# Tag Info

10

The functions $$f_n(x) = \frac{nx}{1+nx^2}$$ are (infinitely often) differentiable on $\Bbb R$. Each $f_n$ is bounded: $$|f_n(x)| = \frac{\sqrt{nx^2}}{1+nx^2} \sqrt n \le \frac 12 \sqrt n \, ,$$ and $f_n(x) \to f(x)$ pointwise for all $x$. Graph of $f_5, f_{10}, f_{40}$ together with the graph of $1/x$:

7

How about: $$f_n(x) = \begin{cases} 1/x, & \mbox{if } x>\frac{1}{n} \\ n^2\cdot x, & \mbox{if } 0\leq n\leq \frac{1}{n}\end{cases}$$ This is continous, though not differentiable, but you can get the idea on how to do it(What I did is just join the point $(\frac{1}{n},n)$ with $(0,0)$ in a continous manner, sure you can come about with a ...

7

For an alternative approach, notice that $$f(2(x-1))+2f(y+1)=f(f(x+y))=f(2x)+2f(y)$$ so that $$2(f(y+1)-f(y))=f(2x)-f(2x-2).$$ Setting $x=0$ say gives $f(y+1)-f(y)=C$ is a constant. Therefore $f(x)=Cx+D$ for constants $C$ and $D$. Now put this into the original equation to find all possible pairs $(C,D)$, etc.

6

Hint: Consider $X = \{(1,0), (1/2,0), (1/3,0), ..., (0,0)\}$. What can we say about the behaviour of $f(x)$ as $x\to (0,0)$?

6

Let me just show you a few tricks so the next time you'll be able to figure such things out without pen or paper, not to mention MSE. First, drop the common part. We need to show that $$x\mapsto \frac{\int_{[x-b,x+b]}e^{-z^2/2}dz}{\int_{[x-c,x+c]\setminus[x-b,x+b]}e^{-z^2/2}dz}$$ is decreasing for $x\ge 0$. Next, use symmetry and the fact that the ...

5

Take$$\begin{array}{rccc}f\colon&\mathbb R&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}x+1&\text{ if }x\in\mathbb Z\\x&\text{ otherwise.}\end{cases}\end{array}$$Then $(\forall x\in\mathbb R):f(x+1)>f(x)$. However, $f$ is not monotonically increasing. It is more natural to deduce that your function is monotonically ...

5

By C-S $$\sqrt{400+2x-x^2}+\sqrt{x^2-20x}\leq\sqrt{(1+1)(400+2x-x^2+x^2-20x)}=20\sqrt2.$$ Also, since $$\sqrt{a}+\sqrt{b}=\sqrt{a+b+2\sqrt{ab}}\geq\sqrt{a+b},$$ we obtain: $$\sqrt{400+2x-x^2}+\sqrt{x^2-20x}\geq\sqrt{400+2x-x^2+x^2-2x}=20.$$ We got the maximal and the minimal value of $f$ and $f$ is a continuous function. Can you end it now?

4

There exist discontinous functions satisfying the given conditions. Such functions are not bounded in any neighborhood of $0$ so they do not satisfy $\lim _{x \to 0} \frac {f(x)} x=1$. So the result, as stated, is false. However, if $f$ is continuous then the first condition implies that $f(x)=cx$ for some $c$ and the second condition implies that $c=1$ so ...

4

Take the function $f(x)={1\over{x-2}}$

4

Let $f(x)=f(y)$. Then $x+1 = f(f(x)) - f(x) = f(f(y))-f(y) = y+1$ and so $x=y$.

3

Let $f(x) = ax+b$ and let $g(x) = cx + d$. Then let $h(x)$ denote the function $$h(x) = (f\circ g)(x) - (g\circ f)(x) = a(cx+d) + b - c(ax+b) - d = ad + b - bc - d.$$ So you're right that this is a constant, and hence can't be surjective (since the codomain $\mathbb{R}$ has more than one point). You're confused about the definition of injectivity though. A ...

3

If $f'(x) = m$ for every $x\in\mathbb R$ then it's not possible, as that implies $f(x) = mx+c$ and is in contradiction that $f(x)\in(-1,1)$. If you make the condition a bit weaker, for example $f'(x) = m$ almost everywhere, then it is possible. For example $$f(x) = \left\{\begin{array}{ll} x-\ln n+ 1 & \text{for }x\in[\ln (n-1),\ln n), \quad n\in\... 3 Summary of results. f(x)=1 for u\lt x\le 1 andf(x)=\frac{u-1}{u} for 0\le x\le u. To maximize \int_0^1f^3(x)dx=1-u+\frac{(u-1)^3}{u^2}=\frac{(1-u)(2u-1)}{u^2}=-\frac{2u^2-3u+1}{u^2}=I(u) Use elementary calculus to find the maximum. I'(u)=-\frac{u^2(4u-3)-2u(2u^2-3u+1)}{u^4} Set I'(u)=0 and get u=\frac{2}{3} and I(u)=.25. Added note f(... 3 Let f: [0,1] \to [-1,1] be any integrable function satisfies the constraint \int_0^1 f(x) = 0. For any x \in [0,1],$$f(x) \in [-1,1]\quad\implies\quad (1 - f(x))\left(f(x) + \frac12\right)^2 \ge 0$$Integrating RHS gives us$$\int_0^1 \left[\frac14 + \frac34 f(x) - f(x)^3\right] dx \ge 0 \implies \int_0^1 f(x)^3 dx \le \frac14 $$Since this value \... 3$$y=\left\lfloor\frac{nx-1}{n-1}\right\rfloor$$where \lfloor\alpha\rfloor denotes the integer part of \alpha (i.e. the largest integer that is \le\alpha). 3 For 1\leq a< b and 1\leq c < d, suppose: a + \frac{b(b-1)}{2} = c + \frac{d(d-1)}{2} and (a,b) \neq (c,d). Wlog assume d>b (if they are equal, then a=c and we have a contradiction), then:$$a -c = \frac{d(d-1)-b(b-1)}{2} = \frac{(d-b)(d+b-1)}{2}\geq \frac{(d+b-1)}{2}> \frac{2b-1}{2}=b-\frac{1}{2}>a$$Therefore c < 0, and ... 3 The property you are asking for is continuity. More succinctly written, continuity for \operatorname{Ord}\to\operatorname{Ord} functions is when$$f(\gamma)=\sup_{\alpha\in\gamma}f(\alpha)=\sup\{f(\alpha)~|~\alpha\in\gamma\}$$for every limit ordinal \gamma. It is notable that the functions you are talking about are increasing functions, so if \gamma\... 3 Your claim is not true if X is an infinite set. Consider the function f(x)=2x mapping the natural numbers to itself. It is injective but not surjective. 3 By odd parity,$$\lim_{x\to0^-}f(x)=-\lim_{x\to0^+}f(x).$$If those limits exist, they must be equal for the ordinary limit to exist, hence 2. 3 This is just a Taylor expansion$$b(r) = \sqrt{\frac{r^3}{r - 2M}}=r \sqrt{\frac{r}{r - 2M}}$$Use long division or Taylor series$$\frac{r}{r - 2M}=1+\frac{2 M}{r}+\frac{4 M^2}{r^2}+O\left(\frac{1}{r^3}\right)\sqrt{\frac{r}{r - 2M}}=1+\frac{M}{r}+\frac{3 M^2}{2 r^2}+O\left(\frac{1}{r^3}\right)b(r)=r+M+\frac{3 M^2}{2 r}+O\left(\frac{1}{r^2}\right)=...

2

Set theory does not allow one to form a set out of all sets with cardinality $3$, say. Nor even a set of all sets with any given cardinality (except for the cardinality $0$, for which $\{\{\}\}$ does the trick). The existence of such a set would contradict the axioms of set theory, in a manner similar to Russell's paradox. So no, it is not true that "...

2

a) true (but what you are saying is not a proof of it). Let $s,s'\in S$ with $f(s)=f(s')$. Then also $(g\circ f)(s)=g(f(s))=g(f(s'))=(g\circ f)(s')$ so that we are allowed to conclude that $s=s'$ because $g\circ f$ is injective. b) false in general. Let $U=\{u\}$ and $S\neq\varnothing$ so that automatically $g\circ f$ is surjective. If however $T$ ...

2

First notice that from $(1)$ you can conclude that $f(\frac{1}{n}) = \frac{f(1)}{n}$ which by $(2)$ equals $\frac{1}{n}$. From this you can conclude that if the limit in the question exists, it must be $1$. Unfortunately, the limit may not exists without further assumptions on $f$. Consider the set $\{1,\pi,\pi^2,...\}$. This is a set of independent real ...

2

(Posted because the one-line comment was not enough...) This is obviously false. For example, suppose that $\{n_i:i\in I\}$ is the collection of all unit vectors. Then for any $n$ there exists $i$ such that $n\cdot n_i=0$. So there does not exist $n$ such that $n\cdot n_i\ne0$ for all $i \in I$.

2

Hint: It is a) What can you say from $f(x_1)=f(x_2)$? Does for each $b$ in $X$ exists $a\in X$ so that $f(a)=b$? Can you finish?

2

Hint: In the ring $\Bbb{R}\oplus\Bbb{R}$ you have $(1,0)\times(0,1)=0$ and $(1,0)+(0,1)=1$. Details: Let $f$ and $g$ denote the images of $(1,0)$ and $(0,1)$ in $C(\Bbb{R})$, respectively. Then $fg=0$ and $f+g=1$, so $g=1-f$, meaning that $$0=fg=f(1-f)=f-f^2.$$ It follows that $f=f^2$, and so $f(x)\in\{0,1\}$ for all $x\in\Bbb{R}$. Because $f$ is continuous ...

2

Another way without using induction. Let $h(x)=f(x)-f(a)$, then $h(a)=0$, so $(x-a)|h(x)$, i.e. there exists $g(x)$ such that $f(x)=(x-a)g(x)+f(a)$. Here $h$ and $g$ are polynomials.

2

Is there any way I can [combine] $h(x)$ into a single function? You just did. What you just wrote is a completely acceptable definition of one single function $h$. However, you may be wondering something. Namely, your function is written in a piecewise form (it's defined separately for $x < 5$ and $x > 5$), and you may be wondering if it's possible ...

2

Hint: Can you show $f(x) = \sqrt{-y+500}+\sqrt{y-100}$ is at least $20,$ and it has a maximum at $20\sqrt2$ when $y=300$? If so, you can see the $9$ integer elements in the range are $20, 21, ..., 28$.

2

You can prove this via contradiction. Suppose $\pi(a,b)=\pi(a',b')$. First case $b=b'$, show this implies $a=a'$. Second case, $b < b'$. Hence $b' \ge b+1$. This gives you an inequality for $a-a'$ which yields a contradiction with the assumption $a < b$. Edit: As $b' \ge b+1$ we have \$\frac{b'(b'-1)}{2}-\frac{b(b-1)}{2}\ge \frac{(b+1)b}{2}-\frac{b(...

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