On Feb 4, 5:05 pm, Neal Becker <[email protected]> wrote: > On Wednesday 04 February 2009, Vinzent Steinberg wrote: > > > Please note that "log" means "ln" in sympy. > > > Solving by hand and using the natural logarithm I get > > > z = -ln(10**(-x) + 10**(-y)) / ln(10) > > > which is the same as your solution. > > > Furthermore > > > ln(1/(10**x) + 1/(10**y)) = ln((10**x + 10**y) / 10**(x + y)) = ln > > (10**x + 10**y) - ln(10**(x+y)) > > > so sympy's result is correct, although one could argue that it might > > be expressed simpler. > > That's what I meant, that it should be expressed simpler. I forgot, though, > that I meant log10 when I said log. > > > Vinzent > > > On Feb 4, 1:32 am, Neal Becker <[email protected]> wrote: > > > solve(Eq(1/(10**(-x) + 10**(-y)), 10**z), z) > > > Out[10]: > > > ⎡ ⎛ x y⎞ ⎛ x + y⎞ ⎤ > > > ⎢- log⎝10 + 10 ⎠ + log⎝10 ⎠ ⎥ > > > ⎢───────────────────────────────⎥ > > > ⎣ log(10) ⎦ > > > > That's not much of an answer. Should be: > > > > z = -log10(10**-x + 10**-y) > > > > What am I doing wrong? > >
What exactly would be simpler in your opinion? Not turning the sum 1/(10**x) + 1/(10**y) into the fraction (10**x + 10**y) / 10**(x + y)? Not expanding the log afterwards? Using log10(...) instead of log(...) / log(10)? I personally think it's better to have only one log. However, sympy's behavior could be more configurable. Vinzent --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "sympy" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/sympy?hl=en -~----------~----~----~----~------~----~------~--~---
