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Measure the width of a sharp peak?

Luca Stadtmueller
Posted 5 years ago

Hi, I need the width of the sharp peak that you can see in the picture below. For a given n this would be Nmax - Nmin as marked in the picture. The plot is a 3D listplot over the natural numbers n and N.

I think this is important because it defines the peak like Nmax >> Nmax + 1 and Nmin >> Nmin-1. So basically I need Nmax - Nmin for a given n and the Programm should choose Nmax and Nmin automatically in a way that Nmax >> Nmax + 1 and Nmin >> Nmin - 1.

Is there some kind of build-in-function wich is able to do it or do I have to implement it myself with If-statements?

thanks in advance,

Best regards

Luca

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POSTED BY: Luca Stadtmueller
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Ok, thats a good idea. I will try it. Thank you
POSTED BY: Luca Stadtmueller

But in this example there is no function. The plot is a plot of data that I got from a stochastic simulation.

Is this really a problem? Do a ListInterpolation with your data - and you will have a function!

data2D = Table[1/(x^2 + 2 y^2), {y, -2.05, 2, .1}, {x, -2.05, 2, .1}];
(*  data2D is just an example *)
func = ListInterpolation[data2D, {{-2.05, 2}, {-2.05, 2}}];
reg = ImplicitRegion[func[x, y] > 1, {x, y}];
CoordinateBounds[reg]
POSTED BY: Henrik Schachner

Thank you Henrik, this might be helpful at a later point. But in this example there is no function. The plot is a plot of data that I got from a stochastic simulation. Sorry, I really should have been more precise.

However, I think I found a solution that works for my problem.

POSTED BY: Luca Stadtmueller

You did not give an example function, so I am making one up; say your "function with a peak" is:

g[x_, y_] := 1/(x^2 + 2 y^2)

and you want to know the dimensions at g=1, then this is maybe giving what you are asking for:

reg = ImplicitRegion[g[x, y] > 1, {x, y}];
CoordinateBounds[reg]
POSTED BY: Henrik Schachner

Sorry, I made a mistake there. The kriteria should be nBan-nBsim(Nmax) >> nBan-nBsim(Nmax + 1) and nBan-nBsim(Nmin) >> nBan-nBsim(Nmin - 1)
POSTED BY: Luca Stadtmueller
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