Interpolation with Finite differences

Numerical Methods - Finite Differences
Dr. N. B. Vyas
Department of Mathematics,
Atmiya Institute of Tech. and Science,
Rajkot (Guj.)
niravbvyas@gmail.com
Dr. N. B. Vyas Numerical Methods - Finite Differences

Finite Differences
Forward diﬀerence
Suppose that a function y = f(x) is tabulated for the equally
spaced arguments x0, x0 + h, x0 + 2h, ..., x0 + nh giving the
functional values y0, y1, y2, ..., yn.

Finite Differences
Forward difference
The constant difference between two consecutive values of x is
called the interval of differences and is denoted by h.

Finite Differences
Forward diﬀerence
The operator ∆ deﬁned by
∆y0 = y1 − y0

Finite Differences
Forward diﬀerence
∆y0 = y1 − y0
∆y1 = y2 − y1

Finite Differences
Forward diﬀerence
∆y0 = y1 − y0
∆y1 = y2 − y1
.......
.......
∆yn−1 = yn − yn−1

Finite Differences
Forward diﬀerence
∆y0 = y1 − y0
∆y1 = y2 − y1
.......
.......
∆yn−1 = yn − yn−1
is called Forward diﬀerence operator.

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

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
∆n
yi = ∆n−1
yi+1 − ∆n−1
yi

Finite Differences
Forward Diﬀerence Table:
x y ∆y ∆2y ∆3y ∆4y ∆5y
x0 y0
∆y0
x1 y1 ∆2y0
∆y1 ∆3y0
x2 y2 ∆2y1 ∆4y0
∆y2 ∆3y1 ∆5y0
x3 y3 ∆2y2 ∆4y1
∆y3 ∆3y2
x4 y4 ∆2y3
∆y4
x5 y5

Finite Differences
The operator ∆ satisﬁes the following properties:
1 ∆[f(x) ± g(x)] = ∆f(x) ± ∆g(x)

Finite Differences
1 ∆[f(x) ± g(x)] = ∆f(x) ± ∆g(x)
2 ∆[cf(x)] = c∆f(x)

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
Backward diﬀerence

Finite Differences
The operator deﬁned by
y1 = y1 − y0

Finite Differences
y1 = y1 − y0
y2 = y2 − y1

Finite Differences
y1 = y1 − y0
y2 = y2 − y1
.......
.......
yn = yn − yn−1

Finite Differences
y1 = y1 − y0
y2 = y2 − y1
.......
.......
yn = yn − yn−1
is called Backward diﬀerence operator.

Finite Differences
The ﬁrst backward diﬀerence is yn = yn − yn−1

Finite Differences
The second backward difference are obtain by the difference
of 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
The second backward difference are obtain by the difference
of 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
n
yi = n−1
yi − n−1
yi−1

Finite Differences
Backward Diﬀerence Table:
x y y 2y 3y 4y 5y
x0 y0
y1
x1 y1
2y2
y2
3y3
x2 y2
2y3
4y4
y3
3y4
5y5
x3 y3
2y4
4y5
y4
3y5
x4 y4
2y5
y5
x5 y5

Finite Differences
Central diﬀerence
The operator δ deﬁned by
δy1
2
= y1 − y0

Finite Differences
Central diﬀerence
δy1
2
= y1 − y0
δy3
2
= y2 − y1

Finite Differences
Central diﬀerence
δy1
2
= y1 − y0
δy3
2
= y2 − y1
.......
.......
δyn−1
2
= yn − yn−1

Finite Differences
Central difference
δy1
2
= y1 − y0
δy3
2
= y2 − y1
.......
.......
δyn−1
2
= yn − yn−1
is called Central difference operator.
Similarly, higher order central differences are defined as
δ2y1 = δy3
2
− δy1
2

Finite Differences
Central diﬀerence
δy1
2
= y1 − y0
δy3
2
= y2 − y1
.......
.......
δyn−1
2
= yn − yn−1
δ2y1 = δy3
2
− δy1
2
δ2y2 = δy5
2
− δy3
2

Finite Differences
Central diﬀerence
δy1
2
= y1 − y0
δy3
2
= y2 − y1
.......
.......
δyn−1
2
= yn − yn−1
δ2y1 = δy3
2
− δy1
2
δ2y2 = δy5
2
− δy3
2
.......
δ3y3
2
= δ2y2 − δ2y1

Finite Differences
Central Diﬀerence Table:
x y δy δ2y δ3y δ4y δ5y
x0 y0
δy1
2
x1 y1 δ2y1
δy3
2
δ3y3
2
x2 y2 δ2y2 δ4y2
δy5
2
δ3y5
2
δ5y5
2
x3 y3 δ2y3 δ4y3
δy7
2
δ3y7
2
x4 y4 δ2y4
δy9
2
x5 y5

Finite Differences
NOTE:
From all three diﬀerence tables, we can see that only the
notations changes not the diﬀerences.
y1 − y0 = ∆y0 = y1 = δy1
2

Finite Differences
NOTE:
y1 − y0 = ∆y0 = y1 = δy1
2
Alternative notations for the function y = f(x).

Finite Differences
NOTE:
y1 − y0 = ∆y0 = y1 = δy1
2
For two consecutive values of x diﬀering by h.

Finite Differences
NOTE:
y1 − y0 = ∆y0 = y1 = δy1
2
∆yx = yx+h − yx = f(x + h) − f(x)

Finite Differences
NOTE:
y1 − y0 = ∆y0 = y1 = δy1
2
∆yx = yx+h − yx = f(x + h) − f(x)
yx = yx − yx−h = f(x) − f(x − h)

Finite Differences
NOTE:
y1 − y0 = ∆y0 = y1 = δy1
2
∆yx = yx+h − yx = f(x + h) − f(x)
yx = yx − yx−h = f(x) − f(x − h)
δyx = yx+h
2
− yx−h
2
= f(x + h
2 ) − f(xh
2 )

Finite Differences
Evaluate the following. The interval of diﬀerence being h.
1 ∆nex

Finite Differences
1 ∆nex
2 ∆logf(x)

Finite Differences
1 ∆nex
2 ∆logf(x)
3 ∆(tan−1x)

Finite Differences
1 ∆nex
2 ∆logf(x)
3 ∆(tan−1x)
4 ∆2cos2x

Finite Differences
Other Diﬀerence Operator
1. Shift Operator E:
E does the operation of increasing the argument x by h so that
Ef(x) = f(x + h);

Finite Differences
Ef(x) = f(x + h);
E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);

Finite Differences
Ef(x) = f(x + h);
E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);
E3f(x) = f(x + 3h);

Finite Differences
Ef(x) = f(x + h);
E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);
E3f(x) = f(x + 3h);
Enf(x) = f(x + nh);

Finite Differences
Ef(x) = f(x + h);
E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);
E3f(x) = f(x + 3h);
Enf(x) = f(x + nh);
The inverse operator E−1 is deﬁned as

Finite Differences
Ef(x) = f(x + h);
E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);
E3f(x) = f(x + 3h);
Enf(x) = f(x + nh);
E−1f(x) = f(x − h);

Finite Differences
Ef(x) = f(x + h);
E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);
E3f(x) = f(x + 3h);
Enf(x) = f(x + nh);
E−1f(x) = f(x − h);
E−nf(x) = f(x − nh);

Finite Differences
Ef(x) = f(x + h);
E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);
E3f(x) = f(x + 3h);
Enf(x) = f(x + nh);
E−1f(x) = f(x − h);
If yx is the function f(x), then

Finite Differences
Ef(x) = f(x + h);
E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);
E3f(x) = f(x + 3h);
Enf(x) = f(x + nh);
E−1f(x) = f(x − h);
Eyx = yx+h;

Finite Differences
Ef(x) = f(x + h);
E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);
E3f(x) = f(x + 3h);
Enf(x) = f(x + nh);
E−1f(x) = f(x − h);
Eyx = yx+h;
E−1yx = yx−h;

Finite Differences
Ef(x) = f(x + h);
E2f(x) = E(Ef(x)) = Ef(x + h) = f(x + 2h);
E3f(x) = f(x + 3h);
Enf(x) = f(x + nh);
E−1f(x) = f(x − h);
Eyx = yx+h;
E−1yx = yx−h;
Enyx = yx+nh;

Finite Differences
2. Averaging Operator µ:
It is deﬁned as
µf(x) =
1
2
f x +
h
2
+ f x −
h
2
;

Finite Differences
It is deﬁned as
µf(x) =
1
2
f x +
h
2
+ f x −
h
2
;
i.e. µyx =
1
2
yx+h
2
+ yx−h
2
;

Finite Differences
It is deﬁned as
µf(x) =
1
2
f x +
h
2
+ f x −
h
2
;
i.e. µyx =
1
2
yx+h
2
+ yx−h
2
;
3. Diﬀerential Operator D:

Finite Differences
It is deﬁned as
µf(x) =
1
2
f x +
h
2
+ f x −
h
2
;
i.e. µyx =
1
2
yx+h
2
+ yx−h
2
;
It is deﬁned as

Finite Differences
It is deﬁned as
µf(x) =
1
2
f x +
h
2
+ f x −
h
2
;
i.e. µyx =
1
2
yx+h
2
+ yx−h
2
;
It is deﬁned as
Df(x) =
d
dx
f(x) = f (x);

Finite Differences
Relation between the operators
1 ∆ = E − 1

Finite Differences
1 ∆ = E − 1
2 = 1 − E−1

Finite Differences
1 ∆ = E − 1
2 = 1 − E−1
3 δ = E
1
2 − E−1
2

Finite Differences
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
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
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
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
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 ∆ − = ∆ = δ2

Finite Differences
9 1 + µ2δ2 = 1 +
1
2
δ2
2

Finite Differences
9 1 + µ2δ2 = 1 +
1
2
δ2
2
10 µ2 = 1 +
1
4
δ2

Finite Differences
9 1 + µ2δ2 = 1 +
1
2
δ2
2
10 µ2 = 1 +
1
4
δ2
11 E
1
2 = µ +
1
2
δ

Finite Differences
9 1 + µ2δ2 = 1 +
1
2
δ2
2
10 µ2 = 1 +
1
4
δ2
11 E
1
2 = µ +
1
2
δ
12 E−1
2 = µ −
1
2
δ

Finite Differences
9 1 + µ2δ2 = 1 +
1
2
δ2
2
10 µ2 = 1 +
1
4
δ2
11 E
1
2 = µ +
1
2
δ
12 E−1
2 = µ −
1
2
δ
13 ∆ =
1
2
δ2 + δ 1 +
δ2
4

Finite Differences
9 1 + µ2δ2 = 1 +
1
2
δ2
2
10 µ2 = 1 +
1
4
δ2
11 E
1
2 = µ +
1
2
δ
12 E−1
2 = µ −
1
2
δ
13 ∆ =
1
2
δ2 + δ 1 +
δ2
4
14 µδ =
1
2
∆E−1 +
1
2
∆

Gregory - Newton Forward Interpolation Formula
To estimate the value of a function near the beginning a table,
the forward diﬀerence interpolation formula in used.

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

x0, x0 + h, x0 + 2h, . . . of x.
Suppose we want to evaluate yx when x = x0 + ph, where p is
any real number.

Let it be yp. For any real number n, we have deﬁned operator E
such that Enf(x) = f(x + nh).
∴ yx = yx0+ph = f (x0 + ph) = Epyx0 = (1 + ∆)p
y0

y0
= 1 + p∆ +
p(p − 1)
2!
∆2 +
p(p − 1)(p − 2)
3!
∆3 + ... y0

y0
= 1 + p∆ +
p(p − 1)
2!
∆2 +
p(p − 1)(p − 2)
3!
∆3 + ... y0
yx = y0 + p∆y0 +
p(p − 1)
2!
∆2
y0 +
p(p − 1)(p − 2)
3!
∆3
y0 + ...
is called Newton’s forward interpolation formula.

Example
Ex. For the data construct the forward diﬀerence formula. Hence,
ﬁnd f(0.5).
x -2 -1 0 1 2 3
f(x) 15 5 1 3 11 25

Example
Ex. The population of the town in decennial census was as given
below estimate the population for the year 1895.
Year: 1891 1901 1911 1921 1931
Population(in thousand): 46 66 81 93 101

Example
Ex. Estimate the value of production for the year 1984 using
Newton’s forward method for the following data:
Year: 1976 1978 1980 1982
Production: 20 27 38 50

Gregory - Newton Backward Interpolation Formula
To estimate the value of a function near the end of a table, the
backward diﬀerence interpolation formula in used.

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

x0, x0 + h, x0 + 2h, . . . of x.
Suppose we want to evaluate yx when x = xn + ph, where p is
any real number.

∴ yx = yxn+ph = f (xn + ph) = Epyxn = (1 − )−p
yn

yn
= 1 + p +
p(p + 1)
2!
2 +
p(p + 1)(p + 2)
3!
3 + ... yn

yn
= 1 + p +
p(p + 1)
2!
2 +
p(p + 1)(p + 2)
3!
3 + ... yn
yx = yn + p yn +
p(p + 1)
2!
2
yn +
p(p + 1)(p + 2)
3!
3
yn + ...
is called Newton’s backward interpolation formula

Example
Ex. Using Newtons backward diﬀerence interpolation, interpolate at
x = 1 from the following data.
x 0.1 0.2 0.3 0.4 0.5 0.6
f(x) 1.699 1.073 0.375 0.443 1.429 2.631

Example
Ex. The table gives the distance in nautical miles of the visible
horizon for the given heights in feet above the earth’s surface.
Find the value of y when x=390 ft.
Height(x): 100 150 200 250 300 350 400
Distance(y): 10.63 13.03 15.04 16.81 18.42 19.90 21.47

Stirling’s Interpolation Formula
To estimate the value of a function near the middle a table, the
central 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, . . . and
the corresponding values of y are y−2, y−1, y0, y1, y2 . . . then we
can form a central diﬀerence table as follows:

x y
1st
difference
2nd
difference
3rd
difference
x0 − 2h y−2
∆y−2(= δy−3/2)
x0 − h y−1 ∆2y−2(= δ2y−1)
∆y−1(= δy−1/2) ∆3y−2(= δ3y−1/2)
x0 y0 ∆2y−1(= δ2y0) ∆
∆y0(= δy1/2) ∆3y−1(= δ3y1/2)
x0 + h y1 ∆2y0(= δ2y1)
∆y1(= δy3/2)
x0 + 2h y2

The Stirling’s formula in forward diﬀerence notation is

yp = y0 + p
∆y0 + ∆y−1
2
+
p2
2!
∆2y−1

yp = y0 + p
∆y0 + ∆y−1
2
+
p2
2!
∆2y−1
+
p(p2 − 12)
3!
∆3y−1 + ∆3y−2
2
+
p2(p2 − 12)
4!
∆4y−2 + . . .

Example
Ex.
Using Stirling’s formula ﬁnd y35
x: 10 20 30 40 50
y: 600 512 439 346 243

Example
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 0.05
y: 98.4342 48.4392 31.7775 23.4492 18.4542

Central Diﬀerence
Gauss’s Forward interpolation formula:

Central Diﬀerence
Pn(x) = y0 + p∆y0 +
p(p − 1)
2!
∆2
y−1 +
(p + 1)p(p − 1)
3!
∆3
y−1+
(p + 1)p(p − 1)(p − 2)
4!
∆4
y−2 + ...

Central Diﬀerence
Pn(x) = y0 + p∆y0 +
p(p − 1)
2!
∆2
y−1 +
(p + 1)p(p − 1)
3!
∆3
y−1+
(p + 1)p(p − 1)(p − 2)
4!
∆4
y−2 + ...
Gauss’s Backward interpolation formula:

Central Diﬀerence
Pn(x) = y0 + p∆y0 +
p(p − 1)
2!
∆2
y−1 +
(p + 1)p(p − 1)
3!
∆3
y−1+
(p + 1)p(p − 1)(p − 2)
4!
∆4
y−2 + ...
Gauss’s Backward interpolation formula:
Pn(x) = y0 + p∆y−1 +
p(p + 1)
2!
∆2
y−1 +
(p + 1)p(p − 1)
3!
∆3
y−2+
(p + 2)(p + 1)p(p − 1)
4!
∆4
y−2 + ...

Example
Ex. Estimate the value of y(2.5) using Gauss’s forward formula given
that:
x: 1 2 3 4
y: 1 4 9 16

Example
Ex. Interpolate by means of Gauss’s backward formula the
population for the year 1936 given the following table:
Year : 1901 1911 1921 1931 1941 1951
Population(in 1000s): 12 15 20 27 39 52

Interpolation with Finite differences

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