# Difference between revisions of "User:Wsjradha/Cotangent Sum Problem"

(→Problem:) |
m (→Solution:) |
||

Line 21: | 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> |

## Revision as of 17:09, 19 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