Wolfram Alpha says that the domain of $$f(x) = \large \ln(1+x)^{\sin(x)}$$
is $x \gt 0$. I can't figure out how we come up with this restrictions. Please help.
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If Wolfram Alpha told you that, it is wrong. – Git Gud Jan 21 '13 at 18:43
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How come you say that? Is there something I should know about?! – George F-bot Jan 21 '13 at 18:45
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2By $\ln^{\sin(x)}(1+x)$, do you mean $$\big(;\ln(1+x);\big)^{\sin(x)}\qquad ?$$ – Zev Chonoles Jan 21 '13 at 18:48
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@GeorgeF-bot Nevermind, apparently I misread the function. I can't even understand what function it is. – Git Gud Jan 21 '13 at 18:49
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@ZevChonoles yes. Any clues? – George F-bot Jan 21 '13 at 18:51
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@George F-bot: I am confused what you mean, see WA Plot. – Amzoti Jan 21 '13 at 18:53
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@GeorgeF-bot: It is incorrect to write powers that way. Even though it is a convention for trigonometric functions, it is a terrible convention (see the confusion it can cause here), and at any rate does not apply to any other functions whatsoever. – Zev Chonoles Jan 21 '13 at 18:55
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@GeorgeF-bot: Are you considering $\mathbb R$ or $\mathbb C$ domains? Regards – Amzoti Jan 21 '13 at 19:02
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@Amzoti, can you please address him by "GeorgeF-bot"? Otherwise, I get the notification instead. :) – George V. Williams Jan 21 '13 at 19:04
1 Answers
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Assuming we agree on having positive bases for exponential functions, it thus must be
$$\log(1+x)>0\Longleftrightarrow 1+x>1\Longleftrightarrow x>0\,$$
just as WA said...
DonAntonio
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What about when $x = -\pi$? Technically, $f(-\pi) = 1$ (because $(a+bi)^0 = 1$). – George V. Williams Jan 21 '13 at 18:53
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2If $x = -\pi$, then $1+x = 1-\pi < 0$, so its logarithm is not even defined. – Javier Jan 21 '13 at 18:55
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1No, then its logarithm is complex ($\ln(1 - \pi) \approx 0.7615 + i\pi$), and $\sin(-\pi) = 0$, and $(0.7615 + i\pi)^0 = 1$. Do we deal with this case when discussing domains? – George V. Williams Jan 21 '13 at 18:57
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That's a stretch, @GeorgeV.Williams: unless specifically asked by the OP, we always assume the "usual" functions are real ones. In particular here, the usual notation for the complex logarithm is "Log" and not "log" or even less, as the OP wrote, "ln". – DonAntonio Jan 21 '13 at 18:59
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