How to prove this equality:
$$\frac{1-e^{-\frac{1}{t}}}{1-e^{-\frac{1}{2t}}}=1+e^{-\frac{1}{2t}}$$
I've no idea how to start so everything is welcome!
Thanks a lot!
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3Set $$e^{-1/2t}=u\implies e^{-1/t}=u^2$$ – lab bhattacharjee Nov 30 '15 at 14:45
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2Or just multiply both sides by the denominator and use the fact that $(1+x)(1-x)=1-x^2$. Also the tag doesn't fit the question too well. – Kuba Nov 30 '15 at 14:46
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Multiplying both sides with $1-e^{-\frac{1}{2t}}$ is really all there is to it: $$\frac{1-e^{-\frac{1}{t}}}{1-e^{-\frac{1}{2t}}}=1+e^{-\frac{1}{2t}} \iff 1-e^{-\frac{1}{t}}=\left(1-e^{-\frac{1}{2t}}\right)\left(1+e^{-\frac{1}{2t}}\right)$$ Now note that $$\left(1-e^{-\frac{1}{2t}}\right)\left(1+e^{-\frac{1}{2t}}\right)=1-e^{-\frac{1}{2t}}+e^{-\frac{1}{2t}}-e^{-\frac{1}{2t}}e^{-\frac{1}{2t}}=1-e^{-2\cdot\frac{1}{2t}}=1-e^{-\frac{1}{t}}$$ And we are done :)
Eric S.
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