I tried it like this:
myEvenFunction[x_] := x^2
Equal[myEvenFunction[x],myEvenFunction[-x]]
Out = x^2
Out = True
myOddFunction[x_] := x^3
Equal[myOddFunction[x], myOddFunction[-x]]
Out = x^3
Out = x^3 == -x^3
Shouldn't it say false here?
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3 Answers
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Rather than imposing x>0 one can also do
FullSimplify[ ForAll[x, myOddFunction[x] == myOddFunction[-x]]]
which yields False.
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2$\begingroup$ Any thoughts about using
Resolvein place ofFullSimplifyhere? (p.s. I am out of votes for the day or I would upvote.) $\endgroup$ May 22, 2015 at 22:20 -
$\begingroup$ i like this answer the most. intuitive, simple.. Thanks :) $\endgroup$– balooMay 22, 2015 at 22:20
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$\begingroup$ *intuitional. Actually i had this in mind, but couldn't transform it into a piece of code. Thanks, again. ^^ I'll try it with Resolve too. $\endgroup$– balooMay 22, 2015 at 22:28
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1$\begingroup$ @Mr.Wizard I don't have any thoughts about
Resolvebecause I didn't know about it until your comment! So IDK :-P Thanks for the accept, @sudo_math! Welcome to Mma.SE! $\endgroup$– evanbMay 22, 2015 at 22:54 -
$\begingroup$ Okay. I think it may be faster but less robust than
FullSimplify. With what I do I useForAllso rarely that I don't have much experience withResolve. Let me know what you find out if you remember, OK? $\endgroup$ May 22, 2015 at 22:56
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evenFQ[f_] := Simplify[f[t] - f[-t]] === 0
oddFQ[f_] := Simplify[f[t] + f[-t]] === 0
Examples:
ef[x_] := x^2
of[x_] := x^3
evenFQ/@ {ef, of}
{True, False}
oddFQ/@ {ef, of}
{False, True}
evenFQ /@ {# &, Im, Sin, Tan, Sinh, Erf}
{False, False, False, False, False, False}
oddFQ /@ {# &, Im, Sin, Tan, Sinh, Erf}
{ True, True, True, True, True, True}
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$\begingroup$ I'm concerned about the assumption that the arguments are real.
evenFQ[Im]andoddFQ[Im]both give true, for example, whileImis really odd. $\endgroup$– evanbMay 22, 2015 at 22:05 -
$\begingroup$ @evanb, thank you; it looks like we don't need the assumption. $\endgroup$– kglrMay 22, 2015 at 22:20
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$\begingroup$ @kglr Hi,
g[x_]:=Sin[x]/(x (x^2-1)); evenFQ[g] (*False*)Why? $\endgroup$ Aug 29, 2023 at 2:26 -
$\begingroup$ @lotus2019,
ClearAll[g]; g[x_] := Sin[x]/(x (x^2 - 1));evenFQ[g]givesTrue(version 13..3.0 - Linux x86 -64bit) $\endgroup$– kglrAug 29, 2023 at 7:05 -
1$\begingroup$ @kglr Thank you, it has been successful now. It may be because I did not use
Clearbefore. $\endgroup$ Aug 29, 2023 at 7:44
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You need Simplify with an assumption:
myOddFunction[x_] := x^3;
Simplify[
Equal[myOddFunction[x], myOddFunction[-x]],
x > 0
]
False
Refine also works in this case, again with the appropriate assumption:
Refine[Equal[myOddFunction[x], myOddFunction[-x]], x > 0]
False
answered May 22, 2015 at 20:55
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$\begingroup$ Please be careful with regards to functions of complex variables. $\endgroup$– evanbMay 22, 2015 at 22:05
Blockin some cases to make it safe; see: (6664). If you have frequent need of this kind of protected evaluation see: (1992) $\endgroup$f[x_] = 5 + Interpolation[{1, 2, 3, 5, 8, 5}][x]. In this case you don't want to execute the expensiveInterpolationfunction every time you call the function. This is just a toy example, I can't come up with a real world scenario right now but I seem to remember that I've seen some. $\endgroup$localSetfrom (1992) makes such uses worry-free; I hope you'll consider using it if you do that often. $\endgroup$