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Thread: The Holy Grail Of Mathematics.

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11-14-2008

#54 (permalink)

Qfwfq

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Re: The Holy Grail Of Mathematics.

Quote:

Originally Posted by modest

$\frac{\left[\left(sin(x^{(1/2)})\right)-1\right]^{-1}}{\pi^2+\left[\ln\left(\ln\left(2\left[\left[2(x^{-1}+1)\right]^{-1}+1\right]\right)\right)\right]^{-1}}-1=0$

Wouldn't it be simpler to write as follows? (I don't know what's gone wrong with the width here)

$(\sin x^{\frac{1}{2}}-1)\left(\pi^2+\frac{1}{\ln{\ln((x^{-1}+1)^{-1}+2)}}\right)=1$

$(\sin x^{\frac{1}{2}}-1)\left(\pi^2+\frac{1}{\ln{\ln(\frac{x}{x+1}+2)}}\right)=1$

Done without pencil & paper, over many edits,

but should be alright now. Now that ought to be simpler to take the derivative of.

----------------
Inutil insegnà al mus, si piart timp, in plui si infastidìs la bestie.

Hypography Forum PITA...... er, Administrator.

Last edited by Qfwfq; 11-14-2008 at 03:33 AM.. Reason: just hackin' around

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