# Closed form for a non-linear recurrence

Does equation $a_n=\sqrt{a_{n-1}+6}$ with $a_1=6$ have a closed form? I've found no linearization method. Any suggestion or hint will be highly appreciated.

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It seems to converge to $3$... – draks ... Aug 3 '12 at 8:43
@draks: It does converge to $3$ (roughly like a geometric series of ratio $1/6$). – Did Aug 3 '12 at 9:07
I would guess not, but it is going to be close to $3 + \frac{k_1}{6^n} + \frac{k_2}{6^{2n}} + \cdots$ for some $k_1,k_2,\ldots$ (in this case it seems $k_1 \approx 16.4558$ and $k_2 \approx 0.614$). – Henry Aug 3 '12 at 9:46
If the question is $a_n=\sqrt{a_{n-1}+2}$ instead I can solve this explicitally. – doraemonpaul Aug 4 '12 at 5:30
@doraemonpaul Then you should post this here (and probably mention that your solution is direct when $|a_1|\leqslant2$ but that it needs some adjustment otherwise). – Did Sep 8 '12 at 13:21

There is no hope to find an explicit formula for $a_n$, but the asymptotics is clear.

One has $a_n=u(a_{n-1})$ where the function $u:x\mapsto\sqrt{x+6}$ has a unique fixed point $a=3$, hence $a_n-a=u(a_{n-1})-a=u(a_{n-1})-u(a)$ and one can suspect that $a_n\to a$. As a matter of fact, $u(a_n)-a=b_n\,(a_{n-1}-a)$ with $b_n=1/(u(a_{n-1})+a)$ hence $0\lt b_n\lt1/a$ hence $|a_n-a|\leqslant a^{-n}\,|a_0-a|$. Since $a\gt1$, this shows that $a_n\to a$.

More is true: since $b_n\to b=1/(2a)=1/6$, $a_n-a=b^{n+o(n)}$. In other words, since $a_n\gt a$ for every $n$, $$\lim\limits_{n\to\infty}\frac{\log(a_n-a)}n=\log(b)=-\log(6),$$ and a little more work shows that $a_n-a=c\,b^n\,(1+o(1))$, where $c$ depends on $a_0$ and has no simple explicit form.

Edit: The algebraic trick used above to compute $b$ might hide the fact that $b=u'(a)$, where $a=u(a)$ is the fixed point of $u$.

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@J.M. About your first Edit: I seem to encounter the construction the asymptotics + is regularly. Could you indicate an authoritative source which forbids this usage? – Did Aug 3 '12 at 16:24
I'll have to get back to you on that, but if memory serves: "asymptotics" the subject matter being studied is singular, but "asymptotics" as in the behavior of the function is plural. A bit like "statistics" in that regard... – J. M. Aug 3 '12 at 16:45
@J.M. Until then, let me stick to my idiom (and to \lim\limits, by the way), if you don't mind. – Did Aug 3 '12 at 16:47
Well, on the matter of \lim\limits, since you have the thing enclosed in $$, whether \limits is there or not, the display is the same, so it'd seem \limits is moot (unless there is some hidden benefit of this idiom of yours). I'd have left the thing if it were enclosed in ... – J. M. Aug 3 '12 at 16:58 @vonbrand Thanks for the explanation on this terminology. – Did Mar 17 '13 at 14:18 Rewrite it as$$ a_n^2=a_{n-1}+6\\ a_{n}-a_{n-1}=a_n-a_n^2+6 $$and approximate it as$$ a_n'-a_n+a_n^2-6=0, $$which is solved by$$a_n = \frac{3 e^{5 n}+2 e^{5 c_1}}{e^{5 n}-e^{5 c_1}},$$with c_1= \frac15 \left(5-3 \log(2)+\log(3)\right)\;\; such that a_1=6. Here's a plot... - I've checked your hidden solution with Mathematica. It didn't satisfy the given recursion. – Christian Blatter Aug 3 '12 at 9:21 i think that a_{n-1}=\sqrt{7} and a_{n-2}=1 – dato datuashvili Aug 3 '12 at 9:27 @ChristianBlatter no, it's an approximation... – draks ... Aug 3 '12 at 10:43 it's an approximation... In which sense? – Did Aug 3 '12 at 10:50 Not an approximation of what? (this is clear) but an approximation in which sense? As you know, approximating discrete iterations by solutions of differential equations is a tricky business. In the case at hand, the limit a_\infty=3 is correct but not much more, for example a_n-a_\infty is not of the order of \mathrm e^{-5n}. – Did Aug 3 '12 at 11:15 I think I found closed-form solution for this equation. We need to find solution for this one (a_0 as a first element):$$ a_n = \sqrt{a_{n-1}+6}, \quad a_0=6. $$We convert equation to this:$$ a_{n-1} = a_n^2 - 6, \quad a_0=6. $$I always can change direction of element index and rewrite last equation in the next form (we call that equation "eq1"):$$ F_{n+1} = F_n^2 - 6, \quad F_0=6. $$If we find solution (as function F(n)) for that equation we can use it to express the closed-form solution for your problem. If F_n = F(n) then a_n = F(-n). How can we found F(n)? Solution for equation b_{n+1} = b_n^2, \ \ b_0=\beta is b_n = \beta^{2^n}. We will seek the solution of the eq1 in this form:$$ F_n = F(n) = f(X^{2^n}) $$where f(x) is some unknown function and X is some number. For f we know that$$ f(x^2) = f^2(x) - 6, $$because we must satisfy the eq1$$ F_{n+1} = f(X^{2^{n+1}}) = f((X^{2^n})^2) = f(X^{2^n})^2 - 6 = F_n^2 - 6. $$We use this property and we find f(x) as a fixed point (without proof of existence)$$ f(x) = \sqrt{6 + f(x^2)}  f(x) = \lim\limits_{k -> \infty} \sqrt{6 + ... \sqrt{6 + f_0(x^{2^k})}}, $$(number 6 in this expression used k times) where f_0(x) is some random function (f_0(x) = x for example). Function f(x) are very interesting (it's not elementary or analytic). Here is some of its properties: 1. f(x^2) = f^2(x) - 6. 2. f(0) = f(1) = 3. f(x^2) = f^2(x) - 6 => f(0) = f^2(0) - 6 => f(0) = \frac{1}{2}(1 + \sqrt{1 + 4 \cdot 6}) => f(0) = 3. (another proof is to using the formula \sqrt{c+\sqrt{c+...}} = \frac{1}{2}(1 + \sqrt{1 + 4 c}).. you can find this on http://en.wikipedia.org/wiki/Nested_radical ) 3. f(a) = 3 for all a : |a|<1. (it's very easy to prove) 4. f(-x) = f(x). 5. f(x) is continuous in all real x. (without proof) Here is a plot of this function: Now we can write the solution of eq1:$$ F_n = F(n) = f(X^{2^n}) $$where X is positive solution of equation f(X) = 6 (because we need satisfy initials F_0 = f(X) = 6). (x \approx 5.46806882358680646837316643) Finally we can write the solution for you equation$$ a_n = F(-n). $$(a_n -> 3 because F(-n) = f(X^{2^{-n}}) -> f(1) = 3) Same method we can use to find closed-form solution in more general cases! If you have Mathematica you can use my code for check solution:  c = 6; (*constant adder in square root*) b = 6; (*initial value (for F_0)*) f0[x_] := x; (*finding f(x)*) N1 = 10; (*precision*) g[x_] := Sqrt[c + x]; f[x_] := Nest[g, f0[x^2^N1], N1]; (*findinf X*) N2 = 10; (*precision*) res1 = FindRoot[f[x] == b, {x, c}, WorkingPrecision -> 1000]; X = x /. res1; F[n_] := f[X^2^n] (* you can use this function for comparing (this for eq1) *) F2[n_] := F2[n - 1]^2 - c; F2[0] := b;   For example:  F[0] F[-1] F[-2]   gets  6.00000000000000000000000000000000000000000000000000000000000000000000... 3.46410161513775458705489268301174473388561050762076125611161395890386... 3.07637800264170309696602586393672241931859085772505962544063421316756...   - Indeed one always can change direction of element index, but, if one does, one obtains a completely different problem. Thus, this answer addresses a, possibly interesting but, quite unrelated question. – Did Sep 8 '12 at 6:40 did: It's not completely different problem. It is a same one problem in some sence. I use is solution to express the solution for initial problem. See below my another answer! Also you always can check my solution. It's really work! – Zegalur Sep 8 '12 at 9:08 For example above some one write that he can find solution for a_n=\sqrt{a_{n-1}+2}. First we find solution for F_n = F^2_{n-1}-2. The solution is F_n = x^{2^n} + x^{-2^n} (where x we find from equation x+\frac{1}{x}=a_0). We also automatic get the solution for a_n. a_n = F_{-n} = x^{2^{-n}} + x^{-2^{-n}}. (x=3+2\sqrt{2}). – Zegalur Sep 8 '12 at 9:23 Also we can use the same general method. In this case we have function f(x) that is equal to g(x)=x+\frac{1}{x} for all x \ge 0 (img193.imageshack.us/img193/1260/nonlinear03.png). We get the same numeric result! – Zegalur Sep 8 '12 at 10:09 As I said, your answer brings zero information about the initial problem. As long as you play with words as when saying that If F_n=F(n) then a_n=F(-n) (??) where f(x) is some unknown function (indeed, f is quite unknown...), I see no way in which one would use i(t)s solution to express the solution for initial problem. Sorry. – Did Sep 8 '12 at 13:17 Zegalur: I followed your interesting post but cannot see the final closed form. My first question is why to "change direction of element index" when the changed recurrence Fn+1=F2n−6 is as hard as the original a2n=an−1+6? Has this backward recurrence, say, something to do with symmmetries in a further elaboration? Would you please clarify a bit? Strange. I can't write any comments.. so I wrote a new answer.. Anyway... I take a square of two parts of equation and move 6 to the left side of equation.$$ a_n = \sqrt{a_{n-1}+6} \ \ \ => \ \ \ a^2_n - 6 = a_{n-1} \ \ \ => \ \ \ a_{n-1}=a^2_{n} - 6. $$If we have a_0 we now can find a_{-1}, a_{-2}, ... (a_{-1}=30). I've changed direction of index because it's more native to work with something like this:$$ F_n = Q(F_{n-1}). $$(when the next element of sequence is expressed through previous one). Now F_{n} = a_{-n}. If I will find solution in a closed-form F_{n} = W(n) then I can use it to express solution for a_n.$$ a_n = F_{-n} = W(-n). $$It is exactly what I did. You also write that you can't find a final result. That is because solution is expressed by non elementary function f(x) (see my answer). But you can't even find a Taylor series for it. But that function exist and it is continuous (see for example the plot of it). Final result is$$ a_n = F_{-n} = f(X^{2^{-n}})  where $f$ and $X$ are defined in my previous answer.

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Zegalur: I followed your interesting post but cannot see the final closed form. My first question is why to "change direction of element index" when the changed recurrence $F_{n+1}=F_n^2-6$ is as hard as the original $a_n^2=a_{n-1}+6$? Has this backward recurrence, say, something to do with symmmetries in a further elaboration? Would you please clarify a bit? – Tavasanis Sep 7 '12 at 15:15
Almost forgot. $f(x)$ is defined as a limit. But we don't need to find it. If we take some big $k$ we can find a very good approximation for it (in my example if I take $k=10$ it is such good approximation that all significant digits for $F_{10}$ is good). We also need to find $X$ with good accuracy. – Zegalur Sep 7 '12 at 17:01