Stack Exchange Network

Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Visit Stack Exchange
5

This is a multivalued function (see especially the first example!), or multifunction, or set-valued function. A set-valued map, taking elements of $X$ and producing subsets of $Y$, is often denoted $f : X \rightrightarrows Y$. It can also be denoted more literally by $f : X \to 2^Y$, as such maps can be thought of as (ordinary, single-valued) functions from ...


5

It is common to use the notation $f^{-1}(A)$, where $A$ is a subset of the value range of $f$, as a shorthand to describe the set $\{x : f(x) \in A \}$. Furthermore, if $A$ is a set with only one value $x$ it is also somewhat common to just write $f^{-1}(x)$ instead of $f^{-1}(\{x\})$. So if in your case the context is clear, it is fine to write $g^{-1}(4) = ...


4

The usual method to obtain the inverse is to let $F(x)=y$, interchange $x$ and $y$, and solve for $y$. Thus we solve $$x=\frac{y}{1-y^2}$$ for $y$: \begin{align*} x&=\frac{y}{1-y^2}\\ \implies x-xy^2&=y\\ \implies xy^2+y-x&=0\\ \implies y^2+\tfrac1xy-1&=0\\ \implies(y+\tfrac1{2x})^2&=\frac{1+4x^2}{4x^2}\qquad\text{(completing the square)}...


4

The hint. Use $$\arctan(n+1)-\arctan{n}=arccot(n^2+n+1)$$ and the telescopic summation.


3

Yes, it is always true. It turns out that $\sin$ is not invertible. The function which is denoted by $\sin^{-1}$ (or, more generally, by $\arcsin$) is the inverse of the restriction of $\sin$ to $\left[-\frac\pi2,\frac\pi2\right]$. And so$$\left(\forall x\in\left[-\frac\pi2,\frac\pi2\right]\right):\arcsin\bigl(\sin(x)\bigr)=x$$and$$\bigl(\forall x\in[-1,1]\...


3

I'll leave it to you to prove by induction that the partial sum is $\arctan\left(1-\frac{1}{n^2+n+1}\right)$, so the limit is $\pi/4$. One approach to obtaining this partial sum is that of @achillehui's telescope, viz. $$\left[\arctan(2r^2+2r+1)\right]_{-1}^n=\arctan\frac{n^2+n}{n^2+n+1}.$$ Edit: just to spell it out, the definition $f(r):=\arctan(2r^2+2r+1)...


3

Let $x$, $y$ and $z$ be positive numbers. We consider a triangle $ABC$ with side lengths $a=BC=y+z$, $b=CA=x+z$ and $c=AB=x+y$. The semi-perimeter $s=x+y+z$ inradius $r$. Now, by Heron’s formula we have $$\eqalign{\cot(A/2)&=\frac{s-a}{r}=\frac{s(s-a)}{{\rm Area}(ABC)}=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}\cr &=\sqrt{\frac{x(x+y+z)}{yz}}}$$ So, $$\...


3

This means $f(0)=1$ or $g(1)=0$. Since $f(x)$ is a bijection $f: R \rightarrow R$ its inverse exists, here it is called $g(x)$. So $$f(g(x))=x \Rightarrow f'(g(x)) g'(x)=1 \Rightarrow g'(1)=\frac{1}{f'(g(1))}= \frac{1}{f'(0)}=2.$$ Note Here the inverse exists but it is not obtainable. Let $y=f(x)$, you can find $g'(y_0)$ only at good points: $(x_0,y_0): (0,...


3

I started by considering the graph of $y=_nC_x$ - illustrated above. I then looked at the natural logarithm - after all, Stirling's formula gives factorials as powers. This curve looks like a quadratic, so I tried to fit a quadratic of the form $y=x(n-x)$. You can see that this is not quite right. I therefore tried variations of the form $y=\left(x(n-x)\...


2

Considering the three different cases $$X = \frac{n! }{ k!(n-k)! }=\frac{ n!}{\Gamma(k+1)\, \Gamma(n+1-k)}$$ $$ X =\frac{n! }{ (n-k)! } =\frac{ n!}{\Gamma(n+1-k)}$$ $$X= \frac{ (n+k-1)! }{ k!(n-1)! }=\frac{ \Gamma(n+k)}{\Gamma(k+1)\, n!}$$ I should look, in the real domain, for the zero of functions $$F(k)=\log (\Gamma (k+1))+\log (\Gamma (n+1-k))-\log \left(...


2

hint Using the fact that $$(\forall x>0) \;\; (g\circ f)(x)=x$$ we get by differentiation $$g'(f(x))f'(x)=1=g'(f(x))(2+\frac 1x)$$ and with $x=1$, $$3g'(2)=1$$


2

Let me call $s_x, s_h$ the sign of $x,h$. $$h = s_x(\sqrt{|x| +1 } -1) + \epsilon x$$ Then $$s_x = s_h \Rightarrow s_h h = s_x (s_x(\sqrt{|x| +1 } -1) + \epsilon x) \Rightarrow \\ |h| = \sqrt{|x| +1 } -1 + \epsilon |x| \Rightarrow |h|+1 = \sqrt{|x| +1 } + \epsilon |x|$$ $$|h|+1-\epsilon |x| = \sqrt{|x|+1} \Rightarrow (|h|+1-\epsilon |x|)^2 = |x|+1 \...


2

That is because $\tan x$ isn't a bijective function. To obtain a bijective function one has to consider its restriction to some relevant interval on which it becomes a bijection – in practice the interval $(-\tfrac\pi2,\tfrac\pi 2)$. So by definition $$y=\arctan x\iff \tan y =x\quad\textbf{and}\quad -\tfrac\pi 2<y<\tfrac\pi 2$$ Thus, we have $$\...


2

Since $\tan x$ is periodic, any function of the form $f(\tan x)$ is also periodic, so isn't the identity function. Consider $\arctan\tan\frac{5\pi}{4}=\frac{\pi}{4}$.


2

To invert the function $y=F(x)$, you need to solve for $y$ in the equation $x=F(y)$. So solve for $y$ in $$ x = \frac{y}{1-y^2} \implies 0 = y^2 x + y - x \implies y = F^{-1}(x) = \frac{-1 \pm \sqrt{1 + 4x^2}}{2x} $$ where the last equality is quadratic formula


2

We have $$y=\frac{x}{1-x^2}$$ so we get $$y-yx^2=x$$ or $$x^2+\frac{x}{y}-1=0$$ using the quadratic formula we get $$x_{1,2}=-\frac{1}{2y}\pm \sqrt{\frac{1}{4y^2}+1}$$


2

Always keep in mind the principal values of https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions $-\dfrac1{\sqrt2} <x<\dfrac1{\sqrt2}$ actually implies $\dfrac{3\pi}4>\arccos x=y>\dfrac\pi4$ as $\arccos(x)$ is decreasing in $[-\dfrac\pi2,\dfrac\pi2]$ $\implies?>2y>\dfrac\pi2$ But $-\dfrac\pi2\le u=\arcsin(\sin2y)\le\dfrac\...


2

As you want to compute the inverse, let us start from that end: Consider $$g(x)=\begin{cases}3x^5-10x^3+30x&-1\le x\le 1\\ 15x+8&x\ge1\\ 15x-8&x\le -1\end{cases} $$ Then $g$ is $C^2$ and has no critical points, hence has a $C^2$ inverse $f$ that looks like a line plus a sigmoidal. To match your demands for $x\ll 0$ and $x\gg 0$, we consider a ...


1

Put $f(x)=\exp\cot^{-1}\cos x$ and $g(x)=\exp\cot^{-1}\sin x$. $$\int_{\pi/2}^{5\pi/2}\frac{f(x)\,dx}{f(x)+g(x)}=\int_0^{2\pi}\frac{f(x)\,dx}{f(x)+g(x)}=\int_0^{2\pi}\frac{g(x)\,dx}{f(x)+g(x)}$$ (the first equality holds because of $2\pi$-periodicity of the integrand; the second is obtained by substituting $x=5\pi/2-y$ (and replacing $y$ by $x$) in the first ...


1

Here are the graphs of $\arctan(\tan(x))$(green) and $x$(blue). Notice that they only overlap in the region $\frac{-\pi}{2}$ and $\frac{\pi}{2}$


1

$\arctan(\tan x)=x$ $\mbox{ }\ \ \ $if $-\dfrac{\pi}2<x<\dfrac{\pi}2$. Otherwise its not.


1

Derivative of inverse function : $g(f(y))=f^{-1}\circ f (y)=y$ so that chain rule is $$ g'(f(y)) f'(y) =1 $$ Hence $$ f(0)=1,\ g'(1)=g'(f(0))= \frac{1}{f'(0)} =2$$


1

We know $g$ is inverse of $f$ so we have $g(f(x))=x$ . Differentiating we have $g'(f(x)).f'(x)=1$ we want $g'(1)$ so in above equation we want $f(x)=1$. Observation suggests that $f(x)=1$ at $x=0$ thus $g'(f(0)).f'(0)=1$ thus $g'(1)\frac{1}{2}=1 $ $g'(1)=2$.


1

Let us construct the inverse function. First, fix the parameter $ q=\lfloor (z'-1)/(n/4)\rfloor $. Analyzing the terms $\Big(1.\mathbf{1}[n/2<x] + 2\cdot \mathbf{1}[n/2<y]\Big)$ on the definition of $z'$, we conclude that: (i) $q=0 \Rightarrow$ $x, y\leq n/2$; (ii) $q=1 \Rightarrow$ $x> n/2$ and $y\leq n/2$; (iii) $q=2 \Rightarrow$ $x\leq n/2$ ...


Only top voted, non community-wiki answers of a minimum length are eligible