Difference between revisions of "User:Wsjradha/Cotangent Sum Problem"
(New page: '''Cotangent Sum Problem''' AIME-level of difficulty Problem: Let <math>z_1, z_2, \ldots, z_{20}</math> be the twenty (complex) roots of the equation: <math>z^{20} - 4z^{19} + 9z^{18...) |
m (Cotangent Sum Problem moved to User:Wsjradha/Cotangent Sum Problem: Sorry, but this doesn't really belong in the main space.) |
||
(5 intermediate revisions by one other user not shown) | |||
Line 1: | Line 1: | ||
− | + | == Problem: == | |
− | + | Let <math>z_1, z_2, \ldots, z_{20}</math> be the twenty (complex) roots of the equation <cmath>z^{20} - 4z^{19} + 9z^{18} - 16z^{17} + \cdots + 441 = 0.</cmath> | |
+ | Calculate the value of <cmath>\cot{\left(\sum_{k = 1}^{20} \cot^{ - 1}{z_k}\right)}.</cmath> | ||
− | + | == Solution: == | |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | Solution: | ||
For the purpose of this solution <math>k_n</math> will be the sum of the roots of the 20th degree polynomial, taken <math>n</math> at a time. | For the purpose of this solution <math>k_n</math> will be the sum of the roots of the 20th degree polynomial, taken <math>n</math> at a time. | ||
Line 26: | Line 15: | ||
Also, <math>c(k_n)</math> will be the sum of the cotangent inverses of the roots, taken <math>n</math> at a time. The cotangent inverses will be multiplied as necessary, then added. | Also, <math>c(k_n)</math> will be the sum of the cotangent inverses of the roots, taken <math>n</math> at a time. The cotangent inverses will be multiplied as necessary, then added. | ||
+ | |||
Also, <math>tc(k_n)</math> will be the sum of the tangents of the cotangent inverses of the roots, taken <math>n</math> at a time. Basically, this is the same as <math>c(k_n)</math> except that the tangents are taken right after the cotangent inverses. | Also, <math>tc(k_n)</math> will be the sum of the tangents of the cotangent inverses of the roots, taken <math>n</math> at a time. Basically, this is the same as <math>c(k_n)</math> except that the tangents are taken right after the cotangent inverses. | ||
For example, | For example, | ||
Line 31: | Line 21: | ||
<math>tc(k_1) = tan(cot^{ - 1}{z_1}) + tan(cot^{ - 1}{z_2}) + \cdots + tan(cot^{ - 1}{z_{20}})</math> | <math>tc(k_1) = tan(cot^{ - 1}{z_1}) + tan(cot^{ - 1}{z_2}) + \cdots + tan(cot^{ - 1}{z_{20}})</math> | ||
− | Let <math>S = \cot{\left(\sum_{k = 1}^{20} \cot^{ - 1}{z_k}\right)}</math> | + | Let <math>S = \cot{\left(\sum_{k = 1}^{20} \cot^{ - 1}{z_k}\right)}</math>. |
− | This equals <math>cot (c(k_1))</math> | + | This equals <math>cot (c(k_1))</math>. There is a formula that states the following, where, for the purposes of this formula only, <math>\sum_{a_1}^n(b_m)</math>, is the sum of <math>a_1</math> through <math>a_n</math>, taken <math>m</math> at a time, in the fashion described above: |
− | There is a formula that states the following, where, for the purposes of this formula only, <math>\sum_{a_1}^n(b_m)</math>, is the sum of <math>a_1</math> through <math>a_n</math>, taken <math>m</math> at a time, in the fashion described above: | ||
<math>tan{\left(a_1+a_2+a_3+\cdots+a_n\right)}=\dfrac{1-\sum_{a_1}^n(b_2)+\sum_{a_1}^n(b_4)-\sum_{a_1}^n(b_6)+\cdots}{\sum_{a_1}^n(b_1)-\sum_{a_1}^n(b_3)+\sum_{a_1}^n(b_5)-\cdots}</math> | <math>tan{\left(a_1+a_2+a_3+\cdots+a_n\right)}=\dfrac{1-\sum_{a_1}^n(b_2)+\sum_{a_1}^n(b_4)-\sum_{a_1}^n(b_6)+\cdots}{\sum_{a_1}^n(b_1)-\sum_{a_1}^n(b_3)+\sum_{a_1}^n(b_5)-\cdots}</math> | ||
Line 54: | Line 43: | ||
<math>S = \dfrac{21^2 - 19^2 + 17^2 - 15^2 + \cdots - 3^2 + 1}{20^2 - 18^2 + \cdots - 2^2}</math> | <math>S = \dfrac{21^2 - 19^2 + 17^2 - 15^2 + \cdots - 3^2 + 1}{20^2 - 18^2 + \cdots - 2^2}</math> | ||
− | This simplifies to <math>\boxed{\dfrac{241}{220}}</math> | + | This simplifies to <math>\boxed{\boxed{\dfrac{241}{220}}}</math> |
Latest revision as of 23:59, 20 December 2008
Problem:
Let be the twenty (complex) roots of the equation
Calculate the value of
Solution:
For the purpose of this solution will be the sum of the roots of the 20th degree polynomial, taken
at a time.
For example,
Also, will be the sum of the cotangent inverses of the roots, taken
at a time. The cotangent inverses will be multiplied as necessary, then added.
Also, will be the sum of the tangents of the cotangent inverses of the roots, taken
at a time. Basically, this is the same as
except that the tangents are taken right after the cotangent inverses.
For example,
Let .
This equals
. There is a formula that states the following, where, for the purposes of this formula only,
, is the sum of
through
, taken
at a time, in the fashion described above:
When applied to this problem, it yields:
Taking the reciprocal of either side, one gets:
Multiple the numerator and the denominator of the right hand side by .
can be determined, from the original 20th degree equation using Vieta's Formulas, to be
Therefore,
This simplifies to