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[edit] mathematica

In[13]:= Simplify[x^2==9,x>0]

Out[13]= x^2 == 9

In mathematica 5.2 , why does mathematica not output the answer " x == 3 "

211.28.238.164 05:01, 12 March 2007 (UTC)

You misstate the request. What you want is

Reduce[x^2 == 9 && x > 0]

This does produce

x == 3

as you wish. --KSmrq^T 05:36, 12 March 2007 (UTC)

As to why, this is because Simplify simplifies sets of equations, not inequations. Reduce simplifies sets of inequations and equations.

That's not really . Simplify works with inequalities just fine (try Simplify[2x>2]) but only uses a very limited set of tools. Also, it doesn't actually simplify a set of equations - it simplifies a single expression (which could be an equation) under a set of assumptions. FullSimplify is slower but gives better results - for example, FullSimplify[x^2==9, x>0] works. -- Meni Rosenfeld (talk) 20:02, 12 March 2007 (UTC)

Mathematica is a large complicated beast, with many ways to approach the same problem. Simplification is known to be a tough problem for several reasons. One is that we have no universal definition of "simpler"; another is computational complexity. Both Simplify and FullSimplify use sets of heuristics — guesses at potentially helpful transformations — hoping to get somewhere. By contrast, Reduce uses the idea of cylindrical algebraic decomposition, which can be horrendously slow but effectively thorough. For example,

• Reduce[x^2 == 9]

produces

• x == -3 || x == 3

which not even FullSimplify will do. Another tool is Solve, and indeed

• Solve[x^2 == 9]

produces

• {{x -> -3}, {x -> 3}}

much like Reduce; however, we have no way to insist on the positive solution. Can we conclude that Reduce is the best tool? No; for consider that

• Reduce[z == Cos[x]^2 + Sin[x]^2]

does nothing, producing

• z == Cos[x]^2 + Sin[x]^2

in contrast to

• Simplify[z == Cos[x]^2 + Sin[x]^2]

which produces

• z == 1

as we would hope. Also, we may sometimes settle for FindInstance. Note: Obviously we are exploring trivial examples, as practice for the real challenges. --KSmrq^T 23:40, 12 March 2007 (UTC)

Is there a way to say

• Solve[x^2 == 9] // SelectPositive

produces

• {{x -> 3}}

202.168.50.40 01:36, 13 March 2007 (UTC)

Use Select with a suitable anonymous function, then apply it to the solution list. —The preceding unsigned comment was added by 129.78.64.102 (talk) 02:39, 13 March 2007 (UTC).

Create a function called SelectPositive and then use it to filter the result of Solve

In[1]:= SelectPositive[soln_]:= Select[ soln, Part[FullForm[#], 1, 1, 2] ≥ 0 & ]

In[2]:= Solve[x^2 == 9, x] // SelectPositive

Out[2]:= {{x->3}}

211.28.123.23 06:19, 13 March 2007 (UTC)

This one is slightly better because it can handle complex numbers

In[3]:= SelectPositive2[soln_] := Select[ soln, Function[X, Module[{val}, val = Part[FullForm[X], 1, 1, 2]; Abs[val] == val ]]]

In[4]:= Solve[x^2 == 16, x] // SelectPositive2

Out[4]:= {{x->4}}

211.28.123.23 07:25, 13 March 2007 (UTC)

You can't consistently assign an ordering to the complexes.

You dont need to assign an ordering, merely get a True / False response

In[1]:= Abs[ 3+4I ] == ( 3+4I ) (* Abs[ 3 + 4i ] == 3 + 4i *)

Out[1]= False

In[2]:= Abs[-5] == -5

Out[2]= False

In[3]:= Abs[3] == 3

Out[3]= True

In[4]:= Abs[0] == 0

Out[4]= True

Call me old-fashioned, but I'm uncomfortable with zero being considered positive. Now might be the time to break out Element[x,Reals]. Also, this Select approach may force us to work finding many roots that we then discard; for some problems, not such a good idea. --KSmrq^T 23:38, 13 March 2007 (UTC)

If you prefer you can always change it to

Abs[val] == val && Abs[val] > 0

and filter out zero as well.

211.28.123.23 00:15, 14 March 2007 (UTC)

Still, you need to have defined some sense of how a complex number is "positive" or not, and there are a few ways of doing this.