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If f has a zero of order m at z 0 then for every small enough ρ > 0 and for sufficiently large k ∈ N (depending on ρ), f k has precisely m zeroes in the disk defined by |z − z 0 | < ρ, including multiplicity. Furthermore, these zeroes converge to z 0 as k → ∞. [1]
In the multiset {a, a, b}, the element a has multiplicity 2, and b has multiplicity 1. In the multiset {a, a, a, b, b, b}, a and b both have multiplicity 3. These objects are all different when viewed as multisets, although they are the same set, since they all consist of the same elements.
The fundamental theorem of algebra shows that any non-zero polynomial has a number of roots at most equal to its degree, and that the number of roots and the degree are equal when one considers the complex roots (or more generally, the roots in an algebraically closed extension) counted with their multiplicities. [3]
Theorem — The number of strictly positive roots (counting multiplicity) of is equal to the number of sign changes in the coefficients of , minus a nonnegative even number. If b 0 > 0 {\displaystyle b_{0}>0} , then we can divide the polynomial by x b 0 {\displaystyle x^{b_{0}}} , which would not change its number of strictly positive roots.
The proof of the theorem is thus a variant of the method of infinite descent [8] and relies on the repeated application of Euclidean divisions on E: let P ∈ E(Q) be a rational point on the curve, writing P as the sum 2P 1 + Q 1 where Q 1 is a fixed representant of P in E(Q)/2E(Q), the height of P 1 is about 1 / 4 of the one of P (more ...
We construct an isomorphism from X minus the fiber over 0 to E×C minus the fiber over 0 by mapping (c,s) to (c-log(s)/2πi,s 2). (The two fibers over 0 are non-isomorphic elliptic curves, so the fibration X is certainly not isomorphic to the fibration E×C over all of C.) Then the fibration X has a fiber of multiplicity 2 over 0, and otherwise ...
After blowing up at its singular point it becomes the ordinary cusp y 2 = x 3, which still has multiplicity 2. It is clear that the singularity has improved, since the degree of defining polynomial has decreased. This does not happen in general. An example where it does not is given by the isolated singularity of x 2 + y 3 z + z 3 = 0 at the ...
This proves Bézout's theorem, if the multiplicity of a common zero is defined as the multiplicity of the corresponding linear factor of the U-resultant. As for the preceding proof, the equality of this multiplicity with the definition by deformation results from the continuity of the U -resultant as a function of the coefficients of the f i ...