I want to set a symbol (or variable) for an expression. Here is an example:

A x^2 + C x + D

I want to reduce this expression to

x^2 + G x + D/A

but Mathematica can't do this way

A x^2 + C x + D /. C/A -> G

The above was just an simple example, but my problem is in complex expression like:

(c Kϕ Sin[(i π z)/l])/hw + (c EIw i^4 π^4 Sin[(i π z)/l])/(hw l^4) - 
  (c i^2 π^2 (-GJ + r0^2 P[z]) Sin[(i π z)/l])/(hw l^2) + 
  d Dw (6/hw^2 + β μ) Sin[(π z)/λ] + (c i π r0^2 Cos[(i π z)/l] Derivative[1][P][z])/(hw l)

and I want to replace some part like this:

enter image description here

and first expression convert to:

enter image description here

How can I do that?

  • 1
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    – user9660
    Jan 4 '17 at 17:34
  • $\begingroup$ I am marking this as a duplicate of mathematica.stackexchange.com/q/3822/121 -- unfortunately there is no robust built-in solution for this kind of thing but it has been brought up many times before and there are many partial solutions. See my answer to that question for links to many more related questions. Please try the solutions therein, and if they do not work edit this question to describe what you tried and what the result was, including a link to the source post. $\endgroup$
    – Mr.Wizard
    Jan 7 '17 at 20:41
A x^2 + C x + D /. {A -> 1, C -> G, D -> D/A}

gives what you expect

  • $\begingroup$ tnx but i said that very simple example to find a way for that complex expression. $\endgroup$
    – Amir
    Jan 4 '17 at 18:15

The question is ambiguous, in that what variables (if any) are to be eliminated when introducing H1 and F is not specified. One course of action is to name the original expression ex and then perform,

Solve[H1 == (EIy β^2 λ)/2, EIy][[1, 1]]
Solve[F == (3 Dw λ)/hw^3 - Dw*(2 - μ)/(2 hw) λ β, Dw][[1, 1]]
ex /. {%, %%}
(* -((2 c F hw^3 (-(6/hw^3) + (β (2 - μ))/hw) Sin[(π z)/λ])/(λ (6 - 2 hw^2 β + 
   hw^2 β μ))) + d Kx Sin[(i π z)/λ] + (2 d H1 i^4 π^4 Sin[(i π z)/λ])/(β^2 λ^5) - 
   (d i^2 π^2 P[z] Sin[(i π z)/λ])/λ^2 + (d i π Cos[(i π z)/λ] Derivative[1][P][z])/λ *)
  • $\begingroup$ I edited my question. please look again. tnx $\endgroup$
    – Amir
    Jan 4 '17 at 20:02
  • $\begingroup$ Please give your definitions of B1, etc in Mathematica format. $\endgroup$
    – bbgodfrey
    Jan 4 '17 at 20:24
  • $\begingroup$ Also, F and Q are not in your final expression. Is that intentional? $\endgroup$
    – bbgodfrey
    Jan 4 '17 at 20:32
  • $\begingroup$ B1 = (GJ [Pi]^2)/(2 [Lambda] hw^2); N1 = (Af r02 [Pi]^2)/(2 [Lambda] hw^2); S1 = (Af [Pi]^2)/(2 [Lambda]); H1 = (EIy [Beta]^2 [Lambda])/2; F = (3 Dw [Lambda])/hw^3 - Dw*(2 - [Mu])/(2 hw) [Lambda] [Beta]; Q = (3 Dw [Lambda])/hw^3 + Dw*[Mu]/(2 hw) [Lambda] [Beta]; $\endgroup$
    – Amir
    Jan 4 '17 at 20:35
  • $\begingroup$ Because I have 2 expression and i just posted one expression. Is it necessary to send exact formula? $\endgroup$
    – Amir
    Jan 4 '17 at 20:37

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