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

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m (Cotangent Sum Problem moved to User:Wsjradha/Cotangent Sum Problem: Sorry, but this doesn't really belong in the main space.)
 
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<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>
  
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Let <math>S = \cot{\left(\sum_{k = 1}^{20} \cot^{ - 1}{z_k}\right)}</math>
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Let <math>S = \cot{\left(\sum_{k = 1}^{20} \cot^{ - 1}{z_k}\right)}</math>.
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This equals <math>cot (c(k_1))</math>
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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:
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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>

Latest revision as of 22:59, 20 December 2008

Problem:

Let $z_1, z_2, \ldots, z_{20}$ be the twenty (complex) roots of the equation \[z^{20} - 4z^{19} + 9z^{18} - 16z^{17} + \cdots + 441 = 0.\]

Calculate the value of \[\cot{\left(\sum_{k = 1}^{20} \cot^{ - 1}{z_k}\right)}.\]

Solution:

For the purpose of this solution $k_n$ will be the sum of the roots of the 20th degree polynomial, taken $n$ at a time. For example,

$k_1 = z_1 + z_2 + \cdots + z_{20}$

$k_2 = z_1z_2 + z_1z_3 + \cdots + z_{19}z_{20}$

Also, $c(k_n)$ will be the sum of the cotangent inverses of the roots, taken $n$ at a time. The cotangent inverses will be multiplied as necessary, then added.

Also, $tc(k_n)$ will be the sum of the tangents of the cotangent inverses of the roots, taken $n$ at a time. Basically, this is the same as $c(k_n)$ except that the tangents are taken right after the cotangent inverses. For example,

$tc(k_1) = tan(cot^{ - 1}{z_1}) + tan(cot^{ - 1}{z_2}) + \cdots + tan(cot^{ - 1}{z_{20}})$

Let $S = \cot{\left(\sum_{k = 1}^{20} \cot^{ - 1}{z_k}\right)}$. This equals $cot (c(k_1))$. There is a formula that states the following, where, for the purposes of this formula only, $\sum_{a_1}^n(b_m)$, is the sum of $a_1$ through $a_n$, taken $m$ at a time, in the fashion described above:

$tan{\left(a_1+a_2+a_3+\cdots+a_n\right)}=\dfrac{1-\sum_{a_1}^n(b_2)+\sum_{a_1}^n(b_4)-\sum_{a_1}^n(b_6)+\cdots}{\sum_{a_1}^n(b_1)-\sum_{a_1}^n(b_3)+\sum_{a_1}^n(b_5)-\cdots}$

When applied to this problem, it yields:

$tc(k_1) = \dfrac{1 - tc(k_2) + tc(k_4) + \cdots + tc(k_{20})}{tc(k_1) + tc(k_3) + \cdots + tc(k_{19})}$

Taking the reciprocal of either side, one gets:

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

Multiple the numerator and the denominator of the right hand side by $tc(k_{20})$.

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

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

$S = \dfrac{21^2 - 19^2 + 17^2 - 15^2 + \cdots - 3^2 + 1}{20^2 - 18^2 + \cdots - 2^2}$

This simplifies to $\boxed{\boxed{\dfrac{241}{220}}}$

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