New answers tagged

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 line $y=x$. 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,...


Top 50 recent answers are included