[Update 3:] I gave a new partial answer following the ansatz in question Q3. I leave the other parts of the question untouched, they are also partially answered in specialized other questions in MSE.

I'm trying to understand the concept of fractional derivatives and am fiddling with the examples at wikipedia. The a'th derivative of a monomial in x, where a can be fractional is accordingly $$ {d^a \over dx^a} x^m = { \Gamma(1+m) \over \Gamma (1+m-a) } x^{m-a} $$.

Q1: But what happens for some function $f(x)$ if I want to evaluate the half-derivative at zero?

Originally I'm interested in the fractional derivatives of the zeta at zero. Because I thought that the monomial-halfiterative is the most easy one to understand I tried first the power-series expression of the zeta-function $$ \zeta(x) = - {1 \over 1-x} + \sum_{k=0}^\infty w_k x^k $$ where $w_k$ are some coefficients beginning with $w_0=0.5, w_1=0.081... , w_2=-0.0031... , \cdots $
But if I want to find the (1/2)'th derivative at $x=0$ I need definitions how I should handle the fractional powers of zero.

Q2: How can I evaluate the fractional derivative of the leading fraction ${1 \over 1-x}$? Can I do better than to express the fraction by its power series and do the derivations termwise at the monomials?

Q3: Or can I do something like with the integer derivatives of the zeta at $s=0$ where I express it as the Dirichlet-series having the logarithms in the numerators?

[update]: concerning Q3, I've now used the alternating-zeta version and assumed, that $$ {d^{1/2} \over dx^{1/2}} {1 \over k^x}={d^{1/2} \over dx^{1/2}} \exp(x(-\log(k))) =(-\log(k))^{1/2} {1 \over k^x} = i \cdot (\log(k))^{1/2} {1 \over k^x} $$ and then I set $x=0$ and evaluate the alternating series $$ \eta(0)^{(1/2)} \underset{\mathfrak E}{=} \sum_{k=0}^\infty \left((-1)^k log(1+k)^{1/2}\cdot i \right)\sim - 0.347006596200 \cdot i $$ where $\mathfrak E $ means Euler-summation of the divergent series. However, even if that result is meaningful this does not yet help much because I've now no further idea how I could use the Euler's zeta/eta-conversion-formula here. (I have just developed the conversion scheme for the integer derivatives, but that transforms to an infinite series if this is at all generalizable to fractional indexes)

[update2]: I tried the Riemann-Liouville-formula for the half-derivative as given by @J.M. in MSE, but the result is inconclusive. First, I need to handle zeros in the denominators, and second, if I replace them by limiting expressions with $\epsilon \to 0$ the numerical integration seems to diverge either to $-\infty$ or $- i \infty$ depending on whether I approach zero from positive or from negative values. So I need some help even for this...

  • $\begingroup$ Just found another very nice (and much compact) treatize at mathpages.com/home/kmath616/kmath616.htm It focuses nicely at the exponential function... but still I'm not farther than in my updated Q3... $\endgroup$ – Gottfried Helms Apr 29 '13 at 12:19
  • $\begingroup$ Here is how you derive this formula. $\endgroup$ – Mhenni Benghorbal May 15 '13 at 0:56
  • $\begingroup$ I'll link to a previous thread of yours, where I derived expressions that will pop up in the Maclaurin series for $\zeta(s)$. Semidifferentiate this series, replace $s$ with $0$, and you now have a series to start with. I might write something up late if I find the time. $\endgroup$ – J. M. ain't a mathematician Dec 12 '15 at 8:14
  • $\begingroup$ @j.m.: yes, that would be very nice. I had no problem do semi-differentiate the terms of the power series, only the leadign term (containing seemingly a fractional power of zero) was the remaining problem when I tried the ansatz with the power series $\endgroup$ – Gottfried Helms Dec 12 '15 at 9:59
  • $\begingroup$ A very silly question, but why isn't (-1)^0,5 not -i? $\endgroup$ – Gerben Nov 18 '19 at 3:48

Here is a tip: Google "super calculus" I just did it, it's the second hit. It deals with fractional calculus and it enables you for example to calculate sums of rediculous series. Not for the everyday mathematician though.

  • $\begingroup$ Hmm, thanks! But I've plenty "sponsored links" at the top of my search-results (from google.de), and many others seem to provide only pdf's. Would you mind to provide an exact link? $\endgroup$ – Gottfried Helms Apr 28 '13 at 23:49
  • $\begingroup$ fractional-calculus.com $\endgroup$ – imranfat Apr 28 '13 at 23:54
  • $\begingroup$ Thank you again. Very nice, indeed! It is much general and I did not find an easy answer for my question so I delay the reading to tomorrow. Meanwhile, I think there should be some relatively simple solution - I were also happy to have some numerical approximation beforehand... $\endgroup$ – Gottfried Helms Apr 29 '13 at 11:20

Here is possibly an answer. I followed the path at Q3 in my above question with a final procedure of divergent summation. The ansatz is to express the problem as a sum of a function evaluated at consecutive arguments, just in the sense of the Euler-MacLaurin-summation method, which requires a power series for the function, which gives the "general term" for the series.

While we don't have a Taylor series for the general term $ \sqrt{-\ln(1+x)} $ around $x=0$ we can rewrite this a bit and get a related series: $$ \sqrt{\ln(1+x)} = \sqrt{x \ln(1+x)/x} = \sqrt x \sqrt{ {\ln(1+x) \over x}} $$ and for the second root-expression we have a nice Taylor-series

$ \displaystyle \qquad \small f(x)=\sqrt{ {\ln(1+x) \over x}} = \sum_{k=0}^{\infty} a_k x^k \\ \\ \small \qquad \qquad = 1 - 1/4 x + 13/96 x^2 - 35/384 x^3 + 6271/92160 x^4 - 2211/40960 x^5 + O(x^6) $

We get then

$ \displaystyle \qquad \small \sqrt{\ln(1+x)} = \sum_{k=0}^{\infty} a_k x^{0.5+k } $

Now I assume, that it is possible to express formally the sum of that function for k over all consecutive integers by the double sum

$ \displaystyle \qquad \small \sum_{k=0}^\infty \sqrt{ - \ln(1+k)} = \sqrt{-1} \cdot \sum_{k=0}^{\infty} \left( a_k \sum_{x=0}^\infty x^{0.5+k } \right) = i \sum_{k=0}^{\infty} a_k \zeta(-0.5-k) $

and the goal to evaluate the fractional half-derivative of $\zeta$ at zero were then, using a procedure $\small \mathfrak N$ for strongly divergent series,

$ \displaystyle \qquad \small \zeta(0)^{(1/2)} \underset{\mathfrak N}{=} i \sum_{k=0}^{\infty} \left( a_k \zeta(-0.5-k) \right) \qquad \sim - 0.200824660892 \ i $

Now conjectured: $$ \zeta(0)^{(1/2)} \underset{\mathfrak N}{=} i \sum_{k=0}^{\infty} \left( a_k \zeta(-0.5-k) \right) \\ \zeta(0)^{(1/2)} \sim - 0.200824660892 \ i $$

The applicability and the correctness of the above sequence is not sure/not proven. Heuristics show, that the method applied to the fractional derivatives of the $\eta$ gives the expected results but for instance for the non-alternating zeta series we might need some integral as summand like in the Ramanujan-summation-method.

So far I can only hope that someone else does crosscheck the method and the recieved value.


Sorry I don't have a rep > 50 so this is just a comment but have you noticed that $\zeta(0)^{(1/2)}$ is eerily close to $\frac{\\i\zeta(0)^{(2)}}{10}$ in value?

  • $\begingroup$ No, I just didn't find reliably a value for the half derivative, so I couldn't say whether it is close to something. What is its value? $\endgroup$ – Gottfried Helms Mar 12 '14 at 22:27
  • $\begingroup$ the 2nd derivative of zeta @ 0 ~ -2.006356455 (per mathworld) . Was thinking there could be a connection between the nth and the 1/nth derivative of zeta $\endgroup$ – gobrewers14 Mar 12 '14 at 22:41
  • $\begingroup$ I'd highly doubt of it. $\endgroup$ – Simply Beautiful Art Nov 27 '16 at 17:10

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.