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Do the roots of $$P_{2n+1}(x)=1+\frac x{1!}+\frac{x^2}{2!}+\cdots+\frac{x^{2n+1}}{(2n+1)!}$$ decrease to $-\infty$?

Can we show this? Indeed, $P_{2n+1}(0)=1$, $P_{2n+1}(-(2n+1))<0$. And $P_{2n+1}'(x)=P_{2n}(x)>0$. So $P_{2n+1}$ has only one root $x_n$. Can we show that $x_n\to-\infty$?

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  • $\begingroup$ I don't think so: wolframalpha.com/input/… $\endgroup$ – Yves Daoust Oct 12 '18 at 13:26
  • $\begingroup$ @YvesDaoust What is the root of $P_{11}$ supposed to demonstrate? $\endgroup$ – Arthur Oct 12 '18 at 13:27
  • $\begingroup$ @Arthur: that some roots even have a positive real part. $\endgroup$ – Yves Daoust Oct 12 '18 at 13:28
  • $\begingroup$ @YvesDaoust Who said anything about complex numbers? $\endgroup$ – Arthur Oct 12 '18 at 13:28
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    $\begingroup$ @YvesDaoust I mean, you can probably find lots of questions here which do not explicitly mention that they restrict themselves to the real numbers. There are many small pieces of evidence, mostly using the variable name $x$, and refering to $<$ and $-\infty$. Those all pull toward real numbers. I see nothing pointing towards complex numbers. $\endgroup$ – Arthur Oct 12 '18 at 13:32
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By using, for instance, the Weierstrass $M$-test, we see that your sequence converges uniformly to $e^x$ on any bounded domain. In other words, for any $X< 0$, there is an $N\in \Bbb N$ such that $$|P_{2n+1}(x) - e^x|<\frac{1}{2}e^X \leq \frac12e^x$$ for any $n> N, x\in [X, -X]$. By the triangle inequality, this implies that $P_{2n+1}$ doesn't have any roots on $[X, -X]$ for such $n$. So yes, $x_n\to -\infty$, as for any finite bound we can pick an $N$ which forces $x_n$ to be below that bound whenever $n>N$.


The Weierstrass $M$-test is a very fancy name for a quite simple idea. Basically, say you have a collection of functions $f_n$ for $n\in \Bbb N$ on some domain, and on that domain the absolute value $|f_n|$ of each of those functions is bounded by some number $M_n$. If $\sum M_n < \infty$, then $\sum f_n$ converges uniformly.

In this case, we have $f_n(x) = \frac{x^n}{n!}$, which on $[X, -X]$ is bounded by $M_n = \frac{(-X)^n}{n!}$. Then $\sum M_n = e^{-X}<\infty$, so $\sum f_n$ converges uniformly.

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  • $\begingroup$ Isn't it standard to define the exponential function $e^x$ as the power series $\sum_{i=0}^\infty \frac{x^n}{n!}$ which has radius of convergence $= \infty$ and thus converges uniformly on each compact set? $\endgroup$ – Paul Frost Oct 12 '18 at 13:28
  • $\begingroup$ @PaulFrost "and thus converges uniformly on each compact set" How would you prove that? Isn't that exactly what I've done here with the $M$-test? Besides, that's just the first half of my argument. The second is actually using that uniform convergence to show the result. $\endgroup$ – Arthur Oct 12 '18 at 13:32
  • $\begingroup$ You are right, it requires a proof. I only wanted to remark that this is a characteristic property of power series. $\endgroup$ – Paul Frost Oct 12 '18 at 13:44

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