a =a+aq + aq3 + aq3 +aq1 + Now, by making, a=1, q=2, this equality becomes an equation of which the first member is negative whilst the second is positive and greater in proportion to the magnitude of q. To interpret this result, we will observe that, when in the a equation =a+aq+aq2 + aq3 + 1-q ---, we stop at a certain term of the series, it is necessary to complete the quotient in order that the equality may subsist. Thus, in stopping, for example, at the fourth term, aq3. It is necessary to add the fractional expression quotient, which gives rigorously, If in this exact equation we make a=1, q=2, it becomes - -1=1+2+4+8+ 16 -1 =1+2+4+8-16, which verifies itself. In general, when an expression involving x, designated by f(x), (which is called a function of x), is developed into a series of the form a+bx+cx2+dx3+. . . ., we have not rigorously f(x)=a+bx+cx2+dx3 +....., unless we conceive that (in stopping at a certain term in the second member) the series is completed by a certain expression involving a. When, in particular applications, the series is decreasing, the expression which serves to complete it may be conceived as small as we please, by prolonging the series; but the contrary is the case when the series is increasing, and it must not be neglected. This is the reason why increasing series cannot be used for approximating to the value of numbers. It is for this reason, also, that algebraists have called those series which go on diminishing from term to term, converging series, and those in which the terms go on increasing, diverging series. In the first, the greater the number of terms taken, the nearer the sum approximates numerically to the expression of which this series is the development; whilst in the others, the more terms we take, the more their sum differs from the numerical value of this expression. 206. The consideration of the five quantities a, q, n, l and S, which enter in the two formulas l=bq”—', S= lq-a 9-1 (Nos. 200 and 201), gives rise to ten particular problems, the enunciations of which do not differ from those relative to progressions by differences, (196), except that the letter r is replaced by q. We will then, as in progressions by differences, determine q and S, knowing a, l and n. tuting this value in the second formula, the value of S will be obtained. The expression q= following question, viz. furnishes the means for resolving the To find m mean proportionals between two given numbers a and b; that is, to find a number m of quantities, which will form with a and b, considered as extremes, a progression by quotients. For this purpose, it is only necessary to know the ratio; now the required number of terms being m, the total number of terms r is equal to m+2. Moreover, we have l=b, there fore the value of q becomes q m+ 1 a ; that is, we must divide one of the given numbers (b) by the other (a), then extr root of the quotient which is denoted by one more than the required number of terms. Thus, to insert six mean proportionals between the numbers 3 and 384, we make m=6, whence |= whence we deduce the progression 7 1384 7 = √128=2, 3:6:12: 24: 48: 96: 192: 384. We will hereafter explain the most expeditious means of m+i calculating numerically the number expressed by q=. When the same number of mean proportionals are inserted between all the terms of a progression by quotients, taken two and two, all the progressions thus formed will constitute a single progression. The demonstration is analogous to that of No. 197. 207. Of the ten principal problems that may be proposed upon progressions, four are susceptible of being easily resolved. The following are the enunciations, with the formulas relating to them. 1st. a, q, n, being given, to find l and S. 4th. q, n, S, being given to find a and l. S(q-1) Sq-1(q-1) a= Two other problems depend upon the resolution of equations of a degree superior to the second; they are those in which the unknown quantities are supposed to be a and q, or 1 and q. or, For, from the second formula we deduce a=lq-Sq+S ; Whence, by substituting this value of a in the first l=aq”—', l=(lq—Sq+S)q”—1, (S—l)q”—Sq11 +7=0. . an equation of the nth degree. In like manner, in determining and q, we would obtain the equation aq"-Sq+S—a=0. 208. Finally, the other four problems lead to the resolution of equations of a peculiar nature; they are those in which n and one of the other four quantities are unknown. From the second formula it is easy to obtain the value of one of the quantities a, q, l, and S, in functions of the other three; hence the problem is reduced to finding n by means of the formula l=aq"1. Now this equality becomes q"= lq an equation of the form a a2=b, a and b being known quantities. Equations of this kind are called exponential equations, to distinguish them from those previously considered, in which the unknown quantity is raised to a power denoted by a known number. § II. Of Exponential Quantities and Logarithms. 209. Resolution of the equation a=b. The object of the question is, to find the exponent of the power to which it is necessary to raise a given number a, in order to produce another given number b. We will first consider some particular cases. Suppose it is required to resolve the equation 2=64. By raising 2 to its different powers, we find that 2° =64; hence x=6 will satisfy the conditions of the question. Again, let there be the equation 3243. The solution is x=5. In fact, so long as the second member b is a perfect power of the given number a, x will be an entire number which may be obtained by raising a to its successive powers, commencing at the first. Suppose it is required to resolve the equation 26. By making x=2, and c=3, we find 22 =4 and 23=8; from which we perceive that x has a value comprised between 2 and 3. Suppose then, that x=2+ 1 (x' is then >1). x' Substituting this value in the proposed equation, it becomes To determine x', make successively x'=1, x'=2; we find therefore a' is comprised between 1 and 2. By substituting this value in the equation (3) The two hypotheses a′′=1 and 2"=2, give (4) -2. therefore x" is comprised between 1 and 2. ing successively x""=1, 2, 3, we find for the two last |