Solve-for-a-and-explain-your-working-a-e-a-

Question Number 1828 by Filup last updated on 07/Oct/15

Solve for a and explain your working

a = e^{a}

Commented by Alejandro Prieto last updated on 14/Oct/15

Gr \overset{´}{a f i c a m e n t e} podr \overset{´}{i a} verse que las funciones

f (x) = e^{x} y g (x) = x

no intersectan en ning \overset{´}{u n} punto .

Por tanto, puede pensarse que no existen soluciones en R .

Commented by 123456 last updated on 16/Oct/15

howeve i suspect that its have solution into C

z = e^{z} (z = x + y ı, x \in R, y \in R)

x + y ı = e^{x} e^{y ı}

x + y ı = e^{x} \cos y + ı e^{x} \sin y

{\begin{cases} x = e^{x} \cos y \\ y = e^{x} \sin y \end{cases}

Answered by 123456 last updated on 08/Oct/15

f (x) = e^{x} - x

f (x) = 0 \Leftrightarrow e^{x} - x = 0 \Leftrightarrow e^{x} = x

x \in R

e^{x} ⩾ 0

x ⩽ 0 ⩽ e^{x}

so if it as a real solution it would be at

[0, + \infty)

f (0) = e^{0} - 0 = 1 \neq 0

f^{'} (x) = e^{x} - 1

x ⩾ 0, e^{x} ⩾ 1 \Rightarrow f^{'} (x) ⩾ 0

so f is increasing into [0, + \infty)

lets x > 0, by mean value theorem into

[0, x], \forall ξ \in [0, x] with g (x) = e^{x}

g^{'} (ξ) = \frac{e^{x} - 1}{x - 0}

g^{'} (ξ) = e^{ξ} ⩾ 1, \forall ξ ⩾ 0

\frac{e^{x} - 1}{x} ⩾ 1

e^{x} - 1 ⩾ x

e^{x} - x - 1 ⩾ 0

f (x) - 1 ⩾ 0

f (x) ⩾ 1

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