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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1887 Excerpt: … being odd, t–1 is even, and the number of trinomial factors in (452) exclusive of (2–l)3, is even; but so that the first and last of these factors are equal. In the same manner it is show a that any two of these factors equally distant from the first and last are equal, therefore, uniting these equal factors and extracting the square root of both members, we have, when m is odd, zm–1 = (z–1) X ( 2–2 z cos—-+ 1) X ( z–2 z cos—-+ 1) X. xfs’-eos-+l) (453 m / 228. To find the quadratic factors of z 1–1, when m is even. When m is even, m–1 is odd, the number of factors in (452), exclusive of (z–1), is odd, and the middle factor will not combine with any other. This factor is the (?) and contains z + 2 z + 1 = z + l)a BO that uniting the remaining factors, and extracting the square root, we have, when m is even. X (-2 Cob J+ l) (454) 229. To find the factors of zm-f-1, when m is odd. In (450) let p = Jt, it gives (rf +l) = (i’-2oosi+l) X ( –2zCOS–+ l). X and it is easily shown, as in the preceding articles, that the factors equally distant from the first and last are equal, and that the middle term is z + 2 z + 1 = (z-f-l)a# M Hence we find, when m is odd, zTM + 1 = (z + 1) X (?–2 z cos–+ l) 230. To find the factors of zTM + 1 wAe w is even. The same process gives zTM + 1 = (V–22 cos–+ l) x (z_2cos JLL+l) X.. X (–2z cos m 1)?r + l) (456) 231. The simple factors of (453) and (454) are obtained from (451) by putting 4 = 0, and those of (455) and (456) by putting / = w. There will be found pairs of equal factors as in the preceding articles, but all the different simple factors will be found by taking only the positive sign of the radical /–1. 232. Any function of the form zam–2p zm-f-q may also be resolve…
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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1887 Excerpt: … being odd, t–1 is even, and the number of trinomial factors in (452) exclusive of (2–l)3, is even; but so that the first and last of these factors are equal. In the same manner it is show a that any two of these factors equally distant from the first and last are equal, therefore, uniting these equal factors and extracting the square root of both members, we have, when m is odd, zm–1 = (z–1) X ( 2–2 z cos—-+ 1) X ( z–2 z cos—-+ 1) X. xfs’-eos-+l) (453 m / 228. To find the quadratic factors of z 1–1, when m is even. When m is even, m–1 is odd, the number of factors in (452), exclusive of (z–1), is odd, and the middle factor will not combine with any other. This factor is the (?) and contains z + 2 z + 1 = z + l)a BO that uniting the remaining factors, and extracting the square root, we have, when m is even. X (-2 Cob J+ l) (454) 229. To find the factors of zm-f-1, when m is odd. In (450) let p = Jt, it gives (rf +l) = (i’-2oosi+l) X ( –2zCOS–+ l). X and it is easily shown, as in the preceding articles, that the factors equally distant from the first and last are equal, and that the middle term is z + 2 z + 1 = (z-f-l)a# M Hence we find, when m is odd, zTM + 1 = (z + 1) X (?–2 z cos–+ l) 230. To find the factors of zTM + 1 wAe w is even. The same process gives zTM + 1 = (V–22 cos–+ l) x (z_2cos JLL+l) X.. X (–2z cos m 1)?r + l) (456) 231. The simple factors of (453) and (454) are obtained from (451) by putting 4 = 0, and those of (455) and (456) by putting / = w. There will be found pairs of equal factors as in the preceding articles, but all the different simple factors will be found by taking only the positive sign of the radical /–1. 232. Any function of the form zam–2p zm-f-q may also be resolve…