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The Summation of Series
By Harold T. Davis Dover Publications, Inc.
Copyright © 2015 Dover Publications, Inc.
All rights reserved.
ISBN: 978-0-486-80134-6
CHAPTER 1
THE CALCULUS OF FINITE DIFFERENCES
1. Finite Differences. The demands not only of statistics, but of physical science as well, have made desirable some knowledge of what is called the calculus of finite differences to distinguish it from the infinitesimal calculus. The calculus of finite differences can be divided into two parts, one concerned with differences, called the difference calculus, and the other concerned with summations, called the summation calculus. These resemble in many details the differential and integral calculus.
Since the elements of the calculus of finite differences are readily understood by any one who has mastered the principles of the infinitesimal calculus, it will be possible to develop readily many of the important features of this very useful discipline.
By the difference of a function f(x), designated by Δf(x), or simply Δf, we mean the expression
Δf(x) = f(x + d) - f(x),
where d is the difference interval.
That the differential calculus is in a sense the limiting form of the difference calculus is at once observed from the following limit:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
In most of the discussion which follows it will simplify the analysis to assume that the difference interval is 1, that is to say, we shall concern ourselves with the special difference defined by
Δf(x) = f(x + 1) - f(x).
Examples of differences are given by the following:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
The difference calculus can now be developed exactly as the differential calculus was developed by obtaining a set of fundamental formulas. Some of these are obtained by inspection. Thus, denoting by u, v, and w given functions of x, we obtain at once the following:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
2. The Factorial Symbols. If we form the differences of x2 and x2 we obtain 2x + 1 and 3x2 + 3x + 1 respectively, that is, x2 = 2x + 1 and Δx2 = 3x2 + 3x + 1. These formulas do not have the same simplicity as the corresponding formulas in the differential calculus where d/dx x2 = 2x and d/dx x3 = 3x2. But this same simplicity can be restored to the symbols of the difference calculus if xn is replaced by what is called the nth factorialx, represented by the symbol x(n) and defined as follows :
x(n) = x(x - 1)(x - 2) ··· (x - n + 1), n > 0.
Thus we have,
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],
and in general,
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
If the factorial index is a negative number, then the nth reciprocal factorialx, represented by the symbol x(-n), is denned as follows:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
For n = 2 and 3, we get
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII];
and in general,
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
Example 1. Find the difference of
u = 2x(2) + 3x(4) - 5x(-3)
Solution:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
Example 2. Show that x4 = x; + 7x(2) + 6x(3) + x(4) and from this identity compute the difference of x4.
Solution: Since both sides of the given equation are polynomials of fourth degree they can be shown to be identical if they coincide for five different values of x. For x = 0 and 1 the agreement is obvious. For x = 2, we get 16 = 2 + 7·2 = 16; for x = 3, 81 = 3 + 7·6 + 6·6 = 81; and for x = 4, 256 = 4 + 7·4·3 + 6·4·3·2 + 4·3·2·1 = 4 + 84 + 144 + 24 = 256. The identity is thus established.
The difference of x4 is thus found to be
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
PROBLEMS.
Find the differences of the following functions :
1. x2
Ans. 2x + 1.
2. 1/x2.
3. x + 1/x + 2
Ans. 1/[(x + 2)(x + 3)]
4. x(2) - x(-2)
5. x + [1/x].
Ans. [x2 + x – 1]/x(x + 1)
6. 1/[x(x + 1)].
7. 1/[x2 - 1].
Ans. [-(1 + 2x)]/[(x - 1) x(x + 1) (x + 2)]
8. Express x3 as the sum of factorials.
Ans. x3 = x(3) + 3x(2) + x.
9. Use the result of Problem 8 to evaluate Δx3.
Ans. 3x(2) + 6x + 1.
10. Show that Δ x(2)/[x + 1] = [x(x + 3)]/[(x + 1)(x + 2)]
3. Table of Differences and Its Application. For convenience of application the following table of differences has been formed. This table is analogous to the one given in most works on the differential calculus.
TABLE OF DIFFERENCES.
(1)Δc = 0, where c is a constant.
(2)Δ(cu) = cΔu.
(3)Δx = 1.
(4)Δ(u + v + w) = Δu + Δv + Δw.
(5)Δuxvx = vx+1Δux + uxΔvx = ux+1Δux + uxΔux.
(6) Δ(ux/vx) = [vxΔux - uxΔvx]/vxvx+1
(7)Δx(n) = nx(n-1).
(8)Δx(-n) = -nx(-n-1).
(9)Δ2x = 2x.
(10)Δax = (a - 1)ax.
(11)Δsin (ax + b) = 2 sin 1/2a cos (ax + b + 1/2a).
(12)Δcos (ax + b) = -2 sin 1/2a sin (ax + b + 1/2a).
(13)Δtan (ax + b) = sin a sec (ax + b) sec (ax + a + b).
(14)Δcot (ax + b) = -sin a csc (ax + b) csc (ax + a + b).
(15) Δcsc (ax + b) = [-2 sin 1/2a cos (ax + b + 1/2a)]/ [sin (ax + b) sin (ax + a + b)]
(16) Δcsc (ax + b) = [-2 sin 1/2a cos (ax + b + 1/2a)]/ [cos (ax + b) cos (ax + a + b)]
Formulas (1) through (4) were discussed in Section 1 and formulas (7) and (8) were derived in Section 2. The others are similarly obtained as shown below.
The difference of the product of two functions as given in formula (5) is derived as follows:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
Since ux and vx can be interchanged in this formula, it is clear that we can also write the difference as follows:
Δuxvx = ux+1Δvx + vxΔux.
The difference of the quotient of two functions is similarly obtained. Thus to establish formula (5) we write
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
The function 2x plays the same role in the difference calculus that ex does in the differential calculus, for we see from formula (9) that 2x is unchanged by differencing just as ex is unchanged by taking its derivative. Thus we have,
Δ2x = 2x+1 - 2x = (2 -1) 2x = 2x.
This is a special case of formula (10) which is readily established as follows:
Δax = ax+1 - ax = (a - 1) ax.
Formula (11) is derived from the following analysis:
Δsin (ax + b) = sin (ax + a + b) - sin (ax + b).
Referring to tables of trigonometric formulas, we see that the difference of two sines can be written,
sin θ - sin φ = 2 sin 1/2 (θ - φ) cos 1/2(θ + φ).
Applying this formula to the difference Δsin (ax + b), we get
Δsin (ax + b) = 2 sin 1/2a cos (ax + b + 1/2a).
Formula (12) is at once derived from this result if we observe that sin (ax + b + 1/2π) = cos (ax + b). If, therefore, in formula (11) b is replaced by b + 1/2π, formula (12) is obtained.
The derivation of formula (13) comes from an application of (6), since we can write
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
Formula (14) is obtained from this result by replacing b by b + 1/2π and observing that cot (ax + b) = -tan (ax + b + 1/2π).
To obtain formula (15) we write
csc (ax + b) = 1/sin (ax + b),
and apply (6). Formula (16) then follows by replacing b by b + 1/2π and observing that sec (ax + b) = csc (ax + b + 1/2π).
The following examples will illustrate the application of these formulas:
Example 1. Compute the difference of the function
u = [x(x - 1) (x - 2)]/[(x + 1) (x + 2)]
Solution: Since we can write the function in the form u = x(3)x(-2), we apply formula (5) and thus obtain,
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
Example 2. Show that Δu2z = (uz+1 + uz) Δuz.
Solution: Writing u2z as the product ux ux we apply formula (5) as follows:
Δu2x = Δux ux = ux+1 Δux + ux Δux = (ux+1 + ux) Δux.
Example 3. Evaluate Δsin2 4x.
Solution: Making use of the result of Example 2, and then applying formula (11), we have
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
The same result could be obtained also by writing sin2 4x = 1/2 - 1/2 cos 8x and applying formula (12). We thus get
Δsin2 4x = Δ (1/2 - 1/2 cos 8x) = sin 4 sin (8x + 4)
PROBLEMS.
Compute the differences of the following functions:
1. 2x(-2) + 3x(2).
Ans. -4x(-3) + 6x.
2. 1/5 x(5) + 1/x + 1 - 1/3x(-3)
3. x cos πx.
Ans. -(1 + 2x) cos πx.
4. x sin πx.
5. x(2) sin πx.
Ans. -2x2 sin πx.
6. x(2) cos πx.
7. 2x x(2).
Ans. 2x [x(2) + 4x].
8. 3x x(3).
9. sin (πx + 5).
Ans. -2 sin (πx + 5).
10. sin (2πx + 5).
11. 4x + 4-x.
Ans. 3(4x - 4-x-1). 12. 2x - 2-x.
13. 2x cos πx.
Ans. - 3 · 2x cos πx.
14. 2x sin πx.
15. 1/sin πx
Ans. -2 csc πx.
16. sec πx.
17. cos πx/sin πx
Ans. 0.
18. sin x cos x.
19. cos 1/2πx/sin 1/2 πx
Ans. -2 csc πx.
20. [(x - 1) (x - 2)]/[(x + 2) (x + 3)].
21. cos2 4x.
Ans. -sin 4 sin (8x + 4).
22. tan2 2x.
23. Show that Δ(sin 2πx + sin 4πx + sin 6πx + · · + sin 2nπx) = 0.
24. Prove that Δu3x (u2x+1 + ux ux+1 + u2x) Δux.
25. Use the formula in Problem 24 to compute Δsin3 πx.
4. Differences of Higher Order. The quantity Δnf(x) is called the nth difference of f(x), or the difference of order n. It is defined as the difference of Δn-1f(x), that is to say, by the equation
Δnf(x) = Δ[Δn-1f(x)].
Thus we have
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],
and in general,
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
It is convenient in connection with differences of higher order to introduce the new symbol: E f(x), defined by the equation
E f(x) = (1 + Δ) f(x) = f(x) + Δf(x) = f(x + 1).
We then have
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],
and, in general,
Enf(x) = (1 + Δ)nf(x) = f(x + n).
The foregoing formulas can be extended without difficulty to the difference Δf(x) = f(x + d) - f(x). We thus have
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (1)
and the companion formula,
Enf(x) = (1 + Δ)n f(x) = f(x + nd). (2)
5. The Gregory-Newton Interpolation Formula. Let us now generalize formula (2) of the preceding section by replacing n by p, where p is assumed to have any positive or negative value. We can then write
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],
that is to say,
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1)
This famous expression is the so-called Gregory-Newton interpolation formula, first announced by James Gregory (1638-1675) in 1670 and later published by Newton in his Principia Mathematica (1687). This formula assumes the same role in the difference calculus as that assumed by Taylor's theorem in the differential calculus.
Introducing factorial polynomials, the analogous character of the two formulas is immediately observed, for we have
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (2)
in the case of the Gregory-Newton formula, and
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (3)
where D, D2, D3, etc. denote derivatives of first, second, third, etc. orders, in the case of Taylor's series.
(Continues...)
Excerpted from The Summation of Series by Harold T. Davis. Copyright © 2015 Dover Publications, Inc.. Excerpted by permission of Dover Publications, Inc..
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