Résoudre une équation à 1 inconnue

• 09/28/2024 at 7:30 AM #238176
  Bodaris

  Bonjour,

  j’aimerais tester un truc, mais je ne sais pas si c’est possible. Je voudrais savoir à quel niveau de moyenne mobile simple correspond un prix.

  Grosse modo je voudrais résoudre cette équation :

  Close = summation[x](close)/X

  X=???

   

  Je sais pas si c’est assez clair ? Possible ou pas ?

  Merci d’avance pour votre réponse :).

  

  Bodaris

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  09/28/2024 at 1:05 PM #238186
  JS

  Salut,

  Je l’ai essayé pour vous, mais la chance que la moyenne soit exactement égale au prix est négligeable…

  Matching Moving Average

  Found=0 N=1000 //Maximum Period CloseTarget = Close // The closing price you're comparing against FOR x = 2 TO N // You can adjust to any maximum period you want SMAX = average[x](Close) // Calculate the moving average for X periods IF (SMAX = CloseTarget) THEN Found = 1 break // Stop the loop when a match is found ENDIF NEXT DrawText("x=#x#", BarIndex, High+5) DrawText("Found=#Found#",BarIndex, Low-5) RETURN

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  Found=0 N=1000 //Maximum Period CloseTarget = Close // The closing price you're comparing against   FOR x = 2 TO N  // You can adjust to any maximum period you want SMAX = average[x](Close) // Calculate the moving average for X periods IF (SMAX = CloseTarget) THEN Found = 1 break // Stop the loop when a match is found ENDIF NEXT DrawText("x=#x#", BarIndex, High+5) DrawText("Found=#Found#",BarIndex, Low-5)   RETURN

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  Iván, Bodaris

  

  JS

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  09/28/2024 at 2:41 PM #238196
  Bodaris

  Bonjour,

  Un grand merci !!! après quelques modifications, j’y suis arrivé. C’est déjà pas mal comme résultat, mieux que rien ^^

   

  j’ai rajouté un screenshot pour une idée du visuel.

  Found=0 N=100 //Maximum Period CloseTarget = high // The closing price you're comparing against FOR x = 2 TO N // You can adjust to any maximum period you want SMAX = average[x](Close) // Calculate the moving average for X periods IF (SMAX => CloseTarget-5 and SMAX<=closetarget+5) THEN Found = 1 //DrawText("x=#x#", BarIndex, High+5) //DrawText("Found=#Found#",BarIndex, Low-5) //ELSE //found=0 break // Stop the loop when a match is found ENDIF NEXT c1= found=1 and smax=Average[7](close) if c1 THEN DrawText("x=#x#", BarIndex, High+5) endif RETURN c1//smax

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  Found=0 N=100 //Maximum Period CloseTarget = high // The closing price you're comparing against     FOR x = 2 TO N // You can adjust to any maximum period you want SMAX = average[x](Close) // Calculate the moving average for X periods IF (SMAX => CloseTarget-5 and SMAX<=closetarget+5) THEN Found = 1 //DrawText("x=#x#", BarIndex, High+5) //DrawText("Found=#Found#",BarIndex, Low-5) //ELSE //found=0 break // Stop the loop when a match is found ENDIF NEXT   c1= found=1 and smax=Average[7](close) if c1 THEN DrawText("x=#x#", BarIndex, High+5) endif   RETURN c1//smax

   

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  Iván, JS

  

  Bodaris

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  09/28/2024 at 2:47 PM #238198
  Bodaris

  Mais vous avez raison, c’est compliqué de tomber sur la valeur d’une moyenne mobile. Je vais voir si je m’en sors. Je cherche à identifier à quel moyenne mobile un pic et un creux peut correspondre.

  

  Bodaris

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  09/28/2024 at 3:04 PM #238199
  JS

  Bonjour,

  Vous pouvez également regarder la régression linéaire (méthode des moindres carrés)…

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  Bodaris

  

  JS

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  12/14/2024 at 10:07 PM #241429
  Bodaris

  J’ai ouvert un nouveau sujet ici https://www.prorealcode.com/topic/utiliser-une-variable-dynamique-comme-periode-pour-une-sma/

  

  Bodaris

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