Re: [math-fun] transcendental puzzles
As a teen, I made up the PUZZLE: Given that x^2 + log_10 (2x) = 3, solve for x > 0 in closed form (which I think I mentioned on math-fun a long time ago). Although it's not hard to solve for x explicitly, it seems to require human intuition rather than any algorithm. How would computer algebra systems handle this??? Since I don't have one handy, would someone out there be willing to use Macsyma, Maple, or Mathematica and ask it for the solution to this equation (without giving the software any prompts other than that x is real) ? I'm very curious to know what happens. --Dan
For eq:=x^2 + a*log(2*x) = 3 Maple gives x = 1/2*exp(-1/2*(a*LambertW(1/2/a*exp(6*1/a))-6)/a) Emeric
Mathematica vs Dan Asimov:
PUZZLE: Given that x^2 + log_10 (2x) = 3, solve for x > 0 in closed form
In[1]:= Solve[ x^2 + Log[10,2x] == 3, x ] InverseFunction::ifun: Inverse functions are being used. Values may be lost for multivalued inverses. Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found. ProductLog[500000 (Log[2] + Log[5])] Out[1]= {{x -> Sqrt[------------------------------------]}} 2 (Log[2] + Log[5]) Hmm, let's see if it can do any better than that... In[2]:= Simplify[%] 5 Log[10] Out[2]= {{x -> Sqrt[---------]}} Log[100] Aah, much better. I have no idea why it fails to recognize Log[10]/Log[100]. I'm not a Mma expert; there's probably some argument you can pass to Simplify to remind it to think about some extra transformation rules... --Michael Kleber kleber@brandeis.edu
FullSimplify[Solve[x^2 + Log[10, 2*x] == 3, x]] Gives x->Sqrt[5/2] in Mathematica Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.qxmail.com/home.htm ----- Original Message ----- From: "Michael Kleber" <kleber@brandeis.edu> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Friday, April 04, 2003 4:59 PM Subject: Re: [math-fun] transcendental puzzles
Mathematica vs Dan Asimov:
PUZZLE: Given that x^2 + log_10 (2x) = 3, solve for x > 0 in closed form
In[1]:= Solve[ x^2 + Log[10,2x] == 3, x ]
InverseFunction::ifun: Inverse functions are being used. Values may be lost for multivalued inverses.
Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found.
ProductLog[500000 (Log[2] + Log[5])] Out[1]= {{x -> Sqrt[------------------------------------]}} 2 (Log[2] + Log[5])
Hmm, let's see if it can do any better than that...
In[2]:= Simplify[%]
5 Log[10] Out[2]= {{x -> Sqrt[---------]}} Log[100]
Aah, much better. I have no idea why it fails to recognize Log[10]/Log[100]. I'm not a Mma expert; there's probably some argument you can pass to Simplify to remind it to think about some extra transformation rules...
--Michael Kleber kleber@brandeis.edu
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participants (4)
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asimovd＠aol.com -
Emeric Deutsch -
Michael Kleber -
Thane Plambeck