2
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By manual manipulation, it's not too difficult to show that the value of the following expression is exactly $1.$ How to convince Mathematica to do the simplification?

2/3 - ProductLog[-(Log[2]/(3 2^(1/3)))]/Log[2]
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4
  • $\begingroup$ The crux of the problem is that -(Log[2]/(3 2^(1/3))) is not easily recognized to be the same as -(Log[2]/3) Exp[-(Log[2]/3)]. $\endgroup$
    – J. M.'s torpor
    Sep 10 '20 at 11:23
  • $\begingroup$ @J.M. the expression is not created by me, but (more or less) the output of Solve[2^(n - 1) == 3 n - 2] $\endgroup$
    – enzotib
    Sep 10 '20 at 11:30
  • $\begingroup$ @J.M. even with your substitution, the simplification fails to give $1$ $\endgroup$
    – enzotib
    Sep 10 '20 at 11:46
  • $\begingroup$ I was explaining why it was failing. ProductLog[-1/8 Exp[-1/8]] evaluates as expected, but Mathematica is not sufficiently smart to figure out that a similar operation applies for ProductLog[-(Log[2]/(3 2^(1/3)))]. Consider sending Support a message. $\endgroup$
    – J. M.'s torpor
    Sep 10 '20 at 18:21

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