Document Sample

Solving the cubic • Three people were involved in the solution of the cubic: del Ferro (1465-1526), Niccolò Tartaglia (1499-1557) and Girolamo Cardano (1501-1557). Del Ferro and Tartaglia. There were thought to be two cases x3+cx=d called the cosa and x3=cx+d with positive c and d. • Del Ferro solved the cosa and kept his result secret. • Tartaglia then first claimed to be able to solve equations of the kind x3+bx2=d, but then also solved all cubic equations. • Cardano was given the formula by Tartaglia and found and published proofs for the formulas for all cubic equations. He was also aware of del Ferro’s solution. Girolamo Cardano (1501-1557) Ars Magna (The great art) 1. Let the cube be equal to the first 1. x3=cx+d power and constant 2. Let DC, DF be cubes with sides AB 2. and let DC and DF be two cubes and BC. 3. the product of the sides of which, AB 3. s.t. (a) AB BC=1/3c. and BC, is equal to one-third the 4. and (b) DC+DF=AB3+BC3=d. coefficient of x, 5. Then AC=x! 4. and let the sum of these cubes be Proof: equal to the constant. 5. I say that AC is the value of x. 6. AB BC=1/3c 3AB BC=c 6. Now since AB x BC equals one-third 7. and under the assumption AC=x, the coefficient of x, 3(AB x BC) will AC(3ABxBC)=cx. equal the coefficient of x, 8. Claim: 7. and the product of AC and 3(AB x BC) AC(3ABxBC)=3ABxBC2+3BCxAB2. is the whole first power, AC having Notice that (AC=AB+BC) so that the been assumed to be x. equation is actually clear algebraically. 8. But AC x 3(AB x BC) makes six 9. By 7. AC(3ABxBC)=cx and by 8. bodies, three of which are AB x BC2 AC(3ABxBC)=3ABxBC2+3BCxAB2 so and the other three, BC x AB2. 3ABxBC2+3BCxAB2=cx 9. Therefore these six bodies are equal to the whole first power, F E D A C B Girolamo Cardano (1501-1557) Ars Magna (The great art) 10. and these [six bodies] plus the cubes 10. 3ABxBC2+3BCxAB2+CD+DF=AE or DC and DF constitute the cube AE, (u+v)(3uv)=3u2v+3uv2+u3+v3=(u+v)3. according to the first proposition of 11. DC+DF=d, this was assumption 3. Chapter VI. 12. So from 9,10,12 AE=cx+d and if 11. The cubes DC and DF are also equal AC=x then x3=cx+d. to the given number. 13. Remains to prove the claim 8. 12. Therefore the cube AE is equal to the 14. Enough to show given first power and number, which AB(BC x AC) = AB x BC2+ BC x AB2 was to be proved. [by eliminating the common factor 13. It remains to be shown that 3AC(AB x 3]. BC) is equal to the six bodies. 15. Since BE=ACxBC: 14. This is clear enough if I prove that AC(ABxBC)= ABxBE AB(BC x AC) equals the two bodies 16. Since BE=CD+DE: ABxBC2 and BC x AB2, ABxBE=AB(CD+DE) 15. for the product of AC and (AB x BC) is 17. Since GE=AB & DG=CB: equal to the product of AB and the AB x DE= ABxGExDG=CB x surface BE — since all sides are equal AB 2 to all sides — 16. but this [i.e., AB x BE] is equal to the 18. So AC(ABxBC) = AB(CD+DE) = product of AB and (CD + DE); ABxBC2+ BC x AB2 [since CD=BC2] F E 17. the product AB x DE is equal to the product CB x AB2, since all sides are equal to all sides; D G 18. and therefore AC(AB x BC) is equal to AB x BC2 plus BC x AB2, as was proposed. A C B The Great Art Short version and the rule: As Cardano says: 1. Set x=u+v The rule, therefore, is: 2. x3=(u+v)3=u3+3u2v+3uv2+v3. When the cube of one-third 3. If (1) u3+v3=d and (2) 3uv=c, so the coefficient of x is not 3u2v+3uv2=(u+v)3uv=cx. greater than the square of 4. Then x3=cx+d. one-half the constant of the equation, subtract the former 5. From (2) v=c/(3u). from the latter and add the 6. Substituted in (1) yields square root of the remainder u3+(c/(3u))3=d u6-du3+(c/3)3=0 to one-half the constant of 7. Set U=u3, then U2-dU+(c/3)3=0 and the equation and, again, subtract it from the same U d ( d ) 2 ( c )3 2 2 3 half, and you will have, as was said, a binomium and its 8. Let V=v3, then V=d-U, so apotome, the sum of the cube roots of which V d ( d ) 2 ( c )3 2 2 3 constitutes the value of x 9. Finally x 3 U 3 V 3 d 2 ( d ) 2 ( c )3 3 2 3 d 2 ( d ) 2 ( c )3 2 3 Binomial Theorem for Exponent 3 in Terms of Solids • (u+v)3=u3+3u2v+3uv2+v3=u3+3uv(u+v)+v3 u3 v u v3 uv(u+v) Examples • x3=6x+40 • Problems with x3=15x+4, – c=6, (c/3)3=23=8 the formula yields – d=40, (d/2)2=202=400 – c=15, (c/3)3=53=125 – 400-8=392 and – d=4, (d/2)2=22=4 – 125-4=121 and se x 3 20 392 3 20 392 x 3 2 121 3 2 121 • x3=6x+6 which contains complex numbers. But actually there – c=6, (c/3)3=23=8 are three real solutions: – d=6, (d/2)2=32=9 4,-2+ √3 and -2-√3. – 9-8=1 and thus x 3 3 1 3 3 1 3 4 3 2 • Also what about the other possible solutions? Rafaeleo Bombelli (1526-1572) started to calculate cube roots of complex numbers

DOCUMENT INFO

Shared By:

Categories:

Tags:
cubic equation, solving cubic equations, the roots, quadratic equation, cubic equations, Quadratic Formula, complex numbers, how to, quartic equations, the cubic formula

Stats:

views: | 40 |

posted: | 4/17/2010 |

language: | English |

pages: | 6 |

OTHER DOCS BY wulinqing

How are you planning on using Docstoc?
BUSINESS
PERSONAL

By registering with docstoc.com you agree to our
privacy policy and
terms of service, and to receive content and offer notifications.

Docstoc is the premier online destination to start and grow small businesses. It hosts the best quality and widest selection of professional documents (over 20 million) and resources including expert videos, articles and productivity tools to make every small business better.

Search or Browse for any specific document or resource you need for your business. Or explore our curated resources for Starting a Business, Growing a Business or for Professional Development.

Feel free to Contact Us with any questions you might have.