937 INTERPOLATION. 



the preceding question on the supposition that the rates are 2, 3, 4, 5, 

 and 6 per cent., in which from 2 to 4 ia 2J intervals, or n = 2J. 



INTERPOLATION. 



933 



672971 

 553676 

 456387 



- 119295 



+ 22006 



97289 4215 



+ 17791 + 833 

 - 79498 3377 



376889 + 14414 



65084 

 311805 



-1 = 15 1 n 2 _ _5 

 ~ ~a~' o ' a i ft' 



-1 n 2 n 3 _ 



5 



128" 

 672971 



- x 119295 = 298238 

 2 



I! x + 22006 = + 41261 

 8 



X 



16 



ns x + 



4215 = 1317 

 838 = 33 



Answer 

 Correct Answer 



414644 

 414643 



The most simple interpolation is that which takes first differences 

 only into account, and ia perfectly well known to every one who can 

 use a table of logarithms, or interpose arithmetical means. 



We now give some instances in which the intermediate terms are 

 expressed by means, not of the differences of given terms, but of the 

 terms themselves. The case which most often occurs is that in which 

 it is required to bisect the interval of the tables, and in this case a rule 

 may be given which amounts to using the third differences, and is 

 extremely simple. Let p, q, r, and be successive terms of a table, 

 and let it be required to find the term intermediate between q and r, 

 that is, if q and r stand opposite to x and x + y, required the term 

 answering to x + Jy. The following formula shows the process : 



.. < . 



Thus in the preceding instance 



;> = 553676 q + r= 833276 

 q = 456387 p +-865481 

 r= 376889 4)- 32205 

 =311805 4)- 8051 

 - 2013 

 i(? + r) =16638 

 Answer 414625 



This more simple rule is equivalent to the use of the preceding 

 method with p and its first three differences. It requires that two 

 terms should lie on each side of the term sought; but if it were 

 required to bisect the interval between p and q by means of p, q, r, 

 and i, the formula is as follows : 



Thus, to find the value at 34 per cent, from the preceding data, we 

 have 



r-p=- 176787 



_ 6 



-883935 



f-q= -144582 



5 (r-p)-(- ? ) = 4)- 739353 



4)-184838 



46210 



q= 456387 



Answer 502597 



Correct Answer 502566 



The formula for the bisection of the interval of r and i by means of 

 p, q, r, and i, is 



16 



Where extreme accuracy of bisection is required, the following rule 



1 be equivalent to going as far as fifth differences, and taking from 



the tabie three terms on each side of the intermediate term required. 



Let the terms of the table be p, j, r, s, t, and u, the intermediate term 

 of r and s being required. 



P 

 u 



t+q=y 



Intermed. 



256 



As an instance, take from the article INTEREST the logarithms to ten 

 places of 1-01, 1-02, 1'03, 1'04, 1-05, and 1'06, for p, q, r, s, t, and u, 

 neglecting decimal points and preliminary ciphers. 



43213738 



253058653 



1=296272391 



298705640 



2433249=z-s 



3 



- 7299747 



86001718 

 211892991 

 y = 297894709 



1*= 



149352820 

 50678 



128372247 

 170333393 

 2=298705640 

 297894709 

 y= 4)81093100 

 20273275 

 7299747 

 MlL'i7;;.v.'S 

 8) 1621691 



Ana. -01 49403498 = log. 1-035 



4) 202711-4 

 50677-9 



The formula (B) is sufficient to bisect the intervals given in the 

 article cited. 



The following is the method by which the formula (A) may be 

 expressed in terms, not of p, A P, &c., but of p, q, r, &c. Suppose 

 this is to be done as far as the third difference, or in terms of p, q, r, 

 and . Assume for the function in question 



A (n-1) (-2) (n-3)+B(n-2) (n-3)+cn(n-l) (-3) + D n (-l) (-2). 

 When = 0, this should be p ; but it then becomes 6 A, whence A = 

 kP- When n=l,thia should bey ; but it then becomes 2 B, whence 

 B =J q. Similarly c = J r, and D = j s : or the function tabulated, 

 within the given limit, is, so far as third differences can determine it, 



2 n-S 



This method may be extended to the interpolation of intermediate 

 values, when the given values are not equally distant. Suppose that 

 according as n is a, i, or c, a function is p, q, or r. Assume for the 

 function 



A(i-6) ( c) +B(-O) (n c) +o(n o) (n 4). 



Then when n=o, we must have A (o 4) (o c)=p, or 



p q r 



~ U-4) (o-c) ' B = (4-o) (4-7)' 0= (c-o) (c-4)' 



The following results will serve as an instance of the application of 

 the last method but one. Suppose it required to interpose four equi- 

 distant values between q and r in the series p, q, r, a, using third 

 differences inclusive. First interpose four arithmetical means between 

 q and r, and let them be A, B, o, D : then interpose four arithmetical 

 means between 85 2p i, and 3 r 2 s p, and let these be A', 

 B', c', and D'. Then the four terms intermediate between q and r 

 must be 



A + sj,A',B + ifoB', C + ifoC', D 



If it be required to interpose three equidistant values between q and 

 r, using third differences, take p, q, r, s, and between q and r interpose 

 three arithmetical means, A, B, and c ; also between 3 y 2 p a and 

 3r 2 p interpose three means, A', B', and c'. Then the three 

 terms required are 



A+jfcA',B + &B', + ,fco'. 



To interpose two terms, still with third differences, find two arith- 

 metical means between 30 y 2p < and 30 r 2 s p ; the twenty- 

 seventh parts of these means are the intermediate terms required. 

 The interpolation of one term has already been given (B). 



When second differences only are used, no material simplification of 

 the fundamental rule can be given. To place it intermediate terms 

 between q and r by means of q, r, and t, interpose k arithmetical means, 

 and correct them as follows. Calculate (q + t 2r)-i-2 (& + !)*, and 

 call this A ; then subtract from the several means 



*A, 2(i !)A, 3(i 2) A, . . . (k 1)2 A, it A. 



It would be a little more correct to let A be (p + r q i) -f- 4 (k + 

 1)*, which in the case of a single intermediate term would amount to 

 using third differences. 



INTERPRETATION. (Mathematics.) This word is coming into 

 use as descriptive of a process which it has long been customary to 

 employ, though without any express name. Whea an algebraical 



