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Show that if $n \geq 0$ and $x>0$, then

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

Not sure where to get started with this induction proof.

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  • $\begingroup$ Yeah, from an exercise set in an induction chapter in a number theory text. $\endgroup$ – liedora Feb 8 '14 at 0:52
  • $\begingroup$ When $x=1$, $e^x=1+x+\frac{x^2}{2}+\mathcal{O}(x^3)$, not $e^x>1+x+\frac{x^2}{2}+\mathcal{O}(x^3)$. I think your $>$ should be changed to $=$. $\endgroup$ – user122283 Feb 8 '14 at 0:53
  • $\begingroup$ That would be the case if the $\dots$ were after the $\frac{x^n}{n!}$ term. $\endgroup$ – liedora Feb 8 '14 at 1:00
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    $\begingroup$ How do you defined $e^x$? Because if you define it in terms of the power series, it is sort of obvious. :) $\endgroup$ – Thomas Andrews Feb 8 '14 at 1:37
  • $\begingroup$ Happy with the answer given by Goos below. But feel free to provide another method of solving it using induction :) $\endgroup$ – liedora Feb 8 '14 at 1:42
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Base case: $e^x > 1$ for $x > 0$.

Induction: suppose we are given $k$ such that for all $x > 0$ $$ e^x > 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^k}{k!} $$ Change variables to $t$ and integrate both sides: $$ \int_0^x e^t \; dt > \int_0^x \left( 1 + t + \frac{t^2}{2!} + \cdots + \frac{t^k}{k!} \right) \; dt $$ $$ e^x - 1 > x + \frac{x^2}{2!} + \cdots + \frac{x^{k+1}}{(k+1)!} $$ $$ e^x > 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^{k+1}}{(k+1)!} $$

Note that if $f > g$ on $(a,b)$ and both are integrable, then $\int_a^b f > \int_a^b g$ (not just $\ge$).

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    $\begingroup$ I wish I could favorite answers. $\endgroup$ – Robert Wolfe Feb 8 '14 at 2:31
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It rather depends on your definition of $e^x$.

Suppose you know $\frac{d}{dx} e^x = e^x$ and $e^0=1$.

Call $f_n(x) = 1 + x + \frac{x^2}{2!} + \dots + \frac{x^n}{n!}$. Then your inductive step could say $f_{n+1}(0) = 1$ and $\frac{d}{dx} f_{n+1}(x) = f_n(x) \lt e^x$ for $x \gt 0$ so $f_{n+1}(x) \lt e^x$.

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