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. 1911 Excerpt: ...of points (sec. 22) that if = a + 6+c+-, then the distance of x cannot exceed the sum of the distances of a, by c, ...; that is, a + b+c+- a+)6+c +-; it is also obvious, by the definition of the subtraction of points, that a--b will denote the distance between the two points a and b. As a further matter of notation, we ...
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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. 1911 Excerpt: ...of points (sec. 22) that if = a + 6+c+-, then the distance of x cannot exceed the sum of the distances of a, by c, ...; that is, a + b+c+- a+)6+c +-; it is also obvious, by the definition of the subtraction of points, that a--b will denote the distance between the two points a and b. As a further matter of notation, we denote the left-hand side of the given equation byf(x): f(x)=poXn+p1Xn1+p2Xn-2+ ' ' +Pn-lX-t-pn, the value of fix) when x = a is then denoted by /(a), and our problem is to show that there is at least one point x = a such that fia) = 0. The function fix) is called a polynomial of the nth degree in #. 44. In order to simplify the proof, we first establish the following properties of the function fix). (1) Given, any distance R, we can find a distance G such that f(x) G whenever x R. For, take Gn-p'Sn, where p is the most distant of the given points p0, ph..., pn and S is a point such that $E and also 1. Then whenever x R, we shall have xk Rk Sh Sny (&=1, 2, ..., ri), and therefore, 1/001 = bo+PiS-1-+Pn-l-+Pn b - + p - +--+p5- + piS- w.p./S-, which is less than G, as required. (2) By taking x sufficiently large, we can make f(x) as large as we please. That is, given any distance g, we can find a distance h such that /0e) g whenever x h. containing x--a as a factor, and if F(a) were zero, then F(x) itself would contain a;--a as a factor, which is not the case. (4) The polynomial f (x) is continuous at every point x = a. Roughly speaking, this means that a small change in the position of x will produce a correspondingly small change in the position of / (x). More precisely, given any radius R about the point f(a), we can always find a radius r about the point a, such that whenever x--a r, f(x)...
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