# Upper bound on exponential function

1. Dec 2, 2007

### phonic

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!

2. Dec 2, 2007

### phonic

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!

3. Dec 2, 2007

### Xevarion

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.

4. Dec 2, 2007

### EnumaElish

What are the conditions on f? ("f(x) = exp(x)" trivially satisfies your inequality.)

5. Dec 2, 2007

### phonic

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, $$x_i$$s are negative real numbers far from zero: $$x_i < 0$$, and all $$x_i$$s 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!

Last edited: Dec 2, 2007
6. Dec 8, 2007

### Gib Z

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.

Last edited: Dec 8, 2007
7. Dec 8, 2007

### Office_Shredder

Staff Emeritus
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

8. Dec 8, 2007

### Xevarion

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).