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Binomial Theorem

283 Indeterminate Analysis, though a

The demonstration

289 variety of forms

324

Multiplication and Division of Compound Indeterminate Equa-

Series

291 tions

350

Multinomial Theorem

293 Continued Fractions

368

Reversion of Series

294 The differential method of Series 379

Demonstration do.

297 Summation of the Infinite Series 382

Permutations and Combinations 298 Diophantine Analysis

390

Logarithms

301 Double and Triple Equalities 402

The method of finding the increase Solutions of a great variety of ques.

of Population in any country,

tions in this refined analysis 404,420

under given circumstances of Miscellaneous Questions, with

births and mortality

306 their solutions and without

Figurate and Polygonal Numbers 311 Some curious Questions with their

Interest and Annuities

312 Solutions

432–6

Compound Interest

314 Deophantine Questions and Solu.

Vanishing Fractions and other par- tions

446 to 474 and 478.

ticular expressions 319 to 323 Four hundred and twenty-four
Decomposition of Rational Frac- Questions for Practice 498

tions into their equivalent Sim- Further Investigation of Com.
ple Fractions

324 pound Interest and Annuity,
Recurring Series
329 twenty pages

541
Table, page 572. Which shews what one dollar will amount to, being

forebone or increase at Compound Interest, in any number of years not

exceeding 21; and being computed yearly at any of the rates 3, 31, 4, 44, 5

and 6

per cent per annum.

Table 1, page 573. Exhibiting the period in which the population of a

country has a tendency to double itself, from an estimate of its increase per

cent., taken at the end of every ten years.

A Table of Reciprocals, squares, cubes and roots.

Page 374.

Page 575, top Table, which shows what $1, payable at the end of any term

of years to come, under 21, is worth in ready money. Discounting or rebate

being yearly computed at any of the following rates: 3, 31, 4, 41, 5 and 6 per

cent. per annum.

Page 575. Table Second, at the bottom, which shows what $1 annuity,

payable by yearly payments, and foreborn any number of years under 21,

will amonnt to at the end of the term, Compound Interest, being computed at

any of the rates to wit, 3, 31, 4, 44, 5, and 6 per cent per annum.

Page 576, the top Table. Which shews the present worth of 1$, Annuity

to continue any term of years under 21, and payable by yearly payments,

Compound Interest being computed at any of these rates, to wit, 3, 31, 4, 41,

5 and 6 per cent. per annum.

Bottom Table on page 576, will show what Annuity payable by yearly pay.

ments to continue any term of years under 21, 1$ will purchase, Compound

Interest being computed at any of these rates, to wit, 3, 31, 4, 41, 5 and 6

per cent. per annum.

The Construction of Geometrical Problems,

Analysis and demonstration of Theorems,

Application of Algebra to Geometry,

It is to be distinctly understood that I am indebted to the following

works, viz. Simson's Algebra, Emerson, Bonnycastle, Euler, Saunderson,

Maclaurin, Bland, Bridge, Wood, Nicholson, Newton's Universal Arithmetic,

Ward, Doddridge, De Moivre's, Clairaut, Wolfius, Peacock, Muller, Hayes,

Fermat, Pierse's, Diophantus, Bachet, Waring, and others, which my limits

will not permit me to name.

1. A and B purchase 900 (a) acres of land at the rate of $2 pe acre, which they paid equally between them; but on dividin the same, A got that part of the farm which contained the house and agreed to pay $tor or $% per acre more than B;:. 45 cts how many acres had A and B, and at what price ? Let X,

and a-a denote the number of acres A and B each had; then

45

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and tur; will be the price A and B each paid per acre

2 and the equation. Reduced I have 20a-200x20ax+9ax—9.2 •.2?_400x=a?, by Art. 70, page

141. =2490-2090—400 =A's share, and 900—400—500=B's share. And 28848885 225_180=45 cents. See Flint's Surveying.

Let x, and 900~« be the acres A and B each bought, and let y and y+45 represent the cents per acre given by B and A respectively, then xy+45x=900yy, (1) and xy+-45x+900yXY=

900y 180000, (2) and by (1) and (2) I have =4000—20y ;..y—

2y+45 155y=4500, by Art. 70. y=180 cents the price of one acre of B's, and 180+45=2.25 cents, the price of one acre of A's land. Otherwise let x, and y, denote the number of acres A and B each bought, then the price of one acre of each man's land will be ex

90000 90000 pressed by and

.. by the question xty=900,

y 90000 90000

90000 90000 and

+45, or y

900—4 y y?+3100y=18&0000, and y=500, as before.

= 2. A man buys. 80 pounds of pepper and 36 pounds of saffron, so that for $8, he had 14 pounds of pepper more than he had of saffron for $26, and what he laid out amounted to $188. How many pounds of pepper had he for $8, and how many of saffron for $26.

Ans. 20 pounds of pepper and 6 of saffron. Let x denote number of pounds of pepper that he bought for $8, and y=number of pounds of saffron for $26, then by the question if x pounds cost $8 what will 80 pounds cost, and if y pounds of saffron cost 26 dollars what will 36 pounds of saffron

8.80 , 36.26 cost, and consequently their sum must =188, that is +

y 160, 234

160

234 =188 or

=47, (1) and x-y=14, (2) or y

y =47, 160y+234y+3276=47y*+658y, or 47y +264y=3276, part 70; .:: y=6 and z=14+6=20=the number of pounds he ..

=+= bought of pepper, and 6 of saffron. Ans .

+45; or

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Investigation of the Rules of Compound Interest.

annuity, rent, or pension, n = number of times that interest is to be paid for the annuity or sum lent, r = rate of interest of 1 dollar for 1 time, m = amount of the annuity, or sum lent, for n times, at r interest, p principal, sum used, or present worth of a sum before it is due, (of compound interest.) First in amounts let q=ltr = amount of one dollar for 1 time. Now a= last year's amount, and 1:9::a: aq= last year but 1 amt. 1:9:: aq: aq last but 2 year's amount. 1:9:: aq : aq = last but 3, and so on to aqn-1 = first year's amount; therefore a t aq + aq + ago, &c., + aq-1 a: aq ::m aq"-?: m

then ma= mg aq”, or -a=m+ mraq?, .. aq” – a= mr, then put A : q",

(A-1)a (A-1)a (A-1)a= mr, i.a=

*A-1
i m=

-;=
2d. In discounts, q:1:: 2 : = first year's present worth,

9 and q:1:: = 2d year's present worth; q:1:: :

9 9 3d year, and

=nth year's present worth, therefore I have

q" ++

: P :p

; 9

q"
9

go

2 pg" (Euc. 12, V.) .. =Pq-a, or

pt pr -a; .. q"

q aq" -a=pro", or (A-1)a=prA; ..A ; hence (A-la

a-TP

mrta =mr=prA; .. A=

(1+r)".

a-rp р
Case 1. Given a, m, and p, to find r.
(A-1)a

mrta
Since A =
Case 2. Given p, m, and n, to find r.
Since (r+1)"

;.

-1. р Case 3. Given a, m, and n, to find r.

(A-1) Since (A-1)a=mr; therefore

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Case 10. Given m, n, and r; required p.

p(1+r)"=m in Logarithms Since A ; , :P

PEM-n XL(1+r).
р

A'
Case 11. Given a, m, and r, to find p.

mrta Since

i ::P р

mrta Case 12. Given a, m, and n, to find p.

2 Find D= ; E

;F={2(D-1)+E}XE.

:

(n+1) Now FE=r, by 3d, but (1+r)".

р Case 13. Given p, n, and r, to find a. Since (A - 1)a = prA;

A-1
Case 14. Given p, m, and n, to find a.
Since A=(r+1)"

=
=

Hence A -1. р

mX (B-1) and B-1=r, are known; a. a=

A-1 Case 15. Given m, n, and r, to find a. Since (A-1)a= mr, therefore a=

A-1 Case 16. Given p, m, and r, to find a. Since A=

A-I is given, р

A-1 Case 17. Given m, p, and r, to find n. Put L for the logarithm, and L' for the arith. comp. of a log

L.m +L'.p Since (1+r)"=

therefore n = P

L.(r+1) L(1+r) Case 18. Given a, p, and r, to find the value of n.

L.a-L.(a-pr) Since (1+r)" =

therefore n= apr

L.(r+1) Case 19. Given a, p, and m, to find n. Since A=(r+1)"=; then L.(r+1)"=L.m-L.p=L.

P Hence A, and A-1, are known; Also L.(A-1)+L.a+L'.m = L.(B-1) by Case 14th hence r = B-1 and r+1, are kno

L.A. therefore n =

L.(r+1) Case 20. Given a, m, and r, to find n. mrta

L.(mrta)—L.C Since A =

(1+r)"; .in=

L.(1+r)

m

mr

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m

n+ 1

an

rem will become =1+-+*+ne nearly.

(%)== =l+r+7,2 ;

nti: r+E=W{(2X[D—"137E) ()

'n+1

=

m

na

n+1

_1=(1+r)^;

па

nr

nr

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&c. ;

net

+ met +xmeter

, nearly; then (%) +=(1–1 + 1 x ***) **; which, by bi2.,

nt1

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na

n1 rt

. 12

12r Let D

1

-god; then gul +
12
12

nti (D—1) X ; let 2E=

=w]+ X E} ; therefore r = F-E. In this solution, 1st find

6 D= -1; 2d, find E=

;

3d, find F=N{2(D—1)+E}•E; 4th, find r= EF --E. Case 4. Given a, p, and n, to find r.

A-1

P Since (A-1)a=Apr, ..

1-A-1

Anr now (1+r)-*=1

n+1 n+in+2 mr + 7 x - nx

Х
2

2
therefore
P

n+2
1

2 3 P

n+ n2 +

nti; 3

n-1 nomial theorem, will become =1+r- -god, nearly. Now,

12

n-1 n+1 ; let G= n+1=1

-gol;thereр р

12 12

12

12 fore pe

(G — 1)'X Let 2H: n-1

-1' then r-H=W{H-2G-1)XH=)K;..r=H-K. Case 5. Given a, n and r, to find m.

(A-1)a 1)a=

=mr, therefore m= Case 6. Given p, n, and r, to find m. Since A:

ni. pA.

р
Case 7. Given a, p, and r; find m.
Since

ap
; .. m =
р -TP

op Case 8. Given a, p, and n, to find m.

3 Find G = n+1; H= ; K=^{H—(2G-1)}XH.

(n-1) Now H-K=r, by the 4th, but (r+1)"=;..m=

.:m=(r+1)"Xp.

= Case 9. Given a, n, and

1, Since (A-lla=prA; therefore p=

(A-1) Xa

Ar

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