Tag Info

Hot answers tagged

24

A function is smooth is it has derivatives of infinite order. $f(x) = x$ is smooth because it has infinitely many derivatives which are all 0, except for the first one. Polynomials are smooth because eventually their derivatives are 0.


20

Yes, the identity function has derivatives of every finite order, and is therefore smooth. It doesn't matter that most of the derivatives are $0$ everywhere -- being $0$ is a perfectly cromulent way to exist.


11

$g(n)=n+2$ will do the trick. Since this $g$ is injective, $f$ must be injective too, so under $f$ each natural number will have one successor and zero or one predecessor. Also there are exactly two numbers that have no pre-predecessor, and that is only possible if there is exactly one number with no predecessor. There also can be no cycles in the action ...


9

The source is likely referring to the English word "homeomorphic" and has nothing to do with mathematics. Notice how closely the definitions map to dictionary.com's homeomorphism: noun similarity in crystalline form but not necessarily in chemical composition. Mathematics. a function between two topological spaces that is continuous, ...


9

b) is incorrect. Consider the function $f(x) = \tan^{-1}(x)$. c) is incorrect. Consider the function $f(x) = \sin(x)$.


7

Probably the most standard term of the sort you're looking for is "equipotent", though it isn't used particularly often (more often, people will just say two sets "have the same cardinality" or "are in bijection"). People who think about categories a lot sometimes say "isomorphic as sets".


7

For each number $x\in[1,\phi)$, let either $$f(x)=\phi x,f(-x)=-\phi x\text{ or else }\\f(x)=-x/\phi,f(-x)=x/\phi$$ The choice is independent for each $x$ in that interval. Then, for all other numbers, $f(x\phi^k)=f(x)\phi^k$. For each ladder $x\phi^{-2},x\phi^{-1},x,x\phi,x\phi^2,...$, $f$ consistently shifts numbers up the scale, or down the scale.


7

Claim: $f(0)=0$. Assume $f(0) = k\neq 0 $, then $f(k)=f(k)+k$, a contradiction. Claim: If $f(x)$ is an analytic function, $f(x)=\phi x$ or $f(x)=-x/\phi$. If $f$ is an analytic function, it has a unique polynomial expansion with zero constant term, i.e. $f(x) = a_1x+a_2x^2 +\cdots$. Then by comparing powers we get that: $$a_1=\phi, -1/\phi$$ And ...


7

$$\implies x^2(y-1)+x(3-3y)+4y+4=0$$ The discriminant $$=(3-3y)^2-4(y-1)(4y+4)=-(y-1)(7y+25)$$ which needs to be $\ge0$ Now $(x-a)(x-b)\le0, a\le b\implies a\le x\le b$ Alternatively, $$\dfrac{x^2-3x-4}{x^2-3x+4}=1+\dfrac{x^2-3x-4}{x^2-3x+4}-1=1-\dfrac8{x^2-3x+4}$$ Now $x^2-3x+4=\dfrac{4x^2-12x+16}4=\dfrac{(2x-3)^2+7}4$ Now ...


7

$$\begin{align}\frac{\sin 80^\circ \sin 20^\circ\sin 100^\circ }{\sin 50^\circ}&=\frac{\sin 80^\circ \sin 20^\circ\cdot 2\sin 50^\circ \cos 50^\circ }{\sin 50^\circ}\\&=2\sin 80^\circ\sin 20^\circ \cos 50^\circ\\&=2\left(-\frac 12(\cos 100^\circ-\cos 60^\circ)\right)\cos 50^\circ\\&=(\cos 60^\circ-\cos 100^\circ)\cos 50^\circ\\&=\frac ...


6

Two topological spaces are homeomorphic iff there exists a mapping as described in (2) between them, a continuous 1-1 map $f$ from one onto the other space whose inverse $f^{-1}$ is also continuous. Because of this condition, both $f$ and $f^{-1}$ are called homeomorphisms, i.e. maps that preserve the underlying topological structure of a space. Thus the ...


5

You may be having issues with the difference between existence and triviality. If $f(x)=x$ then $f(x)=x$ is continuouss $f'(x)=1$ is continuous $f''(x)=0$ is continuous $f'''(x)=0$ is continuous etc... So all its derivatives are continuous. On the other hand, take $g(x)=x\times|x|$ $g(x)=x\times|x|$ is continuous $g'(x)=\frac{|x|}2$ is continuous ...


5

No. Let $f: x \mapsto x$ and $g: x \mapsto 2x$ on $\mathbb{R}$. Then $f,g$ are bijective and $f(0) = g(0) = 0$. But $f \neq g$.


5

For the sake of simplification, set $a=\log (x+3)$ and $b=\log (5x+9)$ then the equation you obtained writes as $2a^2-3ab+b^2=0$. Using standard factoring this is $(a-b)(2a-b)=0$. I think you can do the rest!


5

I assume your friend and you both know that the expression $(-1)^x$ only makes sense if $x$ is an integer. That said, you have $$(-1)^{(-n)} = \frac{1}{(-1)^n} = \left(\frac{1}{-1}\right)^n = (-1)^n$$ I believe all steps only use equalities you learnt in school: For all $a, b$ we have $a^{-b} = \frac{1}{a^b}$ If $\frac{c}{d}$ is a fraction, then ...


5

Denote $g(x)=\frac{x-1}{x+1}$. Note that $g(g(x))=-\frac{1}{x}$ and hence $g(g(g(g(x))))=x$ is the identity function. Now can you compute $f(x+12)$ in terms of $f(x)$ and then compute the desired term?


4

If $f\circ g=1_X$, then you know that $f$ is surjective and $g$ is injective. Aslo, on finite sets, you know that a function is surjective if and only if it is injective... This does not hold for infinite sets, however. For example, define $g(n)=n+1$ for all values $n\in\mathbb N$. Also, define $f(n)=n-1$ for all values $n\in\mathbb N$ except $1$, and ...


4

Hint. I would begin by constructing the function explicitly at the lower end. You may find that it is fairly well constrained by the fact of being strictly increasing. For instance, we know that $f(1) = 2$: It cannot be $1$, because then $f(f(1)) = f(1) = 1 \not= 3$, and it cannot be $3$, because then $f(f(1)) = f(3) > f(1) = 3$. Therefore $f(2) = 3$. ...


3

HINT: $$z\cdot S=z\sum_{k=0}^\infty \frac{2^{2k}z^{2k-1}}{(2k)!}=z\sum_{k=0}^\infty\dfrac{(2z)^{2k}}{(2k)!}$$ Now $e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$ $2\cosh(y)=e^y+e^{-y}=?$


3

Substituting $x = -4$ means all instances of $x$ should be swapped with $-4$. That means $$ f(\color{red}{x}+5) = (\color{red}{x} - 1)^2\\ f(\color{red}{-4}+5) = (\color{red}{-4}-1)^2\\ f(1) = (-5)^2 = 25 $$


3

A fiber of a function $f:A \to B $ is $\{ c \in A : f(c)=f(a) \}$ for any $a \in A$. In your example the SET $ \{... -3,2,7,12,... \}$ is a fiber.


3

We say "$A$ injects into $B$", and may write $f: A \hookrightarrow B$.


3

Here is a specific example. Take $$ f(x) = \left\{\begin{array}{ll} \arctan(x) & \mbox{if $x \neq \pm 1$} \\ \pi/2 & \mbox{if $x = -1$} \\ -\pi/2 & \mbox{if $x = 1$} \end{array}\right. $$


3

Certainly not: if $\lvert f(x)\rvert<1$, on the contrary $\lvert f(x)\rvert>\lvert f^2(x)\rvert$. Same answer for any power $>1$. (I supposed you meant the square of the function, not the function composed with itself). It's the inverse situation for roots: if $\lvert f(x)\rvert<1$,$\lvert \sqrt{\lvert f(x)\rvert}>\lvert f(x)\rvert$.


2

Yes, you are correct. We can "make" a linear transformation onto by restricting the codomain to the image of the transformation.


2

This question is badly defined when you think about it. You can specify a periodic function simply by giving the function on the period and specifying it's periodically continued. Nobody says every function has to be given in the way you are used to from school (a sum or product of existing functions). So, it's perfectly reasonable to turn any function ...


2

This function has period 2. It equals $\lfloor x\rfloor\pmod2$ $$f(x)=\left\{\begin{array}{l}0,\,\, 2n\leq x<2n+1\\1,\,\, 2n+1\leq x<2n+2\end{array}\right.$$


2

It is not even possible to have $f$ continuous on all of $\mathbb C^2$. That is because the condition requires $f(0,1)\neq 0$ and $f(1/n,1)=0$ for all natural $n$.


2

It’s the second part of the proof that’s wrong. If $y\in F[A]\cap F[B]$, there is an $x_A\in A$ such that $F(x_A)=y$, and there is an $x_B\in B$ such that $f(x_B)=y$, but there’s no reason to think that $x_A=x_B$. For a simple example, let $X=\{0,1\}$, and define $F:X\to X$ by $F(0)=F(1)=0$. Let $A=\{0\}$ and $B=\{1\}$; then $F[A\cap ...



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