Question

please help me in solving the eqn.
(x e^x)/(e^x - 1) =1
USES SOME OMEGA FUNCTION??????????

Answer 1

\[\frac{ xe ^{x} }{e ^{x}-1}=1\]

Answer 2

taylor:\[
\frac{x(1+x+x^2/2+x^3/6+\dotsb)}{x+x^2/2+x^3/6+\dotsb} = 1\\
%%\iff \frac{(1+x+x^2/2+x^3/6+\dotsb)}{(x+x^2/2+x^3/6+\dotsb)/x} = 1\\
\iff \frac{e^x}{1+x/2+x^2/6 + \dotsb} = 0\\
%\iff \frac{(1+x+x^2/2+x^3/6+\dotsb)}{1+x/2+x^2/6 + \dotsb} = 1\\
%\iff 1+x+x^2/2+x^3/6+\dotsb = 1+x/2+x^2/6 + \dotsb \\
\iff e^x = 1+x/2+x^2/6 + \dotsb =: g(x)\]

in my opinion, it can be argued that , when \(x<0\), we have \(e^x<g(x)\). (looking at taylors expansion of \(e^x\))
and that \(e^x>g(x)\) when \(x>0\). since \(e^0=g(0)\) this shows that \(x=0\) is the only solution.

Answer 3

Edit: it's a "1" on the RHS of the second equation. not a "0".

Answer 4

x=0
IT DOESNT SATISFY THE EQN. AT ALL!!!!!!!!!!!!!!

Answer 5

x=0.
\(e^0=1\)
and \(1 +0/2 + 0^2/6 + \dotsb= 1\).
I can read small caps better.