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To change an equation of the second class into the form of the first, let x = Substituting this value for ≈, and reducing,

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The value of y can here be found, as in the first class, the reciprocal of which will be the value of x.

Again, by adding and subtracting equal quantities, the equation

of the third class can take the form,

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a

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2, we may find the value of z, as in the first

class, and then determine x from the relation

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Referring now to the investigation in the Mathematical Monthly, Vol. II., p. 85-88, let 7 denote the common logarithm of the quantity before which it is placed; and the following rule is obtained. For the first class of cubics having the form,

23 + ax = =b,

where a, b denote either positive or negative numbers. Regarding a, b as positive numbers, find the quantity

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II. When a is negative, and the index of L less than 10;

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III. When a is negative, and the index of L is 10, or more;

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Here it should be carefully noted to apply to the values of x, thus far, the same algebraic sign with that of b, on the right-hand side of the given cubic. But the two remaining values of x should be taken with a sign contrary to that of b:

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With the foregoing precepts, the auxiliaries v, u must always be taken positively, and not exceeding 90°.

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In this example, a or 15 is positive, showing that the constant L is to be considered a logarithmic tangent, and that the formulas of I.

are to be employed; this has been strictly done in the above operation, as will be seen by comparison. And since b or 4 is positive, the value of x in this case is also positive.

Further examples might easily be multiplied, were it necessary; but the preceding rule, within a short compass, appears sufficiently plain and comprehensive.-L. W. MEECH, Assistant, U. S. Coast Survey, Preston, Ct.

ON THE INDETERMINATE ANALYSIS.

By REV. A. D. WHEELER, Brunswick, Maine.

[Concluded from Page 195.]

PROPOSITION XV. The whole number of possible equations in such a series, or for every integral value of cab+1 is expressed a b − (a + b − 1) ̧

by the formula

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DEMONSTRATION. Dividing ab by a, we have b = the whole number of impossible equations in which c is a multiple of a; and dividing by b, we have a the whole number of impossible equations in which it is a multiple of b. (Prop. IX., Cases 3 and 4.) But it will be seen that the impossible equation ax + by = ab has thus been reckoned twice, when it should have been reckoned but once. Deducting 1 on account of this repetition, we shall have a + b — 1 for the whole number of impossible equations in which c is a multiple of either a or b. If this number be subtracted from ab, there will remain ab-(a + b−1), which, according to Prop. XIV., includes an equal number of possible and of impossible equations. Therefore ab (a+b-1) expresses all the possible equations in which

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c<ab +1.

PROP. XVI. The whole number of equations in such a series, which will admit of one, and only one solution, is denoted by the product a b.

DEM. All values of c > 2ab will admit of two solutions, according to Prop. VIII. It may be shown, as in the preceding proposition, that all the values of c <2ab+1, which will admit of two soluab—(a+b−1); and

tions, may be expressed by the same formula,

2

this, subtracted from a b, representing the whole interval between ab ab (a+b1) and 2ab, gives ab

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2

ab+a+b-1
2

for the whole number of equations within that interval, admitting of but one solution. Adding the expressions for the number of solutions admitting one solution only above a b and below it, we have

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PROP. XVII. Let there be any numbers of independent equations, having the same coefficients to x and y. The sum of all the solutions contained in the whole number of equations may be found by dividing the sum of the values of c by ab, and adding the number of solutions contained in the several remainders, according to the rule which is given under Prop. XIII.

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ax1+by1 = ab (n + n' + n" — 1) +r+r+"&c.

Subtracting the remainders, we have the equation,

a x,, + by,, = a b (n + n' + n"

a b (n + n' + n" — 1) = c ;

and dividing c by ab, we obtain the quotient n + n'+n" — 1, which gives the number of solutions in (D) with reference to the

remainders. But in equation (A) we have n solutions; in equation (B), n' solutions; and in equation (C), n" — 1 solutions 1 solutions; and the

=

sum of these = n + n' + n" — 1, the same expression which we obtained from (D). Now if we ascertain the number of solutions in these several remainders, and add it to the number obtained from (D), we shall have all the possible solutions belonging to the several equations (A), (B), and (C).

Remark. Several of the preceding propositions will be found very convenient in proposing questions which shall have any desired conditions, and several of them will very much facilitate the operations which are necessary to be performed in the solution of problems which are to follow. They may easily be verified by substituting numbers for letters. YOUNG's Algebra contains more than one erroneous proposition relating to this subject, to which demonstrations are professedly given both by the author and the American editors.

PROBLEMS.

I. To find the successive remainders in any Arithmetical Progression, without continuing the operation of division.

1. Suppose the progression to be ascending, and call the first remainder r.

The general expression for any term in this progression is a + (n-1) d; and that for the next succeeding term will be a + (n−1) d+d. Let D be the divisor, and let

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Then will a +(n − 1) d = DQ+r. By substitution, we have

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That is to find the next succeeding remainder, add the difference of

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