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#3: Post edited
by
Derek Elkins
·
2022-08-21T06:56:53Z (about 2 years ago)
Format implications so it not an unreadable blob of text. Correct minor typo/thinko.
Prove $e^x \ge x+1 \\\; \forall x \in \mathbb{R}$ using induction
- > (How) can we prove $e^x \ge x+1 \\; \forall x \in \mathbb{R}$ using induction (without using the derivative of $e^x$ at any stage)? Comments on my attempt are appreciated.
- I stumbled across a very nice proof of $\frac{\mathrm{d}}{\mathrm{d}x}e^x = e^x$ that uses the identity $e^x \ge x+1$. Briefly,
> \begin{align}&e^x \ge x+1 \implies e^{-x} \le 1-x \implies e^x \le \frac{1}{1-x} \implies x+1 \le e^x \le \frac{1}{1-x} \\\\ \implies &x \le e^x - 1 \le \frac{1}{1-x} - 1 \implies x \le e^x - 1 \le \frac{x}{1-x} \implies 1 \le \frac{e^x - 1}{x} \le \frac{1}{1-x} \\\\ \implies &\left(\lim_{x \to 0} 1 = 1 ight) \le \lim_{x \to 0} \frac{e^x-1}{x} \le \left(\lim_{x \to 0} \frac{1}{1-x} = 1 ight) \implies \lim_{h \to 0} \frac{e^h - 1}{h} = 1 \\\\ \implies &\lim_{h \to 0} e^x \cdot \frac{e^h - 1}{h} = e^x \implies \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = e^x \implies \frac{\mathrm{d}}{\mathrm{d}x}e^x = e^x- \end{align}
- I now need to prove $e^x \ge x+1$ while ensuring that my argument is not circular. A lot of the proofs I came across use the derivative of $e^x$ which is not ideal. There were also references to [Bernoulli's inequality](https://en.wikipedia.org/wiki/Bernoulli%27s_inequality) which has some satisfactory proofs. Nonetheless, my first idea was induction, so I wonder if this is possible over the reals. I outline my attempt below, which I'm not very certain of.
- > For the base case $x = 0$,$$e^0 \ge 0 + 1 \implies \lim_{x\to0}e^x \ge \lim_{x\to 0} x + 1$$
- Consider $\epsilon \ge 0 \implies e^{\epsilon} \ge 1$. We now induct as follows:\
- $\underline{x \in \mathbb{R}_{\ge 0}}$
- >
- >$$e^{\epsilon} \cdot e^x \ge e^{\epsilon}(x+1) \implies \lim_{\epsilon \to 0} e^{x + \epsilon} \ge \lim_{\epsilon \to 0} e^{\epsilon}x + \lim_{\epsilon \to 0} e^{\epsilon} \ge x + \lim_{\epsilon\to 0} \epsilon + 1 \tag{1}$$
- $\underline{x \in \mathbb{R}_{< 0}}$
- $$\text{Replacing} \\; \epsilon \\; \text{with} \\; -\epsilon, \\; \text{we follow the same steps as in} \\; (1)$$
- > (How) can we prove $e^x \ge x+1 \\; \forall x \in \mathbb{R}$ using induction (without using the derivative of $e^x$ at any stage)? Comments on my attempt are appreciated.
- I stumbled across a very nice proof of $\frac{\mathrm{d}}{\mathrm{d}x}e^x = e^x$ that uses the identity $e^x \ge x+1$. Briefly,
- > \begin{align}
- e^x \ge x+1
- & \implies e^{-x} \ge 1-x \\\\
- & \implies e^x \le \frac{1}{1-x} \\\\
- & \implies x+1 \le e^x \le \frac{1}{1-x} \\\\
- & \implies x \le e^x - 1 \le \frac{1}{1-x} - 1 \\\\
- & \implies x \le e^x - 1 \le \frac{x}{1-x} \\\\
- & \implies 1 \le \frac{e^x - 1}{x} \le \frac{1}{1-x} \\\\
- & \implies \left(\lim_{x \to 0} 1 = 1 ight) \le \lim_{x \to 0} \frac{e^x-1}{x} \le \left(\lim_{x \to 0} \frac{1}{1-x} = 1 ight) \\\\ & \implies \lim_{h \to 0} \frac{e^h - 1}{h} = 1 \\\\
- & \implies \lim_{h \to 0} e^x \cdot \frac{e^h - 1}{h} = e^x \\\\
- & \implies \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = e^x \\\\
- & \implies \frac{\mathrm{d}}{\mathrm{d}x}e^x = e^x
- \end{align}
- I now need to prove $e^x \ge x+1$ while ensuring that my argument is not circular. A lot of the proofs I came across use the derivative of $e^x$ which is not ideal. There were also references to [Bernoulli's inequality](https://en.wikipedia.org/wiki/Bernoulli%27s_inequality) which has some satisfactory proofs. Nonetheless, my first idea was induction, so I wonder if this is possible over the reals. I outline my attempt below, which I'm not very certain of.
- > For the base case $x = 0$,$$e^0 \ge 0 + 1 \implies \lim_{x\to0}e^x \ge \lim_{x\to 0} x + 1$$
- Consider $\epsilon \ge 0 \implies e^{\epsilon} \ge 1$. We now induct as follows:\
- $\underline{x \in \mathbb{R}_{\ge 0}}$
- >
- >$$e^{\epsilon} \cdot e^x \ge e^{\epsilon}(x+1) \implies \lim_{\epsilon \to 0} e^{x + \epsilon} \ge \lim_{\epsilon \to 0} e^{\epsilon}x + \lim_{\epsilon \to 0} e^{\epsilon} \ge x + \lim_{\epsilon\to 0} \epsilon + 1 \tag{1}$$
- $\underline{x \in \mathbb{R}_{< 0}}$
- $$\text{Replacing} \\; \epsilon \\; \text{with} \\; -\epsilon, \\; \text{we follow the same steps as in} \\; (1)$$
#2: Post edited
by
Carefree Explorer
·
2022-08-20T20:14:57Z (about 2 years ago)
Prove $e^x \ge x+1 \\; \forall x \in \mathbb{R}$ using induction
- Prove $e^x \ge x+1 \\\; \forall x \in \mathbb{R}$ using induction
#1: Initial revision
by
Carefree Explorer
·
2022-08-20T20:14:35Z (about 2 years ago)
Prove $e^x \ge x+1 \\; \forall x \in \mathbb{R}$ using induction
> (How) can we prove $e^x \ge x+1 \\; \forall x \in \mathbb{R}$ using induction (without using the derivative of $e^x$ at any stage)? Comments on my attempt are appreciated.
I stumbled across a very nice proof of $\frac{\mathrm{d}}{\mathrm{d}x}e^x = e^x$ that uses the identity $e^x \ge x+1$. Briefly,
> \begin{align}&e^x \ge x+1 \implies e^{-x} \le 1-x \implies e^x \le \frac{1}{1-x} \implies x+1 \le e^x \le \frac{1}{1-x} \\\\ \implies &x \le e^x - 1 \le \frac{1}{1-x} - 1 \implies x \le e^x - 1 \le \frac{x}{1-x} \implies 1 \le \frac{e^x - 1}{x} \le \frac{1}{1-x} \\\\ \implies &\left(\lim_{x \to 0} 1 = 1\right) \le \lim_{x \to 0} \frac{e^x-1}{x} \le \left(\lim_{x \to 0} \frac{1}{1-x} = 1\right) \implies \lim_{h \to 0} \frac{e^h - 1}{h} = 1 \\\\ \implies &\lim_{h \to 0} e^x \cdot \frac{e^h - 1}{h} = e^x \implies \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = e^x \implies \frac{\mathrm{d}}{\mathrm{d}x}e^x = e^x
\end{align}
I now need to prove $e^x \ge x+1$ while ensuring that my argument is not circular. A lot of the proofs I came across use the derivative of $e^x$ which is not ideal. There were also references to [Bernoulli's inequality](https://en.wikipedia.org/wiki/Bernoulli%27s_inequality) which has some satisfactory proofs. Nonetheless, my first idea was induction, so I wonder if this is possible over the reals. I outline my attempt below, which I'm not very certain of.
> For the base case $x = 0$,$$e^0 \ge 0 + 1 \implies \lim_{x\to0}e^x \ge \lim_{x\to 0} x + 1$$
Consider $\epsilon \ge 0 \implies e^{\epsilon} \ge 1$. We now induct as follows:\
$\underline{x \in \mathbb{R}_{\ge 0}}$
>
>$$e^{\epsilon} \cdot e^x \ge e^{\epsilon}(x+1) \implies \lim_{\epsilon \to 0} e^{x + \epsilon} \ge \lim_{\epsilon \to 0} e^{\epsilon}x + \lim_{\epsilon \to 0} e^{\epsilon} \ge x + \lim_{\epsilon\to 0} \epsilon + 1 \tag{1}$$
$\underline{x \in \mathbb{R}_{< 0}}$
$$\text{Replacing} \\; \epsilon \\; \text{with} \\; -\epsilon, \\; \text{we follow the same steps as in} \\; (1)$$