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Suppose I have the following recurrence relation: $$ E_{n+1} = E_{n} \ \left[ 1 + \left( \frac{ 2 \ \lambda }{ \sqrt{E_{n}} } \right) + \left( \frac{ \lambda }{ \sqrt{E_{n}} } \right)^{2} \right] \tag{1} $$ where $\lambda \in R$ is some known, positive definite constant and we are given $E_{0}$.

Looking at Newton's method, I found a general solution of: $$ f\left( E_{n} \right) = f_{o} \sqrt{ \lambda + 2 \sqrt{E_{n}} } \ e^{-\left(\sqrt{E_{n}}/\lambda\right)} \tag{2} $$ starting from the assumption that we could rewrite Equation 1 in the form: $$ E_{n+1} = E_{n} - \frac{ f\left( E_{n} \right) }{ f'\left( E_{n} \right) } \tag{3} $$

However, I am not sure how this helps me or if it even helps.

Questions
I am curious if there is a way to determine the value of $n$ required to reach some set value for $E_{n+1}$? Is there a general solution (or general approach for this type of recurrence relation) for this example, i.e., a general analytic function that depends upon $E_{0}$, $\lambda$, and $E_{n+1}$ for $n$?

honeste_vivere
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1 Answers1

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Why not expand that all out?

$$E_{n+1}=E_n+2\lambda\sqrt{E_n}+\lambda^2=(\sqrt{E_n}+\lambda)^2$$

Square root both sides,

$$\sqrt{E_{n+1}}=\sqrt{E_n}+\lambda$$

This is an arithmetic sequence with

$$\sqrt{E_n}=\sqrt{E_0}+n\lambda$$

Simply Beautiful Art
  • 74,685
  • Argh... Sorry, I should have seen that. I knew if I committed to asking the question it would be something absurdly obvious like this. Thanks for the answer. – honeste_vivere Aug 02 '17 at 16:59
  • Lol, no problem :D Got lucky I suppose, however, all such quadratic like recurrence relations can be solved by completing the square I believe. – Simply Beautiful Art Aug 02 '17 at 17:01