Remarks on Professor Johnson's Query

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?89? NEW





BY J. B. MOTT, WORTHINGTON, MINNESOTA. In the following example the ordinary method is pursued till the second figure ofthe root is found. We then find the triple product (t. p.) by placing the second figure of the root to the right of three times the preceding

part of the root and multiplying this by said second figure, and so on for all the triple products, finding one from the other. Thus: lst trial divisor + 1st t. p. = 1st complete divisor. lst complete divisor+lst t. p. + b2+2nd t. p. = 2nd complete divisor. = 3rd com. divisor, &c, 2nd complete divisor+2nd t. n.+c2+3rd tp. the to the under t. two each preceding divisor; b, c p. figures right placing the &c. of root. &c, being 2nd, 3rd figures Each complete divisor is used as a trial divisor for the next figure of the The constant left hand figures of any divisor, used as in simple divi? determine as many more figures of the root. will sion,



find the cube root of 2 to fourteen places of decimals.

=3 First trial divisor ? t. = p. 32X2=J>4_ " com. divisor = 364


8625 lst t. p. +4 + 2d t. p., 45025 2nd complete divisor, 2nd t. p+25 + 3d t. p., 218831 4721331 3rd complete divisor,

1_ )1000 728

3731211 3rd t. p.+81 + 4th t. p., 47586431"! 4th complete divisor, 34765144 t. p., 4th t. p.+81+5th 47621196244 5th complete divisor,

)272000 225125_ )46875000 42491979 )4383021000 4282778799

)100242201Q00 95242392488 79375561 5th t. p.+ 4 +6th t. p., 47621989989(51 6th complete divisor, )4999808512000 4762198998961 3779762 6th t. p.+ 0 + 7th t. p., 4762202778723 7th complete divisor, )237609513039 = .04989487 + . This united Dividing, we have 2376095 -s- 47622028 to the root above gives f2 = 1.25992104989487+. by R. J Adcock.?Mr. Query Jonson's paradox involves two questions, the value of u, and the value of du-i-dx = cos ax, when a = co. He says "now if a = co, u = 0 indepen? This I deny. 0 is used to represent both actual zero and dently of x". Remarks

on Professor


?90? infinitesimal

It cannot however be so used unless it can be quantities. In shown that no error will result from such use in the particular case. this case when a = oo, u = actual zero or an infinitesimal, that is u = 0 -r- a or some infinitesimal between l~-a and ? 1-j-a, these infinitesimals

depending for their values upon a and x. And the rate at which these infin? itesimals change their values is du~dx = cos ax, for all values of a and x. When a is infinite, that is greater than any assignable number, then a is indeterminately great, and by consequence indeterminate; cos ax is the co? sine of an are in a given circle, the are being as indeterminate as a, and therefore cos ax is indetermiate both in "form" and value, and from the

given conditions can no more be affirmed to be zero than any other value between +1 and?1. There being no preference or reason for the termina? tion of the are ax in one part of the circumference rather than another. Mathematics

having to deal with truth, like the "Scripture is of no private

interpretation". DIFFERENTIATION







To differentiate the logarithm of a variable, let y = ex; = d(\oge y). . ?. x = loge y; .-.dx = ex(edx-~~l), y + dy = ex+dx, dy = ex+dx?ex or dy eP* =

but dx =

1 +dx+


or edx =

1 + dx;

or dy = exdx = ydx, or dx = dy-*-y, dy = ex(l+dx?1), d (loge 2/); . ?. d(logey) = dy~yf differential of Napierian logarithm. = md(\ogey); ^g10y = m(\ogey); .-. d(log10 y) . ?. d(logl0


7/) =



differential of common log.




of problems in No. 2 have been received as follows: Solutions From Prof. L. G. Barbour, 434; Prof. W. P. Casey, 430, 431, 433; G. E. Curtis, 429, 434; Geo. Eastwood, 431; Wm. Hoover, 429, 430; Prof. P. H. Philbrick, 429, 430, 431, 433, 434; Prof. E. B. Seitz, 430,431, 433, 435; Prof. J. Scheffer, 428, 430, 431, 433.