194

If you're not quite in the market for a full proof: $$a^n=a\times a\times a\times a...\times a$$ $$n!=1\times 2\times 3\times 4...\times n$$ Now what happens as $n$ gets much bigger than $a$? In this case, when $n$ is huge, $a$ will have been near some number pretty early in the factorial sequence. The exponential sequence is still being multiplied by that ...


178

Of course $C e^x$ has the same property for any $C$ (including $C = 0$). But these are the only ones. Proposition: Let $f : \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f(0) = 1$ and $f'(x) = f(x)$. Then it must be the case that $f = e^x$. Proof. Let $g(x) = f(x) e^{-x}$. Then $$g'(x) = -f(x) e^{-x} + f'(x) e^{-x} = (f&#...


134

There are two things I should point out. One is that the arithmetic mean doesn't properly measure annual growth rate. The second is why. The correct calculation for average annual growth is geometric mean. Let $r_1,r_2,r_3,\ldots,r_n$ be the yearly growth of a particular investment/portfolio/whatever. Then if you invest $P$ into this investment, after $n$ ...


127

Another way (not sure if its "simple" though!): $y = x+1$ is the tangent line to $y = e^x$ when $x= 0$. Since $e^x$ is convex, it always remains above its tangent lines.


120

Hint: Let $f(n) = \dfrac{n! }{ a^n}$, for $ a > 1$. What is $\dfrac{f(n+1)}{f(n)}$?


88

"Why is this hard?" I think a different question would be "Why would it be easy?" But there are some things that are known. It is known that $\pi$ and $e$ are transcendental. Thus $(x-\pi)(x-e) = x^2 - (e + \pi)x + e\pi$ cannot have rational coefficients. So at least one of $e + \pi$ and $e\pi$ is irrational. It's also known that at least one of $e \pi$ ...


68

By the fundamental theorem of calculus $$e^{i\theta}-1=\int_0^\theta ie^{it}\mathrm{d}t$$ Hence...


68

Yes. For example \begin{align*} \sinh x &= -i \sin(ix) \\ \cosh x &= \cos(ix) \\ \tanh x &= -i \tan(ix) \\ \end{align*} These identities come from the definitions, $$ \sin x = \frac{e^{xi}-e^{-xi}}{2i} \text{ and } \sinh x = \frac{e^x - e^{-x}}{2} $$ and similar for cosine and tangent.


63

What answer you find most elegant may depend on what definition of $e$ you're starting with, as Dylan suggests, but I find this argument quite short and sweet: $$\begin{align} &\quad 1 + 1 &= 2\\ &< 1 + 1 + \frac12 + \frac1{2\cdot3} + \frac1{2\cdot3\cdot4} + \cdots &= e \\ &< 1 + 1 + \frac12 + \frac1{2\cdot2} + \frac1{2\cdot2\cdot2} ...


63

Analogously, here are several ways to define me: I am the citizen of the US with social security number [XYZ]. This is of primary interest to the government. I am the oldest son of [my mother's name]. This is of primary interest to my family. I am the instructor of [particular course meeting at particular days/times] at [university]. This is of primary ...


60

No. Let $$A=\begin{pmatrix}2\pi i&0\\0&0\end{pmatrix}$$ and note that $e^A=I$. Let $B$ be any matrix that does not commute with $A$.


59

$x^{n+1}=x\cdot x^n$ right? so $x^1=x \cdot x^0$ but $x=x^1$ so for that to hold true, $x^0$ must be $1$. Similarily, $\large x^{-n} = \frac{1}{x^n}$. So $\large x^n \cdot x^{-n} = x^n \frac{1}{x^n} = 1$. But $\large x^n \cdot x^{-n} = x^{n+(-n)} = x^0$, so once more, $x^0=1$. There are really many reasons for that to hold, and all of them are just ...


59

An intuitive way to see this is to consider that you're trying to show $$a^n < n!$$ for sufficiently large $n$. Take the log of both sides, you get $$n\log(a) = \log(a^n) < \log(n!) = \sum_{i = 1}^n\log(i).$$ Now as you increase $n$ you only add $\log(a)$ to the left side, but the $\log(n + 1)$ that you add to the right can be arbitrarily large as $n$ ...


58

How about, \begin{align} \frac{d}{dx}\left(\lim_{n\rightarrow\infty} \left(1+\frac{x}{n}\right)^{n} \right) & \overset{\text{intimidate}}{=} \lim_{n\rightarrow\infty} \frac{d}{dx}\left(1+\frac{x}{n}\right)^{n} \\ & = \lim_{n\rightarrow\infty} \left(1+\frac{x}{n}\right)^{n-1} \\ & = \lim_{n\rightarrow\infty} \frac{\left(1+\frac{x}{n}\right)^{n}}{\...


56

You may write, for $N \geq 2$, $$ \begin{align} e^{3/2}\prod_{n=2}^{N}e\left(1-\dfrac{1}{n^2}\right)^{n^2}&=e^{3/2}\times\prod_{n=2}^{N}e\times\prod_{n=2}^{N}\left(1-\dfrac{1}{n^2}\right)^{n^2}\\\\ &=e^{3/2}\times e^{N-1}\times\prod_{n=2}^{N}\dfrac{(n-1)^{n^2}}{n^{n^2}}\dfrac{(n+1)^{n^2}}{n^{n^2}}\\\\ &=e^{3/2}\times e^{N-1}\times\prod_{n=2}^{N}\...


54

$$ e^x = \lim_{n\to\infty}\left(1+\frac xn\right)^n\ge1+x $$ by Bernoulli's inequality.


53

The shortest proof I could think of: $$1 + x \leq 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = e^x.$$ However, it is not completely obvious for negative $x$. Using derivatives: Take $f(x) = e^x - 1 - x$. Then $f'(x) = e^x - 1$ with $f'(x) = 0$ if and only if $x = 0$. But this is a minimum (global in this case) since $f''(0) = 1 > 0$ (the second ...


52

Yes. They are the same thing. When exponents get really really complicated, mathematicians tend to start using $\exp(\mathrm{stuff})$ instead of $e^{\mathrm{stuff}}$. For example: $e^{x^5+2^x-7}$ is kind of hard to read. So instead one might write: $\exp(x^5+2^x-7)$. Note: For those who use Maple or other computer algebra systems, e^x is not usually the ...


49

Consider the function $x^{\frac{1}{x}}$. Differentiating gives $x^{\frac{1}{x}}(\frac{1}{x^2})(1-\ln x)$, so the function attains its global maximum at $x=e$. Thus $e^{\frac{1}{e}} \geq \pi^{\frac{1}{\pi}}$, and it is clear that the inequality is strict, so $e^{\pi}>\pi^{e}$.


49

The point is that $1-\frac{1}{n}$ is less than $1$, so raising it to a large power will make it even less-er than $1$. On the other hand, $1+\frac{1}{n}$ is bigger than $1$, so raising it to a large power will make it even bigger than $1$. There's been some brouhaha in the comments about this answer. I should probably add that $(1-\epsilon(n))^n$ could go ...


47

If someone asks me, "what is $e$?" I sketch the graph of $y=1/x$, draw a line segment from $(1,1)$ on the curve down to $(1,0)$ on the $x$-axis, and ask, how far to the right do I have to draw another vertical segment to rope off an area of 1? Anyone who is familiar with the idea of graphing a function can appreciate that definition, and it's not surprising ...


46

Besides the connections between hyperbolic and circular functions which arise from substitutions involving imaginary arguments the functions can also be related using only real arguments via the Gudermannian function defined as $$\text{gd}(x)=\int_0^x\text{sech}\,t\,dt$$ This leads to identities such as $\sinh x = \tan (\text{gd}\,x)$ and $\sin x = \tanh( \...


45

$$2^{41}\gt 2^{40}=(2^5)^8\gt (3^3)^8=3^{24}$$


44

Let $f(x)$ be a differentiable function such that $f'(x)=f(x)$. This implies that the $k$-th derivative, $f^{(k)}(x)$, is also equal to $f(x)$. In particular, $f(x)$ is $C^\infty$ and we can write a Taylor expansion for $f$: $$T_f(x) = \sum_{k=0}^\infty c_k x^k.$$ Notice that the fact that $f(x)=f^{(k)}(x)$, for all $k\geq 0$, implies that the Taylor ...


43

If you know Taylor expansion: then $$e^x=1+x+\frac{x^2}{2!}+...$$ We can get (Or you may take derivative to prove it) $$e^{x}>1+x, \forall x>0$$ Then set $$x=\frac{\pi}{e}-1>0$$ We get $$e^{\frac{\pi}{e}-1}>1+\frac{\pi}{e}-1\Leftrightarrow \frac{e^{\frac{\pi}{e}}}{e}>\frac{\pi}{e}\Leftrightarrow e^{\frac{\pi}{e}}>\pi\Leftrightarrow e^{\pi}...


43

Hints. We first show that $2<\mathrm{e}<3$ (see below), and hence $\mathrm{e}$ is not an integer. Next, following up OP's thought, assuming $\mathrm{e}=a/b$, we multiply by $b!$ and we obtain $$ \sum_{k=0}^\infty \frac{b!}{k!}=a\cdot (b-1)! \tag{1} $$ The right hand side of $(1)$ is an integer. The left hand side of $(1)$ is of the form $$ \sum_{k=...


43

After a 50% loss you need a 100% gain to break even. In that scenario the arithmetic average return is 25% and the geometric average return is 0%. It is more important to maximize geometric rather than arithmetic average return -- and this is intimately connected with the concept of risk-adjusted return and mean-variance optimization. Given a set of ...


42

We can confine attention to $b \ge 1$. This is because, if $0<b<1$, then $n^b \le n$. If we can prove that $n/a^n$ approaches $0$, it will follow that $n^b/a^n$ approaches $0$ for any positive $b\le 1$. So from now on we take $b \ge 1$. Now look at $n^b/a^n$, and take the $b$-th root. We get $$\frac{n}{(a^{1/b})^n}$$ or equivalently $$\frac{n}{c^n}$...


41

The number $e$ has many different characterizations. The word "characterization" has a precise meaning in mathematics. An exercise in a textbook may say: (a) Prove that $X$ is enormously purple but not largely purple. (b) Prove that the property of being enormously purple but not largely purple characterizes $X$. The student is expected to understand ...


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