CHAPTER III
FINITE DIFFERENCES. GENERAL FORMULAS
AND SPECIAL CASES
- Starting from the relationship proved in Chapter II, 7, it is now possible to develop two formulas of the utmost importance.
- 2.To express f (x + mh) in terms of f (x) and its leading differences. By definitionf (x + nth) = En' f (x)
= (1 + A)inf (x),since by Chapter II, g 7,E = 1 + A.Expanding the above expression by the Binomial Theorem we have
f(x+nth) =[1+m0+(2)A'+(3)A'+...+(m)0'"]f(x)
=.f(x)+mAf(x)+(2)A'f(x)+(3)A'f(x)+... + 7n) Am f (x) (1).
Bearing in mind that the symbol A obeys the ordinary algebraic laws and that the Binomial Theorem holds for all values of the index, positive, fractional or negative, it will be realised that the above proof is perfectly general.
It is instructive, however, to show how the formula may be deduced for fractional and for negative values of m.
- Fractional value. Let m be a positive proper fraction, and let the interval between the given terms of the series be h. It is required to find the value off (x+nn h).
Also let f(x+h)—f(x)=Af(x),
and f (x+n) —f (x)=f(x).
14 FINITE DIFFERENCES
In the second case, the unit of differencing has been altered to
h and, bearing in mind that n is a positive integer, we may write n
at once from the theorem in the preceding article f(x+h)=(1+S)"`f(x),
therefore (1 + 0) f (x) = (1 + S)"f (x),
and (1 + A)n f (x) = (1 + S) f (x). Since m is also an integer, it follows that (1+S),nf(x)= f (x+mh) =(1+0)"f (x)
=[1+11 0+(n)0'+...~.f(x) n 2
.f(x)+n Af(x)+(2)A'f(x)+....
- 4.Negative value. It is desired to find the value of f (x — mh). Now, from the preceding theorems, it is clear that
(1+0)mf(x—nth) =f(x),
therefore.f(x—mh,)= (1 +0)-"Z f(x)=[1+(—m)0+( 2) 0'+...].f(x) =f(x)+(—m,)Af(x)+( 2)A2f(x)+....
The above proofs illustrate how the principle of Separation of Symbols can be applied whenever the symbols of operation obey the ordinary laws of algebra.
5.To express Am f (x) in terms of f (x) and its successive values.
Amf (x) = (E - f (x)= [Em — m Em—t + (m) Em' — ... + (—1)m] .f (x)=f (x+mh)—mf(x+m—1h)+(2) f(x+m—2h)—...+(—1)m f (x)(2).Alternatively both the above formulas can be easily proved by the ordinary methods of induction.
Formula (1) also follows directly from the ordinary formula of Divided Differences (see Chapter VIII). This method has the advantage of showing directly the application of formula (1) to cases where m has a fractional or a negative value.
GENERAL FORMULAS AND SPECIAL CASES 15
- The above formulas are expressed in a form which applies in the most general way, i.e. when the interval of differencing is h and the leading term is f (x). It is clear, however, that by altering the unit of measurement the formula will be simplified although the result is not affected. Similarly by changing the leading term (which process corresponds to shifting the "origin ") so that the leading term is expressed as f (0) a further simplification in form is made.
If, therefore, the interval of differencing becomes unity and the leading term can be represented by f (0), the first formula can be written.f (n) = f (0) + nAf (0) + (2) A'f (0) + (3).An example will make this clear.Having given the values of.f(10), f(15),.f(20), etc. it is desired to express f (17) in terms of f (10) and its leading differences.The original formula (1) gives the value of f (10 + 1'4h), where h = 5, and therefore we writef(10+ 1.4h)=f(10)+1.4Af(10)+(1-42(4)A'f(10)+...,where 0, 0', ... are taken over the interval h.But the same result is secured if the unit of measurement is changed from 1 to 5 and if at the same time f(10) is made the initial term of the series, for then f (10), f (15), f (20), ... can be written as F(0), F(1), F(2), ... and the required value, viz. f (17), becomes F(1.4) which by formula (3) is equal toF(0)+1.4AF(0)+(142(4)°'F(0)+....
- The above formulas are of general application if sufficient terms of the series are known, but it is convenient at this stage to consider the particular forms taken by the differences of certain special functions.
- f(x)=axn, A (x)=a(x+1)n—axn=a[7(e-1+(;)xn—'+...+1].
The result of differencing has been, therefore, to change the term involving the highest power of x from axn to anxn_1(thus reducing its degree in x).
16 FINITE DIFFERENCES
Similarly a further process of differencing will reduce the degree of x to n — 2 and the coefficient of the highest power of x will be an (n — 1). By repeating the process we arrive at the result that the nth difference of ax'" is independent of x and is equal to a. n!. The (n + 1)th difference is therefore zero.
Corollary. It follows that the nth difference of
ax"+bxn`1+cxn—'+ ... +k is constant and equal to a .n!.
- 9.f(x)=x(x— 1)(x—2)... (x—m+1).
This expression is usually denoted by x('m).Af (x) = (x+1) x (x—1)... (x — m+ 2) — x (x—1) (x—2)... (x—m+l) mx(x—1)...(x—m+2)= mx(m-1).Similarly02 f (x) = m (m — 1) x(m—2).By repeating the process we /arrive at the result
A f(x)=m!,
which is otherwise obvious from the preceding article since f (x) is of the mth degree in x.1
- 10.f(x)=x(x+1)(x+2)...(x+ne—1).
Corresponding to the notation already used, this can be denotedby x(-m).1A f (x) —1(x + 1)(x + 2) ... (x + m) x(x+1)(x-}2)...(x+m—1)—m x(x+1)(x+2)...(x+m) = - mx(-m+1).Similarly02f (x) = m (m + 1) x(-m+4 and so on.
- 11.f (x) = ax,
Af (x) = ax+1— a2
=ax(a—1). Whence Onf (x) = ax (a — 1)".
GENERAL FORMULAS AND SPECIAL CASES 17
- For many purposes it is convenient to have a table of the leading differences of the powers of the natural numbers. These can be represented as the differences of [x"']x=0 and are sometimes known as the "Differences of 0."
The following table gives a number of values of the first term and leading differences of [Omx'1]x_a, which, for convenience, can be written as AmOn:
| n |
f (0) |
0 |
A2 |
A3 |
A4 |
p5 |
pu |
| 1 |
0 |
1 |
0 |
— |
— |
I— |
— |
| 2 |
0 |
1 |
2 |
0 |
— |
— |
— |
| 3 |
0 |
1 |
6 |
6 |
0 |
— |
— |
| 4 |
0 |
1 |
14 |
36 |
24 |
0 |
— |
| 5 |
0 |
1 |
30 |
150 |
240 |
120 |
0 |
- A working formula for constructing a table such as the above by a continuous process may be obtained as follows:
Onf(x)=f(x+n)—nf(x+n—1)+(2n)f(x+n—2)—.... [See formula (2).]Therefore, when f (x) = xm,Anxm=(x+n)m—n(x+n—1)m+(x+n—2)m..., whence, putting x = 0,0"0m=nm—n(n—1)m+(2)(n—2)n''—... =n[nm-1—(n—1)(n—1)m-1+(n21)(n—2)m-1—...] =n [(1+n—1)m-1—(n—1)(1+n—2)m-1n l] = n.. (4).Butf(1)= f(0)+f(0),andAn-if(l) = An-if(0) + Mn f (0). Therefore[On—1x9n—11x=1 =+ [Anxm_1]x_c= An—1 O n—1 + An 0m-1.H. T. B. I.2
18 FINITE DIFFERENCES
Hence An Om = n [An—1Om—i + Om-'] (5).
It follows that the differences of [xm]y_0 can be constructed from
those of [xm-']x=o, and so on.
To take an example from the table given above, 0405 4 [A30' + 0404] = 4 [36 + 24] = 240.
14. By using the result given in § 9, it is possible to expand
f (x) in terms of x(0), x('), x(2), ... .
Let f (x) = A. + A,x(1) + A,x(2) + A,x(o + ... .
Then, putting x = 0, we see that
f (0) = Ao.
Differencing both sides of the equation, we get
Of (x) = A, + 2A,x(') + 3A,x(2) + ... .
Again putting x = 0, we find
Af(0)=A,.
By repeating the above processes, we obtain successively A2f(0)2!A„ 03f (0)=3!A3, ... A"f (0)=n,An, whence
A'f(0) p"f(0)
A0=f(0), A'=Of(0), A,= A"=
and
f (x) =f (0) + x(') A f(0) + 2 : o f (0) + ... + x(n)
Anf (0) + ... (6).