# Prove that the equation has an unique root

Let $$a>0$$. The question is to show that the equation $$ae^{x} - 1 = x +\frac{x^2}{2}$$ has an unique real solution.

My attempt. Define $$f(x) = ae^{x} - 1 - x - \frac{x^2}{2}$$. We have $$\lim_{x \rightarrow + \infty} f(x) = +\infty$$ and $$\lim_{x \rightarrow - \infty} f(x) = -\infty.$$ Thus by the Intermediate value theorem there exist a root for the equation.

Regarding the uniqueness: My feeling is that it can be used the Rolle's Theorem but I am getting anywhere. someone could help me?

• For $a > 1$, you might notice that your $f(x)$ has a strictly positive derivative. For $a=1$, notice that $f'(x) = 0$ only at the origin and is positive everywhere else. For $0<a<1$, you could play around with the relative extrema. – Hyperion Jul 23 '19 at 1:11
• For $0<a<1$, there are two relative extrema. The relative max will always be decreasing as $a$ decreases from $a = 1$ to $a = 0$. Since the "relative max" at $a = 1$ is at $f(x) = 0$, the relative max will always be less than $0$ for $0<a<1$. – automaticallyGenerated Jul 23 '19 at 1:27

Others have shown the uniqueness in the root for $$a \ge 1$$ by looking at the derivative. For $$0, the uniqueness can be found by looking at the relative maximum. The derivative of $$f(x)$$ is $$ae^{x}-1-x$$

Setting this equal to $$0$$ finds the relative maximum to be at $$x = -W(-\frac{a}{e})-1$$ where $$W(x)$$ is the Lambert W function.

The value of $$f(x)$$ at this $$x$$ is then $$-\frac{(-W(-\frac{a}{e})-1)^2}{2}$$. The derivative of the $$y$$ value of the relative maximum with respect to $$a$$ is then $$-\frac{W(-\frac{a}{e})}{a}$$

This is positive for $$0, and thus, the $$y$$ value of the relative maximum is increasing from $$a = 0$$ to $$a = 1$$. This relative maximum never equals $$0$$, only approaches it as $$a$$ approaches $$1$$. This claim is backed by the fact that, when $$a = 1$$, at the $$x$$-value where the derivative of $$f(x)$$ is $$0$$, the $$y$$-value is also $$0$$.

Now that it is shown that the relative maximum is never greater than or equal to $$0$$, it is trivial to show that there is only one distinct real root. Since the function increases until the relative maximum and still does not equal $$0$$, it will not equal $$0$$ at the relative minimum. Then, the Intermediate Value Theorem tells us there is a root and the derivative is positive. Therefore, there is only one real root for all values of $$a$$.

• wow ! I think that my real analysis professor putted the wrong question in the exercises . thanks for your answer! – math student Jul 23 '19 at 2:17

To see the uniqueness, check the sign of its derivative. If $$f'(x)>0$$ (which is clear) then $$f$$ is non-decreasing, which means that if you cut the $$x$$-axis at some point, you cannot go down again.

• We have $f^{'}(x) = ae^{x} - 1 -x .$ why $f^{'}(x) >0$ for all x? – math student Jul 23 '19 at 1:10
• It depends of the choice of $a$, but a well-known inequality is the following : $$e^x \geq 1+x \quad \textrm{for all } x$$ So, the only thing that causes trouble is the case where $0 <a <1$ – Azif00 Jul 23 '19 at 1:12
• This isn't true for $a \leq 1$ – Hyperion Jul 23 '19 at 1:13
• The inequality $e^{x} > 1 + x$ helps in the case $a \geq 1$ . Now the problem is the case $a <1$ – math student Jul 23 '19 at 1:18