True or False Logarithm

[bo198214]

Hey,

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
\(
f_n(b^m) = -\sum_{k=1}^n \left( n \\ k \right) (-1)^{k}\sum_{i=0}^{m-1} b^{ki}
= \sum_{i=0}^{m-1}\left(1-\sum_{k=0}^n \left(n\\k\right)(-1)^{k} b^{ki}\right) \)

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

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 \( \lim_{n\to\infty} f_n(xy)=\lim_{n\to\infty} f_n(x) + f_n(y) \)?

[Gottfried]

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


V(b^0)~
V(b^1)~
V(b^2)~
...

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

Z=[0,1,2,3,...]

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.


Gottfried

Code:

// 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]

(...)
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)

Gottfried

(I forgot it, but we had this already: here )

[bo198214]

Now its out: It is the *true* logarithm. Proof here.

[andydude]

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.

Regards,
Andrew Robbins