4 votes

Why write codomain instead of image when defining a function.

Here is an easy example. If $A$ is a set, then $\mathcal P(A)$ is the power set of $A$, i.e. $\{B\mid B\subseteq A\}$. In many contexts, we want to identify $\mathcal P(A)$ with $2^A$, that is the set ...
Asaf Karagila's user avatar
  • 390k
4 votes

How to prove that $f:B\to \mathbb{N}$ by $f(2^k\cdot 3^n)=nk$ with $n,k\in\mathbb{N}$ is a function.

Uniqueness of prime factorization...
J.D.'s user avatar
  • 81
3 votes

There exists a function that satisfies $\sum_{n=1}^\infty |f^{[n]}(x) - f^{[n]}(y)| < \infty$ but is not a contraction?

let $$ f(x) =\begin{cases} 2x + 1 & 0 \le x < 1 \\ 0 & \text{ otherwise} \end{cases} $$ That's not a contraction (consider $x = 0, 0.5$ being sent to $1, 2$), but $f^{[2]}$ is everywhere $0$...
John Hughes's user avatar
  • 92.5k
3 votes

An Function that turns $\mathbb Z$ into $\mathbb Q$

Suppose $f:\Bbb Z\to\Bbb Q$ is monotonically increasing. If $f$ is constant, then its range is certainly not the whole of $\Bbb Q$. So suppose $f$ is not constant. Then there exists $n\in\Bbb Z$ such ...
TonyK's user avatar
  • 63.4k
3 votes

If $f(x)=\mu x-\sin x, x>0$ an is monotonically increasing function. Then find $\mu$

You found correctly that $f'(x) > 0 \iff \mu > \cos x$. If $f$ is monotonically increasing, then $\mu > \cos x$ for all real $x$. This is only true if $\mu > 1$. It's also true that $$\mu &...
aschepler's user avatar
  • 8,961
3 votes
Accepted

If $f(x)=\mu x-\sin x, x>0$ an is monotonically increasing function. Then find $\mu$

I think that the problem is that you want $$\mu-\cos x>0\quad \forall x$$ and in order to ensure that, you need to use the largest possible value of $\cos x$ (rather than the smallest). My ...
Patricio's user avatar
  • 1,596
2 votes

Why write codomain instead of image when defining a function.

I am not 100% sure if I am going to provide the example you wanted, but let me try. A couple of days ago, I was reminded of Cantor's diagonal argument. It's a well-known proof of the fact that there ...
Dmitry Ivanov's user avatar
2 votes
Accepted

$|f(x)-f(y)| = 2 |x - y| $, find such functions

You didn't prove that the derivative exists. You're right that $\frac{f(x)-f(c)}{x-c} = \frac{\pm 2|x-c|}{x-c}$ but the sign depends on $x$ and $c$. If the derivative exists then it must be equal to $...
Jakobian's user avatar
  • 8,932
2 votes
Accepted

Prove that Gradient points to Steepest Ascent

By the chain rule \begin{align} \frac{\Bbb d}{\Bbb dt}\phi_v(t)\vert_{t=0} &= \nabla f(x)\cdot \frac{\Bbb d}{\Bbb dt}(x+tv)\vert_{t=0} \\ &= \nabla f(x)\cdot v \end{align} So in magnitude we ...
podiki's user avatar
  • 2,011
2 votes
Accepted

Continuous and increasing in every variable does not imply continuous?

It does. For example, consider function on $[0, 1] \times [0, 1]$ that is continuous increasing in both $x$ and $y$, and $f(0, 0) = 0$. We want to show that $\forall \epsilon > 0 \exists \delta >...
mihaild's user avatar
  • 14.9k
2 votes
Accepted

Mapping from set $A$ surjectively onto itself which is not injective.

I assume you mean, can we have a surjection $A\to A$ that is not injective. Then the answer is yes. By counting reasons, you need that $A$ is infinite, but then it is always possible. For a simple ...
SomeCallMeTim's user avatar
1 vote

How do you do this question on powersets and surjective functions?

Hint: For each element $b$ of $B$, there is at least one element $x$ of $X$ such that $f(x)=b$, so pick one for each element of $B$ and make a subset of $X$. Can you justify that $f_p$ of that subset ...
Ross Millikan's user avatar
1 vote

Find $f(2)$ if $f(2 \sin^2(x)+3\sin(x))=1-\tan^2(x)$

Given that: $\begin{aligned} & f\left(2(\sin x)^2+3 \sin x\right)=1-(\tan x)^2 \\ & 2(\sin x)^2+3 \sin x=2 \\ & \Rightarrow 2(\sin x)^2+3 \sin x-2=0 \\ & \Rightarrow 2(\sin x)^2+4 \sin ...
Aditya Naskar's user avatar
1 vote

If $f(x)=\mu x-\sin x, x>0$ an is monotonically increasing function. Then find $\mu$

Observe that $f$ is a smooth function, thus for it to be monotonically increasing we require that $f'(x) \ge 0 \ \forall x$, in particular $f'(0) = \mu -1 \ge 0$, which gives the desired result
Aditya Dwivedi's user avatar
1 vote
Accepted

Continuity at a point versus continuity after algebraic manipulation

In mathematics, what you've done is known as "function extension." By multiplying your original function by a specific expression, you've created a new function that is continuous and ...
Nicholas Simafranca's user avatar
1 vote

Prove that if $f: \mathbb{R} \to \mathbb{R}$ is convex and bounded from above, then $f$ is constant.

The idea here is that $\ x_1\ $ and $\ x_3\ $ are fixed but arbitrary real numbers with $\ x_1\le x_3\ .$ Then for any $\ x_2\ge x_3\ $ (no matter how large) you can always express $\ x_3\ $ in the ...
lonza leggiera's user avatar
1 vote
Accepted

Prove that if $f: \mathbb{R} \to \mathbb{R}$ is convex and bounded from above, then $f$ is constant.

To address what your teacher was doing, he is fixing $x_1$ and $x_3$, and then noting that as long as $x_2 \neq x_1$, we can set $\lambda = \frac{x_3 - x_2}{x_1 - x_2}$ to get that $\lambda x_1 + (1 - ...
stillconfused's user avatar
1 vote

Prove that if $f: \mathbb{R} \to \mathbb{R}$ is convex and bounded from above, then $f$ is constant.

Let $f(x)\leqslant B$ for all $x\in\mathbb{R}$, and suppose $f(x_1)<f(x_2)$ for some $x_1,x_2\in\mathbb{R}$. After possibly replacing $f(x)$ with $f(-x)$, we can assume $x_1<x_2$. Thus the slope ...
BHT's user avatar
  • 2,193
1 vote

How can I create a function that alternates how much it adds each time?

Try the following function f(i) = 50*i + 50*FLOOR(i/2) i f(i) 0 0 1 50 2 150 3 200 4 300 5 350 6 450 7 500 8 600 .. .. i $50\,i + 50\,{\rm floor}(i/2)$
John Alexiou's user avatar
  • 13.2k
1 vote
Accepted

There exists a function that satisfies $\sum_{n=1}^\infty |f^{[n]}(x) - f^{[n]}(y)| < \infty$ but is not a contraction?

Consider the function \begin{align*} f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=\begin{cases} 2^x \pi,& x\in \mathbb{N}, \\ 1,& x\in \mathbb{R} \setminus \mathbb{N}. \end{cases} \end{align*}...
Severin Schraven's user avatar
1 vote
Accepted

Equivalent of sum of arctan(nx)/n²

$f$ is an odd function so that it suffices to investigate its behavior for $x > 0$. One can show that $$ \tag{*} - x \ln (x) - \frac x6 \le f(x) \le - x \ln (x) + 5 x $$ for $0 < x < 1/2$. ...
Martin R's user avatar
  • 108k
1 vote

Equivalent of sum of arctan(nx)/n²

If, for $x>0$, we use $$\tan ^{-1}(n x)=\frac \pi 2 -\tan ^{-1}\left(\frac{1}{n x}\right)$$ and $$\tan ^{-1}\left(\frac{1}{n x}\right)=\sum_{m=0}^\infty\frac {(-1)^m}{(2 m+1)\, x^{2 m+1}\,\, n^{2 m+...
Claude Leibovici's user avatar
1 vote
Accepted

A piecewise function is continuous from a locally compact Hausdorff to metric space.

Clearly $\omega'$ is continuous on $\{f+g > 0\}$. If $x\in \overline{\{f+g > 0\}}$, take a net $x_\alpha \in \{f+g > 0\}$ such that $x_\alpha\to x$. Then from the inequality $|\omega'(x_\...
Jakobian's user avatar
  • 8,932
1 vote

How to define this function so that it is continuous?

I found a definition with floor function that makes the function continuous on entire real line, but this definition is only disguise of a piecewise definition. $$f(x)=\frac{\arctan(2 \tan (x))+\pi \...
azerbajdzan's user avatar
  • 1,116
1 vote

How to define this function so that it is continuous?

Maybe better to leave it in its original implicit form, so that the range is continuous. Converted to its implicit function $$ \tan (x~y)= 2~\tan(x)$$ and directly plotted on ...
Narasimham's user avatar
  • 39.7k
1 vote

How to define this function so that it is continuous?

The question is about the simplest value for the multi-valued function $$ y = f(x) := \text{arctan}(2\tan(x))/x. \tag1 $$ A natural place to start is to define the two-variable function $$ g(x, y) = \...
Somos's user avatar
  • 34.3k

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