- #1

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__http://www.wolframalpha.com/input/?i=x^pi__

same goes for x^e

__http://www.wolframalpha.com/input/?i=x^e__

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- Thread starter cstvlr
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- #1

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same goes for x^e

- #2

Mentallic

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And no it doesn't make pi an odd number.

- #3

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- #4

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why is that true?Those functions are real for [tex]x\geq 0[/tex] only. If you take a look at that interval for both odd and even functions, you'll realize that you can't really tell a difference between their shape.

And no it doesn't make pi an odd number.

- #5

Mentallic

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why is that true?

Because if we take some positive number x, then [tex](-x)^{\pi}=(-1)^{\pi}\cdot x^{\pi}[/tex]

Since x is positive, [tex]x^{\pi}>0[/tex] so we just have to deal with the [tex](-1)^{\pi}[/tex] factor. It is complex, but if you want a proof of this, simply convert it into its complex form:

[tex]e^{i\pi}=-1[/tex] therefore [tex](-1)^{\pi}=e^{i\pi ^2}=cos(\pi ^2)+isin(\pi ^2)[/tex] so if it is to be a real number, then the sin of [tex]\pi ^2[/tex] needs to be equal to 0, but this isn't the case.

In fact we can take any power [tex]x^{\alpha}[/tex], and deduce the circumstances whether it will be real or complex for negative values of x by following a similar process. But be wary, it is a little more complicated dealing with all rational values.

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