I have been trying to solve a series of equations that I know have real positive values for the unknowns (they all have real world analogues and they are not linked to a frequency domain) but despite asking for simplify to only give me real and positive values and stipulating some non-zero values I am still being giving some values that tend toward negative values.

(The Code option seems to be having issues with me using it too so if someone code help with that too, it would be great!) I have attached a link to a screenshot.

screenshot of the code

  • $\begingroup$ Could you post the code as raw text? $\endgroup$ – Feyre Jul 2 '15 at 12:02
  • $\begingroup$ Just copy the Mathematica text (copy as input text from the Edit menu) or convert the cell to InputForm and copy, then paste it here. Try to edit it as good as possible using the Markdown tools and some community member here with the right tools will do the Greek symbols for you. $\endgroup$ – Sjoerd C. de Vries Jul 2 '15 at 12:06
  • $\begingroup$ See also how-to-copy-code-from-mathematica-so-it-looks-good-on-this-site. $\endgroup$ – Sjoerd C. de Vries Jul 2 '15 at 12:20


Simplify[..., Assumptions -> { rho > 0, r > 0 ,....}]


Assuming[ {rho >0, r > 0, ...}, Simplify[...] ]


Simplify[..., rho > 0, r > 0, ...]
  • $\begingroup$ That, and the OP should also not use ...&&Reals&&Positive. $\endgroup$ – Sjoerd C. de Vries Jul 2 '15 at 12:10
  • $\begingroup$ I implement real condition ( x is real ) using : Assumptions -> { x^2 > 0 } $\endgroup$ – Quasar Supernova Jul 2 '15 at 12:27
  • $\begingroup$ That's not the way to do it, use Element[x, Reals] instead $\endgroup$ – Sjoerd C. de Vries Jul 2 '15 at 12:30
  • $\begingroup$ There is nothing wrong with the syntax Simplify[expression, {rho > 0, r > 0}] but note that the assumptions are grouped in a List rather than given as third arguments and beyond. $\endgroup$ – Mr.Wizard Jul 2 '15 at 12:30
  • $\begingroup$ I've tried using Assuming and Simplify but it still gives me negatives i.e. '{R -> (kK)/Subscript[r, u], Subscript[r, b] -> -Subscript[r, u], Subscript[r, c] -> Subscript[r, u], Subscript[V, ut] -> (k^2*Subscript[V, f]*Subscript[[Eta], ut]^2 Subscript[[Rho], u] + 4*Subscript[r, u]^2*(-k^2 + (-1 + k^2)*Subscript[V, f]* Subscript[[Rho], u] - (-1 + k^2)*Subscript[V, u]* Subscript[[Rho], u]))/ (k^2*Subscript[[Eta], ut]^2*Subscript[[Rho], u]), Subscript[[CurlyEpsilon], r] -> -1, Subscript[[Eta], u] -> (2*Subscript[r, u])/k}' $\endgroup$ – Philip Brown Jul 3 '15 at 9:45

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