Re: NMinimize a function of NMaximize

*To*: mathgroup at smc.vnet.net*Subject*: [mg79990] Re: [mg79947] NMinimize a function of NMaximize*From*: Daniel Lichtblau <danl at wolfram.com>*Date*: Fri, 10 Aug 2007 01:48:37 -0400 (EDT)*References*: <200708090924.FAA21421@smc.vnet.net>

sapsi wrote: > I have a function > g[x_]:=1 (Pi - 2 x*Sqrt[1 - x^2] - 2 ArcSin[x]) > d[x_, o_, n_] := Abs[g[x] - Normal[Series[g[y], {y, o, n}]]] /. y -> x > > d[] is the absolute difference of g[] and its Taylor series > approximation > > dm[p_] := NMaximize[{d[x, p, 3], 0 <= x <= 1}, x][[1]] > > dm[] is the maximum absolute difference between the Taylor series > approximation(3rd order) around p and g[]. > I wish to find that value of 'p' that minimizes this maximum > difference - i tried plotting and can see where the minima occurs but > would like the exact value. So, it thought this would work > > NMinimize[{NMaximize[{ d[x, p, 3], 0 <= x <= 1}, x], 0 <= p <= 1}, p]. > Instead i get errors (briefly) > > NMaximize::nnum: The function value -Abs[-2.41057-(2 (<<19>>-p)^2 \ > p)/Sqrt[1-p^2]+2 p Sqrt[1-p^2]+(2 (<<1>>)^3)/(3 Sqrt[1-<<1>>] \ > (-1+p^2))-(4 (0.652468-p) (-1+p^2))/Sqrt[1-p^2]+2 ArcSin[p]] is not a > \ > number at {x} = {0.652468}. >> > NMaximize::nnum: "The function value \ > -Abs[-2.41057-(2\(<<19>>-p)^2\p)/Sqrt[1-p^2]+2\ p\ Sqrt[1-p^2]+(2\(<<\ > 1>>)^3)/(3\Sqrt[1-<<1>>]\(-1+p^2))-(4\(0.652468-p)\(-1+p^2))/Sqrt[1-p^ > \ > 2]+2\ ArcSin[p]] is not a number at {x} = {0.6524678079740285`}." > NMaximize::nnum: The function value -Abs[-2.41057-(2 (<<19>>-p)^2 \ > p)/Sqrt[1-p^2]+2 p Sqrt[1-p^2]+(2 (<<1>>)^3)/(3 Sqrt[1-<<1>>] \ > (-1+p^2))-(4 (0.652468-p) (-1+p^2))/Sqrt[1-p^2]+2 ArcSin[p]] is not a > \ > number at {x} = {0.652468}. >> > NMinimize::nnum: > > Can anyone provide any pointers on how to find the minimum of dm[]? > Thank you for your time > Saptarshi > Getting the code to handle the two-level optimization is relatively easy. You could do, for example: g[x_] := (Pi - 2*x*Sqrt[1 - x^2] - 2*ArcSin[x]) d[x_, o_, n_] := Abs[g[x] - Normal[Series[g[y], {y,o,n}]]] /. y->x maxp[x_,p_Real,n_] := First[NMaximize[{d[x,p,3], 0<=x<=1}, x, Method->DifferentialEvolution]] Timing[{min,pbest} = NMinimize[{maxp[x,p,3], 0<=p<=1}, p]] Problem is it will be quite slow, and moreover not give a reliable result. The reason for the latter deficiency is that the inner optimization will often go to the "wrong" side of the x interval, and not find the correct max. A simple way around this is to realize (by computations, if you like, which is what I did) that for given p the max is always at one of the endpoints, either x=0 or x=1. So you can save both time and mistakes by checking those directly. I recast so that some computations do not get repeated. In particular I'll start with the generic Taylor series. d[x_, o_] = Abs[2*o*Sqrt[1 - o^2] - (4*(-1 + o^2)*(-o + x))/Sqrt[1 - o^2] - (2*o*(-o + x)^2)/Sqrt[1 - o^2] + (2*(-o + x)^3)/ (3*Sqrt[1 - o^2]*(-1 + o^2)) - 2*x*Sqrt[1 - x^2] + 2*ArcSin[o] - 2*ArcSin[x]] maxp[p_Real] := Max[d[0,p],d[1,p]] In[13]:= Timing[{min,pbest} = NMinimize[{maxp[p], 0<=p<=1}, p]] Out[13]= {1.6321, {0.0635301, {p -> 0.593819}}} If you check d[...,p] at this p value you will find that the values are equal at endpoints x = {0,1}. So we find the min p just where the endpoints balance. By the way, this is distantly related to a bilevel optimization used for Trefethen's 2002 SIAM challenge problem #5. There is a zip file at http://library.wolfram.com/infocenter/Conferences/5353/ It contains a Mathematica notebook with code to handle that optimization. Daniel Lichtblau Wolfram Research

**References**:**NMinimize a function of NMaximize***From:*sapsi <saptarshi.guha@gmail.com>