Showing that $ae^x=1+x+\frac{x^2}{2}$ has exactly one real root (part 2)

Question

I made this question here Show $a e^x=1+x+\frac{x^2}{2}$ has exactly one real root but I wrote the equation wrong. I reproduce the post with the correct equation:

I'm struggling with the following problem:

Show that the equation $a e^x=1+x+\frac{x^2}2$, where $a$ is a positive constant, has exactly one real root.

Looking at the graphs of both $ae^x$ and $1+x+\frac{x^2}2$, they will intersect on the left half of the plane when $a>1$ and on the right half when $a<1$ (because a makes $e^x$ grow faster or slower). So if I could find a point $x_1$ where $a e^{x_1}-(1+x_1+\frac{{x_1}^{2}}2)<0$ and a point $x_2$ where $a e^{x_2}-(1+x_2+\frac{{x_2}^2}2)>0$, by the intermediate value theorem, the equation would have a real root. Also,since the derivative of $a e^x-(1+x+\frac{x^2}2)$ is always positive, I could colclude that there can't exist another real root.

So my question is, how can I find $x_1$ and $x_2$? I can't figure out how to express $x$ in terms of $a$.

Answer 1

I will deal with the case $0<a<1$ as the other one is much simpler:

Let $$f(x)=ae^x-1-x-\frac{x^2}{2}$$ Observe $f^{\prime}(x)=ae^x-1-x$, $f^{\prime\prime}(x)=ae^x-1$.

We have $f^{\prime\prime}(x)<0$ for $x<-\ln a$ and $f^{\prime\prime}(x)>0$ for $x>-\ln a$. Therefore, $f^{\prime}$ decreases in $(-\infty,-\ln a]$ and then increases in $[-\ln a,+\infty)$. The minimum is $f^{\prime}(-\ln a)=-a^2+\ln a-1<0$ as $a<1$. Therefore $f^{\prime}$ has only two roots $\xi\in (-\infty,-\ln a)$ and $\eta\in (-\ln a,+\infty)$.

Therefore, $f^{\prime}(x)>0$ in $(-\infty,\xi)$, $f^{\prime}(x)<0$ in $(\xi,\eta)$ and $f^{\prime}(x)>0$ in $(\eta,+\infty)$. This means that $f$ increases in $(-\infty,\xi)$, has a local maximum in $\xi$, decreases in $(\xi,\eta)$, has a local minimum in $\eta$ and finally increases in $(\eta,+\infty)$. Because $f(\xi)<0$ (*), $f$ has only one root in $(\eta,+\infty)$

(*) $f^{\prime}(\xi)=0\implies ae^{\xi}-1-\xi=0$ and so $f(\xi)=\frac{-\xi^2}2<0$ ($\xi\neq 0$ as $a<1$)

Answer 2

With $f(x)=a e^x-(1+x+\frac{x^2}2)$ we have $f'(x)=ae^x-(1+x)$, $f''(x)=ae^x-1$ and $f'''(x)=ae^x$. Assume $f(x_1)=f(x_2)=0$ with $x_1<x_2$. Then $f'(\xi)=0$ for some $\xi\in(x_1,x_2)$. From $f'(x_1)=f(x_1)+\frac{x_1^2}2\ge 0$ and $f'(x_2)=f(x_2)+\frac{x_2^2}2\ge 0$, we conclude that $f'$ has a local minimum in the interior of $[x_1,x_2]$, hence $f''(\zeta)=0$ for some $\zeta\in(x_1,x_2)$. This implies $\zeta=-\ln a$. Therefore, $e^{x_1}<\frac1a<e^{x_2}$ and $ x_1+\frac{x_1^2}2<0<x_2+\frac{x_2^2}2$. This implies $0<x_1<2$ and then $x_2>2$. Consequently, $f$ has at most two zeroes (because otherwise two of them would be $\ge 2$ or two of them would be $\le 2$). Also, $f$ can change sign only at its zeroes. We have $f(x)\to \pm\infty$ as $x\to\pm\infty$, hence $f$ changes signs an odd number of times. Therefore, $f$ has either exactly one zero, or it does not change signs at one of the zeroes. But the latter is impossible as ist would imply $f'(x_i)=0$, hence $f'(x_i)-f(x_i)=\frac{x_i^2}2=0$ and finally $x_i=0$, contradicting $0<x_1<2<x_2$.