Interpolation with Finite differences

Numerical method, Interpolation with finite differences, forward difference, backward difference, central difference, Gregory Newton Forward difference interpolation formula, Gregory Newton Backward difference interpolation formula, Stirlings interpolation formula, Gauss Forward interpolation formula, Gauss Backward interpolation formula

1. Numerical Methods - Finite Differences
Dr. N. B. Vyas
Department of Mathematics,
Atmiya Institute of Tech. and Science,
Rajkot (Guj.)
[email protected]
Dr. N. B. Vyas Numerical Methods - Finite Differences

2. Finite Differences
Forward dierence
Suppose that a functiony=f(x) is tabulated for the equally
spaced argumentsx0; x0+h; x0+ 2h; :::; x0+nhgiving the
functional valuesy0; y1; y2; :::; yn.
The constant dierence between two consecutive values ofxis
called theinterval of dierencesand is denoted byh.
The operator dened by
y0=y1y0
y1=y2y1:::::::
:::::::
yn1=ynyn1
is calledForward dierenceoperator.
Dr. N. B. Vyas Numerical Methods - Finite Differences

3. Finite Differences
Forward dierence
Suppose that a functiony=f(x) is tabulated for the equally
spaced argumentsx0; x0+h; x0+ 2h; :::; x0+nhgiving the
functional valuesy0; y1; y2; :::; yn.
The constant dierence between two consecutive values ofxis
called theinterval of dierencesand is denoted byh.
The operator dened by
y0=y1y0
y1=y2y1:::::::
:::::::
yn1=ynyn1
is calledForward dierenceoperator.
Dr. N. B. Vyas Numerical Methods - Finite Differences

4. Finite Differences
Forward dierence
Suppose that a functiony=f(x) is tabulated for the equally
spaced argumentsx0; x0+h; x0+ 2h; :::; x0+nhgiving the
functional valuesy0; y1; y2; :::; yn.
The constant dierence between two consecutive values ofxis
called theinterval of dierencesand is denoted byh.
The operator dened by
y0=y1y0
y1=y2y1:::::::
:::::::
yn1=ynyn1
is calledForward dierenceoperator.
Dr. N. B. Vyas Numerical Methods - Finite Differences

5. Finite Differences
Forward dierence
Suppose that a functiony=f(x) is tabulated for the equally
spaced argumentsx0; x0+h; x0+ 2h; :::; x0+nhgiving the
functional valuesy0; y1; y2; :::; yn.
The constant dierence between two consecutive values ofxis
called theinterval of dierencesand is denoted byh.
The operator dened by
y0=y1y0
y1=y2y1:::::::
:::::::
yn1=ynyn1
is calledForward dierenceoperator.
Dr. N. B. Vyas Numerical Methods - Finite Differences

6. Finite Differences
Forward dierence
Suppose that a functiony=f(x) is tabulated for the equally
spaced argumentsx0; x0+h; x0+ 2h; :::; x0+nhgiving the
functional valuesy0; y1; y2; :::; yn.
The constant dierence between two consecutive values ofxis
called theinterval of dierencesand is denoted byh.
The operator dened by
y0=y1y0
y1=y2y1:::::::
:::::::
yn1=ynyn1
is calledForward dierenceoperator.
Dr. N. B. Vyas Numerical Methods - Finite Differences

7. Finite Differences
Forward dierence
Suppose that a functiony=f(x) is tabulated for the equally
spaced argumentsx0; x0+h; x0+ 2h; :::; x0+nhgiving the
functional valuesy0; y1; y2; :::; yn.
The constant dierence between two consecutive values ofxis
called theinterval of dierencesand is denoted byh.
The operator dened by
y0=y1y0
y1=y2y1:::::::
:::::::
yn1=ynyn1
is calledForward dierenceoperator.
Dr. N. B. Vyas Numerical Methods - Finite Differences

8. Finite Differences
Therst forward dierenceis yn=yn+1yn
Thesecond forward dierenceare dened as the dierence of
the rst dierences.

2
y0= (y0) = (y1y0) = y1y0
=y2y1(y1y0) =y22y1+y0

2
y1= y2y1

2
yi= yi+1yi
In general nth forward dierence offis dened by

n
yi=
n1
yi+1
n1
yi
Dr. N. B. Vyas Numerical Methods - Finite Differences

9. Finite Differences
Therst forward dierenceis yn=yn+1yn
Thesecond forward dierenceare dened as the dierence of
the rst dierences.

2
y0= (y0) = (y1y0) = y1y0
=y2y1(y1y0) =y22y1+y0

2
y1= y2y1

2
yi= yi+1yi
In general nth forward dierence offis dened by

n
yi=
n1
yi+1
n1
yi
Dr. N. B. Vyas Numerical Methods - Finite Differences

10. Finite Differences
Forward Dierence Table:
x y y
2
y
3
y
4
y
5
y
x0y0
y0
x1y1
2
y0
y1
3
y0
x2y2
2
y1
4
y0
y2
3
y1
5
y0
x3y3
2
y2
4
y1
y3
3
y2
x4y4
2
y3
y4
x5y5
Dr. N. B. Vyas Numerical Methods - Finite Differences

11. Finite Differences
The operator satises the following properties:
1 f(x)g(x)] = f(x)g(x)
2 cf(x)] =cf(x)3
m

n
f(x) =
m+n
f(x),m; nare positive integers
4
n
ynis a constant,
n+1
yn= 0 ,
n+2
yn= 0,: : :
i.e. (n+ 1)
th
and higher dierences are zero.
Dr. N. B. Vyas Numerical Methods - Finite Differences

12. Finite Differences
The operator satises the following properties:
1 f(x)g(x)] = f(x)g(x)
2 cf(x)] =cf(x)3
m

n
f(x) =
m+n
f(x),m; nare positive integers
4
n
ynis a constant,
n+1
yn= 0 ,
n+2
yn= 0,: : :
i.e. (n+ 1)
th
and higher dierences are zero.
Dr. N. B. Vyas Numerical Methods - Finite Differences

13. Finite Differences
The operator satises the following properties:
1 f(x)g(x)] = f(x)g(x)
2 cf(x)] =cf(x)3
m

n
f(x) =
m+n
f(x),m; nare positive integers
4
n
ynis a constant,
n+1
yn= 0 ,
n+2
yn= 0,: : :
i.e. (n+ 1)
th
and higher dierences are zero.
Dr. N. B. Vyas Numerical Methods - Finite Differences

14. Finite Differences
Backward dierence
Suppose that a functiony=f(x) is tabulated for the equally
spaced argumentsx0; x0+h; x0+ 2h; :::; x0+nhgiving the
functional valuesy0; y1; y2; :::; yn.
The operatorrdened by
ry1=y1y0
ry2=y2y1:::::::
:::::::
ryn=ynyn1
is calledBackward dierenceoperator.
Dr. N. B. Vyas Numerical Methods - Finite Differences

15. Finite Differences
Backward dierence
Suppose that a functiony=f(x) is tabulated for the equally
spaced argumentsx0; x0+h; x0+ 2h; :::; x0+nhgiving the
functional valuesy0; y1; y2; :::; yn.
The operatorrdened by
ry1=y1y0
ry2=y2y1:::::::
:::::::
ryn=ynyn1
is calledBackward dierenceoperator.
Dr. N. B. Vyas Numerical Methods - Finite Differences

16. Finite Differences
Backward dierence
Suppose that a functiony=f(x) is tabulated for the equally
spaced argumentsx0; x0+h; x0+ 2h; :::; x0+nhgiving the
functional valuesy0; y1; y2; :::; yn.
The operatorrdened by
ry1=y1y0
ry2=y2y1:::::::
:::::::
ryn=ynyn1
is calledBackward dierenceoperator.
Dr. N. B. Vyas Numerical Methods - Finite Differences

17. Finite Differences
Backward dierence
Suppose that a functiony=f(x) is tabulated for the equally
spaced argumentsx0; x0+h; x0+ 2h; :::; x0+nhgiving the
functional valuesy0; y1; y2; :::; yn.
The operatorrdened by
ry1=y1y0
ry2=y2y1:::::::
:::::::
ryn=ynyn1
is calledBackward dierenceoperator.
Dr. N. B. Vyas Numerical Methods - Finite Differences

18. Finite Differences
Backward dierence
Suppose that a functiony=f(x) is tabulated for the equally
spaced argumentsx0; x0+h; x0+ 2h; :::; x0+nhgiving the
functional valuesy0; y1; y2; :::; yn.
The operatorrdened by
ry1=y1y0
ry2=y2y1:::::::
:::::::
ryn=ynyn1
is calledBackward dierenceoperator.
Dr. N. B. Vyas Numerical Methods - Finite Differences

19. Finite Differences
Therst backward dierenceisryn=ynyn1
Thesecond backward dierenceare obtain by the dierence
of the rst dierences.
r
2
y2=r(ry2) =r(y2y1) =ry2 ry1
=y2y1(y1y0) =y22y1+y0
r
2
y3=ry3 ry2
r
2
yn=ryn ryn1
In general nth backward dierence offis dened by
r
n
yi=r
n1
yi r
n1
yi1
Dr. N. B. Vyas Numerical Methods - Finite Differences

20. Finite Differences
Therst backward dierenceisryn=ynyn1
Thesecond backward dierenceare obtain by the dierence
of the rst dierences.
r
2
y2=r(ry2) =r(y2y1) =ry2 ry1
=y2y1(y1y0) =y22y1+y0
r
2
y3=ry3 ry2
r
2
yn=ryn ryn1
In general nth backward dierence offis dened by
r
n
yi=r
n1
yi r
n1
yi1
Dr. N. B. Vyas Numerical Methods - Finite Differences

21. Finite Differences
Therst backward dierenceisryn=ynyn1
Thesecond backward dierenceare obtain by the dierence
of the rst dierences.
r
2
y2=r(ry2) =r(y2y1) =ry2 ry1
=y2y1(y1y0) =y22y1+y0
r
2
y3=ry3 ry2
r
2
yn=ryn ryn1
In general nth backward dierence offis dened by
r
n
yi=r
n1
yi r
n1
yi1
Dr. N. B. Vyas Numerical Methods - Finite Differences

22. Finite Differences
Backward Dierence Table:
x y ryr
2
yr
3
yr
4
yr
5
y
x0y0
ry1
x1y1 r
2
y2
ry2 r
3
y3
x2y2 r
2
y3 r
4
y4
ry3 r
3
y4 r
5
y5
x3y3 r
2
y4 r
4
y5
ry4 r
3
y5
x4y4 r
2
y5
ry5
x5y5
Dr. N. B. Vyas Numerical Methods - Finite Differences

23. Finite Differences
Central dierence
The operatordened by
y1
2
=y1y0
y3
2
=y2y1:::::::
:::::::
y
n
1
2
=ynyn1
is calledCentral dierenceoperator.
Similarly, higher order central dierences are dened as

2
y1=y3
2
y1
2

2
y2=y5
2
y3
2
:::::::

3
y3
2
=
2
y2
2
y1
Dr. N. B. Vyas Numerical Methods - Finite Differences

24. Finite Differences
Central dierence
The operatordened by
y1
2
=y1y0
y3
2
=y2y1:::::::
:::::::
y
n
1
2
=ynyn1
is calledCentral dierenceoperator.
Similarly, higher order central dierences are dened as

2
y1=y3
2
y1
2

2
y2=y5
2
y3
2
:::::::

3
y3
2
=
2
y2
2
y1
Dr. N. B. Vyas Numerical Methods - Finite Differences

25. Finite Differences
Central dierence
The operatordened by
y1
2
=y1y0
y3
2
=y2y1:::::::
:::::::
y
n
1
2
=ynyn1
is calledCentral dierenceoperator.
Similarly, higher order central dierences are dened as

2
y1=y3
2
y1
2

2
y2=y5
2
y3
2
:::::::

3
y3
2
=
2
y2
2
y1
Dr. N. B. Vyas Numerical Methods - Finite Differences

26. Finite Differences
Central dierence
The operatordened by
y1
2
=y1y0
y3
2
=y2y1:::::::
:::::::
y
n
1
2
=ynyn1
is calledCentral dierenceoperator.
Similarly, higher order central dierences are dened as

2
y1=y3
2
y1
2

2
y2=y5
2
y3
2
:::::::

3
y3
2
=
2
y2
2
y1
Dr. N. B. Vyas Numerical Methods - Finite Differences

27. Finite Differences
Central dierence
The operatordened by
y1
2
=y1y0
y3
2
=y2y1:::::::
:::::::
y
n
1
2
=ynyn1
is calledCentral dierenceoperator.
Similarly, higher order central dierences are dened as

2
y1=y3
2
y1
2

2
y2=y5
2
y3
2
:::::::

3
y3
2
=
2
y2
2
y1
Dr. N. B. Vyas Numerical Methods - Finite Differences

28. Finite Differences
Central dierence
The operatordened by
y1
2
=y1y0
y3
2
=y2y1:::::::
:::::::
y
n
1
2
=ynyn1
is calledCentral dierenceoperator.
Similarly, higher order central dierences are dened as

2
y1=y3
2
y1
2

2
y2=y5
2
y3
2
:::::::

3
y3
2
=
2
y2
2
y1
Dr. N. B. Vyas Numerical Methods - Finite Differences

29. Finite Differences
Central dierence
The operatordened by
y1
2
=y1y0
y3
2
=y2y1:::::::
:::::::
y
n
1
2
=ynyn1
is calledCentral dierenceoperator.
Similarly, higher order central dierences are dened as

2
y1=y3
2
y1
2

2
y2=y5
2
y3
2
:::::::

3
y3
2
=
2
y2
2
y1
Dr. N. B. Vyas Numerical Methods - Finite Differences

30. Finite Differences
Central Dierence Table:
x y y
2
y
3
y
4
y
5
y
x0y0
y1
2
x1y1
2
y1
y3
2

3
y3
2
x2y2
2
y2
4
y2
y5
2

3
y5
2

5
y5
2
x3y3
2
y3
4
y3
y7
2

3
y7
2
x4y4
2
y4
y9
2
x5y5
Dr. N. B. Vyas Numerical Methods - Finite Differences

31. Finite Differences
NOTE:
From all three dierence tables, we can see that only the
notations changes not the dierences.
y1y0= y0=ry1=y1
2
Alternative notations for the functiony=f(x).For two consecutive values ofxdiering byh.yx=yx+hyx=f(x+h)f(x)ryx=yxyxh=f(x)f(xh)yx=y
x+
h
2
y
x
h
2
=f(x+
h
2
)f(x
h
2
)
Dr. N. B. Vyas Numerical Methods - Finite Differences

32. Finite Differences
NOTE:
From all three dierence tables, we can see that only the
notations changes not the dierences.
y1y0= y0=ry1=y1
2
Alternative notations for the functiony=f(x).For two consecutive values ofxdiering byh.yx=yx+hyx=f(x+h)f(x)ryx=yxyxh=f(x)f(xh)yx=y
x+
h
2
y
x
h
2
=f(x+
h
2
)f(x
h
2
)
Dr. N. B. Vyas Numerical Methods - Finite Differences

33. Finite Differences
NOTE:
From all three dierence tables, we can see that only the
notations changes not the dierences.
y1y0= y0=ry1=y1
2
Alternative notations for the functiony=f(x).For two consecutive values ofxdiering byh.yx=yx+hyx=f(x+h)f(x)ryx=yxyxh=f(x)f(xh)yx=y
x+
h
2
y
x
h
2
=f(x+
h
2
)f(x
h
2
)
Dr. N. B. Vyas Numerical Methods - Finite Differences

34. Finite Differences
NOTE:
From all three dierence tables, we can see that only the
notations changes not the dierences.
y1y0= y0=ry1=y1
2
Alternative notations for the functiony=f(x).For two consecutive values ofxdiering byh.yx=yx+hyx=f(x+h)f(x)ryx=yxyxh=f(x)f(xh)yx=y
x+
h
2
y
x
h
2
=f(x+
h
2
)f(x
h
2
)
Dr. N. B. Vyas Numerical Methods - Finite Differences

35. Finite Differences
NOTE:
From all three dierence tables, we can see that only the
notations changes not the dierences.
y1y0= y0=ry1=y1
2
Alternative notations for the functiony=f(x).For two consecutive values ofxdiering byh.yx=yx+hyx=f(x+h)f(x)ryx=yxyxh=f(x)f(xh)yx=y
x+
h
2
y
x
h
2
=f(x+
h
2
)f(x
h
2
)
Dr. N. B. Vyas Numerical Methods - Finite Differences

36. Finite Differences
NOTE:
From all three dierence tables, we can see that only the
notations changes not the dierences.
y1y0= y0=ry1=y1
2
Alternative notations for the functiony=f(x).For two consecutive values ofxdiering byh.yx=yx+hyx=f(x+h)f(x)ryx=yxyxh=f(x)f(xh)yx=y
x+
h
2
y
x
h
2
=f(x+
h
2
)f(x
h
2
)
Dr. N. B. Vyas Numerical Methods - Finite Differences

37. Finite Differences
Evaluate the following. The interval of dierence beingh.
1
n
e
x
2logf(x)3 tan
1
x)4
2
cos2x
Dr. N. B. Vyas Numerical Methods - Finite Differences

38. Finite Differences
Evaluate the following. The interval of dierence beingh.
1
n
e
x
2logf(x)3 tan
1
x)4
2
cos2x
Dr. N. B. Vyas Numerical Methods - Finite Differences

39. Finite Differences
Evaluate the following. The interval of dierence beingh.
1
n
e
x
2logf(x)3 tan
1
x)4
2
cos2x
Dr. N. B. Vyas Numerical Methods - Finite Differences

40. Finite Differences
Evaluate the following. The interval of dierence beingh.
1
n
e
x
2logf(x)3 tan
1
x)4
2
cos2x
Dr. N. B. Vyas Numerical Methods - Finite Differences

41. Finite Differences
Other Dierence Operator
1. Shift OperatorE:
Edoes the operation of increasing the argumentxbyhso that
Ef(x) =f(x+h);
E
2
f(x) =E(Ef(x)) =Ef(x+h) =f(x+ 2h);E
3
f(x) =f(x+ 3h); E
n
f(x) =f(x+nh);The inverse operatorE
1
is dened asE
1
f(x) =f(xh);E
n
f(x) =f(xnh);Ifyxis the functionf(x), thenEyx=yx+h; E
1
yx=yxh;E
n
yx=yx+nh; Dr. N. B. Vyas Numerical Methods - Finite Differences

42. Finite Differences
Other Dierence Operator
1. Shift OperatorE:
Edoes the operation of increasing the argumentxbyhso that
Ef(x) =f(x+h);
E
2
f(x) =E(Ef(x)) =Ef(x+h) =f(x+ 2h);E
3
f(x) =f(x+ 3h); E
n
f(x) =f(x+nh);The inverse operatorE
1
is dened asE
1
f(x) =f(xh);E
n
f(x) =f(xnh);Ifyxis the functionf(x), thenEyx=yx+h; E
1
yx=yxh;E
n
yx=yx+nh; Dr. N. B. Vyas Numerical Methods - Finite Differences

43. Finite Differences
Other Dierence Operator
1. Shift OperatorE:
Edoes the operation of increasing the argumentxbyhso that
Ef(x) =f(x+h);
E
2
f(x) =E(Ef(x)) =Ef(x+h) =f(x+ 2h);E
3
f(x) =f(x+ 3h); E
n
f(x) =f(x+nh);The inverse operatorE
1
is dened asE
1
f(x) =f(xh);E
n
f(x) =f(xnh);Ifyxis the functionf(x), thenEyx=yx+h; E
1
yx=yxh;E
n
yx=yx+nh; Dr. N. B. Vyas Numerical Methods - Finite Differences

44. Finite Differences
Other Dierence Operator
1. Shift OperatorE:
Edoes the operation of increasing the argumentxbyhso that
Ef(x) =f(x+h);
E
2
f(x) =E(Ef(x)) =Ef(x+h) =f(x+ 2h);E
3
f(x) =f(x+ 3h); E
n
f(x) =f(x+nh);The inverse operatorE
1
is dened asE
1
f(x) =f(xh);E
n
f(x) =f(xnh);Ifyxis the functionf(x), thenEyx=yx+h; E
1
yx=yxh;E
n
yx=yx+nh; Dr. N. B. Vyas Numerical Methods - Finite Differences

45. Finite Differences
Other Dierence Operator
1. Shift OperatorE:
Edoes the operation of increasing the argumentxbyhso that
Ef(x) =f(x+h);
E
2
f(x) =E(Ef(x)) =Ef(x+h) =f(x+ 2h);E
3
f(x) =f(x+ 3h); E
n
f(x) =f(x+nh);The inverse operatorE
1
is dened asE
1
f(x) =f(xh);E
n
f(x) =f(xnh);Ifyxis the functionf(x), thenEyx=yx+h; E
1
yx=yxh;E
n
yx=yx+nh; Dr. N. B. Vyas Numerical Methods - Finite Differences

46. Finite Differences
Other Dierence Operator
1. Shift OperatorE:
Edoes the operation of increasing the argumentxbyhso that
Ef(x) =f(x+h);
E
2
f(x) =E(Ef(x)) =Ef(x+h) =f(x+ 2h);E
3
f(x) =f(x+ 3h); E
n
f(x) =f(x+nh);The inverse operatorE
1
is dened asE
1
f(x) =f(xh);E
n
f(x) =f(xnh);Ifyxis the functionf(x), thenEyx=yx+h; E
1
yx=yxh;E
n
yx=yx+nh; Dr. N. B. Vyas Numerical Methods - Finite Differences

47. Finite Differences
Other Dierence Operator
1. Shift OperatorE:
Edoes the operation of increasing the argumentxbyhso that
Ef(x) =f(x+h);
E
2
f(x) =E(Ef(x)) =Ef(x+h) =f(x+ 2h);E
3
f(x) =f(x+ 3h); E
n
f(x) =f(x+nh);The inverse operatorE
1
is dened asE
1
f(x) =f(xh);E
n
f(x) =f(xnh);Ifyxis the functionf(x), thenEyx=yx+h; E
1
yx=yxh;E
n
yx=yx+nh; Dr. N. B. Vyas Numerical Methods - Finite Differences

48. Finite Differences
Other Dierence Operator
1. Shift OperatorE:
Edoes the operation of increasing the argumentxbyhso that
Ef(x) =f(x+h);
E
2
f(x) =E(Ef(x)) =Ef(x+h) =f(x+ 2h);E
3
f(x) =f(x+ 3h); E
n
f(x) =f(x+nh);The inverse operatorE
1
is dened asE
1
f(x) =f(xh);E
n
f(x) =f(xnh);Ifyxis the functionf(x), thenEyx=yx+h; E
1
yx=yxh;E
n
yx=yx+nh; Dr. N. B. Vyas Numerical Methods - Finite Differences

49. Finite Differences
Other Dierence Operator
1. Shift OperatorE:
Edoes the operation of increasing the argumentxbyhso that
Ef(x) =f(x+h);
E
2
f(x) =E(Ef(x)) =Ef(x+h) =f(x+ 2h);E
3
f(x) =f(x+ 3h); E
n
f(x) =f(x+nh);The inverse operatorE
1
is dened asE
1
f(x) =f(xh);E
n
f(x) =f(xnh);Ifyxis the functionf(x), thenEyx=yx+h; E
1
yx=yxh;E
n
yx=yx+nh; Dr. N. B. Vyas Numerical Methods - Finite Differences

50. Finite Differences
Other Dierence Operator
1. Shift OperatorE:
Edoes the operation of increasing the argumentxbyhso that
Ef(x) =f(x+h);
E
2
f(x) =E(Ef(x)) =Ef(x+h) =f(x+ 2h);E
3
f(x) =f(x+ 3h); E
n
f(x) =f(x+nh);The inverse operatorE
1
is dened asE
1
f(x) =f(xh);E
n
f(x) =f(xnh);Ifyxis the functionf(x), thenEyx=yx+h; E
1
yx=yxh;E
n
yx=yx+nh; Dr. N. B. Vyas Numerical Methods - Finite Differences

51. Finite Differences
Other Dierence Operator
1. Shift OperatorE:
Edoes the operation of increasing the argumentxbyhso that
Ef(x) =f(x+h);
E
2
f(x) =E(Ef(x)) =Ef(x+h) =f(x+ 2h);E
3
f(x) =f(x+ 3h); E
n
f(x) =f(x+nh);The inverse operatorE
1
is dened asE
1
f(x) =f(xh);E
n
f(x) =f(xnh);Ifyxis the functionf(x), thenEyx=yx+h; E
1
yx=yxh;E
n
yx=yx+nh; Dr. N. B. Vyas Numerical Methods - Finite Differences

52. Finite Differences
Other Dierence Operator
1. Shift OperatorE:
Edoes the operation of increasing the argumentxbyhso that
Ef(x) =f(x+h);
E
2
f(x) =E(Ef(x)) =Ef(x+h) =f(x+ 2h);E
3
f(x) =f(x+ 3h); E
n
f(x) =f(x+nh);The inverse operatorE
1
is dened asE
1
f(x) =f(xh);E
n
f(x) =f(xnh);Ifyxis the functionf(x), thenEyx=yx+h; E
1
yx=yxh;E
n
yx=yx+nh; Dr. N. B. Vyas Numerical Methods - Finite Differences

53. Finite Differences
2. Averaging Operator:
It is dened as
f(x) =
1
2

f

x+
h
2

+f

x
h
2

;
i.e.yx=
1
2
h
y
x+
h
2
+y
x
h
2
i
;
3. Dierential OperatorD:
It is dened asDf(x) =
d
dx
f(x) =f
0
(x);
Dr. N. B. Vyas Numerical Methods - Finite Differences

54. Finite Differences
2. Averaging Operator:
It is dened as
f(x) =
1
2

f

x+
h
2

+f

x
h
2

;
i.e.yx=
1
2
h
y
x+
h
2
+y
x
h
2
i
;
3. Dierential OperatorD:
It is dened asDf(x) =
d
dx
f(x) =f
0
(x);
Dr. N. B. Vyas Numerical Methods - Finite Differences

55. Finite Differences
2. Averaging Operator:
It is dened as
f(x) =
1
2

f

x+
h
2

+f

x
h
2

;
i.e.yx=
1
2
h
y
x+
h
2
+y
x
h
2
i
;
3. Dierential OperatorD:
It is dened asDf(x) =
d
dx
f(x) =f
0
(x);
Dr. N. B. Vyas Numerical Methods - Finite Differences

56. Finite Differences
2. Averaging Operator:
It is dened as
f(x) =
1
2

f

x+
h
2

+f

x
h
2

;
i.e.yx=
1
2
h
y
x+
h
2
+y
x
h
2
i
;
3. Dierential OperatorD:
It is dened asDf(x) =
d
dx
f(x) =f
0
(x);
Dr. N. B. Vyas Numerical Methods - Finite Differences

57. Finite Differences
2. Averaging Operator:
It is dened as
f(x) =
1
2

f

x+
h
2

+f

x
h
2

;
i.e.yx=
1
2
h
y
x+
h
2
+y
x
h
2
i
;
3. Dierential OperatorD:
It is dened asDf(x) =
d
dx
f(x) =f
0
(x);
Dr. N. B. Vyas Numerical Methods - Finite Differences

58. Finite Differences
Relation between the operators
1 E1
2r= 1E
1
3=E
1
2E

1
2
4=
1
2
fE
1
2+E

1
2g
5 Er=rE=E
1
2
6E=e
hD
7 r) = 18 r= r=
2
Dr. N. B. Vyas Numerical Methods - Finite Differences

59. Finite Differences
Relation between the operators
1 E1
2r= 1E
1
3=E
1
2E

1
2
4=
1
2
fE
1
2+E

1
2g
5 Er=rE=E
1
2
6E=e
hD
7 r) = 18 r= r=
2
Dr. N. B. Vyas Numerical Methods - Finite Differences

60. Finite Differences
Relation between the operators
1 E1
2r= 1E
1
3=E
1
2E

1
2
4=
1
2
fE
1
2+E

1
2g
5 Er=rE=E
1
2
6E=e
hD
7 r) = 18 r= r=
2
Dr. N. B. Vyas Numerical Methods - Finite Differences

61. Finite Differences
Relation between the operators
1 E1
2r= 1E
1
3=E
1
2E

1
2
4=
1
2
fE
1
2+E

1
2g
5 Er=rE=E
1
2
6E=e
hD
7 r) = 18 r= r=
2
Dr. N. B. Vyas Numerical Methods - Finite Differences

62. Finite Differences
Relation between the operators
1 E1
2r= 1E
1
3=E
1
2E

1
2
4=
1
2
fE
1
2+E

1
2g
5 Er=rE=E
1
2
6E=e
hD
7 r) = 18 r= r=
2
Dr. N. B. Vyas Numerical Methods - Finite Differences

63. Finite Differences
Relation between the operators
1 E1
2r= 1E
1
3=E
1
2E

1
2
4=
1
2
fE
1
2+E

1
2g
5 Er=rE=E
1
2
6E=e
hD
7 r) = 18 r= r=
2
Dr. N. B. Vyas Numerical Methods - Finite Differences

64. Finite Differences
Relation between the operators
1 E1
2r= 1E
1
3=E
1
2E

1
2
4=
1
2
fE
1
2+E

1
2g
5 Er=rE=E
1
2
6E=e
hD
7 r) = 18 r= r=
2
Dr. N. B. Vyas Numerical Methods - Finite Differences

65. Finite Differences
Relation between the operators
1 E1
2r= 1E
1
3=E
1
2E

1
2
4=
1
2
fE
1
2+E

1
2g
5 Er=rE=E
1
2
6E=e
hD
7 r) = 18 r= r=
2
Dr. N. B. Vyas Numerical Methods - Finite Differences

66. Finite Differences
9
2

2
=

1 +
1
2

2

2
10
2
= 1 +
1
4

211E
1
2=+
1
2
12E

1
2=
1
2

13
1
2

2
+
r
1 +

2
4
14=
1
2
E
1
+
1
2

Dr. N. B. Vyas Numerical Methods - Finite Differences

67. Finite Differences
9
2

2
=

1 +
1
2

2

2
10
2
= 1 +
1
4

211E
1
2=+
1
2
12E

1
2=
1
2

13
1
2

2
+
r
1 +

2
4
14=
1
2
E
1
+
1
2

Dr. N. B. Vyas Numerical Methods - Finite Differences

68. Finite Differences
9
2

2
=

1 +
1
2

2

2
10
2
= 1 +
1
4

211E
1
2=+
1
2
12E

1
2=
1
2

13
1
2

2
+
r
1 +

2
4
14=
1
2
E
1
+
1
2

Dr. N. B. Vyas Numerical Methods - Finite Differences

69. Finite Differences
9
2

2
=

1 +
1
2

2

2
10
2
= 1 +
1
4

211E
1
2=+
1
2
12E

1
2=
1
2

13
1
2

2
+
r
1 +

2
4
14=
1
2
E
1
+
1
2

Dr. N. B. Vyas Numerical Methods - Finite Differences

70. Finite Differences
9
2

2
=

1 +
1
2

2

2
10
2
= 1 +
1
4

211E
1
2=+
1
2
12E

1
2=
1
2

13
1
2

2
+
r
1 +

2
4
14=
1
2
E
1
+
1
2

Dr. N. B. Vyas Numerical Methods - Finite Differences

71. Finite Differences
9
2

2
=

1 +
1
2

2

2
10
2
= 1 +
1
4

211E
1
2=+
1
2
12E

1
2=
1
2

13
1
2

2
+
r
1 +

2
4
14=
1
2
E
1
+
1
2

Dr. N. B. Vyas Numerical Methods - Finite Differences

72. Finite Differences
9
2

2
=

1 +
1
2

2

2
10
2
= 1 +
1
4

211E
1
2=+
1
2
12E

1
2=
1
2

13
1
2

2
+
r
1 +

2
4
14=
1
2
E
1
+
1
2

Dr. N. B. Vyas Numerical Methods - Finite Differences

73. Gregory - Newton Forward Interpolation Formula
To estimate the value of a function near thebeginninga table,
the forward dierence interpolation formula in used.
Letyx=f(x) be a function which takes the values
yx0
; yx0+h; yx0+2h; : : :corresponding to the values
x0; x0+h; x0+ 2h; : : :ofx.
Suppose we want to evaluateyxwhenx=x0+ph, wherepis
any real number.
Dr. N. B. Vyas Numerical Methods - Finite Differences

74. Gregory - Newton Forward Interpolation Formula
To estimate the value of a function near thebeginninga table,
the forward dierence interpolation formula in used.
Letyx=f(x) be a function which takes the values
yx0
; yx0+h; yx0+2h; : : :corresponding to the values
x0; x0+h; x0+ 2h; : : :ofx.
Suppose we want to evaluateyxwhenx=x0+ph, wherepis
any real number.
Dr. N. B. Vyas Numerical Methods - Finite Differences

75. Gregory - Newton Forward Interpolation Formula
To estimate the value of a function near thebeginninga table,
the forward dierence interpolation formula in used.
Letyx=f(x) be a function which takes the values
yx0
; yx0+h; yx0+2h; : : :corresponding to the values
x0; x0+h; x0+ 2h; : : :ofx.
Suppose we want to evaluateyxwhenx=x0+ph, wherepis
any real number.
Dr. N. B. Vyas Numerical Methods - Finite Differences

76. Gregory - Newton Forward Interpolation Formula
To estimate the value of a function near thebeginninga table,
the forward dierence interpolation formula in used.
Letyx=f(x) be a function which takes the values
yx0
; yx0+h; yx0+2h; : : :corresponding to the values
x0; x0+h; x0+ 2h; : : :ofx.
Suppose we want to evaluateyxwhenx=x0+ph, wherepis
any real number.
Dr. N. B. Vyas Numerical Methods - Finite Differences

77. Gregory - Newton Forward Interpolation Formula
Let it beyp. For any real numbern, we have dened operatorE
such thatE
n
f(x) =f(x+nh).
)yx=yx0+ph=f(x0+ph) =E
p
yx0
= (1 + )
p
y0
=

1 +p +
p(p1)
2!

2
+
p(p1)(p2)
3!

3
+:::

y0
yx=y0+py0+
p(p1)
2!

2
y0+
p(p1)(p2)
3!

3
y0+:::
is calledNewton's forward interpolationformula.
Dr. N. B. Vyas Numerical Methods - Finite Differences

78. Gregory - Newton Forward Interpolation Formula
Let it beyp. For any real numbern, we have dened operatorE
such thatE
n
f(x) =f(x+nh).
)yx=yx0+ph=f(x0+ph) =E
p
yx0
= (1 + )
p
y0
=

1 +p +
p(p1)
2!

2
+
p(p1)(p2)
3!

3
+:::

y0
yx=y0+py0+
p(p1)
2!

2
y0+
p(p1)(p2)
3!

3
y0+:::
is calledNewton's forward interpolationformula.
Dr. N. B. Vyas Numerical Methods - Finite Differences

79. Gregory - Newton Forward Interpolation Formula
Let it beyp. For any real numbern, we have dened operatorE
such thatE
n
f(x) =f(x+nh).
)yx=yx0+ph=f(x0+ph) =E
p
yx0
= (1 + )
p
y0
=

1 +p +
p(p1)
2!

2
+
p(p1)(p2)
3!

3
+:::

y0
yx=y0+py0+
p(p1)
2!

2
y0+
p(p1)(p2)
3!

3
y0+:::
is calledNewton's forward interpolationformula.
Dr. N. B. Vyas Numerical Methods - Finite Differences

80. Example
Ex.
ndf(0:5).
x-2-10123
f(x)155131125
Dr. N. B. Vyas Numerical Methods - Finite Differences

81. Example
Ex.
below estimate the population for the year 1895.
Year: 18911901191119211931
Population(in thousand):46 66 81 93101
Dr. N. B. Vyas Numerical Methods - Finite Differences

82. Example
Ex.
Newton's forward method for the following data:
Year: 1976197819801982
Production:20 27 38 50
Dr. N. B. Vyas Numerical Methods - Finite Differences

83. Gregory - Newton Backward Interpolation Formula
To estimate the value of a function near theendof a table, the
backward dierence interpolation formula in used.
Letyx=f(x) be a function which takes the values
yx0
; yx0+h; yx0+2h; : : :corresponding to the values
x0; x0+h; x0+ 2h; : : :ofx.
Suppose we want to evaluateyxwhenx=xn+ph, wherepis
any real number.
Dr. N. B. Vyas Numerical Methods - Finite Differences

84. Gregory - Newton Backward Interpolation Formula
To estimate the value of a function near theendof a table, the
backward dierence interpolation formula in used.
Letyx=f(x) be a function which takes the values
yx0
; yx0+h; yx0+2h; : : :corresponding to the values
x0; x0+h; x0+ 2h; : : :ofx.
Suppose we want to evaluateyxwhenx=xn+ph, wherepis
any real number.
Dr. N. B. Vyas Numerical Methods - Finite Differences

85. Gregory - Newton Backward Interpolation Formula
To estimate the value of a function near theendof a table, the
backward dierence interpolation formula in used.
Letyx=f(x) be a function which takes the values
yx0
; yx0+h; yx0+2h; : : :corresponding to the values
x0; x0+h; x0+ 2h; : : :ofx.
Suppose we want to evaluateyxwhenx=xn+ph, wherepis
any real number.
Dr. N. B. Vyas Numerical Methods - Finite Differences

86. Gregory - Newton Backward Interpolation Formula
Let it beyp. For any real numbern, we have dened operatorE
such thatE
n
f(x) =f(x+nh).
)yx=yxn+ph=f(xn+ph) =E
p
yxn= (1 r)
p
yn
=

1 +pr+
p(p+ 1)
2!
r
2
+
p(p+ 1)(p+ 2)
3!
r
3
+:::

yn
yx=yn+pryn+
p(p+ 1)
2!
r
2
yn+
p(p+ 1)(p+ 2)
3!
r
3
yn+:::
is calledNewton's backward interpolationformula
Dr. N. B. Vyas Numerical Methods - Finite Differences

87. Gregory - Newton Backward Interpolation Formula
Let it beyp. For any real numbern, we have dened operatorE
such thatE
n
f(x) =f(x+nh).
)yx=yxn+ph=f(xn+ph) =E
p
yxn= (1 r)
p
yn
=

1 +pr+
p(p+ 1)
2!
r
2
+
p(p+ 1)(p+ 2)
3!
r
3
+:::

yn
yx=yn+pryn+
p(p+ 1)
2!
r
2
yn+
p(p+ 1)(p+ 2)
3!
r
3
yn+:::
is calledNewton's backward interpolationformula
Dr. N. B. Vyas Numerical Methods - Finite Differences

88. Gregory - Newton Backward Interpolation Formula
Let it beyp. For any real numbern, we have dened operatorE
such thatE
n
f(x) =f(x+nh).
)yx=yxn+ph=f(xn+ph) =E
p
yxn= (1 r)
p
yn
=

1 +pr+
p(p+ 1)
2!
r
2
+
p(p+ 1)(p+ 2)
3!
r
3
+:::

yn
yx=yn+pryn+
p(p+ 1)
2!
r
2
yn+
p(p+ 1)(p+ 2)
3!
r
3
yn+:::
is calledNewton's backward interpolationformula
Dr. N. B. Vyas Numerical Methods - Finite Differences

89. Example
Ex.
x= 1 from the following data.
x 0.1 0.2 0.3 0.40.50.6
f(x)1.6991.0730.3750.4431.4292.631
Dr. N. B. Vyas Numerical Methods - Finite Differences

90. Example
Ex.
horizon for the given heights in feet above the earth's surface.
Find the value of y when x=390 ft.
Height(x):100150200250300350400
Distance(y):10.6313.0315.0416.8118.4219.9021.47
Dr. N. B. Vyas Numerical Methods - Finite Differences

91. Stirling's Interpolation Formula
To estimate the value of a function near the middle a table, the
central dierence interpolation formula in used.
Letyx=f(x) be a functional relation betweenxandy.
Ifxtakes the valuesx02h; x0h; x0; x0+h; x0+ 2h; : : :and
the corresponding values ofyarey2; y1; y0; y1; y2: : :then we
can form a central dierence table as follows:
Dr. N. B. Vyas Numerical Methods - Finite Differences

92. Stirling's Interpolation Formula
x y
1
st
difference
2
nd
difference
3
rd
difference
x02h y2
y2(=y
3/2)
x0h y1
2
y2(=
2
y1)
y1(=y
1/2)
3
y2(=
3
y
1/2)
x0 y0
2
y1(=
2
y0)
4
y2(=
4
y0)
y0(=y
1/2)
3
y1(=
3
y
1/2)
x0+h y1
2
y0(=
2
y1)
y1(=y
3/2)
x0+ 2h y2
Dr. N. B. Vyas Numerical Methods - Finite Differences

93. Stirling's Interpolation Formula
The Stirling's formula in forward dierence notation is
yp=y0+p

y0+ y1
2

+
p
2
2!

2
y1
+
p(p
2
1
2
)
3!


3
y1+
3
y2
2

+
p
2
(p
2
1
2
)
4!

4
y2+: : :
Dr. N. B. Vyas Numerical Methods - Finite Differences

94. Stirling's Interpolation Formula
The Stirling's formula in forward dierence notation is
yp=y0+p

y0+ y1
2

+
p
2
2!

2
y1
+
p(p
2
1
2
)
3!


3
y1+
3
y2
2

+
p
2
(p
2
1
2
)
4!

4
y2+: : :
Dr. N. B. Vyas Numerical Methods - Finite Differences

95. Stirling's Interpolation Formula
The Stirling's formula in forward dierence notation is
yp=y0+p

y0+ y1
2

+
p
2
2!

2
y1
+
p(p
2
1
2
)
3!


3
y1+
3
y2
2

+
p
2
(p
2
1
2
)
4!

4
y2+: : :
Dr. N. B. Vyas Numerical Methods - Finite Differences

96. Example
Ex.
Using Stirling's formula ndy35
x:1020304050
y:600512439346243
Dr. N. B. Vyas Numerical Methods - Finite Differences

97. Example
Ex.
Find y for x=0.0341
x:0.01 0.02 0.03 0.04 0.05
y:98.434248.439231.777523.449218.4542
Dr. N. B. Vyas Numerical Methods - Finite Differences

98. Central Dierence
Gauss's Forward interpolation formula:
Pn(x) =y0+py0+
p(p1)
2!

2
y1+
(p+ 1)p(p1)
3!

3
y1+
(p+ 1)p(p1)(p2)
4!

4
y2+:::
Gauss's Backward interpolation formula:
Pn(x) =y0+py1+
p(p+ 1)
2!

2
y1+
(p+ 1)p(p1)
3!

3
y2+
(p+ 2)(p+ 1)p(p1)
4!

4
y2+:::
Dr. N. B. Vyas Numerical Methods - Finite Differences

99. Central Dierence
Gauss's Forward interpolation formula:
Pn(x) =y0+py0+
p(p1)
2!

2
y1+
(p+ 1)p(p1)
3!

3
y1+
(p+ 1)p(p1)(p2)
4!

4
y2+:::
Gauss's Backward interpolation formula:
Pn(x) =y0+py1+
p(p+ 1)
2!

2
y1+
(p+ 1)p(p1)
3!

3
y2+
(p+ 2)(p+ 1)p(p1)
4!

4
y2+:::
Dr. N. B. Vyas Numerical Methods - Finite Differences

100. Central Dierence
Gauss's Forward interpolation formula:
Pn(x) =y0+py0+
p(p1)
2!

2
y1+
(p+ 1)p(p1)
3!

3
y1+
(p+ 1)p(p1)(p2)
4!

4
y2+:::
Gauss's Backward interpolation formula:
Pn(x) =y0+py1+
p(p+ 1)
2!

2
y1+
(p+ 1)p(p1)
3!

3
y2+
(p+ 2)(p+ 1)p(p1)
4!

4
y2+:::
Dr. N. B. Vyas Numerical Methods - Finite Differences

101. Central Dierence
Gauss's Forward interpolation formula:
Pn(x) =y0+py0+
p(p1)
2!

2
y1+
(p+ 1)p(p1)
3!

3
y1+
(p+ 1)p(p1)(p2)
4!

4
y2+:::
Gauss's Backward interpolation formula:
Pn(x) =y0+py1+
p(p+ 1)
2!

2
y1+
(p+ 1)p(p1)
3!

3
y2+
(p+ 2)(p+ 1)p(p1)
4!

4
y2+:::
Dr. N. B. Vyas Numerical Methods - Finite Differences

102. Example
Ex.
that:
x:1234
y:14916
Dr. N. B. Vyas Numerical Methods - Finite Differences

103. Example
Ex.
population for the year 1936 given the following table:
Year : 190119111921193119411951
Population(in 1000s):12 15 20 27 39 52
Dr. N. B. Vyas Numerical Methods - Finite Differences