# How to find the variable $k$ which minimizes the function $\Omega^2$?

$$\Omega^2(k) = 4(\sum\limits_{m=1}^{\infty}\frac{1-\cos(km)}{m^{1+\alpha}})(\sum\limits_{m=1}^{\infty}\frac{1-\cos(km)}{m^{1+\alpha}} - A^2)$$

Let $$A = 1$$ and $$\alpha = 1$$ and write $$w(k) := \sum\limits_{m=1}^{\infty}\frac{1-\cos(km)}{m^{1+\alpha}}$$. I am told that there exists a unique $$k$$, call it $$k_{max}$$, which minimizes $$\Omega^2$$ where $$k_{max}$$ satisfies $$w(k_{max}) = \frac{A^2}{2}$$ and $$\Omega^2(k_{max}) = -A^4$$.

$$\epsilon=1$$ in the above. However, in my attempt to compute this $$k_{max}$$

\[Alpha] = 0.7;
A = 1;
M = 5;
f[k_, \[Alpha]_, M_] :=
Sum[(1 - Cos[k*m])/m^(1 + \[Alpha]), {m, 1, M}];
Solve[f[k, \[Alpha], M] == A^2/2 &&
4 f[k, \[Alpha], M] (f[k, \[Alpha], M] - A^2) == -A^4, k]


this yields an empty solution. Did I write my Mathematica code correctly?

\$Version

(* "13.3.1 for Mac OS X ARM (64-bit) (July 24, 2023)" *)

Clear["Global*"]

f[k_, α_] := Assuming[0 < k < Pi,
Sum[(1 - Cos[k*m])/m^(1 + α), {m, 1, Infinity}] // Simplify]

kmax = With[{A = 1, α = 1},
ArgMin[{4 f[k, α] (f[k, α] - A^2), 0 < k < Pi}, k] //
Simplify]

(* π - Sqrt[-2 + π^2] *)


Verifying,

{4 f[kmax, α] (f[kmax, α] - A^2) == -A^4, f[kmax, α] == A^2/2,
0 < kmax < Pi} /. {A -> 1, α -> 1} // FullSimplify

(* {True, True, True} *)

• Why was what I wrote not yielding the same answer? Commented Nov 11, 2023 at 0:27
• The text indicates that alpha = 1 but your code used alpha = 0.7; also you truncated the sum with M = 5 rather than using M = Infinity`. Commented Nov 11, 2023 at 0:42
• I've tried plotting the kmax values for a range of alpha, but I see lots of these kind of messages: i.imgur.com/EZkm5cq.png Does that happen because finding the value kmax is difficult for Mathematica to do? Commented Nov 11, 2023 at 17:31