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779. Let there be a geometrical progression of seven terms, whereof the first =3 and the exponent =2. Then a =3, b = 2, and n=7. The first term will = 3 x 26, or 3 x 64 = 192, and the progression will be

3, 6, 12, 24, 48, 96, 192.

Multiplying the last term 192 by the exponent 2 we have 384; subtracting the first term the remainder is 381; and dividing this by b - 1 or by 1, we have 381 for the sum of the whole progression.

780. When the exponent is less than 1, and the terms of the progression consequently diminish, the sum of such a decreasing progression, which would go on to infinity, may nevertheless be accurately expressed. Thus, let the first term = 1, the exponent j, and the

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subtracting the preceding progression the remainder is s =2 for the sum of the proposed infinite progression. In general, suppose the first term = a and the exponent of the pro

gression =# (a fraction less than 1), and consequently c is greater than b, the sum of the

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subtracting, the remainder is (1 –#)--a Hence s = I#. Multiplying both terms of the

fraction by c we have s = ..", The sum, therefore, of the progression is found by dividing the first term a by 1 minus the exponent, or by multiplying a by the denominator of the exponent, and dividing the product by the same denominator, diminished by the numerator of the exponent.

781. So are found the sums of progressions whose terms are alternately affected by the signs + and —. For example:

2 4 s=a-'+'-' : -, &c.

Multiplying by # we have

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dC c-i-b" and c=5, we shall have for the sum of the progression #####, + ###, &c. = 1 : for by subtracting the exponent from 1 there remains #; and by ăivil: and dividing the first term by that remainder the quotient is 1. 782. Suppose the terms were alternately positive and negative, thus –

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In the infinite progression, # +15+15+1 out Túw &c. the first term is il, and the exponent '. Subtract this last from 1, and the remainder is £. If we divide the first term by this fraction the quotient is for the sum of the progression; so that by taking only one term of the progression, namely, i, the error is only 1, But taking two terms, it + ISO =#, there would still be T'u wanting to make the sum = }. We shall conclude with another example in the infinite progression: –

9++, +1%, H Tân i Tu'un, &c. Here the first term is 9 and the exponent 15. Then 1 minus the exponent =# and # = 10, the sum required. This series is thus expressed by a decimal fraction 99999999, &c.

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INFINITE DECIMAL FRACTions.

783. It has already been seen that decimal fractions are used in logarithmic calculations, in which vulgar fractions would be useless and cumbersome. In other calculations they are of such importance that we shall here dwell upon them, and show how to transform a vulgar fraction into a decimal, and the converse. 784. Generally let it be required to change the fraction # into a decimal fraction. Now, as this fraction expresses no more than the quotient of a divided by b, instead of a let us write a-OOOOOOO, whose value does not at all differ from a, since it contains neither tenth nor hundredth parts. Let us divide this by b by the common rules of division, taking care to put in the proper place the point which separates the decimals and the integers, and the operation is performed. Let the fraction, for example, be equivalent to , the division in decimals will then stand thus : 2) I ooooooo 0.5oooooo-k From this it appears that =0.5000000 or 0.5; which is sufficiently manifest, since this decimal fraction represents #, which is the same as '. 785. Let the given fraction be , and we have 3)1-ooooooo o:3333333" 786. From this it is seen that the decimal fraction equivalent to cannot be discontinued, but that the number 3 is repeated ad infinitum. Indeed, it has been already seen, in the preceding article, that the fractions # * 1:54. The 4 Twin added together make #. In the same way, the decimal fraction which expresses the value of 3 is 0.6666666, evidently the double of . 787. Suppose to be the proposed fraction, we have 4)1 ooooooo O'25OCOOO so that 3–0 2500000 or o:25. The proof whereof is that #4 in =#=}. 788. In the same way for the fraction $, we have *)'-l O.75OOOOOT * Thus we see #=0.75, that is, *, +15++3. - 789. The fraction # is changed into a decimal fraction by making 4)5:0000000_s. 1.25ooooo."

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for 1 + #=#.

790. 'So will be found =0-2,3-04, 3–0.6, 3–0.8, #=1,?=12, &c. In the occurrence of the denominator 7, the decimal fractions become a little more complicated; thus we have 1=o 142857142857, &c., in which the six figures are continually repeated. By transforming this decimal fraction into a geometrical progression, we may see that it precisely expresses the value }, the first term of this progression = '#' and the exponent = Toulon, Hence the sum = 1 '-'. (multiplying both terms by 10000000)=}. There is, however, a simpler ": of proving that this decimal fraction is exactly 3, by substituting for its value the letters, as under:

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Now, dividing by 999999, we shall have s=#=}; hences=}. - 791. The same will be seen by trial upon other fractions whose denominator is 7, the decimal fraction being infinite and six figures continually repeated. The reason is, that in continuing the division, we must return to a remainder which has already been had; and in

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Subtracts, and we have 99 s =9; consequently s =# =1, so with #, &c. 793. There are many of these decimal fractions which are called recurring, sometimes with two, and at other times with more, figures. Their values may be found without difficulty. Thus in the case of a single figure constantly repeated, let it be represented by a, so that s=0 aaaaaaa, we have 1Os = a aaaaaaa subtracting s = O'aaaaaaa

we have 9s = a, so that s = #. 794. In the case of two figures, as ab, we shall finds =#. In the case of three figures,

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795. So that if a decimal fraction occurs, it is easy to find its value; for instance, of 0.296296, the value will #1-#. which fraction, it may easily be proved, will give again the decimal fraction required. 796. We shall close this section with a curious example of changing into a decimal l

fraction the vulgar fraction 1:2:3:1:5:7:To the operation whereofis as follows:

2) 1:00000000000000 3)o sooooooooooooo. 4)o 16666666666666 5)oroalgoggggggggg 6)o-oossssssssssss 7)o-ool 38888888888 s)ooool984.126984.1 9)ooooo.24801587so 10)o-ooooo.275578192 o-oooooo.27557319.

CALCULATION OF INTEREST.

797. Interest, or the value of the use of money, is usually expressed per cent., or after the rate per hundred on the principal lent. Thus, if we put out 500 pounds sterling at 5 per cent., it signifies that for every hundred pounds the lender is to receive five pounds per annum during the continuance of the loan. . The solution of this question, which is one merely of simple interest, is so obvious, that it is unnecessary further to detain the reader upon it; and we therefore pass on to compound interest, or interest upon interest, which arises from the principal and interest taken together, as it becomes due at the end of each stated time of payment.

798. In the resolution of this question, we are to consider that 100l. at the end of a year becomes 105l. Let a = principal. Its amount at the end of the year is found by saying, if 100 gives 105, what will a give; and we answer # =#, which may be also expressed #x a, or a + '; x a.

799. Thus, by adding its twentieth part to the original principal, we have the principal at the end of the first year; adding to this last its twentieth, we know the amount of the given principal in two years, and so on. Hence the annual increases to the principal may be easily computed. Suppose, for instance, the principal of 1000l. Expressing the values in decimal fractions, it will be worth –

After 1 year - - - £1050
52.5 One year's interest on £1050.

After 2 years - - - 1102.5

55-125 - 1102.5 After 3 years - - - 1157-625

57.881 - 1157.625 After 4 years - - - 1215.506

60-775 - 1215'506 After 5 years - - - 1276-281 &c.

The method above exhibited would, however, in calculations for a number of years, become very laborious, and it may be abridged in the following manner.

800. Let the present principal = a, now, since a principal of 20l. will amount to 211 at the end of a year, the principal a will amount to #xa at the end of that time. At the end of the following year the same principal will amount to # x a = (#)*x a. This principal of two years will, the year after, amount to (#): x a, which will therefore be the principal of three years; increasing in this manner, at the end of four years the principal becomes (#): x a. After a century it will amount to (#)100 x a, and in general (#)” x a is the amount of the principal after n years; a formula serving to determine the amount of principal after any number of years.

801. The interest of 5 per cent., which has been taken in the above calculation, determined the fraction #3. Had the interest been reckoned at 6 per cent. the principal a would at the end of a year be (#) x a; at the end of two years to (#)2 x a; and at the end of n years to (#)”x a. Again, if the interest be at 4 per cent. the principal a will, after n years, be (#)”x a. Now all these formulae are easily resolved by logarithms; for if, according to the first supposition, the question be (#)”x a, this will be L.(#)” + L.a, and as (£)" is a power, we have L.(#)”=n L. #: so that the logarithm of the principal required is = n x L.344 L.a, and the logarithm of the fraction 3}= L.21 – L.20.

802. We shall now consider what the principal of 1000l. will amount to at compound interest of 5 per cent. at the end of 100 years. Here n=100. Hence the logarithm of the principal required will be = 100L.: + L. 1000, calculated as under: —

. L.21 = 1 .32221.93 Subtracting L.20=1:3010300

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* 5-1189soo= Logarithm of the principal required; from the characteristic whereof the principal must be a number of six figures, and by the tables it will appear to be 131,5011. In the case of a principal of 34521 at 6 per cent for sixty-four years, we have a=3452 and n=64. Principal at the end of the first year therefore ' Hence the logarithm of the principal sought=64 x L.: + L.3452, which will be found to amount to 143,768l. 803. When the number of years is very great, errors of considerable magnitude may arise from the logarithms not being sufficiently extended in the decimal places; but as our object here is only to show the principle on which these calculations are founded, we do not think it necessary further to pursue that subject. $94. There is another case which now requires our consideration; it is that of not only adding the interest annually to the principal, but increasing it every year by a new sum **. The original principal a would then increase in the following manner:After 1 year, 'a + b After 2 years, (#)2a+#b+b After 3 years, (3})*a + (£)°b+#b+b After 4 years, (3)"a +(#)*b + (#)*b + #b+b After n years, (3)"a +(#)"b+(#)"*b*. . . . $45+b This principal evidently consists of two parts, whereof the first =(#)"a, and the other, taken inversely, forms the series b + #54 (#)*b +(#)"b+.... (#)”-'b. This last series is evidently a geometrical progression, whose exponent =#. Its sum, therefore, will be found by first multiplying the last term (£)"b by the exponent 3, which gives (#)"b. Subtract the first term b, and we have the remainder (#)"b-b; and lastly, dividing by the exponent minus 1, that is, by , we have the sum required, = 20($)"b–2Ob. Wherefore the principal sought is (£)"a +20($)"b–2Ob=(#)” x (a +20b)–20b.

805. To resolve this formula we must separately calculate its first term (#!)" x (a +20b), which is nL. #4 L.(a +20b), for the number which answers to this logarithm in the tables will be the first term, and if from this we subtract 20b we have the principal sought.

806. Suppose a principal of 1000l. placed out at 5 per cent. compound interest, and to it there be annually added 100l. besides its compound interest, and it be required to know to what it will amount at the end of 25 years. Here a = 1000, b= 100, n = 25; and the operation is as follows: —

L.# =0.021189299

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=4-0068537885 The first part or number which answers to this logarithm is 10159-11. ; from which if we subtract 20b = 2000 we find the principal in question to be after 25 years 8159:11. 807. If it be required to know in how many years a principal of 1000l. under the above conditions would amount to 1,000,000l. ; let n be the number of years required, and since a=1000, b=100, the principal at the end of n years will be (#)” (3000)-2000, which sum must make 1,000,000l., whence results this equation: – 3000 (£)”–2000–1000000 Adding to both sides 2000 we have 3000 ($)”= 1002000 Dividing both sides by 3000 we have (#)=334

Using logarithms we have n L-35–L.334, and dividing by L.3}, we obtain n= # Now L.334=2:5237465 and L. #=0.021 1893, wherefore n= #. If, lastly, the two terms of this fraction be multiplied by 10000000, we shall have n = #. equal to one hundred and nineteen years one month and seven days, which is the time wherein the principal of 1000l. will be increased to 1,000,000l. In the case of an annual decrease instead of increase of the capital by a certan sum, we shall have the following gradations as the values of a, year after year, the interest being at 5 per cent, and, representing by b the sum annually abstracted from the principal,

After 1 year it would be #a-b

After 2 years – (#)'a-#-b

After 3 years – (#) a-(#)'b-#5–t

After n years - (#)"a-(#)"b-(#)"-*b . . . . -(#)b-b.

This principal evidently consists of two parts, one whereof is (#)"a, and the other to be sub

tracted therefrom, taking the terms inversely, forms a geometrical progression, as follows : –

b+(#)b+(#)"b+(#)'b+ . . . . (#)"b

The sum of this progression has already been found = 20 ($)"b–2Ob; if therefore, this be subtracted from ( #"a, we have the principal required after n years =(#)"(a–20b)+2ob.

808. For a less period than a year, the exponent n becomes a fraction; for example, 1 day ==}s, 2 days==#5, and so on. It often happens that we wish to know the present value of a sum of money payable at the end of a number of years. Thus, as 20 pounds in ready money amount in a twelvemonth to 21 pounds, so, reciprocally, 21 pounds payable at the end of a year can be worth only 20 pounds. Therefore, if a be a sum payable at the end of a year, the present value of it is #a. Hence, to find the present value of a principal a at the end of a year, we must multiply by #; to find its present value at the end of two years, it must be multiplied by (#)*a; and, in general, its value n years before the time of payment will be expressed by (#)"a.

809. Thus, suppose a rent of 100l. receivable for 5 years, reckoning interest at 5 per cent., if we would know its value in present money, we have

For £100 due after 1 year, the present value is £95.239

after 2 years - 90-704
after 3 years - 86-385
after 4 years - 82.272
after 5 years - 78-355

Sum of the five terms £432.955 So that in present money, the value is 4321.19s. 1d. 810. But for a great number of years such a calculation would become laborious. It may be facilitated as follows: — Let the annual rent = a, commencing directly and con

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