EXERCISES. 1. At what rate of interest will money double itself every ter years? Ans. 7.177. 2. At what rate will it treble itself every 15 years? Ans. 7.599. 3. A man having invested $1000, with all the interest it yielded him, for 25 years, finds that it amounts to $3386. What was the Ans. 5 per cent. rate of interest? 4. A life company issued to a man of 20 a paid-up policy for $10,000, the single premium charged being $3150. If he dies at the age of 60, at what rate must the company invest its money to make itself good? Ans. 2.93 per cent. 5. A man who can gain 4 per cent interest wants to invest such a sum that it shall amount to $5000 when his daughter, now 5 years old, attains the age of 20. How much must he invest? Ans. $2776.62. 6. How much must a man leave in order that it may amount to $1,000,000 in 500 years at 2 per cent interest? Ans. $4.361 7. How much if the time is 1000 years, the rate being still 2 per cent, and the amount $1,000,000? Ans. 0.0019 of a cent. 8. A man finds that his investment has increased fivefold in 25 What is the average rate of interest he has gained? years. Ans. 6.65. 9. An endowment of $7500 is payable to a man when he attains the age of 65. What is its value when he is 45, supposing the rate of interest to be 4 per cent? Ans. $3423. 13. Accumulation of an Annuity. It is often necessary to ascertain the present or future value of a series of equal annual payments. Thus it is very common to pay a constant annual premium for a policy of life insurance. The value of such a series of payments at any epoch is found by reducing the value of each one to the epoch, allowing for interest, and taking the sum. Supposing the epoch to be the present time, the problem may be stated as follows: A man agrees to pay p dollars a year for n years, the first payment being due in one year, and the total number of payments n. What is the present value of all n payments? rate of interest Putting, as before, p = 100 the present value of p dollars payable after y years will, by § 12, Case II., be Putting in succession, y = 1, y = 2, . . . y = n, the sum of the present values is By College Algebra, § 212, the sum of this progression will be p (1+p)n - 1 (1) - 1 years, If the first payment is to be made immediately, instead of at the end of a year, the last or nth payment will be due in n and the progression will be p (1 + p)2 Ρ We find the sum of the geometric progression to be Σ, = 1 (1 + p)n — (1 ́+p)n−1° EXERCISES. (2) 1. What is the present value of 15 annual payments of $85 each, of which the first is due in one year, the rate being 5 per cent? 2. The same thing being supposed, what would be the present value if the rate of interest were 4 per cent? Ans. $945.80. 3. What is the present value of 25 annual payments of $1000 Ans. $17,935. each, the first due immediately, if the rate of interest is 3 per cent? 4. A debtor owing $10,000 wishes to pay it in 10 equal annual instalments, the first being payable immediately. If the rate of interest is 6 per cent, how much should each payment be? Ans. $1281.76. NOTE. This problem is the reverse of the given one, since, in the equation (2), we have given Z, 10000, p = 0.06, and n = 10, to find p. 5. The same thing being supposed, what should be the annual payment in case the payments should begin in a year? Ans. $1358.69. Perpetual annuities. If the rate of interest were zero, the present value of an infinity of future payments would be infinite. But with any rate of interest, however small, they are finite. For if, will converge in the first equation (1), we suppose n infinite, toward zero, and we shall have p + p Σ= n (3) This result admits of being put into a concise form, thus: Since is the present value of the perpetual annuity p, the annual interest on this value will be pz. But the equation (3) gives The present value of a perpetual annuity is the sum of which the annuity is the annual interest. Example. If the rate of interest were 3 per cent, the present value of a perpetual annuity of $70 would be $2000. EXERCISES. 1. A government owing a perpetual annuity of $1000 wishes to pay it off by 10 equal annual payments. If the rate of interest is 4 per cent, what should be the amount of each payment? Ans. $3082.30. 2. A government bond of $100 is due in 15 years with interest at 6 per cent. The market rate of interest having meanwhile fallen to 3 per cent, what should be the value of the bond? NOTE. We find, separately, the present value of the 15 annual instalments TABLE II. MATHEMATICAL CONSTANTS. 14. In this table is given a collection of constant quantities which frequently occur in computation, with their logarithms. The logarithms are given to more than five decimals, in order to be useful when greater accuracy is required. When used in fiveplace computations, the figures following the fifth decimal are to be dropped, and the fifth decimal is to be increased by unity in case the figure next following is 5 or any greater one. TABLES III. AND IV. LOGARITHMS OF TRIGONOMETRIC FUNCTIONS. 15. By means of these tables the logarithms of the six trigonometric functions of any angle may be found. The logarithm of the function instead of the function itself is given, because the latter is nearly always used as a factor. We begin by explaining Table IV., because Table III. is used only in some special cases where Table IV. is not convenient. I. Angles less than 45°. If the angle of which a function is sought is less than 45°, we seek the number of degrees at the top of the table and the minutes in the left-hand column. Then in the line opposite these minutes we find successively the sine, the tangent, the cotangent, and the cosine of the angle, as given at the heading of the page. Example. log sin 31° 27′ = 9.717 47; log tan 31° 27′ = 9.786 47; log cotan 31° 27′ = 0.213 53; cos 31° 27′ = 9.931 00. The sine, tangent, and cosine of this angle being all less than unity, the true mantissæ of the logarithm are negative; they are therefore increased by 10, on the system already explained. If the secant or cosecant of an angle is required, it can be found by taking the arithmetical complement of the cosine or sine. It is shown in trigonometry that secant = 1 cosine' After each column, upon intermediate lines, is given the differ |