By Jean-Etienne Montucla
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Extra resources for Histoire des mathematiques... par M. Montucla FRENCH
These districts are by boundary lines along which Q vanishes. To while in others marked some so that in districts it is aid the eye, the positive districts are shaded. An arc (2 7i, 2/4 1) of the circle, along which + Q is positive, a positive district. This district lies partly inside and the outside circle. Designate by / the part of it that partly Several cases may arise. The area / may terminate is inside. lies in + 1) inside, as does (2, 12, 3), in which case (2 h } 2 /? arc of the circle on its boundary.
10, p. 19. proofs of this theorem, the fourth (1849) being a simplification of the first It is (1799). The one given here is in substance Gauss's proof of 1849. geometrical in character, and is open to the objection raised in the foot-note . of 22, , ; SOME ELEMENTARY 1'UOPEUTIES OF EQI/ATIONS "27 polynomial f(z) of the wtli degree has real coefficients only. We wish to prove that there exists always at least one value of z, either real or complex, which causes the polynomial J\z) to vanish. Let = x -f iy, then, by z 22, the variable' represents points in a plane, and the function f(z) has a definite value at each 8, we may write J\z) = I'+iQ, point in the plane.
Evidently, between each pair of distinct successive real roots there must be at least one maximum or minimum value of f(x). But each maximum or minimum point represents a value of x which Kolle's is a root of the equation Theorem f '(#) () ( 44). Hence is proved. From the examination of the figure we see that two successive roots of the derived function may not comprise between them 0, as in case of the roots represented by any real root of f(x) D and E\ they may comprise one distinct root, as in case of the roots at ^1 and J3, B and Cs K and F, but they can never comprise more than one root of f(x) = 0.