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

 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

Best Answer 

 #5
avatar+28188 
+10

Here's an alternative proof:

proof

 Oct 29, 2015
 #1
avatar+105634 
+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. ://

 Oct 29, 2015
 #2
avatar+6045 
+5

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

 

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

 

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

 

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

.
 Oct 29, 2015
 #3
avatar+105634 
0

Thanks Rom,

 

I should of thought of that I guess.    blush

 

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

 Oct 29, 2015
 #4
avatar+28188 
+5

I think the questioner knew how to solve it Rom's way (see the question title).  He/she was looking for a possible alternative method!

 Oct 29, 2015
 #5
avatar+28188 
+10
Best Answer

Here's an alternative proof:

proof

Alan Oct 29, 2015
 #6
avatar
0

Thank you guys alot...

Thank you Alan for your alternative answer, it helped me alot!   smiley

I'm new to this site so excuse my late reply

 Nov 2, 2015

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