Group Abstract Group Abstract

Message Boards Message Boards

0
|
3.7K Views
|
3 Replies
|
3 Total Likes
View groups...
Share
Share this post:

FourierCosCoefficient takes long time with no output

Zhang Yuhan
Posted 3 years ago
POSTED BY: Zhang Yuhan
3 Replies
Sort By:

The expression is very complicated for Mathematica.

For general parameters, Mathematica is struggling to calculate the integral.

If we assume:

Subscript[m, e] = 1; Subscript[p, w] = 2;

Assuming[{L > 0, Subscript[a, e] > 0, I0 > 0, Subscript[c, e] > 0, 
  Subscript[c, w] > 0, Subscript[\[Alpha], e0] > 0, 
  Subscript[\[Alpha], w0] > 0}, 
 FourierCosCoefficient[(I0 + Subscript[a, e]*Cos[\[Theta]])*(1 - (
     Subscript[\[Alpha], e0]*Subscript[c, e]*
      L)/((Subscript[m, e]*Cos[\[Theta]])^2 + 1) - (
     Subscript[\[Alpha], w0]*Subscript[c, w]*
      L)/((Subscript[p, w] + Subscript[m, e]*Cos[\[Theta]])^2 + 
      1)), \[Theta], 2]]

 (*-(1/(20 Sqrt[2] \[Pi]))
  L (40 (-3 + 2 Sqrt[2]) I0 \[Pi] Subscript[c, e] Subscript[\[Alpha], 
      e0] + (I0 (((-42 + 4 I) Sqrt[-1 - 2 I] + 
              80 Sqrt[2] - (6 + 26 I) Sqrt[20 - 10 I]) \[Pi] + (2 + 
              21 I) Sqrt[-1 - 
             2 I] (-2 I ArcTan[1/2] + Log[5] + 
              4 Log[-Sqrt[-(2/5) - I/5]])) + (((88 + 34 I) Sqrt[-1 - 
                2 I] - 320 Sqrt[2] + (38 + 46 I) Sqrt[
               20 - 10 I]) \[Pi] - (44 + 17 I) Sqrt[-1 - 
             2 I] (2 ArcTan[1/2] + 
              I (Log[5] + 4 Log[-Sqrt[-(2/5) - I/5]]))) Subscript[a, 
         e]) Subscript[c, w] Subscript[\[Alpha], w0])*)
POSTED BY: Mariusz Iwaniuk

If:

$\text{FourierCosCoefficient}[f(t),t,2]=\frac{2 \int_0^{\pi } f(t) \cos (2 t) \, dt}{\pi }$

then:

2/Pi*Integrate[((I0 + Subscript[a, e]*Cos[\[Theta]])*(1 - (
       Subscript[\[Alpha], e0]*Subscript[c, e]*
        L)/((Subscript[m, e]*Cos[\[Theta]])^2 + 1) - (
       Subscript[\[Alpha], w0]*Subscript[c, w]*
        L)/((Subscript[p, w] + Subscript[m, e]*Cos[\[Theta]])^2 + 
        1))) Cos[2 \[Theta]], {\[Theta], 0, 
   Pi}]

I abort computation with 1 hour.

For general integral Mathematica can compute:

 f = 2/Pi*Integrate[((I0 + Subscript[a, e]*Cos[\[Theta]])*(1 - (
         Subscript[\[Alpha], e0]*Subscript[c, e]*
          L)/((Subscript[m, e]*Cos[\[Theta]])^2 + 1) - (
         Subscript[\[Alpha], w0]*Subscript[c, w]*
          L)/((Subscript[p, w] + Subscript[m, e]*Cos[\[Theta]])^2 + 
          1))) Cos[2 \[Theta]], \[Theta], 
    Assumptions -> Subscript[m, e] > 0]

(*(1/(3 \[Pi]))(3 I0 Sin[2 \[Theta]] + 
  Sin[3 \[Theta]] Subscript[a, e] - (
  3 L Log[-Sin[\[Theta]] Subscript[m, e] + Sqrt[1 + 
\!\(\*SubsuperscriptBox[\(m\), \(e\), \(2\)]\)]] Subscript[a, e]
    Subscript[c, e] (2 + 
\!\(\*SubsuperscriptBox[\(m\), \(e\), \(2\)]\)) Subscript[\[Alpha], 
   e0])/(\!\(
\*SubsuperscriptBox[\(m\), \(e\), \(3\)]\ 
\*SqrtBox[\(1 + 
\*SubsuperscriptBox[\(m\), \(e\), \(2\)]\)]\)) + (
  3 L Log[Sin[\[Theta]] Subscript[m, e] + Sqrt[1 + 
\!\(\*SubsuperscriptBox[\(m\), \(e\), \(2\)]\)]] Subscript[a, e]
    Subscript[c, e] (2 + 
\!\(\*SubsuperscriptBox[\(m\), \(e\), \(2\)]\)) Subscript[\[Alpha], 
   e0])/(\!\(
\*SubsuperscriptBox[\(m\), \(e\), \(3\)]\ 
\*SqrtBox[\(1 + 
\*SubsuperscriptBox[\(m\), \(e\), \(2\)]\)]\)) - (
  6 I0 L ArcTan[Subscript[m, e] - Sqrt[1 + 
\!\(\*SubsuperscriptBox[\(m\), \(e\), \(2\)]\)]
       Tan[\[Theta]/2]] Subscript[c, e] (2 + 
\!\(\*SubsuperscriptBox[\(m\), \(e\), \(2\)]\)) Subscript[\[Alpha], 
   e0])/(\!\(
\*SubsuperscriptBox[\(m\), \(e\), \(2\)]\ 
\*SqrtBox[\(1 + 
\*SubsuperscriptBox[\(m\), \(e\), \(2\)]\)]\)) + (
  6 I0 L ArcTan[Subscript[m, e] + Sqrt[1 + 
\!\(\*SubsuperscriptBox[\(m\), \(e\), \(2\)]\)]
       Tan[\[Theta]/2]] Subscript[c, e] (2 + 
\!\(\*SubsuperscriptBox[\(m\), \(e\), \(2\)]\)) Subscript[\[Alpha], 
   e0])/(\!\(
\*SubsuperscriptBox[\(m\), \(e\), \(2\)]\ 
\*SqrtBox[\(1 + 
\*SubsuperscriptBox[\(m\), \(e\), \(2\)]\)]\)) + (
  6 L ArcTanh[((I + Subscript[m, e] - Subscript[p, w]) Tan[\[Theta]/
      2])/Sqrt[\!\(
\*SubsuperscriptBox[\(m\), \(e\), \(2\)] - 
\*SuperscriptBox[\((\(-I\) + 
\*SubscriptBox[\(p\), \(w\)])\), \(2\)]\)]] Subscript[c, 
   w] (-I I0 Subscript[m, e] + 
     Subscript[a, e] (1 + I Subscript[p, w])) (\!\(
\*SubsuperscriptBox[\(m\), \(e\), \(2\)] - \(2\ 
\*SuperscriptBox[\((\(-I\) + 
\*SubscriptBox[\(p\), \(w\)])\), \(2\)]\)\)) Subscript[\[Alpha], 
   w0])/(\!\(
\*SubsuperscriptBox[\(m\), \(e\), \(3\)]\ 
\*SqrtBox[\(
\*SubsuperscriptBox[\(m\), \(e\), \(2\)] - 
\*SuperscriptBox[\((\(-I\) + 
\*SubscriptBox[\(p\), \(w\)])\), \(2\)]\)]\)) - (
  6 L ArcTan[((1 + I Subscript[m, e] - 
       I Subscript[p, w]) Tan[\[Theta]/2])/Sqrt[\!\(
\*SubsuperscriptBox[\(m\), \(e\), \(2\)] - 
\*SuperscriptBox[\((I + 
\*SubscriptBox[\(p\), \(w\)])\), \(2\)]\)]] Subscript[c, 
   w] (-I0 Subscript[m, e] + 
     Subscript[a, e] (I + Subscript[p, w])) (\!\(
\*SubsuperscriptBox[\(m\), \(e\), \(2\)] - \(2\ 
\*SuperscriptBox[\((I + 
\*SubscriptBox[\(p\), \(w\)])\), \(2\)]\)\)) Subscript[\[Alpha], 
   w0])/(\!\(
\*SubsuperscriptBox[\(m\), \(e\), \(3\)]\ 
\*SqrtBox[\(
\*SubsuperscriptBox[\(m\), \(e\), \(2\)] - 
\*SuperscriptBox[\((I + 
\*SubscriptBox[\(p\), \(w\)])\), \(2\)]\)]\)) - (
  12 L \[Theta] (I0 Subscript[c, e] Subscript[m, e]
       Subscript[\[Alpha], e0] + 
     Subscript[c, 
      w] (I0 Subscript[m, e] - 
        2 Subscript[a, e] Subscript[p, w]) Subscript[\[Alpha], w0]))/
\!\(\*SubsuperscriptBox[\(m\), \(e\), \(3\)]\) + (
  3 Sin[\[Theta]] Subscript[a, e] (\!\(
\*SubsuperscriptBox[\(m\), \(e\), \(2\)] - \(4\ L\ \((
\*SubscriptBox[\(c\), \(e\)]\ 
\*SubscriptBox[\(\[Alpha]\), \(e0\)] + 
\*SubscriptBox[\(c\), \(w\)]\ 
\*SubscriptBox[\(\[Alpha]\), \(w0\)])\)\)\)))/
\!\(\*SubsuperscriptBox[\(m\), \(e\), \(2\)]\))*)

If we assume function is continuous,we take the limits:

Answer = Limit[f, \[Theta] -> Pi, Direction -> 1, 
    Assumptions -> Subscript[m, e] > 0] - Limit[f, \[Theta] -> 0] // 
  Simplify

  (*(1/
  \!\(\*SubsuperscriptBox[\(m\), \(e\), \(3\)]\))L ((
     2 I0 Subscript[c, e] Subscript[m, e] (2 + 
  \!\(\*SubsuperscriptBox[\(m\), \(e\), \(2\)]\)) Subscript[\[Alpha], 
      e0])/Sqrt[1 + 
  \!\(\*SubsuperscriptBox[\(m\), \(e\), \(2\)]\)] + ((Log[(
         I + Subscript[m, e] - Subscript[p, w])/Sqrt[\!\(
  \*SubsuperscriptBox[\(m\), \(e\), \(2\)] - 
  \*SuperscriptBox[\((\(-I\) + 
  \*SubscriptBox[\(p\), \(w\)])\), \(2\)]\)]] - 
        Log[(-I -cript[m, e] + Subscript[p, w])/Sqrt[\!\(
  \*SubsuperscriptBox[\(m\), \(e\), \(2\)] - 
  \*SuperscriptBox[\((\(-I\) + 
  \*SubscriptBox[\(p\), \(w\)])\), \(2\)]\)]]) Subscript[c, 
      w] (-I I0 Subscript[m, e] + 
        Subscript[a, e] (1 + I Subscript[p, w])) (\!\(
  \*SubsuperscriptBox[\(m\), \(e\), \(2\)] - \(2\ 
  \*SuperscriptBox[\((\(-I\) + 
  \*SubscriptBox[\(p\), \(w\)])\), \(2\)]\)\)) Subscript[\[Alpha], 
      w0])/(\[Pi] Sqrt[\!\(
  \*SubsuperscriptBox[\(m\), \(e\), \(2\)] - 
  \*SuperscriptBox[\((\(-I\) + 
  \*SubscriptBox[\(p\), \(w\)])\), \(2\)]\)]) + ((2 \[Pi] Floor[(
          2 \[Pi] - 2 Arg[-I + Subscript[m, e] - Subscript[p, w]] + 
           Arg[\!\(
  \*SubsuperscriptBox[\(m\), \(e\), \(2\)] - 
  \*SuperscriptBox[\((I + 
  \*SubscriptBox[\(p\), \(w\)])\), \(2\)]\)])/(4 \[Pi])] - 
        2 \[Pi] Floor[(
          2 \[Pi] - 2 Arg[2 I - 2 Subscript[m, e] + 2 Subscript[p, w]] +
            Arg[\!\(
  \*SubsuperscriptBox[\(m\), \(e\), \(2\)] - 
  \*SuperscriptBox[\((I + 
  \*SubscriptBox[\(p\), \(w\)])\), \(2\)]\)])/(4 \[Pi])] - 
        I (Log[-I + Subscript[m, e] - Subscript[p, w]] - 
           Log[I - Subscript[m, e] + Subscript[p, w]])) Subscript[c, 
      w] (-I0 Subscript[m, e] + 
        Subscript[a, e] (I + Subscript[p, w])) (\!\(
  \*SubsuperscriptBox[\(m\), \(e\), \(2\)] - \(2\ 
  \*SuperscriptBox[\((I + 
  \*SubscriptBox[\(p\), \(w\)])\), \(2\)]\)\)) Subscript[\[Alpha], 
      w0])/(\[Pi] Sqrt[\!\(
  \*SubsuperscriptBox[\(m\), \(e\), \(2\)] - 
  \*SuperscriptBox[\((I + 
  \*SubscriptBox[\(p\), \(w\)])\), \(2\)]\)]) - 
     4 (I0 Subscript[c, e] Subscript[m, e] Subscript[\[Alpha], e0] + 
        Subscript[c, 
         w] (I0 Subscript[m, e] - 
           2 Subscript[a, e] Subscript[p, w]) Subscript[\[Alpha], w0]))*)
POSTED BY: Mariusz Iwaniuk
Zhang Yuhan
Zhang Yuhan
Posted 3 years ago

Thanks a million. I have tried the way you said, however, I want to get a general parameter solution.

Best wishes to you all!
POSTED BY: Zhang Yuhan
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Tag limit exceeded
Note: Only the first five people you tag will receive an email notification; the other tagged names will appear as links to their profiles.
Reply Preview
Attachments
Remove
or Discard