Where did the other root go?

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Rabeeb
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Where did the other root go?

Unread post by Rabeeb » Mon Jan 26, 2015 2:39 pm

Let's solve the equation: $\frac{e^{x}+e^{-x}}{2}=1$
$\Rightarrow e^{x}+e^{-x}=2$
$\Rightarrow e^{2x}-2e^{x}+1=0$
$\Rightarrow e^{x}=1\pm \sqrt{2}$
But $e^{x }$ cannot be negative, so $x= ln (1+ \sqrt {2}) \approx 0.881$
But from the equation, it is clearly observable that $-0.881$ is an approx root too, which we did not obtain in the normal process. Where did it go?

SMMamun
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Joined: Thu Jan 20, 2011 6:57 pm

Re: Where did the other root go?

Unread post by SMMamun » Mon Jan 26, 2015 4:09 pm

It didn't go anywhere. You have made an error in the root extraction formula for the quadratic form of equations. Please check again. :)

Nirjhor
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Re: Where did the other root go?

Unread post by Nirjhor » Mon Jan 26, 2015 9:29 pm

Notice that $e^{2x}-2e^x+1=\left(e^x-1\right)^2=0\Rightarrow e^x=1$ so $x=0$ is the only solution.

Another way to observe is that the left side is the definition of hyperbolic cosine. So $\cosh x=1$ gives $x=2\pi i n$ for all $n\in\mathbb{Z}$. Only real solution is $x=0$ for $n=0$.
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


Revive the IMO marathon.

Rabeeb
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Joined: Tue Dec 14, 2010 7:52 pm
Location: Mymensingh, Bangladesh
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Re: Where did the other root go?

Unread post by Rabeeb » Thu Jan 29, 2015 8:06 pm

Sorry, i made a mistake.. :mrgreen: Correcting it below:
$\frac{e^{x}-e^{-x}}{2}=1$
$\Rightarrow e^{x}-e^{-x}=2$
$\Rightarrow e^{2x}-2e^{x}-1=0$
$\Rightarrow e^{x}=1\pm \sqrt{2}$
The rest part follows now...

Nirjhor
Posts: 136
Joined: Thu Aug 29, 2013 11:21 pm
Location: Varies.

Re: Where did the other root go?

Unread post by Nirjhor » Fri Jan 30, 2015 2:00 am

Then how is $-\ln\left(1+\sqrt 2\right)$ another root?
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


Revive the IMO marathon.

Rabeeb
Posts: 25
Joined: Tue Dec 14, 2010 7:52 pm
Location: Mymensingh, Bangladesh
Contact:

Re: Where did the other root go?

Unread post by Rabeeb » Mon Feb 02, 2015 11:57 pm

Nirjhor wrote:Then how is $-\ln\left(1+\sqrt 2\right)$ another root?
Replace x by -x in the equation, and -x by x. Do u see any change?

Nirjhor
Posts: 136
Joined: Thu Aug 29, 2013 11:21 pm
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Re: Where did the other root go?

Unread post by Nirjhor » Wed Feb 04, 2015 12:52 pm

Yes I do. Then $\dfrac{e^x-e^{-x}}{2}$ changes to $\dfrac{e^{-x}-e^x}{2}$. They're obviously not equal. They'd be if the sign was 'plus'. But we've already dealt with that case.
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


Revive the IMO marathon.

Rabeeb
Posts: 25
Joined: Tue Dec 14, 2010 7:52 pm
Location: Mymensingh, Bangladesh
Contact:

Re: Where did the other root go?

Unread post by Rabeeb » Wed Feb 04, 2015 4:33 pm

Sorry didn't see it :mrgreen:

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