Making in (8), first n=1 and then n=2, we find In this expression for the nth term of a recurring series, whose scale of relation is p and q, A1 and A2, are the first two terms of the series, a' and a" are the two roots of the quadratic EXAMPLES. 1. Suppose the scale of relation of the series 1+x+3x2+5x+11x+21x5+&c., to be p⇒x;q=2x2, what will be the nth term, and what will be the sum of n terms? Substituting x and 2x2 for p and q, in the equations (11), we find a'=2x and a"=-x. Now, substituting these values in (A), we obtain for the general term sought. Writing n-1 for n, we find these values of A, and A,-1, together with the above values of p, q, A1 and A2, being substituted in (C), Art. 205, give 1 S= 2(x+1)(2x)"+(1—2x)(−x)”—3 Which, when reduced to its simplest form, becomes 2. Find the nth term, and the sum of n terms of the recurring series 1+x+5x2+13x3 +41x2+121x5+&c., whose scale of relation is p=2x; q=3x2. 3. Find the nth term of the recurring series 1+2x+5x2+13x3 +34x2+89x5+&c., whose scale of relation is p=3x; q=-x2. xn-1 Ans. A 2" √5 xn. n-1. (1-√5)(3-√5)n-1 Referring to question 22, page 240, Higher Arithmetic, we see that if we call x=1, the above expression for A, will give the number of branches of the oak tree at the end of n years, thus the number of branches at the end of 20 years is (1+√5)(3+√5)19 (1−√5)(3−√5)19 220 √5 220 √5 ny CHAPTER VIII. LOGARITHMS. (210.) Logarithms are numbers, by the aid of which maarithmetical operations are greatly simplified. In the following relations : a=b a=e a2=d &c., (1)) (2) (A) (3) x, y, and z, are respectively the logarithms of b, c, and d. (211.) The assumed root a is called the base of the system of logarithms. (212.) If in (1), of equations (A), we make x=0, we shall have ao=b=1, (Art. 41,) for all values of a, therefore the logarithm of 1 is always 0. (213.) If in (1), we suppose the base to be negative, we shall have (-a)=b. If b is positive, then x must be even, if b is negative, then x must be odd; hence we can not represent all values of b by the expression (-a)". Therefore the base of every system of logarithms must be positive. log. log. (214.) If in (1), we suppose 1 to be negative, we shall have q*=-6. Now, since a is always positive, the expression a® is positive for all values of x either positive or negative. Therefore, the logarithm of a negative quantity is impossible. (215.) Each different base must produce a different system of logarithms; the logarithms in common use have 10 for their base. So that we have 10°=1; 101=10; 102=100; 103=1000; &c. Hence, we have log. 1=0 1 -1 log. 10=1 10 log. 100=2 1 -2 log. 1000=3 100 log. 10000=4 1 &c. 3 1000 1 : -4 10000 &c. (216.) If we take the product of equations (1) and (2) of group (A), we shall have a++y=bc, (4) from which we discover that, the logarithm of the product of two quantities is equal to the sum of the logarithms. And in general, the logarithm of a number consisting of any number of factors is equal to the sum of the logarithms of all its factors. (217.) It also follows from the above, that n. times the logarithm of any number is equal to the logarithm of its nth power. log. log. : |