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I found this approximation of which an earlier version I posted in the chat room:

$$7 \pi -\text{Log}\left[\frac{7}{2} e^{-7 \pi /2}+\frac{5}{2} e^{-5 \pi /2}+\frac{3}{2} e^{-3 \pi /2}+e^{5 \pi /2}+2 \pi \right] = 14.13472514154629716253329494571302508888...$$

The first non trivial zeta zero: $$14.13472514173469379045725198356247027078$$

Can you improve on the formula above?

Edit 2.9.2012

Based on the comments below I would like to explain how I reasoned:

Any Taylor series evaluated at $x=1$ is convergent for variants of it when multiplied element wise with rows in this matrix:

$$\begin{bmatrix} 0&0&0&0&0&0&0 \\ 1&-1&1&-1&1&-1&1 \\ 1&1&-2&1&1&-2&1 \\ 1&1&1&-3&1&1&1 \\ 1&1&1&1&-4&1&1 \\ 1&1&1&1&1&-5&1 \\ 1&1&1&1&1&1&-6 \end{bmatrix}$$

Many Taylor series have the second row as part of its coefficients. That is: $$(1,-1,1,-1,1,-1,1,-1,1,-1,...)$$

Such Taylor series are for example $\log 2$, $\sqrt 2$, $\cos 1$, $\sin 1$. The reason for the convergence of such series and divisibility defined variants of thereof, seems to be that in the matrix above, a period sums to zero.

The simplest Dirichlet series that sums to zero and is not a an element wise multiplication of two other Dirichlet series, is the first row:

$$\frac{0}{1}+\frac{0}{2}+\frac{0}{3}+\frac{0}{4}+\frac{0}{5}+... = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$

This suggests that one should try to find an expression for a such sequence.

The definition of a number raised to a complex number is:

$$n^{(a+ib)} = n^{a}(\cos (b \log (n))+i\sin (b \log (n)))$$

and the Riemann zeta function is:


where $s$ is a complex number.

Here I then made a mistake. I started studying the equation: $$\cos (\log (n)) = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$ in order to get something similar to the Dirichlet series with numerators equal to the all zeros sequence in expression $(1)$ above. But if I understand correctly this would be the same as seeking the undefined sequence:


After that I just guessed that by combining values from the solutions to equation $(2)$ one could possibly find an expression for the zeta zeros.

Edit 23.12.2012: For what it is worth. Here is how the actual calculation went:

The first Riemann zeta zero is:

$$\Im(\rho _1)$$ $$=14.1347251417346937904572519836$$

A number close to the first Riemann zeta zero is:

$$\frac{9 \pi }{2}$$ $$=14.1371669411540695730818952248$$

That number can be split up into:

$$\frac{9 \pi }{2} = 7 \pi -\log \left(e^{\frac{5 \pi }{2}}\right)$$

To see what is missing within the logarithm I added an $x$ and solved the equation:

$$\text{Solve}\left[N\left[7 \pi -\log \left(x+e^{\frac{5 \pi }{2}}\right),30\right]=N[\Im(\rho _1),30],x\right]$$

This gives the solution:

$$\{\{x\to 6.297688980465813720589098\}\}$$

which is close to:

$$2\pi = 6.28318530717958647692528676656...$$

Substituting $x$ with $2\pi$:

$$7 \pi -\log \left(e^{\frac{5 \pi }{2}}+2 \pi \right)$$

which is closer:


Some small number seems to be missing, the second harmonic number could be it:

$$7 \pi -\log \left(e^{-\frac{1}{2} (3 \pi )}+e^{\frac{5 \pi }{2}}+2 \pi \right)$$

which again is closer:


Multiplying the added number with $\frac{3}{2}$

$$7 \pi -\log \left(\frac{3}{2} e^{-\frac{1}{2} (3 \pi )}+e^{\frac{5 \pi }{2}}+2 \pi \right)$$

closer still:


continuing the trick with similar numbers:

$$7 \pi -\log \left(\frac{5}{2} e^{-\frac{1}{2} (5 \pi )}+\frac{3}{2} e^{-\frac{1}{2} (3 \pi )}+e^{\frac{5 \pi }{2}}+2 \pi \right)$$



and once more:

$$7 \pi -\log \left(\frac{7}{2} e^{-\frac{1}{2} (7 \pi )}+\frac{5}{2} e^{-\frac{1}{2} (5 \pi )}+\frac{3}{2} e^{-\frac{1}{2} (3 \pi )}+e^{\frac{5 \pi }{2}}+2 \pi \right)$$

it works:


but then I can't get further.

Edit: 5.11.2013:

$$\frac{\sqrt{\frac{\Im(\rho _1)}{\pi }+\frac{1}{2}}}{\sqrt{5}}=0.999922272089659461895288929782$$

$$\frac{\Im(\rho _1)}{\pi }+\frac{1}{2}=4.99922275110473484848654142318$$

$\rho _1$ = first riemann zeta zero = 14.134725141734693790457...

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You might want to add some explanations about this exact-approximate formula (the one with $\pi$ and exponentials). Is it composed of the first terms of an explicit series? –  Did Sep 2 '12 at 17:01
Interesting, but $2171956\over153661$ is closer. I mean, unless the pattern lurking in your exression leads to a convergent sequence (which seems is not the case), this looks quite fruitless to me. –  Hagen von Eitzen Sep 2 '12 at 17:02
Agreed, unless you can come up with a "reason," this could easily just be a coincidence. –  Thomas Andrews Sep 2 '12 at 17:06
I will try to explain by editing the question above, with a Edit section below the original question. –  Mats Granvik Sep 2 '12 at 17:11
I fail to see what the Edit section explains. –  Did Sep 2 '12 at 18:15

2 Answers 2

up vote -5 down vote accepted

I like your question.

But im also a bit skeptical and i will try to explain why.

In short , it may just be curve fitting.

You see you got about a few handful integers in your approximation and a few handful correct decimals.

Thus your amount of correct decimals is not much more than the length of your formula.

I hope you understand that.

On the other hand , you got the numbers e and pi involved and the integers follow a pattern.

That makes it seems nonrandom again contrary to the argument above.

Hence its an interesting question.

I wonder how you arrived at that " result ".

One can express the zero's of meromorphic functions with integrals , especially contour integrals.

Maybe if we convert such an integral in a sum we might make progress with a series for the first zero.

Although this might look different from your approximation. On the other hand if we can transform that series into another one we might get closer to your approximation.



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I wonder how best answers can get so many downvotes , isnt that demotivational for the answerer ? In any case it feels weird to me. I got the impression that votes to answers are related to the votes of the questions which is imho 'emotional'. I guess it gets cancelled a bit by the reputation for best answer but that is only once while downvotes can happen everyday. Its not about reputation btw. Personally i would not never downvote a best answer if it is the only one. Sorry if this is a chatty reply but its not good for chat imho and i cant say more about this math here. –  mick Sep 9 '12 at 17:59
Motivation has nothing to do with it. If people are downvoting your answer that means there is something they don't like about it. Perhaps it is poorly written or even wrong. This is not an emotional issue. –  Qiaochu Yuan Sep 9 '12 at 19:08
Whats wrong with it ? Its the best answer and i dont see what usefull thing could be said more. Nobody explained what is wrong with it , so nobody knows or nobody told us. –  mick Sep 9 '12 at 19:15
The remark that the amount of correct decimals is not much more than the length of [the] formula (5th paragraph) is a good one but it should be made into a comment, not an answer. Furthermore, mick, indeed this answer is the best one (so far). It is also the worst (so far) since there is only one answer. Hence, I fail to see why you invoke (twice) this (completely neutral) fact as an argument against the downvotes. –  Did Sep 9 '12 at 19:57
I cant' believe anyone would accept never mind upvote this answer, so it must be some bug in the software. –  user10389 Sep 13 '12 at 22:50

Can you improve on the formula above? -- Yes: $$ \text{Log}\left[\frac{2}{3} e^{-5 \pi /2}+\frac{e^{7 \pi }}{\frac{7}{2} e^{-7 \pi /2}+\frac{5}{2} e^{-5 \pi /2}+\frac{3}{2} e^{-3 \pi /2}+e^{5 \pi /2}+2 \pi }\right] = 14.1347251417343... $$

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