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I'm looking for a reference for deriving the following commonly used upper and lower bounds for $e^x$:

$1 - x \le e^{-x}$ and, assuming $x \le 1/2$,

$1 - x \ge e^{-2x}. $

1 Answers 1

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First Inequality: Let $f(x)=e^{-x}-(1-x)$. Then $f'(x)=-e^{-x}+1$. This is $0$ at $x=0$, negative if $x \lt 0$, and positive for $x \gt 0$. So $f(x)$ reaches an absolute minimum at $x=0$. The minimum value is $0$, so $f(x)\ge 0$ for all $x$.

Second Inequality: Let $g(x)=(1-x)-e^{-2x}$. Then $g'(x)=-1+2e^{-2x}$. This is $0$ for $x=(\ln 2)/2$. We find that $g'(x)$ is positive for $x\lt (\ln 2)/2$, and negative for $x \gt (\ln 2)/2$. The function $g(x)$ is increasing up to $x=(\ln 2)/2$, then decreasing.

We have $g(0)=0$. Since $g(x)$ is increasing until $(\ln 2)/2$, it follows that $g(x)$ is negative when $x$ is negative, so our inequality fails if $x \lt 0$.

For positive $x$, $g(x)$ increases up to $x=(\ln 2)/2$, and then decreases. After a while it will become negative. Let us check whether $g(x)$ is still positive at $x=\frac{1}{2}$. Calculate. We get $g(1/2)\approx 0.132$. So $g(x) \ge 0$ at least in the interval $[0,1/2]$. In fact $g(x) \ge 0$ from $0$ to a bit beyond $0.79$.

Remark: There will be few standard North American calculus books that don't do the first inequality. Usually it is done in the equivalent version $1+x \le e^x$. It is an early application of the connection between increasing functions and positive derivatives. The inequality (among many others) is mentioned here.

The second inequality uses the same method, but the details involve more work.