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Chapter 6
Finite Differences
6.1 Introduction
For a function ðŠ = ð ð¥ , finite differences refer to
changes in values of ðŠ (dependent variable) for any
finite (equal or unequal) variation in ð¥ (independent
variable).
In this chapter, we shall study various differencing
techniques for equal deviations in values of ð¥ and
associated differencing operators; also their
applications will be extended for finding missing
values of a data and series summation.
6.2 Shift or Increment Operator (ð¬)
Shift (Increment) operator denoted by âðžâ operates on ð(ð¥) as ðžð ð¥ = ð ð¥ + ð
Or ðžðŠð¥ = ðŠð¥+ð , where âðâ is the step height for equi-spaced data points.
Clearly effect of the shift operator ðž is to shift the function value to the next
higher value ð ð¥ + ð or ðŠð¥+ð
Also ðž2ð ð¥ = ðž ðžð(ð¥) = ðžð ð¥ + ð = ð ð¥ + 2ð
⎠ðžðð ð¥ = ð ð¥ + ðð
Moreover ðžâ1ð ð¥ = ð ð¥ â ð , where ðžâ1 is the inverse shift operator.
6.3 Differencing Operators
If ðŠ0, ðŠ1, ðŠ2 , âŠðŠð be the values of ðŠ for corresponding values of ð¥0, ð¥1, ð¥2 ,
âŠð¥ð , then the differences of ðŠ are defined by (ðŠ1 â ðŠ0), (ðŠ2 â ðŠ1), ⊠, (ðŠð â ðŠðâ1) , and are denoted by different operators discussed in this section.
6.3.1 Forward Difference Operator â
Forward difference operator âââ operates on ðŠð¥ as âðŠð¥ = ðŠð¥+1â ðŠð¥
Or âð ð¥ = ð ð¥ + ð â ð ð¥ , where ð is the height of differencing.
⎠âðŠ0 = ðŠ1â ðŠ0
âðŠ1 = ðŠ2â ðŠ1
â®
âðŠð = ðŠð+1â ðŠð
Also â2ðŠ0 = âðŠ1â âðŠ0 = ðŠ2â ðŠ1 â ðŠ1â ðŠ0 = ðŠ2 â 2ðŠ1 + ðŠ0
â®
âððŠ0 = ðŠð â ðð¶1ðŠðâ1 + ðð¶2ðŠðâ2 ââ¯+ â1 ðâ1 ðð¶ðâ1ðŠ1 + â1 ððŠ0
Generalizing âððŠð = ðŠð+ð â ðð¶1ðŠð+ðâ1 + ðð¶2ðŠð+ðâ2 ââ¯+ â1 ððŠð
Here âð is the ðð¡ð order forward difference; Table 6.1 shows the forward
differences of various orders.
Table 6.1 Forward Differences
ð ð â âð âð âð âð ðð ðŠð
âðŠð ðð ðŠ1 â2ðŠð
âðŠ1 â3ðŠð ðð y2 â2ðŠ1 â4ðŠ0
âðŠ2 â3ðŠ1 â5ðŠ0 ðð ðŠ3 â2ðŠ2 â4ðŠ1
âðŠ3 â3ðŠ2 ðð ðŠ4 â2ðŠ3
âðŠ4 ðð ðŠ5
The arrow indicates the direction of differences from top to bottom. Differences in
each column notate difference of two adjoining consecutive entries of the previous
column.
Relation between â and ð¬
â and ðž are connected by the relation â â¡ ðž â 1
Proof: we know that âðŠð = ðŠð+1â ðŠð
= ðžðŠðâ ðŠð
â âðŠð = ðž â 1 ðŠð
â â â¡ ðž â 1 or ðž â¡ 1 + â
Properties of operator âââ
âð¶ = 0 , ð¶ being a constant
âð¶ ð(ð¥) = ð¶ð(ð¥)
â[ðð 𥠱 ðð(ð¥)] = ð âð 𥠱 ð âð(ð¥)
â ð ð¥ ð ð¥ = ð ð¥ + ð â ð ð¥ + ð ð¥ âð ð¥ , ð & ð may be interchanged
â ð ð¥
ð ð¥ =
ð ð¥ âð ð¥ âð(ð¥)âð ð¥
ð ð¥+ð ð ð¥
Result 1: The ððð differences of a polynomial of degree 'n' are constant and all
higher order differences are zero.
Proof: Consider the polynomial ð(ð¥) of ðð¡ð degree
ð(ð¥) = ð0ð¥ð + ð1ð¥
ðâ1 + ð2ð¥ðâ2 + â¯+ ððâ1ð¥ + ðð
First differences of the polynomial ð(ð¥) are calculated as:
â ð ð¥ = ð ð¥ + ð â ð(ð¥)
= ð0 (ð¥ + ð)ð â ð¥ð + ð1 (ð¥ + ð)ðâ1 â ð¥ðâ1 + â¯+ ððâ1 ð¥ + ð â ð¥ = ð0ðð ð¥ðâ1 + ð1
â² ð¥ðâ1 + ð2â² ð¥ðâ2 + â¯+ ððâ1
â² ð + ððâ²
where ð1â² , ð2
â² ,⊠,ððâ1â² , ðð
â² are new constants
â First difference of a polynomial of degree ð is a polynomial of degree (ð â 1)
Similarly â2ð ð¥ = â ð ð¥ + ð â â ð ð¥
= ð0ð(ð â 1)ð2 ð¥ðâ2 + ð1â²â² ð¥ðâ3 + âŠ+ ðð
â²â²
⎠Second difference of a polynomial of degree ð is a polynomial of degree (ð â 2)
Repeating the above process âðð ð¥ = ð0ð(ð â 1)âŠ2.1ðð ð¥ðâð
â âðð ð¥ = ð0ð! ðð which is a constant
⎠ðð¡ð Difference of a polynomial of degree ð is a polynomial of degree zero.
Thus (ð + 1)ð¡ðand higher order differences of a polynomial of ðð¡ð degree are all
zero.
The converse of above result is also true , i.e. if the ðð¡ð difference of a
polynomial given at equally spaced points are constant then the function is
a polynomial of degree âðâ.
6.3.2 Backward Difference Operator ð
Backward difference operator â â â operates on ðŠð as âðŠð = ðŠðâ ðŠðâ1
⎠The differences (ðŠ1 â ðŠ0) , (ðŠ2 â ðŠ1) , ⊠, (ðŠð â ðŠðâ1) when denoted by
âðŠ1, âðŠ2, ⊠,âðŠð are called first backward differences.
Also â2ðŠð = âðŠð â âðŠðâ1 , â3ðŠð = â2ðŠð â â2ðŠðâ1 denote second and third
backward differences respectively.
Table 6.2 shows the backward differences of various orders.
Table 6.2 Backward Differences
ð ð ð ðð ðð ðð ðð ðð ðŠð
âðŠ1 ðð ðŠ1 â2ðŠ2
âðŠ2 â3ðŠ3 ðð y2 â2ðŠ3 â4ðŠ4
âðŠ3 â3ðŠ4 â5ðŠ5 ðð ðŠ3 â2ðŠ4 â4ðŠ5
âðŠ4 â3ðŠ5 ðð ðŠ4 â2ðŠ5
âðŠ5 ðð ðŠ5
The arrow indicates the direction of differences from bottom to top. Differences in
each column notate difference of two adjoining consecutive entries of the previous
column, i.e. âðŠ1
= ðŠ1â ðŠ
ð, â2ðŠ
2= âðŠ
2â âðŠ
1, ⊠,â5ðŠ5 = â4ðŠ5 â â4ðŠ4.
Relation between ð and ð¬
â and ðž are connected by the relation â â¡ 1 â ðžâ1
Proof: we know that âðŠð = ðŠðâ ðŠðâ1
= ðŠðâ ðžâ1ðŠð
â âðŠð = 1 â ðžâ1 ðŠð
â â â¡ 1 â ðžâ1
6.3.3 Central Difference Operator ð
Central difference operator â ÎŽ â operates on ðŠð as ÎŽ ðŠð = ðŠð+
1
2
â ðŠðâ
1
2
⎠The differences (ðŠ1 â ðŠ0) , (ðŠ2 â ðŠ1) , ⊠, (ðŠð â ðŠðâ1) when denoted by
ÎŽðŠ1
2
, ÎŽðŠ3
2
, ⊠, ÎŽðŠðâ
1
2
are called first central differences.
Also ÎŽ2ðŠð = ÎŽðŠð+
1
2
â ÎŽðŠðâ
1
2
, ÎŽ3ðŠð = ÎŽ2ðŠð+
1
2
â ÎŽ2ðŠðâ
1
2
denote second and third
central differences respectively as shown in Table 6.3.
Table 6.3 Central Differences
ð ð ð ð ð ð ð ð ð ð ð ðð ðŠð
ÎŽðŠ12
ðð ðŠ1 ÎŽ2ðŠ1
ÎŽðŠ32 ÎŽ3ðŠ3
2
ðð y2 ÎŽ2ðŠ2 ÎŽ4ðŠ
2
ÎŽðŠ52 ÎŽ3ðŠ5
2
ÎŽ5ðŠ52
ðð ðŠ3 ÎŽ2ðŠ3 ÎŽ4ðŠ
3
ÎŽðŠ72 ÎŽ3ðŠ7
2
ðð ðŠ4 ÎŽ2ðŠ4
ÎŽðŠ92
ðð ðŠ5
Central differences in each column notate difference of two adjoining consecutive
entries of the previous column, i.e. ÎŽðŠ1
2
= ðŠ1â ðŠ
ð, ⊠, ÎŽ5ðŠ5
2
= ÎŽ4ðŠ3 â ÎŽ4ðŠ2.
Relation between ð and ð¬
ÎŽ and ðž are connected by the relation ÎŽ â¡ ðž1
2 â ðžâ 1
2
Proof: we know that ÎŽ ðŠð = ðŠð+
1
2
â ðŠðâ
1
2
= ðž1
2ðŠðâ ðžâ 1
2ðŠð
â ÎŽ ðŠð = ðž1
2 â ðžâ 1
2 ðŠð
⎠Ύ â¡ ðž1
2 â ðžâ 1
2
Observation: It is only the notation which changes and not the difference.
⎠ðŠ1 â ðŠð = âðŠ0 = âðŠ1 = ÎŽðŠ1
2
6.3.4 Averaging Operator ð
Averaging operator â ÎŒ â operates on ðŠð¥ as ÎŒ ðŠð¥ =1
2 ðŠ
ð¥+ð
2
+ ðŠð¥â
ð
2
Or ÎŒ ð ð¥ = 1
2 ð ð¥ +
ð
2 + ð ð¥ â
ð
2 , â h â is the height of the interval.
Relation between ð and ð¬
We know that ÎŒ ðŠð =1
2 ðŠ
ð+ð
2
+ ðŠðâ
ð
2
=1
2 ðž
1
2ðŠð+ ðžâ 1
2ðŠð
â ÎŒ ðŠð =1
2 ðž
1
2 + ðžâ 1
2 ðŠð
⎠Ό â¡1
2 ðž
1
2 + ðžâ 1
2
Result 2: Relation between ð¬ and ð«, where ð« â¡ð
ð ð
We know ðŠ ð¥ + ð = ðŠ ð¥ + ð ðŠâ² ð¥ +ð2
2! ðŠâ²â² ð¥ +. .. By Taylorâs theorem
= ðŠ ð¥ + ð ð·ðŠ(ð¥) +ð2
2! ð·2ðŠ(ð¥)+. ..
= 1 + ðð· +ð2
2! ð·2 + ⯠ðŠ(ð¥)
â ðž ðŠ ð¥ = ððð·ðŠ(ð¥)
⎠ðž = ððð·, ð· â¡ð
ðð¥
Result 3: Relation between â and ð«, where ð« â¡ð
ð ð
We know that â â¡ ðž â 1
â â â¡ ððð· â 1 âµ ðž = ððð·
Result 4: Relation between ðµ and ð«, where ð« â¡ð
ð ð
We know that â â¡ 1 â ðžâ1 = 1 â ðâðð· âµ ðž = ððð·
Result 5: Relation between â and ðµ
We know that ðž â¡ 1 + â â¯â
Also ðžâ1 â¡ 1 â â
â ðž â¡1
1ââ ⯠â¡
â 1 + ââ¡1
1ââ From â and â¡
â ââ¡1
1âââ 1
â ââ¡â
1ââ
Result 6: Relation between ð , ð¹ and ð¬
We have ÎŒ â¡1
2 ðž
1
2 + ðžâ 1
2
Also ÎŽ â¡ ðž1
2 â ðžâ 1
2
â ΌΎ â¡1
2 ðž
1
2 + ðžâ 1
2 ðž1
2 â ðžâ 1
2
â ΌΎ â¡1
2 ðž â ðžâ 1
Result 7: Relation between ð , ð¹ , â and â
We have ΌΎ â¡1
2 ðž â ðžâ 1 =
1
2 1 + â) â (1 â â
â ΌΎ â¡1
2 â + â
Result 8: âððŠð = âððŠð+ð
We have âððŠð = (ðž â 1)ððŠð âµ â= ðž â 1
= ðŠð+ð â ðð¶1ðŠð+ðâ1 + ðð¶2ðŠð+ðâ2 ââ¯+ â1 ððŠð
= ðžð â ðð¶1ðžðâ1 + ðð¶2ðž
ðâ2 ââ¯+ â1 ð ðŠð
= ðžððŠð â ðð¶1ðžðâ1ðŠð + ðð¶2ðž
ðâ2ðŠð ââ¯+ â1 ððŠð
= ðŠð+ð â ðð¶1ðŠð+ðâ1 + ðð¶2ðŠð+ðâ2 ââ¯+ â1 ððŠð
Also âððŠð+ð = (1 â Eâ1)ððŠð+ð âµ â â¡ 1 â ðžâ1
= 1 â ðð¶1ðžâ1 + ðð¶2ðž
â2 ââ¯+ â1 ððžâð ðŠð+ð
= ðŠð+ð â ðð¶1ðŠð+ðâ1 + ðð¶2ðŠð+ðâ2 ââ¯+ â1 ððŠð
⎠âððŠð = âððŠð+ð
Example 1 Evaluate the following:
i. âðð¥ ii. â2ðð¥ iii. â ð¡ððâ1ð¥ iv. â ð¥+1
ð¥2â3ð¥+2 v. âðð
2 = ðð + ðð+1 âðð
Solution: i. âðð¥ = ðð¥+ð â ðð¥ = ðð¥(ðð â 1)
âðð¥ = ðð¥(ð â 1) , if ð = 1
ii. â2ðð¥ = â(âðð¥)
= â ðð¥ ðð â 1
= ðð â 1 âðð¥
= ðð â 1 ðð¥+ð â ðð¥
= ðð â 1 ðð¥(ðð â 1)
= ðð¥ (ðð â 1)2
iii. âð¡ððâ1ð¥ = ð¡ððâ1 ð¥ + ð â ð¡ððâ1ð¥
= ð¡ððâ1 ð¥+ðâð¥
1+(ð¥+ð)ð¥
= ð¡ððâ1 ð
1+(ð¥+ð)ð¥
iv. â ð¥+1
ð¥2â3ð¥+2 = â
ð¥+1
ð¥â1 (ð¥â2)
= â â2
ð¥â1+
3
ð¥â2 = â
â2
ð¥â1 + â
3
ð¥â2
= â2 1
ð¥+1â1â
1
ð¥â1 + 3
1
ð¥+1â2â
1
ð¥â2
= â2 1
ð¥â
1
ð¥â1 + 3
1
ð¥â1â
1
ð¥â2
= â(ð¥+4)
ð¥ ð¥â1 (ð¥â2)
v. âðð2 = ðð+1
2 â ðð2 = ðð+1 + ðð ðð+1 â ðð = ðð + ðð+1 âðð
Example 2 Evaluate the following:
i. âðð¥ log 2ð¥ ii. â ð¥2
cos 2ð¥
Solution: i. Let ð ð¥ = ðð¥ and ð ð¥ = log 2ð¥
We have â ð ð¥ ð ð¥ = ð ð¥ + ð â ð ð¥ + ð ð¥ âð ð¥
⎠âðð¥ log 2ð¥ = ðð¥+ðâ log 2ð¥ + log 2ð¥ âðð¥
= ðð¥+ð log 2 ð¥ + ð â log 2ð¥ + log 2ð¥ ðð¥+ð â ðð¥
= ðð¥ðð log 1 +ð
ð¥ + ðð¥ log 2ð¥ ðð â 1
= ðð¥ ðð log 1 +ð
ð¥ + log 2ð¥ ðð â 1
ii. Let ð ð¥ = ð¥2 and ð ð¥ = cos 2ð¥
We have â ð ð¥
ð ð¥ =
ð ð¥ âð ð¥ âð(ð¥)âð ð¥
ð ð¥+ð ð ð¥
=cos 2ð¥ ð¥+ð 2âð¥2 âð¥2 cos 2 ð¥+ð âcos 2ð¥
cos 2 ð¥+ð cos 2ð¥
= ð2+2ðð¥ cos 2ð¥+2ð¥2 sin 2ð¥+ð sin ð
cos 2 ð¥+ð cos 2ð¥
Example 3 Evaluate â4 1 â 2ð¥ 1 â 3ð¥ 1 â 4ð¥ 1 â ð¥ ,where interval of differencing is one.
Solution: â4 1 â 2ð¥ 1 â 3ð¥ 1 â 4ð¥ 1 â ð¥
= â4 24ð¥4 + â¯+ 1 = 24.4!. 14 = 576
âµ âðð ð¥ = ð0ð! ðð and â4ð¥ð = 0 when ð < 4
Example 4 Prove that â3ðŠ3 = â3ðŠ6
Solution: â3ðŠ3 = (ðž â 1)3ðŠ3 âµ â= ðž â 1
= ðž3 â 1 â 3ðž2 + 3ðž ðŠ3
= ðž3ðŠ3 â ðŠ3 â 3ðž2ðŠ3 + 3ðžðŠ3
= ðŠ6 â ðŠ3 â 3ðŠ5 + 3ðŠ4
Also â3ðŠ6 = (1 â Eâ1)3ðŠ6 âµ â â¡ 1 â ðžâ1
= 1 â Eâ3 â 3ðžâ1 + 3ðžâ2 ðŠ6
= ðŠ6 â ðŠ3 â 3ðŠ5 + 3ðŠ4
Example 5 Prove that â + â = â
âââ
â
Solution: L.H.S. = â + â = ðž â 1 + 1 â ðžâ1
= ðž â ðžâ1
R.H.S. =â
âââ
â
=ðžâ1
1âðžâ1â
1âðž â1
ðžâ1
= ðžâ1 2 â 1â ðž â1
2
1âðž â1 ðžâ1
= ðž2+1â 2ðž â 1+ ðžâ2â2ðžâ1
ðž + ðžâ1â2
=ðž2âðžâ2â2ðž+2ðžâ1
ðž + ðžâ1â2
= ðž+ðžâ1 ðžâðž â1 â 2 ðžâ ðžâ1
ðž + ðžâ1â2
= ðž â ðžâ1 ðž + ðž â1â2
ðž + ðžâ1â2
= ðž â ðžâ1 = R.H.S.
Example 6 Prove that ðž = 1 +1
2ÎŽ2 + ÎŽ 1 +
1
4ÎŽ2
Solution: R.H.S. = 1 +1
2ÎŽ2 + ÎŽ 1 +
1
4ÎŽ2
= 1 +1
2 ðž
1
2 â ðžâ 1
2 2
+ ðž1
2 â ðžâ 1
2 1 +1
4 ðž
1
2 â ðžâ 1
2 2
âµ ÎŽ â¡ ðž1
2 â ðžâ 1
2
= 1 +1
2 ðž + ðžâ1 â 2 + ðž
1
2 â ðžâ 1
2 1 +1
4 ðž + ðžâ1 â 2
= 1 +1
2 ðž + ðžâ1 â 2 + ðž
1
2 â ðžâ 1
2 1
4 ðž + ðžâ1 + 2
= 1 +1
2 ðž + ðžâ1 â 2 + ðž
1
2 â ðžâ 1
2 1
4 ðž
1
2 + ðžâ 1
2 2
= 1 +1
2 ðž + ðžâ1 â 2 +
1
2 ðž
1
2 â ðžâ 1
2 ðž1
2 + ðžâ 1
2
=1
2 ðž + ðžâ1 +
1
2 ðž â ðžâ1 = ðž = L.H.S.
Example 7 Prove that â = â1
2ÎŽ2 + ÎŽ 1 +
1
4ÎŽ2
Solution: R.H.S. = â1
2ÎŽ2 + ÎŽ 1 +
1
4ÎŽ2
= â1
2 ðž
1
2 â ðžâ 1
2 2
+ ðž1
2 â ðžâ 1
2 1 +1
4 ðž
1
2 â ðžâ 1
2 2
âµ ÎŽ â¡ ðž1
2 â ðžâ 1
2
= â1
2 ðž + ðžâ1 â 2 + ðž
1
2 â ðžâ 1
2 1 +1
4 ðž + ðžâ1 â 2
= â1
2 ðž + ðžâ1 â 2 + ðž
1
2 â ðžâ 1
2 1
4 ðž + ðžâ1 + 2
= â1
2 ðž + ðžâ1 â 2 + ðž
1
2 â ðžâ 1
2 1
4 ðž
1
2 + ðžâ 1
2 2
= â1
2 ðž + ðžâ1 â 2 +
1
2 ðž
1
2 â ðžâ 1
2 ðž1
2 + ðžâ 1
2
= â1
2 ðž + ðžâ1 â 2 +
1
2 ðž â ðžâ1 = 1 â ðžâ1 = â= L.H.S.
Example 8 Prove that (i) â â â = ÎŽ2 (ii) ÎŒ = 1 +1
4ÎŽ2 = 1 +
â
2 1 + â â
1
2
Solution: (i) ÎŽ2 = ðž1
2 â ðžâ 1
2 2
= ðž + ðžâ1 â 2 âµ ÎŽ â¡ ðž1
2 â ðžâ 1
2
= ðž â 1 â 1 â ðžâ1 = â â â
âµ ðž â 1 â¡ â and 1 â ðžâ1 = â
(ii) 1 +1
4ÎŽ2 = 1 +
1
4 ðž
1
2 â ðžâ 1
2 2
âµ ÎŽ â¡ ðž1
2 â ðžâ 1
2
= 1 +1
4 ðž + ðžâ1 â 2
= 1
4 ðž + ðžâ1 + 2
= 1
4 ðž
1
2 + ðžâ 1
2 2
=1
2 ðž
1
2 + ðžâ 1
2 = ÎŒ âµ ÎŒ â¡1
2 ðž
1
2 + ðžâ 1
2
Also 1 +â
2 1 + â â
1
2 = 1 +ðžâ1
2 1 + ðž â 1 â
1
2
âµ â â¡ ðž â 1
= ðž+1
2 ðžâ
1
2
=1
2 ðžâ
1
2 + ðž1
2 = Ό
Example 9 Prove that (i) â â¡ ðžâ â¡ âE = ÎŽðž 12 (ii) Er = ÎŒ +
ð¿
2
2ð
Solution: (i) ðžâ = E 1 â ðžâ 1 = ðž â 1 = â âµ â â¡ 1 â ðžâ1
âE = 1 â ðžâ 1 ðž = ðž â 1 = â
ÎŽðž 12 = ðž
1
2 â ðžâ 1
2 ðž 12 = ðž â 1 = â âµ ÎŽ â¡ ðž
1
2 â ðžâ 1
2
(ii) R.H.S. = ÎŒ +ð¿
2
2ð
= 1
2 ðž
1
2 + ðžâ 1
2 +1
2 ðž
1
2 â ðžâ 1
2
2ð
âµ ÎŒ â¡1
2 ðž
1
2 + ðžâ 1
2 and ÎŽ â¡ ðž1
2 â ðžâ 1
2
= 1
2 2ðž
1
2
2ð
= ðž1
2 2ð
= Er = L.H.S
Example 10 Prove that (i) ð· â¡ 1
ðððð ðž (iii) ðð· â¡ ððð 1 + â â¡ âlog(1 â â)
(iii) â2 â¡ ð2ð·2 â ð3ð·3 +7
12ð4ð·4 + â¯
Solution: (i) We know that ðž â¡ ððð·
â ððððž â¡ ððð ððð·
â ððððž â¡ ðð ððð ð
â ð· â¡ 1
ðððð ðž âµ ððð ð = 1
(ii) ð ð· â¡ ððððž From relation (i)
â¡ log(1 + â) âµ ðž â¡ 1 + â
Also ð ð· â¡ ððððž â¡ â logðžâ1
â¡ âlog(1 â â) âµ â â¡ 1 â ðžâ1
(iii) We know that â â¡ 1 â ðžâ1
â â â¡ 1 â1
ðž
â¡ 1 â ðâðð· âµ ðž = ððð·
â ââ¡ 1 â 1 â ðð +ð2ð·2
2!â
ð3ð·3
3!+ â¯
â ââ¡ ðð âð2ð·2
2! +
ð3ð·3
3!+ â¯
⎠â2 â¡ ðð âð2ð·2
2! +
ð3ð·3
3!+ â¯
2
â â2 â¡ ð2ð·2 + ð2ð·2
2!
2
+ â¯â 2 ðð ð2ð·2
2! + 2 ðð
ð3ð·3
3! ââ¯
â â2 â¡ ð2ð·2 â ð3ð·3 + ð4ð·4
4+
ð4ð·4
3 ââ¯
â â2 â¡ ð2ð·2 â ð3ð·3 +7
12ð4ð·4 ââ¯
Remark: In order to prove any relation, we can express the operators (â ,â,ð¿) in
terms of fundamental operator ðž.
Example 11 Form the forward difference table for the function
ð ð¥ = ð¥3 â 2ð¥2 â 3ð¥ â 1 for ð¥ = 0, 1, 2, 3, 4.
Hence or otherwise find â3ð ð¥ , also show that â4ð ð¥ = 0
Solution: ð 0 = â1, ð 1 = â5, ð 2 = â7, ð 3 = â1, ð 4 = 19
Constructing the forward difference table:
ð ð(ð) â âð âð âð ð â1 â4
1 â5 2 â2 6 ð â7 8 0 6 6 ð â1 14 20 ð 19
From the table, we see that â3ð ð¥ = 6 and â4ð ð¥ = 0
Note: Using the formula âðð ð¥ = ð0ð! ðð , â3ð ð¥ = 1.3!. 1ð = 6
Also âð+1ð ð¥ = 0 for a polynomial of degree ð, ⎠â4ð ð¥ = 0
Example 12 If for a polynomial, five observations are recorded as: ðŠ0 = â8,
ðŠ1 = â6, ðŠ2 = 22, ðŠ3 = 148, ðŠ4 = 492, find ðŠ5.
Solution: ðŠ5 = ðž5ðŠ0 = (1 + â)5ðŠ0 âµ ðž â¡ 1 + â
= ðŠ0 + 5ð¶1âðŠ0 + 5ð¶2â2ðŠ0 + 5ð¶3â
3ðŠ0 + 5ð¶4â4ðŠ0 + â5ðŠ0 âŠâ
Constructing the forward difference table:
ð ð â âð âð âð ðð â8
2 ðð â6 26
28 72 ðð 22 98 48
126 120 ðð 148 218
344 ðð 492
From table âðŠ0 = 2, â2ðŠ0 = 26 , â3ðŠ0 = 72 , â3ðŠ0 = 48 ⊠â¡
â ðŠ5
= â8 + 5 2 + 10 26 + 10 72 + 5 48 = 1222 using â¡ in â
6.4 Missing values of Data Missing data or missing values occur when an observation is missing for a
particular variable in a data sample. Concept of finite differences can help to locate
the requisite value using known concepts of curve fitting.
To determine the equation of a line (equation of degree one), we need at least two
given points. Similarly to trace a parabola (equation of degree two), at least three
points are imperative. Thus we essentially require ð + 1 known observations to
determine a polynomial of ðð¡ð degree.
To find missing values of data using finite differences, we presume the degree of
the polynomial by the number of known observations and use the result
âð+1ð ð¥ = 0 for a polynomial of degree ð.
Example 13 Use the concept of missing data to find ðŠ5 if ðŠ0 = â8, ðŠ1 = â6,
ðŠ2 = 22, ðŠ3 = 148, ðŠ4 = 492
Solution: Constructing the forward difference table taking ðŠ5 as missing value
ð ð â âð âð âð âð ðð â8
2 ðð â6 26
28 72 ðð 22 98 48
126 120 ðŠ5 â 1222 ðð 148 218 ðŠ5 â 1174
344 ðŠ5 â 1054 ðð 492 ðŠ5 â 836
ðŠ5 â 492 ðð ðŠ5
Since 5 observations are known, let us assume that the polynomial represented by
given data is of 4ð¡ð degree. ⎠â5ðŠ = 0 â ðŠ5â 1222 = 0 or ðŠ5 = 1222
Example 14 Find the missing values in the following table
ð ð ð ðð ðð ðð ðð
ð(ð) 6 ? 13 17 22 ?
Solution: Since there are 4 known values of ð ð¥ in the given data, let us assume
the polynomial represented by the given data to be of 3ððdegree.
Constructing the forward difference table taking missing values as ð and ð.
ð ð â âð âð âð ð 6
ð â 6 ð ð 19 â 2ð
13 â ð 3ð â 28 ðð 13 ð â 9 38 â 4ð
4 10 â ð 15 17 1 ð + ð â 38
5 ð â 28 2ð 22 ð â 27
ð â 22
25 ð
Since the polynomial represented by the given data is considered to be of
3ððdegree, 4ð¡âand higher order differences are zero i.e. â4ðŠ = 0
⎠38 â 4ð = 0 and ð + ð â 38 = 0 Solving these two equations, we get ð = 9.5 ð = 28.5
6.5 Finding Differences Using Factorial Notation We can conveniently find the forward differences of a polynomial using factorial
notation.
6.5.1 Factorial Notation of a Polynomial
A product of the form ð¥ ð¥ â 1 ð¥ â 2 ⊠ð¥ â ð + 1 is called a factorial
polynomial and is denoted by ð¥ ð
⎠ð¥ = ð¥
ð¥ 2 = ð¥(ð¥ â 1)
ð¥ 3 = ð¥ ð¥ â 1 (ð¥ â 2)
â®
ð¥ ð = ð¥ ð¥ â 1 ð¥ â 2 ⊠ð¥ â ð + 1
In case, the interval of differencing is ð, then
ð¥ ð = ð¥ ð¥ â ð ð¥ â ð ⊠ð¥ â ðâ 1 ð
The results of differencing ð¥ ð are analogous to that differentiating ð¥ð
⎠â ð¥ ð = ð ð¥ ðâ1
â2 ð¥ ð = ð(ð â 1) ð¥ ðâ2
â3 ð¥ ð = ð ð â 1 (ð â 2) ð¥ ðâ3
â®
âð ð¥ ð = ð ð â 1 ð â 2 ⊠3.2.1 = ð!
âð+1 ð¥ ð = 0
Also 1
â ð¥ =
ð¥ 2
2 ,
1
â ð¥ 2 =
ð¥ 3
3 and so on
1
â2 ð¥ =
1
â ð¥ 2
2 =
ð¥ 3
6
â®
Remark:
i. Every polynomial of degree ð can be expressed as a factorial
polynomial of the same degree and vice-versa.
ii. The coefficient of highest power of ð¥ and also the constant term
remains unchanged while transforming a polynomial to factorial
notation.
Example15 Express the polynomial 2ð¥2 + 3ð¥ + 1 in factorial notation.
Solution: 2ð¥2 â 3ð¥ + 1 = 2ð¥2 â 2ð¥ + 5ð¥ + 1
= 2ð¥ ð¥ â 1 + 5ð¥ + 1
= 2 ð¥ 2 + 5 ð¥ + 1
Example16 Express the polynomial 2ð¥3 â ð¥2 + 3ð¥ â 4 in factorial notation.
Solution: 2ð¥3 â ð¥2 + 3ð¥ â 4 = 2 ð¥ 3 + ðŽ ð¥ 2 + ðµ ð¥ â 4
Using remarks i and ii
= 2ð¥ ð¥ â 1 ð¥ â 2 + ðŽð¥ ð¥ â 1 + ðµð¥ â 4
= 2ð¥3 + ðŽ â 6 ð¥2 + âðŽ + ðµ + 4 ð¥ â 4
Comparing the coefficients on both sides
A â 6 = â1, âðŽ + B + 4 = 3
â A = 5, B = 4
⎠2ð¥3 â ð¥2 + 3ð¥ â 4 = 2 ð¥ 3 + 5 ð¥ 2 + 4 ð¥ â 4
We can also find factorial polynomial using synthetic division as shown:
Coefficients ðŽ and ðµ can be found as remainders under ð¥2 and ð¥ columns
ð¥3 ð¥2 ð¥
1 2 â1 3 â4
â 2 1
2 2 1 4 = ðµ
â 4
2 5 = A
Example 17 Find â3ð ð¥ for the polynomial ð ð¥ = ð¥3 â 2ð¥2 â 3ð¥ â 1
Also show that â4ð ð¥ = 0
Solution: Finding factorial polynomial of ð ð¥ as shown:
Let ð¥3 â 2ð¥2 â 3ð¥ â 1 = ð¥ 3 + ðŽ ð¥ 2 + ðµ ð¥ â 1
Coefficients ðŽ and ðµ can be found as remainders under ð¥2 and ð¥ columns
ð¥3 ð¥2 ð¥
1 1 â2 â3 â1
â 1 â1
2 1 â1 â 4 = ðµ
â 2
1 1 = A
⎠ð ð¥ = ð¥3 â 2ð¥2 â 3ð¥ â 1 = ð¥ 3 + ð¥ 2 â 4 ð¥ â 1
â3ð ð¥ = â3 ð¥ 3 + ð¥ 2 â 4 ð¥ â 1
= 3! + 0 = 6 âµ âð ð¥ ð = ð! and âð+1 ð¥ ð = 0
Also â4ð ð¥ = â4 ð¥ 3 + ð¥ 2 â 4 ð¥ â 1 = 0
Note: Results obtained are same as in Example 11, where we have used
forward difference table to compute the differences.
Example 18: Obtain the function whose first difference is 8ð¥3 â 3ð¥2 + 3ð¥ â 1
Solution: Let ð ð¥ be the function whose first difference is 8ð¥3 â 3ð¥2 + 3ð¥ â 1
â âð ð¥ = 8ð¥3 â 3ð¥2 + 3ð¥ â 1
Let 8ð¥3 â 3ð¥2 + 3ð¥ â 1 = 8 ð¥ 3 + ðŽ ð¥ 2 + ðµ ð¥ â 1
Coefficients ðŽ and ðµ can be found as remainders under ð¥2 and ð¥ columns
ð¥3 ð¥2 ð¥
1 8 â3 3 â1
â 8 5
2 8 5 8 = ðµ
â 16
8 21 = A
⎠âð ð¥ = 8ð¥3 â 3ð¥2 + 3ð¥ â 1 = 8 ð¥ 3 + 21 ð¥ 2 + 8 ð¥ â 1
ð ð¥ = 1
â 8 ð¥ 3 + 21 ð¥ 2 + 8 ð¥ â 1
=8 ð¥ 4
4+
21 ð¥ 3
3+
8 ð¥ 2
2â ð¥ âµ
1
â ð¥ =
ð¥ 2
2 ,
1
â ð¥ 2 =
ð¥ 3
3, âŠ
= 2 ð¥ 4 + 7 ð¥ 3 + 4 ð¥ 2 â [ð¥] = 2ð¥ ð¥ â 1 ð¥ â 2 ð¥ â 3 + 7ð¥ ð¥ â 1 ð¥ â 2 + 4ð¥ ð¥ â 1 â ð¥
= ð¥ 2 ð¥ â 1 ð¥ â 2 ð¥ â 3 + 7 ð¥ â 1 ð¥ â 2 + 4 ð¥ â 1 â 1 = ð¥ 2ð¥3 â 5ð¥2 + 5ð¥ â 3 = 2ð¥4 â 5ð¥3 + 5ð¥2 â 3ð¥
â ð ð¥ = 2ð¥4 â 5ð¥3 + 5ð¥2 â 3ð¥
6.6 Series Summation Using Finite Differences The method of finite differences may be used to find sum of a given series by
applying the following algorithm:
1. Let the series be represented by ð¢0, ð¢1 , ð¢2 , ð¢3, âŠ
2. Use the relation ð¢ð = ðžðð¢0 to introduce the operator ðž in the series.
3. Replace ðž by â by substituting ðž â¡ 1 + â and find the sum the series by
any of the applicable methods like sum of a G.P., exponential or logarithmic
series or by binomial expansion and operate term by term on ð¢0 to find the
required sum.
Example 19 Prove the following using finite differences:
i. ð¢0 + ð¢1ð¥
1!+ ð¢2
ð¥2
2!+ ⯠= ðð¥ ð¢0 + ð¥
â ð¢0
1!+ ð¥2 â2 ð¢0
2!+ â¯
ii. ð¢0 â ð¢1 + ð¢2 â ð¢3 + ⯠= 1
2 ð¢0 â
1
4â ð¢0 +
1
8â2 ð¢0 ââ¯
Solution: i. ð¢0 + ð¢1ð¥
1!+ ð¢2
ð¥2
2!+ ⯠= ð¢0 +
ð¥
1!ðžð¢0 +
ð¥2
2!ðž2ð¢0 + â¯
= 1 +ð¥ðž
1!+
ð¥2ðž2
2!ð¢0 + ⯠ð¢0
= ðð¥ðž ð¢0 = ðð¥(1+â) ð¢0
= ðð¥ðð¥â ð¢0
= ðð¥ 1 +ð¥â
1!+
ð¥2â2
2!+ ⯠ð¢0
= ðð¥ ð¢0 + ð¥â ð¢0
1!+ ð¥2 â2 ð¢0
2!+ â¯
ii. ð¢0 â ð¢1 + ð¢2 â ð¢3 + ⯠= ð¢0 â ðžð¢0 + ðž2ð¢0 â ðž3ð¢0 + â¯
= 1 â ðž + ðž2 â ðž3 + ⯠ð¢0
= 1 + ðž â1 ð¢0
= 2 + â â1 ð¢0
= 2â1 1 +â
2 â1
ð¢0
=1
2 1 â
â
2+
â2
4â⯠ð¢0
= 1
2 ð¢0 â
1
4â ð¢0 +
1
8â2 ð¢0 ââ¯
Example 20 Sum the series 12, 22, 32,âŠ, ð2 using finite differences.
Solution: Let the series 12, 22, 32,âŠ, ð2 be represented by ð¢0, ð¢1 , ð¢2 ,âŠ, ð¢ðâ1
⎠ð = ð¢0 + ð¢1 + ð¢2 + â¯+ð¢ðâ1
â ð = ð¢0 + ðžð¢0 + ðž2ð¢0 + â¯+ðžðâ1ð¢0
= 1 + ðž + ðž2 + â¯+ ðžðâ1 ð¢0
=1âðžð
1âðžð¢0 =
ðžðâ1
ðžâ1ð¢0 âµ ðð = ð
1âðð
1âð
â ð =(1+â)ðâ1
(1+â)â1ð¢0
=1
â 1 + ðâ +
ð(ðâ1)
2!â2 +
ð ðâ1 (ðâ2)
3!â3 + ⯠â 1 ð¢0
= ðð¢0 +ð(ðâ1)
2!âð¢0 +
ð ðâ1 (ðâ2)
3!â2ð¢0 + â¯
Now ð¢0 = 12 = 1
âð¢0 = ð¢1 â ð¢0 = 22 â 12 = 3
â2ð¢0 = âð¢1 â âð¢0 = ð¢2 â 2ð¢1 + ð¢0 = 32 â 2( 22) + 12 = 2
â3ð¢0, â4ð¢0 ⊠are all zero as given series is an expression of degree 2
⎠ð = ð + ð(ðâ1)
2! 3 +
ð ðâ1 (ðâ2)
3! 2 + 0
= ð +3ð ðâ1
2+
ð ðâ1 ðâ2
3
=1
6 6ð + 9ð ð â 1 + 2ð ð â 1 ð â 2
=1
6ð 6 + 9ð â 9 + 2ð2 â 6ð + 4
=1
6ð 2ð2 + 3ð + 1 =
1
6ð ð + 1 (2ð + 1)
Example 21 Prove that ð¢0 + ð¢1ð¥ + ð¢2ð¥2 + ⯠=
ð¢0
1âð¥+
ð¥âð¢0
(1âð¥)2+
ð¥2âð¢02
(1âð¥)3+ â¯
and hence evaluate 1.2 + 2.3ð¥ + 3.4ð¥2 + 4.5ð¥3 + â¯
Solution: ð¢0 + ð¢1ð¥ + ð¢2ð¥2 + ⯠= ð¢0 + ð¥ðžð¢0 + ð¥2ðž
2ð¢0 + â¯
= 1 + ð¥ðž + ð¥2ðž2 + ⯠ð¢0
=1
1âð¥ðžð¢0 âµ ðâ =
ð
1âð
=1
1âð¥(1+â)ð¢0 =
1
(1âð¥)âð¥âð¢0
=1
1âð¥
1
1âð¥â
1âð¥ ð¢0
=1
1âð¥ 1 â
ð¥â
1âð¥ â1ð¢0
=1
1âð¥ 1 +
ð¥â
1âð¥ +
ð¥2â2
(1âð¥)2+ ⯠ð¢0
= ð¢0
1âð¥+
ð¥âð¢0
(1âð¥)2+
ð¥2âð¢02
(1âð¥)3+ ⯠= R.H.S.
Now to evaluate the series 1.2 + 2.3ð¥ + 3.4ð¥2 + 4.5ð¥3 + â¯
Let ð¢0 = 1.2 = 2, ð¢1 = 2.3 = 6, ð¢2 = 3.4 = 12, ð¢3 = 4.5 = 20,âŠ
Forming forward difference table to calculate the differences
ð â âð âð âð ðð = ð.ð = ð
4 ðð = ð.ð = ð 2
6 0 ðð = ð.ð = ðð 2 0
8 0 ðð = ð.ð = ðð 2
10 ðð = ð.ð = ðð
⎠1.2 + 2.3ð¥ + 3.4ð¥2 + 4.5ð¥3 + ⯠= ð¢0
1âð¥+
ð¥âð¢0
(1âð¥)2+
ð¥2âð¢02
(1âð¥)3+ â¯
=2
1âð¥+
4ð¥
(1âð¥)2+
2ð¥2
(1âð¥)3+ 0
=2
(1âð¥)3
Exercise 6A
1. Express ðŠ4 in terms of successive forward differences.
2. Prove that âðð3ð¥+5 = ð3 â 1 ðð3ð¥+5
3. Evaluate â2 5ð¥+12
ð¥2+5ð¥+6
4. If ð¢0 = 3, ð¢1 = 12, ð¢2 = 81, ð¢3 = 2000, ð¢4 = 100, calculate â4ð¢0.
5. Prove that ÎŒ =2+â
2 1+â= 1 +
1
4ÎŽ2
6. Find the missing value in the following table
ð¥ 0 5 10 15 20 25
ðŠ 6 10 - 17 - 31
7. Sum the series 13, 23, 33,âŠ, ð3 using finite differences.
Answers
1. ðŠ4 = ðŠ0 + 4âðŠ0 + 6â2ðŠ0 + 4â3ðŠ0 + â4ðŠ0
3. â3(ð¥2+9ð¥+15)
ð¥(ð¥+1)(ð¥+4)(ð¥+5)(ð¥+8)(ð¥+9)
4. â7459
6. 13.25, 22.5
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