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All variables are positive integers.

For:

$$a_1\qquad\frac{\sqrt{x}}{y}$$ $$a_2\qquad\frac{\sqrt{x\!+\!\sqrt{x}}}{y}$$ $$\cdots$$ $$a_n\qquad\frac{\sqrt{x\!+\!\sqrt{\!x+\!\sqrt{\!\cdots\!+\sqrt{x}}}}}{y}$$

Is there a formula of an unconditional form to describe series $a_n$?

I thought of something along the lines of:

$$\sum _{k=1}^{n } \left(\sum _{j=1}^k \frac{\sqrt{x}}{y}\right)$$

but, I quickly realized that it was very incorrect; Then I thought of:

$$\sum _{k=1}^{n} \frac{\sum _{j=1}^k \sqrt{x}}{y}$$

which I also concluded as very incorrect...

I'm blank, but I would like to see an example of something along the lines of:

$$\sum _{k=1}^{n } \frac{\sqrt{x+\sqrt{x+\sqrt{\cdot\cdot\cdot+\sqrt{x}}}}}{y}$$

where each $\sqrt{x+\sqrt{\cdots}}$ addition, repeats $k$ times. (i.e $k=3 \Rightarrow \sqrt{x+\sqrt{x+\sqrt{x}}}$); If it is possible...

Cheers!

JohnWO
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  • What means $(x)_k$ ? – Gaston Burrull May 26 '13 at 06:19
  • @GastónBurrull: $(x)_k$ such that $k$ is the $k$th term. – JohnWO May 26 '13 at 06:21
  • k-th term of what? – Gaston Burrull May 26 '13 at 06:23
  • As the definition implies: $S(a)n=(a)_k+(a){k+1}+\cdots+(a){n-1}+(a){n}, k = 1$ – JohnWO May 26 '13 at 06:25
  • But what means $(a)_k$?, Why you wrote $k=1$, is $k$ a variable or the number $1$? – Gaston Burrull May 26 '13 at 06:26
  • I think he just means $\left(x\right)_k$ is the $k$-th term in the sequence. Consider the arithmetic series $10+20+30+\ldots$. In his notation, one would write $\left(10\right)_1 + \left(20\right)_2 + \left(30\right)_3 + \ldots$. – parsiad May 26 '13 at 06:29
  • @Gastón Burrull: To imply that the first term is denoted as $1$, as opposed to some other $n$. If i wanted terms $3\rightarrow 7$, $k$ would be $= 3$... – JohnWO May 26 '13 at 06:29
  • I apologize for the confusion caused by the notation, but I know of no other form to express the series within these terms. – JohnWO May 26 '13 at 06:31
  • That not answer my question, is $k$ a constant or a variable? – Gaston Burrull May 26 '13 at 06:32
  • Could you use the standart notation for sequences to avoid confusion? http://en.wikipedia.org/wiki/Sequence if you want indexing you're doing it wrong, I can't understand what are you asking. – Gaston Burrull May 26 '13 at 06:35
  • Edited to avoid confusion. – JohnWO May 26 '13 at 06:42
  • I understood what is $a_n$ and @par answered you, your first sum is $\sum _{k=1}^{n } \left(\sum _{j=1}^k \frac{\sqrt{x}}{y}\right)=\frac{\sqrt{x}}{y} \sum _{k=1}^{n }\sum _{j=1}^k 1=\frac{\sqrt{x}}{y} \sum _{k=1}^{n }k=\frac{n(n+1)\sqrt{x}}{2y}$ – Gaston Burrull May 26 '13 at 06:45
  • @GastónBurrull: U trollin' ? – JohnWO May 26 '13 at 06:56
  • @JohnWO, No. I just developed your first sum, and $x,y$ does not depend on $j,k$ (I assumed $x,y$ constant since you never defined who is x or y) can get out the constant $\sqrt{x}/y$ for sum. – Gaston Burrull May 26 '13 at 06:59
  • @GastónBurrull: I quote from my question: "I quickly realized that it was very incorrect." And what do you mean by: "I assumed $x,y$ constant since you never defined who is $x$ or $y$" since defining them, i.e: $a+b = 1+2$, makes $a,b$ by definition constant, and since when they aren't defined, they are by definition variable? – JohnWO May 26 '13 at 07:06
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    @JohnWO do you only want your last sum (I was solved your first and second sum which are the same)? your last sum is just $\sum _{k=1}^n a_k$ and has not closed form. For example $s_n\to\infty$ has a closed form (for limit you can solve $z=(\sqrt{x+\sqrt{z}})/y$) – Gaston Burrull May 26 '13 at 07:12
  • @GastónBurrull: I'm sorry, but your observations are trivial to my problem. – JohnWO May 26 '13 at 07:16
  • @JohnWO I didn't understood your problem. You can't compute $a_n$ in a closer form. – Gaston Burrull May 26 '13 at 07:18
  • $s_k$ approaches its limit very rapidly, so $\sum_{k=1}^n s_k \approx s_{\infty}$ is a very good approximation to your problem (for large enough $k$). For example, in the case where $a=2$, note that $s_0=0.00000$ $s_1\approx 1.41421$ $s_2\approx 1.84776$ $s_3\approx 1.96157$ $s_4\approx 1.99037$ $s_5\approx 1.99759$ $s_6\approx 1.99940$ $s_7\approx 1.99985$ $s_8\approx 1.99996$ $s_9\approx 1.99999$ – parsiad May 26 '13 at 07:25
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    That does not mean that you can get the exact closed-form expression for any $n$. – Gaston Burrull May 26 '13 at 07:29
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    actually, this is kind of neat. When $a=1$, $s_{\infty}$ is the golden ratio. – parsiad May 26 '13 at 07:31
  • It is a very little related problem finding the limit, as well I said the limit has closed form. Sorry but this discussion is more like a chat. – Gaston Burrull May 26 '13 at 07:38
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    @GastónBurrull: added finding the fixed point to the answer. – parsiad May 26 '13 at 08:02

1 Answers1

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If all you are looking for is a compact representation, let $$ s_{k}=\begin{cases} 0 & \text{if }k=0\\ \sqrt{a+s_{k-1}} & \text{if }k>0 \end{cases}. $$ Then \begin{align*} S_n & =\left(\frac{\sqrt{a}}{b}\right)+\left(\frac{\sqrt{a+\sqrt{a}}}{b}\right)+\ldots+\left(\frac{\sqrt{a+\sqrt{a+\ldots+\sqrt{a}}}}{b}\right)\\ & =\frac{1}{b}\left[\sqrt{a}+\sqrt{a+\sqrt{a}}+\ldots+\sqrt{a+\sqrt{a+\ldots+\sqrt{a}}}\right]\\ & =\frac{1}{b}\sum_{k=1}^{n}s_{k}. \end{align*}

Assume $a\in\mathbb{R}$ (you don't have to do this). We can show that the recurrence is stable everywhere (weakly stable at $a=-\frac{1}{4}$). Particularly, the fixed point is given by $$ s^{2}-s-a=0, $$ which has roots $$ \frac{1\pm\sqrt{1+4a}}{2}. $$ Particularly, the locally stable fixed point is the solution with $\pm$ is $+$. So, for large enough $k$, $$ s_k\approx\frac{1+\sqrt{1+4a}}{2}. $$

This is as good an answer as you can hope for, save for error bounds on the above expression.

parsiad
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  • Thanks for a great example of a compact representation, however, I would prefer an unconditional representation, as in for example: $\sum _{k=1}^n \frac{a k}{b}=\frac{a n (n+1)}{2 b}$ – JohnWO May 26 '13 at 06:52
  • @JohnWO $S\neq \sum \frac{ak}{b}$ – Gaston Burrull May 26 '13 at 06:55
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    @GastónBurrull: That was an example... And you are trolling me, right? – JohnWO May 26 '13 at 06:58
  • @JohnWO So.. What do you mean by unconditional representation? – Gaston Burrull May 26 '13 at 07:02
  • Read this answer: http://math.stackexchange.com/questions/147075/why-is-solving-non-linear-recurrence-relations-hopeless – parsiad May 26 '13 at 07:04
  • @par: Thank you very much! I presume that my case is a general one, and as the answer in the question linked states: "(...)hopeless in the most general case." – JohnWO May 26 '13 at 07:18
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    We usually say "closed form" instead of "unconditional representation". – tba May 26 '13 at 08:17