Upper bound on exponential function

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The discussion centers on finding an upper bound for the exponential function or the sum of exponential functions, particularly in the context of error analysis where the variables are negative real numbers approaching negative infinity. Participants note that traditional methods, such as Jensen's inequality, are not applicable due to the convex nature of the exponential function. One suggested approach is to use the maximum of the variables to bound the sum, while another involves using Taylor series expansions, although these methods may not yield satisfactory upper bounds for large negative values. The conversation emphasizes the need for clarity on the function f(x) and the specific requirements for bounding the exponential terms effectively. Overall, the search for a suitable upper bound remains challenging, especially for values far from zero.
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Dear All,

I am searching for an upper bound of exponential function (or sum of experiential functions):

1) \exp(x)\leq f(x)
or:
2) \sum_{i=1}^n \exp(x_i) \leq f(x_1,\cdots,x_n, n) .

Since exponential function is convex, it is not possible to use Jenssen's inequality to get an upper bound like 2). Does anyone has an ideal?

Thanks a lot!
 
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upper bound on exponential function?

Dear All,

I am searching for an upper bound of exponential function (or sum of experiential functions):

1) \exp(x)\leq f(x)
or:
2) \sum_{i=1}^n \exp(x_i) \leq f(x_1,\cdots,x_n, n) .

Since exponential function is convex, it is not possible to use Jenssen's inequality to get an upper bound like 2). Does anyone has an ideal?

Thanks a lot!
 
We need more information.
a) What kind of numbers are x_i?
b) How good do you want your bound to be? For what kind of values of x do you want it to be effective?

I'll assume that your numbers are real. Since the exponential has a fast rate of growth, you can easily reduce 2 to 1 by saying
\sum_{i=1}^n e^{x_i} \le ne^{\max_{i}(x_i)}
but this might not be as good a bound as you want.

If the x_i aren't too big, then you can just use the first few terms of the Taylor series for the exponential:
e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots
This gives you a lower bound usually, but of course you can use a slightly larger value of x or multiply by a constant and get an upper bound. But as I said this won't work for a large range of values.

Another thing you can use is the fact that
\lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n = e^x

This converges pretty quickly, but it's also from below so you'd have to do the trick of using it for a slightly larger value of x than the thing you actually want to bound. But again n might be big and this might not work for a large range of x.

Anyway it would also help if you explained why you wanted a bound like this. That would help us figure out what sort of thing you were looking for. The exponential is one of the nicest functions; why would you want to replace it by something else? Or maybe you want a specific thing, which would be helpful for us to know.
 
What are the conditions on f? ("f(x) = exp(x)" trivially satisfies your inequality.)
 
Xevarion and EnumaElish,

Thanks for helping.

Actually, I am working on 2). It is about error analysis. Each of the exponential function in the summation is an error term. I want to bound the total error by a function f(x), which is preferably an exponential function. In my case, x_is are negative real numbers far from zero: x_i < 0, and all x_is approach -\infty.

Xevarion pointed out two facts for x near zeros. Are there some results related to exponential functions with x far away from zero? Thank for your discussions!
 
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Actually, Xevarion had two methods, both work for all x, only problems being they are both approaching from below and you said you wanted from above. Also, you still haven't given proper restrictions to f(x) for (1), and as EnumaElish points out, exp(x) is a very good upper bound for exp(x) =]

EDIT: e^x \approx \frac{x^x\sqrt{2\pi x}}{\Gamma(x+1)} \cdot \left( 1+\frac{1}{12x} + \frac{1}{288x^2} - \frac{139}{51840x^3} -\frac{571}{2488320x^4} \right) works remarkably good as well.
 
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Xevarion pointed out two facts for x near zeros. Are there some results related to exponential functions with x far away from zero? Thank for your discussions!

Unfortunately no (or at least nothing easy). basically, the reason things near 0 work is because in the taylor series, near zero for larger degrees of x the coefficients and the xn are both really tiny, meaning you can ignore them. Cutting off the Taylor series at a given x, as the magnitude of x gets large, the last term you're looking at grows faster than everything else, and the terms you're ignoring are growing faster, meaning eventually you are ignoring very non-trivial terms
 
Yeah. Luckily if your x_i are far apart in (b), you can definitely replace that sum with the biggest x_i (I mean least negative) times a small constant (like 3/2 or 2 maybe, depends on how much smaller the other terms are).
 

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