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I was tinkering with the question whether the logarithm $\log(x)$ can be expressed by some more useful series than by the Mercator series (in terms of (1+x)) for a certain question.

One idea was to try, what a representation as "iteration-series" would look like, where I write "iteration series" for a series composed by the iterates of a function (not as a series of powers of its argument $x$).
Here I write the iterate of a function $f(x)$ as

$ \qquad \qquad f^{°2}(x)\overset{def}=f(f(x)) \\ \qquad \qquad f^{°k+1}(x) = f^{°k}(f(x)) \\ \qquad \qquad f^{°1}(x)=f(x), f^{°0}(x)=x $

Then I try the following naive ansatz:

$$ \begin{array} {ll} \log(x) &= f(x) + f^{°2}(x) + f^{°3}(x) + ... \\ \log(f(x)) &= f^{°2}(x) + f^{°3}(x) + ... \\ f(x) &= \log(x) - \log(f(x)) \\ e^{f(x)} &= x / f(x) \\ x &= f(x) \cdot e^{f(x)} \end{array}$$ This has the form of the Lambert-W-function, so a conclusion should be $$ W(x)\overset{def}: W(x) \cdot e^{W(x)} = x \to f(x) = W(x) $$ However, the result is not correct; simply insert $x=0$ to see the contradiction; also the lhs in the first equation is likely divergent, and approximations using $x=2$ or similar show wrong results.

Q: So how could I repair this ansatz? Is there a correction, say in modification of the argument $x$ or of the iteration-series itself?


[update] Using an alternating iteration-series and a recentering of the logarithm to write $\log(1+x)$ I get a quickly converging series by $$ \begin{array} {ll} \log(1+x) &= x - f(x) + f^{°2}(x) - f^{°3}(x) + ... \\ \log(1+f(x)) &= f(x) - f^{°2}(x) + f^{°3}(x) - ... \\ x &= \log(1+x) + \log(1+f(x)) \\ e^x &= (1+x) \cdot (1+f(x)) \\ f(x) &= {e^x\over 1+x} -1 \end{array}$$

and for small values $x=2$ and $x=3$ which are of interest for me, the iteration-series made by this converges to many digits with a handful of terms: below are the first few partial sums for $\log(2)$:

   k   partial sums          signed k'th iterate of f(1) 
  [0,  1,                   1    ]
  [1,  0.6408590857704774, -0.3591409142295226]
  [2,  0.6945383440651704,  0.05367925829469305]
  [3,  0.6931462124431120, -0.001392131622058478]
  [4,  0.6931471805604139,  0.0000009681173019793587]
  [5,  0.6931471805599453, -4.686252527398870 E-13]

However, I am searching for a non-alternating series...

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  • $\begingroup$ It sounds like you're looking for something analagous to iterations of Newton's Method (en.wikipedia.org/wiki/Newton%27s_method) for the logarithm - essentially you're doing rootfinding on the equation $e^y=x$. $\endgroup$ – Steven Stadnicki Oct 4 '16 at 21:21
  • $\begingroup$ @StevenStadnicki Newton's method is the linear approximation for a root using some manipulation of the point-slope equation and derivatives. This is also not an iterative type of sequence. To me, it seems he wants an expansion of the $\log$ for some interesting sort of manipulation. $\endgroup$ – Simply Beautiful Art Oct 4 '16 at 21:28
  • $\begingroup$ @SimpleArt is correct. I'm looking at "iteration-series" in various contexts to get more fundamental understanding of this (and possibly find later some useful relations to, say, power series and/or in general useful applications). I have for instance a speculation, that iteration-series are better candidates to express fractional iterates of functions than powerseries and so on... $\endgroup$ – Gottfried Helms Oct 4 '16 at 23:12
  • $\begingroup$ Fractional iterates as in $\sum_{n=1}^{1.5}f^{^\circ n}(x)$? $\endgroup$ – Simply Beautiful Art Oct 4 '16 at 23:26
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    $\begingroup$ @SimpleArt - thanks for your link. This looks much interesting! Unfortunately I'm not much literate in integrals, so possibly this is over my head. I'll see. Note: perhaps you'll get problems with your question - because it is more of the discussion-style and less of (answerable) Q&A for which is MSE made for, so perhaps some people might tend to vote for close and link you to elsewhere, say tetrationforum or so. $\endgroup$ – Gottfried Helms Oct 10 '16 at 13:09
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$$\log(x+1)=\sum_{n=1}^\infty f^{^\circ n}(x)$$

$$\log(f(x)+1)=\sum_{n=1}^\infty f^{^\circ n+1}(x)$$

$$f(x)=\log(x+1)-\log(f(x)+1)$$

$$e^{f(x)}(f(x)+1)=x+1$$

$$g(x)e^{g(x)-1}=x+1\tag{$g(x)=f(x)+1$}$$

$$g(x)e^{g(x)}=e(x+1)$$

$$g(x)=W(e(x+1))$$

$$f(x)=W(e(x+1))-1$$

Trying this for $x=0$, it is correct.

Trying this for $x=-1$, both are beyond their domain. (not really sure what happens to the expansion if one includes complex values)

I can't verify this for any other values... bit much for me. I can see, however, that $W^{^\circ n}(e(x+1))-1\to0$, so there is some hope in the convergence.

For $x=1$, I expanded out the first $3$ terms and found that their sum was $\approx0.618$, where as $\log(2)\approx0.693$. Clearly, the sum is also increasing, so it may converge. Very slowly though...


For $e^x$:

$$e^x-1=\sum_{n=1}^\infty g^{^\circ n}(x)$$

$$g(x)=e^x-e^{g(x)}$$

$$g(x)+e^{g(x)}=e^x$$

$$e^{g(x)}e^{e^{g(x)}}=e^{e^x}$$

$$g(x)=\log\left(W\left(e^{e^x}\right)\right)$$

This holds true for $x=0$.

Seems it might hold for other values of $x$.


In general, when setting up these types of problems, I would start with the following form:

$$f(x+a)+b=\sum_{n=1}^\infty g^{^\circ n}(x)$$

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  • $\begingroup$ Nice idea, thank you. Using $\small f(x)$ as indicated I find with 24 terms for $\small x=1$ the value $\small 0.693147145431$ which is off from $\small \ln2$ by $\small -0.0000000351288341412$ (working with Pari/GP and 200 dec digits precision) . Using 48 terms I get an error of $\small -2.09384166197 E-15$ and using 96 terms an error of $\small -7.43881991373 E-30$ so this is not so bad... $\endgroup$ – Gottfried Helms Oct 4 '16 at 23:19
  • $\begingroup$ @GottfriedHelms Have you confirmed for any other values? I tried a bit, but I'm not sure how accurate my 15 digit scientific calculator allows XD $\endgroup$ – Simply Beautiful Art Oct 4 '16 at 23:23
  • $\begingroup$ Hmm, using $x=2$ or $x=12$ gives similar or even better precision (possibly depending on the default-precision in the W()-function). I don't have much time at the moment and I'll look at this deeper tomorrow! $\endgroup$ – Gottfried Helms Oct 4 '16 at 23:25
  • $\begingroup$ @GottfriedHelms Ah, nice. If only it converged faster, than this would be perfect, huh? $\endgroup$ – Simply Beautiful Art Oct 4 '16 at 23:29
  • $\begingroup$ Yes, that woulöd be perfect. What I'm after next is to find more material in which cases I should try some ansatz with a function $f(x)$ and when with $f(x+1)$ and similar questions (say if I want to express $\exp(x)$ or $\exp(x)-1$ as iteration-series). This is not yet obvious for me. $\endgroup$ – Gottfried Helms Oct 4 '16 at 23:32

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