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I have been using

Manipulate[Plot[{(LogIntegral[x])^(1/2), 
(((x*E^s)/Log[x*E^s]) ((((Log[Log[x*E^s]])^(w - 1))/((w - 1)!))))/
RiemannR[x]}, {x, 2, 5000}, PlotStyle -> {Blue,Red}, ImageSize -> 700], 
{w, 33.34, 40, 0.01}, {s, 43.2, 50, 0.01}]

in Mathematica to play with the plot

enter image description here

I am trying to find the max values of w and s where the red curve is at no point greater in value that the blue curve. Is there a better way of doing this? What would the mathematical approach be?

I am guessing I will have to find the derivative of $$ {{\rm Li}^{1/2}\left(x\right)\,y \left[\log\left(\log\left(y\right)\right)\right]^{z} \over {\rm R}\left(x\right)\log\left(y\right)\left(z!\right)}$$ where ${\rm R}$ is the Riemann counting function, but am at a bit of a loss as to how to proceed.

(I have started w and s at a reasonable estimate - but when values become much higher than 1000, manipulation is not really feasible.)

NB My best guess so far at the relationship between s and w is s=[N[Log[((w/5) + 1)!]], but this is clearly way off.

martin
  • 8,998
  • If you can tell me how to apply that answer in this case, I would be most grateful :) – martin Dec 11 '13 at 16:45
  • I took Ross Millikan's advice & posted it as a separate question. – martin Dec 11 '13 at 16:47
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    @anorton: this is a different function from the one in the question you link to. I think I saw this function recently. – Ross Millikan Dec 11 '13 at 16:48
  • @ anorton - if not, I would appreciate you removing the suggestion that it has already been answered - many thanks. – martin Dec 11 '13 at 16:50
  • You should show the function with $w$ and $s$ in it-changing to $y,z$ is confusing. It would also be good to specify what the blue and red curves are. Taking the derivative of this one looks quite hard-have you asked Mathematica? You can use numeric root finders to find intersection points. NSolve will do this. – Ross Millikan Dec 11 '13 at 16:54
  • Yes - but the point is, I don't really want them to intersect :/ – martin Dec 11 '13 at 17:02
  • I have asked what approaches I should take on Mathematica StackExchange, but haven't had much joy there. – martin Dec 11 '13 at 17:03
  • @martin OH! I cannot believe I did that. :-o Sorry. I'll retract that vote. (I was out to lunch, hence the latish reply.) – apnorton Dec 11 '13 at 17:58
  • @ anorton - thanks, no worries :) – martin Dec 11 '13 at 18:01

0 Answers0