How useful is Abel Ruffini's theorem

Zeros of real polynomials

Transcript

1 introductory seminar for teaching

2 Overview Description of possible solutions for solving real polynomials up to degree 4 Brief explanation of the problem of determining roots in polynomials with degree> 4 = Abel-Ruffini's theorem; Galois' idea Some hints how the position of zeros can be estimated and approximate determination of zeros in practice

3 Overview Description of possible solutions for solving real polynomials up to degree 4 Brief explanation of the problem of determining the roots of polynomials with degree> 4 = Abel-Ruffini's theorem; Galois' idea Some hints how the position of zeros can be estimated and approximate determination of zeros in practice

4 Overview Description of possible solutions for solving real polynomials up to degree 4 Brief explanation of the problem of determining the roots of polynomials with degree> 4 = Abel-Ruffini's theorem; Galois' idea Some hints how the position of zeros can be estimated and approximate determination of zeros in practice

5 1 Introduction 2 Basic terms Definition of polynomials Fundamental theorem of algebra Definition of zero Existence of zeros 3 Polynomials with n 4 Linear polynomials Quadratic polynomials Cubic polynomials Polynomials with n = 4 4 NS with bel. Degree n Abel-Ruffini theorem Galois idea 5 zeros in practice Bisection Newton method Secant method Horner's scheme

6 What are real polynomials? In linear algebra, polynomials are defined as finite sequences (i.e. all but a finite number of a i are equal to zero). For our following considerations, however, suffices: Definition p: CC is called a polynomial of degree n N 0 if a 0, ..., an C: an 0 x C: f (x) = naixi = anxn + an 1 xna 2 x 2 + a 1 x + a 0 i = 0

7 What are real polynomials? In linear algebra, polynomials are defined as finite sequences (i.e. all but a finite number of a i are equal to zero). For our following considerations, however, suffices: Definition p: CC is called a polynomial of degree n N 0 if a 0, ..., an C: an 0 x C: f (x) = naixi = anxn + an 1 xna 2 x 2 + a 1 x + a 0 i = 0

8 What are real polynomials? f (x) = n a i x i = a n x n + a n 1 x n a 2 x 2 + a 1 x + a 0 i = 0 where - the highest exponent n determines the polynomial degree of p. - the numbers a 0, ..., a n are called the coefficients of p. - we call p normalized if the leading coefficient a n = 1. Polynomials are real if the coefficients are a 0, ..., a n R. Danger! Polynomials with real coefficients can also have complex zeros.

9 What are real polynomials? f (x) = n a i x i = a n x n + a n 1 x n a 2 x 2 + a 1 x + a 0 i = 0 where - the highest exponent n determines the polynomial degree of p. - the numbers a 0, ..., a n are called the coefficients of p. - we call p normalized if the leading coefficient a n = 1. Polynomials are real if the coefficients are a 0, ..., a n R. Danger! Polynomials with real coefficients can also have complex zeros.

10 What are real polynomials? f (x) = n a i x i = a n x n + a n 1 x n a 2 x 2 + a 1 x + a 0 i = 0 where - the highest exponent n determines the polynomial degree of p. - the numbers a 0, ..., a n are called the coefficients of p. - we call p normalized if the leading coefficient a n = 1. Polynomials are real if the coefficients are a 0, ..., a n R. Danger! Polynomials with real coefficients can also have complex zeros.

11 What are real polynomials? f (x) = n a i x i = a n x n + a n 1 x n a 2 x 2 + a 1 x + a 0 i = 0 where - the highest exponent n determines the polynomial degree of p. - the numbers a 0, ..., a n are called the coefficients of p. - we call p normalized if the leading coefficient a n = 1. Polynomials are real if the coefficients are a 0, ..., a n R. Danger! Polynomials with real coefficients can also have complex zeros.

12 What are real polynomials? f (x) = n a i x i = a n x n + a n 1 x n a 2 x 2 + a 1 x + a 0 i = 0 where - the highest exponent n determines the polynomial degree of p. - the numbers a 0, ..., a n are called the coefficients of p. - we call p normalized if the leading coefficient a n = 1. Polynomials are real if the coefficients are a 0, ..., a n R. Danger! Polynomials with real coefficients can also have complex zeros.

13 Fundamental Law of Algebra Main Law of Algebra: Let p: CC: If p (x) = nk = 0 akxk = anxn + an 1 xna 2 x 2 + a 1 x + a 0 is a complex polynomial of degree n, then there is unique certain numbers x 1, ..., xn (the zeros of the polynomial), so that p (x) = an (xxn) (xxn 1) ... (xx 2) (xx 1) applies. The total number of zeros, if they are counted with the correct multiplicity, is therefore equal to the degree of the polynomial.

14 Fundamental Law of Algebra Main Law of Algebra: Let p: CC: If p (x) = nk = 0 akxk = anxn + an 1 xna 2 x 2 + a 1 x + a 0 is a complex polynomial of degree n, then there is unique certain numbers x 1, ..., xn (the zeros of the polynomial), so that p (x) = an (xxn) (xxn 1) ... (xx 2) (xx 1) applies. The total number of zeros, if they are counted with the correct multiplicity, is therefore equal to the degree of the polynomial.

15 What is a zero? Definition An element x 0 C of a polynomial p: C C is called the zero of p if p (x 0) = 0. According to the main law of algebra, the numbers x 1, ..., x n are the zeros of a polynomial. One speaks of a simple zero if x i occurs only once in the linear factors (x x i). There is a multiple (k-fold) zero in x i C if the linear factor occurs (x x i) k times.

16 What is a zero? Definition An element x 0 C of a polynomial p: C C is called the zero of p if p (x 0) = 0. According to the main law of algebra, the numbers x 1, ..., x n are the zeros of a polynomial. One speaks of a simple zero if x i occurs only once in the linear factors (x x i). There is a multiple (k-fold) zero in x i C if the linear factor occurs (x x i) k times.

17 What is a zero? In the case of polynomials with real coefficients, genuinely complex zeros always appear as pairs of complex conjugate numbers. That is, if λ = x + iy is a zero, so is λ = x iy. Polynomials of even or odd degrees always have an even or an odd number of real zeros if one counts every zero according to its multiplicity. Polynomials of odd degree over the real numbers always have at least one real zero.

18 What is a zero? In the case of polynomials with real coefficients, genuinely complex zeros always appear as pairs of complex conjugate numbers. That is, if λ = x + iy is a zero, so is λ = x iy. Polynomials of even or odd degrees always have an even or an odd number of real zeros if one counts every zero according to its multiplicity. Polynomials of odd degree over the real numbers always have at least one real zero.

19 What is a zero? In the case of polynomials with real coefficients, genuinely complex zeros always appear as pairs of complex conjugate numbers. That is, if λ = x + iy is a zero, so is λ = x iy. Polynomials of even or odd degrees always have an even or an odd number of real zeros if one counts every zero according to its multiplicity. Polynomials of odd degree over the real numbers always have at least one real zero.

20 Existence of zeros The intermediate value theorem can be used to estimate whether a zero exists between two positions a and b of a continuous function: Bolzano's intermediate value theorem: Let p: R [a, b] R be continuous and γ R with min {f (x ): axb} γ max {f (x): axb}. Then there is (at least) one ˆx [a, b] with f (ˆx) = γ. If f (a) and f (b) have different signs, the intermediate value theorem guarantees the existence of a zero of f in the open interval (a, b).

21 Polynomials with degree n 4 The zeros of polynomials of the first, second, third and fourth degree can be calculated explicitly with formulas. In contrast, higher-degree polynomials can only be factored exactly in special cases. The simple cases will now be dealt with here first.

22 Solving linear polynomials Linear polynomials are polynomials of the form p (x) = ax + b. It should be p (x) = 0. This results in the solution for x: {L: = x R x = b}, a, b R a Geometric description of linear polynomials at p: R R: straight line that intersects the x-axis at ONE point x Figure: 3x - 6

23 Solution of linear polynomials Linear polynomials are polynomials of the form p (x) = ax + b. It should be p (x) = 0. This results in the solution for x: {L: = x R x = b}, a, b R a Geometric description of linear polynomials at p: R R: straight line that intersects the x-axis at ONE point x Figure: 3x - 6

24 Quadratic Polynomials To solve a quadratic equation (p (x) = ax 2 + bx + c = 0) the following formula (midnight formula) can be used: x 1,2 = b ± b 2 4ac 2a For the special case of a normalized quadratic polynomial (ie a = 1 p (x) = x 2 + px + q) the so-called pq formula results: x 1,2 = p 2 ± (p 2) 2 ppq = 2 ± 2 4 q Every quadratic equation can be brought into the normal form by suitable equivalence transformations p = ba, q = approx

25 Quadratic Polynomials To solve a quadratic equation (p (x) = ax 2 + bx + c = 0) the following formula (midnight formula) can be used: x 1,2 = b ± b 2 4ac 2a For the special case of a normalized quadratic polynomial (ie a = 1 p (x) = x 2 + px + q) the so-called pq formula results: x 1,2 = p 2 ± (p 2) 2 ppq = 2 ± 2 4 q Every quadratic equation can be brought into the normal form by suitable equivalence transformations p = ba, q = approx

26 Quadratic Polynomials To solve a quadratic equation (p (x) = ax 2 + bx + c = 0) the following formula (midnight formula) can be used: x 1,2 = b ± b 2 4ac 2a For the special case of a normalized quadratic polynomial (ie a = 1 p (x) = x 2 + px + q) the so-called pq formula results: x 1,2 = p 2 ± (p 2) 2 ppq = 2 ± 2 4 q Every quadratic equation can be brought into the normal form by suitable equivalence transformations p = ba, q = approx

27 Number of solutions to quadratic equations For an equation with real coefficients, the expression under the root (the discriminant D = b 2 4ac) determines how many real-valued solutions the equation has. There are three cases: 1 D> 0: The parabola (A) has two points of intersection with the x-axis, so there are two different real zeros x 1 and x 2, 2 D = 0: The parabola (B) has one point of contact with the x-axis, it is x 1 = x 2 (double zero). The point of contact is also a maximum (a <0) or minimum (a> 0) 3 D <0, the parabola (C) has no intersection with the x-axis, there is no real solution of the quadratic equation, but there is complex yx 2 3 Legend ABC

28 Number of solutions to quadratic equations For an equation with real coefficients, the expression under the root (the discriminant D = b 2 4ac) determines how many real-valued solutions the equation has. A distinction is made between three cases: 1 D> 0: The parabola (A) has two points of intersection with the x-axis, so there are two different real zeros x 1 and x 2, 2 D = 0: The parabola (B) has one point of contact with the x-axis, it is x 1 = x 2 (double zero). The point of contact is also a maximum (a <0) or minimum (a> 0) 3 D <0, the parabola (C) has no intersection with the x-axis, there is no real solution of the quadratic equation, but there is complex yx 2 3 Legend ABC

29 Number of solutions to quadratic equations For an equation with real coefficients, the expression under the root (the discriminant D = b 2 4ac) determines how many real-valued solutions the equation has. A distinction is made between three cases: 1 D> 0: The parabola (A) has two points of intersection with the x-axis, so there are two different real zeros x 1 and x 2, 2 D = 0: The parabola (B) has one point of contact with the x-axis, it is x 1 = x 2 (double zero). The point of contact is also a maximum (a <0) or minimum (a> 0) 3 D <0, the parabola (C) has no intersection with the x-axis, there is no real solution of the quadratic equation, but there is complex yx 2 3 Legend ABC

30 Derivation of the formulas To derive the solution formulas, the quadratic addition is used: x 2 + p x + q = 0 q x 2 + p x = q + `p 2 square of the half of the linear term add. 2 x 2 + p + `p 2` = p 2 q () with binomial. Summarize the formula 2 2 `x + p 2` = p 2 q take the root 2 2 q` x + p = ± p 2 q 2 2 You get the midnight formula from here by using p = b and q = caa: qx 1,2 = p ± p 2 qp, q replace with so 2 q 4 = 1 b ± 1 b 2 c 1 2 aq 4 a 2 a 2a exclude = 1 (b) ± 1 (b 2a 4a 2 4ac) 2 up write a fraction line b ± = b 2 4ac 2a

31 Derivation of the formulas To derive the solution formulas, the quadratic addition is used: x 2 + p x + q = 0 q x 2 + p x = q + `p 2 square of the half of the linear term add. 2 x 2 + p + `p 2` = p 2 q () with binomial. Summarize the formula 2 2 `x + p 2` = p 2 q take the root 2 2 q` x + p = ± p 2 q 2 2 You get the midnight formula from here by using p = b and q = caa: qx 1,2 = p ± p 2 qp, q replace with so 2 q 4 = 1 b ± 1 b 2 c 1 2 aq 4 a 2 a 2a exclude = 1 (b) ± 1 (b 2a 4a 2 4ac) 2 up write a fraction line b ± = b 2 4ac 2a

32 Derivation of the formulas To derive the solution formulas, the quadratic addition is used: x 2 + p x + q = 0 q x 2 + p x = q + `p 2 square of the half of the linear term add. 2 x 2 + p + `p 2` = p 2 q () with binomial. Summarize the formula 2 2 `x + p 2` = p 2 q take the root 2 2 q` x + p = ± p 2 q 2 2 You get the midnight formula from here by using p = b and q = caa: qx 1,2 = p ± p 2 qp, q replace with so 2 q 4 = 1 b ± 1 b 2 c 1 2 aq 4 a 2 a 2a exclude = 1 (b) ± 1 (b 2a 4a 2 4ac) 2 up write a fraction line b ± = b 2 4ac 2a

33 Derivation of the formulas To derive the solution formulas one uses the quadratic extension: x 2 + p x + q = 0 q x 2 + p x = q + `p 2 square of the half of the linear term add. 2 x 2 + p + `p 2` = p 2 q () with binomial. Summarize the formula 2 2 `x + p 2` = p 2 q take the root 2 2 q` x + p = ± p 2 q 2 2 You get the midnight formula from here by using p = b and q = caa: qx 1,2 = p ± p 2 qp, q replace with so 2 q 4 = 1 b ± 1 b 2 c 1 2 aq 4 a 2 a 2a exclude = 1 (b) ± 1 (b 2a 4a 2 4ac) 2 up write a fraction line b ± = b 2 4ac 2a

34 Geometric solution with Vieta's theorem Quadratic equations have an important property: x 1 + x 2 = p and x 1 x 2 = q This is called the Vieta group of sentences after the Latinized form of François Viète, a French. Mathematician (). This property can be used to geometrically determine the solution of quadratic equations. For the case 0 q (p 2) the following applies with the aid of the height theorem: Figure: Christiane solutions Sutter according to zeros Theorem of real vonpolynomials Vieta

35 Geometric solution with Vieta's theorem Quadratic equations have an important property: x 1 + x 2 = p and x 1 x 2 = q This is called the Vieta group of sentences after the Latinized form of François Viète, a French. Mathematician (). This property can be used to geometrically determine the solution of quadratic equations. For the case 0 q (p 2) the following applies with the aid of the height theorem: Figure: Christiane solutions Sutter according to zeros Theorem of real vonpolynomials Vieta

36 Geometric solution with Vieta's theorem Quadratic equations have an important property: x 1 + x 2 = p and x 1 x 2 = q This is called the Vieta group of sentences after the Latinized form of François Viète, a French. Mathematician (). This property can be used to geometrically determine the solution of quadratic equations. For the case 0 q (p 2) the following applies with the aid of the height theorem: Figure: Christiane solutions Sutter according to zeros Theorem of real vonpolynomials Vieta

37 Cubic Polynomials Cubic polynomials are third-degree polynomials. Geometrically, the real variant of the cubic equation describes a cubic parabola in the x-y plane, which always runs from ... + (a> 0) or from + ... (a <0). Therefore there must always be at least one point of intersection with the x-axis, i.e. at least one real zero. Solution formulas to cubic polynomials were first published in 1545 by Girolamo Cardano () in his book Ars magna.

38 Cubic Polynomials Cubic polynomials are third-degree polynomials. Geometrically, the real variant of the cubic equation describes a cubic parabola in the x-y plane, which always runs from ... + (a> 0) or from + ... (a <0). Therefore there must always be at least one point of intersection with the x-axis, i.e. at least one real zero. Solution formulas to cubic polynomials were first published in 1545 by Girolamo Cardano () in his book Ars magna.

39 Cubic Polynomials Cubic polynomials are third-degree polynomials. Geometrically, the real variant of the cubic equation describes a cubic parabola in the x-y plane, which always runs from ... + (a> 0) or from + ... (a <0). Therefore there must always be at least one point of intersection with the x-axis, i.e. at least one real zero. Solution formulas to cubic polynomials were first published in 1545 by Girolamo Cardano () in his book Ars magna.

40 Solution of reduced cubic polynomials The solution of the cubic equation is based on the cubic binomial formula (u + v) 3 = 3uv (u + v) + (u 3 + v 3), which Cardano was able to derive using geometric means. The equation can be interpreted as a cubic equation x 3 + px + q = 0 if the following conditions are met: u + v 3uv u 3 + v 3 = x = p = q

41 Solution of reduced cubic polynomials So that the above equation can be solved, suitable quantities u and v have to be found. Since we know both the product and the sum of the two quantities u 3 and v 3, we can calculate them using Vieta's theorem as the two solutions to the quadratic equation. For clarification: (p) 3 w 2 + qw = 0 3 x 1 + x 2 = -pu 3 + v 3 = -qx 1 x 2 = qu 3 v 3 = (-p / 3) 3 x 2 + px + q = 0 0 = w 2 + qw - (p / 3) 3

42 Solution of reduced cubic polynomials So that the above equation can be solved, suitable quantities u and v have to be found. Since we know both the product and the sum of the two quantities u 3 and v 3, we can calculate them using Vieta's theorem as the two solutions to the quadratic equation. For clarification: (p) 3 w 2 + qw = 0 3 x 1 + x 2 = -pu 3 + v 3 = -qx 1 x 2 = qu 3 v 3 = (-p / 3) 3 x 2 + px + q = 0 0 = w 2 + qw - (p / 3) 3

43 Solution of reduced cubic polynomials So that the above equation can be solved, suitable quantities u and v have to be found. Since we know both the product and the sum of the two quantities u 3 and v 3, we can calculate them using Vieta's theorem as the two solutions to the quadratic equation. For clarification: (p) 3 w 2 + qw = 0 3 x 1 + x 2 = -pu 3 + v 3 = -qx 1 x 2 = qu 3 v 3 = (-p / 3) 3 x 2 + px + q = 0 0 = w 2 + qw - (p / 3) 3

44 Solution of reduced cubic polynomials Using the pq formula, this leads to the two values: u = 3 q (q) 2 (p) and v = q (q) 2 (p) For the sought solution x = u + v of the cubic Equation x 3 + px + q = 0 one obtains the so-called Cardano formula x = 3 q (q) 2 (p) q (q) 2 (p)

45 Solution of reduced cubic polynomials Using the pq formula, this leads to the two values: u = 3 q (q) 2 (p) and v = q (q) 2 (p) For the sought solution x = u + v of the cubic Equation x 3 + px + q = 0 one obtains the so-called Cardano formula x = 3 q (q) 2 (p) q (q) 2 (p)

46 Cardano's solution for the general cubic case But how do you solve the general case x 3 + ax 2 + bx + c = 0? Cardano solved the problem by using a generally applicable procedure to eliminate the square limb. First, the summand a / 3 is added to the sought solution x: x 3 + ax 2 = x 3 + ax 2 + (x a2 3 + a3 27 a2 3 x a3 27 = + a) 3 ax 1 27 a3 = (x + a) 3 a 2 (x + a) a3 Then substitute the unknowns x within the entire equation by x = ya 3

47 Cardano's solution for the general cubic case But how do you solve the general case x 3 + ax 2 + bx + c = 0? Cardano solved the problem by using a generally applicable procedure to eliminate the square limb. First, the summand a / 3 is added to the sought solution x: x 3 + ax 2 = x 3 + ax 2 + (x a2 3 + a3 27 a2 3 x a3 27 = + a) 3 ax 1 27 a3 = (x + a) 3 a 2 (x + a) a3 Then substitute the unknowns x within the entire equation by x = ya 3

48 Cardano's solution for the general cubic case But how do you solve the general case x 3 + ax 2 + bx + c = 0? Cardano solved the problem by using a generally applicable procedure to eliminate the square limb. First, the summand a / 3 is added to the sought solution x: x 3 + ax 2 = x 3 + ax 2 + (x a2 3 + a3 27 a2 3 x a3 27 = + a) 3 ax 1 27 a3 = (x + a) 3 a 2 (x + a) a3 Then substitute the unknowns x within the entire equation by x = ya 3

49 Cardano's solution for the general cubic case But how do you solve the general case x 3 + ax 2 + bx + c = 0? Cardano solved the problem by using a generally applicable procedure to eliminate the square limb. First, the summand a / 3 is added to the sought solution x: x 3 + ax 2 = x 3 + ax 2 + (x a2 3 + a3 27 a2 3 x a3 27 = + a) 3 ax 1 27 a3 = (x + a) 3 a 2 (x + a) a3 Then substitute the unknowns x within the entire equation by x = ya 3

50 Cardano's solution to the general cubic case After sorting the terms according to powers of y, one obtains the identity x 3 + ax 2 + bx + c = y 3 + py + q with p = 1 3 a2 + b and q = 2 27 a3 1 3 ab + c The solution for y is obtained using Cardano's formula. The original equation can then be solved with the transformation x = y - a / 3. In the general case: 2 x = 3 27 a3 1 3 from + c (2 27 a3 1 3 from + c 2) a3 1 3 from + c 2 (27 a3 1 3 from + c) 2 ((1 3 a2 + b 3 3 a2 + b 3) 3) 3 a 3

51 Cardano's solution to the general cubic case After having sorted the terms according to powers of y, one obtains the identity x 3 + ax 2 + bx + c = y 3 + py + q with p = 1 3 a2 + b and q = 2 27 a3 1 3 ab + c The solution for y is obtained using Cardano's formula. The original equation can then be solved with the transformation x = y - a / 3. In the general case: 2 x = 3 27 a3 1 3 from + c (2 27 a3 1 3 from + c 2) a3 1 3 from + c 2 (27 a3 1 3 from + c) 2 ((1 3 a2 + b 3 3 a2 + b 3) 3) 3 a 3

52 Cardano's solution to the general cubic case After having sorted the terms according to powers of y, one obtains the identity x 3 + ax 2 + bx + c = y 3 + py + q with p = 1 3 a2 + b and q = 2 27 a3 1 3 ab + c The solution for y is obtained using Cardano's formula. The original equation can then be solved with the transformation x = y - a / 3. In the general case: 2 x = 3 27 a3 1 3 from + c (2 27 a3 1 3 from + c 2) a3 1 3 from + c 2 (27 a3 1 3 from + c) 2 ((1 3 a2 + b 3 3 a2 + b 3) 3) 3 a 3

53 Type of solution of cubic equations Depending on the sign of the discriminant D = 4p q 2, the cubic polynomial x 3 + px + q for D> 0 has one real and two complex zeros (A) D <0 three real zeros (B). D = 0 a triple real zero if p = q = 0 (C) a double and a single real zero if 4p 3 = 27p 2 0 (D)

54 Kind of solutions for cubic equations Depending on the sign of the discriminant D = 4p q 2, the cubic polynomial x 3 + px + q for D> 0 has one real and two complex zeros (A) D <0 three real zeros (B). D = 0 a triple real zero if p = q = 0 (C) a double and a single real zero if 4p 3 = 27p 2 0 (D)

55 Kind of solutions to cubic equations Depending on the sign of the discriminant D = 4p q 2, the cubic polynomial x 3 + px + q for D> 0 has one real and two complex zeros (A) D <0 three real zeros (B). D = 0 a triple real zero if p = q = 0 (C) a double and a single real zero if 4p 3 = 27p 2 0 (D)

56 Kind of solutions to cubic equations (A) D> 0 one real and 2 compl. NS; (B) D <0 three real NS; (C) D = 0 a 3-fold real NS if p = q = 0 or (D) a 2-fold and a 1-fold real NS if 4p 3 = 27q 2 0,

57 Casus irreducibilis A special case is D <0: When determining the three real solutions with the above formula, negative roots must be taken into account. That is why this case is called casus irreducibilis. When Cardano carried out this calculation, it was, so to speak, the birth of complex numbers.

58 Quartic and biquadratic equations Quartic equations: A polynomial equation of the 4th degree (also called quartic equation) has the form: Ax 4 + Bx 3 + Cx 2 + Dx + E = 0. Biquadratic equations: Is B = 0 and D = 0, then the equation can be reduced to a quadratic equation by substitution. This special form is sometimes referred to as the biquadratic equation in textbooks. But: You often find this designation for the general form of the equation of the 4th degree! This is due to the fact that with the help of the fundamental theorem of algebra all quartic equations can be written as the product of two quadratic terms.

59 Quartic and biquadratic equations Quartic equations: A polynomial equation of the fourth degree (also called quartic equation) has the form: Ax 4 + Bx 3 + Cx 2 + Dx + E = 0. Biquadratic equations: Is B = 0 and D = 0, then the equation can be reduced to a quadratic equation by substitution. This special form is sometimes referred to as the biquadratic equation in textbooks. But: You often find this designation for the general form of the equation of the 4th degree! This is due to the fact that with the help of the fundamental theorem of algebra all quartic equations can be written as the product of two quadratic terms.

60 Quartic and biquadratic equations Quartic equations: A polynomial equation of the 4th degree (also called quartic equation) has the form: Ax 4 + Bx 3 + Cx 2 + Dx + E = 0. Biquadratic equations: Is B = 0 and D = 0, then the equation can be reduced to a quadratic equation by substitution. This special form is sometimes referred to as the biquadratic equation in textbooks. But: You often find this designation for the general form of the equation of the 4th degree! This is due to the fact that with the help of the fundamental theorem of algebra all quartic equations can be written as the product of two quadratic terms.

61 Solution of the reduced form quart. Equations In addition to third-degree equations, Cardano also published a general formula for solving fourth-degree equations in his Ars magna. He made use of the knowledge of his student Ludovico Ferrari (). Quartic equations of the form x 4 + px 2 + qx + r = 0 can be transformed in such a way that a square is created on both sides. To this, one adds to both sides of the equation 2zx 2 + z 2 using a value z that is still to be selected appropriately and thus obtains x 4 + 2zx 2 + z 2 = (2z p) x 2 qx + (z 2 r). On the left we already have: (x 2 + z) 2. How should p be chosen on the right? 2 2z p z 2 r = q

62 solution of the reduced form quart. Equations In addition to third-degree equations, Cardano also published a general formula for solving fourth-degree equations in his Ars magna. He made use of the knowledge of his student Ludovico Ferrari (). Quartic equations of the form x 4 + px 2 + qx + r = 0 can be transformed in such a way that a square is created on both sides. To this, one adds to both sides of the equation 2zx 2 + z 2 using a value z that is still to be selected appropriately and thus obtains x 4 + 2zx 2 + z 2 = (2z p) x 2 qx + (z 2 r). On the left we already have: (x 2 + z) 2. How should p be chosen on the right? 2 2z p z 2 r = q

63 Solution of the reduced form quart. Equations In addition to third-degree equations, Cardano also published a general formula for solving fourth-degree equations in his Ars magna. He made use of the knowledge of his student Ludovico Ferrari (). Quartic equations of the form x 4 + px 2 + qx + r = 0 can be transformed in such a way that a square is created on both sides. To this, one adds to both sides of the equation 2zx 2 + z 2 using a value z that is still to be selected appropriately and thus obtains x 4 + 2zx 2 + z 2 = (2z p) x 2 qx + (z 2 r). On the left we already have: (x 2 + z) 2. How should p be chosen on the right? 2 2z p z 2 r = q

64 solution of the reduced form quart. Equations In addition to third-degree equations, Cardano also published a general formula for solving fourth-degree equations in his Ars magna. He made use of the knowledge of his student Ludovico Ferrari (). Quartic equations of the form x 4 + px 2 + qx + r = 0 can be transformed in such a way that a square is created on both sides. Using a value z that has yet to be suitably selected, one adds to this on both sides of the equation 2zx 2 + z 2 and thus obtains x 4 + 2zx 2 + z 2 = (2z p) x 2 qx + (z 2 r). On the left we already have: (x 2 + z) 2. How should p be chosen on the right? 2 2z p z 2 r = q

65 solution of the reduced form quart. Equations In addition to third-degree equations, Cardano also published a general formula for solving fourth-degree equations in his Ars magna. He made use of the knowledge of his student Ludovico Ferrari (). Quartic equations of the form x 4 + px 2 + qx + r = 0 can be transformed in such a way that a square is created on both sides. Using a value z that has yet to be suitably selected, one adds to this on both sides of the equation 2zx 2 + z 2 and thus obtains x 4 + 2zx 2 + z 2 = (2z p) x 2 qx + (z 2 r). On the left we already have: (x 2 + z) 2. How should p be chosen on the right? 2 2z p z 2 r = q

66 Solution of the reduced form quart. Equations By squaring, multiplying and transforming this condition on both sides, one finally obtains a cubic equation of the form z 3 p 2 z2 rz + pr 2 q2 8 = 0, with the solution z which can be determined by the Cardano formula. zq = p3 1 6 pr q2 + 1 pp 4 r + 12p 3 qp 2 r 2 432prq q 4 768r 3 + q p3 1 6 pr q2 1 pp 4 r + 12p 3 qp 2 r 2 432prq q 4 768r p Solutions for x is obtained by using the equation transformed into two-sided squares: (x 2 + z) 2 = (2z p) x 2 + (z 2 r) Taking the roots and reshaping on both sides leads to the quadratic equation ppx 2 ± 2z px + z2 r + z = 0

67 solution of the reduced form quart. Equations By squaring, multiplying and transforming this condition on both sides, one finally obtains a cubic equation of the form z 3 p 2 z2 rz + pr 2 q2 8 = 0, with the solution z which can be determined by the Cardano formula. zq = p3 1 6 pr q2 + 1 pp 4 r + 12p 3 qp 2 r 2 432prq q 4 768r 3 + q p3 1 6 pr q2 1 pp 4 r + 12p 3 qp 2 r 2 432prq q 4 768r p Solutions for x is obtained by using the equation transformed into two-sided squares: (x 2 + z) 2 = (2z p) x 2 + (z 2 r) Taking the roots and reshaping on both sides leads to the quadratic equation ppx 2 ± 2z px + z2 r + z = 0

68 Solution of the reduced form quart. Equations By squaring, multiplying and transforming this condition on both sides, one finally obtains a cubic equation of the form z 3 p 2 z2 rz + pr 2 q2 8 = 0, with the solution z. zq = p3 1 6 pr q2 + 1 pp 4 r + 12p 3 qp 2 r 2 432prq q 4 768r 3 + q p3 1 6 pr q2 1 pp 4 r + 12p 3 qp 2 r 2 432prq q 4 768r p Solutions for x is obtained by using the equation transformed into two-sided squares: (x 2 + z) 2 = (2z p) x 2 + (z 2 r) Taking the roots and reshaping on both sides leads to the quadratic equation ppx 2 ± 2z px + z2 r + z = 0

69 Solution of the reduced form quart. Equations By squaring, multiplying and transforming this condition on both sides, one finally obtains a cubic equation of the form z 3 p 2 z2 rz + pr 2 q2 8 = 0, with the solution z which can be determined by the Cardano formula. zq = p3 1 6 pr q2 + 1 pp 4 r + 12p 3 qp 2 r 2 432prq q 4 768r 3 + q p3 1 6 pr q2 1 pp 4 r + 12p 3 qp 2 r 2 432prq q 4 768r p Solutions for x is obtained by using the equation transformed into two-sided squares: (x 2 + z) 2 = (2z p) x 2 + (z 2 r) Taking the roots and reshaping on both sides leads to the quadratic equation ppx 2 ± 2z px + z2 r + z = 0

70 solution of the reduced form quart. Equations Because of the sign variants, four solutions are obtained using the pq formula: x 1,2 = 1 2z p ± z 1 4 p + z 2 rx 3,4 = 1 2 2z p ± 1 2 z 1 4 pz 2 r representation of the Solution only depending on p, q, r see Maple x1x2x3x4allgemein.mws

71 Quartic case (general) So far we have not discussed any quartic equations in which the unknown x appears to the third power. So how can a general equation of the form x 4 + ax 3 + bx 2 + cx + d = 0 be transformed into an equation of the reduced form y 4 + py 2 + qy + r = 0? The unknown x is substituted by x = ya 4, whereby the resulting terms to the power y 3 cancel each other out: x 4 + ax 3 + bx 2 + cx + d = y 4 + py 2 + qy + r where the coefficients are the reduced equation can be calculated using polynomial expressions.

72 Quartic Case (General) So far we have not discussed any quartic equations in which the unknown x appears to the third power. So how can a general equation of the form x 4 + ax 3 + bx 2 + cx + d = 0 be transformed into an equation of the reduced form y 4 + py 2 + qy + r = 0? The unknown x is substituted by x = ya 4, whereby the resulting terms to the power y 3 cancel each other out: x 4 + ax 3 + bx 2 + cx + d = y 4 + py 2 + qy + r where the coefficients are the reduced equation can be calculated using polynomial expressions.

73 Quartic case (general) So far we have not discussed any quartic equations in which the unknown x appears to the third power. So how can a general equation of the form x 4 + ax 3 + bx 2 + cx + d = 0 be transformed into an equation of the reduced form y 4 + py 2 + qy + r = 0? The unknown x is substituted by x = ya 4, whereby the resulting terms to the power y 3 cancel each other out: x 4 + ax 3 + bx 2 + cx + d = y 4 + py 2 + qy + r where the coefficients are the reduced equation can be calculated using polynomial expressions.

74 Type of solutions to quartic equations If all coefficients are real, three case distinctions can be given for the possible solutions, since every polynomial with real coefficients can be decomposed into linear and quadratic factors with real coefficients, regardless of its degree. (Fundamental theorem of algebra) The equation has four real solutions. It breaks down into four linear factors with real coefficients. (A) The equation has two real and two complex conjugate solutions. It is divided into two linear factors and a quadratic factor with real coefficients. (B) The equation has two pairs of complex conjugate solutions. It breaks down into two quadratic factors with real coefficients. (C)

75 Type of solutions to quartic equations If all coefficients are real, three case distinctions can be given for the possible solutions, since every polynomial with real coefficients can be broken down into linear and quadratic factors with real coefficients, regardless of its degree. (Fundamental theorem of algebra) The equation has four real solutions. It breaks down into four linear factors with real coefficients. (A) The equation has two real and two complex conjugate solutions. It is divided into two linear factors and a quadratic factor with real coefficients. (B) The equation has two pairs of complex conjugate solutions. It breaks down into two quadratic factors with real coefficients. (C)

76 Kind of solutions to quartic equations If all coefficients are real, three case distinctions can be given for the possible solutions, since every polynomial with real coefficients can be decomposed into linear and quadratic factors with real coefficients, regardless of its degree. (Fundamental theorem of algebra) The equation has four real solutions. It breaks down into four linear factors with real coefficients. (A) The equation has two real and two complex conjugate solutions. It is divided into two linear factors and a quadratic factor with real coefficients. (B) The equation has two pairs of complex conjugate solutions. It breaks down into two quadratic factors with real coefficients. (C)

77 Kind of solutions to quartic equations If all coefficients are real, three case distinctions can be given for the possible solutions, since every polynomial with real coefficients can be decomposed into linear and quadratic factors with real coefficients, regardless of its degree. (Fundamental theorem of algebra) The equation has four real solutions. It breaks down into four linear factors with real coefficients. (A) The equation has two real and two complex conjugate solutions. It is divided into two linear factors and a quadratic factor with real coefficients. (B) The equation has two pairs of complex conjugate solutions. It breaks down into two quadratic factors with real coefficients. (C)

78 Types of solutions to quartic equations

79 Abel-Ruffini theorem Abel-Ruffini theorem General polynomials of the fifth degree or higher cannot be resolved by radicals, i.e. mathematical expressions that only use roots and basic arithmetic operations. The first proof of this theorem was published by Paolo Ruffini () in 1799. However, this evidence was sketchy and largely ignored. Complete proof was provided by Niels Henrik Abel () in 1824. In addition to Galois, who generalized Abel's investigations into the unsolvability of equations to special equations (so-called Galois theory), Abel was an important co-founder of group theory.

80 Abel-Ruffini theorem Abel-Ruffini theorem General polynomials of the fifth degree or higher cannot be resolved by radicals, i.e. mathematical expressions that only use roots and basic arithmetic operations. The first proof of this theorem was published by Paolo Ruffini () in 1799. However, this evidence was sketchy and largely ignored. Complete proof was provided by Niels Henrik Abel () in 1824. In addition to Galois, who generalized Abel's investigations into the unsolvability of equations to special equations (so-called Galois theory), Abel was an important co-founder of group theory.

81 Abel-Ruffini theorem Abel-Ruffini theorem General polynomials of the fifth degree or higher cannot be resolved by radicals, i.e. mathematical expressions that only use roots and basic arithmetic operations. The first proof of this theorem was published by Paolo Ruffini () in 1799. However, this evidence was sketchy and largely ignored. Complete proof was provided by Niels Henrik Abel () in 1824. In addition to Galois, who generalized Abel's investigations into the unsolvability of equations to special equations (so-called Galois theory), Abel was an important co-founder of group theory.

82 Galois' idea A generalization of Abel's approaches, which can also be used for special equations, was found a few years later by Evariste Galois (), who was only twenty at the time. On the eve of a fatal duel for him, he summarized his results in a letter. This includes criteria that allow each individual equation to be examined to determine whether or not its solutions can be represented with the help of root expressions. For example, the solutions to the fifth degree equation x 5 x 1 = 0 cannot be represented by nested root expressions, whereas for the equation x x 44 = 0, for example, x 1 = a solution.

83 Galois' idea A generalization of Abel's approaches, which can also be used for special equations, was found a few years later by Evariste Galois, who was only twenty at the time (). On the eve of a fatal duel for him, he summarized his results in a letter. This includes criteria that allow each individual equation to be examined to determine whether or not its solutions can be represented with the help of root expressions. For example, the solutions to the fifth degree equation x 5 x 1 = 0 cannot be represented by nested root expressions, whereas for the equation x x 44 = 0, for example, x 1 = a solution.

84 Galois' idea A generalization of Abel's approaches, which can also be used for special equations, was found a few years later by Evariste Galois, who was only twenty at the time (). On the eve of a fatal duel for him, he summarized his results in a letter. This includes criteria that allow each individual equation to be examined to determine whether or not its solutions can be represented with the help of root expressions. For example, the solutions to the fifth degree equation x 5 x 1 = 0 cannot be represented by nested root expressions, whereas for the equation x x 44 = 0, for example, x 1 = a solution.

85 Galois' idea A generalization of Abel's approaches that can also be used for special equations was found a few years later by Evariste Galois (), who was only twenty at the time. On the eve of a fatal duel for him, he summarized his results in a letter. This includes criteria that allow each individual equation to be examined to see whether or not its solutions can be represented with the help of root expressions. For example, the solutions to the fifth degree equation x 5 x 1 = 0 cannot be represented by nested root expressions, whereas for the equation x x 44 = 0, for example, x 1 = a solution.

86 Idea of ​​Galois Using the more general results of Galois theory, only two points need to be shown to prove the Abel-Ruffini theorem: The general equation of the fifth degree (i.e. the equation with variables as coefficients) has as Galois group the symmetrical group S 5 The symmetrical Group S 5 cannot be resolved. Main theorem of Galois theory: Each equation solution corresponds to nested number ranges (which are formed by the coefficients and the solutions x 1, x 2, ...) and these can all be found by analyzing the Galois group. Such an analysis by the Galois group therefore already answers the question of whether solutions can be represented in the form of nested root expressions.

87 Idea of ​​Galois Using the more general results of Galois theory, only two points need to be shown to prove the Abel-Ruffini theorem: The general equation of the fifth degree (i.e. the equation with variables as coefficients) has as Galois group the symmetrical group S 5 The symmetrical Group S 5 cannot be resolved. Main theorem of Galois theory: Each equation solution corresponds to nested number ranges (which are formed by the coefficients and the solutions x 1, x 2, ...) and these can all be found by analyzing the Galois group. Such an analysis by the Galois group therefore already answers the question of whether solutions can be represented in the form of nested root expressions.

88 Idea of ​​Galois Using the more general results of Galois theory, only two points need to be shown to prove the Abel-Ruffini theorem: The general equation of the fifth degree (i.e. the equation with variables as coefficients) has the symmetrical group S 5 as the Galois group Group S 5 cannot be resolved. Main theorem of Galois theory: Each equation solution corresponds to nested number ranges (which are formed by the coefficients and the solutions x 1, x 2, ...) and these can all be found by analyzing the Galois group. Such an analysis by the Galois group therefore already answers the question of whether solutions can be represented in the form of nested root expressions.

89 Idea of ​​Galois Using the more general results of Galois theory, only two points have to be shown to prove the Abel-Ruffini theorem: The general equation of the fifth degree (i.e. the equation with variables as coefficients) has the symmetrical group S 5 as the Galois group Group S 5 cannot be resolved. Main theorem of Galois theory: Each equation solution corresponds to nested number ranges (which are formed by the coefficients and the solutions x 1, x 2, ...) and these can all be found by analyzing the Galois group. Such an analysis by the Galois group therefore already answers the question of whether solutions can be represented in the form of nested root expressions.

90 Idea of ​​Galois Symmetric Group The symmetric group S n is a group that consists of all permutations of a set with n elements. Group operation is the concatenation of permutations. The neutral element is the identity id. Definition A group is called solvable if there is a descending sequence G = G 0 G 1 ... G n = 1 of normal divisors whose quotients G k / G k + 1 (factor groups) are Abelian. In group theory, a subgroup N of a group G is called a normal subgroup (or normal subgroup) if the following applies to all a G and b N: aba 1 N. Notation: N GDh for each element a G the left minor class of N is equal to right, so on = Well.

91 Idea of ​​Galois Symmetric Group The symmetric group S n is a group that consists of all permutations of a set with n elements. Group operation is the concatenation of permutations. The neutral element is the identity id. Definition A group is called solvable if there is a descending sequence G = G 0 G 1 ... G n = 1 of normal divisors whose quotients G k / G k + 1 (factor groups) are Abelian. In group theory, a subgroup N of a group G is called a normal subgroup (or normal subgroup) if the following applies to all a G and b N: aba 1 N. Notation: N GDh for each element a G the left minor class of N is equal to right, so on = Well.

92 Idea of ​​Galois Symmetric group The symmetric group S n is a group that consists of all permutations of a set with n elements. Group operation is the concatenation of permutations. The neutral element is the identity id. Definition A group is called solvable if there is a descending sequence G = G 0 G 1 ... G n = 1 of normal divisors whose quotients G k / G k + 1 (factor groups) are Abelian. In group theory, a subgroup N of a group G is called a normal subgroup (or normal subgroup) if the following applies to all a G and b N: aba 1 N. Notation: N GDh for each element a G the left minor class of N is equal to right, so on = Well.

93 Numerics In numerical practice today, Cardano's formulas are of little importance. In an age in which the computing power of computers is de facto unlimited, an explicit formula (and the root towers, as they appear in quartic equations, for example) is dispensable in practical applications, since the solutions are completely sufficient for such applications to be approximately determined by numerical methods. The most important of these procedures are briefly explained below.

94 Bisection We are looking for the zero of a strictly monotonically increasing function f in the interval [a, b]. This should be specified with an accuracy ε. (The subinterval of [a, b] that contains the zero has a maximum length of ε.) Idea: Estimation of the position of a zero with a ZW-record. Then halve the interval and modify the interval limits (function values ​​of the interval limits must have different VZ). This leads to the following algorithm: 1. Set l = a and r = b. 2. Test whether [l, r] contains a zero. If not: abort. 3. Test whether r l <ε. If so, we've found our interval! 4. Otherwise divide [l, r] in the middle and continue the process recursively from 2. with both partial intervals.