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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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    $f(x) = \log(\sqrt{x^2+1}+x) = \text{ArcSinh}(x) $2012-12-26

7 Answers 7

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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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    easy! nice answer +12012-12-27
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Hint: $(\sqrt{x^2+1}+x)(\sqrt{x^2+1}-x) = 1$.

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    that's why I love $m$ath2012-12-26
18

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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    Thanks julien for your kind words. +1002013-02-19
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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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    @BabakSorouh: Offcourse, we are taking about odd functions.2012-12-26
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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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    The link is not valid2017-02-03
5

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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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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    It is also crucial that the integral start at $0$ and not someplace else.2012-12-26