ppt explaning Newton's Divided Difference.

1. Newton’s Divided
Difference Interpolation

2. Group members
22BCE10175 – Arya Shah
22BCE10463 – Harsh Kumar Shukla
22BCE10620 – Devansh Tiwari
22BCE10628 – Ikshu Patni
22BCE10739 – Abhishek Bhardwaj

3. Introduction
In this method, we do not need the different values of X to be spaced
evenly like forward and backward interpolation.
We can work with uneven spaced values of X and still get the value of
Y at any X.
We first have to make the divided difference table based on the
values of X and Y provided.

4. Xi fi
1st div. diff.
f [ Xi , Xj ]
2nd div. diff.
f [ Xi , Xj , Xk ]
3rd div. diff.
f [ Xi , Xj , Xk , Xl ]
X0 f0
f [X0 , X1]
=
(f1 – f0) / (X1 – X0)
X1 f1
f [ X0 , X1 , X2]
=
f [ X1 , X2 ]- f [ X0 , X1 ]
/ (X2 - X0)
f [X1 , X2]
=
(f2 – f1) / (X2 – X1)
f [X0 , X1 , X2 , X3]
=
f [ X1 , X2 , X3]– f [ X0 , X1 , X2 ]
/ (X3 – X0)
X2 f2
f [ X1 , X2 , X3]
=
f [ X2 , X3 ]- f [ X1 , X2 ]
/ (X3 - X1)
f [X2 , X3]
=
(f3 – f2) / (X3 – X2)
X3 f3

5. To get the value of f(X) at a particular X, we use the Newton’s
divided difference interpolation formula :
f(X) = f[X0] +(X – X0) f[X0 , X1] + (X – X0)(X – X1) f[X0 , X1 , X2]
+ (X – X0)(X – X1)(X – X2) f[X0 , X1 , X2 , X3]
based on the previous divided difference table.
After this we will substitute the values in the formula with the values
from the table and solve the equation to get f(X).

6. What is the need of Newton's
interpolation formula?
It is to furnish some mathematical tools that are used in
developing methods in the areas of approximation theory,
numerical integration, and the numerical solution of differential
equations.
It is the use of nested multiplication and the relative easiness
to add more data points for higher-order interpolating
polynomials.

7. This can be used for any interval.
The value of the function can be found at any point in the interval.
For different points also the same table can be used.
For new data, we can use this method by slightly changing the divide
difference table.
The value of y can be found regardless of the nature of x and of units
in which x is expressed.
It is less efficient than Lagrange’s interpolation when several
data sets are interpolated on the same data points
Advantages
Disadvantage

8. Solved Questions :
1) For the given table, use Newton divided difference to derive an
expression of a polynomial and get the value of f (2.7)
X
1 2 3
f (X)
3 5 8

9. Solution
First we will create the divided difference table :
Xi fi
1st div. diff.
f [ Xi , Xj ]
2nd div. diff.
f [ Xi , Xj , Xk ]
1 3
(f1 – f0) / (X1 – X0)
= 2/1
2 5 f [ X1 , X2 ]- f [ X0 , X1 ]
/ (X2 - X0)
= 1/2
(f2 – f1) / (X2 – X1)
= 3/1
3 8

10. Newton’s divided difference interpolation formula :
P(X) = f[X0] +(X – X0) f[X0 , X1] + (X – X0)(X – X1) f[X0 , X1 , X2]
After substituting every value from the table in the previous slide, we get
P(X) = 3 + (X – 1)(2) + (X – 1)(X – 2)(
1
2
)
= 3 + 2X – 2 + (X2 – 3X + 2)(
1
2
)
= 3 + 2X – 2 +
X2
2
–
3𝑋
2
+ 1
=
X2
2
+
X
2
+2
The polynomial is : f(X) =
X2
2
+
X
2
+2
The value of f(X) at X = 2.7
f(2.7) =
7.29
2
+
2.7
2
+2 = 3.645 + 1.35 + 2
f(2.7) = 6.995

11. 2) By means of Newton's divided difference formula, find f(8) and f(9) from
the following data
Solution
We need to find the values of Y at X=8 and X=9
First we will create divided difference table
X 4 5 7 10 11 13
f (X) 48 100 294 900 1210 2028

12. X f (X) 1st div.
diff.
2nd div.
diff.
3rd div.
diff.
4th div.
diff.
5th div.
diff.
4 48
52
5 100 45/3=
15
194/2=9
7
6/6=1
7 294 105/5=2
1
0
606/3=2
02
6/6=1 0
10 900 108/4=2
7
0
310 6/6=1
11 1210 99/3=
33
818/2=
409
13 2028

13. Newton’s divided difference interpolation formula :
P(X) = f[X0] +(X – X0) f[X0 , X1] + (X – X0)(X – X1) f[X0 , X1 , X2]
+ (X – X0)(X – X1)(X – X2) f[X0 , X1 , X2 , X3]
+ (X – X0)(X – X1)(X – X2)(X – X3) f[X0 , X1 , X2 , X3 , X4]
+ (X – X0)(X – X1)(X – X2)(X – X3)(X – X4) f[X0 , X1 , X2 , X3 , X4 , X5]
After substituting every value from the table in the previous slide, we get
P(X) = 48 + (X – 4)(52) + (X – 4)(X – 5)(15) + (X – 4)(X – 5)(X – 7)(1)
+ (X – 4)(X – 5)(X – 7)(X – 10)(0)
+ (X – 4)(X – 5)(X – 7)(X – 10)(X – 11)(0)
= 48 + 52X – 208 + 15X2 – 135X + 300 + X3 – 16X2 + 83X - 140
= X3 – X2
The polynomial is : f(X) = X3 – X2

14. The value of f(X) at X = 8
f (8) = 83 – 82
= 512 – 64
= 448
The value of f(X) at X = 9
f (9) = 93 – 92
= 729– 81
= 648

15. Application of Newton Divided Difference
Interpolation
• Interpolation of tabulated data: When we have a set of tabulated data that does not
have a continuous functional form, we can use Newton divided difference interpolation
to estimate the value of the function at a point between the tabulated values.
• Construction of polynomial models: Newton divided difference interpolation can be used
to construct polynomial models that approximate a given data set. This can be useful in a
variety of applications, such as curve fitting and data analysis.
• Numerical integration: Newton divided difference interpolation can be used to
approximate the value of an integral by constructing a polynomial that interpolates the
integrand at a set of points, and then integrating this polynomial.

16. Application of Newton Divided Difference
Interpolation
• Solving differential equations: Newton divided difference
interpolation can be used as a tool in numerical methods for solving
differential equations. In particular, it can be used to approximate the
solution of a differential equation at intermediate points, given a set
of initial conditions.
• Overall, Newton divided difference interpolation is a powerful
numerical method that has many applications in science, engineering,
and mathematics.

17. Practice Questions with solutions
1) Find Solution using Newton's Divided Difference Interpolation
formula at X = 301
Solution : f(301) = 2.4785
2) Find Solution using Newton's Divided Difference Interpolation
formula at X = 2.7
Solution : f(2.7) = 0.9941
X 300 304 305 307
f (X) 2.4771 2.4829 2.4843 2.4871
X 2 2.5 3
f (X) 0.69315 0.91629 1.09861

18. 3) Find Solution using Newton's Divided Difference
Interpolation formula at X = 3.8
Solution : Y(3.8) = 52.072
4) Find Solution using Newton's Divided Difference
Interpolation formula at X = 3.8
Solution : Y(3.8) = 95.544
5) Compute f(0.3) for the data given below using Newton's divided
difference formula
Solution : f(0.3) = 1.831
X 2 2.5 3 3.5 4
Y 7 14.125 25 40.375 61
X 2 2.5 3 3.5 4
Y 9 22.25 43 72.75 113
X 0 1 3 4 7
f (X) 1 3 49 129 813