True or False Logarithm

the following function \( f_n \) iapproaches - if the limit \( n\to\infty \) exists - the intuitive logarithm to base \( b \), i.e. the intuitive Abel function of \( bx \) developed at 1:

\( f_n(x)=-\sum_{k=1}^n \left(n\\k\right)(-1)^{k}\frac{1-x^k}{1-b^k} \)

The question is whether this is indeed the logarithm, i.e. if \( \lim_{n\to\infty} f_n(x) = \log_b(x) \) for \( \left|1-\frac{x}{b}\right|<1 \), provided that the limit exists at all.

It has a certain similarity to Euler's false logarithm series (pointed out by Gottfried here) as it can indeed be proven that \( f(b^m) = m \) for natural numbers \( m \) (even for \( m=0 \) in difference to Euler's series):

\( f_n(b^m) = -\sum_{k=1}^n \left(n\\k\right)(-1)^{k}\frac{1-b^{mk}}{1-b^k} \)
if we now utilize that \( \frac{1-y^m}{1-y}=\sum_{i=0}^{m-1} y^i \) for \( y=b^k \) then we get
f_n(b^m) = -\sum_{k=1}^n \left( n \\ k \right) (-1)^{k}\sum_{i=0}^{m-1} b^{ki}
= \sum_{i=0}^{m-1}\left(1-\sum_{k=0}^n \left(n\\k\right)(-1)^{k} b^{ki}\right) \)

\( f_n(b^m)=\sum_{i=0}^{m-1} 1-(1-b^i)^n \)
\( \lim_{n\to\infty} f_n(b^m) = m \)

But is this true also for non-integer \( m \)? Do we have some rules like \( \lim_{n\to\infty} f_n(x^n)=n \lim_{n\to\infty} f_n(x) \), or even \( \lim_{n\to\infty} f_n(xy)=\lim_{n\to\infty} f_n(x) + f_n(y) \)?
Here is another view into a polynomial interpolation for the logarithm which does not provide the true mercator-series for the logarithm. It is just a q&d reproduction of some analysis I've done recently but which I didn't yet document properly. It deals much with that (b-1)^m and (b^m-1) terms - so maybe you can see a/the relation to your own procedure...

We want, with some vector of coefficients X, a quasi-logarithmic solution for a base b,

V(b^m)~ * X = m

We generate a set of interpolation-points for the b^m-parameters


Make this a matrix, call it VV. Note that this matrix is symmetric!

VV = matrix{r,c=0..inf}(b^(r*c))

We generate a vector for the m-results


Then the classical ansatz to find the coefficientsvector X by vandermonde interpolation is:

VV * X = Z
X = VV^-1 * Z

But VV cannot be inverted in the case of infinite size. So we factorize VV into triangular and diagonal factors and invert that factors separately

[L,D,U] = LU(VV)

Here VV is symmetric, thus U is the transpose of L. So actually

[L,D,L~] = LU(VV)

Moreover, we have the remarkable property, that L is simply the q-analoguous of the binomial-matrix to base b and D contains q-factorials. Thus we neither have actually to calculate the LU-factorization nor the inversion - the entries of the inverted factors can directly be set.

So we have LI = L^-1, DI = D^-1 just by inserting the known values for the inverse q-binomial-matrix (see description of entries at end)
Then, formally, the coefficientsvector X could be computed by the product

X = (LI~ * DI * LI) * Z

But LI~ * DI * LI could imply divergent dot-products, (I didn't actually test this here) so we leave it with two separate factors:

W = LI~ * DI // upper triangular
WI = LI * Z // columnvector, see explicte description of entries at end

At this point -since we know explicitely the entries of W and WI- we could dismiss all the matrix-stuff and proceed to the usual notation with the summation-symbol and the known coefficients and have very simple formulae...

But well, since we are just here, let's proceed that way... :-)

I'll denote a formal matrix-product, which cannot be evaluated, by the operator <*>. Then we expect (at least for *integer m*) this to be a correct formula:

V(b^m)~ * W <*> WI = m

We compute the leftmost product first, and actually the result-vector Y~ in

V(b^m)~ * W = Y~

becomes rowfinite - it just contains the q-binomials (m:k)_b for k=0..m

So in the formula

( V(b^m)~ * W ) <*> WI = m

we have actually

( [1,(m:1)_b, (m:2)_b,...,1,0,0,0,...]) * WI = m

thus the product with WI can be done and we get an exact (and correct) solution for integer m.

So far, so good.
However, this does not apply for fractional m. The vector Y is no more finite and approximations suggest, that for all fractional values the formula is false.

Additional remark: because by the factor DI we get the same denominators as shown in the article on Euler's false logarithms and the overall-structure is very similar I assume, that this procedure provides simply the taylor-coefficients of that Eulerian series.


// description of entries in LI,DI and WI    
A quick inspection of an actual example gives the following (please crosscheck this!);

the symbol (r:c) means the binomial r over c
the symbols x!_b and (r:c)_b denote the according q-analogues to base b

     LI = matrix {r=0..inf,c=0..r} ( (-1)^(r-c)*b^(r-c:2)*(r:c)_b)        
     DI = diagonal(vector{r=0..inf}( 1/ ( r!_b*(b-1)^r * b^(r:2)  ))                      
     WI = vector{r=0..inf}(  if(r==0)  : 0                                 )
                          (  if(r >0)  : (-1)^(r-1) (r-1)!_b *(b-1)^(r-1)  )
Gottfried Helms, Kassel
(07/27/2010, 01:12 PM)Gottfried Wrote: (...)
Then, formally, the coefficientsvector X could be computed by the product

X = (LI~ * DI * LI) * Z

But LI~ * DI * LI could imply divergent dot-products, (I didn't actually test this here) so we leave it with two separate factors:
Hmm, I just tried this with base 2 and 3, and actually the entries of X seem to be computable. I get

for the row r=0 \( X[0] = - \sum_{k=0}^{\infty} \frac1{b^k - 1}

for a row r>0 \( X[r] = - (-1)^r \frac{b^r }{b^r -1}* \prod_{k=1}^r \frac 1{b^k-1} \)

Of course the prod-expression can be rewritten as q-factorial multiplied by powers of (b-1)

\( X[r] = - (-1)^r * \frac{b^r}{b^r-1}*\frac 1{(b -1)^r * r !_b } \)
and then

\( falselog(b^m) = -\sum_{k=1}^{\infty}\frac1{b^k-1}
- \sum_{r=1}^{\infty} (-1)^r \frac{b^r}{(b^r-1) * r!_b * (b-1)^r}*(b^m)^r \)

correct for positive integer m and wrong for other m.
(But now it seems that I drifted far away from Henryk's formula, sorry)


(I forgot it, but we had this already: here )
Gottfried Helms, Kassel
Now its out: It is the *true* logarithm. Proof here.
(08/11/2010, 02:37 AM)bo198214 Wrote: Now its out: It is the *true* logarithm. Proof here.

For some reason, it took me over a year to find this post/paper. It is brilliantly written. Good job Henryk.

Andrew Robbins

Possibly Related Threads…
Thread Author Replies Views Last Post
Question Derivative of the Tetration Logarithm Catullus 1 638 07/03/2022, 07:23 AM
Last Post: JmsNxn
Question Iterated Hyperbolic Sine and Iterated Natural Logarithm Catullus 2 951 06/11/2022, 11:58 AM
Last Post: tommy1729
  Is bugs or features for super-logarithm? Ember Edison 10 18,969 08/07/2019, 02:44 AM
Last Post: Ember Edison
  Can we get the holomorphic super-root and super-logarithm function? Ember Edison 10 19,889 06/10/2019, 04:29 AM
Last Post: Ember Edison
  Principal Branch of the Super-logarithm andydude 7 22,925 06/20/2011, 09:32 PM
Last Post: tommy1729
  Logarithm reciprocal bo198214 10 35,111 08/11/2010, 02:35 AM
Last Post: bo198214
  Kneser's Super Logarithm bo198214 16 70,870 01/29/2010, 06:43 AM
Last Post: mike3
  A false interpolation paradigm (?); a reconsideration Gottfried 4 13,264 09/17/2009, 08:17 AM
Last Post: bo198214
  Unique Holomorphic Super Logarithm bo198214 3 11,042 11/24/2008, 06:23 AM
Last Post: Kouznetsov
  Jabotinsky's iterative logarithm bo198214 21 40,981 06/14/2008, 12:44 AM
Last Post: andydude

Users browsing this thread: 1 Guest(s)