1
-2
4
1
1.5
0.25
5
2
3
𝒙𝒊 − 𝒙𝒊 − 𝟏
𝟏. 𝟓 − 𝟏 = 𝟎. 𝟓
Newton-Raphson Method
• EXP V: Find the approximate root of x3+3x-7 up to 4 iterations
Secant Method – Derivation
The Geometric Similar Triangles
f(x)
f(xi)
AB DC

AE DE
B
can be written as
f ( xi )
f ( xi 1 )

xi  xi 1 xi 1  xi 1
C
f(xi-1)
xi+1
E D
xi-1
A
xi
X
xi 1  xi 
f ( xi )( xi  xi 1 )
f ( xi )  f ( xi 1 )
Secant Method – Derivation
Secant Method
 The open method which may converge or may not converge
 If it converges, it is faster than the bisection and Newton-Raphson method.
 Use two initial guesses 𝑥𝑖 𝑎𝑛𝑑 𝑥𝑖−1
The formula for the next 𝑥𝑖+1 = 𝑥𝑖 − 𝑓(𝑥𝑖 )
𝑥𝑖 −𝑥𝑖−1
𝑓 𝑥𝑖 −𝑓(𝑥𝑖−1 )
Stop the iteration when 𝑥𝑖+1 = 𝑥𝑖 (the point is repeated)
𝑓 𝑥𝑖 ≈ 0
A secant line is the straight points line joining two points on a function
Secant Method – Derivation
Exp 8: Solve x2- 3 = 0 using secant method with initial value x0 = 1 and x1 = 2 use 4 decimal
place in all calculations
𝑖
0
𝑥𝑖
1
𝑓(𝑥𝑖 )
𝒙𝟐 = 𝟐 − 𝟏
𝟐−𝟏
𝟏− −𝟐
= 𝟏. 𝟔𝟔𝟔𝟕
𝒙𝟑 = 𝟏. 𝟔𝟔𝟔𝟕 − −𝟎. 𝟐𝟐𝟐𝟏
𝟏.𝟔𝟔𝟔𝟕−𝟐
−𝟎.𝟐𝟐𝟐𝟏−𝟏
-2
The root of 𝒇(𝒙) is approximate 1.7321
1
2
1
2
1.6667
-0.2221
3
1.7273
-0.0164
4
1.7321
0.0002
5
1.7321
0.0002
= 𝟏. 𝟕𝟐𝟕𝟑
The fixed-point interaction method
 Fixed point of a function = element of the function’s domain that is mapped to itself by the
function
 The number p is a fixed point for a given function 𝑔 𝑥 𝑖𝑓 𝑔 𝑝 = 𝑝
 Reformulate a nonlinear equation 𝑓 𝑥 = 𝑜 𝑡𝑜 𝑥 = 𝑔(𝑥)
 Start with one initial guess
 Stop the interaction when 𝑥𝑖+1 − 𝑥1 = 0 (the point is repeated)
The fixed-point interaction method
Convergence condition
If 𝑔 𝑥 𝑎𝑛𝑑 𝑔′(𝑥) are continuous for all in the interval (a, b), then the iteration process defined
by 𝑥𝑖+1 = 𝑔(𝑥) will converge to the solution s if 𝑔′(𝑥0 ) < 1 with 𝑥0 is an initial value where s
and 𝑥0 belong to the interval (a, b)
The fixed-point interaction method
Example IX: Solve 2x2-6x+3 = 0 using fixed point iteration method with initial value x0 =1 and
𝑥=𝑔 𝑥 =
2𝑥 2 +3
6
Use 4 decimal places in all calculations
Solution
𝑔 𝑥 =
2𝑥 2 +3
6
, 𝑔′ 𝑥 =
When X0 = 1 𝑔 1 =
Hence 𝑥𝑖+1 =
2𝑥 2 +3
6
2
3
2𝑥
3
1 = 0.6667 < 1 , it converges
The fixed-point interaction method
Example IX:
𝑖
𝑥𝑖
𝑥𝑖+1 − 𝑥𝑖
−
0
1.0000
1
0.8333
0.1667
2
0.7315
0.1018
3
0.6784
0.0531
4
0.6534
0.0250
5
0.6423
0.0111
6
0.6375
0.0048
7
0.6355
0.0020
8
0.6346
0.0009
9
0.6342
0.0004
10
0.6341
0.0001
11
0.6340
0.0001
12
0.6340
0.0000
The root of 𝑓 𝑥 = 2𝑥 2 − 6𝑥 + 3 is approximately 0.6340
Systems of nonlinear equations
So far, the roots of a single equation have been handled , now it is time to focus on solving the
roots of a set of simultaneous equations .
Newton-Raphson method
Consider a system of two equation
𝑓 𝑥, 𝑦 = 0 𝑎𝑛𝑑 𝑔 𝑥, 𝑦 = 0
Let (𝑥0 , 𝑦0 ) are initial approximation to the root of equation
If (𝑥0 + ℎ, 𝑦0 + 𝑘) is the root of the system then
𝑓 𝑥0 + ℎ, 𝑦0 + 𝑘 = 0
g 𝑥0 + ℎ, 𝑦0 + 𝑘 = 0
Systems of nonlinear equations
Assuming 𝑓 𝑎𝑛𝑑 𝑔 are sufficiently differential then by Taylor's series we obtain
𝑓0 + ℎ
𝑑𝑓
𝑑𝑥0
+𝑘
𝑑𝑓
𝑑𝑦0
+ ⋯……… = 0
𝑔0 + ℎ
𝑑𝑔
𝑑𝑥0
+𝑘
𝑑𝑔
𝑑𝑦0
+ ⋯……… = 0
Where
𝑑𝑓
𝑑𝑥0
=
𝑑𝑓
,
𝑑𝑥 𝑥=𝑥0
𝑓0 = 𝑓(𝑥0 , 𝑦0 ) etc
Neglecting the second and higher order derivative terms, we obtain the following system of linear equations
𝑑𝑓
𝑑𝑓
ℎ
+𝑘
= −𝑓0
𝑑𝑥0
𝑑𝑦0
𝑑𝑔
𝑑𝑔
ℎ
+𝑘
= −𝑔0
𝑑𝑥0
𝑑𝑦0
Systems of nonlinear equations
The above equations solution of the determinant of the Jacobian matrix
𝑑𝑓
𝑑𝑥0
𝑑𝑔
𝑑𝑥0
D=
𝑑𝑓
𝑑𝑦0
𝑑𝑔
𝑑𝑦0
≠0
By Cramer's rule , the solution of equation is given by
h=
1
𝐷
−𝑓0
−𝑔0
𝑑𝑓
𝑑𝑦0
𝑑𝑔
𝑑𝑦0
and k
𝑑𝑓
1 𝑑𝑥
= 𝑑𝑔0
𝐷
𝑑𝑥0
−𝑓0
−𝑔0
The new approximations are therefore
𝑥1 = 𝑥0 + ℎ
𝑦1 = 𝑦0 + 𝑘
The process is to be repeated until we obtain the roots to desired accuracy .
Systems of nonlinear equations
Example: Solve the following system using Newton-Raphson method
3𝑦𝑥2 − 10𝑥 + 7 = 0
𝑦2 − 5𝑦 + 4 = 0
𝑓 𝑥, 𝑦 = 3𝑦𝑥2 − 10𝑥 + 7 = 0
𝑔 𝑥, 𝑦 = 𝑦2 − 5𝑦 + 6 = 0
𝑑𝑓
𝑑𝑥
= 6xy − 10 ,
𝑑𝑓
𝑑𝑦
= 3x2,
𝑑𝑔
𝑑𝑥
=0,
Taking 𝑥0 = 𝑦0 = 0.5 as initial approximate
𝑓0 = 3 0.5 0.5
𝑔0 = 0.5
2
2
− 10 0.5 + 7 = 2.375
− 5 0.5 + 4 = 1.75
𝑑𝑔
𝑑𝑦
= 2y − 5 ,
Systems of nonlinear equations
𝑑𝑓
𝑑𝑥 0
= 6 0.5 0.5 − 10 = −8.5
𝑑𝑔
𝑑𝑥 0
=0
D=
𝑑𝑓
𝑑𝑥0
𝑑𝑔
𝑑𝑥0
𝑑𝑔
𝑑𝑦 0
,
𝑑𝑓
𝑑𝑦0
= 3 0.5 2 = 0.75
= 2 0.5 − 5 = 4
𝑑𝑓
𝑑𝑦0
𝑑𝑔
𝑑𝑦0
−8.5 0.75
= 34
0 −4
=
h =
𝑑𝑓
𝑑𝑦0
1
𝑑𝑔
𝐷
−𝑔0 𝑑𝑦
0
K =
𝑑𝑓
1 𝑑𝑥0
𝐷 𝑑𝑔
𝑑𝑥0
−𝑓0
1
= 34
−𝑓0
−𝑔0
−2.375 0.75
= 0.3180
−1.75 − 4
1
= 34
−8.5 − 2.375
= 0.4375
−0 − 1.75
𝑥1 = 𝑥0 + ℎ = 0.5 + 0.3180 = 0.818
𝑦1 = 𝑦0 + 𝑘 = 0.5 + 0.4375 = 0.9375
Systems of nonlinear equations
For 2nd approximation
𝑓1 = 3 0.9375 0.818 2 − 10 0.818 + 7 = 0.7019
𝑔1 = 0.9375 2 − 5 0.9375 + 4 = 0.1914
𝑑𝑓
𝑑𝑥1
= 6 0.9375 0.818 − 10 = −5.3988
𝑑𝑔
𝑑𝑥1
=0
D=
𝑑𝑓
𝑑𝑥1
𝑑𝑔
𝑑𝑥1
𝑑𝑔
𝑑𝑦 2
𝑑𝑓
, 𝑑𝑦1 = 3 0.818
= 2 0.9375 − 5 = -3.125
𝑑𝑓
𝑑𝑦1
𝑑𝑔
𝑑𝑦1
−5.3988 2.0074
= 16.8712
0
− 3.125
=
h =
𝑑𝑓
𝑑𝑦1
1
𝑑𝑔
𝐷
−𝑔1 𝑑𝑦
1
K =
𝑑𝑓
1 𝑑𝑥1
𝐷 𝑑𝑔
𝑑𝑥1
−𝑓1
1
= 16.8712
−𝑓1
−𝑔1
1
−0.7019
−0.1914
= 16.8712
2.0074
= 0.1528
− 3.125
−5.3988 − 0.7109
= 0.0612
−0
− 0.1914
𝑥2 = 𝑥1 + ℎ = 0.818 + 0.1528 = 0.9708
𝑦2 = 𝑦1 + 𝑘 = 0.9375 + 0.0612 = 0.9987
2
= 2.0074
Systems of nonlinear equations
For 3nd approximation
𝑓2 = 3 0.9987 0.9708
𝑔2 = 0.9987
2
2
− 10 0.9708 + 7 = 0.1156
− 5 0.9987 + 4 = 0.0039
𝑑𝑓
𝑑𝑥 2
= 6 0.9987 0.9708 − 10 = −4.1827
𝑑𝑔
𝑑𝑥 2
=0
D=
𝑑𝑓
𝑑𝑥2
𝑑𝑔
𝑑𝑥2
𝑑𝑔
𝑑𝑦 2
𝑑𝑓
, 𝑑𝑦2 = 3 0.9708
2
= 2.8273
= 2 0.9987 − 5 = -3.0026
𝑑𝑓
𝑑𝑦2
𝑑𝑔
𝑑𝑦2
−4.1827 2.8273
= 12.55897
0
− 3.0026
=
h =
𝑑𝑓
𝑑𝑦2
1
𝑑𝑔
𝐷
−𝑔2 𝑑𝑦
2
K =
𝑑𝑓
1 𝑑𝑥2
𝐷 𝑑𝑔
𝑑𝑥2
−𝑓2
1
= 12.55897
−𝑓2
−𝑔2
1
−0.1156
−0.0039
= 12.55897
2.8273
= 0.0285
− 3.0026
−4.1827 − 0.1156
= 0.00129
−0
− 0.0039
𝑥3 = 𝑥2 + ℎ = 0.9708 + 0.0285 = 0.9993
𝑦3 = 𝑦2 + 𝑘 = 0.9987 + 0.00129 = 0.9999
Therefore x = 1 and y = 1
Systems of nonlinear equations
EXERCISE
1. Solve the system of nonlinear by Newton Raphson method
x2 + y2 -1
y = x2 Initial approximation x0 = 0.5 and y0 = 0.5
2. Solve the following equation using Newton Raphson’s method
sin 𝑥 − 𝑦 = −0.9793
cos 𝑦 − 𝑥 = −0.6703 with xo = 0.5 , y0 = 1.5 as initial approximation
3. Solve the following system of nonlinear equations
-2x3 +3y2+42 = 0
5x2 +3y3-69 = 0 use Newton’s method , start x = 1 ,y = 1 and carryout the first five iterations
Systems of nonlinear equations
EXERCISE
1. Solve the system of nonlinear by Newton Raphson method
x2 + y2 -1
y = x2 Initial approximation x0 = 0.5 and y0 = 0.5
2. Solve the following equation using Newton Raphson’s method
sin 𝑥 − 𝑦 = −0.9793
cos 𝑦 − 𝑥 = −0.6703 with xo = 0.5 , y0 = 1.5 as initial approximation
3. Solve the following system of nonlinear equations
-2x3 +3y2+42 = 0
5x2 +3y3-69 = 0 use Newton’s method , start x = 1 ,y = 1 and carryout the first five iterations
Interpolation polynomials
Interpolation
 Process of constructing a polynomial /mathematical equation based on set of data points (𝑥0, 𝑦0), (𝑥𝑛 −
Number of point n
Order /degree of polynomial
2
P1(x) first order
3
P2(x) Second order
4
P3(x) third order
n+1
Pn(x) n order
Interpolation polynomials
Interpolation polynomial methods
 Lagrange interpolation
 Newton divided interpolation
 Neville’s method
Lagrange Interpolation
Definition
Given data points (x0, yo), (x1,y1)……………………….(xn-1, yn-1) (xn,yn), the interpolation polynomial represented by
Lagrange interpolation is
Pn(x) = L0(x)yo+ L1(x)y1+ L2(x)y2+ …………………………………………..Ln(x)yn
Where
𝑥 − 𝑥0 𝑥 − 𝑥1 … … (𝑥 − 𝑥𝑖−1 )(𝑥 − 𝑥𝑖+1 )(𝑥 − 𝑥𝑛)
𝐿𝑖 =
(𝑥𝑖 − 𝑥0) 𝑥𝑖 − 𝑥1 … . . 𝑥𝑖 − 𝑥𝑖−1 … 𝑥𝑖 − 𝑥𝑖+1 . . (𝑥𝑖 − 𝑥𝑛)
Interpolation polynomials
Example One: Form an interpolation polynomial using Lagrange methods for the following data sets . Then
find y for x= 2.5
x
1
2
3
y
2
5
7
Example TWO :Apply Lagrange interpolation polynomial to evaluate P3( 3.5) based on the following data set
x
1
2
3
5
y
2
5
7
3
Interpolation polynomials
Example Three: Given the following data
x
y
2
7.2
4.25
7.1
5.25
6.0
7.81
5.0
9.20
3.5
10.60
5.0
What is the value of y at x = 4 Using the first order Lagrange interpolation polynomial.
Interpolation polynomials
Newton divided difference interpolation
 Process of constructing a polynomial /mathematical equation based on set of data points (𝑥0, 𝑦0), (𝑥𝑛 −
Interpolation polynomials
x
y/f(x)
X0
f0
X1
f1
First d.d
𝑓 𝑥0 , 𝑥1
𝑓1 − 𝑓𝑜
𝑥1 − 𝑥0
𝑓 𝑥1, 𝑥2
X2
x3
f2
f3
𝑓2−𝑓1
𝑥2−𝑥1
𝑓 𝑥2, 𝑥3
𝑓3 − 𝑓2
𝑥3 − 𝑥2
Second d.d
𝑓 𝑥0, 𝑥1, 𝑥2
𝑓 𝑥1, 𝑥2 − 𝑓(𝑥0, 𝑥1)
𝑥2 − 𝑥0
𝑓 𝑥1, 𝑥2, 𝑥3
𝑓 𝑥2, 𝑥3 − 𝑓(𝑥1, 𝑥2)
𝑥3 − 𝑥1
Third d.d
𝑓 𝑥0, 𝑥1, 𝑥2, 𝑥3
𝑓 𝑥1, 𝑥2, 𝑥3 − 𝑓(𝑥0, 𝑥1, 𝑥2)
𝑥3 − 𝑥0
Interpolation polynomials
Example: Form an interpolation polynomial using Newton divided difference methods for the following data sets .
Then find y for 2.5
x
1
2
3
y
2
5
7
P2 x = f x0 + x − x0 f x0, x1 + x − x0
x − x1 f x0, x1, x2
= 2 + 𝑥 − 1 𝑓 𝑥0, 𝑥1 + 𝑥 − 1 𝑥 − 2 𝑓(𝑥0, 𝑥1, 𝑥2)
x
Y=f(x)
1
2
First d.d
Second d.d
5−2
=3
2−1
2
3
5
7
7−5
3−2
=2
2−3
= −0.5
3−1
= 2 + 𝑥 − 1 3 + 𝑥 − 1 𝑥 − 2 (−0.5) = -0.5x2 +4.5x-2
P2(2.5) = -0.5(2.5)2+4.5(2.5) = 6.125
Interpolation polynomials
Neville’s Method
 Newton's method of interpolation involves two steps: computation of the coefficients, followed
by evaluation the polynomial.
 This works well if the interpolation is carried out repeatedly at different values of x using the
same polynomial.
 If only one point is to be interpolated, a method that computes the interpolant in a single step,
such as Neville’s algorithm, is a better choice.
Interpolation polynomials
Let f be defined at (n+1)distinct points x0, xi……xn
Then for each 0≤ 𝑖 < 𝑗 ≤ 𝑛
Then 𝑃𝑖,𝑖+1…..𝑗(𝑥) =
𝑥−𝑥𝑖 𝑃𝑖+1,𝑖+2….𝑗 𝑥 −(𝑥−𝑥 )𝑃
𝑗 𝑖,𝑖+1,…𝑗−1(𝑥)
𝑥𝑗 −𝑥𝑖
is the 𝑛𝑡ℎ Lagrange
polynomial
NB : The above theorem implies that the interpolating polynomial can be generated recursively.
for Example
Interpolation polynomials
Interpolation polynomials
Neville’s Method Example
Consider the following table of x and corresponding y values.
x
8.1
8.3
8.6
y
16.9446
17.56492
18.50515
8.7
18.82091
Suppose we are interested in interpolating a polynomial that passes through these points to approximate the resulting y value
from an x value of 8.4.
Interpolation polynomials
x
8.1
8.3
8.6
8.7
y
16.9446
17.56492
18.50515
18.82091