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. 1861 Excerpt: ...how many roots lie within a given interval, and we may then divide that interval into smaller intervals within which the roots lie singly. Suppose then that we know that an equation has one root and only one between two given quantities a and /?, and we wish to approximate to the value of this root. If we substitute ...
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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. 1861 Excerpt: ...how many roots lie within a given interval, and we may then divide that interval into smaller intervals within which the roots lie singly. Suppose then that we know that an equation has one root and only one between two given quantities a and /?, and we wish to approximate to the value of this root. If we substitute any quantity y which is intermediate between a and /? for x in f(x), we shall know by the sign of f(y) whether the root lies between a and y or between y and /3. Suppose it to lie between a and y; then we may substitute for x a quantity 8 which lies between o and y, and we shall know by the sign of f(S) whether the root lies between a and 8 or between 8 and y. This process may be continued to any extent, and we may approximate as closely as we please to the numerical value of the root; for by each operation we can thus halve the interval within which the root must lie. The operation here described would however be very laborious, and methods have been proposed for attaining the required result with less calculation. We shall first explain Lagrange's method. 212. Lety(ai) = 0 be an equation which is known1 to have one root, and only one, between two consecutive positive integers a and a + 1. Put x = a +-, and substitute this value of a: in the y proposed equation; thus /fa +-J = 0. If we clear this equation of fractions, we obtain an equation in y of the same degree as the original equation in x; denote it by $l(y) = 0. This equation in y has only one positive root, because the original equation in x has only one root between a and a + 1. We may then determine the consecutive integers between which the value of y must lie, by substituting in f, (y) successively the values 1, 2, 3, ... until two consecutive results are obtained which are of contrary.
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PLEASE NOTE, WE DO NOT SHIP TO DENMARK. New Book. Shipped from UK in 4 to 14 days. Established seller since 2000. Please note we cannot offer an expedited shipping service from the UK.
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PLEASE NOTE, WE DO NOT SHIP TO DENMARK. New Book. Shipped from UK in 4 to 14 days. Established seller since 2000. Please note we cannot offer an expedited shipping service from the UK.
