- #126

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I had someone show me this; it'll prove that for even ##n##, ##p_n(x)=\sum_{k=0}^n\frac{x^n}{n!}>0## for ##x<0##, thus proving 2) and 3).Not yet a full solution; missing a detail. I'd be grateful if someone could give me a hint for 3) of the first part.

We know that ##p_n(0)=1## for all ##n##, ##p_0(x)=1## has no zeroes, and ##p_1(x)## has only one zero (linear function). For the rest, I suppose that ##n\geq2##.

Let ##n## be an even natural number.

1) ##\forall x\geq0,\,p_n(x)>0## since we have a sum of positive numbers.

2) ##\forall x\in[-1,\,0),\,p_n(x)>0##.

I can write ##p_n(x)## like this:

$$p_n(x)=1+\sum_\underset{\text{step}=2}{k=1}^{n-1}\left(\frac{x^k}{k!}+\frac{x^{k+1}}{(k+1)!}\right)$$

I am going to prove that the parenthesis is always positive:

$$\begin{align*}

x^k\geq |x|^{k+1}&\Leftrightarrow\frac{x^k}{k!}>\frac{x^k}{(k+1)!}\geq\frac{|x|^{k+1}}{(k+1)!} \\

&\Leftrightarrow\frac{x^k}{k!}>\frac{|x|^{k+1}}{(k+1)!}\\

&\Leftrightarrow\frac{x^k}{k!}>\frac{x^{k+1}}{(k+1)!}>-\frac{x^k}{k!}\\

&\Leftrightarrow-\frac{x^k}{k!}<-\frac{x^{k+1}}{(k+1)!}<\frac{x^k}{k!}\\

&\Leftrightarrow0<\frac{x^k}{k!}+\frac{x^{k+1}}{(k+1)!}

\end{align*}$$

3) ##\forall x<-1,\,p_n(x)>0##. (WIP)

Since ##p_n(x)>0## for all real values of ##x##, it has no real zeroes when ##n## is even.

Let ##n## be an odd natural number.

I notice that ##p'_n(x)=p_{n-1}(x)##. We have that ##n-1## is even, so ##p_{n-1}>0## for ##x\in\mathbb{R}##. From this I can say that ##p_n(x)## is a bijection, since it is constantly growing, with no absolute/local maximum or minimum. ##(*)##

We also have that:

$$\lim_{x\to\pm\infty}p_n(x)=\lim_{x\to\pm\infty}\frac{x^n}{n!}=\pm\infty$$

1) This tells me that there exists a real number ##N##, such that ##p_n(N)p_n(-N)<0##.

2) Since ##p_n(x)## is a polynomial, it is continuous over all ##\mathbb{R}##.

Using 1) and 2), I conclude from the intermediate value theorem that there exists a real number ##c## such that ##p_n(c)=0##. Moreover, using ##(*)##, I also conclude that ##c## is unique.

Conclusion:

If ##n## is even, then ##p_n(x)## has no real zeroes. Else, it has only one.

$$e^x=p_n(x)+E(x)\Leftrightarrow p_n(x)=e^x-E(x)$$

Where ##E(x)## is the remainder, and is given by:

$$E(x)=\frac{e^\zeta}{(n+1)!}x^{n+1},\text{ with }x<\zeta<0$$

Since ##x<0## and ##n## is even, then ##-x^{n+1}=|x|^{n+1}##:

$$p_n(x)=e^x-E(x)=e^x+\frac{|x|^{n+1}e^\zeta}{(n+1)!}>0$$

I think that my proof is now complete.

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