18
$\begingroup$

This is also related to an older thread in MSE ("what is the half derivative of zeta at zero?") .

One of the possible steps in the problem of that thread was to evaluate the series

$$s_a=\eta^{(0.5)}(0) = î \small{\left(\sqrt{\ln(1)}-\sqrt{\ln(2)}+\sqrt{\ln(3)}-\sqrt{\ln(4)} \cdots \right) \underset{\mathfrak E}{\approx}} - 0.347006596200 \, î $$ as the (regularized by Euler-summation $\mathfrak E$) value for the half-derivative of the alternating zeta (or "Dirichlet's eta"). Note, that the additional imaginary factor $î$ is due to the fact, that we had originally negative values under the square-root terms. In the following I'll leave this factor out for convenience.

Q: But how can I do the nonalternating series $$ s_p=\zeta^{(0.5)}(0)= î \small{ \left( \sqrt{\ln(1)}+\sqrt{\ln(2)}+\sqrt{\ln(3)}+\sqrt{\ln(4)}+ \cdots \right)} \underset{\mathfrak ???}{\approx} ??? $$ $\qquad \qquad$ I don't see so far any possibility for instance in the sense of L. Euler's famous $\eta() \to \zeta() $ - conversion.


[update]: what I've additionally just done was to apply a procedure to find the/a formal power series for the problem of finite sums of consecutive terms.

The procedure is that of the approximation of the Neumann-series of the Carleman-matrix for the function $f(x) = \sqrt{\ln(\exp(x^2)+1)}$ which is the iterative transfer function which produces the term of the series for index $n+1$ from the term at index $n$.

I'll explain this now in more detail:

For the procedure which is also known with the name "indefinite summation" we need first a function, which generates the terms of the series to be summed as iterates of itself. What function can transfer $\sqrt{\ln(x)}$ to $\sqrt{\ln(x+1)}$? This is $f(x)$ as given above, because for instance it gives for $z= \sqrt{\ln(5)}$ the result $f( z) = \sqrt{\ln(6)}$ and $f°^2( z) = \sqrt{\ln(7)}$ and so on.
So we can formally write the series $$ s_p =î \cdot ( z + f(z) + f°^2(z) + f°^3(z) + ... ) \qquad \qquad \text{ with } z=\sqrt{\ln(1)}=0 $$ An approach which I've exercised a couple of times is to implement $f()$ by a Carlemanmatrix based on $f()$'s formal powerseries. That powerseries is $$ \mathcal {\text{Taylor}} (f(x)) \approx \small 0.83255461 + 0.30028060 x^2 + 0.020918484 x^4 - 0.0075447481 x^6 + ... $$

Let now $ C = \text{carleman}(f) $ be the Carlemanmatrix for $f(x)$ then the dotproduct of a vector $V(x)=[1,x,x^2,x^3,x^4,...]$ with $C$ gives $V(x) \cdot C= V(f(x))$ by definition, and if we look at the second column of $C$ only we have $V(x) \cdot C_{0..\infty,1} = f(x)$ at least as formal powerseries, and if it is convergent for small $x$ we can also evaluate numerically.

Now the idea of the Neumann-series comes into play. As $V(x) \cdot C_{0..\infty,1} = f(x)$ we should formally also have $V(x) \cdot C^2_{0..\infty,1} = f°^2(x)$, $V(x) \cdot C^3_{0..\infty,1} = f°^3(x)$ and so on, such that we make the ansatz: $$ V(z) \cdot ( C^0 + C^1 + C^2 + C^3 + ... )_{0..\infty,1} \overset?= z + f°(z) + f°^2(z) + f°^3(z) + ... $$ and the key-observation is here, that we have in the parenthese of the lhs the geometric series of the matrix $C$ (such a construct is also called "Neumann-series"). We can surely expect, that this is no proper sum, but with some examples of alternating geometric series instead I could get meaningful approximations when using empirical approximations to $B = (I+C)^{-1}$ and then approximate for instance $V(z) \underset{\mathfrak E}\cdot B_{0..\infty,1} \approx z - f(z) + f°^2(z) - ... + ...$ where $\mathfrak E$ means Eulersummation in the dotproduct if needed.

This is not so simple and straightforward for the non-alternating geometric series. As the Carlemanmatrix of any function has the eigenvalue $1$ (at least once) by construction we would run into $\frac 10$ and cannot immediately try the approximation with finitely truncated matrices $A = ( I - C)^{-1}$. One of the workarounds, which gives sometimes meaningful results is, to omit the first column of $(I - C)$ which gives then some result, but where the first row is then systematically missing/unknown.

Remark: a case where this workaround is successful is the problem of the sum-of-like-powers where the Carlemanmatrix is the upper-triangular Pascalmatrix $P$ . The removal of the empty first column in $ Q = (I - P)_{0..(n-1),1..n} $ allows inversion and provides the matrix $Q^{-1}$ of coefficients, with which Faulhaber had solved the summing-of-like-powers-problem. The same ansatz was also tried as far as I know by two authors for solving the extension of tetration to real iteration heights. I've similarly attempted some other series-problems with such "iteration-series" of iterated functions $f°^h(x)$ and their according Neumann-series with meaningful approximations

So I've now tested $_n Q = (I - C)_{0..n-1,1..n}$ with increasing $n$ and computed $\,_nA^* = \,_n Q^{-1} $ and made $\,_nA$ by inserting the unknown first row in $\,_nA^*$ The top-left of the heuristically approximated matrix $\,_nA$ is $$\small \,_nA_{0..16,0..1}=\begin{bmatrix} ?? & ?? \\ 0 & 1 \\ -1 & -0.662055270527 \\ 0 & 0 \\ -\frac1{2!} & -0.561866397242 \\ 0 & 0 \\ -\frac1{3!} & -0.249581408503 \\ 0 & 0 \\ -\frac1{4!} & -0.0755503260124 \\ 0 & 0 \\ -\frac1{5!} & -0.0172091887343 \\ 0 & 0 \\ -\frac1{6!} & -0.00315760368955 \\ 0 & 0 \\ -\frac1{7!} & -0.000499047959470 \\ 0 & 0 \\ -\frac1{8!} & -0.0000687607442729 \\ ... & ... \end{bmatrix} $$ Using the coefficients of the first column for a power series in $x$ building the function $a_0(x)$ we get $$ a_0( \sqrt{\ln(n_1)})-a_0( \sqrt{\ln(n_2)}) = n_2 - n_1 $$ which indicates the sum of the $(f°^k(n_1))^0$ - and which is just counting the terms.

Using the coefficients $a_{k,1}$ of the second column for a power series in $x$ building the function $a_1(x)$ we get the finite sum of the function at consecutive arguments: $$ a_1( \sqrt{\ln(n_1)})-a_1( \sqrt{\ln(n_2)}) = \sum_{k=1}^\infty (\sqrt{\ln(n_1)}^k - \sqrt{\ln(n_2)}^k )a_{k,1} = \sum_{k=n_1}^{n_2-1} \sqrt{\ln (k)} $$ which indicates the sum of the $f°^k(n_1)$ which is the desired finite sums $s_p$ for the required terms to a very good (and seemingly arbitrary) numerical approximation.
------ End of lengthy explanation

Q: However - I'm missing the first coefficient for the second power series $a_1(x)$ . That should just contain the representative value for the infinite sum of $\sqrt{\ln(k)} $ with $k=1 ... \infty$

$\endgroup$
  • $\begingroup$ Very interesting sum, but I can only half understand what is going on. $\endgroup$ – Simply Beautiful Art Jun 16 '16 at 15:50
  • $\begingroup$ What do you need? What a "half-derivative" is? $\endgroup$ – Gottfried Helms Jun 16 '16 at 19:41
  • $\begingroup$ No, I understand what fractional derivatives are, but familiar with anything starting from the UPDATE section, which is half the post. $\endgroup$ – Simply Beautiful Art Jun 17 '16 at 11:42
  • $\begingroup$ I see... Well, this is a method which needs some more background I'm afraid. The idea is, to derive a solution by converting the problem of a powerseries into one containing iterates of a function, which I've applied several times to similar problems. A related link is go.helms-net.de/math/divers/BernoulliForLogSums.pdf where I try to explain what I'm also doing here. Perhaps a more basic introduction is this which deals with the sums-of-like-powers-problem with this method: go.helms-net.de/math/binomial_new/04_3_SummingOfLikePowers.pdf , the $\eta()$-function. $\endgroup$ – Gottfried Helms Jun 17 '16 at 12:23
  • $\begingroup$ @simpleArt : This is not a complete "private" method, one can find application of this under the term "indefinite summation" and there are question&answers here in MSE as well as in MO which cover this subject. Unfortunately I've no good idea at the moment for a good historical text, maybe the Euler-Maclaurin-formula is a usable example, but I've nothing at the top of my head in the moment... $\endgroup$ – Gottfried Helms Jun 17 '16 at 12:25
1
$\begingroup$

One form of the Ramanujan summation is done as follows:

$$\sum_{n\ge1}^\Re f(n)=\lim_{N\to\infty}\sum_{n=1}^Nf(n)-\int_1^Nf(t)\ dt$$

For our case,

$$\sum_{n\ge1}^\Re\sqrt{\ln(n)}=\lim_{N\to\infty}\sum_{n=1}^N\sqrt{\ln(n)}-\int_1^N\sqrt{\ln(t)}\ dt\approx1.6$$

$\endgroup$
  • $\begingroup$ Thank you for the input. I'll look at it next week. ( Hmm, doing this in Pari/GP gives a likely result $\gt 1.6$ by N=1000000;sum(k=1,N,sqrt(log(k)))-intnum(x=1,N,sqrt(log(x))) (with 200 digits internal precision) giving = 1.65764... and from $N=10^2$ to $N=10^6$ giving around 0.2 more for each step in the exponent with slowly decreasing tendency) $\endgroup$ – Gottfried Helms Jan 19 '17 at 18:03
  • $\begingroup$ @GottfriedHelms with my low tech, I could only manage like $n=60$k or so... $\endgroup$ – Simply Beautiful Art Jan 19 '17 at 18:08
  • $\begingroup$ I think it grows around the rate of about $\sqrt{\ln(n)}$, so I might fix this to the constant term in the Euler-Maclaurin expansion. $\endgroup$ – Simply Beautiful Art Jan 19 '17 at 18:13
  • $\begingroup$ Sorry, forgot completely this question/your answer after the courses... I'll look at your answer again from tomorrow (It's night already here, my pillow is waiting...) $\endgroup$ – Gottfried Helms Mar 13 '17 at 0:01
  • $\begingroup$ XD Same, forgetting and looking at my bed. :D $\endgroup$ – Simply Beautiful Art Mar 13 '17 at 0:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.