View source for User:Wsjradha/Cotangent Sum Problem

== 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: ==

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 example,

<math>k_1 = z_1 + z_2 + \cdots + z_{20}</math>

<math>k_2 = z_1z_2 + z_1z_3 + \cdots + z_{19}z_{20}</math>

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.
For example,

<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>.
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:

<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>

When applied to this problem, it yields:

<math>tc(k_1) = \dfrac{1 - tc(k_2) + tc(k_4) + \cdots + tc(k_{20})}{tc(k_1) + tc(k_3) + \cdots + tc(k_{19})}</math>

Taking the reciprocal of either side, one gets:

<math>S = \dfrac{tc(k_1) + tc(k_3) + \cdots + tc(k_{19})}{1 - tc(k_2) + tc(k_4) + \cdots + tc(k_{20})}</math>

Multiple the numerator and the denominator of the right hand side by <math>tc(k_{20})</math>.

<math>S = \dfrac{k_{20} - k_{18} + k_{16} - k_{14} + k_{12} - k_{10} + k_8 - k_6 + k_4 - k_2 + 1}{k_{19} - k_{17} + k_{15} - k_{13} + k_{11} - k_9 + k_7 - k_5 + k_3 - k_1}</math>

<math>k_n</math> can be determined, from the original 20th degree equation using Vieta's Formulas, to be <math>(n+1)^2</math>
Therefore,

<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{\boxed{\dfrac{241}{220}}}</math>