which are given, other similar numbers which shall correspond to intermediate values of the function. For example, suppose p, q, r, s, &c., to be a series of tabulated numbers corresponding to, and written opposite the functions a, a + b, a + 2b, a + 3b, &c., and it were required to find the tabulated number corresponding to the function a + 216. This is a question of interpolation, and is resolved by taking the successive differences of the tabulated numbers, thus : From the above equations, we have q=p+dp, r=q+dq=p+ dp + dq = p + 2dp + d2p; in which notation it should be observed, that d2, d3, &c., denote the second, third, &c. differences of the successive tabulated numbers. It is plain, that the above law from which the numerical coefficients for any term may be derived, is similar to that for the co-efficients of a binomial: hence, if T denote the n+1 term of the tabulated numbers, reckoning from p inclusive, we shall have Let it be required to find the tabulated number corresponding to a + 3b. We then have, n = 3: hence, T = p + 3dp+ 3d2p + d3p, the same value as that found above for r. Next, let it be required to find the tabulated value answering to the functions a + b. Then, n = 1, and if we know the tabulated number p, and the successive differences d, d2, &c., the approximate value of T can easily be found. It is plain from the series that the interpolated values are but approximations, since no order of difference can reduce to zero; and hence, the series will contain an infinite number of terms. Generally, however, the tabulated values are themselves but approximations, and the successive differences decrease so rapidly in value, that the series becomes very converging. Let us suppose for example, that we have the logarithms of 12, 13, 14, 15, &c., and that it is required to find the logarithm of 12 and a half. Then, INTEREST. 273. The solution of all questions relating to interest, may be greatly simplified by employing the algebraic formulas. In treating of this subject, we shall employ the following notation: Let p = the amount bearing interest, called the principal; r = the part of $1, which expresses its interest for one year, called the rate per cent.; t = the time that p draws interest; i = the interest of p dollars for t years; S=p+ the interest which accrues in the time t, which is called the amount. Simple Interest. To find the interest of a sum p for t years, at the rate r, and the amount then due. Sincer denotes the part of a dollar which expresses its interest for a single year, the interest of p dollars for the same time will be expressed by pr; and for t years it will be t times as nuch: hence, 1. What is the interest, and what the amount of $365 for three years and a half, at the rate of 4 per cent. per annum. Here, Present Value and Discount at Simple Interest. The present value of any sum S, due t years hence, is the principal p, which put at interest for the time t, will produce the amount S. The discount on any sum due t years hence, is the difference between that sum and the present value. To find the present value of a sum of dollars denoted by S, due t years hence, at simple interest, at the rate r; also, the discount. We have, from formula (2), S = p + ptr; and since p is the principal which in t years will produce the sum S, we have and for the discount, which we will denote by D, we have 1. Required the discount on $100, due 3 months hence, at the rate of 5 per cent. per annum. Compound interest is when the interest on a sum of money becoming due, and not paid, is added to the principal, and the interest then calculated on this amount as on a new principal. To find the amount of a sum p placed at interest for t years, compound interest being allowed annually at the rate r. At the end of one year the amount will be S=p+pr = p (1 + r). Since compound interest is allowed, this sum now becomes the principal, and hence at the end of the second year the amount will be S' = p (1 + r) + pr (1+r) = p (1 + r)2. Regard p (1+r)2 as a new principal; we have, at the end of the third year, S" = p (1 + r)2 + pr (1 + r)2= p (1 + r)3; and at the end of t years, S = p (1 + r)..... And from Article 260 we have log. S = log. p + t log. (1 + r); (5). and if any three of the four quantities S, p, t, and r, are given, the remaining one can be determined. Let it be required to find the time in which a sum p will double itself at compound interest, the rate being 4 per cent. per annum. We have from equation (5), The discount being the difference between the sum S and p, we have D=S-1)=(1-1) : |