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It is well known $1+x\leq e^x$ for all $x\in\mathbb R$. See for example this post for several proofs of it.

I wanted to show for every $n\in\mathbb N$ there is a constant $C_n>0$ such that $$e^x\leq C_n(1+|x|)^n,$$ forall $x\in\mathbb R$.

I did a proof (hope it is right):

If $x\in (-\infty, 0]$ then $e^x<1\leq (1+|x|)^n$ for every $n\in\mathbb N$;

If $x\in (0, 1)$ then $e^x<e\leq e(1+|x|)^n$ for every $n\in\mathbb N$;

If $x\geq 1$ then by Taylor's formula there is $c\in [0, x]$ such that: \begin{align*} \displaystyle e^x&=|e^x|\\ &=\left| e^x-\sum_{j=0}^{n-1} \frac{1}{j!}x^j+\sum_{j=0}^{n-1} \frac{1}{j!}x^j\right|\\ &\leq \left| e^x-\sum_{j=0}^{n-1} \frac{1}{j!}x^j\right|+\left|\sum_{j=0}^{n-1} \frac{1}{j!}x^j\right|\\ &\leq \frac{e^c}{n!}|x|^{n}+\sum_{j=0}^{n-1} \frac{1}{j!}|x|^j\\ &\leq \frac{e^c}{n!}|x|^n+\sum_{j=0}^{n-1} \frac{1}{j!}|x|^{n}\quad (\textrm{since}\ |x|\geq 1)\\ &\leq \left(\frac{e^c}{n!}+\sum_{j=0}^{n-1} \frac{1}{j!}\right) (1+|x|)^n\\ &=C_n (1+|x|)^n, \end{align*} for every $n\in\mathbb N$.

The problem is that $c$ is depending on $x$ (isn't it?) and I'd like that not to happen.

Sketching the graph of the functions it is clear for me one can always find $C_n$ not depending on $x$, but I wasn't able to prove it.

So, is my above proof right? Is there really a dependence on $x$ in my constant $C_n$? Can I take drop this dependence?

Thanks

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PtF
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  • Since $e^x$ is continuous, it is bounded for all $x$. So, what does this tell you? – Mark Viola Oct 28 '15 at 23:53
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    Do you mean it is bounded on compact subsets containing $x$, right? The function $x\longmapsto e^x$ is not bounded on $\mathbb R$.. – PtF Oct 28 '15 at 23:56
  • Yes. But the order of statement is important here. Here you first fix $x$ and then for any $n$ find the $C_n$, which depends on $x$. If you try to do it the other way, you will fail since we can always find a value of $x$ for which $e^x$ exceeds a given $C(1+|x|)^n$ for any $n$ and $C$. – Mark Viola Oct 28 '15 at 23:59
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    Agreed, but how do I use your first comment? – PtF Oct 29 '15 at 00:15
  • It was a question. For any $x$, we can find a $C$ such that $C>e^x$. Of what added value is the term $(1+|x|)^n\ge1$ and the indexing of $C$, which seems superfluous? – Mark Viola Oct 29 '15 at 00:23
  • The statement you want to show is false. You can not take $C_n$ independent of $x$. – PhoemueX Oct 29 '15 at 21:42

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