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I am using Mathematica 10, FunctionPeriod which is supposed to return period of a given function returns 0 if the input trigonometric function have a Real phase

In[1]:= FunctionPeriod[Cos[t+1.2], t]
Out[1]= 0
In[2]:= FunctionPeriod[Cos[t+1], t]
Out[2]= 2 π
In[3]:= FunctionPeriod[Cos[t+1.0], t]
Out[3]= 0

Why the period is zero if the phase is a real number instead of 2 π and why Cos[t+1] and Cos[t+1.0] are not having the same period

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  • $\begingroup$ I suppose the reason is related to the result of Cos[t+1.2+2 Pi] == Cos[t+1.2] $\endgroup$
    – LLlAMnYP
    Commented Oct 4, 2016 at 13:57
  • $\begingroup$ What is the question? $\endgroup$
    – QuantumDot
    Commented Oct 4, 2016 at 14:03
  • $\begingroup$ @QuantumDot I'd guess that the implicit question is either "how do I get the expected behavior" i.e. a result of 2 Pi or "why am I not getting the expected behavior". $\endgroup$
    – LLlAMnYP
    Commented Oct 4, 2016 at 14:05
  • 2
    $\begingroup$ See what happens when you evaluate Cos[1.] - Cos[1. + 2 Pi]. $\endgroup$
    – John Doty
    Commented Oct 4, 2016 at 14:20
  • 1
    $\begingroup$ Yeah, I think @JohnDoty has the right of it. Further evidence: FunctionPeriod[Rationalize@Cos[t + 1.2], t] yields 2 π. $\endgroup$
    – march
    Commented Oct 4, 2016 at 15:55

1 Answer 1

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These problems are fixed in V 14.1 (or earlier)

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