Mathematics

Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra states that every non-constant polynomial equation with complex coefficients has at least one complex root. In other words, every polynomial of degree n has n roots, counting multiplicities. This theorem is a fundamental result in algebra and has many applications in various fields of mathematics.

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6 Key excerpts on "Fundamental Theorem of Algebra"

  • Book cover image for: Mathematical Proofs and Theorems
    ______________________________ WORLD TECHNOLOGIES ______________________________ Chapter 12 Fundamental Theorem of Algebra and Fundamental Theorem of Calculus Fundamental Theorem of Algebra In mathematics, the Fundamental Theorem of Algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. Equivalently, the field of complex numbers is algebraically closed. Sometimes, this theorem is stated as: every non-zero single-variable polynomial with complex coefficients has exactly as many complex roots as its degree, if each root is counted up to its multiplicity. Although this at first appears to be a stronger statement, it is a direct consequence of the other form of the theorem, through the use of successive polynomial division by linear factors. In spite of its name, there is no purely algebraic proof of the theorem, since any proof must use the completeness of the reals (or some other equivalent formulation of completeness), which is not an algebraic concept. Additionally, it is not fundamental for modern algebra; its name was given at a time in which algebra was mainly about solving polynomial equations with real or complex coefficients. History Peter Rothe (Petrus Roth), in his book Arithmetica Philosophica (published in 1608), wrote that a polynomial equation of degree n (with real coefficients) may have n solutions. Albert Girard, in his book L'invention nouvelle en l'Algèbre (published in 1629), asserted that a polynomial equation of degree n has n solutions, but he did not state that they had to be real numbers. Furthermore, he added that his assertion holds “unless the equation is incomplete”, by which he meant that no coefficient is equal to 0.
  • Book cover image for: Theoretical Introduction to Mathematical Theorems, A
    ________________________ WORLD TECHNOLOGIES ________________________ Chapter- 3 Fundamental Theorem of Algebra and Fundamental Theorem of Calculus Fundamental Theorem of Algebra In mathematics, the Fundamental Theorem of Algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. Equivalently, the field of complex numbers is algebraically closed. Sometimes, this theorem is stated as: every non-zero single-variable polynomial with complex coefficients has exactly as many complex roots as its degree, if each root is counted up to its multiplicity. Although this at first appears to be a stronger statement, it is a direct consequence of the other form of the theorem, through the use of successive polynomial division by linear factors. In spite of its name, there is no purely algebraic proof of the theorem, since any proof must use the completeness of the reals (or some other equivalent formulation of com-pleteness), which is not an algebraic concept. Additionally, it is not fundamental for modern algebra; its name was given at a time in which algebra was mainly about solving polynomial equations with real or complex coefficients. History Peter Rothe (Petrus Roth), in his book Arithmetica Philosophica (published in 1608), wrote that a polynomial equation of degree n (with real coefficients) may have n solu-tions. Albert Girard, in his book L'invention nouvelle en l'Algèbre (published in 1629), asserted that a polynomial equation of degree n has n solutions, but he did not state that they had to be real numbers. Furthermore, he added that his assertion holds “unless the equation is incomplete”, by which he meant that no coefficient is equal to 0.
  • Book cover image for: Elements of Mathematics
    eBook - PDF

    Elements of Mathematics

    From Euclid to Gödel

    The mere existence of written symbolic computation made all this possible, without much conceptual input from algebra. Calculus in fact surpassed algebra, to the extent that it was able to solve the main algebraic problem of the time: proving the Fundamental Theorem of Algebra. The solution was not what had been hoped for in the sixteenth century—a formula giving the roots of any equation in terms of the coefficients—but a new kind of proof: a proof of existence . We say more about this theorem in the next section, and we give a proof in section 10.3, but the idea of an existence proof can be illustrated with the cubic equation. Any cubic polynomial, say x 3 − x − 1, corresponds to a cubic curve, in this case y = x 3 − x − 1, whose graph for real values is shown in figure 4.5. This picture confirms something that can also be seen algebraically, that the polynomial function x 3 − x − 1 has large positive values for large positive x and large negative values for large negative x . We can also see something that is more subtle to explain algebraically, that the curve is continuous , and hence that it meets the x axis somewhere , at an x value that is a root of the equation x 3 − x − 1 = 0. Algebra • 141 x y O 1 2 -2 Figure 4.5: Graph of the curve y = x 3 − x − 1. Thus the Fundamental Theorem of Algebra turns out to involve a concept from outside algebra: 3 the concept of continuous function . As we will see in chapter 6, continuity is a fundamental concept of calculus, though its importance was not properly understood until the nineteenth century. For many reasons, continuity is an advanced concept, despite its apparent simplicity in cases like the graph of y = x 3 − x − 1. One might hope, perhaps, for a proof of the Fundamental Theorem of Algebra that does not involve the continuity concept. However, there is another reason why the theorem falls short of what the Italians were hoping for.
  • Book cover image for: A MatLab® Companion to Complex Variables
    • A. David Wunsch(Author)
    • 2018(Publication Date)
    • CRC Press
      (Publisher)
    5 Polynomials, Roots, the Principle of the Argument, and Nyquist Stability 5.1  Introduction
    The question of where an equation like f(z) = 0 is satisfied in the complex plane, when f(z) is an analytic function, is a major subject in complex variable theory. This problem often arises in engineering and scientific applications. A value of z for which the equation is satisfied is said to be a “zero” of f(z). Similarly, we must pay attention to the related problem of where g(z) = 1/f(z) possesses pole singularities because of the vanishing of f(z). Some of these questions are directly resolved in MATLAB® , while others require the more sophisticated technique called the principle of the argument.
    The Nyquist stability criterion, based directly on the principle of the argument, which we discuss, is a favorite technique of engineers; it is accessed directly in a MATLAB toolbox. 5.2  The Fundamental Theorem of Algebra
    One of the great pleasures of studying functions of a complex variable is to see how easily one can prove the Fundamental Theorem of Algebra (see W section 4.6 and S problem 5.10). Here is a way to state the theorem: We have a polynomial
    P
    ( z )
    =
    a n
    z n
    +
    a
    n 1
    z
    n 1
    +
    a 0
    (5.1)
    in the variable z, where a0 , a1 , … are constants (real and/or complex); n is a positive integer; and a
    n
    ≠ 0. We say that the polynomial is of degree n, although some authors prefer the term order instead of degree. The theorem states that
    P
    ( z )
    = 0
    (5.2)
    has exactly n solutions, or roots, in the complex z plane.* Equivalently, the fundamental theorem can be stated that if z1 , z2 , … z
    n
    are the roots, then we can factor the left side of Equation 5.1 so that
    P
    ( z )
    =
    a n
    (
    z
    z 1
    )
    (
    z
    z 2
    )
    (
    z
    z n
    )
    (5.3)
    In applying the Fundamental Theorem of Algebra, we may have to count a root more than once. For example, the equation z3 + (2 − i)z2 + z(1 − 2i) − i = 0 has roots at z = i and z = −1. If you try any other values of z, you will find that the equation is not satisfied. We seem to have two roots, not the expected n = 3 from our theorem. But if we factor the left side, we get (z + 1)2 (zi) = z3 + (2 − i)z2 + z(1 − 2i) − i = 0. The root at −1 is of multiplicity two—this follows from the exponent in (z + 1)2
  • Book cover image for: Numbers and Proofs
    Do note, however, that there is no trichotomy law, as there is for the real numbers. If there were, we would (since i φ 0) have either i > 0 or i < 0, i.e. — i > 0. But then one could, by squaring up, presumably 'prove' that i 2 > 0, i.e. — 1 > 0, or ( —i) 2 > 0, i.e. -1 > 0. The Fundamental Theorem of Algebra That the complex numbers are quite remarkable is emphasized by the Fundamental Theorem of Algebra. Let f(z) = c n z n + ο η _ γ ζ η -1 + ... + c x z + c 0 be a polynomial (in z) with complex coefficients c n ,c n _ l ,...,c 0 . Then,/(z) factorizes into a product c n (z — a x )(z — a 2 )...(z — a M ) of n linear factors (i.e. factors of degree 1) for suitable complex numbers a l5 a 2 ,...,a„. In particular, a 1? a 2 ,..., cc n (which may not all be distinct) are the n roots of f(z). Example 2 (a) z 4 + 2z 2 + 1 = (z - i)(z - i)(z + i)(z + i) (b) z 3 - 7z 2 + 25z - 39 = (z - 3)(z - {2 - 3i})(z - {2 + 3i}) (c) z 3 - (2 + 2i)z 2 + (-2 + 3i)z - (5 + 5i) = (z - {3 + i})(z - {-1 + 2i})(z + i) The Fundamental Theorem of Algebra is too tricky to prove here. The first adequate proof was given by Gauss when aged about 22, Euler and others having previously failed in the attempt. Note that the corresponding result for real polynomials is false: polynomials with real coefficients need not factorize into products of linear factors x — (Xi with real numbers a¿. Indeed, there are no real numbers α ΐ5 α 2 such that the polynomial x 2 + 1 factorizes as (x — a x )(x — a 2 ). Put another way: to 162 NUMBERS AND PROOFS factorize a real polynomial into linear factors we may have to call upon complex numbers. Therefore, to factorize a polynomial with complex coefficients, it would not seem unreasonable to expect to have to call upon some new 'supercomplex' numbers. The remarkable fact is that we don't need to 'invent' any new numbers to allow the factorization to go ahead: the complex numbers we already have to hand will suffice.
  • Book cover image for: Topics In Polynomials: Extremal Problems, Inequalities, Zeros
    eBook - PDF
    • Gradimir V Milovanovic, Themistocles M Rassias, D S Mitrinovic(Authors)
    • 1994(Publication Date)
    • World Scientific
      (Publisher)
    C H A P T E R 1 General Concepts of Algebraic Polynomials 1.1. P O L Y N O M I A L S A N D E Q U A T I O N S 1.1.1. Preliminaries Let F be a field. The ring of polynomials in x over F, denoted by F[i], is defined to be the set of all formal expressions P(x) = aa + atx + + a n x n , where the a;, the coefficients of the polynomial P in x, are elements in F. If a n ^ 0, then the degree of P, denoted by dg P, is n. Therefore, the degree of a polynomial P is the highest power of x that occurs in the expression for P(x) with a non-zero coefficient. In F[i] we define equality, sum, and product of two polynomials as follows. Given two polynomials P(x) = a 0 + a 1 x + --- +a n x n and Q(x) = b 0 + b i x + ---+b m x m , then the polynomials are equal if, and only if, a; = 6; for all i > 0, that is when their corresponding coefficients are equal. We define the sum of P and Q by {P + Q)(x) = P(x) + Q(x) = c 0 + ax + • • • + c r x r , l 2 GENERAL CONCEPTS OF POLYNOMIALS where for each t, C; = a; + (>;, and the product of P and Q by (PQ)(x) - P(x)<30r) = co + cia; + • + c,*', where c,- = a,6 0 + <*i-i&i + h ai&i_i + a 0 b„ for every i . Then, one can show that F[x] is a commutative ring with unit. It is easy to prove that if P(x) and Q(x) are non-zero elements of V[x], then dg(PQ) = dgP + dgt?. Also, if P{x),Q{x) e F[x] and P(x) + Q(x) ± 0, then dgtP + g j ^ m a x I d g ^ d g Q } . We do not assign a degree to 0{x) = Ox + 0 x n _ 1 + • • • + Ox + 0. The polynomials of degree 0 are called the constants (these are the elements of the field F). Let k ,... , k m be non-negative integers and fc = ( * i , . . . , A m ) , = + + x k = »f» • * * » .
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