Mathematical functions power

• 03/31/2017 at 3:06 PM #30535
  Broc

  Is there a Prorealtime POWER math function e.g.

  y = x^3  as in 4^3 = 64?

   

  

  Broc

  Participant

  Topics: 2

  Replies: 2

  Been thanked: 8 times
  03/31/2017 at 3:23 PM #30539
  Nicolas

  Unfortunately no, this is a already a query for new instruction in the users wish list: https://www.prorealcode.com/topic/centralization-of-queries-and-suggestions-on-prorealtime/

  The best you could do to get the result of this math function is by making a loop like this:

  number=4 pow=3 result=0 i=1 while i<pow do if i=1 then result=number endif result=result*number i=i+1 wend return result

  1 2 3 4 5 6 7 8 9 10 11 12 13 14

  number=4 pow=3   result=0 i=1 while i<pow do if i=1 then   result=number endif result=result*number i=i+1 wend   return result

  Let me know if it’s clear for you.

  

  Nicolas

  Keymaster

  Topics: 81

  Replies: 20863

  Been thanked: 4773 times
  03/31/2017 at 3:43 PM #30547
  Broc

  Hi Nicolas

  Thanks for the prompt reply!

  I have found a mathematical equivalent if the Prorealtime LOG function (called Napierian log in the manual) is actually the common LN function:

  y = x^p = e^(p*LN(x))

  This could then be coded as:

  y = EXP(p*LOG(x))

  This works fine.

  W

   

  6 users thanked author for this post.

  Balmora74, Francesco78, Wing, Maz, zilliq, Nicolas

  

  Broc

  Participant

  Topics: 2

  Replies: 2

  Been thanked: 8 times
  04/02/2017 at 10:32 AM #30680
  zilliq

  Thanks for your answer wessel 😉

  How did you know that ?

  1 user thanked author for this post.

  laurenzo

  

  zilliq

  Participant

  Topics: 58

  Replies: 683

  Been thanked: 152 times
  04/03/2017 at 7:30 AM #30778
  Broc

  I had to do a bit of math research…  Prorealtime could use this formula to create an easy power math function.

   

  B-)

  2 users thanked author for this post.

  Wing, zilliq

  

  Broc

  Participant

  Topics: 2

  Replies: 2

  Been thanked: 8 times
  05/03/2017 at 1:34 PM #34416
  Maz

  facepalm

  

  Maz

  Participant

  Topics: 14

  Replies: 121

  Been thanked: 138 times
  05/03/2017 at 2:14 PM #34426
  Maz

  I have a power function like this?

  // n = 2 // x = 3 // r = n^x once j = x-1 r = n if x > 0 then for i = 1 to j do r = (r * n) next else for i = 1 to j do r = (r / n) next endif return r

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

  // n = 2 // x = 3 // r = n^x   once j = x-1   r = n if x > 0 then for  i = 1 to j do r = (r * n) next else for  i = 1 to j do r = (r / n) next endif   return r

   

  1 user thanked author for this post.

  Nicolas

  

  Maz

  Participant

  Topics: 14

  Replies: 121

  Been thanked: 138 times
  05/03/2017 at 2:15 PM #34428
  Maz

  Here’s tanh (subsequently sinh and cosh)

   

  // x = 2 // result = tanh(x) once e = exp(1) mx = x * -1 eX = CALL fnPower[e, x] eXm = CALL fnPower[e, mx] sinh = 0.5* (eX - eXm) cosh = 0.5* (ex + eXm) tanh = sinh / cosh return tanh

  1 2 3 4 5 6 7 8 9 10 11 12 13

  // x = 2 // result = tanh(x)   once e  = exp(1) mx = x * -1 eX  = CALL fnPower[e, x] eXm = CALL fnPower[e, mx]   sinh = 0.5* (eX - eXm) cosh = 0.5* (ex + eXm) tanh = sinh / cosh   return tanh

   

  See above for powers

  2 users thanked author for this post.

  Wing, Nicolas

  

  Maz

  Participant

  Topics: 14

  Replies: 121

  Been thanked: 138 times
  05/03/2017 at 2:52 PM #34439
  Nicolas

  Great job Maz!

  1 user thanked author for this post.

  Maz

  

  Nicolas

  Keymaster

  Topics: 81

  Replies: 20863

  Been thanked: 4773 times
  05/03/2017 at 9:51 PM #34483
  Maz

  You’re welomce; unfortunately it’s very slow but that’s all we have for now 🙂

  

  Maz

  Participant

  Topics: 14

  Replies: 121

  Been thanked: 138 times
• Author
  Posts