43
$\begingroup$

$$f(x) = \log(\sqrt{x^2+1}+x)$$ 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 $-\log(\sqrt{x^2+1}+x)$.

$\endgroup$
1
  • 4
    $\begingroup$ $f(x) = \log(\sqrt{x^2+1}+x) = \text{ArcSinh}(x) $ $\endgroup$
    – S L
    Commented Dec 26, 2012 at 18:29

7 Answers 7

98
$\begingroup$

If$$f(x) = \log(\sqrt{x^2+1}+x)$$ then $$f(-x) = \log \left(\sqrt{(-x)^2+1}-x\right)=$$ $$= \log \left((\sqrt{x^2+1}-x)\cdot\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}\right)=$$ $$= \log \left(\frac{1}{\sqrt{x^2+1}+x}\right)=- \log({\sqrt{x^2+1}+x})=-f(x)$$

$\endgroup$
1
  • 2
    $\begingroup$ easy! nice answer +1 $\endgroup$ Commented Dec 27, 2012 at 7:10
91
$\begingroup$

Hint: $(\sqrt{x^2+1}+x)(\sqrt{x^2+1}-x) = 1$.

$\endgroup$
3
  • 5
    $\begingroup$ I am so surprised by your nice, simple complete answer Dan. :-) $\endgroup$
    – Mikasa
    Commented Dec 26, 2012 at 18:33
  • $\begingroup$ @BabakSorouh Wow, this is unexpected. Thank you very much! $\endgroup$
    – Dan Shved
    Commented Dec 26, 2012 at 18:39
  • 10
    $\begingroup$ that's why I love math $\endgroup$
    – nicolas
    Commented Dec 26, 2012 at 19:37
19
$\begingroup$

We have $f(-x)=\log \left(\frac{(\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x)}{(\sqrt{x^2+1}+x)}\right)=\log\left(\frac{x^2+1-x^2}{\sqrt{x^2+1}+x}\right)=-\log(\sqrt{x^2+1}+x)=-f(x)$.

$\endgroup$
2
  • 4
    $\begingroup$ 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
    Commented Feb 16, 2013 at 20:57
  • $\begingroup$ Thanks julien for your kind words. +100 $\endgroup$
    – Mikasa
    Commented Feb 19, 2013 at 15:23
8
$\begingroup$

Another thing to add is that the Taylor series (of odd functions, if it exists) has only odd powers

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

$\endgroup$
2
  • $\begingroup$ Odd answer for the odd function. ;-) $\endgroup$
    – Mikasa
    Commented Dec 26, 2012 at 18:40
  • 1
    $\begingroup$ @BabakSorouh: Offcourse, we are taking about odd functions. $\endgroup$ Commented Dec 26, 2012 at 18:51
5
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ +1 for your nice alternative approach, and for your kind words. $\endgroup$
    – Julien
    Commented Feb 19, 2013 at 15:31
  • $\begingroup$ I like! Helpful, too! $\endgroup$
    – amWhy
    Commented Mar 2, 2013 at 0:49
  • $\begingroup$ The link is not valid $\endgroup$
    – miracle173
    Commented Feb 3, 2017 at 9:57
4
$\begingroup$

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.

$\endgroup$
1
  • 4
    $\begingroup$ It is also crucial that the integral start at $0$ and not someplace else. $\endgroup$
    – whuber
    Commented Dec 26, 2012 at 20:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .