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 Post subject: Sequences
PostPosted: Fri, 14 Aug 2009 20:09:24 UTC 
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Joined: Thu, 26 Jul 2007 19:08:57 UTC
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Let a_n,b_n,c_n,d_n be integer sequences such that
(1+\sqrt p + \sqrt q )^n = a_n + b_n \sqrt p + c_n \sqrt q + d_n \sqrt {pq}
where p,q are prime numbers.

Find the following limits:
$ \mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}
{{{b_n}}},\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}
{{{c_n}}},\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}
{{{d_n}}}


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 Post subject: Re: Sequences
PostPosted: Sat, 15 Aug 2009 04:40:07 UTC 
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
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Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
M.A wrote:
Let a_n,b_n,c_n,d_n be integer sequences such that
(1+\sqrt p + \sqrt q )^n = a_n + b_n \sqrt p + c_n \sqrt q + d_n \sqrt {pq}
where p,q are prime numbers.

Find the following limits:
$ \mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}
{{{b_n}}},\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}
{{{c_n}}},\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}
{{{d_n}}}


Spoiler:
Because
$
\begin{aligned}
4a_n
&=\mathrm{Tr}_{\mathbb{Q}(\sqrt{p},\sqrt{q})/\mathbb{Q}}
((1+\sqrt{p}+\sqrt{q})^n)\\
&=(1+\sqrt{p}+\sqrt{q})^n+(1+\sqrt{p}-\sqrt{q})^n
+(1-\sqrt{p}-\sqrt{q})^n+(1-\sqrt{p}+\sqrt{q})^n\\
&\sim(1+\sqrt{p}+\sqrt{q})^n
\end{aligned}
and similarly,
$
\begin{aligned}
4b_n
&=\mathrm{Tr}_{\mathbb{Q}(\sqrt{p},\sqrt{q})/\mathbb{Q}}
((1+\sqrt{p}+\sqrt{q})^n/\sqrt{p})\\
&=\dfrac{(1+\sqrt{p}+\sqrt{q})^n+(1+\sqrt{p}-\sqrt{q})^n
-(1-\sqrt{p}-\sqrt{q})^n-(1-\sqrt{p}+\sqrt{q})^n}{\sqrt{p}}\\
&\sim\dfrac{(1+\sqrt{p}+\sqrt{q})^n}{\sqrt{p}}
\end{aligned}
So \dfrac{a_n}{b_n}\sim\sqrt{p}. By the same reasoning, \dfrac{a_n}{c_n}\sim\sqrt{q} and \dfrac{a_n}{d_n}\sim\sqrt{pq}.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject:
PostPosted: Wed, 9 Sep 2009 18:13:38 UTC 
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Joined: Thu, 26 Jul 2007 19:08:57 UTC
Posts: 56
Correct and nice !

This is another idea for a more lengthy approach:
Spoiler:
Write:
\[\begin{array}{rcl}    {{a_{n + 1}} + {b_{n + 1}}\sqrt p  + {c_{n + 1}}\sqrt q  + {d_{n + 1}}\sqrt {pq} } &  =  & {{{(1 + \sqrt p  + \sqrt q )}^{n + 1}} \\ ~ & = & (1 + \sqrt p  + \sqrt q )({a_n} + {b_n}\sqrt p  + {c_n}\sqrt q  + {d_n}\sqrt {pq} )}  \\ {} &  =  & { ({a_n} + p{b_n} + q{c_n}) + ({a_n} + {b_n}q{d_n})\sqrt p   \\ ~ & ~ & +   ({a_n} + {c_n} + p{d_n})\sqrt p  + ({b_n} + {c_n} + {d_n})\sqrt {pq} }  \end{array} \]

by equating the coefficients we get a system of difference equations that can be solved using standard methods (e.g. diagonalization of the system matrix)
\left[ {\begin{array}{*{20}{c}}
   {{a_{n + 1}}}  \\
   {{b_{n + 1}}}  \\
   {{c_{n + 1}}}  \\
   {{d_{n + 1}}}  \\

 \end{array} } \right] = \left[ {\begin{array}{*{20}{c}}
   1 & p & q & 0  \\
   1 & 1 & 0 & q  \\
   1 & 0 & 1 & p  \\
   0 & 1 & 1 & 1  \\

 \end{array} } \right]\left[ {\begin{array}{*{20}{c}}
   {{a_n}}  \\
   {{b_n}}  \\
   {{c_n}}  \\
   {{d_n}}  \\

 \end{array} } \right], with the initial condition:\,\,\left[ {\begin{array}{*{20}{c}}
   {{a_^\circ }}  \\
   {{b_^\circ }}  \\
   {{c_^\circ }}  \\
   {{d_^\circ }}  \\

 \end{array} } \right] = \left[ {\begin{array}{*{20}{c}}
   1  \\
   1  \\
   1  \\
   0  \\

 \end{array} } \right]


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