# Tag Info

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 meansf(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 \Rightarrowx, y\leq n/2$; (ii)$q=1 \Rightarrowx> n/2$and$y\leq n/2$; (iii)$q=2 \Rightarrowx\leq n/2\$ ...

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