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I would like to find the general solution of the difference equation $y(n+1) = \alpha y(n)^2 $. I know that the general solution to $y(n+1) = y(n)^2$ is $y(n) = \exp({c\cdot 2^{n}})$. However, I've not yet been able to rewrite it to a general solution of $y(n+1) = \alpha y(n)^2$. Is this a efficient approach to finding the general solution. If so, what steps do I need to take to get to the general solution of $y(n+1) = \alpha y(n)^2 $. If not, what would be a better approach to solve this difference equation?

abiessu
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reinout-g
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  • $$\alpha y(n)=(\alpha y(n-1))^2\implies\frac{\log(\alpha y(n))}{2^{n}}=\frac{\log(\alpha y(n-1))}{2^{n-1}}=\cdots=\log(\alpha y(0))$$ – Did Jan 19 '15 at 20:24

3 Answers3

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$$ y(n)\rightarrow \beta \bar{y}(n) $$ thus $$ \beta \bar{y}(n+1) = \beta^2\alpha\bar{y}(n)^2\implies \bar{y}(n+1) = \alpha\beta\bar{y}(n)^2 $$ let $\alpha\beta = 1\implies \beta = \frac{1}{\alpha}$ thus $$ \bar{y}(n+1) = \bar{y}(n)^2\implies \bar{y}(n) = \mathrm{e}^{c2^n} $$ so $$ y(n) = \frac{1}{\beta}\mathrm{e}^{c2^n} = \alpha \mathrm{e}^{c2^n} $$

$\textbf{edit}$ As pointed out by the OP..i made a mistake in the last line. $$ y(n) = \beta \bar{y}(n) = \beta \mathrm{e}^{c2^n} = \frac{1}{\alpha}\mathrm{e}^{c2^n} $$

Chinny84
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    Thank you for your answer, I have some trouble trying to fully understand it, but I believe there must be a small mistake somewhere since $y(n) = \alpha e^{c2^{n}}$ is not the correct answer since $y(n+1) = \alpha y(n) = \alpha^2 e^{c2^{n+1}} \neq \alpha e^{c2^n+1}$. Your answer did however bring me to the correct answer $y(n) = \frac{1}{\alpha}e^{c2^{n}}$. Thank you for your help – reinout-g Jan 19 '15 at 16:38
  • I do have a mistake here .. ha. $y(n) = \beta \bar{y}(n) = \beta \mathrm{e}^{c2^n} = \frac{1}{\alpha}\mathrm{e}^{c2^n}$ – Chinny84 Jan 19 '15 at 16:43
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Let $z(n)=\alpha\,y(n)$. Then $$ z(n+1)=\alpha\,y(n+1)=\alpha^2\,y(n)^2=z(n)^2. $$

Julián Aguirre
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I like this methods:

since $$y_{n+1}=a(y_{n})^2\Longrightarrow \ln{(y_{n+1})}=\ln{a}+2\ln{(y_{n})}$$ let $$\ln{(y_{n})}=x_{n}$$ so $$x_{n+1}=\ln{a}+2x_{n}\Longrightarrow x_{n+1}+\ln{a}=2\left(x_{n}+\ln{a}\right)$$ so $$\Longrightarrow x_{n}+\ln{a}=(\ln{x_{1}}+\ln{a})2^{n-1}$$ so $$y_{n}=e^{x_{n}}=e^{(\ln{x_{1}}+\ln{a})2^{n-1}-\ln{a}}=e^{\ln{\left(\dfrac{(ax_{1})^{2^{n-1}}}{a}\right)}}=\dfrac{(ax_{1})^{2^{n-1}}}{a}$$ where $x_{1}=\ln{y_{1}}$

math110
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  • Could you please explain to me why $x_{n+1} + \ln{a} = 2(x_n + \ln{a}) \Rightarrow x_n +\ln{a} = (\ln{x_1} + \ln {a}) 2^{n-1}$? – reinout-g Jan 19 '15 at 20:22