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Hi,

For a given $\theta < 1$, and $N$ a positive integer, I am trying to find an $x > 0$ (preferably the smallest such $x$) such that the following inequality holds:

$$\sum_{k=0}^{N} \frac{x^k}{k!} \leq \theta e^{x}$$

In my application, even $N$ is an integer function of $x$, i.e. $N = N(x)$, but for simplicity sake, let's assume $N$ is given for now.

Any ideas?

Thanks for reading

Fred

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  • $\begingroup$ You won't find an exact answer. However, the difference between the left side and $e^x$ is the tail of the Taylor series so can be expressed in many ways including as an integral. Bounding that integral will give you bounds on $x$. $\endgroup$
    – Brendan McKay
    Commented Apr 12, 2013 at 3:32
  • $\begingroup$ Thanks Brendan. I am going to play with this idea and see where I end up. Cheers $\endgroup$
    – Fred
    Commented Apr 12, 2013 at 5:39
  • $\begingroup$ and in fact this idea will lead exactly to the inequality that William suggested, see her remark below. So I am still on the square one! :-( $\endgroup$
    – Fred
    Commented Apr 12, 2013 at 6:13

1 Answer 1

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The approach that looks most promising to me here is to use the incomplete Gamma function, $$ \Gamma(n, x) = \int_x ^{\infty} t^{n-1}e^{-t}dt = (n-1)!e^{-x}\sum_{k=0} ^{\infty} \frac {x^{k}} {k!} $$ for $n \in \mathbb{Z}^+$. From this and your inequality, we get that $\frac{\Gamma(N+1, x)}{\Gamma(N+1, 0)} \le \theta$, or, equivalently, that your inequality holds wherever $\int_0^xt^Ne^{-t}dt \ge N!(1-\theta)$. Hope this helps.

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  • $\begingroup$ Thanks William, This was a great observation. In fact, this inequality came out of a more general inequality involving incomplete Gamma function. I was, specifically, trying to bound the CDF of a Chi-Square random variable. I was hoping to make the problem simpler by going to the series format, rather than integral. Cheers. $\endgroup$
    – Fred
    Commented Apr 12, 2013 at 5:42
  • $\begingroup$ The upper limit of the sum should be $n-1$ I believe. $\endgroup$
    – Brendan McKay
    Commented Apr 12, 2013 at 8:21
  • $\begingroup$ yes that is true. But the last inequality is OK....have not got any further with this one yet? Does anyone know of any tight upper/lower bound for the incomplete Gamma function? $\endgroup$
    – Fred
    Commented Apr 12, 2013 at 21:12
  • $\begingroup$ Tight upper/lower bound will be difficult without further information. For instance, if you are interested in $\theta \ll 1$, then $x$ will be large and you could try to use asymptotic approximations like dlmf.nist.gov/8.11 $\endgroup$
    – André Schlichting
    Commented Apr 26, 2013 at 12:14

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