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17

This is possible if and only if $\tilde x$ is an isolated point of the domain. Example 1: For example, the domain could be $$(-\infty,\tilde x - \epsilon)\cup \{\tilde x\}\cup (\tilde x +\epsilon,\infty)$$ if it is a subset of $\mathbb R$ (which, incidentally, is not specified) and the function rule could be $$f(x) = \begin{cases} 0 & \textrm{ if }x ...


12

This function cannot exist, if it must be continuous. Suppose such a function $f:R \to R$ exists, and for some $x \in R$, $f(x) = c \ne 0$. Let $y \ne x$. By the Intermediate Value Theorem, then $\exists z \in [x, y]$ (or $[y, x]$) such that $0 < f(z) < c$.


7

Good question. The functions $\frac{x^3-8}{x-2}$ and $x^2+2x+4$ are equal for all $x$ except at $x=2$, where the first function is undefined, and the second function is $12$. However, if you are taking the limit of the first function as $x \to 2$, all you care about are the values of the function at $x$ near $2$; the value of $x$ at $2$ is irrelevant when ...


5

The difference is that $\mapsto$ denotes the function itself. Thus you need not name the function. $a\mapsto b$ fully describes the action of the function. $\to$, on the other hand, describes only the domain and codomain. Thus one might say $x\mapsto x+1$ is equivalent to $f(x)=x+1$, in which case we would say $f:\mathbb{R}\to\mathbb{R}$.


5

The $\to$ arrow points from the domain of a function to its codomain or target set. The $\mapsto$ arrow (fittingly called \mapsto in TeX) shows what an individual element of the domain will be mapped to, i.e. it shows what the function does while $\to$ shows where it operates (so to say). So, the elaborate way to write a function definition is ...


5

The answer is no. Take a look at this paper. The corollary on p. 353 ensures that for any two disjoint countable dense subsets $A, B \subset \mathbb{R}$, there exists an everywhere differentiable function $H$ such that $H'>0$ on $A$ and $H'<0$ on $B$. Note that this function is monotone on no subinterval of $\mathbb{R}$. I should give credit to user ...


5

We say $f:A\to B$ is a function if, for any $a\in A$ there exists exactly one $b\in B$ such that $f(a)=b$. Since $\infty$ is not an element of $\mathbb Z^+$, if you want $f:\mathbb Z^+\to \mathbb Z^+$, you can't have $f(n)=\infty$ for any positive integer $n$.


4

The following is maybe too fancy. Let $a=-1$ and $b=1$. Let $$f(x)= \begin{cases} x^2\sin(1/x) & \text{if } x \ne 0, \\ 0 & \text{if } x=0. \end{cases}$$ The derivative of this function exists at $0$, but is not continuous there. If you want greater smoothness at $0$, replace $x^2$ by $x^{77}$ If you want the same sort of thing on a general ...


4

Letting $t=2^{1/3}\Rightarrow t^3=2$, we have $$(x-2)^3=(t^2+t)^3$$$$\Rightarrow x^3-6x^2+12x-8=t^6+3t^5+3t^4+t^3=4+6t^2+6t+2.$$ Then, we have $$\begin{align}x^3-6x^2+6x&=(6t^2+6t+6)+8-6x\\&=6t^2+6t+14-6(t^2+t+2)\\&=14-12\\&=2.\end{align}$$


4

You are right, $f$ and $g$ are different functions since they don't have the same domain. And, yes, $f$ is not continuous at $2$ because it is not defined at $2$. If we talk about limits then you can use the fact that if $f(x) = g(x)$ except at $x=a$, then $$ \lim_{x\to a} f(x) = \lim_{x\to a} g(x). $$ So now you note that for any $x\neq 2$, you indeed ...


4

There is no continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(x)=0$ for all $x\not=x_0$ and $f(x_0)\not=0$, since by definition of continuity $$f(x_0)=\lim_{x\rightarrow x_0,x\not=x_0} f(x)=0$$


3

As @Josh Keneda points out, a characteristic function won't work in general if $L$ is infinite. But we can use the following slight modification: $$ f(x) := \sum_{n=1}^\infty \frac{1}{n} \cdot \chi_{x_n} (x). $$ It is clear that $f$ is discontinuous at every $x_n$, because the set where $f(x) = 0$ is dense. Below is a proof that $f$ does what you want ...


3

I'm guessing that $\;x=(x_1,x_2)\;$ , and then $$f(x)\ge 0\;\;\forall\;x\in\Bbb R^2\;,\;\;\;and\;\;\;\; f(2,1)=0$$ so...


3

Yes, since differentiability is sufficient for continuity. Not necessarily. Consider the function \begin{equation} f(x,y)=\left\{ \begin{array}{lr} 0 & x=0\textrm{ or }y=0\\ 1 & otherwise \end{array} \right. \end{equation} Since $\frac{\partial f}{\partial x}(0,0)= \lim_{h \to 0} \frac{f(h,0)-f(0,0)}{h}=\lim_{h \to 0} ...


3

As stated in the comments, we have $x=y=0$ implies $2f(0)=4f(0)$ so $f(0)=0$. Also we must have an even function since with $x=0$ we see that $f(y)+f(-y)=2f(y)$ implying $f(y)=f(-y)$. Now for $k\in\mathbb N$ we can prove by induction that $f(ky)=k^2f(y)$. This is clearly true for $k=1$. So if we assume this holds for $k$ and proceed to $k+1$ we see that ...


3

You are close. I think some people use $B^A$ to denote the set of all functions from $A$ to $B$. So you could write $$\left\{f \in B^A: \left|f^{-1}(\{k\})\right|=c_k\right\}$$ to denote the set of all functions such that $c_k$ elements of $A$ are sent to $k \in B$.


3

Is this what you're looking for? $$\min(a,\,b) \; = \; \frac{a\,+\,b\; - \; |a-b|}{2} $$ $$\max(a,\,b) \; = \; \frac{a\,+\,b\; + \; |a-b|}{2} $$


3

I'll just count the number of times the sum=A[i]+A[j]+A[k] statement gets executed: \begin{align*} \sum_{i=1}^n \sum_{j=1}^i \sum_{k=1}^j 1 &= \sum_{i=1}^n \sum_{j=1}^i j \\ &= \sum_{i=1}^n \frac{i(i+1)}{2} \\ &= \frac{1}{2}\left[\sum_{i=1}^n i^2 + \sum_{i=1}^n i\right] \\ &= \frac{1}{2}\left[\frac{n(n + 1)(2n + 1)}{6} + ...


3

If you say $x^2+y=\text{constant}$, you're defining a curve in the $xy$-plane, which is a parabola $y=-x^2+\text{constant}$. If you move along that curve, both $x$ and $y$ are changing. But if you write $z=y^2+x$, defining $z$ as a function of those two variables, both of which can vary freely, then the expression $\partial z/\partial x$ means the rate of ...


3

In this context, $\mathbb{R}-\{x\}$ means the set of all real numbers that are different from $x$. Injective: if $f(s)=f(t)$ then $\frac{5s+1}{s-2}=\frac{5t+1}{t-2}$ so that $$ (5s+1)(t-2)=(5t+1)(s-2)\implies 5st-10s+t-2=5st-10t+s-2 $$ which simplifies to imply that $s=t$. Surjective: suppose that $y$ is any number in $\mathbb{R}-\{5\}$, let us demonstrate ...


2

Using L'Hospital rule for $\infty/\infty$ format, if $\lim_{x\to\infty}f(x)=\infty$: $$\lim_{x\to\infty}\frac{f(x)}{x}=0\implies\lim_{x\to\infty}\frac{f'(x)}{1}=0$$ See if you can use the same for second part similiary.


2

$$ (2k + 1)^2 + (2l + 1)^2 = 4\left(k^2 + l^2 + k + l\right) + 2 $$ This means that $\left(a^2 + b^2\right) = 4 \lambda + 2$, meaning that $4$ could not possibly divide $\left(a^2 + b^2\right)$, since there's always a remainder of $2$. ...by the way, this means that for $4$ to divide $a^2 + b^2$, both $a$ and $b$ must be even, since certainly if one is odd ...


2

Continuity of $f$ can be seen as "if you move just a bit, the image through $f$ will move a bit too". In your case $f^{-1}$ being not continuous would imply that you can find two points on the codomain "close enough" such that their $f$-counter image are actually far away. Think of a rectangle and glue two opposite sides together, and think of the obvious ...


2

It reads as: "Show that if $f$ of $x$ equals $f$ of $y$ for arbitrary $x$ and $y$ in $A$ with $x$ different than $y$, then $x$ equals $y$". The comma as nothing to do with pairs, it is just a comma from common language. Moreover $f \colon A \to B$ should be read as: $f$ is a function from the set $A$ into the set $B$.


2

The first part should be read "Suppose that $f$ is a function from the set $A$ into the set $B.$" The second part doesn't make sense, as written. It should say "Show that if $f(x)=f(y)$ for arbitrary $x,y\in A,$ then $x=y.$" That is, if $f(x)=f(y)$ implies that $x=y$ for any elements $x$ and $y$ of $A$ (not necessarily different elements, just with ...


2

Assume you can do it for $n=2^k$, specifically $n=4$, then $f(\frac{(x_1+x_2+x_3+(x_1+x_2+x_3)/3)}{4})\leq \frac{f(x_1)+f(x_2)+f(x_3)+f((x_1+x_2+x_3)/3)}{4}$ rearrange to get n=3 case. Same approach can be generalized to any non $2^k$ number.


2

Let me answer this by way of a different problem since there is no way to give a hint without giving the answer directly. Suppose we want to find out how far the ball falls in $.5$ seconds. Our distance function $d(t)$ tells us this. In order to find out what the distance is, we need only to plug in the time (in seconds). Thus the distance the object fell ...


2

As this is a question about notation allow me a few remarks: You won't wind up with a "four-element set". As $B$ has only two elements, $f[A]$ - i.e. the image of $A$ - will have at most two elements. This also means that your $\{5,5,6,5\}$ and $\{5,6,6,5\}$ are the same set, namely just $\{5,6\}$. The functions will differ, though. If you write your ...


2

Your operative definition of $\longrightarrow$ and $\longmapsto$ are correct. In your example you are dealing with vectors in $\mathbb{R}^n$, which are elements of a set, so you should use $\longmapsto$. see also this wikipedia article: http://en.wikipedia.org/wiki/List_of_mathematical_symbols


2

You have $$ A\cap B\subset A\implies f(A\cap B)\subset f(A),\\ A\cap B\subset B\implies f(A\cap B)\subset f(B) $$ so it follows that $f(A\cap B)\subset f(A)\cap f(B)$.



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