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I try to maximise a following function:

 x[t_] := 
 1/4 (-1 + a) + a/4 + 1/2 Sqrt[a - a^2] Cos[d] - 
  1/4 (1 - a) Cos[d + 2*t] + 1/4 a Cos[d - 2*t] + 
  1/2 Sqrt[a - a^2] Cos[2*t]

with respect to a and d. What is more, $t \in (0,\frac{\pi}{4})$. From NMaximize I know what the range of a and d should be (see figures, where, however, $t \in (0,\frac{\pi}{2})$),

Numerical maximisation with respect to a

Numerical maximisation with respect to d

but I am looking for the analytical formula.

I have written something like this

Maximize[{x[t], 0 < t < Pi/4, 4/5 < a < 1, 0 < d < Pi/2}, {a, d}]

(I have hoped that limiting the possible values of a and d will help Mathematica), but the output is the same as the input.

Does it mean that the Mathematica is not able to give the analytical form of the maximisation?

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  • 2
    $\begingroup$ "analytical formula" - considering the moderate complexity of your (transcendental!) function, expecting an analytic answer seems a little unreasonable. $\endgroup$
    – J. M.'s persistent exhaustion ♦
    May 16, 2020 at 13:00
  • $\begingroup$ Can one say something more rather than it is unreasonable? I mean if one can exactly state, that, according to some theorems, such an analytical solution does not exist? $\endgroup$
    – Agnieszka
    May 16, 2020 at 13:11
  • 1
    $\begingroup$ If you recall that you essentially want to find the zero of a derivative, and then recall that there is usually no systematic method for finding roots of a transcendental equation... $\endgroup$
    – J. M.'s persistent exhaustion ♦
    May 16, 2020 at 13:15
  • $\begingroup$ Hm, ok, this seems reasonable. Oh, damn it :(. $\endgroup$
    – Agnieszka
    May 16, 2020 at 13:17
  • $\begingroup$ Note that I didn't say there surely isn't a closed-form solution. Just that it's unlikely. $\endgroup$
    – J. M.'s persistent exhaustion ♦
    May 16, 2020 at 13:32

1 Answer 1

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A non-rigorous approach.

Clear["Global`*"]

x[t_] := 1/4 (-1 + a) + a/4 + 1/2 Sqrt[a - a^2] Cos[d] - 
  1/4 (1 - a) Cos[d + 2*t] + 1/4 a Cos[d - 2*t] + 1/2 Sqrt[a - a^2] Cos[2*t]

{max, arg} =
 (NMaximize[{x[t], 0 < t < Pi/4, 4/5 < a < 1, 0 < d < Pi/2},
      {a, d, t}, WorkingPrecision -> 20] // N // Chop) /.
  v_?NumericQ :> RootApproximant[v]

(* {1/Sqrt[2], {a -> 1/4 (2 + Sqrt[2]), d -> 0, t -> 0}} *)

The derivatives are all identically zero at arg

Grad[x[t], {t, a, d}] /. arg // Simplify

(* {0, 0, 0} *)

nmax[t_?(0 <= # <= Pi/4 &)] := Module[
  {tt = SetPrecision[t, 20]},
  NMaxValue[{x[tt], 4/5 < a < 1, 0 < d < Pi/2},
   {a, d}, WorkingPrecision -> 15]]

data = Table[{t, nmax[t]}, {t, 0, Pi/4, Pi/100}] // N;

ListLinePlot[data, Frame -> True, FrameLabel -> {"t", "nmax"}]

enter image description here

EDIT: Maximize finds the global maximum if the constraints are modified slightly.

{max, arg} = 
 Maximize[{x[t], 0 <= t < Pi/4, 0 < a < 1, 0 <= d < Pi/2}, {t, a, d}]

(* {1/Sqrt[2], {t -> 0, a -> 1/2 (1 + 1/Sqrt[2]), d -> 0}} *)

Setting d == 0, then the maximum for this special case as a function of t is

{maxd0, argd0} = Maximize[
   {x[t] /. d -> 0 // Simplify, 0 <= t < Pi/4, 0 < a < 1}, a] // 
  Simplify[#, 0 <= t < Pi/4] &

(* {Cos[t]^2/Sqrt[2], {a -> 1/4 (2 + Sqrt[2])}} *)

Plot[maxd0, {t, 0, Pi/4},
 PlotLabel -> StringForm["d = 0 && ``", Equal @@ argd0[[1]]],
 AxesLabel -> (Style[#, 12, Bold] & /@ {"t", "maxd0"})]

enter image description here

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  • $\begingroup$ Thanking for using RootApproximant, as I didn't know it yet, but these a and d do not maximise the function for all t: 0<t<Pi/4. E.g. a choice a -> 9/10, d -> 9/10*Pi/2 get greater max from ca. 0.5... $\endgroup$
    – Agnieszka
    May 16, 2020 at 14:17
  • $\begingroup$ 0.529791 < 1/Sqrt[2] $\endgroup$
    – Bob Hanlon
    May 16, 2020 at 14:26
  • $\begingroup$ Oh, sorry, your maximisation is only for t=0, and there it is true. The problem is that I want to maximise the time function x[t] with respect to a and d, but not t. $\endgroup$
    – Agnieszka
    May 16, 2020 at 14:40
  • $\begingroup$ Hm, ok, bit this is not an analytical solution. I do not have problems to get the numerical one. $\endgroup$
    – Agnieszka
    May 16, 2020 at 15:09

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