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How to transpose one term to the front of expression?

Jacques Ou
Posted 4 years ago

The expression is

1/(2 (-1 + 
    A6) A7 v0) ((-1 + 2 A6 - A7) (1 + ks) v0 SuperStar[(Subscript[c, 
     m])] - (-1 + A7) (Imax - A6 Imax + 
      A6 Km (1 + 
         ks) v0 + \[Sqrt](((-1 + A6) Imax - 
           A6 Km (1 + ks) v0)^2 + (1 + ks) v0 SuperStar[(Subscript[c, 
           m])] (2 (-1 + A6) Imax + 
            2 A6 Km (1 + ks) v0 + (1 + ks) v0 SuperStar[(Subscript[c, 
              m])]))))

Now I need transpose 

(1 + ks) v0 SuperStar[(Subscript[c, m])]

from inside the brackets to the outside of brackets.

Thanks.
POSTED BY: Jacques Ou
4 Replies
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Updating Name
Updating Name
Posted 4 years ago

It works both inside and outside of the radical symbol,

eq1 = 1/(2 A7 (1 + 
     ks) v0) ((-1 + A7) Imax - (1 + ks) v0 (A7 Km - 
       SuperStar[(Subscript[c, m])]) + Sqrt[
    4 A7 Km (1 + ks)^2 v0^2 SuperStar[(Subscript[c, 
       m])] + ((-1 + A7) Imax - 
       A7 Km (1 + ks) v0 + (1 + ks) v0 SuperStar[(Subscript[c, 
         m])])^2])

eq2 = (1 + ks) v0 SuperStar[(Subscript[c, m])]

eq3 = eq1/eq1[[5]]*eq2*
   Map[Divide[#, eq2] &, eq1[[5]]] /. {a_*Sqrt[b_] :> 
    Sqrt[Map[Divide[#, 1/a] &, b]]}
POSTED BY: Updating Name 
expr = (1 + ks) v0 SuperStar[(Subscript[c, m])];
Sqrt[((-1 + A3) Imax - A3 Km v0)^2 + 
   v0 SuperStar[(Subscript[c, m])] (2 (-1 + A3) Imax + 2 A3 Km v0 + 
      v0 SuperStar[(Subscript[c, m])])] /. 
 Sqrt[stuff_] :> Sqrt[expr*Map[Divide[#, expr] &, stuff]]
POSTED BY: Gianluca Gorni

I don't understand what you mean, but perhaps this goes some way:

Collect[%, (1 + ks) v0 SuperStar[(Subscript[c, m])], Factor]

There are SuperStars inside the square root. How do you expect to "transpose" them?
POSTED BY: Gianluca Gorni
Jacques Ou
Jacques Ou
Posted 4 years ago

It doesn't work inside the square root symbol.

Sqrt[((-1 + A3) Imax - A3 Km v0)^2 + 
 v0 SuperStar[(Subscript[c, 
   m])] (2 (-1 + A3) Imax + 2 A3 Km v0 + 
    v0 SuperStar[(Subscript[c, m])])]
POSTED BY: Jacques Ou
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