# How to deal with alternating series when index $\to +\infty$ and $\to - \infty$?

This question stems from some exercises in summing the alternating series of iterations of (simple) functions. I get some sort of paradox...

Assume $f(x) = x^2 - 0.5$ which has the inverse $g(x) = \sqrt{0.5+x}$. Let's write the iterates using $h$ as "iteration-height" as second parameter: $f(x,h+1)= f(f(x,h)) \qquad f(x,0)=x$ and $\qquad g(x,h)$ analoguously.

The iterations, if started at $x_0=1$ have two fixpoints: $x_\omega=\lim_{k\to\infty} f(x_0,h)={1-\sqrt 3 \over 2} \approx -0.366$ and $x_{-\omega} = \lim_{k\to\infty} f(x_0,-h) = g(x_0,h) ={1+\sqrt 3 \over 2} \approx 1.366$.

Let's denote the iterates from $x_0$ by the index $x_1 = f(x_0,1), x_2=f(x_0,2),\ldots,x_\omega$ and $x_{-1} = f(x_0,-1), x_{-2}=f(x_0,-2),\ldots, x_{-\omega}$

Now I compute the alternating sums: \begin{align} \operatorname{sp}(x_0)&= \sum_{k=0}^\infty (-1)^k x_k \\\ \operatorname{sn}(x_0)&= \sum_{k=0}^\infty (-1)^k x_{-k} \\\ y(x_0) &= \operatorname{sp}(x_0) + \operatorname{sn}(x_0) - x_0 \\\ &= x_\omega \cdots +x_2 -x_1 + x_0 - x_{-1}+ x_{-2} - \cdots + \cdots x_\omega \end{align}

Because the iterations converge to the fixpoint and the sum is alternating each series can be Abel- or Euler summed and I can simply use the sumalt-function in Pari/GP to get numerical values, for instance,

\begin{align} \operatorname{sp}(1)&= \sum_{k=0}^\infty (-1)^k f(1,k) &\approx 0.7098 \\\ \operatorname{sn}(1)&= \sum_{k=0}^\infty (-1)^k f(1,-k)&\approx 0.41976 \\\ y(1) &= \operatorname{sp}(1) + \operatorname{sn}(1) - 1 &\approx\ 0.1296 \\\ \end{align}

Intuitively I'd assume, that taking $x_{-4},x_{-2},x_0,x_2,x_4,\ldots$ as central values for the two alternating sums (in steps of two iterations) the results $y(x_{0+2k})$ should be identical for integer k - but apparently this is not true: Pari/GP as well as some checks with Euler-summation give a nonconclusive empirical result. Here is a table of some alternating sums centered at the iterates from $x_0=1$
$\qquad \qquad \small \begin{array} {rrr} x & \operatorname{sp}(x) & \operatorname{sn}(x) & y(x) \\ 1 & 0.709801988103 & 0.419756033790 & 0.129558021893 \\ 0.500000000000 & 0.290198011897 & 0.0802439662097 & -0.129558021893 \\ -0.250000000000 & 0.209801988103 & -0.330243966210 & 0.129558021893 \\ -0.437500000000 & -0.459801988103 & -0.361385096486 & -0.383687084589 \\ -0.308593750000 & 0.0223019881028 & -0.348680460029 & -0.0177847219265 \\ -0.404769897461 & -0.330895738103 & -0.364204951981 & -0.290330792623 \\ -0.336161330109 & -0.0738741593581 & -0.355478984338 & -0.0931918135864 \\ -0.386995560139 & -0.262287170751 & -0.363410316203 & -0.238701926815 \\ -0.350234436433 & -0.124708389388 & -0.358352789732 & -0.132826742687 \\ -0.377335839537 & -0.225526047045 & -0.362477669053 & -0.210667876562 \\ \vdots & \vdots & \vdots & \vdots \\ -0.366025403784 & -0.183012701892 & -0.361005561798 & -0.177992859906 \\ -0.366025403784 & -0.183012701892 & -0.361005561798 & -0.177992859906 \\ -0.366025403784 & -0.183012701892 & -0.361005561798 & -0.177992859906 \\ -0.366025403784 & -0.183012701892 & -0.361005561798 & -0.177992859906 \end{array}$

This confuses me much: I think, that the summation is valid, but I also think, that the two-way-infinite sum should be invariant to the centering in steps of two iterations.

Where do I go wrong here?

[Update] I've seen one source of the problem: this is the two-valuedness of the inverse $f(x)$-function (the taking of squareroot). From this follows the infinite multifoldness of the trajectory if going with negative height. As simple as it is: on the trajectory is $x_0=1, x_1=0.5, x_2=-0.25, x_3=-0.4375,\ldots$ but if we go back from $x_3$ to $x_2$ by $x_2 = f(x_3,-1)= g(x_3)$ we need a correction of sign after taking the squarerroot in g(x). This inherits to the full trajectory as far as it is in the negative domain.
There is some more interesting stuff following from this - indefinities or better: multivaluedness of iteration height of the values on the curve y=x^2-0.5=f(x) near the negative fixpoint - we have a continuous set mapped to itself (with contraction), which may have more consequences also in general, which I cannot yet fully recognize.

Hmm, while this solves at least the "practical" or "obvious" problem before the eyes here, I'm still scared/insecure about the more general problem of summing series if they are assumed with infinite index at both ends (even if things are generally easier if the series have alternating signs). But this question may then be too broad/unspecified for math.SE so maybe I'll later formally "accept" some null-answer or I'll retract the whole question.

[end update]

For convenience, here is Pari/GP-code to reproduce the behaviour:

\\ f can be positively or negatively be iterated:
f(x,h=0) = for(k=1,h,x=x^2-0.5);  for(k=1,-h,x=sqrt(0.5+x));  x

\\ helper functions for sumalt. sumalt asks for consecutive iterations
\\ so we do not need to compute the full iteration at each function call
\\ instead we use a global variable and do only one step of iteration
fsa(k,x)=if(k==0,  gl_f=x,   gl_f =      gl_f^2 - 0.5  );  return(gl_f);
gsa(k,x)=if(k==0,  gl_f=x,   gl_f = sqrt(gl_f   + 0.5) );  return(gl_f);

fmt(400)
\\ check one solution using central value x0=1.0
x=1.0
sp=sumalt(k=0,(-1)^k*fsa(k,x))
sn=sumalt(k=0,(-1)^k*gsa(k,x))
sp+sn-x

\\ check for sequence of solutions for central values in steps of 1 iteration
vsumaltp=vectorv(n,r,[x=f(1,r-1),sp=sumalt(k=0,(-1)^k*fsa(k,x)),sn=sumalt(k=0,(-1)^k*gsa(k,x)),sp+sn-x])
vsumaltn=vectorv(n,r,[x=f(1,1-r),sp=sumalt(k=0,(-1)^k*fsa(k,x)),sn=sumalt(k=0,(-1)^k*gsa(k,x)),sp+sn-x])
Mat(vsumaltp) \\ list as documented , central values towards f(x,+inf)
Mat(vsumaltn) \\                      central values towards f(x,-inf)

• Just a quick question,Are you working with absolutley convergent series? if not then the sum will vary depending on the order of summation. sorry for the pedestrain question but had to ask the simplest thing first. – jimjim Jun 2 '11 at 23:19
• @Arjang: the sums are not absolutely converging; the terms approach a constant <>0 to the left as well as to the right when the index goes to infinity. – Gottfried Helms Jun 2 '11 at 23:23

This is simply due to the fact, that $\small f(x)$ is not bijective in the full range between the two fixed points. So for a subrange of $\small x$ it is $\small f(f(x,h),-h) \ne x$ for some h and the shift of the initial point in the different $\small fsa$-calls (see the Pari/GP-code) introduces errors by unresolved ambiguity.
An example, where this does not occur and the function is in fact bijective in the range between the two fixpoints is $\small f(x) = 1/16 x^2 + x - 1$; here the described problem does not occur.