1 FINITE DIFFERENCES AND NUMERICAL

METHODS
1.1 THE FORWARD DIFFERENCES OPERATOR ∆
Let
y = f (x).
The values, which the independent variable x takes, are called arguments and the
corresponding values of f (x) are called entries. The difference between consecutive
values of x called the interval of differecing. If the interval of differencing be h and
the first argument be a, then

Arguments x : a, a + h, a + 2h, a + 3h, . . .

Entries f (x) : f (a), f (a + h), f (a + 2h), f (a + 3h), . . .

For brevity, these entries are denoted by

y0 , y1 , y2 , y3 , . . .

y1 − y0 = f (a + h) − f (a) is called the first forward differences of y0 and is denoted

by △y0 or △f (a). Thus

△y0 = y1 − y0 or △ f (a) = f (a + h) − f (a)

Similarly, △ y1 = y2 − y1 , △y2 = y3 − y2
In general, △ yn = yn+1 − yn or △ f (x) = f (x + h) − f (x)

The differences of the first forward differences are called second forward differences.
Thus △(△y0 ) = △(y1 − y0 ) or △2 y0 = △y1 − △y0 is calld the second forward
differences of y0 . Similarly,

△2 y1 = △y2 − △y1 , △2 y1 = △y3 − △y2

In general, △2 yn = △yn+1 − △yn

Similarly, we can define differences of higher order.
The table showing the forward differences is called forward differences table and is
given below
Argument Entry First Diff. Second Diff. Third Diff.
x y △y △2 y △3 y
a y0
y − y0 = △y0
a+h y1 △y1 − △y0 = △2 y0
y2 − y1 = △y1 △2 y1 − △2 y0 = △3 y0
a + 2h y2 △y2 − △y1 = △2 y1
y3 − y2 = △y2
a + 3h y3

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y0 is called the leading term and △y0 , △2 y0 , △3 y0 , . . . are called the leading
differences.

The operator △ has the following properties :

(i) △c = 0, c begin a constant.

(ii) △ f (x) = c, △f (x)
(iii) △[af (x) + bg(x)] = a △ f (x) + b △ g(x)
(iv) The nth difference of an nth degree polynomial is a constant = (co−ef f. of xn )n!.h
and hence higher order differences are zero.
Example 1.1.1 Prove that :
(i) △[f (x)g(x)] = f (x + h) △ g(x) + g(x) △ f (x)
 
f (x) g(x) △ f (x) − f (x) △ g(x)
(ii) △ =
g(x) g(x + h)g(x)
Solution 1.1.2
(i) △ [f (x)g(x)] =f (x + h)g(x + h) − f (x)g(x)
=f (x + h)g(x + h) − f (x + h)g(x) + f (x + h)g(x) − f (x)g(x)
=f (x + h)[g(x + h) − g(x)] + g(x)[g(x + h) − f (x)]
=f (x + h) △ g(x) + g(x) △ f (x)
 
f (x) f (x + h) f (x) f (x + h)g(x) − f (x)g(x + h)
(ii) △ = − =
g(x) g(x + h) g(x) g(x + h)g(x)
f (x + h)g(x) − f (x)g(x) + f (x)g(x) − f (x)g(x + h)
=
g(x + h)g(x)
g(x)[f (x + h) − f (x)] − f (x)[g(x + h) − g(x)] g(x) △ f (x) − f (x) △ g(x)
= =
g(x + h)g(x) g(x + h)g(x)
Example 1.1.3 Evaluate the following, interval of differencing being unity.
x x
   
2 e
(i) △ tan−1 ax (ii) △ (iii) △
(x + 1)! ex + e−x

Solution 1.1.4
a(x + 1) − ax a
(i) △ tan−1 ax = tan−1 a(x + 1) − tan−1 ax = tan−1 = tan−1 .
1 + a(x + 1).ax 1 + a x + a2 x 2
2

2x 2x+1 2x 2x+1 − 22 (x + 2) 2x+1 − x.2x − 2x+1 x.2x

 
(ii) △ = − = = =−
(x + 1)! (x + 2)! (x + 1)! (x + 2)! (x + 2)! (x + 2)!

ex ex+1 ex e2x+1 + e − e2x+1 − e−1 e − e−1

 
(iii) △ = x+1 − = x+1 −(x+1) x = x+1
ex + e−x e + e−(x+1) ex + e−x [e e ](e + e−x ) (e + e−x−1 )(ex +

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Example 1.1.5 Evaluate the following interval of differencing being h:

(i) △ (x2 + sin x) (ii) △ (sin 2x cos 4x) (iii) △ cot ax

Solution 1.1.6

(i) △ (x2 + sin x) = △ x2 + △ sin x

 
2 2 2 h h
=[(x + h) − x ] + [sin(x + h) − sin x] = 2hx + h + 2 cos x + sin
2 2
 
h h
=h(h + 2x) + 2 sin cos x +
2 2

1
(ii) △ (sin 2x cos 4x) = △ ( .2 cos 4x sin 2x)
2
1 1
= △ (sin 6x − sin 2x) = (△ sin 6x − △ sin 2x)
2 2
1
= [{sin 6(x + h) − sin 6x} − {sin 2(x + h) − sin 2x}]
2
1
= [2 cos(6x + 3h) sin 3h − 2 cos(2x + h) sin h]
2
= sin 3h cos 3(2x + h) − sin h cos(2x + h)

cot ax+h cot ax

(iii) △ cot ax = cot ax+h − cot ax = −
sin ax+h sin ax
x x+h x x+h
sin a cot a − cot a sin a sin(ax − ax+h ) sin ax (1. − ah )
= = =
sin ax+h sin ax sin ax+h sin ax sin ax+h sin ax

1.2 DIFFERENCES OF FACTORIAL POLYNOMIAL

If n is a positive integer, then the expression

x(x − h)(x − 2h) . . . (x − (n − 1)h)

Involving n factors, beginning with x and decreasing by h every time is called a

factorial polynomial of degree n and is denoted by xn .

For example, x(1) = x.x(2) = x(x − h), x3 = x(x − h)(x − 2h)

△x(n) =(x + h)(n) − x(n)

=[(x + h)(x)(x + h) . . . (x − n − 2h)] − [x(x − h) . . . (x − n − 2h)(x − n − 1h)]
=x(x − h) . . . (x − n − 2h)[(x + h) − (x − n − 1h)]
=nh.x(x − h) . . . (x − n − 2h) = nh.xn−1

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Similarly,
△2 xn = △ (x(n) ) = nh △ x(n−1)
=nh.(n − 1)hx(n−2) = n(n − 1)h2 x(n−2)
△3 x(n) =n(n − 1)(n − 2)h3 x(n−3)
..........................................................
△n xn =n(n − 1)(n − 2) . . . 2.1hn = n!hn
=constant
△n+1 x(n) = 0
Note. If h = 1, △xn = nxn−1
⇒ Differencing is analogous in differentiation.
⇒ The process of getting the function whose first differences are given is analogous
to integration.

Example 1.2.1 Express the function f (x) = 2x3 + 3x2 − 5x + 4 and its successive
differeces in factorial notation. Also obtain a function whose first difference is f (x).
.
Solution 1.2.2 We first express f (x) in factorial notation.
.

1 2 3 -5 4=A3
2 5
2 2 5 0=A2
4
2 9=A1
2=A0

NOTES

Thus A3 , A2 , A1 , A0 are the successive remainders in the divition of f (x) by

x, x − 1, x − 2

∴ f (x) =2x(3) + 9x(2) + 4

△f (x) =6x(2) + 18x(1)
△2 f (x) =12x(1) + 181
△3 f (x) =12

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Differences of higher order are zero.

Now let F (x) be the function whose first differences is f (x). Then

△F (x) =f (x)
1 1 (3) 2 x(4) x(3)
⇒ F (x) = f (x) = [2x + 9x + 4] = 2. + 9. + 4.x(1) + c
△ △ 4 3
1
= .x(x − 1)(x − 2)(x − 3) + 3x(x − 1)(x − 2) + 4x + c
2
1 1
= [x(x3 − 6x2 + 11x − 6) + 6x(x2 − 3x + 2) + 8x] + c = (x4 − 7x2 + 14x) + c.
2 2

1.3 BACKWARD DIFFERECES OPERATOR ▽

The backward difgferences operator ▽ is defined by
▽yn =yn − yn−1 or ▽ f (a) = f (a) − f (a − h)
T hus ▽ y0 =y0 − y−1 , ▽y1 = y1 − y0 ▽ y2 = y2 − y1 etc.
▽2 y0 = ▽ (▽y0 ) ▽ (y0 − y−1 ) = ▽y0 − ▽y−1
▽2 y1 = ▽ y1 − ▽y0 , ▽2 y2 = ▽y2 − ▽y1 etc.
The table showing the various backward differences is called backward differencestable
given below.

x y ▽y ▽2 y ▽3 y
a − 3h y−3
y−2 − y−3 = ▽y−2
a − 2h y−2 ▽y−1 − ▽y−2 = ▽2 y−1
y−1 − y−2 = ▽y−1 ▽2 y0 − ▽2 y−1 = ▽3 y0
a−h y−1 ▽y0 − ▽y−1 = ▽2 y0
y0 − y−1 = ▽y0
a y0

1.4 THE DISPLACEMENT (OR SHIFT) OPERATOR E

The operator E increases the value of the arguments by one interval.
If
x : a, a + h, a + 2h, . . .

and y0 =f (a), y1 = f (a + h), y2 = f (a + 2h), . . .

then Ef (a) =f (a + h) or Ey0 = y1
E, (a + h) =f (a + 2h) or Ey1 = y2

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When the operator E is applied twice, the value of the argument increases by two
intervals.

E 2 y0 = y2 , E 2 y1 = y3 , E −2 yn = yn−2

In general E r yn = yn+r E −r yn = yn−r

The operator E has the following properties:
(i) Ecf (x) = cEf (x)

(ii) E[af (x) + bg(x)] = aEf (x) + bEg(x)

(iii) E m [E n f (x)] = E m+n f (x)

(iv) E and △ are commutative,i.e, E △ f (x) = △Ef (x).

(a) 1.4.1 RELATIONS BETWEEN △, ▽ AND E

(a) E = 1 + △ and △ = E − 1

△yn =yn−1 − yn = Eyn − yn = (E − 1)yn

⇒ △ =E − 1 and E = 1 + △

Note. In general E n = (1 + △)n

(b) ▽ = 1 − E −1

▽yn =yn − yn−1 = yn − E −1 yn = (1 − E −1 )yn ⇒ ▽ = 1 − E −1

(c) ▽ = △E −1

▽yn =yn − yn−1 = △yn−1 = △E −1 yn ⇒ ▽ = △E −1

Example 1.4.2 Prove that ▽E = E▽ = △ = E − 1.

.
Solution 1.4.3

▽Eyn = ▽ yn+1 = yn+1 − yn = △yn

E ▽ yn =E(yn − yn−1 =) = yn+1 − yn = △yn
(E − 1)yn =yn+1 − yn = △yn
Hence ▽E =E▽ = △ = E − 1

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Example 1.4.4 Evaluate (▽ + △)2 (x2 + x), h = 1.

Solution 1.4.5

(▽ + △)2 (x2 + x) =(1 − E −1 + E − 1)2 (x2 + x)

=(1 − E −1 )2 (x2 + x) = (E 2 − 2 + E 2 )(x2 + x)
=E 2 (x2 + x) − 2(x2 + x) + E −2 (x2 + x)
=[(x + 2)2 + (x + 2)] − 2(x2 + 2) + [(x − 2)2 + (x − 2)]
=(x2 + 5x + 6) − (2x2 + 2x) + (x2 − 3x + 2) = 8.

△2 △2 u(x)
 
Example 1.4.6 Explain the difference between u(x) and .
E Eu(x)
Solution 1.4.7
 2
(E − 1)2
   2 
△ E − 2E + 1
u(x) = u(x) = u(x)
E E E
=(E − 2 + E −1 )u(x) = u(x + 1) − 2u(x) + (x − 1)
△2 u(x) (E − 1)2 u(x) (E 2 − 2E + 1)u(x) u(x + 2) − 2u(x − 1) + u(x)
= = =
Eu(x) u(x + 1) u(x + 1) u(x + 1)

The difference is evident.

Example 1.4.8 Find (△ − ▽)x2 , where h is the interval of differencing.

Solution 1.4.9

(△ − ▽)x2 =[(E − 1) − (1 − EE −1 )]x2

=(E − 2 + E −1 )x2 = Ex2 − 2x2 + E −1 x2 = ((x + h)2 − 2x2 + (x − h)2 = 2h2 ).

Example 1.4.10 Prove that △▽ = △ − ▽

Solution 1.4.11

△▽ = △ (1 − E −1 ) = △ − △E −1 = △ − ▽.

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(b) 1.4.12 TWO MORE OPERATIONS

(i) Central Difference Operator δ

The central difference operator δ is difined as
   
h h
δf (x) = f x + −f x−
2 2

(ii) Averaging Operator µ

The averaging operator µ is defined as

    
1 h h
µf (x) = f x + +f x−
2 2 2

1.4.13 RELATIONS BETWEEN THE OPERATORS

1
(i) δ =E 1/2 − E −1/2 (ii) µ = (E 1/2 + E −1/2 )
2
(iii) δ = △ E −1/2 (vi) δ = ▽ E 1/2 .
Proofs.
   
h h
(i) δf (x) =f x + −f x− = E 1/2 f (x) − E −1/2 f (x) = (E 1/2 − E −1/2 )f (x)
2 2
∴ δ =E 1/2 − E −1/2
    
1 h h 1
(ii) µf (x) = f x + −f x− = [E 1/2 f (x) + E −1/2 f (x)]
2 2 2 2
1
= (E 1/2 + E −1/2 )f (x)
2
1
∴ µ = (E 1/2 + E −1/2 )
2     
h h h
(iii) δf (x) =f x + −f x− = △f x − = △E −1/2 f (x)
2 2 2
∴ δ = △ E −1/2
     
h h h
(iv) δf (x) =f x + −f x− = ▽f x − = ▽E 1/2 f (x)
2 2 2
∴ δ = △ E 1/2

(c) Example 1.4.14 Prove that

(i) hD = log(1 + △) = −log(△) = sinh−1 (µδ) q
(ii) µ2 = 1 + 41 δ 2 (iii) △ = 21 δ 2 + δ 1 + 14 δ 2 .

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 Solution 1.4.15
(i) We know that
ehD =E = 1 + △
∴ hD =log(1 − △) . . . (1)
−1
Also hD =logE = −log = −log(1 − ▽) . . . (2)
1
and µδ = (E 1/2 + E −1/2 )(E 1/2 − E −1/2 )
2
1 1
= (E − E −1 ) = (ehD − e−hD ) = sinh(hD)
2 2
⇒ hD =sinh−1 (µδ) . . . (3)
F rom (1), (2) and (3) , we have
hD =log(1 + △) = −log(1 − ▽) = sinh−1 (µδ).
1 1
(ii) 1 + δ 2 =1 + (E 1/2 − E −1/2 )2
4 4
1 1
=1 + [(E 1/2 + E −1/2 )2 − 4] = 1 + [(2µ)2 − 4] = µ2 .
r 4 4r
1 2 1 1 1
(iii) δ + δ 1 + δ 2 = (E 1/2 − E −1/2 ) + (E 1/2 − E −1/2 ) 1 + (E 1/2 − E −1/2 )2
2 4 2 4
r
1 1
= (E + E −1 − 2) + (E 1/2 − E −1/2 ) (4 + E + E −1 − 2)
2 4

1 1
= (E + E −1 − 2) + (E 1/2 − E −1/2 )(E 1/2 + E −1/2 )
2 2
1 1
= [(E + E −1 − 2) + (E − E −1 )] = (2E − 2) = E − 1 = △.
2 2
Example 1.4.16 Prove that:

 m  m
m 1 2 m 1.3 3 m 1.3.5 3 m 1 1
(i) △x − △ x + △ x − △ x + ... = x + − x−
2 2.4 2.4.6 2 2
 2  3
x x x
(ii) xy1 + x2 y2 + x3 y3 + · · · = 1−x y1 + △ y1 + △2 y 1 + . . .
1−x 1−x
u1 x u2 x2 u3 x3 x2 2 x3 3
 
x
(iii) u0 + + + + · · · = e u0 + x △ u0 + △ u0 + △ u0 + . . .
1! 2! 3! 2! 3!
Solution 1.4.17
 
1 1.3 2 1.3.5 3
(i) L.H.S = △ 1 − △ + △ − △ + . . . xm
2 2.4 2.4.6
= △ (1 + △)−1/2 xm = △E −1/2 xm
 m  m  m
1 1 1
=△ x− = x+ − x− = R.H.S
2 2 2

Page 9 of 11
   2 
x x x 2
(ii) R.H.S. = 1+ △+ △ + . . . y1
1−x 1−x 1−x
 −1  −1
x x x 1 − x(1 + △)
= 1− △ y1 = y1
1−x 1−x 1−x 1−x
   
1 1
=x y1 = x y1
1 − x(1 + △) 1 − xE
=x(1 − xE)−1 y1 = x(1 + xE + x2 E 2 + . . . )y1 = xy1 + x2 y2 + x3 y3 + . . .
=L.H.S.

x x2 x3
(iii) L.H.S. =u0 + Eu0 + E 2 u0 + E 3 u0 + . . .
1! 2! 3!
xE x2 E 2 x3 E 2
 
= 1+ + + + . . . u0 = exE u0 = ex(1+△) u0 = ex .ex△ u0
1! 2! 3!
x 2 △2 x 3 △3
 
x
=e 1 + x △ + + + . . . u0
2! 3!
x2 2 x3 3
 
x
=e u0 + x △ u0 + △ u0 + △ u0 + . . .
2! 3!
=R.H.S.

TEST YOUR KNOWLEDGE

 
ux vx △ ux − ux △ vx
(1) Prove that △ =
vx vx vx+1
(2) Prove that
(i) E −1 = 1 − ▽ (ii) ▽2 = 1 − 2E −1 + E −2

(iii) (1 + △)(1 − ▽) = 1 (iv) △ ▽ = ▽△ = △ − ▽ = δ 2

△ ▽
(v) △ +▽ = ▽
− △
(vi) ▽ E = E▽ = △ = E −1

d
(vii) e−hD = 1 − ▽, where D = .
dx

(3) Evaluate the following, the interval of differencing

 2  being unity:
△
(i) △ log x (ii) (E −1 △)x (iii) x2
E
△2 x 3 ▽2 x 2
(iv) (v) (vi) △2 E 3 x2
Ex3 E[x + log x]
(vii) (2 △ +1)2 (x + 2)2 (viii) (E + 2)(E − 1)(ex + x)

(ix) (▽ + △)2 (x2 + x) (x) △3 (1 − x)(1 − 2x)(1 − 3x)

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 (4) Evaluate △x log x, the interval of differencing being h.

△2 x Eex
 
x
(5) Prove that e = e . 2 x : the interval of differencing being unity.
E △e

(6) If f (E) is a polynomial in E. prove that f (E)ex = ex f (e).

(7) Prove that y4 = y3 + △y2 + △2 y1 + △3 y1 .

△2 △2 sin(x + h)
(8) Prove that sin(x + h) + = 2(cosh −1)[sin(x + h) + 1].
E E sin(x + h)

(9) If u0 = 3, u1 = 12, u2 = 81, u3 = 200, u4 = 100, u5 = 8, find the value of

△5 u0

(10) Express u = x4 − 12x3 + 24x2 − 30x + 9 and its successive differences in

factorial notation. Hence show that

△5 u = 0.

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