On Sat, 28 Nov 2009, Dmitry Vinokurov wrote:
< Hello,
<
< I've got an equation, for example 'x_1 + x_2 = 0' (of course real case
< is not so simple, but it's enough for demonstration) and I need to get
< function F(x_2) from it.
<
< I tried to use solve for it 'solve(x_1 + x_2 = 0, x_1)' and get 'x_1 = -
< x_2'. Now I couldn't find, how to extract right part of solve result
< into function. Just 'F(x_2) := solve(x_1 + x_2 = 0, x_1)' of course
< doesn't work, it results to 'x_1 = -x_2' instead of wanted '- x_2'.
<
< Can anybody help with it? And maybe I'm on the wrong path and shouldn't
< use 'solve' for it?
<
< Thanks in advance.
How about
(%i5) eq : sin(x)=u; define(F(u),rhs(part(solve(eq,x),1)));
(%o5) sin(x) = u
solve: using arc-trig functions to get a solution.
Some solutions will be lost.
(%o6) F(u):=asin(u)
You can tweak this if you have multiple solutions, etc.
Leo
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