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$$f(x) = \log(\sqrt{x^2+1}+x)$$ I can't figure out, why this function is odd. I mean, of course, its graph shows, it's odd, but when I investigated $f(-x)$, I couldn't find way to $-\log(\sqrt{x^2+1}+x)$.

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    $\begingroup$ $f(x) = \log(\sqrt{x^2+1}+x) = \text{ArcSinh}(x) $ $\endgroup$ – Santosh Linkha Dec 26 '12 at 18:29
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If$$f(x) = \log(\sqrt{x^2+1}+x)$$ then $$f(-x) = \log \left(\sqrt{(-x)^2+1}-x\right)=$$ $$= \log \left((\sqrt{x^2+1}-x)\cdot\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}\right)=$$ $$= \log \left(\frac{1}{\sqrt{x^2+1}+x}\right)=- \log({\sqrt{x^2+1}+x})=-f(x)$$

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    $\begingroup$ easy! nice answer +1 $\endgroup$ – Grijesh Chauhan Dec 27 '12 at 7:10
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Hint: $(\sqrt{x^2+1}+x)(\sqrt{x^2+1}-x) = 1$.

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    $\begingroup$ I am so surprised by your nice, simple complete answer Dan. :-) $\endgroup$ – mrs Dec 26 '12 at 18:33
  • $\begingroup$ @BabakSorouh Wow, this is unexpected. Thank you very much! $\endgroup$ – Dan Shved Dec 26 '12 at 18:39
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    $\begingroup$ that's why I love math $\endgroup$ – nicolas Dec 26 '12 at 19:37
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We have $f(-x)=\log \left(\frac{(\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x)}{(\sqrt{x^2+1}+x)}\right)=\log\left(\frac{x^2+1-x^2}{\sqrt{x^2+1}+x}\right)=-\log(\sqrt{x^2+1}+x)=-f(x)$.

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    $\begingroup$ Interestingly, I had exactly the same answer as Adi Dani one minute before, and I get 9 upvotes while he/she gets 76... $\endgroup$ – Julien Feb 16 '13 at 20:57
  • $\begingroup$ Thanks julien for your kind words. +100 $\endgroup$ – mrs Feb 19 '13 at 15:23
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Another thing to add is that the Taylor series (of odd functions, if it exists) has only odd powers

$$ x-{\frac {1}{6}}{x}^{3}+{\frac {3}{40}}{x}^{5}+\dots.$$

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  • $\begingroup$ Odd answer for the odd function. ;-) $\endgroup$ – mrs Dec 26 '12 at 18:40
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    $\begingroup$ @BabakSorouh: Offcourse, we are taking about odd functions. $\endgroup$ – Mhenni Benghorbal Dec 26 '12 at 18:51
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To add another hint to Dan's answer, consider that $-\log a = \log \frac1a$ and then simplify the radical out of the denominator for your function.

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This function is another command for showing $f(x)=\text{arcsinh}(x)$ being increasing continous in $[0,\infty]$. We know that $f(x)=\sinh(x)$ is an odd one-one function. $$f(x)=\text{arcsinh(x)}\to \sinh(f(x))=x$$ so if $x\to -x$ then $$\sinh(f(-x))=-x\longrightarrow -\sinh(f(-x))=x\longrightarrow\sinh(-f(-x))=x=\sinh(f(x))$$ so $f(-x)=-f(x)$. This means $f(x)$ is an odd function. There is another different approach for this. See this link http://ddmf.msr-inria.inria.fr/1.8/ddmf

enter image description here

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  • $\begingroup$ +1 for your nice alternative approach, and for your kind words. $\endgroup$ – Julien Feb 19 '13 at 15:31
  • $\begingroup$ I like! Helpful, too! $\endgroup$ – amWhy Mar 2 '13 at 0:49
  • $\begingroup$ The link is not valid $\endgroup$ – miracle173 Feb 3 '17 at 9:57
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Note by differentiating that $$f(x)=\int_0^x \frac{dt}{\sqrt{t^2+1}}.$$ The result now follows from the fact that $\dfrac{1}{\sqrt{t^2+1}}$ is even.

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    $\begingroup$ It is also crucial that the integral start at $0$ and not someplace else. $\endgroup$ – whuber Dec 26 '12 at 20:53

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