The general term formula of such a recursion formula is what? I wanted to know how to prove it has the general term formula and what it is? $$ \frac{1}{a_n^2}+a_n^2=a_{n+1}^2$$ $$a_0=1$$
The general term of such a recursion formula
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sequences-and-series
recurrence-relations
closed-form
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0What is $a_{1}$? – 2017-01-21
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0sorry, i forgot,it's 1 – 2017-01-21
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0Start by putting $b_{n} = a_{n}^2$. The recursion becomes $$ {1 \over b_{n}} + b_{n} = b_{n+1}, \quad b_{1} = 1. $$ – 2017-01-21
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0@Mclalalala -- What makes you think a closed form exists? – 2017-01-21
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0i am not sure, so i'm thinking if there exists an expression of the infinite series or something else, and I want to prove whether or not there exists a close form. – 2017-01-21
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0@Mclalalala -- I doubt there is a closed form. Moreover, if no closed form exists, proving such non-existence would most likely be a very hard problem, requiring advanced methods. – 2017-01-21
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0advanced methods are ok if you can get one – 2017-01-21
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0@Mclalalala -- The advanced methods I alluded to are certainly beyond my range of knowledge. Typically, proving that a closed form _doesn't_ exist involves first developing a customized version of Galois Theory, tailored to your class of equations -- a nontrivial task, and possibly years of work (even after the time spent learning the prerequisites). – 2017-01-21
1 Answers
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The numerators and denominators of $a_n^2$ are sequences A073833 and A073834 in the OEIS. There doesn't seem to be a (known) closed form for your sequence. The OEIS gives a few references that you could try looking at:
H. L. Montgomery, Ten Lectures on the Interface Between Analytic Number Theory and Harmonic Analysis, Amer. Math. Soc., 1996, p. 187.
D. J. Newman, A Problem Seminar, Springer; see Problem #60.
J. H. Silverman, The arithmetic of dynamical systems, Springer, 2007, see p. 113 Table 3.1
Edit: Neither Newman nor Silverman gives an explicit formula, but Newman does give very good bounds.
Proposition: Let $x_0 = 1$ and $x_{n+1} = x_n + 1/x_n$. Then $$ \sqrt{2n + 2} \lt x_n \lt \sqrt{2n + \frac52 + \frac12 \ln(n-1)}$$ for all $n \ge 2$.
Proof. Let $y_n = x_n^2 - 2n$. A short calculation shows that $$y_{n+1} = y_n + \frac{1}{y_n + 2n}$$ for all $n \ge 0$. It follows that $$y_n \lt y_{n+1} < y_n + \frac{1}{2n}$$ for all $n \ge 1$. Thus $$2 \lt y_n \lt 2 + \frac{1}{2} \sum_{k=1}^{n-1} \frac{1}{k} < 2 + \frac{1}{2}(\ln(n-1) + 1) = \frac52 + \frac12 \ln(n-1)$$ for all $n \ge 2$. Rewriting this in terms of $x_n$ yields the desired inequality. $\square$
A final remark: I have found one paper where the recurrence $x_{n+1} = x_n + 1/x_n$ is mentioned explicitly. I suppose you could try emailing the author, but I wouldn't get my hopes up if I were you.
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0Can it be expresses in an infinite series with the advanced methods? – 2017-01-21
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0Did you look at the links I gave? They give a series formula for $a_n^2$ and also some references to the litterature. – 2017-01-21
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0I have seen, but I still don't know what exactly it is and where is the formula. – 2017-01-21
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0What is it that you don't know exactly? The formula is in the "Comments" section of the first link, but it is provided without explanations. You should probably look at the books in the "References" section. – 2017-01-21
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0sorry, I really can't find the books you mentioned – 2017-01-21
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0i didn't find the pdf of such books,how can I do? – 2017-01-21