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The following equation holds:

\begin{align} & \frac{9}{2}\pi \\[8pt] & = x \\[8pt] & {}+\cos (x) \\[8pt] & {}+\cos (x+\cos (x)) \\[8pt] & {}+\cos (x+\cos (x)+\cos (x+\cos (x))) \\[8pt] & {}+\cos (x+\cos (x)+\cos (x+\cos (x))+\cos (x+\cos (x)+\cos (x+\cos (x)))) \\[8pt] & {}+\cos (x+\cos (x)+\cos (x+\cos (x))+\cos (x+\cos (x)+\cos (x+\cos (x)))+\cos (x+\cos (x)+\cos (x+\cos (x))+\cos (x+\cos (x)+\cos (x+\cos (x))))) \\[8pt] & {}+\cdots \end{align}

for $x=\Im(\rho _1)$

where $\rho _1 = \frac{1}{2}+i14.1347251417346937904572519836\ldots$ is the first Riemann zeta zero.

Is it possible to solve this equation analytically? Or does the non-existence of a general solution form for higher order polynomials set the barrier? Or is this type of identity meaningless?

The right hand expression in the equation above is defined recursively so that each line includes recursively the previous line. With the right hand side is meant all the lines below the equal sign.

The following Mathematica program illustrates the equation:

Clear[x]
N[9/2*Pi, 90]
x = N[Im[ZetaZero[1]], 90];
x + Cos[x] + Cos[x + Cos[x]] + Cos[x + Cos[x] + Cos[x + Cos[x]]] + 
 Cos[x + Cos[x] + Cos[x + Cos[x]] + 
   Cos[x + Cos[x] + Cos[x + Cos[x]]]] + 
 Cos[x + Cos[x] + Cos[x + Cos[x]] + 
   Cos[x + Cos[x] + Cos[x + Cos[x]]] + 
   Cos[x + Cos[x] + Cos[x + Cos[x]] + 
         Cos[x + Cos[x] + Cos[x + Cos[x]]]]]

with the output:

14.137166941154069573081895224757762978887262297187976194387250665385173... 14.137166941154069573081895224757762978887262297187976194387250665385173...

which is the numerical value for $\frac{9}{2}\pi$.


Edit 2.11.2012: As suggested in the answer below the plot could be interesting. Turns out to be a step function. Replacing the cosine with the sine function gives a more symmetric plot:

Step function from cosine expression

The Mathematica code for the plot is:
Clear[x, y, n, a, xmin, xmax, ymin, ymax]
xmin = -5*Pi;
xmax = +5*Pi;
x = N[Range[xmin, xmax, 1/(25*Pi)], 90];
y = (x + Sin[x] + Sin[x + Sin[x]] + 
    Sin[x + Sin[x] + Sin[x + Sin[x]]] + 
    Sin[x + Sin[x] + Sin[x + Sin[x]] + 
      Sin[x + Sin[x] + Sin[x + Sin[x]]]] + 
    Sin[x + Sin[x] + Sin[x + Sin[x]] + 
      Sin[x + Sin[x] + Sin[x + Sin[x]]] + 
      Sin[x + Sin[x] + Sin[x + Sin[x]] + 
        Sin[x + Sin[x] + Sin[x + Sin[x]]]]]);
a = Table[{x[[n]], y[[n]]}, {n, 1, Length[x]}];
ymin = Min[y];
ymax = Max[y];
ymin = -10/2*Pi;
ymax = 10/2*Pi;
ListPlot[a, 
 Ticks -> {Flatten[{xmin, xmax, Table[n, {n, xmin, xmax, Pi/2}]}], 
   Flatten[{ymin, ymax, Table[k, {k, ymin, ymax, Pi/2}]}]}, 
 ImageSize -> Full]    
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1  
Spooky. Again I started reading the question and knew it was yours :-) –  joriki Nov 2 '12 at 13:28
    
So the equation is of the form $c = y + cos(y)$, and if I'm seeing clearly(?), it suffices to be able solve equations of that form. –  hardmath Nov 2 '12 at 13:35
    
Could be, but how do you solve that then? Will a series expansion of the cosine do it? –  Mats Granvik Nov 2 '12 at 13:41
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1 Answer

up vote 4 down vote accepted

For any $x$ whatsoever, the cosine of the right hand side should be zero, if the expression on the right hand side is to converge. So I think you should get an odd multiple of $\frac{\pi}{2}$ for anything you plug in, zeta-value or no. Now, I don't understand which multiple you will get, or whether the right hand side always converges. It might be nice to plot it.

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Yes, you are right. I see my mistake now. I did this a long time ago, and concluded something similar. But not as clearly as you put it. Any value of x will do. –  Mats Granvik Nov 2 '12 at 13:53
    
It's still really interesting - it's worth figuring out when the RHS converges and where it jumps between different odd multiples of $\pi/2$. –  user29743 Nov 2 '12 at 13:53
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