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Let sequence $$a_{1}=1,a_{n+1}=\dfrac{2}{2a_{n}+1}$$

show that $$\left|a_{2n}-a_{n}\right|<\dfrac{10}{27}$$ and the constant $\frac{10}{27}$ A smaller number instead

$$a_{n+1}+\dfrac{1+\sqrt{17}}{4}=2(1+\sqrt{17})\cdot\dfrac{a_{n}+\frac{1+\sqrt{17}}{4}}{4(2a_{n}+1)}\tag{1}$$ $$a_{n+1}-\dfrac{-1+\sqrt{17}}{4}=-2(-1+\sqrt{17})\cdot\dfrac{a_{n}-\frac{-1+\sqrt{17}}{4}}{4(2a_{n}+1)}\tag{2}$$ $\dfrac{(1)}{(2)}$,that $$\dfrac{a_{n+1}+\dfrac{1+\sqrt{17}}{4}}{a_{n+1}-\dfrac{-1+\sqrt{17}}{4}}=\dfrac{1+\sqrt{17}}{1-\sqrt{17}}\cdot\left(\dfrac{a_{n}+\dfrac{1+\sqrt{17}}{4}}{a_{n}-\dfrac{-1+\sqrt{17}}{4}}\right)$$ so we I get ugly even more can't solve this problem.can you some recommended or solve this?

math110
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    Prove by induction – Tesla Jun 05 '16 at 14:58
  • can you post full solution,because I found it's nod easy,even more ugly – math110 Jun 05 '16 at 15:08
  • @Sigma,Assmue that $ |a_{2n}-a_{n}|<\dfrac{27}{10}$,then consider $|a_{2(n+1)}-a_{n+1}|?$ or other? – math110 Jun 05 '16 at 15:12
  • I think this problem exist without induction methods and simple,and we consider How to replace the smaller numbers? because this absult,use induction maybe can't works? – math110 Jun 05 '16 at 15:23
  • Aren't all the terms between 1 and 2/3? – davik Jun 05 '16 at 16:20
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    @Sigma "Prove by induction" I do not understand this comment. Did you have a specific idea in mind when you posted this or is it a shot in the dark? (Apparently, 5 users not only know how to apply your hint (I don't) but think that it yields the best solution. Beats me.) – Did Jun 06 '16 at 04:30
  • @Did a shot in the dark. After that I tried to do so but I failed which I thought is most likely because I'm lacking approximation skills..apparently 5 other people were agreeing, so I didn't delete the comment :-D – Tesla Jun 06 '16 at 04:48
  • @Sigma Posting shots in the dark is allright but only if you mention them as such. You did quite the opposite here since your comment reads as "I did it by induction, I am absolutely sure this is the best way, just follow me". Indeed I would delete the comment if I were you (and please do not get me started on upvotes on mse... :-)). – Did Jun 06 '16 at 04:57

3 Answers3

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The function $f(x)=\frac{1}{x+1/2}$ maps $[2/3,1]$ to $[2/3,6/7] \subset [2/3,1]$, since $1/3 < 10/27$ the result follows.

davik
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  • you mean $a_{n}\in [2/3,1]$ for all $n?$ – math110 Jun 05 '16 at 16:27
  • Yes, that is what i mean – davik Jun 05 '16 at 16:28
  • I think this methods is very nice +1 – math110 Jun 05 '16 at 16:30
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    so $|a_{2n}-a_{n}|\le 1/3$ the $\dfrac{1}{3}$ is best constant for $n$,because $a_{2}-a_{1}=\dfrac{1}{3}$, – math110 Jun 05 '16 at 16:32
  • @functionsug Indeed, and, for each $N$, a constant such that $|a_{2n}-a_n|\leqslant c_N$ for every $n\geqslant N$ is $c_N=|a_{N+1}-a_N|$. Note that $(c_N)$ is decreasing and that $c_N\to0$ when $N\to\infty$. – Did Jun 06 '16 at 05:05
  • @Did,Nice ,and How to prove $c_{N}\to 0,N\to+\infty$? – math110 Jun 06 '16 at 05:12
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    @functionsug Contraction property... Let $\ell$ in $[\frac23,1]$ solve $f(\ell)=\ell$. Since $|f'(\ )|\leqslant r$ on $[\frac23,1]$, with $r=\frac67$, one knows that $|a_n-\ell|\leqslant r^{n-1}\cdot|a_1-\ell|$ for every $n$, qed. – Did Jun 06 '16 at 05:18
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For convenience, let $p=\frac{1+\sqrt{17}}{4}$ and $q=\frac{-1+\sqrt{17}}{4}$

Let $b_{n}=\frac{a_{n}+p}{a_{n}-q}$, then your last equation can be rewritten as follows:

$b_{n+1}=\frac{p}{q}b_{n}$

Since the above is a geometric sequence, we have $b_{n}=b_{1}r^{n-1}$, where $r=\frac{p}{q}$.

Unsubstitting, we have $\frac{a_{n}+p}{a_{n}-q}=b_{1}r^{n-1}$.

So $a_{n}=\frac{qb_{1}r^{n-1}+p}{b_{1}r^{n-1}-1}$

I think you can derive your desired inequality using extra conditions such as $a_{1}=1$.

Sah-moo Kim
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    Thanks, I known this,But How to prove inequality? – math110 Jun 05 '16 at 15:15
  • I don't understand why this anwer has been downvoted (I upvote immediately). It is perfectly economical in its explanations : the auxiliary sequence $b_n$ being a geometric sequence is very easy to manipulate. see for example this – Jean Marie Jun 05 '16 at 16:17
  • @JeanMarie Can you perform the missing step? If you answer "no", you have an idea about why people may downvote this and why you may want to revise your immediate-upvote procedure. (To prevent misunderstandings, I didn't vote yet, but only because I did not fully check this is a deadend although I suspect it is.) – Did Jun 06 '16 at 05:11
  • @Did I just think that Sah-moo Kim is fully in the spirit of this site by 1) taking into account where the OP has met a difficulty, 2) abstract it by using economical notations and giving the name "geometrical sequence" for the auxiliary sequence he uses. I agree on the fact that there is a missing step. I would fill it like this : as the terms of $b_n$ can be confined in an arbitrarily small vicinity of $0$, this property is transferred by continuity of a certain $f(x)=(rx+t)/(ux+v)$ to the $a_n \rightarrow L$ which thus can be confined for ex. in $(L-5/27,L+5/27)$ for all $n > n_0.$ – Jean Marie Jun 06 '16 at 07:45
  • @JeanMarie The spirit of the site is to downvote posts that do not answer the questions. Re the "filling in" you suggest: indeed, various methods show that the sequence $(a_n)$ converges hence, for every $n$ large enough, yes, $|a_{2n}-a_n|\leqslant0.00001$. The question is quite different since it asks to show the upper bound $\frac{10}{27}$ for every $n$. So it seems we are back at my "deadend" conjecture (sorry). – Did Jun 06 '16 at 07:47
  • Yes. But one must recognize that to such an artificial question (it could have been as well $|a_{3^n}-a_{23n}|$...), some answers concentrate only on the important fact, the convergence... – Jean Marie Jun 06 '16 at 07:53
  • @JeanMarie Again: "the important fact" here is to answer the question as it is. If one does not like a question but one wants to post an answer, then one can write a post explaining why the question is bad and answering it, not only the former. – Did Jun 06 '16 at 08:03
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This is not a proof, only an indication that this issue is situated in $|a_2-a_1|$ only (because afterwards, it is a classical oscillatory convergence). When you look at the first values of the sequence :

$a_1=1, a_2=2/3=0.66667, a_3=6/7=0.85714, a_4=14/19=0.73684, a_5=38/47=0.80851, a_6=94/123=0.76423, a_7=246/311=0.79100, a_8=622/803=0.77460, 0.78456, 0.77847, 0.78218, 0.77992, 0.78130, 0.78046, 0.78097, 0.78066, 0.78085, 0.78073, 0.78080, 0.78076, ....$

you understand that the "hard core" of what has to be proved is essentially in the first cases and in $|a_2-a_1|$ mainly.

$|a_2-a_1| = \dfrac{1}{3} < \dfrac{10}{27}=0.37037...$, rather tight, but

$|a_4-a_2| = \dfrac{4}{57}=0.07017$ is much smaller than $\dfrac{10}{27}$

$|a_8-a_4| = \dfrac{576}{15257}=0.0377... << 10/27$, and so on...

We are far from being with a tight bound... So, may I ask to the author of this question, what has motivated her/him (homework, or personal research) for being interested in $a_{2n}-a_n$ and not in $a_{3n}-a_{2n}$ or...

Jean Marie
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