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The expression $\sqrt{L^2 M+\sqrt{L^2 (-J^2 + L^2 M^2)}}/\sqrt{2}$,can be simplified into $\sqrt{L}(\sqrt{LM + J}+\sqrt{LM - J})/2$, but how can I obtain this result with Mathematica?

That is, how can I simplify

Sqrt[L^2 M + Sqrt[L^2 (-J^2 + L^2 M^2)]]/Sqrt[2]

into

Sqrt[L]/2 (Sqrt[L M + J] + Sqrt[L M - J])
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  • $\begingroup$ Have You tried Simplify? $\endgroup$ – Wojciech Nov 13 '13 at 9:34
  • $\begingroup$ Please correct your expressions clarify the layout. Note: LM is not equivalent to L M. $\endgroup$ – Yves Klett Nov 13 '13 at 10:13
  • $\begingroup$ sometimes the best you can do is get mathematica verify that your expression is the same, have you done that? Show that code... $\endgroup$ – george2079 Nov 13 '13 at 13:11
  • $\begingroup$ Equal[{Sqrt[L^2 M + Sqrt[L^2 (-J^2 + L^2 M^2)]]/Sqrt[2], Sqrt[L]/2 (Sqrt[L M + J] + Sqrt[L M - J])}] returns true, but I'm not sure what are we trying to achieve - to verify the statement or to get Mathematica to return the simplified form? $\endgroup$ – Wojciech Nov 13 '13 at 13:23
  • $\begingroup$ @WojciechSitkiewicz that is incorrect syntax for equal. The expressions are in fact only equal under certain conditions (eq J=1,M=1,L=-1) they are not equal. $\endgroup$ – george2079 Nov 13 '13 at 14:56
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Are you sure that the first expression can be simplified to the second?

o1 = Sqrt[L^2 M + Sqrt[L^2 (-J^2 + L^2 M^2)]]/Sqrt[2];
o2 = Sqrt[L]/2 (Sqrt[L M + J] + Sqrt[L M - J]);

and

ru = {L -> 5, M -> 2, J -> 3.};

Then just

o1 /. ru
o2 /. ru
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  • $\begingroup$ Sorry, it is " plus " in the first equation. $\endgroup$ – user10537 Nov 13 '13 at 9:43
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    $\begingroup$ Given that the equation was incorrect I suggest you delete your answer, as it actually doesn't answer the question. $\endgroup$ – Sjoerd C. de Vries Nov 13 '13 at 13:12
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By squaring both expressions and making some assumptions we can show them equal..

     Simplify[
            (Sqrt[L^2 M + Sqrt[L^2 (-J^2 + L^2 M^2)]]/Sqrt[2] )^2 
              ==
            ( Sqrt[L]/2 (Sqrt[L M + J] + Sqrt[L M - J]))^2  , 
       Assumptions -> { L > 0, M > Abs[J/L]}] 

     (* True *)

As near as I can tell under those assumptions this equality should still hold without squaring the expressions, but Simplify fails to see that for some reason.

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