- #1
daniel6874
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The following came up in the context of a genetics question.
\begin{align*}
a_{n+1} &= 2a_{n}d_{n} + b_{n}d_{n}\\
b_{n+1} &= 2a_{n}e_{n} + b_{n}d_{n} + b_{n}e_{n} + 2c_{n}d_{n} + c_{n}e_{n}\\
c_{n+1} &= b_{n}e_{n} + c_{n}e_{n}\\
d_{n+1} &= 2a_{n}d_{n} + 2a_{n}e_{n} + b_{n}d_{n} + b_{n}e_{n} + c_{n}e_{n}\\
e_{n+1} &= b_{n}d_{n} + b_{n}e_{n} + 2c_{n}d_{n} + c_{n}e_{n}
\end{align*}
If the initial values for the variables is ##a_0=b_0=d_0=e_0=1, c_0=0## does the ratio
##e_{n}/(a_{n}+b_{n}+c_{n}+d_{n}+e_{n})## converge as ##n## gets large? Can the ratio be calculated without a computer?
Mathematica gives that ##e_{n}/(a_{n}+b_{n}+c_{n}+d_{n}+e_{n})## is about ##0.109## for ##n=20##, FWIW.
\begin{align*}
a_{n+1} &= 2a_{n}d_{n} + b_{n}d_{n}\\
b_{n+1} &= 2a_{n}e_{n} + b_{n}d_{n} + b_{n}e_{n} + 2c_{n}d_{n} + c_{n}e_{n}\\
c_{n+1} &= b_{n}e_{n} + c_{n}e_{n}\\
d_{n+1} &= 2a_{n}d_{n} + 2a_{n}e_{n} + b_{n}d_{n} + b_{n}e_{n} + c_{n}e_{n}\\
e_{n+1} &= b_{n}d_{n} + b_{n}e_{n} + 2c_{n}d_{n} + c_{n}e_{n}
\end{align*}
If the initial values for the variables is ##a_0=b_0=d_0=e_0=1, c_0=0## does the ratio
##e_{n}/(a_{n}+b_{n}+c_{n}+d_{n}+e_{n})## converge as ##n## gets large? Can the ratio be calculated without a computer?
Mathematica gives that ##e_{n}/(a_{n}+b_{n}+c_{n}+d_{n}+e_{n})## is about ##0.109## for ##n=20##, FWIW.
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