The Fundamental Theorem of Algebra:
If P(z) = Σnk=0 akzk where ak Є C, n ≥ 1, and an ≠ 0, then P(z0) = 0 for some z0 Є C. Descriptively, this says that any nonconstant polynomial over the complex number space, C, can be written as a product of linear factors.
Proof:
First off, we need to apply the Heine-Borel theorem to C. The Heine-Borel theorem states that if S is a closed and bounded set in an m-dimensional Euclidean space (written as Rm), then S is compact.
From above, P(z) = Σnk=0 akzk where ak Є C, n ≥ 1, and an ≠ 0. Let m = inf{|P(z)| : z Є C} where inf is the infinum, or the greatest lower bound of the set.
From the triangle inequality, |P(reit)| ≥ rn(|an| - r-1|an-1| - … - r-n|a0|),
so limr --> ∞ |P(reit)| = ∞. Therefore there is a real number R that |P(reit)| > m + 1 whenever r > R.
If S = {reit : r ≤ R}, then S is compact in C, by the Heine-Borel Theorem; and let m = inf{|P(z)| : z Є S}. |P| is a continuous and real-valued function in S, so, using the result from another proof not done here, it has a minimum value on S; i.e., there is a value for z0 Є S that makes |P(z0)| = m. So, if m = 0 then the theorem is proved.
We're going to show that m = 0 by proving that m can't equal anything else, and since we know m exists, it has no choice but to be zero. So, suppose m ≠ 0 and let Q(z) = P(z + z0)/P(z0), z Є C.
Q is therefore a polynomial with degree n and |Q(z)| ≥ 1 for all z Є C.
Q(0) = 1 so Q(z) can be expressed via P's series as:
Q(z) = 1 + bkzk + … + bnzn where k is the smallest positive integer ≤ n such that bk ≠ 0.
Since |-|bk|/bk| = 1, there exists a t0 Є [0, 2π/k] such that eikt0 = -|bk|/bk.
Then Q(reit0) = 1 + bkrkeikt0 + bk+1rk+1ei(k+1)t0 + … + bnrneint0
= 1 - rk|bk| + bk+1rk+1ei(k+1)t0 + … + bnrneint0.
So, if rk|bk| < 1 then |Q(reit0)| ≤ 1 - rk(|bk| - r|bk+1| - … - rn-k|bn|).
That means that if we pick a small enough r, we can make |Q(reit0)| ≤ 1 which contradicts the statement above that |Q(z)| ≥ 1 for all z Є C. Therefore m ≠ 0 doesn't hold and P(z0) = 0
Q.E.D.
Another proofSuppose P has no zeroes. Then we can define the function f(z) = 1 / P(z), and f is analytic. By the proof above, P(z) tends to infinity as z tends to infinity; hence f(z) tends to 0 as z tends to infinity. So there is a disc S such that f, restricted to the outside of S, is bounded. Also by the proof above, f is bounded inside the disc as well; therefore f is bounded. Now we apply a theorem called Liouville's Theorem, which says that any analytic function which is defined on all of C and is bounded must be a constant. So f is a constant; therefore P is constant. But we were assuming that P is not constant, so this is a contradiction.(To prove Liouville's Theorem: Suppose M is a bound for the function f, i.e. |f(z)| < M for all z. Suppose a and b are complex numbers, and we want to show f(a) = f(b). Use the theorem that f(a) = integral of f(z)/(z-a) / (2 * pi * i) around the circle of radius R and centre 0. Then, if R is sufficiently large:
|f(b) - f(a)|
= | integral, around circle, of (f(z) * (1/(z-b) - 1/(z-a))) | / (2*pi)
= | integral around circle of (f(z) * (b-a) / ((z-a)(z-b)) ) | / (2*pi)
<= (M * |b-a| / ((R-|a|)(R-|b|)) ) * (2*pi*R) / (2*pi)
The last line uses the formula |integral| <= |pathlength| * |maximum value|. Then we get |f(b) - f(a)| <= M * |b-a| * R / ((R-|a|)(R-|b|)). Letting R tend to infinity, we can prove that |f(b)-f(a)| is as small as we like; therefore f(a) = f(b).
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The Liouville theorem states that every bounded entire function must be constant and the consequences of which are that it proves the fundamental proof of Algebra.
The fundamental theorem of algebra was proved by Carl Friedrich Gauss in 1799. His proof demonstrated that every polynomial equation with complex coefficients has at least one complex root. This theorem laid the foundation for the study of complex analysis and was a significant contribution to mathematics.
Parts of formal proof of theorem?
the best mathematician at algebra are Pythagoras because of him, there is pythagorean theorem
He proved the "fundamental theorem of algebra" and developed a method of minimizing statistical error called "the method of least squares" which is still used today.
look in google if not there, look in wikipedia. fundamental theorem of algebra and their proofs
The Liouville theorem states that every bounded entire function must be constant and the consequences of which are that it proves the fundamental proof of Algebra.
Algebra is used for mathematics
Carl Friedrich Gauss...
The fundamental theorem of algebra was proved by Carl Friedrich Gauss in 1799. His proof demonstrated that every polynomial equation with complex coefficients has at least one complex root. This theorem laid the foundation for the study of complex analysis and was a significant contribution to mathematics.
Parts of formal proof of theorem?
the best mathematician at algebra are Pythagoras because of him, there is pythagorean theorem
He proved the "fundamental theorem of algebra" and developed a method of minimizing statistical error called "the method of least squares" which is still used today.
The Fundamental theorem of arithmetic states that every naturalnumber is either prime or can be uniquely written as a productof primes.
No. A corollary goes a little bit further than a theorem and, while most of the proof is based on the theorem, the extra bit needs additional proof.
Theory_of_BPT_theorem
Although the Pythagorean theorem (sums of square of a right angled triangle) is called a theorem it has many mathematical proofs (including the recent proof of Fermats last theorem which tangentially also prooves Pythagorean theorem). In fact Pythagorean theorem is an 'axiom', a kind of 'super law'. It doesn't matter if anyone does oppose it, it is one of the few fundamental truths of the universe.