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 April 18th, 2017, 05:56 AM #1 Member   Joined: Sep 2013 Posts: 83 Thanks: 0 Help to put Expression equal to zero I have this expression: $\displaystyle g = \sqrt {2\ln {{{\sigma ^2}} \over {{{2r} \over {\sqrt a }}x\log x{e^{ - a{{(r + (1/2){\sigma ^2})}^2}\tau /(2{\sigma ^2})}}}}}$ I want to put this expression equal to zero so it would be something like $\displaystyle 0 = {e^{{{{g^2}} \over 2}}} - { \ldots \over \ldots }$ $\displaystyle {e^{{{{g^2}} \over 2}}}$ i put just to remove squared and 2ln from the right side How should i express this correctly?
April 18th, 2017, 08:00 AM   #2
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Quote:
 Originally Posted by Ku5htr1m I have this expression: $\displaystyle g = \sqrt {2\ln {{{\sigma ^2}} \over {{{2r} \over {\sqrt a }}x\log x{e^{ - a{{(r + (1/2){\sigma ^2})}^2}\tau /(2{\sigma ^2})}}}}}$ I want to put this expression equal to zero so it would be something like $\displaystyle 0 = {e^{{{{g^2}} \over 2}}} - { \ldots \over \ldots }$
No, it would be $\displaystyle \sqrt {2\ln {{{\sigma ^2}} \over {{{2r} \over {\sqrt a }}x\log x{e^{ - a{{(r + (1/2){\sigma ^2})}^2}\tau /(2{\sigma ^2})}}}}}= 0$. I don't know where you got $\displaystyle e^{g^2/2}$

Quote:
 $\displaystyle {e^{{{{g^2}} \over 2}}}$ i put just to remove squared and 2ln from the right side How should i express this correctly?
So are you trying to "set g equal to 0", as you said, or are you trying to solve for r for any value of g?

 April 18th, 2017, 08:24 AM #3 Math Team   Joined: Jul 2011 From: Texas Posts: 2,751 Thanks: 1401 $g = \sqrt{2\ln\left(\dfrac{\sigma^2}{\text{big mess}}\right)}$ $g = 0 \implies \dfrac{\sigma^2}{\text{big mess}} = 1$
 April 18th, 2017, 08:36 AM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,421 Thanks: 832 Hmmm...your problem statement is: "Help to put Expression equal to zero" Sooo.... g = sqrt(bigmess) g^2 = bigmess g^2 - bigmess = 0 There..."equal to zero" What's a big mess Skeeter? The Folies Bergeres' dressing room?

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