$$x-\frac{1}{x} - \frac{1}{x-\frac{1}{x}} - \frac{1} {x-\frac{1}{x} - \frac{1}{x-\frac{1}{x}}} - \dots$$ this expression is the result of recursively subtracting $1/x$ from $x$,


$$x_{n+1} = x_n - 1/x_n$$

when executing the previous expression multiple times it does not converge, as the number of executions increase the value of $x$ tends to change slower and slower.

I wonder if this expression can be represented by some integral or summation

  • $\begingroup$ The infinite series diverges similar to the Harmonic series. $\endgroup$
    – Somos
    Commented Mar 9, 2021 at 13:56
  • 1
    $\begingroup$ Hard to decipher your math. Please edit both the title and body of the query with MathJax. $\endgroup$ Commented Mar 9, 2021 at 13:57
  • $\begingroup$ For $x_1>1$, $x_n$ first becomes negative when $n\approx x_1^2/2$ $\endgroup$
    – PM 2Ring
    Commented Mar 9, 2021 at 14:47
  • 1
    $\begingroup$ In short, the expression does not converge to anything and hence does not mean anything, represent it whatever you like. $\endgroup$ Commented Mar 9, 2021 at 14:48
  • $\begingroup$ Also, I've come to regret my tag edit, because this is not really about continued fractions, but the original was no better. $\endgroup$ Commented Mar 9, 2021 at 15:06

2 Answers 2


If $x_{n+1} = x_n - \frac{1}{x_n}$ were to converge to a limit $L$ for some value of $x_1$, taking limits in the recurrence relation we would have $L = L - \frac{1}{L}$ which is absurd.

  • 1
    $\begingroup$ I don't think this addresses the question > I wonder if this expression can be represented by some integral or summation $\endgroup$ Commented Mar 9, 2021 at 15:06
  • 3
    $\begingroup$ The expression does not have a meaning. There is nothing to represent. $\endgroup$ Commented Mar 9, 2021 at 15:10
  • 1
    $\begingroup$ @Chickenmancer The reasonable way to interpret the expression the OP wrote is as a limit, but such limit doesn't exist for any $x$, so that actually doesn't define an expression. $\endgroup$
    – jjagmath
    Commented Mar 9, 2021 at 15:47
  • $\begingroup$ jjagmath - you are discussing the existence of a fixpoint (point of convergence) . But the non-existence of a fixpoint (1-periodic point) does not necessarily say something about the existence of $p$-periodic points with $p>1$. See my answer to have an example for a $4$-periodic point $x_{k+4}=x_k$ and $x_{k+2}=-x_k$ $\endgroup$ Commented Mar 10, 2021 at 13:13
  • $\begingroup$ A $2$-periodic sequence is $(a,b)_k=(-1,1.41421356237),(-1,-1.41421356237),(-1,1.41421356237),...$. A $3$-periodic sequence is $(a,b)_k=\{(-1.39285043931,0.952765813768),$ $(1.03226965040 -1.32706028227), (-0.695508361630,-1.36988405364),$ $(-1.39285043931,0.952765813768),...\}$ (The actual decimal precision can be made arbitrary, depending on software. While the exact expression for the $2$-periodic case looks obvious, W/A might be able to decode the numbers for the $3$-periodic sequence. Likely it also gives a solution as exact roots of the involved polynomial. $\endgroup$ Commented Mar 10, 2021 at 13:23

As user @jjagmath has already noticed, there is no (convergence to a) fixpoint, or one might say, the only fixpoint is infinity.
But there can be said more: there are periodic points of any period length, and they can be numerically approximated to any precision. Since the iteration-process in the question produces divergence, any small error in these approximations shall escalate after a certain finite number of iterations. However, such periodic points can be found as roots of polynomials, and are thus algebraic numbers.

It came out to be helpful to decode the variable $x$ as fractional representation, for instance $x = \frac ab$. I'll use that below.

The iterative approach to find periodic points

The iteration $\frac ab \to { a^2-b^2\over ab}$ can be inverted; however the results in the reverse iteration are ambiguous having either $ \frac ab $ or $ \frac {-a}{-b}$ . So with the route of reverse iterations we must supply a vector containing the signs for each iteration. If the vector is finite of length, say, $h$ and the provided signs are taken periodically, then the reverse iteration converges to a set of $h$-periodic points. I've done the calculation of iterations in $(a,b)$ in terms of a 2x2-matrix $A = \begin{bmatrix}a&-b\\b&a \end{bmatrix} $ and $D = \small \begin{bmatrix}1&2\\2&1 \end{bmatrix} $ and $D_r = \small \begin{bmatrix}1&1/2\\1/2&1 \end{bmatrix} $ to iterate $ A \to A^2 \star D_r $ where the "$\star$" means Hadamard-multiplication (=elementwise multiplication), and the reverse iteration $ A \to_\pm \sqrt{D \star A} $ where -if we iterate- the signs are taken from a predefined vector or list.

So for instance I got with the Pari/GP-routine (containing a simple matrix-root-routine which also includes the Hadamard-multiplication with constant matrix $D$)

 il=0;list=vectorv(1000);  \\ takes the protocol
     M2*=[1,-1][k % 2+1];   \\ vector of signs is [1,-1] and taken as infinitely repeating
     il++;list[il]=M2[1,];  \\ document current values of [a,b]
 list=Mat(VE(list,il)); }  \\ reshape list to get it as matrix

This got me the protocol:

    a                   b
   ...              ...             (a couple of hundred lines of 
   ...              ...                     not yet well approximates)
 -1.34150376263   -1.02674252883    (approximated to more than 170 dec digits)
  0.745432124647   1.37737896568
 -1.34150376263    1.02674252883
  0.745432124647  -1.37737896568
 -1.34150376263   -1.02674252883   (repeating previous values precisely)
    ...                ...

where, after $800$ iterations the difference from $ |a_{k+4}-a_k| \lt 1e-170$. The last entry cutted to $40$ visible dec digits protocol gives
a =-1.341503762630577719675693514485165986352 and b=-1.026742528828304264759792215760033619885 .
You might try iteration starting at this values (with high enough internal precision) to see the periodicity.

Periodic points via solving polynomials

Periodic points in the reals can be found solving polynomials as they occur from 2- or more steps iteration.

The assumed fixpoint iteration at $L$ leads to $L=L-\frac 1L$ as mentioned correctly by user @jjagmath: if $x$ is expressed as $\frac ab$ then we find the fixpoint $$ \frac ab = \frac ab - \frac ba = {a^2-b^2\over ab } \\ \text{solving} \\ \begin{array}{} (1)& a &= a^2 - b^2 \\ (2)& b &= ab &\qquad \to & a=1 \\ \to & 1 &= 1^2 - b^2 &\qquad \to & b=0\\ \end{array} $$ which represents of course the singularity for the expression $x = \frac ab$.

However, this does not exhaust all interesting properties of the iteration in the OP.

Using the polynomials in $(a,b)$ which occur if we iterate 2 or more times symbolically (with vector of signs provided) allows to get infinitely many solutions over algebraic numbers defining periodic iterations.

For instance the above numerical approximation, initially been found by reverse iteration, gives the ratio $$ {a \over b} = \sqrt{1+\sqrt{1/2}} = x $$ as value, which defines a 4-periodic set of values. We find this value if we write $ (a_0,b_0) \to (a_1,b_1)=(a_0^2-b_0^2,a_0 b_0) \to (a_2,b_2)=(a_1^2-b_1^2, a_1 b_1)$ and solve $(a_0,b_0)=(a_2,-b_2)$

The above heuristical protocol more detailed:

   a_k              b_k          x_k               x_k symbolic
-1.34150376263 ,-1.02674252883, 1.30656296488,  sqrt(1+sqrt(1/2))
 0.745432124647, 1.37737896568, 0.541196100146, sqrt(1-sqrt(1/2))
-1.34150376263 , 1.02674252883,-1.30656296488, -sqrt(1+sqrt(1/2))

P.S. Wolfram Alpha decomposes the following values (from second row in the protocol)

 [a_1 ≈] 0.74543212464725619 ≈ 1/(1 + sqrt(2))^(1/3) `        
 [b_1 ≈] 1.377378965676004   ≈ 2^(1/4) (1 + sqrt(2))^(1/6) `       

Final remark: since we have periodic points for those points the infinite iteration-series (as intended in the OP) might be regularizable by the $\zeta(0)$ or $\eta(0)$ application.

Example (hopefully:simple) and possibly regularization

If we do not assume a fixpoint $L$ to which the partial sums converge in the sense that in the limit $L=L - 1/L$ (which can only hold for infinite $L$) but a 2- periodic point, then we get for the limit: $$L = M - 1/M \text{ and } M= L - 1/L \tag 1 $$ Let's see, whether we can find some $L$: $$ \begin{array} {} L &= L-1/L - 1/(L-1/L) \\ &=(L^2-1)/L - L/(L^2-1) \\ &=((L^2-1)^2-L^2)/(L(L^2-1)) \\ L^2(L^2-1)&=(L^2-1)^2-L^2 \\ L^4&=(L^2-1)^2 \\ L^4&=L^4-2L^2+1 \\ 2L^2&=1 \\ L&=\sqrt{1/2} \\ M &= \sqrt{1/2}-\sqrt2 \\ &=\sqrt{1/2}(1-2)=-\sqrt{1/2}=-L \end{array} \tag 2 $$ With this we find the partial sums of the OP's expression, when $x=L$ as $$ \begin{array}{r|r|l} \text{ps} & \text{ps} & \text{partial expression in OP ; ps:partial sum} \\ \hline L & L & x \\ M & -L & x - 1/x \\ L & L & x - 1/x - 1/(x-1/x)\\ M & -L & x - 1/x - 1/(x-1/x) - 1/(...)\\ \vdots & \vdots & \vdots \end{array}\tag 3 $$

One might consider to assign a regularized value in the sense of the alternating $\zeta(0)$-series; if I didn't mess it up we should get the (Euler-) regularized value for $f(x)$ at $x=L=1/\sqrt 2$ as $$f(L) \underset{\mathfrak E}=0 \tag 4$$

Final remarks

In 2020 I've done a (formally) similar analyis (without the final attempt to regularization) on the question of periodic points in the iteration of the (complex-valued) function $z_{k+1} =\exp(z_k)$ and have discussed this to some extent here in MSE because I had some open questions with this. I find it quite funny to find here such a familarity with the derivations...

Form my earlier comment at the OP one may take the external references discussing interesting aspects of the OP's problem:

See oeis.org/A147985 for more information. Clark Kimberling has done an article on this in JIS see: cs.uwaterloo.ca/journals/JIS/VOL12/Kimberling/kimberling56.pdf


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