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

$e^x when $ \;x<0$ or

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

Thanks a lot!

  • 0
    Sorry, "x is discrete" was a mistake.2012-10-10

3 Answers 3

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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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    Yes, and so does the OP. However, I stumbled upon this question while looking for such a bound for x>0 and your answer pleased me so I just wanted to complete it in that case.2016-10-13
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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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How about $e^{-x}\leq c(\gamma)(\frac{1}{1+x})^\gamma$ for non-negative $x$?