# A Treatise on the Dynamics of a System of Rigid Bodies. with Numerous Examples

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This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1884 edition. Excerpt: ...292 that the signs of their terms are alternately positive and negative, and since their roots are real every one of those roots is positive. Hence all the subsequent auxiliary functions f3(y), ft(y), &o. have their roots real and positive. The signs therefore of all their terms are alternately positive and negative, and by Art. 297 tho coefficient of the highest power is in every case positive. In this way we are led to an extension of the theorem in Art. 297. Supposing p0 to have heen made positive, we see by the preceding reasoning that though it is necessary and sufficient that all the terms in the first column should be positive, yet it is also true that the terms in every column must be positive. Hence as we perform the process indicated in that article we may stop as soon as we find any negative term, and conclude at once that / (z) has some roots with their real parts negative. 300. Ex. 1. Express the condition that the real roots and the real parts of the imaginary roots of the cubic z3 + p1z2+p2z+p3 = 0 should be all negative. By Art. 296 /, y) = y3-p2y, /a (y)=PiVi-Pi Using the method of cross-multiplication given in Art. 297 and omitting the divisors as shown in Art. 299 we have fay) = PiPi-Pa)y fiy) = (PiP2-Pa)P3The necessary conditions are that pv p2-p3, and p3 should be all positive. We have retained the powers of y in order to separate the terms, and also the negative signs in the second column, but both these are unnecessary and in accordance with Art. 297 might have been omitted. In both this and the next example all the numerical calculations are shown. Ex. 2. Express the corresponding conditions for the biquadratic Z +pxZ3 +PnZ +P3Z + p4 = 0, /i(/) = 2/4-PtiP+Pv f3(y)=(PiPsl-p3)y;-pip fi (y) = (PiPa-Pi) Pa-Pi Pi i...