exp(x) inequality [Solution]

The problem was to show that

$$e^x \ge 1 + x + \frac{x^2}{2!} + \dots + \frac{x^n}{n!}$$

for all $x \ge 0$ and integer $n \ge 0$.

Solution

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Let $f_{n}(x) = 1 + x + \frac{x^2}{2!} + \dots + \frac{x^n}{n!}$ with $f_{0}(x) = 1$

Now $e^x \ge f_{0}(x)$ for all $x \ge 0$

Now suppose $e^x \ge f_{n}(x)$ for all $x \ge 0$.

Let $h(x) = e^x - f_{n+1}(x)$

Now $h(0) = 0$.

We also have that the derivative of $h$, $h'(x) = e^x - f_n(x)$

By induction assumption $h'(x) \ge 0$ for all $x \ge 0$.

Thus $h$ is increasing and so $h(x) \ge h(0) = 0$ for all $x \ge 0$.

Thus by induction, we are done.