For the function $$f(x)=b^x-1 = x_1 \qquad g(x)=\log(1+x)/\log(b) $$ and its iterative notation $$ x_0=x \qquad x_h=f(x_{h-1})=g(x_{h+1}) \qquad x_{-1}=g(x_0) $$ with b from the interval $1 \lt b \lt e$ we can define alternating series with infinitely many terms with negative as well with positive indexes. Let us denote the idea $$A(x) = \ldots + x_{-2} - x_{-1} + x - x_1 + x_2 - \ldots $$ and define it by $$ \begin{eqnarray}A(x) &=& A_n(x) + A_p(x) - x \\ \text{ where } \qquad A_n &=& \sum_{k=0}^\infty (-1)^k x_{-k} \\ \text{ and } \qquad A_p &=& \sum_{k=0}^\infty (-1)^k x_{k} \\ \end{eqnarray} $$ With, say, $b=1.3$ and $x=3$ the series $A_p$ is convergent and $A_n$ must be evaluated using Cesaro/Euler-summation. It is obvious that this is 2-periodic, that means $$A(x)=A(x_2)=A(x_4)=...=A(x_{-2})=... $$
such that it is irrelevant where I define the center $x$ as long as I choose the iterates in steps of 2.

I expect, that this should be true for the derivatives in the same way, but with some examples I find, that this is not the case. $$ ... \ne A'(x_{-2}) \ne A'(x) \ne A'(x_2) \ne ... \\ ... \ne A''(x_{-2}) \ne A''(x) \ne A''(x_2) \ne ... $$
where I compute the derivatives in various ways, for instance simply by $$A'(x)=\lim_{h\to0} {A(x+h/2)-A(x-h/2) \over h} $$ Q: I think I must have some basic misunderstanding here (which, I think, is not specific to the given example function $f(x)$ and $g(x)$ here). Can someone explain this to me?
[Added:] It seems, I've found one hint myself: checking numerically I found that $$A(x_0)' = - x_1' \cdot A(x_1)' \quad \text{ where } \quad x_1' = \lim_{h\to0} {f(x_0+h/2)-f(x_0-h/2)\over h}$$ So at least the result is not completely random (and in particular not depending on problems with the Cesaro/Euler-summation), but my intuition is still unable to grasp this conceptually...
Numerical example using $ b=1.3 , x_0=3.2 $ $$ \small \begin{array} {rclll} & x_0 & A(x_0) & A'(x_0) & A''(x_0) \\ & 3.20000000000 & -0.00119822450167& 0.0175377529574& 0.000817628416425 \\ & x_1 & A(x_1) & A'(x_1) & A''(x_1) \\ &1.31536107268& 0.00119822450167& -0.0288702496568& 0.0102533145478 \\ & x_2 & A(x_2) & A'(x_2) & A''(x_2) \\ &0.412136407584& -0.00119822450167& 0.0779236358328& -0.129878114856 \end{array} $$

  • $\begingroup$ Just to check, with Cesaro Summation, are you allowed to take derivatives? $\endgroup$ – Calvin Lin Jan 3 '13 at 8:31
  • $\begingroup$ @cal: hmm, good question. I didn't think about it; perhaps I can check this with another example-function, where the terms decrease to zero on both sides infinite indexes - however, at the moment I don't have an idea for such a function... $\endgroup$ – Gottfried Helms Jan 3 '13 at 8:35
  • $\begingroup$ @calvinLin : It seems, that the Cesaro-summation does not disturb; it seems there is some systematic effect which I'm simply lacking to understand correctly; please see my added remark in the question $\endgroup$ – Gottfried Helms Jan 3 '13 at 14:03
  • $\begingroup$ Before I try to understand this any deeper, are you familiar with this? Conditionally convergent series sum ambiguously depending on their ordering, and a 2-sided series would seem to aggravate this. $\endgroup$ – Zook Apr 23 '14 at 15:01
  • $\begingroup$ @Zook: yes, I'm aware of that; but I think this does not apply here because we do not reorder. It's more that I think I have some "fiddling-error" while I'm trying to discuss the formal derivative of a function and then the derivative at some x. I think it must have something to do with this - but I'm not sure... $\endgroup$ – Gottfried Helms Apr 23 '14 at 17:19

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