Dr. Nirav Vyas numerical method 2.pdf

8/20/2019 Dr. Nirav Vyas numerical method 2.pdf

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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


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Finite Differences

Forward difference

Suppose that a function y = f (x) is tabulated for the equallyspaced arguments x0, x0 + h, x0 + 2h,...,x0 + nh giving thefunctional values y0, y1, y2,...,yn.



3/103

Finite Differences

Forward difference


The constant difference between two consecutive values of x is

called the interval of differences and is denoted by h.



4/103

Finite Differences

Forward difference



called the interval of differences and is denoted by h.The operator ∆ defined by

∆y0 = y1 − y0



5/103

Finite Differences

Forward difference




∆y0 = y1 − y0∆y1 = y2 − y1



6/103

Finite Differences

Forward difference




∆y0 = y1 − y0∆y1 = y2 − y1.......

.......

∆yn−1 = yn − yn−1


F D


7/103

Finite Differences

Forward difference




∆y0 = y1 − y0∆y1 = y2 − y1.......

.......

∆yn−1 = yn − yn−1

is called Forward difference operator.


F D


8/103

Finite Differences

The first forward difference is ∆yn = yn+1 − yn


F D


9/103

Finite Differences

The first forward difference is ∆yn = yn+1 − yn

The second forward difference are defined as the difference of the first differences.

∆2y0 = ∆(∆y0) = ∆(y1 − y0) = ∆y1 − ∆y0= y2 − y1 − (y1 − y0) = y2 − 2y1 + y0

∆2y1 = ∆y2 − ∆y1∆2yi = ∆yi+1 − ∆yi

In general nth forward difference of f is defined by

∆nyi = ∆n−1yi+1 − ∆

n−1yi


F D


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Finite Differences

Forward Difference Table:

x y ∆y ∆2y ∆3y ∆4y ∆5y

x0 y0∆y0

x1 y1 ∆2y0

∆y1 ∆3

y0x2 y2 ∆2y1 ∆

4y0∆y2 ∆

3y1 ∆5y0

x3 y3 ∆2y2 ∆

4y1∆y3 ∆

3y2

x4 y4 ∆2y3∆y4

x5 y5


Finite Differences


11/103

Finite Differences

The operator ∆ satisfies the following properties:

1 ∆[f (x) ± g(x)] = ∆f (x) ± ∆g(x)


Finite Differences


12/103

Finite Differences


1 ∆[f (x) ± g(x)] = ∆f (x) ± ∆g(x)

2 ∆[cf (x)] = c∆f (x)


Finite Differences


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Finite Differences


1 ∆[f (x) ± g(x)] = ∆f (x) ± ∆g(x)

2 ∆[cf (x)] = c∆f (x)

3 ∆m∆nf (x) = ∆m+nf (x), m, n are positive integers

4 Since ∆nyn is a constant, ∆n+1yn = 0 , ∆

n+2yn = 0, . . .i.e. (n + 1)th and higher differences are zero.


Finite Differences


14/103

Finite Differences

Backward differenceSuppose that a function y = f (x) is tabulated for the equallyspaced arguments x0, x0 + h, x0 + 2h,...,x0 + nh giving thefunctional values y0, y1, y2,...,yn.


Finite Differences


15/103

Finite Differences


The operator ∇ defined by

∇y1 = y1 − y0


Finite Differences


16/103

Finite Differences



∇y1 = y1 − y0∇y2 = y2 − y1


Finite Differences


17/103

Finite Differences



∇y1 = y1 − y0∇y2 = y2 − y1.......

.......

∇yn = yn − yn−1


Finite Differences


18/103

Finite Differences



∇y1 = y1 − y0∇y2 = y2 − y1.......

.......

∇yn = yn − yn−1

is called Backward difference operator.


Finite Differences


19/103

Finite Differences

The first backward difference is ∇yn = yn − yn−1


Finite Differences


20/103

Finite Differences

The first backward difference is ∇yn = yn − yn−1The second backward difference are obtain by the differenceof the first differences.

∇2y2 = ∇(∇y2) = ∇(y2 − y1) = ∇y2 −∇y1= y2 − y1 − (y1 − y0) = y2 − 2y1 + y0

∇2y3 = ∇y3 −∇y2∇2yn = ∇yn −∇yn−1


Finite Differences


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Finite Differences

The first backward difference is ∇yn = yn − yn−1The second backward difference are obtain by the differenceof the first differences.

∇2y2 = ∇(∇y2) = ∇(y2 − y1) = ∇y2 −∇y1= y2 − y1 − (y1 − y0) = y2 − 2y1 + y0

∇2y3 = ∇y3 −∇y2∇2yn = ∇yn −∇yn−1

In general nth backward difference of f is defined by

∇nyi = ∇n−1yi −∇

n−1yi−1


Finite Differences


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Backward Difference Table:

x y ∇y ∇2y ∇3y ∇4y ∇5yx0 y0

∇y1x1 y1 ∇

2y2∇y

2 ∇3y

3x2 y2 ∇2y3 ∇

4y4∇y3 ∇

3y4 ∇5y5

x3 y3 ∇2y4 ∇

4y5∇y4 ∇

3y5

x4 y4 ∇2

y5∇y5

x5 y5


Finite Differences


23/103

Central difference

The operator δ defined by

δy 12 = y1 − y0


Finite Differences


24/103

Central difference


δy 12 = y1 − y0δy 3

2

= y2 − y1


Finite Differences


25/103

Central difference


δy 12 = y1 − y0δy 3

2

= y2 − y1

.......

.......

δyn−

1

2

= yn − yn−1


Finite Differences


26/103

Central difference


δy 12 = y1 − y0δy 3

2

= y2 − y1

.......

.......

δyn−

1

2

= yn − yn−1

is called Central difference operator.Similarly, higher order central differences are defined as

δ 2y1 = δy 32

− δy 12


Finite Differences


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Central difference


δy 12 = y1 − y0δy 3

2

= y2 − y1

.......

.......

δyn−

1

2

= yn − yn−1


δ 2y1 = δy 32

− δy 12

δ 2y2 = δy 52

− δy 32


Finite Differences


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Central difference


δy 12 = y1 − y0δy 3

2

= y2 − y1

.......

.......

δyn−

1

2

= yn − yn−1


δ 2y1 = δy 32

− δy 12

δ 2y2 = δy 52

− δy 32

.......

δ 3y 32

= δ 2y2 − δ 2y1


Finite Differences


29/103

Central difference


δy 12 = y1 − y0δy 3

2

= y2 − y1

.......

.......

δyn−

1

2

= yn − yn−1


δ 2y1 = δy 32

− δy 12

δ 2y2 = δy 52

− δy 32

.......

δ 3y 32

= δ 2y2 − δ 2y1


Finite Differences


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Central Difference Table:

x y δy δ 2

y δ 3

y δ 4

y δ 5

yx0 y0

δy 12

x1 y1 δ 2y1

δy 32

δ 3y 32

x2 y2 δ 2y2 δ 4y2δy 5

2

δ 3y 52

δ 5y 52

x3 y3 δ 2y3 δ

4y3δy 7

2

δ 3y 72

x4 y4 δ 2y4δy 9

2

x5 y5


Finite Differences


31/103

NOTE:From all three difference tables, we can see that only thenotations changes not the differences.

y1 − y0 = ∆y0 = ∇y1 = δy 12


Finite Differences


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y1 − y0 = ∆y0 = ∇y1 = δy 12

Alternative notations for the function y = f (x).


Finite Differences


33/103


y1 − y0 = ∆y0 = ∇y1 = δy 12

Alternative notations for the function y = f (x).For two consecutive values of x differing by h.


Finite Differences


34/103


y1 − y0 = ∆y0 = ∇y1 = δy 12


∆yx = yx+h − yx = f (x + h) − f (x)


Finite Differences


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y1 − y0 = ∆y0 = ∇y1 = δy 12


∆yx = yx+h − yx = f (x + h) − f (x)

∇yx = yx − yx−h = f (x) − f (x − h)


Finite Differences


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y1 − y0 = ∆y0 = ∇y1 = δy 12


∆yx = yx+h − yx = f (x + h) − f (x)

∇yx = yx − yx−h = f (x) − f (x − h)

δyx = yx+h2− yx−h

2= f (x + h2 ) − f (xh2 )


Finite Differences


37/103

Evaluate the following. The interval of difference being h.

1 ∆nex


Finite Differences


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1 ∆nex

2 ∆logf (x)


Finite Differences


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1 ∆nex

2 ∆logf

(x

)3 ∆(tan−1x)


Finite Differences


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1 ∆nex

2 ∆logf

(x

)3 ∆(tan−1x)

4 ∆2cos2x


Finite Differences


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Other Difference Operator

1. Shift Operator E :

E does the operation of increasing the argument x by h so thatEf (x) = f (x + h);


Finite Differences


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E 2f (x) = E (Ef (x)) = Ef (x + h) = f (x + 2h);


Finite Differences


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E 2f (x) = E (Ef (x)) = Ef (x + h) = f (x + 2h);

E 3f (x) = f (x + 3h);


Finite Differences


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E 2f (x) = E (Ef (x)) = Ef (x + h) = f (x + 2h);

E 3f (x) = f (x + 3h);

E n

f (x) = f (x + nh);


Finite Differences

O h Diff O


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E 2f (x) = E (Ef (x)) = Ef (x + h) = f (x + 2h);

E 3f (x) = f (x + 3h);

E n

f (x) = f (x + nh);The inverse operator E −1 is defined as


Finite Differences

Oth Diff O t


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E 2f (x) = E (Ef (x)) = Ef (x + h) = f (x + 2h);

E 3f (x) = f (x + 3h);

E n


E −1f (x) = f (x − h);


Finite Differences

Oth Diff O t


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E 2f (x) = E (Ef (x)) = Ef (x + h) = f (x + 2h);

E 3f (x) = f (x + 3h);

E n


E −1f (x) = f (x − h);

E −nf (x) = f (x − nh);


Finite Differences

Oth Diff O t


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E 2f (x) = E (Ef (x)) = Ef (x + h) = f (x + 2h);

E 3f (x) = f (x + 3h);

E n


E −1f (x) = f (x − h);

E −nf (x) = f (x − nh);

If yx is the function f (x), then


Finite Differences



49/103




E 2f (x) = E (Ef (x)) = Ef (x + h) = f (x + 2h);

E 3f (x) = f (x + 3h);

E n


E −1f (x) = f (x − h);

E −nf (x) = f (x − nh);

If yx is the function f (x), thenEyx = yx+h;


Finite Differences



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E 2f (x) = E (Ef (x)) = Ef (x + h) = f (x + 2h);

E 3f (x) = f (x + 3h);

E n


E −1f (x) = f (x − h);

E −nf (x) = f (x − nh);


E −1yx = yx−h;


Finite Differences



51/103




E 2f (x) = E (Ef (x)) = Ef (x + h) = f (x + 2h);

E 3f (x) = f (x + 3h);

E n


E −1f (x) = f (x − h);

E −nf (x) = f (x − nh);


E −1yx = yx−h;

E nyx = yx+nh;


Finite Differences



52/103




E 2f (x) = E (Ef (x)) = Ef (x + h) = f (x + 2h);

E 3f (x) = f (x + 3h);

E n


E −1f (x) = f (x − h);

E −nf (x) = f (x − nh);


E −1yx = yx−h;

E nyx = yx+nh;


Finite Differences


53/103

2. Averaging Operator µ:

It is defined as

µf (x) = 1

2

f

x +

h

2

+ f

x −

h

2

;


Finite Differences


54/103


It is defined as

µf (x) = 1

2

f

x +

h

2

+ f

x −

h

2

;

i.e. µyx =

1

2

yx+h2 + yx−h2

;


Finite Differences


55/103


It is defined as

µf (x) = 1

2

f

x +

h

2

+ f

x −

h

2

;

i.e. µyx =

1

2

yx+h2 + yx−h2

;

3. Differential Operator D:


Finite Differences


56/103


It is defined as

µf (x) = 1

2

f

x +

h

2

+ f

x −

h

2

;

i.e. µyx =

1

2

yx+h2 + yx−h2

;


It is defined as


Finite Differences


57/103


It is defined as

µf (x) = 1

2

f

x +

h

2

+ f

x −

h

2

;

i.e. µyx =

1

2

yx+h2 + yx−h2

;


It is defined as

Df (

x) =

d

dxf

(x

) = f

(x

);


Finite Differences


58/103

Relation between the operators

1 ∆ = E − 1


Finite Differences


59/103


1 ∆ = E − 1

2 ∇ = 1 − E −1


Finite Differences


60/103


1 ∆ = E − 1

2 ∇ = 1 − E −1

3 δ = E 1

2 − E −1

2


Finite Differences


61/103


1 ∆ = E − 1

2 ∇ = 1 − E −1

3 δ = E 1

2 − E −1

2

4 µ =

1

2{E

1

2

+ E −

1

2

}


Finite Differences


62/103


1 ∆ = E − 1

2 ∇ = 1 − E −1

3 δ = E 1

2 − E −1

2

4 µ =

1

2{E

1

2

+ E −

1

2

}

5 ∆ = E ∇ = ∇E = δE 1

2


Finite Differences


63/103


1 ∆ = E − 1

2 ∇ = 1 − E −1

3 δ = E 1

2 − E −1

2

4 µ =

1

2{E

1

2

+ E −

1

2

}

5 ∆ = E ∇ = ∇E = δE 1

2

6 E = ehD


Finite Differences


64/103


1 ∆ = E − 1

2 ∇ = 1 − E −1

3 δ = E 1

2 − E −1

2

4 µ =

1

2{E

1

2

+ E −

1

2

}

5 ∆ = E ∇ = ∇E = δE 1

2

6 E = ehD

7 (1 + ∆)(1 −∇) = 1


Finite Differences


65/103


1 ∆ = E − 1

2 ∇ = 1 − E −1

3 δ = E 1

2 − E −1

2

4 µ =

1

2{E

1

2

+ E −

1

2

}

5 ∆ = E ∇ = ∇E = δE 1

2

6 E = ehD

7 (1 + ∆)(1 −∇) = 1

8 ∆ −∇ = ∆∇ = δ 2


Finite Differences


66/103

9 1 + µ2

δ

2

=

1 +

1

2 δ 2

2


Finite Differences


67/103

9 1 + µ

2

δ

2

=

1 +

1

2 δ

22

10 µ2 = 1 + 1

4δ 2


Finite Differences


68/103

9 1 + µ

2

δ

2

=

1 +

1

2 δ

22

10 µ2 = 1 + 1

4δ 2

11 E 1

2 = µ + 1

2δ


Finite Differences


69/103

9 1 + µ2δ 2

=

1 +

1

2δ 2

2

10 µ2 = 1 + 1

4δ 2

11 E 1

2 = µ + 1

2δ

12 E −1

2 = µ − 1

2δ


Finite Differences


70/103

9 1 + µ2δ 2

=

1 +

1

2δ 2

2

10 µ2 = 1 + 1

4δ 2

11 E 1

2 = µ + 1

2δ

12 E −1

2 = µ − 12

δ

13 ∆ = 1

2δ 2 + δ

1 +

δ 2

4


Finite Differences


71/103

9 1 + µ2δ 2 = 1 + 12

δ 22

10 µ2 = 1 + 1

4δ 2

11 E 1

2 = µ + 1

2δ

12 E −1

2 = µ − 12

δ

13 ∆ = 1

2δ 2 + δ

1 +

δ 2

4

14 µδ = 12

∆E −1 + 12

∆


Finite Differences


72/103

9 1 + µ2δ 2 = 1 + 12

δ 22

10 µ2 = 1 + 1

4δ 2

11 E 1

2 = µ + 1

2δ

12 E −1

2 = µ − 12

δ

13 ∆ = 1

2δ 2 + δ

1 +

δ 2

4

14 µδ = 12

∆E −1 + 12

∆


Gregory - Newton Forward Interpolation Formula


73/103

To estimate the value of a function near the beginning a table,the forward difference interpolation formula in used.




74/103


Let yx = f (x) be a function which takes the valuesyx0, yx0+h, yx0+2h, . . . corresponding to the valuesx0, x0 + h, x0 + 2h , . . . of x.




75/103



Suppose we want to evaluate yx when x = x0 + ph, where p isany real number.




76/103



Suppose we want to evaluate yx when x = x0 + ph, where p isany real number.




77/103

Let it be y p. For any real number n, we have defined operator E such that E nf (x) = f (x + nh).

∴ yx = yx0+ ph = f (x0 + ph) = E pyx0 = (1 + ∆)

py0




78/103


∴ yx = yx0+ ph = f (x0 + ph) = E pyx0 = (1 + ∆)

py0

=

1 + p∆ + p( p − 1)2! ∆2 + p( p − 1)( p − 2)3! ∆

3 + ...

y0




79/103


∴ yx = yx0+ ph = f (x0 + ph) = E pyx0 = (1 + ∆)

py0

=

1 + p∆ + p( p − 1)2! ∆2 + p( p − 1)( p − 2)3! ∆

3 + ...

y0

yx = y0 + p∆y0 + p( p − 1)

2! ∆2y0 +

p( p − 1)( p − 2)

3! ∆3y0 + ...

is called Newton’s forward interpolation formula.


Example


80/103

Ex. For the data construct the forward difference formula. Hence,find f (0.5).

x -2 -1 0 1 2 3f(x) 15 5 1 3 11 25


Example


81/103

Ex. The population of the town in decennial census was as givenbelow estimate the population for the year 1895.

Year: 1891 1901 1911 1921 1931Population(in thousand): 46 66 81 93 101


Example


82/103

Ex. Estimate the value of production for the year 1984 usingNewton’s forward method for the following data:

Year: 1976 1978 1980 1982Production: 20 27 38 50


Gregory - Newton Backward Interpolation Formula


83/103

To estimate the value of a function near the end of a table, thebackward difference interpolation formula in used.




84/103


Let yx = f (x) be a function which takes the valuesyx0, yx0+h, yx0+2h, . . . corresponding to the values

x0, x0 + h, x0 + 2h , . . . of x.




85/103


Let yx = f (x) be a function which takes the valuesyx0, yx0+h, yx0+2h, . . . corresponding to the values

x0, x0 + h, x0 + 2h , . . . of x.

Suppose we want to evaluate yx when x = xn + ph, where p isany real number.




86/103


∴ yx = yxn+ ph = f (xn + ph) = E pyxn = (1 −∇)

− pyn




87/103



− pyn

=

1 + p∇ +

p( p + 1)

2! ∇2

+

p( p + 1)( p + 2)

3! ∇3

+ ...

yn




88/103



− pyn

=

1 + p∇ +

p( p + 1)

2! ∇2

+

p( p + 1)( p + 2)

3! ∇3

+ ...

yn

yx = yn + p∇yn + p( p + 1)

2! ∇2yn +

p( p + 1)( p + 2)

3! ∇3yn + ...

is called Newton’s backward interpolation formula


Example


89/103

Ex. Using Newtons backward difference interpolation, interpolate atx = 1 from the following data.

x 0.1 0.2 0.3 0.4 0.5 0.6f(x) 1.699 1.073 0.375 0.443 1.429 2.631

Dr N B Vyas Numerical Methods - Finite Differences

Example


90/103

Ex. The table gives the distance in nautical miles of the visiblehorizon for the given heights in feet above the earth’s surface.Find the value of y when x=390 ft.

Height(x): 100 150 200 250 300 350 400Distance(y): 10.63 13.03 15.04 16.81 18.42 19.90 21.47


Stirling’s Interpolation Formula


91/103

To estimate the value of a function near the middle a table, thecentral difference interpolation formula in used.

Let yx = f (x) be a functional relation between x and y.

If x takes the values x0 − 2h, x0 − h, x0, x0 + h, x0 + 2h , . . . andthe corresponding values of y are y−2, y−1, y0, y1, y2 . . . then wecan form a central difference table as follows:




92/103

x y 1st

difference2nd

difference3rd

difference

x0 − 2h y−2∆y−2(= δy−3/2)

x0 − h y−1 ∆2y−2(= δ

2y−1)

∆y−1(= δy−1/2) ∆3y−2(= δ 3y−1/2)x0 y0 ∆

2y−1(= δ 2y0)

∆y0(= δy1/2) ∆3y−1(= δ

3y1/2)

x0 + h y1 ∆2y0(= δ

2y1)∆y1(= δy3/2)

x0 + 2h y2




93/103

The Stirling’s formula in forward difference notation is




94/103


y p = y0 + p

∆y0 + ∆y−1

2

+

p2

2!∆2y−1




95/103


y p = y0 + p

∆y0 + ∆y−1

2

+

p2

2!∆2y−1

+ p( p2 − 12)

3!

∆3y−1 + ∆3y−2

2

+ p

2( p2 − 12)4!

∆4y−2 + . . .


Example


96/103

Ex.

Using Stirling’s formula find y35

x: 10 20 30 40 50y: 600 512 439 346 243


Example


97/103

Ex. The function y is given in the table below:

Find y for x=0.0341

x: 0.01 0.02 0.03 0.04 0.05y: 98.4342 48.4392 31.7775 23.4492 18.4542


Central Difference

Gauss’s Forward interpolation formula:


98/103


Central Difference



99/103

P n(x) = y0 + p∆y0 + p( p − 1)

2! ∆2y−1 + (

p + 1) p( p − 1)3!

∆3y−1+

( p + 1) p( p − 1)( p − 2)

4! ∆4y−2 + ...


Central Difference



100/103

P n(x) = y0 + p∆y0 + p( p − 1)

2! ∆2y−1 + (

p + 1) p( p − 1)3!

∆3y−1+

( p + 1) p( p − 1)( p − 2)

4! ∆4y−2 + ...

Gauss’s Backward interpolation formula:


Central Difference



101/103

P n(x) = y0 + p∆y0 + p( p − 1)2!

∆2y−1 + ( p + 1) p( p − 1)3!

∆3y−1+

( p + 1) p( p − 1)( p − 2)

4! ∆4y−2 + ...

Gauss’s Backward interpolation formula:

P n(x) = y0 + p∆y−1 + p( p + 1)

2! ∆2y−1 +

( p + 1) p( p − 1)

3! ∆3y−2+

( p + 2)( p + 1) p( p − 1)

4! ∆4y−2 + ...


Example


102/103

Ex. Estimate the value of y(2.5) using Gauss’s forward formula giventhat:

x: 1 2 3 4

y: 1 4 9 16


Example


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Ex. Interpolate by means of Gauss’s backward formula thepopulation for the year 1936 given the following table:

Year : 1901 1911 1921 1931 1941 1951

Population(in 1000s): 12 15 20 27 39 52


Dr. Nirav Vyas numerical method 2.pdf

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