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## New answers tagged inverse-function

1

Using $\arcsin x=\arctan\dfrac x{\sqrt{1-x^2}}$ for $-1<x<1$ $$(2\sqrt2+i)(\sqrt2+i)^2=(2\sqrt2+i)(1+2\sqrt2i)=i(1+8)$$ Take argument in both sides See also: atan2

4

$$\sin x=\frac13\implies \tan x=\frac{\frac13}{\sqrt{1-(\frac13)^2}}=\frac1{2\sqrt2}\implies x=\arctan\frac1{2\sqrt2}\tag1$$ $$\tan(y/2)=\frac1{\sqrt2}\implies\tan y=\frac{2\frac1{\sqrt2}}{1-(\frac1{\sqrt2})^2}=2\sqrt2\implies y=\arctan(2\sqrt2)\tag2$$ $$(1)\& (2) \implies x+y=\frac\pi2.$$ Somewhat shorter aproach: $$\cos y=\frac {1-\tan^2\frac y2}{... 3 It's enough to prove that$$\cos\left(\arcsin|x|+\arcsin|y|\right)<\cos\arcsin\left|\frac{x+y}{1+xy}\right|$$or$$\sqrt{(1-x^2)(1-y^2)}-|xy|<\sqrt{1-\left(\frac{x+y}{1+xy}\right)^2}$$or$$|xy|>\sqrt{(1-x^2)(1-y^2)}-\frac{\sqrt{(1-x^2)(1-y^2)}}{1+xy}$$or$$|xy|(1+xy)>xy\sqrt{(1-x^2)(1-y^2)},$$for which it's enough to prove that$$(1+xy)^2>(...

0

In practice, the first efficient formula was John Machin's, discovered in 1706: $$\frac \pi 4=4\arctan\frac 15-\arctan\frac 1{239}.$$ When you expand the series defining $\arctan x$ with these values up to the $n^\text{th}$ term, you obtain $15$ exact decimal digits. Other formulæ in a similar vein were found out later by various mathematicians (Euler, ...

1

To make use of Leibniz formula you need to express function $\arctan x$ in the form of power series. That is $$\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \dots = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}$$ Here I show how it is derived, but you don't know this to use it. You have, for $|x|<1$: $$(\arctan x)' = \frac{1}{1+... 2 Your function f(x)=\frac{x}{x^2+1} is of course continuous, since it is a ratio of polynomials and the denominator has no real roots. If a continuous function is a bijection (i.e. it has an inverse), then it must be monotonic (see this question for a proof). But f(x) is clearly not monotonic, since f(0)<f(1) but f(2)<f(1). 0 So consider the function$$ f(x,y) = (2x-y, x- 2y) $$It is a function of two inputs and two outputs. Let's write the outputs as x_{\text{out}} and y_{\text{out}}, so we have$$ \tag{1} \left\{ \begin{array}{ccc} x_{\text{out}} &=& 2x&-y \\ y_{\text{out}} &=& x & -2y \end{array} \right. $$In order to invert the process, we assume ... 0 To give a function is the same to give the sets where the function is constant: c\mapsto f^{-1}(c). This would be a map F:\mathbb{R}\to \text{power}(\mathbb{R}). The function F is almost injective (since f^{-1}(c)\cap f^{-1}(c')=\emptyset for c\neq c'). It might not be injective if f^{-1}(c)=f^{-1}(c')=\emptyset, that is, if c and c' are not ... 0 It is possible to have a set-valued function just like your f. The only 'but' is that it is not a function f: \mathbb R\to \mathbb R but actually f:\mathbb R\to\mathcal P(\mathbb R), i.e. its target is the set of parts (or power set) of \mathbb R. What you are actually describing by your f is the level sets of g. Any more I would add is ... 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) = ... 0 It is always true. Technically a function is not invertible except on certain intervals. f:X->Y is invertibel iff f is a bijection between X and Y. A bijection is into(injective) and onto(surjective). Into means that for every a and b in X, f(a)=f(b) necessarily implies a=b. f is onto iff for every y \in Y \exists x \in X that f(x)... 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]\...

4

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

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

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)}...

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 $$x_{1,2}=-\frac{1}{2y}\pm \sqrt{\frac{1}{4y^2}+1}$$

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 $f(r):=\arctan(2r^2+2r+1)... 0 Like Inverse trigonometric function identity doubt:$\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when$x<0$,$y<0$, and$xy>1, $$\tan^{-1}\sqrt{\dfrac{x(x+y+z)}{yz}}+\tan^{-1}\sqrt{\dfrac{y(x+y+z)}{zx}}$$ $$=\begin{cases} \tan^{-1}\left(\dfrac{\sqrt{\dfrac{x(x+y+z)}{yz}}+\sqrt{\dfrac{y(x+y+z)}{zx}}}{1-\sqrt{\dfrac{x(x+y+z)... 0 Hint:-$$ \tan^{-1}a + \tan^{-1}b + \tan^{-1}c= \pi$$Only and only if$$a+b+c=abc$$0 Use https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values, -\dfrac\pi2\le \tan^{-1}a\le\dfrac\pi2 Now \sqrt b\ge0 for real \sqrt b \implies0\le\tan^{-1}\sqrt b\le\dfrac\pi2 So, here the sum will lie in \in[0,3\pi/2] Now the sum will be =0 only if each term under is individually =0 i.e. if x+y+z=0 Otherwise ... 3 Let x, y and z be positive numbers. We consider a triangle ABC with side lengths a=BC=y+z, b=CA=x+z and c=AB=x+y. The semi-perimeter s=x+y+z inradius r. Now, by Heron’s formula we have$$\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}}}$$So,$$\... 0 Inverse functions are symmetric with respect to the liney=x$. They don't necessarily contain a point such as$(1,1)$on that line, but if they do (e.g.,$y=x^2, x^3, ...$), then they intersect there. In fact,$y=-x^3$intersects its inverse at$(0,0)$on that line. 0 Consider the curve$y=1-x$. It's inverse is$y=1-x$, i.e. it is self inverse. This means it intersects all along its curve, despite only intersecting$y=x$once. Now suppose a curve$y=f(x)$intersects the line$y=x$at$x_0$. This means that $$y_0=f(x_0)=x_0.$$ Applying$f$to both sides yields $$f(y_0)=f(x_0)=x_0,$$ and hence the inverse of the curve ... 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 ... 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 ...

0

Not an answer to the question asked This doesn't provide a function $s$ such that it's sigmoid, but $f(x) = x + s(x)$ is easily invertible. But it still may be of use: Have you considered an alternative like rejection sampling? That doesn't require the inverse. For something like a sigmoid, this should result in only about a 50% rejection rate, which isn't ...

0

I will denote the function by $\Phi$ and make some remarks. 1) It is better to go for $u\in(0,1)$. This because $\Phi(u)\in\mathbb R$ for every $u\in(0,1)$ which is not necessarily true if $u=0$ or $u=1$. 2) It is not necessary to demand that $F$ is continuous. 3) Characteristic is the relation:$$F(x)\geq u\iff x\geq \Phi(u)$$Based on that relation it ...

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 $\frac{\pi}{2}$

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 $$\... 2 Since \tan x is periodic, any function of the form f(\tan x) is also periodic, so isn't the identity function. Consider \arctan\tan\frac{5\pi}{4}=\frac{\pi}{4}. 1 \arctan(\tan x)=x \mbox{ }\ \ \ if -\dfrac{\pi}2<x<\dfrac{\pi}2. Otherwise its not. 2 Let me call s_x, s_h the sign of x,h.$$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 \...

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 $g'(1)\frac{1}{2}=1$ $g'(1)=2$.

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,...

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