Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

Divide the propofed Stock (viz. 634,4) by the given Principal (viz. 375,5) and the Quotient will fhew the tabular Number that ftands over against the Time fought. Thus 375,5) 634,4 (1,689479 &c. this Number being fought in the Table, will be found to ftand against 9 Years, which is the Time required.

But if the Quotient cannot be truly found in the Table of Amounts for Years, as above; then take out of that Table the neareft Number that is lefs, and make it a Divifor, by which you muft divide the first Quotient; and then feek the fecond Quotient in the Table of Amounts for Days (which is inferted a little further on) and it will affign the Number of Days: as in this Example.

In what Time will 5631. amount to 8601. at 6 per Cent. per Annum, Compound Intereft?

Anfwer. In 7 Years and 99 Days.. Thus 563) 860 (1,52753 which fhews the Time to be more (or above) feven Years; for over againft 7 Years is 1,50363 which being made the new Divifor: Viz. 1,50363) 1,52753 (1,01589 &c. this Number is the nearest Amount to 99 Days.

Note, If the Stock, Principal, and Time be given; the Rate of Intereft will be beft found by extracting the Root, &c. as before in the fourth Question.

The next Thing that I fhall here propofe, is to make this Table (which is only calculated for the Rate of 6 per Cent.) univerfally useful for all the Rates of Compound Intereft, which I may prefume to say, is a new Improvement of my own, being well fatisfied it never was published before; and not only fo, but I have heard feveral very good Artists affirm it was impoffible to be done.

The Method of performing it is briefly thus, Let x= the Difference between 1,06 R, the Amount of 17, for one Year (in the Table) and any other proposed Amount of 1 l. for one Year; which admits of two Cafes.

Cafe 1. If the propofed Rate be greater than the 1,06 = R, then will R+the true Amount of 1 I, for one Year at that Rate.

Cafe 2. But if the propofed Rate be lefs than 1,06R, then it will be Rx the Amount of 1 1. &c.

Make

{t-1=b, t-2 = c, t = 3 =d, t—4=f, &c. 2} + b = 8,468 m, 4 dmn, ±ƒn=s&c.

LI

Then

[ocr errors]

the

Then will R+ t Rb x + g Rc x2 + m Rd x* &c. = the Amount of 11. at the given Rate, for any Time denoted by t, in Cafe 1. And R − t Rb x + g R¢ x3 — m Rd x3 &c. Amount of 11. in Cafe 2. Which is no more but this: Let R+x or R. — * (which foever it is) be involved (as directed in Sect. 5. Chap. 2.) to the fame Power or Height as the Index t the given Time in the Que. ftion denotes: rejecting all the Powers of x above x ×× or ×××× at moft, as ufelefs. Then multiply that Power of R+x or Rx into the given Principal, and their Product will be the Amount required.

7

An Example or two in each Cafe will render all easy.

[blocks in formation]

Suppofe it were required to find what 2561. would amount to in fifteen Years, at 81. per Cent. per Annum Compound Intereft? Here t=15.

First 100: 108 :: : 1,08 the Amount of 11. at 8 per Cent. Next 1,08-1,06=0,02=x. And R+x=1,08 as in Cafe 1. Then R15 R14x + 105 R13 x x + 455 R12 xxx &c, the Amount of i. for 15 Years, at 8 per Cent.

12

Here x 0,02. xx 0,0004.. and x x x

15 R14 x

And 105 R13 xx

,000008 By the Table R15 = 2,396558

2,260904 x 15 x,02

2,132928 x 105 x,0004

0,678271 =0,089583

Sum 3,171736

455 R12 xxx= 2,012196 x 455 ×,0000080,007324

Then 3,171736 x 256 811,964416 A

That is, 811. 95. 3 d. fere. Which is the Anfwer required.

[blocks in formation]

What will 3651. amount to in feven Years at four and a half per Cent, &c.

First 100 104,5 : 1,045 the Amount of 1 l. at 4% per Cent.

Next 1,06-1,045=0,015=x. Confequently R-x=1,045 as in Cafe 2.

The R-7R6x + 21 R3 xx— 35 R4xxx &c. the Amount of 1 1. for 7 Years, at 4 per Cent.

Here

Here x =,015; xx=,000225; and xxx000003375
By the Table R' =+1,503630

7 R6 x

[blocks in formation]

=0,148944

And+21 R3 x x = +0,006323

[blocks in formation]

R' — 7 R6 x + 21 R3 x x — 35 R+ xxx=1,360868 Then 1,360868 x 365 496,71682=A.

That is, 496. 145. 34 d. is the Anfwer required.

If the Reason of these two Operations be but well understood, it will be very easy to conceive how to find P, the Principal, by having A, t, and x given (because R and it's Powers are always given by the Table.)

For R2±t Rbx+g R¢ x x + m Rd x x x × P = A (as above.)

Therefore

A

R2 + t Rb x + g R¢ x x + m R4 x x x Or if A, P, and t, be given x may be found.

A

P

For R2 ± 1 Rb x + 8 R‹ × × ±m Rdxxx=2. This Equation

g

P

being folved (as in Chap. 10.) the Value of x will be found; and then either R+x, or R -x will fhew the Rate of Intereft, &c.

But I fhall leave the numerical Operations to the Learner's Practice, fuppofing enough done to fhew how all Questions of this Kind that are limited by whole Years may be computed.

And if the Time given or fought be not terminated by whole Years, but by Weeks, Months, Quarters, or Half-Years, &e. for refolving fuch Questions, the beft Way will be to reduce those Parts of a Year into Days; that done, find an Anfwer according to the Demand of the Queftion (and agreeing to 1 l. as before) for that Number of Days; and in order to that, it will be requifite to find the Amount of 1. for one Day (as in my Compendium of Algebra, Page 110) which I fhall here infert.

Put a the Amount fought, then it will be

1:a :: a: aa¦ ¦ aa: aaa :: aaa : à a a a÷to a365.

As one Pound is to it's Amount for one That is Amount to the Amount of two Days of two Days to that of three Days. to 365 Days.

L12

Doy :: fo is that

and fo is that And fo on in

Then

[ocr errors]

Then the laft of the Terms will be a365 — 1,06
Putra. And let r = 1

1365

z in Numb.

3- 1

2μ365 +365r36 +e+66430r363ee=a365=1,06
31365e+ 66430e e = 1,06

4 365 e66430 ee = 0,06

466430 5,00549 e +ee = 0,0000009032 D

5 ÷ 6e=

D

[blocks in formation]

,00549 +e

ift Divifor ,00559) 0,0000009032

+e=,00015

2d Divifor,00574

[blocks in formation]
[blocks in formation]
[blocks in formation]

2870

and only too much by

[blocks in formation]

New r1,00016 for a fecond Operation. Then

2 in Numb. 71,06013401407 +386,887 e +70402,172 €8 1,06. Hence it appears that r — e = a. 81,06013401407 — 386,887 e +70402,172ee

Therefore

1,06

-

89386,887 e — 70402,172 e e = 0,0001 3401407
910,0054953 e e =,0000000019035503

[merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

Which being further pursued to a third Operation will give

a = 1,000159653587453 &c.

This

I

This Value of a is the Amount of 11. for one Day, from which, if 1. be fubtracted, the Remainder =,000159653587 &c. will be the Intereft of 11. for one Day. Confequently, if any proposed Principal be multiplied into either of these, the respective Product will be the Amount or Intereft of that Principal for one Day, at 6 per Cent, &c.

And that the Amount (or Intereft) of any Principal or Sum may be eafily computed for any Number of Days lefs than a Year; I have here inferted the following Table, which with a great deal of Care (and I believe Exactnefs) is calculated from the laft found (1,000159653587453) Amount of 11. for one Day. To which alfo is annexed a Table of the Amounts of 1 l for Months.

[blocks in formation]

1.0019175262 37 1.0059241901 62 1.0099468767

15
16 1.0025575184 41

13 1.0020774859 38
14 1.0022374712 39 1.0062454146 64
1.0023974820 40 1.0064060653 65

1.0060847895 63

1.0101081184

1.0102693858

1.0104306789

1.0065667416 66

1.0105919978

17

19

1.0027175803 42 1.0067274436 67 1.0028776677 43 1.0030377808 44 1.0070489245 69 1.0031979193 45 21 1.0033580850 46 22 1.0035182732 47 23 1.0036784885 48 24 1.0038387294 49 25 1.0039989958

1.0107533424

1.0078881712 68

1.0109147128

1.0110761090

1.0072097035 70

1.0112375309

1.0115604521

1.0073705082 71 1.0113989786
1.0075313385 72
1.0076921945 73 1.0117219513
1.0078530762 74

1.0118834764

50

1.0080139835 75

1.0120450272

Days

« ΠροηγούμενηΣυνέχεια »