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Could someone please explain what the following would look like in standard math notation (or explain to me what this means):

InverseFunction[(
   z Log[1 - #1] + (1 - 2 z) Log[#1] + (-1 + z) Log[z - #1 + z #1])/(
   z (-1 + 2 z)) &][-(t/2) + (
  z Log[1 - a0] + Log[a0] - 2 z Log[a0] - Log[z - a0 + z a0] + 
   z Log[z - a0 + z a0])/(z (-1 + 2 z))]

I'm new to Mathematica so the use of the pure functions is a little confusing, especially since it's been combined with the inverse.

Also is there a way to get Mathematica to print this without using pure functions?

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closed as off-topic by user9660, MarcoB, Öskå, Bob Hanlon, ubpdqn Jun 21 '16 at 8:41

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – MarcoB, Öskå, Bob Hanlon, ubpdqn
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ "Also is there a way to get Mathematica to print this without using pure functions?" - not really. To interpret something like InverseFunction[Tan[#] &][x], it just says it is the inverse function of Tan[] applied to its argument x; that is, ArcTan[]. Since most functions do not admit an easy representation for the inverse, we sometimes cannot avoid InverseFunction[]. $\endgroup$ – J. M. will be back soon May 2 '15 at 15:39
  • $\begingroup$ @J.M. that;s a valid answer from my point of view :) $\endgroup$ – Kuba Dec 7 '15 at 13:55
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One way to write it is to let $$f(x) = \frac{z \log (1-x)+(1-2 z) \log (x)+(z-1) \log (x z-x+z)}{z (2 z-1)}$$ and write $f^{(-1)}\left(\frac{z \log (1-{a_0})-2 z \log ({a_0})+z \log ({a_0} z-{a_0}+z)-\log ({a_0} z-{a_0}+z)+\log ({a_0})}{z (2 z-1)}-\frac{t}{2}\right)$ for InverseFunction[..][..]. So in regular mathematical notation, the expression is done in two pieces (or sometimes more). In Mathematica, one needs a single expression (to compute with). I doubt there's a simpler way than InverseFunction[..][..].

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  • $\begingroup$ I doubt there's a simpler way to answer. +1 $\endgroup$ – user9660 Jun 19 '16 at 5:19

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