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CUBIC EQUATIONS SECTION to be made into a website... History [citation needed] Cubic equations were known to the ancient Indians and ancient Greeks since the 5th century [citation BC, and even earlier to the ancient Babylonians who were able to solve certain cubic equations, needed] and also to the ancient Egyptians. Doubling the cube is the simplest and oldest studied cubic [1] equation, and one which the ancient Egyptians considered to be impossible. Hippocrates reduced this problem to that of finding two mean proportionals between one line and another of twice its length, but [2] could not solve this with a compass and straightedge construction, a task which is now known to be impossible. Hippocrates, Menaechmus and Archimedes are believed to have come close to solving the [2] problem of doubling the cube using intersecting conic sections, though historians such as Reviel Netz dispute whether the Greeks were thinking about cubic equations or just problems that can lead to cubic equations. Some others like T. L. Heath , who translated all Archimedes' works, disagree, putting forward evidence that Archimedes really solved cubic equations using intersections of two cones, but also [3] discussed the conditions where the roots are 0, 1 or 2. Two-dimensional graph of a cubic, the polynomial ƒ(x) = 2x3 − 3x2 − 3x + 2. In the 11th century, the Persian poet-mathematician, Omar Khayyám (1048–1131), made significant progress in the theory of cubic equations. In an early paper he wrote regarding cubic equations, he discovered that a cubic equation can have more than one solution and stated that it cannot be solved using compass and straightedge constructions. He also found a geometric solution which could be used to get a numerical answer by consulting trigonometric tables. In his later work, the Treatise on Demonstration of Problems of Algebra, he wrote a complete classification of cubic equations with general [4][5] geometric solutions found by means of intersecting conic sections. In the 12th century, another Persian mathematician, Sharaf al-Dīn al-Tūsī (1135–1213), wrote the Al- Mu'adalat (Treatise on Equations), which dealt with eight types of cubic equations with positive solutions and five types of cubic equations which may not have positive solutions. He used what would later be known as the "Ruffini-Horner method" to numerically approximate the root of a cubic equation. He also developed the concepts of a derivative function and the maxima and minima of curves in order to solve [6] cubic equations which may not have positive solutions. He understood the importance of the [7] discriminant of the cubic equation and used an early version of Cardano's formula to find algebraic [8] solutions to certain types of cubic equations. Leonardo de Pisa, also known as Fibonacci (1170–1250), was able to find the positive solution to the 3 2 cubic equation x +2x +10x = 20, using the babylonian numerals. He gave the result as 1,22,7,42,33,4,40 2 3 4 5 6 [9] which is equivalent to: 1+22/60+7/60 +42/60 +33/60 +4/60 +40/60 . In the early 16th century, the Italian mathematician Scipione del Ferro (1465–1526) found a method for 3 solving a class of cubic equations, namely those of the form x + mx = n. In fact, all cubic equations can be reduced to this form if we allow m and n to be negative, but negative numbers were not known to him at that time. Del Ferro kept his achievement secret until just before his death, when he told his student Antonio Fiore about it. In 1530, Niccolò Tartaglia (1500–1557) received two problems in cubic equations from Zuanne da Coi and announced that he could solve them. He was soon challenged by Fiore, which led to a famous contest between the two. Each contestant had to put up a certain amount of money and to propose a number of problems for his rival to solve. Whoever solved more problems within 30 days would get all the 3 money. Tartaglia received questions in the form x + mx = n, for which he had worked out a general 3 2 method. Fiore received questions in the form x + mx = n, which proved to be too difficult for him to solve, and Tartaglia won the contest. Later, Tartaglia was persuaded by Gerolamo Cardano (1501–1576) to reveal his secret for solving cubic equations. In 1539, Tartaglia did so only on the condition that Cardano would never reveal it and that if he did reveal a book about cubics, that he would give Tartaglia time to publish. Some years later, Cardano learned about Ferro's prior work and published Ferro's method in his book Ars Magna in 1545, meaning Cardano gave Tartaglia 6 years to publish his results (with credit given to Tartaglia for an independent solution). Cardano's promise with Tartaglia stated that he not publish Tartaglia's work, and Cardano felt he was publishing del Ferro's, so as to get around the promise. Nevertheless, this led to a challenge to Cardano by Tartaglia, which Cardano denied. The challenge was eventually accepted by Cardano's student Lodovico Ferrari (1522–1565). Ferrari did better than Tartaglia in the competition, and Tartaglia [10] lost both his prestige and income . Cardano noticed that Tartaglia's method sometimes required him to extract the square root of a negative number. He even included a calculation with these complex numbers in Ars Magna, but he did not really understand it. Rafael Bombelli studied this issue in detail and is therefore often considered as the discoverer of complex numbers. [edit] Roots of a cubic function [edit] The nature of the roots Every cubic equation with real coefficients has at least one solution x among the real numbers; this is a consequence of the intermediate value theorem. We can distinguish several possible cases using the discriminant, [11] The following cases need to be considered: If Δ > 0, then the equation has three distinct real roots. If Δ = 0, then the equation has a multiple root and all its roots are real. If Δ < 0, then the equation has one real root and two nonreal complex conjugate roots. See also: multiplicity of a root of a polynomial [edit] General formula of roots For the general cubic equation ax3 + bx2 + cx + d = 0, if (2b3 − 9abc + 27a2d)2 − 4(b2 − 3ac)3 > 0, the general [12] formulas for the roots, in terms of the coefficients, are as follows: However this formula is wrong if the operand of the square root is negative: When it is positive, the cubic roots are real and well defined. On the other hand, if the operand of the square root is negative, the square root is not real and one has to choose, once for all a determination for it, for example the one with positive imaginary part. For extracting the cubic roots we have also to choose a determination for the cubic roots, and this gives nine possible values for the first root of an equation which has only three roots. A correct solution may be obtained by remarking that the proof of above formula shows that the product of the two cubic roots is rational. This gives the following formula in which or stand for any determination of the square or cubic root. [edit] Monic formula of roots For a monic polynomial (dividing by the leading coefficient), renaming the coefficients as x3 + ax2 + bx + c = 0, this reduces to Introducing the following notation simplifies the above equations to In this form, they can be identified as the inverse discrete Fourier transform of the triple: which is the perspective taken in the method of Lagrange resolvents. Since there are 3 possible values for each cube root, and must be taken so that they satisfy αβ = k. [edit] Cardano's method The solutions can be found with the following method due to Scipione del Ferro and Tartaglia, published [13] by Gerolamo Cardano in 1545. We first divide the standard equation by the leading coefficient to arrive at an equation of the form The substitution eliminates the quadratic term, giving the so-called depressed cubic where We introduce two variables u and v linked by the condition and substitute this in the depressed cubic (2), giving . At this point Cardano imposed a second condition for the variables u and v 3 which, combined with (3) (the first parenthesis vanishes, then multiply by u and substitute uv) gives 3 This can be seen as a quadratic equation in u . When we solve this equation, we find that and thus Since t = v + u, t = x + a/3, and v = −p/3u, we find Note that there are six possibilities in computing u with (4), since there are two possibilities for the square root ( ), and three for the cubic root (the principal root and the principal root multiplied by ). The sign of the square root however does not affect the resulting t (a simple calculation shows that −p/3u = v), although care must be taken in three special cases to avoid divisions by zero: First, if p = q = 0, then we have the triple real root Second, if p = 0 and q ≠ 0, then Third, if p ≠ 0 and q = 0 then in which case the three roots are where [edit] Summary In summary, for the cubic equation the solutions for x are given by where The expression above for u can generate up to three values (there are three cubic roots related by a factor which is one of the two non-real cubic roots of one, and two square roots of any sign ; but these 6 expressions can generate only 3 pairs). This also applies to the final solutions for x. [edit] Alternate method An alternate method to obtain the same results is as follows. We know that Since u and v must satisfy , it can be shown that Writing out the three cube roots we get Remembering t = u + v we get only three possible values for t because only three combinations of u and v are possible if is to remain valid as it must — so and x is obtained from The above methods do apply if p and q are complex. This solution avoids the addition of an inverted cubic radical in the solution, and also resolves the ambiguity of signs for the square roots in the first solution given above. Summary To simplify the expressions above, it is customary to define this resolution in several steps by defining intermediate variables. Let and Then the discriminant of the quadratic equation of or is , Let's also define a constant that represents a generator for the three cubic roots of unity: Then the solutions for x = t - A can be simply defined for k in 0, 1, 2 in : where are the two possible values for or . In the case p and q are both real, the following cases can be distinguished, according to the sign of the discriminant. 1. If D is strictly positive then there are one real root and two non-real, conjugate roots. 2. If D is zero then there is one real root (a triple root) or two real roots (a single root and a double root.) 3. If D is strictly negative then there are three real roots (Casus irreducibilis). [edit] Lagrange resolvents [edit] Description Lagrange resolvents, introduced by Joseph Louis Lagrange in his paper Réflexions sur la résolution algébrique des équations, reduced the solution of a cubic equation to the solution of an auxiliary quadratic [14][15] polynomial, the "resolving equation" or resolvent quadratic of the original equation, by a change of variables from the roots x0,x1,x2 to the Lagrange resolvents r0,r1,r2. In modern terms, the resolvents are the discrete Fourier transform of the original roots. In the modern language of Galois theory, Lagrange exploited the fact that the symmetric group S3 has the cyclic group of order three C3 as a normal subgroup, thus allowing one to decompose the permutation group of the roots as – the C3 corresponds to the discrete Fourier transform (of order 3), while the C2 = S2 corresponds to the quadratic resolving polynomial. In the same way, the symmetric group of order four S4 has a Klein four-group as normal subgroup, with quotient a symmetric group of order three S3, which allows one to solve a quartic in terms of a cubic resolving polynomial. However, this method does not work for polynomials of degree five or greater, as the resolving polynomial [14] has higher degree than the original polynomial (for a quintic the resolving polynomial has degree 24 ). This is explained by the Abel–Ruffini theorem, which proves that such polynomials cannot be solved by radicals. [edit] Procedure Suppose that r0, r1 and r2 are the roots of equation (1), and define , so that ζ is a primitive third root of unity. We now set This is the discrete Fourier transform of the roots: observe that while the coefficients of the polynomial are symmetric in the roots, in this formula an order has been chosen on the roots, so these are not symmetric in the roots. The roots may then be recovered from the three si by inverting the above linear transformation via the inverse discrete Fourier transform, giving We already know the value s0 = −a, so we only need to seek values for the other two. The si are not symmetric in the roots – s0 is invariant, while the two non-trivial cyclic permutations of the roots send s1 to ζs1 and s2 to ζ2s2, or s1 to ζ2s1 and s2 to ζs2 (depending on which permutation), while transposing r1 and r2 switches s1 and s2 – other transpositions switch these roots and multiplying them by a power of ζ. Thus, if one takes the cubes, the factors of ζ become factors of ζ3 = 1, so every cyclic permutation leaves 3 3 the cubes invariant, and a transposition of two roots exchanges s1 and s2 – in other words, S3 (permutation of the roots) acts as a permutation group on specifically by the sign permutation exchanging the last two, meaning that the action factors through the sign map corresponding to the cyclic group of order 3 being a normal subgroup. Hence the polynomial is invariant under permutations of the roots, and so has coefficients expressible in terms of (1). Using calculations involving symmetric functions or alternatively field extensions, we can calculate (5) to be The roots of this quadratic equation are where D is the discriminant. Taking cube roots give us s1 and s2, from which we can recover the roots ri of (1). [edit] Factorization If r is any root of (1), then we may factor using r to obtain Hence if we know one root we can find the other two by solving a quadratic equation, giving for the other two roots. [edit] Root-finding formula The formula for finding the roots of a cubic function, based on Cardano's method, is fairly complicated. Therefore, it is common to use the rational root test or a numerical solution instead. If we have with and , let and we define the discriminant: There are two distinct cases: in which case there are one real root and two non-real roots that are conjugate. We define: and in which case we have 3 real roots. We express the complex quantity in polar form: and we define: and In both cases, the solutions are [edit] Solution in terms of Chebyshev radicals If we have a cubic equation which is already in depressed form, we may write it as . Substituting we obtain or equivalently From this we obtain solutions to our original equation in terms of the Chebyshev cube root as If now we start from a general equation and reduce it to the depressed form under the substitution x = t − a/3, we have and , leading to This gives us the solutions to (1) as [edit] The case of a cubic equation with real coefficients 2 Suppose the coefficients of (1) are real. If s is the quantity q/r from the section on real roots, then s = t ; hence 0 < s < 4 is equivalent to −2 < t < 2, and in this case we have a polynomial with three distinct real roots, expressed in terms of a real function of a real variable, quite unlike the situation when using cube roots. If s > 4 then either t > 2 and is the sole real root, or t < −2 and is the sole real root. If s < 0 then the reduction to Chebyshev polynomial form has given a t which is a pure imaginary number; in this case is the sole real root. We are now evaluating a real root by means of a function of a purely imaginary argument; however we can avoid this by using the function which is a real function of a real variable with no singularities along the real axis. If a polynomial can be reduced to the form x3 + 3x − t with real t, this is a convenient way to solve for its roots. [edit] Derivative Through the quadratic formula the roots of the derivative , are given by 2 and provide the critical points where the slope of the cubic equation is zero. If b -3ac>0, then the cubic 2 function has a local maximum and a local minimum. If b -3ac=0, then the cubic's inflection point is the 2 2 only critical point. If b -3ac<0, then there are no critical points. In the cases where b -3ac≤0, the cubic is strictly monotonic. [edit] Bipartite cubics The graph of where 0<a<b , is called a bipartite cubic. This is from the theory of elliptic curves. One can graph a bipartite cubic on a graphing device by graphing the function corresponding to the upper half of the bipartite cubic. It is defined on [edit] See also CUBIC EQUATIONS SECTION to be made into a website... History [citation needed] Cubic equations were known to the ancient Indians and ancient Greeks since the 5th century [citation BC, and even earlier to the ancient Babylonians who were able to solve certain cubic equations, needed] and also to the ancient Egyptians. Doubling the cube is the simplest and oldest studied cubic [1] equation, and one which the ancient Egyptians considered to be impossible. Hippocrates reduced this problem to that of finding two mean proportionals between one line and another of twice its length, but [2] could not solve this with a compass and straightedge construction, a task which is now known to be impossible. Hippocrates, Menaechmus and Archimedes are believed to have come close to solving the [2] problem of doubling the cube using intersecting conic sections, though historians such as Reviel Netz dispute whether the Greeks were thinking about cubic equations or just problems that can lead to cubic equations. Some others like T. L. Heath , who translated all Archimedes' works, disagree, putting forward evidence that Archimedes really solved cubic equations using intersections of two cones, but also [3] discussed the conditions where the roots are 0, 1 or 2. Two-dimensional graph of a cubic, the polynomial ƒ(x) = 2x3 − 3x2 − 3x + 2. In the 11th century, the Persian poet-mathematician, Omar Khayyám (1048–1131), made significant progress in the theory of cubic equations. In an early paper he wrote regarding cubic equations, he discovered that a cubic equation can have more than one solution and stated that it cannot be solved using compass and straightedge constructions. He also found a geometric solution which could be used to get a numerical answer by consulting trigonometric tables. In his later work, the Treatise on Demonstration of Problems of Algebra, he wrote a complete classification of cubic equations with general [4][5] geometric solutions found by means of intersecting conic sections. In the 12th century, another Persian mathematician, Sharaf al-Dīn al-Tūsī (1135–1213), wrote the Al- Mu'adalat (Treatise on Equations), which dealt with eight types of cubic equations with positive solutions and five types of cubic equations which may not have positive solutions. He used what would later be known as the "Ruffini-Horner method" to numerically approximate the root of a cubic equation. He also developed the concepts of a derivative function and the maxima and minima of curves in order to solve [6] cubic equations which may not have positive solutions. He understood the importance of the [7] discriminant of the cubic equation and used an early version of Cardano's formula to find algebraic [8] solutions to certain types of cubic equations. Leonardo de Pisa, also known as Fibonacci (1170–1250), was able to find the positive solution to the 3 2 cubic equation x +2x +10x = 20, using the babylonian numerals. He gave the result as 1,22,7,42,33,4,40 2 3 4 5 6 [9] which is equivalent to: 1+22/60+7/60 +42/60 +33/60 +4/60 +40/60 . In the early 16th century, the Italian mathematician Scipione del Ferro (1465–1526) found a method for 3 solving a class of cubic equations, namely those of the form x + mx = n. In fact, all cubic equations can be reduced to this form if we allow m and n to be negative, but negative numbers were not known to him at that time. Del Ferro kept his achievement secret until just before his death, when he told his student Antonio Fiore about it. In 1530, Niccolò Tartaglia (1500–1557) received two problems in cubic equations from Zuanne da Coi and announced that he could solve them. He was soon challenged by Fiore, which led to a famous contest between the two. Each contestant had to put up a certain amount of money and to propose a number of problems for his rival to solve. Whoever solved more problems within 30 days would get all the 3 money. Tartaglia received questions in the form x + mx = n, for which he had worked out a general 3 2 method. Fiore received questions in the form x + mx = n, which proved to be too difficult for him to solve, and Tartaglia won the contest. Later, Tartaglia was persuaded by Gerolamo Cardano (1501–1576) to reveal his secret for solving cubic equations. In 1539, Tartaglia did so only on the condition that Cardano would never reveal it and that if he did reveal a book about cubics, that he would give Tartaglia time to publish. Some years later, Cardano learned about Ferro's prior work and published Ferro's method in his book Ars Magna in 1545, meaning Cardano gave Tartaglia 6 years to publish his results (with credit given to Tartaglia for an independent solution). Cardano's promise with Tartaglia stated that he not publish Tartaglia's work, and Cardano felt he was publishing del Ferro's, so as to get around the promise. Nevertheless, this led to a challenge to Cardano by Tartaglia, which Cardano denied. The challenge was eventually accepted by Cardano's student Lodovico Ferrari (1522–1565). Ferrari did better than Tartaglia in the competition, and Tartaglia [10] lost both his prestige and income . Cardano noticed that Tartaglia's method sometimes required him to extract the square root of a negative number. He even included a calculation with these complex numbers in Ars Magna, but he did not really understand it. Rafael Bombelli studied this issue in detail and is therefore often considered as the discoverer of complex numbers. [edit] Roots of a cubic function [edit] The nature of the roots Every cubic equation with real coefficients has at least one solution x among the real numbers; this is a consequence of the intermediate value theorem. We can distinguish several possible cases using the discriminant, [11] The following cases need to be considered: If Δ > 0, then the equation has three distinct real roots. If Δ = 0, then the equation has a multiple root and all its roots are real. If Δ < 0, then the equation has one real root and two nonreal complex conjugate roots. See also: multiplicity of a root of a polynomial [edit] General formula of roots For the general cubic equation ax3 + bx2 + cx + d = 0, if (2b3 − 9abc + 27a2d)2 − 4(b2 − 3ac)3 > 0, the general [12] formulas for the roots, in terms of the coefficients, are as follows: However this formula is wrong if the operand of the square root is negative: When it is positive, the cubic roots are real and well defined. On the other hand, if the operand of the square root is negative, the square root is not real and one has to choose, once for all a determination for it, for example the one with positive imaginary part. For extracting the cubic roots we have also to choose a determination for the cubic roots, and this gives nine possible values for the first root of an equation which has only three roots. A correct solution may be obtained by remarking that the proof of above formula shows that the product of the two cubic roots is rational. This gives the following formula in which or stand for any determination of the square or cubic root. [edit] Monic formula of roots For a monic polynomial (dividing by the leading coefficient), renaming the coefficients as x3 + ax2 + bx + c = 0, this reduces to Introducing the following notation simplifies the above equations to In this form, they can be identified as the inverse discrete Fourier transform of the triple: which is the perspective taken in the method of Lagrange resolvents. Since there are 3 possible values for each cube root, and must be taken so that they satisfy αβ = k. [edit] Cardano's method The solutions can be found with the following method due to Scipione del Ferro and Tartaglia, published [13] by Gerolamo Cardano in 1545. We first divide the standard equation by the leading coefficient to arrive at an equation of the form The substitution eliminates the quadratic term, giving the so-called depressed cubic where We introduce two variables u and v linked by the condition and substitute this in the depressed cubic (2), giving . At this point Cardano imposed a second condition for the variables u and v 3 which, combined with (3) (the first parenthesis vanishes, then multiply by u and substitute uv) gives 3 This can be seen as a quadratic equation in u . When we solve this equation, we find that and thus Since t = v + u, t = x + a/3, and v = −p/3u, we find Note that there are six possibilities in computing u with (4), since there are two possibilities for the square root ( ), and three for the cubic root (the principal root and the principal root multiplied by ). The sign of the square root however does not affect the resulting t (a simple calculation shows that −p/3u = v), although care must be taken in three special cases to avoid divisions by zero: First, if p = q = 0, then we have the triple real root Second, if p = 0 and q ≠ 0, then Third, if p ≠ 0 and q = 0 then in which case the three roots are where [edit] Summary In summary, for the cubic equation the solutions for x are given by where The expression above for u can generate up to three values (there are three cubic roots related by a factor which is one of the two non-real cubic roots of one, and two square roots of any sign ; but these 6 expressions can generate only 3 pairs). This also applies to the final solutions for x. [edit] Alternate method An alternate method to obtain the same results is as follows. We know that Since u and v must satisfy , it can be shown that Writing out the three cube roots we get Remembering t = u + v we get only three possible values for t because only three combinations of u and v are possible if is to remain valid as it must — so and x is obtained from The above methods do apply if p and q are complex. This solution avoids the addition of an inverted cubic radical in the solution, and also resolves the ambiguity of signs for the square roots in the first solution given above. Summary To simplify the expressions above, it is customary to define this resolution in several steps by defining intermediate variables. Let and Then the discriminant of the quadratic equation of or is , Let's also define a constant that represents a generator for the three cubic roots of unity: Then the solutions for x = t - A can be simply defined for k in 0, 1, 2 in : where are the two possible values for or . In the case p and q are both real, the following cases can be distinguished, according to the sign of the discriminant. 1. If D is strictly positive then there are one real root and two non-real, conjugate roots. 2. If D is zero then there is one real root (a triple root) or two real roots (a single root and a double root.) 3. If D is strictly negative then there are three real roots (Casus irreducibilis). [edit] Lagrange resolvents [edit] Description Lagrange resolvents, introduced by Joseph Louis Lagrange in his paper Réflexions sur la résolution algébrique des équations, reduced the solution of a cubic equation to the solution of an auxiliary quadratic [14][15] polynomial, the "resolving equation" or resolvent quadratic of the original equation, by a change of variables from the roots x0,x1,x2 to the Lagrange resolvents r0,r1,r2. In modern terms, the resolvents are the discrete Fourier transform of the original roots. In the modern language of Galois theory, Lagrange exploited the fact that the symmetric group S3 has the cyclic group of order three C3 as a normal subgroup, thus allowing one to decompose the permutation group of the roots as – the C3 corresponds to the discrete Fourier transform (of order 3), while the C2 = S2 corresponds to the quadratic resolving polynomial. In the same way, the symmetric group of order four S4 has a Klein four-group as normal subgroup, with quotient a symmetric group of order three S3, which allows one to solve a quartic in terms of a cubic resolving polynomial. However, this method does not work for polynomials of degree five or greater, as the resolving polynomial [14] has higher degree than the original polynomial (for a quintic the resolving polynomial has degree 24 ). This is explained by the Abel–Ruffini theorem, which proves that such polynomials cannot be solved by radicals. [edit] Procedure Suppose that r0, r1 and r2 are the roots of equation (1), and define , so that ζ is a primitive third root of unity. We now set This is the discrete Fourier transform of the roots: observe that while the coefficients of the polynomial are symmetric in the roots, in this formula an order has been chosen on the roots, so these are not symmetric in the roots. The roots may then be recovered from the three si by inverting the above linear transformation via the inverse discrete Fourier transform, giving We already know the value s0 = −a, so we only need to seek values for the other two. The si are not symmetric in the roots – s0 is invariant, while the two non-trivial cyclic permutations of the roots send s1 to ζs1 and s2 to ζ2s2, or s1 to ζ2s1 and s2 to ζs2 (depending on which permutation), while transposing r1 and r2 switches s1 and s2 – other transpositions switch these roots and multiplying them by a power of ζ. Thus, if one takes the cubes, the factors of ζ become factors of ζ3 = 1, so every cyclic permutation leaves 3 3 the cubes invariant, and a transposition of two roots exchanges s1 and s2 – in other words, S3 (permutation of the roots) acts as a permutation group on specifically by the sign permutation exchanging the last two, meaning that the action factors through the sign map corresponding to the cyclic group of order 3 being a normal subgroup. Hence the polynomial is invariant under permutations of the roots, and so has coefficients expressible in terms of (1). Using calculations involving symmetric functions or alternatively field extensions, we can calculate (5) to be The roots of this quadratic equation are where D is the discriminant. Taking cube roots give us s1 and s2, from which we can recover the roots ri of (1). [edit] Factorization If r is any root of (1), then we may factor using r to obtain Hence if we know one root we can find the other two by solving a quadratic equation, giving for the other two roots. [edit] Root-finding formula The formula for finding the roots of a cubic function, based on Cardano's method, is fairly complicated. Therefore, it is common to use the rational root test or a numerical solution instead. If we have with and , let and we define the discriminant: There are two distinct cases: in which case there are one real root and two non-real roots that are conjugate. We define: and in which case we have 3 real roots. We express the complex quantity in polar form: and we define: and In both cases, the solutions are [edit] Solution in terms of Chebyshev radicals If we have a cubic equation which is already in depressed form, we may write it as . Substituting we obtain or equivalently From this we obtain solutions to our original equation in terms of the Chebyshev cube root as If now we start from a general equation and reduce it to the depressed form under the substitution x = t − a/3, we have and , leading to This gives us the solutions to (1) as [edit] The case of a cubic equation with real coefficients 2 Suppose the coefficients of (1) are real. If s is the quantity q/r from the section on real roots, then s = t ; hence 0 < s < 4 is equivalent to −2 < t < 2, and in this case we have a polynomial with three distinct real roots, expressed in terms of a real function of a real variable, quite unlike the situation when using cube roots. If s > 4 then either t > 2 and is the sole real root, or t < −2 and is the sole real root. If s < 0 then the reduction to Chebyshev polynomial form has given a t which is a pure imaginary number; in this case is the sole real root. We are now evaluating a real root by means of a function of a purely imaginary argument; however we can avoid this by using the function which is a real function of a real variable with no singularities along the real axis. If a polynomial can be reduced to the form x3 + 3x − t with real t, this is a convenient way to solve for its roots. [edit] Derivative Through the quadratic formula the roots of the derivative , are given by 2 and provide the critical points where the slope of the cubic equation is zero. If b -3ac>0, then the cubic 2 function has a local maximum and a local minimum. If b -3ac=0, then the cubic's inflection point is the 2 2 only critical point. If b -3ac<0, then there are no critical points. In the cases where b -3ac≤0, the cubic is strictly monotonic. [edit] Bipartite cubics The graph of where 0<a<b , is called a bipartite cubic. This is from the theory of elliptic curves. One can graph a bipartite cubic on a graphing device by graphing the function corresponding to the upper half of the bipartite cubic. It is defined on [edit] See also

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