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I am looking for a tight upper bound of exponential function (or sum of exponential functions):

$e^x<f(x)$ when $x<0$ or

$\sum_{i=1}^n e^{x_i} < g(x_1,...,x_n)$ when $x_i<0$

Thanks a lot!

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4  
If $x < 0$ then $e^x < 1$. Is this what you need? –  Pragabhava Oct 10 '12 at 18:07
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What do you mean with $x$ is discrete? –  Hagen von Eitzen Oct 10 '12 at 18:42
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Thanks Pragabhava, but I am looking for a tight upper bound of exponential function. –  Mehran Oct 10 '12 at 19:32
    
Thanks Pragabhava, but I am looking for a tight upper bound for exponential function. I know a close upper bound ($1/(1-x)$) , but I am looking for polynomial form! –  Mehran Oct 10 '12 at 19:39
    
Sorry, "x is discrete" was a mistake. –  Mehran Oct 10 '12 at 19:48

2 Answers 2

Since you suggest in the comments you would like a polynomial bound, you can use any even Taylor polynomial for $e$.

Proposition. $\boldsymbol{1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!}}$ is an upper bound for $\boldsymbol{e^x}$ when $\boldsymbol{n}$ is even and $\boldsymbol{x \le 0}$.

Proof. We wish to show $f(x) \ge 0$ for all $x$, where $f: (-\infty, 0] \to \mathbb{R}$ is the function defined by $f(x) = 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} - e^x.$ Since $f(x) \to \infty$ as $x \to - \infty$, $f$ must attain an absolute minimum somewhere on the interval $(-\infty, 0]$.

  • If $f$ has an absolute minimum at $0$, then for all $x$, $f(x) \ge f(0) = 1 - e^0 = 0$, so we are done.

  • If $f$ has an absolute minimum at $y$ for some $y < 0$, then $f'(y) = 0$. But differentiating, $$ f'(y) = 1 + y + \frac{y^2}{2!} + \cdots + \frac{y^{n-1}}{(n-1)!} - e^y = f(y) - \frac{y^n}{n!}. $$ Therefore, for any $x$, $$ f(x) \ge f(y) = \frac{y^n}{n!} + f'(y) = \frac{y^n}{n!} > 0, $$ since $n$ is even. $\square$


Keep in mind that any polynomial upper bound will only be tight up to a certain point, because the polynomial will blow up to infinity as $x \to -\infty$. Also note, the same proof shows that the Taylor polynomial is a lower bound for $e^x$ when $n$ is odd and $x \le 0$.

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For $x < 0$ and $n \in \mathbb{N}$,

$$ 0 < \sum_{k=0}^{n} \frac{(-x)^k}{k!} < e^{-x} $$

so

$$ e^x < 1\left/\sum_{k=0}^{n} \frac{(-x)^k}{k!}\right.. $$

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