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First-order homogeneous recurrence relation with variable coefficient

  • Thread starter Robin04
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  • #1
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Homework Statement


I need to find the explicit formula for the following recursive sequence:
##v_n=\frac{2}{1+q^n}v_{n-1}## where ##0<q<1## is a constant

Homework Equations


I found the following method to solve it:
https://en.wikipedia.org/wiki/Recurrence_relation#Solving_first-order_non-homogeneous_recurrence_relations_with_variable_coefficients

The Attempt at a Solution


Referring to Wikipedia, first I need to find
, where

My sequence is homogenous so ##g_n = 0##

I'm already in a problem at the beginning. In order to find ##A_n## I need to calculate##\prod_{k=0}^{n-1} \frac{2}{1+q^k} = 2^{n-1}\prod_{k=0}^{n-1} \frac{1}{1+q^k}##
I don't really have any idea how to do this. Can you give me some hint?
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


I need to find the explicit formula for the following recursive sequence:
##v_n=\frac{2}{1+q^n}v_{n-1}## where ##0<q<1## is a constant

Homework Equations


I found the following method to solve it:
https://en.wikipedia.org/wiki/Recurrence_relation#Solving_first-order_non-homogeneous_recurrence_relations_with_variable_coefficients

The Attempt at a Solution


Referring to Wikipedia, first I need to find
, where

My sequence is homogenous so ##g_n = 0##

I'm already in a problem at the beginning. In order to find ##A_n## I need to calculate##\prod_{k=0}^{n-1} \frac{2}{1+q^k} = 2^{n-1}\prod_{k=0}^{n-1} \frac{1}{1+q^k}##
I don't really have any idea how to do this. Can you give me some hint?
You are done; that is about as simple a formula as can possibly be found.
 
  • #3
SammyS
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You are done; that is about as simple a formula as can possibly be found.
Don't you also get the result that ##\displaystyle \ A_{n} = A_{n-1} \ \dots \ ## so that ##\displaystyle \ A_{n} = A_{0} \,?##
 
  • #4
Ray Vickson
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Don't you also get the result that ##\displaystyle \ A_{n} = A_{n-1} \ \dots \ ## so that ##\displaystyle \ A_{n} = A_{0} \,?##
Using the recursion directly (and not bothering to look up formulas in Wikipedia) we have
$$ \begin{array}{rcl}
v_1 &=& \frac{2}{1+q} v_0 \\
v_2&=& \frac{2}{1+q^2} v_1 = \frac{2^2}{(1+q)(1+q^2)} v_0 \\
v_3 &=& \frac{2}{1+q^3} v_2 = \frac{2^3}{(1+q)(1+q^2)(1+q^3)} v_0 \\
\vdots & \vdots &\hspace{4em} \vdots
\end{array} $$ So
$$v_n = 2^n \left( \prod_{k=1}^n \frac{1}{1+q^k} \right) \: v_0,$$ and it does not get simpler than that.
 
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