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April 18th, 2017, 05:56 AM   #1
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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?
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April 18th, 2017, 08:00 AM   #2
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Quote:
Originally Posted by Ku5htr1m View Post
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?
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April 18th, 2017, 08:24 AM   #3
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$g = \sqrt{2\ln\left(\dfrac{\sigma^2}{\text{big mess}}\right)}$

$g = 0 \implies \dfrac{\sigma^2}{\text{big mess}} = 1$
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April 18th, 2017, 08:36 AM   #4
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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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