Re: how to plot nminimized result
- To: mathgroup at smc.vnet.net
- Subject: [mg113521] Re: how to plot nminimized result
- From: Daniel Lichtblau <danl at wolfram.com>
- Date: Mon, 1 Nov 2010 04:59:48 -0500 (EST)
----- Original Message -----
> From: "tarun dutta" <tarunduttaz at gmail.com>
> To: mathgroup at smc.vnet.net
> Sent: Sunday, October 31, 2010 2:09:22 AM
> Subject: [mg113502] how to plot nminimized result
> p = 0.01;
> q = 1;
> n = 5;
> CompNum[i_] := a[i] + I b[i];
> TCompNum[i_] := a[i] - I b[i];
> f = Sum[SetPrecision[
> Sqrt[i + 1] p CompNum[i] TCompNum[i + 1] +
> p CompNum[i + 1] TCompNum[i] + (q*i + i (i - 1)) CompNum[
> i] TCompNum[i], Infinity], {i, 0, n}];
> c = \!\(
> \*UnderoverscriptBox[\(\[Sum]\), \(i = 0\), \(n\)]
> \*SuperscriptBox[\(Abs\ [CompNum[i]]\), \(2\)]\) ==
> 1; c // TraditionalForm;
> dp = {a[n + 1] -> 0, b[n + 1] -> 0};
> var = Table[CompNum[i], {i, 0, n}] /. a_ + I b_ -> {a, b} // Flatten;
> prob = Join[ComplexExpand[f] /. a_ + I b_ -> {a^2 + b^2} /. dp, {c}];
> {val, res} =
> NMinimize[prob, var, MaxIterations -> 10000,
> AccuracyGoal -> 30]; // AbsoluteTiming
>
> this is my main program.for fixed value of p and q.I got the minimized
> value and also the variable like a[o],a[1]..etc for example(from above
> program)
>
> In[12]:= val
>
> Out[12]= 5.25767*10^-6
> In[13]:= res
>
> Out[13]= {a[0] -> -0.805443, b[0] -> 0.591251, a[1] -> -0.00824279,
> b[1] -> 0.0402093, a[2] -> 0.0000547477, b[2] -> -0.000202243,
> a[3] -> -1.11872*10^-7, b[3] -> 3.84618*10^-7,
> a[4] -> 6.22738*10^-10, b[4] -> 8.88902*10^-11,
> a[5] -> 3.02522*10^-10, b[5] -> 3.02899*10^-10}
>
>
> now I want to vary the value of p from 0 to 2 for a fixed value of q.q
> will also vary from 0 to 5.so, program will start taking the first
> value of q as 0 and scan p from 0 to 2 in steps 0.01.every time i will
> get the corresponding {val,res}.for example
> q=1 p=0.01 val 5.25767*10^-6 res a[0] -> -0.805443,
> b[0] -> 0.591251, a[1] -> -0.00824279,
>
> b[1] -> 0.0402093, a[2] -> 0.0000547477, b[2] ->
> -0.000202243,
> a[3] -> -1.11872*10^-7, b[3] -> 3.84618*10^-7,
>
> a[4] -> 6.22738*10^-10, b[4] -> 8.88902*10^-11,
>
> a[5] -> 3.02522*10^-10, b[5] -> 3.02899*10^-10}
>
>
>
> q=1 p=0.02 val= res =
>
>
> now i want to check the result of res----if any a[i] and b[i] of all
> a[i] and b[i] is nearly equal to 1(or > o.1) and other a[i] are zero
> then print 'true' and also print the corresponding value of p and q
> if more than one a[i] and b[i] have value nearly equal to 1 then print
> 'false'
> from above example..
> q=1 p=0.01 res=True because only a[0] and b[0] have nearly = to 1.we
> ignore all other a[i] and b[i] cause they have value in order of 10^-2
> or more.
>
> so there will be some kind of table as
> q=1 p=0.01 res true
> q=1 p=0.02 res true
> q=1 p=0.03 res true
> q=1 p=0.04 res false
> now we only consider only the last value of p for which we get
> res=true after that point we get false
> from above example we note the value of p=0.03 for q=1
> similarly
> q=1.1 res true p=0.01
> q=1.1 res true p=0.02
> q=1.1 res false p=0.03
> here we note the value p=0.02,q=1.1
> in this way we get a table of true value like
> q=1 p=0.03
> q=1.1 p=0.02
> ....
> .....
> now I want to plot(contour) q vs p.......
> this is my problem....so how will I do all this in mathematica...
> help
> if the problem is not clear to you people just mail me...
> with regards,
> tarun
If it helps anyone, below is code I sent to the original poster in private email several days ago. It makes a function of the parameters (p,q) and also uses a faster minimization (FindMinimum with interior point).
n = 5;
x[6] = 0;
x[i_] = re[i] + I*im[i];
conj[a_] := ComplexExpand[Conjugate[a]]
f[p_, q_] =
Together[Sum[
Sqrt[i + 1]*(p x[i]*conj[x[i + 1]] +
p*x[i + 1] conj[x[i]]) + (q*i + i (i - 1)) x[i]*
conj[x[i]], {i, 0, n}]];
c = Expand[Sum[x[i]*conj[x[i]], {i, 0, n}]] == 1;
v = Join[Array[re, n + 1, 0], Array[im, n + 1, 0]];
fmin[p_?NumericQ, q_?NumericQ] :=
FindMinimum[{f[p, q], c}, v, Method -> "InteriorPoint"]
Even using FindMinimum instead of NMinimize, this can take time. With increments of 1/10 instead of 1/1000, we have
In[72]:= Timing[
pts = Flatten[
Table[{p, q, fmin[p, q]}, {p, 0, 2, 1/10}, {q, 0, 5, 1/10}], 1];]
Out[72]= {75.598, Null}
So a run over the full requested grid might be over two hours. Possibly using sensible explicit initial values for the variables would help; I have not tried that.
One probably would do better to use explicit loops on {q,p}, and terminate the inner loop whenever the condition above goes to False. I had not been addressing that issue, hence the full table.
Daniel Lichtblau
Wolfram Research