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## e^x + e^-x - 2 = 0

This is a discussion on e^x + e^-x - 2 = 0 within the General Off Topic+ forums, part of the QJ.net Forum Miscellaneous category; e^x + e^-x - 2 = 0 The answer is x=0. However, I'm having a problem explaining it algebraically. When ...

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1. ## e^x + e^-x - 2 = 0

e^x + e^-x - 2 = 0
The answer is x=0. However, I'm having a problem explaining it algebraically. When I try to solve for x, I usually get ln(2) = x, ln(2) = 0, or something along those lines.:Argh::Argh::Argh:
Anybody willing to help? Its not my homework, I was just curious.

2. e^x = 1 + x/1! + x^2/2! + x^3/3! ...

e^-x = 1 - x/1! + x^2/2! - x^3/3! ...

therefore

e^x + e^-x - 2 = x^2/2! + x^4/4! + x^6/6! ... = 0

the term in the middle is only ever 0 for x = 0 (otherwise > 0)

3. Originally Posted by kuroneko
e^x = 1 + x/1! + x^2/2! + x^3/3! ...

e^-x = 1 - x/1! + x^2/2! - x^3/3! ...

therefore

e^x + e^-x - 2 = x^2/2! + x^4/4! + x^6/6! ... = 0

the term in the middle is only ever 0 for x = 0 (otherwise > 0)
It makes sense, but only if I can figure out how e^x=1 + x/1 +...
How is that possible?

4. Originally Posted by Gam3freQ
It makes sense, but only if I can figure out how e^x=1 + x/1 +...
How is that possible?
http://en.wikipedia.org/wiki/E_(math...omplex_numbers

To be honest, I just looked it up. I remember from school that series can be helpful sometimes.

5. Originally Posted by kuroneko
http://en.wikipedia.org/wiki/E_(math...omplex_numbers

To be honest, I just looked it up. I remember from school that series can be helpful sometimes.
Ah, thank you.
I wouldn't have figured that out on my own :)

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