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I have posed a specific question yesterday but it may be too trivial to answer.

I think the question could be asked in a more general way so it is easier to answer. Then I could solve the original problem by myself.

Usually, we define a function as

F[x_] := x + 10 

where the right-hand side is an expression.

Now, I would like define a function where the right-hand side is the solution of a set of equations. Like

optV[V_, r_, λ_, μ_, η_, σ_] := 
  V /. 
      A V^θ[r, λ, μ, η, σ] h[V, c, r, λ, μ, η, σ] == V - c, 
      D[A*V^θ*h[V, c, r, λ, μ, η, σ], V] == 1
      {A, 0}, {V, 1}]; 

a[V_, r_, λ_, μ_, η_, σ_] := 
  A /. 
      A V^θ[r, λ, μ, η, σ] h[c, r, λ, μ, η, σ] == V - c, 
      D[A V^θ[r, λ, μ, η, σ]  h[V, c, r, λ, μ, η, σ], V] == 1
      {A, 0}, {V,1}]

This may not be correct code, but I think you can understand my question from it. optV and a are both parameters for another final function which would be manipulated. So you could regard optV and a as intermediate parameters. So the solution of the set of equations, correspondingly optV and a, varies along with c, r, λ, μ, η, σ, which are parameters input for manipulating plot.

share|improve this question
Since NSolve[] will return a list of lists, you'll likely want to put Part[] in there somewhere... – J. M. Jun 17 '13 at 17:45
@0x4A4D – Dr. belisarius Jun 17 '13 at 17:49
@bel, FWIW: it's one of the songs I listen to after a hard day... precisely that medley. – J. M. Jun 17 '13 at 17:51
Just yesterday I posted an answer where this is illustrated. There are countless other answers on this site that do similar things. Maybe just searching for NDSolve will turn up other good examples. – Jens Jun 17 '13 at 17:56
you can easily do what you asked, eg.f[a_] := x /. NSolve[{x + y == a, x^2 + y^2 == 1}, {x, y}]. I cant understand what your intent is with those examples though. The arg x on input does what? – george2079 Jun 17 '13 at 18:37

If I understand you right, here is the answer, but it is in a form of a more transparent example.

Let this

eq=x^2 - b*x - 1 == 0

be an equation depending upon a parameter b with b>0. Let us define a=a[b] which is its smaller solution of the equation eq:

a[b_] := Solve[eq, x][[1, 1, 2]]

It seems that this (or something alike) is what you need. You can check that a is indeed a function of b by, say, plotting a[b]. Evaluate this:

Plot[a[b], {b, 0, 3}]

With some care one can do the same with the FindRoot statement:

a[b_] := FindRoot[eq, {x, 0}][[1, 2]]
Plot[a[b], {b, 0, 3}]
share|improve this answer
While this works, I would be careful about explicitely making eq a function of b: 'eq[b_] = x^2 - b*x - 1 == 0' and I would force the definition of a[b] to take only numeric values: a[b_?NumberQ]. In this example it is fine, but you have to be careful of the order of evaluation when you are mixing both symbolic and numerical values. – Jonathan Shock Jun 18 '13 at 8:41
Isn't it inefficient to call Solve every time the function a is evaluated? If you used Set instead of SetDelayed, the equation would only be solved once. – nikie Jun 18 '13 at 9:07
@nikie yes in this simple example. There are however situations where you need to Solve (or especially NSolve ) for the specific value of the parameters. – george2079 Jun 18 '13 at 19:15

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