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Fixed in Version 14.0


According to Exponent:

  • Exponent[expr, form] gives the maximum power with which form appears in the expanded form of expr.
  • Exponent[expr, form, h] applies h to the set of exponents with which form appears in expr.
  • The default taken for h is Max.

So Exponent[..., …, Max] is equivalent to Exponent[..., …]. But how to explain the following output?

In[1]:= Exponent[x, Sqrt[x]]

Out[1]= 2

In[2]:= Exponent[x, Sqrt[x], Max]

Out[2]= 1

In[3]:= Exponent[Sqrt[x], Sqrt[x], Defer]

              1
Out[3]= Defer[-]
              2

Isn't this a bug?

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  • $\begingroup$ It seems weird, indeed. For example, you can use List as the 3rd argument to get all the exponents. But Exponent[x, Sqrt[x]] gives 2 and Exponent[x, Sqrt[x], List] gives {1}. $\endgroup$
    – Domen
    Sep 12, 2023 at 7:40
  • $\begingroup$ The documentation says under Possible Issues that Exponent is purely syntactical and does no analysis. $\endgroup$
    – Roland F
    Sep 12, 2023 at 8:51
  • $\begingroup$ @RolandF If so, shouldn't Exponent[x, Sqrt[x]] return 0 (instead of 2)? And shouldn't Exponent[Sqrt[x], Sqrt[x], Identity] return 1 (rather than 1/2)? $\endgroup$
    – user688486
    Sep 12, 2023 at 10:21
  • $\begingroup$ By experiment, Exponent[x^a, x^b] works reasonably with numbers as exponents of symbols, but not with exponent symbols. Even specifying everything] as positive integers or expecting simplifications as in Exponent[(1-x^2)/(1+x),x] returns input. Using a non-symbol expression as variable in polynomial reduction may be termed 'abuse of powers'. $\endgroup$
    – Roland F
    Sep 12, 2023 at 10:58

1 Answer 1

1
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Fixed in Version 14.0, now Exponent[Sqrt[x], Sqrt[x], Defer] returns 1/2

enter image description here

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