问题
If I evaluate Solve[f[x,y]==0,x]
, I get a bunch of solutions like:
{{x -> something g[y]}, {x -> something else}}
, etc.
Now I want to convert each of those x->somethings
into a function. Typically, my requirements are low, and my function f[x]
is at the most a cubic, with straightforward solutions for x
. So I've always just defined g1[y_]:=something
, g2[y_]:=...
etc, manually.
However, for a function that I have now, Solve
outputs a complicated polynomial running 4 pages long, and there are 4 such solutions. I've tried reducing to simpler forms using Simplify
, Collect
, Factor
etc, but it just seems irreducible.
Is there a way I can automatically assign them to functions? (It's extremely hard to scroll through pages and copy each one... and I have to look for where the next one begins!)
Something like: {g1[y_], g2[y_], g3[y_]} = output of Solve
?
回答1:
It appears Simon beat me to an answer (I am glad that StackOverflow gives me a pop-up to let me know!), therefore I will take a different approach. You should know how to use the output of Solve directly, as quite a few times it will be convenient to do that.
Starting with
ClearAll[a, x, sols]
sols = Solve[x^2 + a x + 1 == 0, x]
Here are some things you can do.
Find the solutions to x
for a == 7
x /. sols /. a -> 7
Plot the solutions
Evaluate
is used here not out of necessity for basic function, but to allow the Plot function to style each solution separately
Plot[Evaluate[x /. sols], {a, 1, 4}]
Define a new function of a
for the second solution
Notice the use of =
rather than :=
here
g[a_] = x /. sols[[2]]
Here is an alternative to Simon's method for defining functions for each solution
MapIndexed[(gg[#2[[1]]][a_] := #) &, x /. sols]
The function is then used with the syntax gg[1][17]
to mean the first solution, and a == 17
Plot[gg[1][a], {a, 1, 4}]
gg[2] /@ {1, 2, 3}
These uses do generally require that a
(in this example) remain unassigned.
回答2:
Here's a simple solution that could be cleaned up
In[1]:= solns = Solve[x^2+a x+b==0, x]
Out[1]= {{x -> 1/2 (-a-Sqrt[a^2-4 b])}, {x -> 1/2 (-a+Sqrt[a^2-4 b])}}
In[2]:= Table[Symbol["g"<>ToString[i]][a_,b_] := Evaluate[x/.solns[[i]]],
{i,Length[solns]}];
In[3]:= DownValues/@{g1,g2}
Out[3]= {{HoldPattern[g1[a_,b_]]:>1/2 (-a-Sqrt[a^2-4 b])},
{HoldPattern[g2[a_,b_]]:>1/2 (-a+Sqrt[a^2-4 b])}}
回答3:
The following function will automatically convert the output of Solve
to a list of functions (assuming Solve
finds solutions of course):
solutionFunctions[expr_, var_] :=
Check[Flatten @ Solve[expr, var], $Failed] /.
(_ -> x_) :>
Function[Evaluate[Union @ Cases[x, _Symbol?(!NumericQ[#]&), Infinity]], x]
Here is an example:
In[67]:= g = solutionFunctions[x^2+a x+1==0, x]
Out[67]= {Function[{a},1/2(-a-Sqrt[-4+a^2])],Function[{a},1/2(-a+Sqrt[-4+a^2])]}
The functions can be called individually:
In[68]:= g[[1]][1]
Out[68]= 1/2 (-1-I Sqrt[3])
In[69]:= g[[2]][1]
Out[69]= 1/2 (-1+I Sqrt[3])
Or, all of the functions can be called at once to return all solutions:
In[70]:= Through[g[1]]
Out[70]= {1/2 (-1-I Sqrt[3]),1/2 (-1+I Sqrt[3])}
The function will fail if Solve
cannot find any solutions:
In[71]:= solutionFunctions[Log[x]==Sin[x],x]
During evaluation of In[71]:=
Solve::nsmet: This system cannot be solved with the methods available to Solve.
Out[71]= $Failed
Variables are automatically identified:
In[72]:= solutionFunctions[a x^2 + b x + c == 0, x]
Out[72]= { Function[{a, b, c}, (-b - Sqrt[b^2 - 4 a c])/(2 a)],
Function[{a, b, c}, (-b + Sqrt[b^2 - 4 a c])/(2 a)] }
回答4:
Here's the simplest way:
In[1]:= f = Solve[x^2 + ax + 1 == 0, x]
Out[1]= {{x -> -Sqrt[-1 - ax]}, {x -> Sqrt[-1 - ax]}}
In[2]:= g1[y_] := x /. f[[1]] /. a -> y
g2[y_] := x /. f[[2]] /. a -> y
In[4]:= g1[a]
g2[a]
Out[4]= -Sqrt[-1 - ax]
Out[5]= Sqrt[-1 - ax]
回答5:
This is really cool. Thanks. By converting Solve results into functions I could use Manipulate in a Plot. Something like
In[73]:= g = solutionFunctions[x^2 + a x + b == 0, x]
Out[73] = {Function[{a, b}, 1/2 (-a - Sqrt[a^2 - 4 b])],
Function[{a, b}, 1/2 (-a + Sqrt[a^2 - 4 b])]}
In[74]:= Manipulate[Plot[g[[1]][a, b], {a, 0, 4}], {{b, 1}, 0, 10}]
And you get a plot where you can manipulate parameter b
来源:https://stackoverflow.com/questions/5319771/how-to-create-a-function-directly-from-the-output-of-solve