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I have the function $f(x,b,t,m)$ for $\{x>0,t,b\}$ all reals and $m=\{1,2,3,4,5,6\}$. For each particular $m\in\{1,2,3,4,5,6\}$, I see that $f(x,b,t,m)$ has a multiplicative factor $e^{7 i x}$ times another real function, for example, for $m=1$, we have

$$e^{7 i x} \left(\cos 7 b+\cos t+448 \cos 5 x+128 \cos 7 x+14 \cos x)(251-342 \sin \frac{\pi }{14}+326 \sin \frac{3 \pi }{14}-346 \cos \frac{\pi }{7})-112 \cos 3 x(17 \sin \frac{\pi }{14}-16 (1+\sin \frac{3 \pi }{14})+17 \cos \frac{\pi }{7})\right)$$

I want to get a similar result for the general $m$; in other words, to show that $f(x,b,t,m)$ can be expressed in terms of this multiplicative factor $e^{7 i x}$ times another real function which is $m$ dependent.

f[x_,b_,t_,m_]:=64 + 336 E^(2 I x) + 1008 E^(4 I x) + 1757 E^(6 I x) + 
 1757 E^(8 I x) + 1008 E^(10 I x) + 336 E^(12 I x) + 64 E^(14 I x) + 
 E^(7 I x) Cos[7 b] + 
 8 E^(2 I x) (24 + 110 E^(2 I x) + 215 E^(4 I x) + 215 E^(6 I x) + 
    110 E^(8 I x) + 12 E^(10 I x)) Cos[(2 m \[Pi])/7] + 
 4 E^(2 I x) (40 + 202 E^(2 I x) + 201 E^(4 I x) + 201 E^(6 I x) + 
    202 E^(8 I x) + 40 E^(10 I x)) Cos[(4 m \[Pi])/7] + 
 4 E^(2 I x) (32 + 82 E^(2 I x) + 335 E^(4 I x) + 335 E^(6 I x) + 
    82 E^(8 I x) + 32 E^(10 I x)) Cos[(6 m \[Pi])/7] + 
 2 E^(2 I x) (48 + 128 E^(2 I x) + 527 E^(4 I x) + 527 E^(6 I x) + 
    128 E^(8 I x) + 48 E^(10 I x)) Cos[(8 m \[Pi])/7] + 
 2 E^(2 I x) (32 + 176 E^(2 I x) + 183 E^(4 I x) + 183 E^(6 I x) + 
    176 E^(8 I x) + 32 E^(10 I x)) Cos[(10 m \[Pi])/7] + 
 4 E^(2 I x) (8 + 50 E^(2 I x) + 111 E^(4 I x) + 111 E^(6 I x) + 
    50 E^(8 I x) + 32 E^(10 I x)) Cos[(12 m \[Pi])/7] + 
 2 E^(4 I x) (20 + 59 E^(2 I x) + 59 E^(4 I x) + 20 E^(6 I x)) Cos[(
   16 m \[Pi])/7] + 
 2 E^(4 I x) (8 + 13 E^(2 I x) + 13 E^(4 I x) + 8 E^(6 I x)) Cos[(
   18 m \[Pi])/7] + 
 8 E^(4 I x) (74 + 3 E^(2 I x) + 3 E^(4 I x) + 74 E^(6 I x)) Cos[(
   20 m \[Pi])/7] + 4 E^(6 I x) (1 + E^(2 I x)) Cos[(22 m \[Pi])/7] + 
 1198 E^(6 I x) (1 + E^(2 I x)) Cos[(24 m \[Pi])/7] + 
 E^(7 I x) Cos[t] - 96 I E^(12 I x) Sin[(2 m \[Pi])/7] - 
 804 I E^(6 I x) (1 + E^(2 I x)) Sin[(4 m \[Pi])/7] + 
 328 I E^(4 I x) (1 + E^(6 I x)) Sin[(6 m \[Pi])/7] - 
 256 I E^(4 I x) (1 + E^(6 I x)) Sin[(8 m \[Pi])/7] + 
 366 I E^(6 I x) (1 + E^(2 I x)) Sin[(10 m \[Pi])/7] - 
 96 I E^(12 I x) Sin[(12 m \[Pi])/7] - 
 26 I E^(6 I x) (1 + E^(2 I x)) Sin[(18 m \[Pi])/7] - 
 584 I E^(4 I x) (1 + E^(6 I x)) Sin[(20 m \[Pi])/7] - 
 1196 I E^(6 I x) (1 + E^(2 I x)) Sin[(24 m \[Pi])/7]

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  • 1
    $\begingroup$ Sounds like you want $f(x,b,t,m)/e^{7 i x}$, since the function is $e^{7 i x}$ times this factor; but you probably mean something else. $\endgroup$
    – Michael E2
    Commented Oct 29, 2022 at 1:42
  • $\begingroup$ @MichaelE2 Thanks for your comment. No, the problem is exactly this; I want to show that for a general $m$, the function $f(x,b,t,m)$ has (as it must) this coefficient $e^{7ix}$. I tried FullSimplify for the general case, but maybe my computer is slow and it takes a lot of time. I thought maybe there is a simple way to simplify the function. $\endgroup$
    – math2021
    Commented Oct 29, 2022 at 1:59
  • $\begingroup$ But it seems to be what I said then, no? For instance f[x, b, t, m]/E^(7 I x) // ReIm // ComplexExpand // {1, I}.Simplify[#, m \[Element] Integers && 1 <= m <= 6] & $\endgroup$
    – Michael E2
    Commented Oct 29, 2022 at 2:30

1 Answer 1

1
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$Version

(* "13.1.0 for Mac OS X x86 (64-bit) (June 16, 2022)" *)

Clear["Global`*"]

f[x_, b_, t_, m_] := 
  64 + 336 E^(2 I x) + 1008 E^(4 I x) + 1757 E^(6 I x) + 1757 E^(8 I x) + 
   1008 E^(10 I x) + 336 E^(12 I x) + 64 E^(14 I x) + E^(7 I x) Cos[7 b] + 
   8 E^(2 I x) (24 + 110 E^(2 I x) + 215 E^(4 I x) + 215 E^(6 I x) + 
      110 E^(8 I x) + 12 E^(10 I x)) Cos[(2 m π)/7] + 
   4 E^(2 I x) (40 + 202 E^(2 I x) + 201 E^(4 I x) + 201 E^(6 I x) + 
      202 E^(8 I x) + 40 E^(10 I x)) Cos[(4 m π)/7] + 
   4 E^(2 I x) (32 + 82 E^(2 I x) + 335 E^(4 I x) + 335 E^(6 I x) + 
      82 E^(8 I x) + 32 E^(10 I x)) Cos[(6 m π)/7] + 
   2 E^(2 I x) (48 + 128 E^(2 I x) + 527 E^(4 I x) + 527 E^(6 I x) + 
      128 E^(8 I x) + 48 E^(10 I x)) Cos[(8 m π)/7] + 
   2 E^(2 I x) (32 + 176 E^(2 I x) + 183 E^(4 I x) + 183 E^(6 I x) + 
      176 E^(8 I x) + 32 E^(10 I x)) Cos[(10 m π)/7] + 
   4 E^(2 I x) (8 + 50 E^(2 I x) + 111 E^(4 I x) + 111 E^(6 I x) + 
      50 E^(8 I x) + 32 E^(10 I x)) Cos[(12 m π)/7] + 
   2 E^(4 I x) (20 + 59 E^(2 I x) + 59 E^(4 I x) + 
      20 E^(6 I x)) Cos[(16 m π)/7] + 
   2 E^(4 I x) (8 + 13 E^(2 I x) + 13 E^(4 I x) + 
      8 E^(6 I x)) Cos[(18 m π)/7] + 
   8 E^(4 I x) (74 + 3 E^(2 I x) + 3 E^(4 I x) + 
      74 E^(6 I x)) Cos[(20 m π)/7] + 
   4 E^(6 I x) (1 + E^(2 I x)) Cos[(22 m π)/7] + 
   1198 E^(6 I x) (1 + E^(2 I x)) Cos[(24 m π)/7] + E^(7 I x) Cos[t] - 
   96 I E^(12 I x) Sin[(2 m π)/7] - 
   804 I E^(6 I x) (1 + E^(2 I x)) Sin[(4 m π)/7] + 
   328 I E^(4 I x) (1 + E^(6 I x)) Sin[(6 m π)/7] - 
   256 I E^(4 I x) (1 + E^(6 I x)) Sin[(8 m π)/7] + 
   366 I E^(6 I x) (1 + E^(2 I x)) Sin[(10 m π)/7] - 
   96 I E^(12 I x) Sin[(12 m π)/7] - 
   26 I E^(6 I x) (1 + E^(2 I x)) Sin[(18 m π)/7] - 
   584 I E^(4 I x) (1 + E^(6 I x)) Sin[(20 m π)/7] - 
   1196 I E^(6 I x) (1 + E^(2 I x)) Sin[(24 m π)/7];

Restructuring f

(expr = f[x, b, t, m] // ComplexExpand // Simplify) // Short[#, 2] &

enter image description here

The initial factor of expr is E^(7*I*x)

(expr2 = ReplacePart[expr, 1 -> TrigToExp[expr[[1]]]]) // Short

enter image description here

Verifying that expr2 is equivalent to f[x, b, t, m]

f[x, b, t, m] == expr2 // Simplify

(* True *)
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  • $\begingroup$ Thanks. I need the complete form of the function in your brackets (which should be $m$ dependent); can we obtain it? $\endgroup$
    – math2021
    Commented Oct 29, 2022 at 2:10
  • 1
    $\begingroup$ The expressions are complete. Short is used to suppress their full display (note that the parentheses are used to isolate the Short from the expressions definitions). Remove each Short and you will see the full expressions. $\endgroup$
    – Bob Hanlon
    Commented Oct 29, 2022 at 2:17

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