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e

How is



?
Original post by Banny Dyrne
How is



?


e−2=1e2 e^{-2} = \frac{1}{e^{2}} . Same thing for e−4 e^{-4}

So

e−2−e−42=(1e2−1e4)⋅12 \frac{e^{-2} - e^{-4}}{2} = \left(\frac{1}{e^2} - \frac{1}{e^4} \right) \cdot \frac{1}{2}

Multiply top and bottom of 1e2 \frac{1}{e^2} by e2 e^2 (to get a common denominator)

(e2e2⋅e2=e4−1e4)⋅12 \left( \frac{e^2}{e^2 \cdot e^2 = e^4} - \frac{1}{e^4} \right) \cdot \frac{1}{2}

e2−1e4⋅12=e2−12e4 \frac{e^2 - 1}{e^4} \cdot \frac{1}{2} =\frac{e^2 - 1}{2e^4}



EDIT: Why did I get negged?
(edited 11 years ago)
Reply 2
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Star-girl
12
Original post by Banny Dyrne
How is



?


Just multiply numerator and denominator of LHS by e4e^4
Reply 3
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Banny Dyrne
OP
3
Oh ok. Is it necessary to do this simplification though?
Reply 4
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TenOfThem
18
Original post by Banny Dyrne
Oh ok. Is it necessary to do this simplification though?


necessary for what?
Reply 5
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Star-girl
12
Original post by Banny Dyrne
Oh ok. Is it necessary to do this simplification though?


Not really but the RHS looks prettier.
Original post by Banny Dyrne
How is



?

using the facts that:
x*1=x
and e4e4=1 \frac{e^4}{e^4} = 1
you should be able to do it.

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