# Inequality of Partial Taylor Series

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?

Fred

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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 $x$. –  Brendan McKay Apr 12 '13 at 3:32
Thanks Brendan. I am going to play with this idea and see where I end up. Cheers –  Fred Apr 12 '13 at 5:39
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! :-( –  Fred Apr 12 '13 at 6:13

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.
The upper limit of the sum should be $n-1$ I believe. –  Brendan McKay Apr 12 '13 at 8:21
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 –  André Schlichting Apr 26 '13 at 12:14