Prove that f(x)=ln(x + (x^2+1)^0.5 ) is an odd function... :/

Guest Oct 28, 2015

edited by
Guest
Oct 28, 2015

edited by Guest Oct 28, 2015

edited by Guest Oct 28, 2015

edited by Guest Oct 28, 2015

edited by Guest Oct 28, 2015

edited by Guest Oct 28, 2015

edited by Guest Oct 28, 2015

#1**+5 **

Prove that f(x)=ln(x + (x^2+1)^0.5 ) is an odd function... :/

If this is an odd function then

f(-x)=-f(x)

and since x^2= (-x)^2

that means you are suppose to prove that

\(ln(-x+\sqrt{x^2+1})=-ln(x+\sqrt{x^2+1})\)

Now

\(ln(x+\sqrt{x^2+1})\) \(=sinh^{-1}(x)\)

is the inverse hyperbolic sine function so it is definitely odd. But I do not know how to prove this.

I shall flag other mathematicians to take a look for you. You should get an answer but it meight take a day or so. ://

Melody Oct 29, 2015

#2**+5 **

\(-\ln(x+\sqrt{x^2+1}) = \ln\left(\dfrac{1}{x+\sqrt{x^2+1}}\right) =\)

\(\ln\left(\dfrac{1}{x+\sqrt{x^2+1}} \dfrac{x-\sqrt{x^2+1}}{x-\sqrt{x^2+1}}\right) = \ln\left( \dfrac{x-\sqrt{x^2+1}}{x^2-(x^2+1)}\right)=\)

\(\ln\left(-(x-\sqrt{x^2+1})\right) =\ln\left(-x + \sqrt{x^2+1}\right)\)

\(\mbox{so }f(-x)=-f(x) \mbox{ and f is odd.}\)

.Rom Oct 29, 2015

#3**0 **

Thanks Rom,

I should of thought of that I guess.

I suppose things often look easy when someone shows you how to do it.

Melody Oct 29, 2015