![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](https://image.slidesharecdn.com/newtonsdivideddifferenceinterpolation18-230511171805-df557b0f/85/Newton-s-Divided-Difference-Interpolation-18-pptx-4-320.jpg)
![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).](https://image.slidesharecdn.com/newtonsdivideddifferenceinterpolation18-230511171805-df557b0f/85/Newton-s-Divided-Difference-Interpolation-18-pptx-5-320.jpg)
![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](https://image.slidesharecdn.com/newtonsdivideddifferenceinterpolation18-230511171805-df557b0f/85/Newton-s-Divided-Difference-Interpolation-18-pptx-9-320.jpg)
![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](https://image.slidesharecdn.com/newtonsdivideddifferenceinterpolation18-230511171805-df557b0f/85/Newton-s-Divided-Difference-Interpolation-18-pptx-10-320.jpg)
![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](https://image.slidesharecdn.com/newtonsdivideddifferenceinterpolation18-230511171805-df557b0f/85/Newton-s-Divided-Difference-Interpolation-18-pptx-13-320.jpg)
Newton’s Divided Difference Interpolation 18.pptx
- 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