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I have the following function $h_1(x)$:

h1[x : _] = (1/(8 a))Sech[a x]^2 (-12 a^3 + 8 a d + 8 a c x + 96 a^3 Log[Cosh[a x]] + 
  12 a^3 Cosh[3 a x] Sech[a x] + 
  c Sech[a x] Sinh[
  3 a x] - (c + 4 a^2 (4 b + x (12 a^2 + 2 d - 12 a^3 x + c x)) + 
  96 a^4 x (-Log[1 + E^(-2 a x)] + Log[Cosh[a x]]) + 
  48 a^3 PolyLog[2, -E^(-2 a x)]) Tanh[a x])

I want to choose $b$, $c$ and $d$ to make $h_1$ go to zero at infinity.

Is there a way to do this with asymptotic series in Mathematica?

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  • $\begingroup$ @user64494, please thoroughly and carefully read instructions for editing posts. Especially pay attention to: Tiny, trivial edits are discouraged. $\endgroup$
    – Domen
    Jan 8 at 18:24
  • $\begingroup$ @Domen: Did you pay your attention to "To fix grammatical or spelling mistakes"? I did. $\endgroup$
    – user64494
    Jan 8 at 19:37
  • $\begingroup$ @user64494, again: Tiny, trivial edits are discouraged ... correcting all problems that you observe. And also, you've been warned before that "How to ..." should not end with a question mark. $\endgroup$
    – Domen
    Jan 8 at 19:43
  • $\begingroup$ @Domen: Sorry, there is no description of "Tiny, trivial edits" in the instruction. I am sure that the edit of grammar mistakes in titles makes forum better. BTW, all the titles in the instruction end with a question mark. $\endgroup$
    – user64494
    Jan 8 at 19:59

1 Answer 1

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Here a solution using Asymptotic

h[x_] := (1/(8 a)) Sech[a x]^2 (-12 a^3 + 8 a d + 8 a c x + 
    96 a^3 Log[Cosh[a x]] + 12 a^3 Cosh[3 a x] Sech[a x] + 
    c Sech[a x] Sinh[
      3 a x] - (c + 4 a^2 (4 b + x (12 a^2 + 2 d - 12 a^3 x + c x)) + 
       96 a^4 x (-Log[1 + E^(-2 a x)] + Log[Cosh[a x]]) + 
       48 a^3 PolyLog[2, -E^(-2 a x)]) Tanh[a x])

find asymptote

asy = Asymptotic[h[x], x -> Infinity] (*ConditionalExpression[6 a^2 + c/(2 a), (b | c | d) \[Element] Reals && a > 0]*)

should be zero

solc = Solve[asy[[1]] == 0, c][[1]] (* {c -> -12 a^3} *)

solution h[x]

h[x]/.solc//Simplify (* depends on parameter `a>0, b,d`)

enter image description here

higher order approximations follow step by step

asy2 = Asymptotic[h[x] /. solc, x -> Infinity ] // Simplify
(* ConditionalExpression[4 a E^(-2 a x)x (-d + 6 a^2 (-1 + Log[4])),(b | d) \[Element] Reals && a > 0] *)

h[x] /. solc /. d -> 6 a^2 (-1 + Log[4]) // Simplify

enter image description here

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    $\begingroup$ +1. I found the same condition by hand with help of Mathematica. The first step was Expand[(TrigToExp[(1/(8 a)) Sech[a x]^2 (-12 a^3 + 8 a d + 8 a c x + 96 a^3 Log[Cosh[a x]] + 12 a^3 Cosh[3 a x] Sech[a x] + c Sech[a x] Sinh[ 3 a x] - (c + 4 a^2 (4 b + x (12 a^2 + 2 d - 12 a^3 x + c x)) + 96 a^4 x (-Log[1 + E^(-2 a x)] + Log[Cosh[a x]]) + 48 a^3 PolyLog[2, -E^(-2 a x)]) Tanh[a x])] /. {Exp[-a*x] -> 0, Exp[-2*a*x] -> 0, Exp[-3*a*x] -> 0}) /. Log[E^(a x)/2] -> a*x - Log[2]]. $\endgroup$
    – user64494
    Jan 8 at 18:11
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    $\begingroup$ @simon Hope it helps! You are welcome! $\endgroup$ Jan 8 at 18:21
  • $\begingroup$ Expand[(TrigToExp[h1[x]] /. {Exc = -12 a^3p[-ax] -> 0, Exp[-2*ax] -> 0, Exp[-3*ax] -> 0}) /. Log[E^(a x)/2] -> ax - Log[2]] Thank you so c = -12 a^3 but the following still draws an echo not a limit of 0 Evaluate[Limit[h1[x], x -> Infinity]] $\endgroup$
    – simon
    Jan 8 at 18:57
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    $\begingroup$ @simon See the comments of the first two commands. I'm unsure whether Asymptotic is known in v9 $\endgroup$ Jan 8 at 19:27
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    $\begingroup$ @simon: Did you pay you attention to the words "The first step was ..." in my comment? $\endgroup$
    – user64494
    Jan 8 at 19:39

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