TITLE: Hyperbolic trigonometric functions identity
QUESTION [1 upvotes]: $$\cosh(\sinh^{-1}x) = \sqrt{x^{2}+1}$$
I used the fact that $\cosh(x) = \frac{1}{2}(e^{x}+e^{-x})$
and that $\sinh^{-1}(x) = \ln(x + \sqrt{x^{2}+1})$
Eventually I simplified to
$$\frac{x^{2}+x\sqrt{x^{2}+1}+1}{x+ \sqrt{x^{2}+1}}$$
I verified and so far I am correct, but I can't see how I can simplify further.

REPLY [4 votes]: Note that the top is $\sqrt{x^2+1}\left(x+\sqrt{x^2+1}\right)$.
Remark: You may have worked too hard by first obtaining a formula for $\sinh^{-1}x$, though such a formula is in fact useful.
For we have $\cosh^2 t-\sin^2 t=1$ for all $t$. Let $t=\sinh^{-1}(x)$. We get $\cosh^2 t=1+x^2$. But $\cosh t$ is always positive, so $\cosh t=\sqrt{1+x^2}$.


TITLE: Prove inequality $\sum_{i=1}^n\left(\frac{n-1}{n}\right)^{\lceil n\log n+ cn\rceil}\leq n e^{-(n\log(n)+cn)/n}$
QUESTION [0 upvotes]: Can anyone help me see why this inequality holds?
$$\sum_{i=1}^n\left(\frac{n-1}{n}\right)^{\lceil n\log n+ cn\rceil}\leq n e^{-(n\log(n)+cn)/n}$$
Any help is appreciated

REPLY [1 votes]: Because $\; e^x\ge 1+x\ge 0 \;$ for $\; x\ge-1 \; \;$ $\textbf{(*)}$
$ e^{rx}\ge (1+x)^r \; $ for $\; r\ge 0  $ $\\$
Using the inequality gives us for $x=\frac{-1}{n} \;$ and $\; r=\lceil n\log n+ cn\rceil\ge 0 \; \;$  (I assumed $c\ge 0$ here)
$$\sum_{i=1}^n\left(\frac{n-1}{n}\right)^{\lceil n\log n+ cn\rceil}=\sum_{i=1}^n\left(1-\frac{1}{n}\right)^{\lceil n\log n+ cn\rceil}\leq\sum_{i=1}^ne^{-(\lceil n\log n+ cn\rceil)/n}\\$$
$$\le\sum_{i=1}^ne^{-( n\log n+ cn)/n}= ne^{-( n\log n+ cn)/n}$$
$\textbf{(*):}$ $\;$ For $\; x=0 \; \;$ and $\; x=-1 \; $ the inequality is true. $ \; \; \frac{\partial [e^x-(1+x)]}{\partial x}\ge 0 \;$ for  $\; x\ge 0 \; \;$ and $\; \; \frac{\partial [e^x-(1+x)]}{\partial x}\le 0 \; $ for  $\; x\le 0 \; \;$


TITLE: What does it mean for a point to be "inside" a plane curve?
QUESTION [3 upvotes]: Suppose that we have a simple closed curve in the $xy$-plane: $f(x,y)=0$ (for example, $f(x,y) = x^2 + y^2 - 4$. If I give you the point $(x,y)$, is there an analytical method to determine whether $(x,y)$ is "inside" $f$? (besides looking at a graph, etc.)
That raises the following question: what exactly does it mean to be "in" a curve? The closest definition my friends and I achieved involved a new space where the line integral of the curve from a given point to a second point is mapped to the distance from that second point to $(x,y)$.
Of course, I feel like this must have been solved a long time ago, and my explanation was rather naïve. Still, I'm having a hard time navigating Wikipedia and the such (MRH).
What does it mean to be inside a simple closed curve?

REPLY [1 votes]: Assume your simple closed curve is the zero set of a function
$$ C: f(x,y) = 0$$
and $f$ also takes positive and negative values. (thanks @Federico Poloni for drawing attention to the need of this last condition).
Now the complement of the curve has two connected components by Jordan's theorem so none of these components can intersects both of the non-void sets $\{f<0\}$ and $\{f>0\}$. It follows that the components of $\mathbb{R}^2 \backslash C$ are  $\{f<0\}$  and $\{f>0\}$. Now you have to figure out what is the bounded component (the inside). Look at the sign of $f$ for large values of $(x,y)$ ( at infinity). The inside is the region of opposite sign.
Inside means not with the infinite.