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The value of x is evidently between 3 and 4, since 33=27 and 4=256; hence, taking the logarithms of both sides of the equation, we have
x log. r= log. 100=2.* First, let xi= 3.5; then
Second, let x,= 3.6; then 3.5 log. 3.5= 1.9042380
3.6 log. 3•6= 2.0026890 true no. = 2:0000000
true no.= 2.0000000 error = –.0957620
error = +.0026890 Then, as the difference of the results is to the difference of the assumed numbers, so is the least error to a correction of the assumed number corresponding to the least error ; that is,
•098451 ::1:: •002689 : •00273; hence r=3.6—.00273=3.59727, nearly.
Again, by forming the value of x* for x=3.5972, we find the error to be -.0000841, and for r=3.5973, the error is +.0000149;
hence, as .000099 : .0001 ::.0000149 : .0000151 ; therefore, x=3.5973–.0000151=3.5972849, the value nearly.
EXAMPLES FOR PRACTICE. (1) Find x from the equation r=5.
Ans. 2.129372. (2) Solve the equation r=123456789.
Ans. 8.6400268. (3) Find x from the equation x=2000.
Ans. 4.8278226. DEMONSTRATION OF THE BINOMIAL THEOREM.
CASE 1. 326. If, at Prop. V., Art. 245, we suppose the second terms an, Az, az, &c., of the binomials to be all positive instead of negative, and all equal to a, then the products two and two will all become a? ; those three and three, a?, and so on; and, by recurring to Art. 203, we perceive that the number of combinations or products two and two, if we suppose that there are n binomials, will . n(n-1)
n(n-1)(n-2) be expressed by 10, the number three and three by
1.2.3 80'on. Hence, where n is a whole number,
(r+a)"=r" + naro-1+" Ja*xn-, &c., +a",
will be a
Applying the rule at Art. 113 for extracting the root of a polynomial, the first term of the root will be a ñ; the divisor of the second term of the given polynomial, n(a) =na" 7; and the quotient or second term of the root mm--(m-m) mm
Pr=max. When the two terms of the root thus found are raised to the nth power, and subtracted from the given polynomial according to the rule, the first two terms of the latter will be canceled, and the next highest power of a to be divided by the constant divisor na" 7 will be a"multiplied by z?, and the quotient, which is the third term of the root, will contain a to the power n–2–
"-2 with z?, and so on, so that the root may be written under the form
añ+ -+r+ A'zā *r+B'aä *2+, &c., the same form, so far as regards the exponents, as when the exponent is a whole number. The coefficients will be examined for this and the next case together.
CASE II. When the exponent is negative, either entire or fractional, as a consequence of what has just been demonstrated, we have
1 latu) (a + x)" a" + man-r+, &c. But if the division be effected according to the ordinary rules, the quotient will be indefinite, and of the form
a-m-ma-m-'r+A"a--mor? +, &c.; then, whatever be the exponent of a binomial, its development, as to the coefficients of the first two terms and the exponents of all, is of the same form, viz., that indicated by equation (1).
Now, to examine the coefficients of the other terms, for the sake of generality, I shall consider two consecutive terms of any rank whatever, and I shall write
(a +x)"="+mam-ır... + Mam-ox" + Nam-h-12.+1+, &c.
tain neither a nor z, the above expression becomes
(a +r+y)"=a" + mam-1.+y)....
.:+Mam-"(+y)" + Nam-n-Hr+y)"++, &c. By changing a into a +y, we should have found
.... +M(a+y)"="x" + N(a+y)--txutit, &c. In the two preceding equalities the first members are equal, therefore the second members must be equal also ; and this is the case whatever values z and y may have. Then, if they be arranged according to the powers of y, they must be identical. It is true, they contain binomials, but we know the first two terms of each of these binomials, so that we can form the part which, in each second member, contains y to the first degree, and that will suffice for our purpose. Designating it by Yy in the one and by Y'y in the other, it is easy to find
Y=mam-1.... + Mnam-2-1 + N(n+1)am-r-x"....
Y'=ma"-1.... + M(m-nam-n-12" + N(m-1-1)a".--22271..... These two quantities must be equal, whatever be the value of r; the coefficients, therefore, of the same powers of x must be equal. Considering only those which pertain to am---"x", we have
• M(m-n) N(n+1)=M(m-n).. N=
n+1 We see by this according to what law, in the development (1), any coefficient whatever is formed from the preceding. It is the same that we have found for the case of a positive exponent (Art. 107, IV.); and as we have seen that the first two terms are composed in the same manner, whatever be the exponent m, it will be so also with all the other terms.
An abbreviate notation, sometimes employed to express the coefficients of the binomial formula, is the initial letter B of the word binomial, with the exponent of the power of the binomial before it, and the order of the coefficient above. Thus, the coefficient of the 1° term, if the exponent be n, is expressed by B ; of the 20, oB; of the 3o, B, &c.; of the kib term m(n-1)...(n=k+1).
? by B, or otherwise simply nk. 1.2.3...k
327. To develop the expression in a series, place
atbr=A+BX+Cr2+, &c., and proceeding by the method of undetermined coefficients, explained at Art. 209, we find
A= BEA, C==}B, D=-c, &e. From which we perceive that each coefficient is obtained by multiplying the preceding by - Here the series is a simple geometrical progression. Proceeding in a similar manner with the fraction
a' + b.r
Ihrer=A+Br+Cr2+, &c., we obtain
A=%, B=LZA, CE-A-B, D=-B-C, &c. Here each coefficient from the 3o is the sum of the two preceding, multiplied respectively by --and -, or each term is the sum of the two pre
ceding multiplied by - and ca:
Again, in the development of
at brtcr: +dx=A+Br+, &c., each term will be composed of the three preceding, multiplied respectively by
dx3 cx? bx
a' to'rtc'r? ... th'rm-1
a +bx+cx?...+kx" produces a series, each term of which from the (m+1)th is composed of the
b m preceding, multiplied respectively by — 72", -22. , ..
Series of this form are called recurrent, and the assemblage of quantities by which it is necessary to multiply several consecutive terms to obtain the following term, is called the scale of relation of the terms.
328. PROBLEM.--A recurring series being given, to return to the generating fraction.
In this enunciation it is supposed that the recurring series is arranged with respect to an indeterminate x. Let
S=A+BX+Cx?+... be such a series, having for a scale of relation (pro, qx?, rx]. Since this scale contains three terms, the generating fraction is of the form
a+bx+cx+d.2.3 If this fraction had been given, we have seen that the scale of relation would
a b c
For example, let S=1-22-2° — 523 +424 — ... be a recurring series, whose scale of relation is [+2, +4r”, — 2x]. Taking the above formula, we shall have
A=1, B=-2, с=-1, p=1, q=4, r=-2, and we shall find
i +2.—4r? — 233 329. PROBLEM.—A series being given, to determine whether it be recurring, and, in this case, to return to the generating fraction. Let the given series be
and place S=10. From this equation we derive
1 a+bx a b
g=a atami the quotient, therefore, of (1), divided by the series, ought to be exact, and of the form p+qr. Then the generating fraction will be expressed thus :
p+qr If the division does not stop at the second term this series will not be recurring, or else it will arise from a more complicated fraction.
a'+b'r Place S=
Tatortekli we shall have
g= a't b'r =p+qx+a'th'ri that is to say, dividing (1) by the series S, if we stop the division after we have obtained as a quotient terms of the form p+qr, the series Siro, which is the remainder that we then bave, and which is always divisible by zo, will be
S a' such that, after we have removed this factor, we must have == Hence we derive
S a' to'r
Si= a =Pi +913; that is to say, the new division ought to terminate at the second term in the quotient; and then, to find the generating fraction, we shall have the two equations
g=p+ t", 5.=pit917, whence
Consequently, the generating fraction will be