Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 4"
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<math>\zeta_1, \zeta_2,</math> and <math>\zeta_3</math> are [[complex number]]s such that | <math>\zeta_1, \zeta_2,</math> and <math>\zeta_3</math> are [[complex number]]s such that | ||
| − | < | + | <cmath>\zeta_1+\zeta_2+\zeta_3=1</cmath> |
| − | \zeta_1^2+\zeta_2^2+\zeta_3^2 | + | <cmath>\zeta_1^2+\zeta_2^2+\zeta_3^2=3</cmath> |
| − | \zeta_1^3+\zeta_2^3+\zeta_3^3 | + | <cmath>\zeta_1^3+\zeta_2^3+\zeta_3^3=7</cmath> |
Compute <math>\zeta_1^{7} + \zeta_2^{7} + \zeta_3^{7}</math>. | Compute <math>\zeta_1^{7} + \zeta_2^{7} + \zeta_3^{7}</math>. | ||
| Line 10: | Line 10: | ||
==Solution== | ==Solution== | ||
We let <math>e_1 = \zeta_1 + \zeta_2 + \zeta_3,\ e_2 = \zeta_1\zeta_2 + \zeta_2\zeta_3 + \zeta_3\zeta_1,\ e_3 = \zeta_1\zeta_2\zeta_3</math> (the [[elementary symmetric sums]]). Then, we can rewrite the above equations as | We let <math>e_1 = \zeta_1 + \zeta_2 + \zeta_3,\ e_2 = \zeta_1\zeta_2 + \zeta_2\zeta_3 + \zeta_3\zeta_1,\ e_3 = \zeta_1\zeta_2\zeta_3</math> (the [[elementary symmetric sums]]). Then, we can rewrite the above equations as | ||
| − | < | + | <cmath>\zeta_1+\zeta_2+\zeta_3=e_1 = 1</cmath> |
| − | \zeta_1^2+\zeta_2^2+\zeta_3^2 | + | <cmath>\zeta_1^2+\zeta_2^2+\zeta_3^2= e_1^2 - 2e_2 = 3</cmath> |
from where it follows that <math>e_2 = -1</math>. The third equation can be factored as | from where it follows that <math>e_2 = -1</math>. The third equation can be factored as | ||
| − | < | + | <cmath>7 =\zeta_1^3+\zeta_2^3+\zeta_3^3 = (\zeta_1+\zeta_2+\zeta_3)(\zeta_1^2+\zeta_2^2+\zeta_3^2-\zeta_1\zeta_2-\zeta_2\zeta_3 -\zeta_3\zeta_1)+3\zeta_1\zeta_2\zeta_3\\ = e_1^3 - 3e_1e_2 + 3e_3,</cmath> |
from where it follows that <math>e_3 = 1</math>. Thus, applying [[Vieta's formulas]] backwards, <math>\zeta_1, \zeta_2,</math> and <math>\zeta_3</math> are the roots of the polynomial | from where it follows that <math>e_3 = 1</math>. Thus, applying [[Vieta's formulas]] backwards, <math>\zeta_1, \zeta_2,</math> and <math>\zeta_3</math> are the roots of the polynomial | ||
| − | < | + | <cmath>x^3 - x^2 - x - 1 = 0 \Longleftrightarrow x^3 = x^2 + x + 1</cmath> |
Let <math>s_n = \zeta_1^n + \zeta_2^n + \zeta_3^n</math> (the [[power sums]]). Then from <math>(1)</math>, we have the [[recursion]] <math>s_{n+3} = s_{n+2} + s_{n+1} + s_n</math>. It follows that <math>s_4 = 7 + 3 + 1 = 11, s_5 = 21, s_6 = 39, s_7 = \boxed{071}</math>. | Let <math>s_n = \zeta_1^n + \zeta_2^n + \zeta_3^n</math> (the [[power sums]]). Then from <math>(1)</math>, we have the [[recursion]] <math>s_{n+3} = s_{n+2} + s_{n+1} + s_n</math>. It follows that <math>s_4 = 7 + 3 + 1 = 11, s_5 = 21, s_6 = 39, s_7 = \boxed{071}</math>. | ||
| − | ==See | + | ==See Also== |
{{Mock AIME box|year=Pre 2005|n=3|num-b=3|num-a=5}} | {{Mock AIME box|year=Pre 2005|n=3|num-b=3|num-a=5}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
Latest revision as of 01:49, 19 March 2015
Problem
and 
are complex numbers such that
![\[\zeta_1+\zeta_2+\zeta_3=1\]](http://latex.artofproblemsolving.com/2/0/f/20f1cdf3667c7b392addc8b1658aef13e0a65918.png)
![\[\zeta_1^2+\zeta_2^2+\zeta_3^2=3\]](http://latex.artofproblemsolving.com/1/9/0/1905da989b086b38a25f8d01de7119ca8dc675d1.png)
![\[\zeta_1^3+\zeta_2^3+\zeta_3^3=7\]](http://latex.artofproblemsolving.com/f/c/a/fcaa6db09874632947ab41ccd40e1f90e3311b44.png)
Compute 
.
Solution
We let 
(the elementary symmetric sums). Then, we can rewrite the above equations as
![\[\zeta_1+\zeta_2+\zeta_3=e_1 = 1\]](http://latex.artofproblemsolving.com/8/3/e/83e28ea00ae7b53d77e12ff22900374d431cc2de.png)
![\[\zeta_1^2+\zeta_2^2+\zeta_3^2= e_1^2 - 2e_2 = 3\]](http://latex.artofproblemsolving.com/2/d/b/2dbd6dece2117341b07590d218f1b74ef7a2cce3.png)
from where it follows that 
. The third equation can be factored as
![\[7 =\zeta_1^3+\zeta_2^3+\zeta_3^3 = (\zeta_1+\zeta_2+\zeta_3)(\zeta_1^2+\zeta_2^2+\zeta_3^2-\zeta_1\zeta_2-\zeta_2\zeta_3 -\zeta_3\zeta_1)+3\zeta_1\zeta_2\zeta_3\\ = e_1^3 - 3e_1e_2 + 3e_3,\]](http://latex.artofproblemsolving.com/7/6/4/7648d05b4643d4570678215d3ebc4a31b7ae6d2d.png)
from where it follows that 
. Thus, applying Vieta's formulas backwards, and are the roots of the polynomial
![\[x^3 - x^2 - x - 1 = 0 \Longleftrightarrow x^3 = x^2 + x + 1\]](http://latex.artofproblemsolving.com/2/2/6/226f7e0b22c5aa7b367db210135911e418bb4b4c.png)
Let 
(the power sums). Then from 
, we have the recursion 
. It follows that 
.
See Also
| Mock AIME 3 Pre 2005 (Problems, Source) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
