CHAPTER XV
RELATION OF DIFFERENTIAL CALCULUS TO
FINITE DIFFERENCES
DIFFERENTIAL CALCULUS AND FINITE DIFFERENCES 89 Thus, taking Stirling's formula, we have f(x)=f(0)+x°f(0)+2Af(-1)+ of(-1)
+x(x'--1) 0 f(1)+A3f(-2)+....
3! 2 Differentiating with regard to x,
f (x) __Of(o)+Af(1)A3f(1)+O$f(2)
2 12 +
+ terms involving x and its powers.
Putting x = 0 we obtain
0 =AJ(0)+Af(1)A3f(1)+A3f(2)+... (4).
f, (0) 2 12
Taking the first term only, we have
f,(0)_A.f(0)+Of(1)f(1)f(1) (5),
2 2
a very useful approximation, which, by altering the origin and the unit of measurement, may be expressed more generally as
f, (x) f(x+h) _f(x-h)
If h = 1 we arrive at the approximate result
f' (x) _ A f (x 2) (7)
=A(1+0) lf(x)
=[0 A'+$A3 ifiA'+issA'...]f(x). Comparing this with the value given by formula (1) we see that the error involved in taking f ' (x) = A f (x 2) is
[ Q403+1',;04 4470A5+...]f(x),
which is approximately equal to } 03 f (x D.
Hence we may write
f'(x)=of(x-2)-2~4A'f(x-i) (8).
3. The introduction of the second term in equation (4) gives the useful approximation
f, (0) 8 {.f (1) -f(-1)} If (2) -f( 2)} (9).
12
The other formulas of central differences, when treated as above, give these and other expressions for f' (x). The above, however, are the most accurate and useful for general use.
90 DIFFERENTIAL CALCULUS AND FINITE DIFFERENCES
- Using the central difference formula, we shall obtain for the differential coefficients of higher orders,
f"(0)=0'f(1) 0'f(2) (10)andf,,. (0) = off(1) j 03f ( 2)
= Q3 f( i) approx (11).
- 5.An example of the working of the above formulas is now given.
Example. Find the first three differential coefficients of logx, when x = 2. having given the values of loge 1'80, loge 1.90, ....
| Number |
N log al |
A |
Aa |
|
|
p5 |
| 1.80 |
587787 |
054067 |
1)02774 |
000271 |
'000038 |
000007 |
|
1'90 |
'641854 |
051293 |
002503 |
000233 |
'000031 |
000005 |
| 2.00 |
'693147 |
'048790 |
002270 |
'000202 |
'000026 |
'000004 |
| 2.10 |
741937 |
046520 |
002068 |
'000176 |
'000022 |
000004 |
| 2.20 |
'788457 |
'044452 |
001892 |
000154 |
'000018 |
|
| 2.30 |
832909 |
042560 |
'001738 |
'000136 |
|
|
| 2.40 |
'875469 |
'040832 |
'001602 |
|
|
|
| 2.50 |
'916291 |
'039220 |
|
|
|
|
| 2.60 |
'955511 |
|
|
|
|
Using the advancing difference formula (1), starting at the term 2.00, and remembering that the interval of differencing is not unity but '1, we get
'00.2270 '000202 '000026 '000004 f' (x) = 10 (048790 + 2 + 3 + 4 + 5 )
= '50000.
Similarly
f" (x) = 100 ( '002270 000202 x 000026 x '000004) '2499,
f "' (x) = 1000 ('000202 + x '000026 +'-~ x '000004)
= '248.
The true values are, of course, given by the differential coefficients
of log x, which are respectively 1, 1 2 When x = 2 these x x3'
become ,}, 4 , i; the closeness of the above approximations is apparent.
Formulas (6) and (9) give respectively for f' (x) values of '50042 and '50000.