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Good evening to everyone. I am trying to solve the following problem. Let $ (a_n) $ be a sequence defined recursively as: $\displaystyle a_{n+1}=\frac{2a_n^2+1}{2a_n}$ and $a_1=11$ . I know there is a theorem in order to find a closed formula if the number 2 on the numerator did not exist, but here it seems complicated. Actually this was an exercise on a problem sheet that gave to me one of my students. The ''highlight'' theorem in his course was the following: If $\displaystyle a_{n+1}=\frac{a_n^2 + k^2}{2a_n}$ , then $\displaystyle \frac{a_n - k}{a_n +k} = \left({\frac{a_1-k}{a_1+k}}\right)^{2^{n-1}}$ . On the problem sheet there were many exercises that were exactly applications of the above theorem, nothing really difficult. But the LAST exercise was the one I mentioned in my original post,which was given,I suppose, either to show that it is in general hard to solve general recursions or there is some trick here I cannot think by myself . Does anybody know how to solve this? I am asking because it is possible there is a typo here, so I would like to know if indeed this exercise cannot be solved. Thanks a lot!

Petros Karajan
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  • Keeping in mind my lack of formal training in this area, my standard approach to your question would be to compute $a_n$ for each $n \in {1,2,3,4,5,6}$. Then, I would look for a pattern in the data, form a hypothesis based on this data, and then try to prove the hypothesis (plausibly through induction). Here, there is a potential shortcut. Since it is reasonable to surmise that the problem is an application of the theorem, and yet the problem doesn't seem to directly coorelate with the theorem's conclusion, there is an alternative. ...see next comment – user2661923 Feb 08 '22 at 22:25
  • It could be that while the problem is not a direct application of the theorem, the problem may be an application of the analysis in the proof to the theorem. That is, if you study the theorem's proof, you may be able to alter the proof, so that it applies to the problem that you have posed. – user2661923 Feb 08 '22 at 22:27
  • Can you show the proof about the theorem? Nice problem. We can see for example $\frac{1}{2}a_{n+1}=\frac{a_{n}^{2}+\frac{1}{2}}{2a_{n}}$. –  Feb 08 '22 at 22:28
  • Well, induction seems work with the theorem. –  Feb 08 '22 at 22:35
  • @Alex, re my 2nd comment, the difficulty is that induction might not provide the theorem-analysis needed to conquer the posed problem. If a direct proof of the theorem can be derived, then (perhaps) that derivation can be altered to suit the needs of the posed problem. – user2661923 Feb 08 '22 at 22:40
  • For the theorem's proof: notice that $ a_{n+1}-k=\frac{(a_n-k)^2}{2a_n}$ and similarly $ a_{n+1}+k=\frac{(a_n+k)^2}{2a_n}$ . Now divide the two equations and get: $\frac{a_{n+1}-k}{a_{n+1}+k}=(\frac{a_n-k}{a_n+k})^2$. Inductively, you see that the pattern goes the same way until you reach $\frac{a_1-k}{a_1+k}$ which has the power $2^{n-1}$. But the initial trick we did here worked well because the factor in front of $a_n^2$ was 1 .If there is a diferent factor, it does not work, cause you cannot create a perfect square in the right without changing the pattern on the left side. – Petros Karajan Feb 08 '22 at 22:55
  • If we write $\frac{1}{2} a_{n+1}=\frac{a_n^2+\frac{1}{2}}{2a_n}$ you cannot apply the theorem, because on the left you have the factor $\frac{1}{2}$. Even if you call $b_{n+1}=\frac{1}{2}a_{n+1}$,you again cannot use the theorem, because you do not have on the right $\frac{b_n^2+\frac{1}{2}}{2b_n}$ . – Petros Karajan Feb 08 '22 at 22:59
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    Normally, I would suggest a direct approach derivation, or an induction derivation. With induction, you assume that the formula is given by a polynomial and deduce what constraints the polynomial must satisfy. However, I regard either of these two approaches as outside the spirit of the question, which is: What was the problem composer's intent? How did the problem composer intend that either the theorem, or the proof to the theorem should be used to solve the problem? – user2661923 Feb 09 '22 at 02:45
  • I don't understand where exactly is you question. The "highlight" theorem result you give is rediscovered in many different places, like here or there for example (found with formula searching tool https://approach0.xyz/search/) – Jean Marie Feb 09 '22 at 09:36
  • Thank all for your comments. The question was to find a closed formula for $a_n$ (something like $a_n=\frac{5n}{n^3+2}$ ) . I mentioned this theorem cause it COULD give an explicit formula, if there was no 2 on the numerator. So I supposed there were 3 scenarios: i)there was a typo in the exercise (which is by far the most possible scenario) , ii) there was an extra trick someone could use together with the theorem's proof,and deduce the closed formula , iii) the lecturer just wanted to show to the students that in general it is difficult to find closed formulas for recursive sequences. – Petros Karajan Feb 09 '22 at 11:01

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Based on previous work in a related MSE question 3489102 I found a similar approach that works in this case. There is no closed form solution for these that I know of.

Given the function $\,f(x) := x + \frac1{2x}\,$ define the sequence $\,a_{n+1} := f(a_n).\,$ Numerical experiments suggest to define the sequence $\,b_n := a_n^2.\,$ Now define $\,g(x) := x + 1 + \frac1{4x}\,$ and $\,b_{n+1} := g(b_n) .\,$ The asymptotic expansion of $\,b_n\,$ depends on the solution to the functional equation $\,B(x+1) = g(B(x))\,$ and is given by $$ B(x) := x + \frac{2+y}{4} + \frac{1+y}{16x} + \frac{1-3y^2}{384x^2} + \frac{7-6y-9y^2+6y^3}{4608x^3} + \cdots $$ where $\,y:=\log(x).\,$ Then $\,b_n \approx B(n+c_0)\,$ where $\,c_0\,$ depends only on $\,b_0.\,$

For example, if $\,b_0=2, a_0=\sqrt{2}\,$ then $\,c_0\approx 1.36157\,$ and $\,a_{100} \approx 10.14986 \approx \sqrt{B(100+c_0)}.\,$

By the way, the "highlight" theorem you refer to is the Babylonian or Newton's method for computing square roots. That is, define the iteration $\,a_{n+1}=\frac{a_n^2 + N}{2a_n}.\,$ Then the sequence converges to $\,k=\sqrt{N}\,$ and the formula with the $\,2^{n-1}\,$ exponent gives the precise convergence of the iteration sequence.

Somos
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  • Thank you for your help. Could you explain in a little bit more detail why we cannot find a closed form solution ? What do you mean by ''asymptotic expansion of $b_n$ '' and where did the function $B(x)$ come from? – Petros Karajan Feb 09 '22 at 11:19
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    @PetrosKarajan I don't have any proof that it has no closed form for a precisely defined class of "closed form" solutions. That would be very difficult. Also, I don't have any proof that the series for $,B(x),$ converges. It may be similar to Stirling's asymptotic expansion of the gamma function. The function $B(x)$ comes from my original work in 1999 which I present a summary of in the MSE question 3489102 I linked to. – Somos Feb 09 '22 at 12:15
  • Ok,thank you a lot! Yes, probably there is a typo in this exercise,the tutor just typed an extra 2 on the numerator. – Petros Karajan Feb 10 '22 at 22:46