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Je veux savoir pourquoi $x=e^x$ n'a aucune solution dans $\Bbb R$. Lorsque j'ai essayé de tracer le graphe de la fonction $e^x$, j'ai trouvé en fait qu'elle est une fonction strictement croissante mais je ne sais pas quoi faire après.

Merci mon ami.


I would like to know why $x=e^x$ has no solution in $\Bbb R$? I tried to plot the function $e^x$ and I found that it is an increasing function but I do not know what to do next.

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    $\begingroup$ Parlez-vous anglais? $\endgroup$ – Pedro Tamaroff Jun 30 '14 at 21:05
  • $\begingroup$ Oui je peux parler. $\endgroup$ – Brika Jun 30 '14 at 21:05
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    $\begingroup$ If it's not too much trouble, could you rephrase in English? $\endgroup$ – James47 Jun 30 '14 at 21:06
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This is easily checked, by definition of $e^x$. For $x > 0$ we have $$ e^x = 1 + x + \sum_{k = 2}^\infty\frac{x^k}{k!}>x$$ and for $x \leq 0$ we have $$x \leq 0 < e^x.$$

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    $\begingroup$ True, and a nice approach if $e^x$ is defined using the power series. In many first calculus books, it is not. (One defines $\ln x$, and then defines $\exp(x)$ as the inverse function.) There are other possibilities. $\endgroup$ – André Nicolas Jun 30 '14 at 21:20
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By Bernoulli's inequality, $$ \left(1+\frac xn\right)^n \ge 1+x \qquad\text{if $n > |x|$.} $$ Letting $n\to\infty$ yields $e^x \ge 1+x > x$.

(I like this argument because it doesn't need to treat $x<0$ specially.)

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  • $\begingroup$ The hard part is to show that $e^x \ge (1+x/n)^n$. Also, the usual elementary version of Bernoulli's inequality is shown for integer $n$; the general version is a good deal harder. $\endgroup$ – marty cohen Jun 30 '14 at 22:06
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    $\begingroup$ @martycohen, it's a good thing, then, that neither the inequality $e^x\ge(1+x/n)^n$, nor the general case of Bernoulli's inequality, are invoked in this answer. $\endgroup$ – user21467 Jun 30 '14 at 22:46
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If we look at the initial value for each function $f(x)=x$ and $g(x)=e^x$ we see that $f(0)=0 < g(0)=1$.

Now for $x>0$ we have $f'(x)=1$ and $g'(x) = e^x$ which tells us that $f'(x) < g'(x)$ for all $x > 0$, and so $$x \le e^x$$ for all $x \ge 0$. For negative $x$ we know that $e^x$ is strictly positive, so this tells us that $$x \le e^x$$ for all $x \in \mathbb{R}$.

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The exponential function is a convex function, since the second derivative equals the function, so it is non-negative. This implies that the graphics of $f(x)=\exp(x)$ lies above the tangent line in $x=0$, i.e.: $$\forall \in\mathbb{R},\quad e^x \geq x+1.$$

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If $x \leq 0$, then $e^x > 0 \geq x$.

Suppose $x > 0$ and let $f(x) = e^x - x $. So $f'(x) = e^x - 1 > e^0 - 1 = 0$. Hence $f$ is increasing and since $f(0) > 0$, $f(x) = e^x - x > 0$, finally $e^x > x$.

The inequality is strict.

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