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Factor out known multipliers in algebraic expressions?

André Barbosa

Posted 8 years ago

Hi there, I have this function of a[t], and it's a long function of other variables in time and some constant parameter (Ixx, Iyy, Izz), which I really want to explicit in a somatory of any combination of those constant parameters. I've tried to Simplify and then Apart, but Mathematica doesn't group similar coefficients. I wonder if I could get this result applying some kind of mapping (...)? Well, below is the function a[t]: a[t_] := -((Cos[?[t]]* Izz[t](Ty[t] - Cos[?[t]]Iyy[t]Derivative[1][?][t] Derivative[1][?][t] + Cos[?[t]](-Ixx[t] + Izz[t]) (Sin[?[t]]Derivative[1][?][t] + Derivative[1][?][t])((-Cos[?[t]])* Tan[?[t]]Derivative[1][?][t] + Derivative[1][?][t]) + Iyy[t]Derivative[1][?][ t](Cos[?[t]]Sin[?[t]]* Derivative[1][?][t] + Cos[?[t]]Sin[?[t]] Derivative[1][?][t])) - Iyy[t]Sin[?[t]](Tz[t] + Izz[t]Sin[?[t]]Derivative[1][?][t]* Derivative[1][?][ t] + (Ixx[t] - Iyy[t])(Sin[?[t]]Derivative[1][?][t] + Derivative[1][?][t])* (Cos[?[t]]Cos[?[t]] Derivative[1][?][t] + Sin[?[t]]Derivative[1][?][t]) + Izz[t]Derivative[1][?][ t](Cos[?[t]]Cos[?[t]]* Derivative[1][?][t] - Sin[?[t]]Sin[?[t]] Derivative[1][?][t])))/((-Cos[?[t]]^2)* Cos[?[t]]Iyy[t]Izz[t] - Cos[?[t]]Iyy[t]Izz[t]Sin[?[t]]^2)); Summarizing, what I want is a way to say : rewrite a[t] factorizing in a sum by these terms (Ixx, Iyy, Izz) on any combination between them.* ( In other way, I could ask the same by rewriting a[t] factorizing in terms that don't vary in time? Would this be possible/easier? ) Can anybody suggest a good thing to try? Thanks. Attachments:

Hi there,

I have this function of a[t], and it's a long function of other variables in time and some constant parameter (Ixx, Iyy, Izz), which I really want to explicit in a somatory of any combination of those constant parameters. I've tried to Simplify and then Apart, but Mathematica doesn't group similar coefficients. I wonder if I could get this result applying some kind of mapping (...)?

Well, below is the function a[t]:

a[t_] := -((Cos[?[t]]*
        Izz[t]*(Ty[t] - 
          Cos[?[t]]*Iyy[t]*Derivative[1][?][t]*
           Derivative[1][?][t] + 
          Cos[?[t]]*(-Ixx[t] + Izz[t])*
                    (Sin[?[t]]*Derivative[1][?][t] + 
             Derivative[1][?][t])*((-Cos[?[t]])*
              Tan[?[t]]*Derivative[1][?][t] + 
             Derivative[1][?][t]) + 

          Iyy[t]*Derivative[1][?][
            t]*(Cos[?[t]]*Sin[?[t]]*
              Derivative[1][?][t] + 
             Cos[?[t]]*Sin[?[t]]*
              Derivative[1][?][t])) - 

       Iyy[t]*Sin[?[t]]*(Tz[t] + 
          Izz[t]*Sin[?[t]]*Derivative[1][?][t]*
           Derivative[1][?][
            t] + (Ixx[t] - 
             Iyy[t])*(Sin[?[t]]*Derivative[1][?][t] + 
             Derivative[1][?][t])*
                    (Cos[?[t]]*Cos[?[t]]*
              Derivative[1][?][t] + 
             Sin[?[t]]*Derivative[1][?][t]) + 
          Izz[t]*Derivative[1][?][
            t]*(Cos[?[t]]*Cos[?[t]]*
              Derivative[1][?][t] - 

             Sin[?[t]]*Sin[?[t]]*
              Derivative[1][?][t])))/((-Cos[?[t]]^2)*
        Cos[?[t]]*Iyy[t]*Izz[t] - 
       Cos[?[t]]*Iyy[t]*Izz[t]*Sin[?[t]]^2));

Summarizing, what I want is a way to say : rewrite a[t] factorizing in a sum by these terms (Ixx, Iyy, Izz) on any combination between them. ( In other way, I could ask the same by rewriting a[t] factorizing in terms that don't vary in time? Would this be possible/easier? )

Can anybody suggest a good thing to try?

Thanks.

POSTED BY: André Barbosa

6 Replies

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Hans Dolhaine

Hans Dolhaine, retired

Posted 8 years ago

Another method, perhaps easier to be made more general is: Find the "variables" - meaning the expressions not being dependet of A or B vv = Complement[Variables[y], {A, B}] Now find the coefficients of these variables cc = Coefficient[y, #] & /@ vv and then find the factors of these Collect[y, cc]

POSTED BY: Hans Dolhaine

André Barbosa

Posted 8 years ago

Thank you! This idea worked pretty neat! Precisely what I intended to obtain as an answer form of the equation. Very easy to understand. Now I will try to apply those methods into a[t]. Because I need to simplify redundant functions for applying linear solving methods (non singular matrix). I'm grateful! Thanks again.

POSTED BY: André Barbosa

Hans Dolhaine

Hans Dolhaine, retired

Posted 8 years ago

hmmm - ok. perhaps you should disguise the mixed terms. Seems that this Collect[y /. {A B -> ab, A/B -> adb, B/A -> bda}, {A, B, ab, adb,bda}] /. {ab -> A B, adb -> A/B, bda -> B/A} works in your simple example, but for sure it will get tedious in more complicated cases. And what is wrong with the method I mentioned first? This gives num1 as a polynomial in Ixx, Ixx^2, Ixx * Iyy and so on with coefficients as functions depending only of t. Same for the denominator and then you can write your a [ t ] in the form you want.

POSTED BY: Hans Dolhaine

André Barbosa

Posted 8 years ago

Attachments:

POSTED BY: André Barbosa

Hans Dolhaine

Hans Dolhaine, retired

Posted 8 years ago

I am not quite sure what you want to do, but first of all in a [ t ] your "constant parameter (Ixx, Iyy, Izz)" are functions of t as well. So I remove this num = Numerator[a[t]] /. {Ixx[t] -> Ixx, Iyy[t] -> Iyy, Izz[t] -> Izz} // Expand den = Denominator[a[t]] /. {Ixx[t] -> Ixx, Iyy[t] -> Iyy, Izz[t] -> Izz} // FullSimplify and Collect terms num1 = Collect[num, {Ixx, Iyy, Izz, Ixx Iyy, Ixx Izz, Iyy Izz}] Check it (a[t] - num1/den /. {Ixx[t] -> Ixx, Iyy[t] -> Iyy, Izz[t] -> Izz}) // Simplify You may well replace Ixx for Ixx [ t ] ans so on in num1 / den again.... Hope this helps.

I am not quite sure what you want to do, but first of all in a [ t ] your "constant parameter (Ixx, Iyy, Izz)" are functions of t as well.

So I remove this

num = Numerator[a[t]] /. {Ixx[t] -> Ixx, Iyy[t] -> Iyy,  Izz[t] -> Izz} // Expand
den = Denominator[a[t]] /. {Ixx[t] -> Ixx, Iyy[t] -> Iyy,  Izz[t] -> Izz} // FullSimplify

and Collect terms

num1 = Collect[num, {Ixx, Iyy, Izz, Ixx Iyy, Ixx Izz, Iyy Izz}]

Check it

(a[t] - num1/den /. {Ixx[t] -> Ixx, Iyy[t] -> Iyy,     Izz[t] -> Izz}) // Simplify

You may well replace Ixx for Ixx [ t ] ans so on in num1 / den again....

Hope this helps.

POSTED BY: Hans Dolhaine

André Barbosa

Posted 8 years ago

Thanks for the help. Check the new file that I've uploaded below.

POSTED BY: André Barbosa

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