Hot answers tagged

8

This is very interesting. Apparently the interpretations of $\arccot(x)$ vary! While your math textbook assumes $\arccot(x)$ as the inverse of $\cot(x)$ on $\left(0, \pi\right)$, Symbolab (and Wolfram Alpha and other mathematics software) use $\left(-{\pi \over 2}, {\pi \over 2}\right]-\{0\}$. The result are two different versions of $\arccot(x)$, of ...


7

Solve the system of questions: $$ \begin{align} F(x) + F\left(\frac{x-1}{x}\right) &= 1+x \\ F\left(\frac{x-1}{x}\right) + F\left(\frac{1}{1-x}\right) &= \frac{2x-1}{x} \\ F\left(\frac{1}{1-x}\right) + F(x) &= \frac{2-x}{1-x} \end{align} $$


6

Notice, $$\lim_{x\to 1}\frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3}$$ $$=\lim_{x\to 1}\frac{(\sqrt{x^2+3}-2)(\sqrt{x^2+3}+2)}{(\sqrt{x^2+8}-3)(\sqrt{x^2+8}+3)}\cdot \frac{(\sqrt{x^2+8}+3)}{(\sqrt{x^2+3}+2)}$$ $$=\lim_{x\to 1}\frac{x^2+3-4}{x^2+8-9}\cdot \frac{(\sqrt{x^2+8}+3)}{(\sqrt{x^2+3}+2)}$$ $$=\lim_{x\to 1}\frac{(x^2-1)}{(x^2-1)}\cdot ...


5

Take the log of both sides and examining the 2 sides of the limits yields $$\ln L^+=\lim \limits_{x \to 0^+} \frac{\ln (\sin x)-\ln(x)}{x}$$ $$\ln L^-=\lim \limits_{x \to 0^-} \frac{\ln (\sin (-x))-\ln(-x)}{x}$$ which can be solved by L'Hopital's to both equal $0$, so the limit is 1.


4

The graph of $y=2x(1-x)$ is a parabola opening down. It has its vertex halfway between the zeroes at $x=0$ and $x=1$, i.e., at $x=\frac12$, and $2\cdot\frac12\left(1-\frac12\right)$ is only $\frac12$. Thus, no value of the function is greater than $\frac12$, and it cannot map $\Bbb R$ onto $[0,1]$. If you prefer an algebraic approach, note that ...


4

The integral diverges. To see this, we can write $$\int_0^n \left(1-\frac{3x}n\right)^ne^{x/2}\,dx=\int_0^{n/3} \left(1-\frac{3x}n\right)^ne^{x/2}\,dx+\int_{n/3}^n \left(1-\frac{3x}n\right)^ne^{x/2}\,dx \tag 1$$ We will present two parts. In Part $1$, we will show that the first integral on the right-hand side of $(1)$ converges. In Part $2$, we will ...


4

We see $$3y-2=-4\sqrt{1-y^2}$$ so $$(3y-2)^2=16-16y^2$$ therefore $$9y^2-12y+4=16-16y^2$$ rearranging gives $$25y^2-12y-12=0$$ And so the solutions are $$y_0,y_1=\frac{12}{50}\pm\frac{1}{50}\sqrt{144+1200}$$ and simplifying yields $$y_0,y_1=\frac{6\pm4\sqrt{21}}{25}$$ Checking these solutions will give us the unique solution $$y_0=\frac{6-4\sqrt{21}}{25}$$


4

Hint: $$ \frac{n^3}{n!}=\frac{n^2}{(n-1)!}=\frac{1}{(n-1)!}+\frac{n+1}{(n-2)!}=\frac{1}{(n-1)!}+\frac{3}{(n-2)!}+\frac{1}{(n-3)!} $$


4

The text book is correct. Note that the limit of the argument is $$\lim_{x\to 1^{\pm}}\frac{x^2+1}{x^2-1}=\pm \infty$$ and that for the Principal Values of the arccotangent, we have $$\lim_{x\to \infty}\arccot(x)=0$$ and $$\lim_{x\to -\infty}\arccot(x)=\pi$$


4

You should see if the two functions (comprising the given function) have derivatives at $x = 0$ and if so, if these two are equal. If they are equal, then the main function also has a derivative at $x = 0$. Think about the geometric meaning of the derivative: imagine the two functions both have derivatives (at $x=0$) but they are not equal. This would ...


3

Clearly $\dfrac{x^2+1}{2x}$ can never be $0$ so the range cannot be $(-\infty,+\infty)$. We have $$ \frac{x^2+1}{2x} = \frac x 2 + \frac 1 {2x}. $$ Notice that $\dfrac x 2$ goes up to $\infty$ as $x\to+\infty$ and $\dfrac 1 {2x}$ goes up to $+\infty$ as $x\downarrow0$, and the sum is positive everywhere in between, so the range of the restriction of this ...


3

Hints: it converges normally on any $I_\delta \stackrel{\rm def}{=}(-\infty, \delta)\cup(\delta, \infty)$ (for any fixed $\delta > 0$. Indeed, for all $n\geq 0$ the function $f_n$ is even, non-negative, and decreasing on $(\delta,\infty)$, so that $$ \sup_{x\in I_\delta} \lvert f_n(x)\rvert = \sup_{x\in I_\delta} f_n(x) = \frac{1}{1+n^a\delta^4}. $$ ...


3

For $a>1$ and $x\ge \delta>0$, we have $$\sum_{n=1}^\infty\frac{1}{1+n^ax^4}\le \sum_{n=1}^\infty\frac{1}{n^a\delta^4}=\frac{1}{\delta^4}\zeta(a)$$ which exists for $a>1$, which one can show using, say, the integral test. However, we can choose a number $\epsilon=\frac12$, and a number $x=1/n^{a/4}$ such that for any $n$ ...


3

Hint: try a power function $f(x)=x^{p}$ and see what $p$ would have to be.


3

Any function defined from $\Bbb R$ (the set of real numbers) to $\Bbb R$ is monotonic iff its derivative never changes sign, yes. But $\tan(x)$ is not a function from $\Bbb R$ to $\Bbb R$, since it's not defined on all real numbers. In fact, the thing about derivatives is only true when the domain is a connected set (i.e. an interval)! (Remember that $\Bbb ...


3

Let $f_n(x)=\left(1-\frac{3x}{n}\right)^ne^{x/2}$. Then $$\lim_{n\to\infty }f_n(x)\lambda_{[0,n]}(x)=\lim_{n\to\infty }f_n(x)\cdot \underbrace{\lim_{n\to\infty }\lambda_{[0,n]}(x)}_{=\lambda_{[0,\infty [}(x)}=\lambda_{[0,\infty [}(x)\lim_{n\to\infty }f_n(x).$$ Therefore $$\int \lim_{n\to\infty }f_n(x)\lambda_{[0,n]}(x)\mathrm d x=\int \lambda_{[0,\infty ...


3

$$a\ge b\iff\dfrac am\ge\dfrac bm$$ only if $m>0$ If $m<0,$ $$a\ge b\iff\dfrac am\le\dfrac bm$$ $$\dfrac{2x-5}{3x-1}\ge1\iff2x-5\ge3x-1$$ only if $3x-1>0$


3

To get an answer, you have to specify the codomain. $f:[0,1]\to[0,1]$ given by $f(x)=4x(1-x)$ is onto; $f:[0,1]\to[0,1]$ given by $f(x)=2x(1-x)$ is not onto; $f:[0,1]\to[0,2]$ given by $f(x)=4x(1-x)$ is not onto; $f:[0,1]\to[0,1/2]$ given by $f(x)=2x(1-x)$ is onto.


3

Given $\varepsilon >0$ there exist $\delta_{\varepsilon}>0$ such that $$-\delta_{\varepsilon}<x<\delta_{\varepsilon}\qquad \implies \qquad\left|\frac{f(x)}{x^2}-L\right|<\varepsilon$$ So $$|f(x)-Lx^2|<\varepsilon x^2$$ Now, take $\delta=\min(\delta_{\varepsilon},1)$, then $$-\delta<x<\delta\qquad \implies ...


3

The functions $x \mapsto \arctan 2x$ and (some restriction of) $x \mapsto \tan 2x$ are not inverses: The inverse of $x \mapsto \arctan 2x$ is $$x \mapsto \tfrac{1}{2} \tan x, \qquad -\tfrac{\pi}{2} < x < \tfrac{\pi}{2} .$$


2

Your (original) proof doesn't really concerns the case when $-1<x<0$, because in that case the remainder might be negative. Actually the remainder is nonnegative, which can indeed be proved with the Lagrange remainder. You can consider, instead, the function $f(x)=x-\ln(1+x)$; its derivative is $$ f'(x)=1-\frac{1}{1+x}=\frac{x}{1+x} $$ that only ...


2

With Taylor expansions: (just for reference) We will use $$\begin{align} \sin u &= u + o(u^2) \\ \ln(1+x) &= u + o(u) \end{align}$$ when $u\to0$. (In particular, $\ln(1+o(u)) = o(u)$.) Write $$ \left(\frac{\sin x}{x}\right)^{\frac{1}{x}} = \left(\frac{x+o(x^2)}{x}\right)^{\frac{1}{x}} = \left(1+o(x)\right)^{\frac{1}{x}} = e^{ ...


2

Notice, $$\lim_{x\to 0}\left(\frac{\sin x}{x}\right)^{1/x}$$ $$=\lim_{x\to 0}\exp \left(\frac{1}{x}\ln\left(\frac{\sin x}{x}\right)\right)$$ $$=\lim_{x\to 0}\exp \left(\frac{\ln(\sin x)-\ln(x)}{x}\right)$$ using L'Hosptal's rule for $\frac 00$ form, $$=\lim_{x\to 0}\exp \left(\frac{\frac{\cos x}{\sin x}-\frac1x}{1}\right)$$ $$=\lim_{x\to 0}\exp ...


2

HINT.-$$ (\frac{\sin x}{x})^{1/x}=\left([1+(\frac{\sin x}{x}-1)]^{\frac{1}{\frac{\sin x}{x}-1)}}\right)^{\frac{\sin x-x}{x^2}}$$ It follows $$\lim \limits_{x \to 0} (\frac{\sin x}{x})^{1/x}=e^{\lim {x\to 0}\frac{\sin x -x}{x^2}}=e^0=1$$


2

By the product rule $$\frac{f'(1)}{f(1)}=\sum_{k=1}^n \frac{1}{a_k+1}$$ The inequality you want to prove is: $$ \left(\sum_{k=1}^n \frac{1}{a_k+1} \right) (1+a_1)(1+a_2)...(1+a_n)\geq n\left ( 1+ \sqrt[n]{a_1...a_n} \right)^{n-1}$$ Since by AM-GM you have $$\left(\sum_{k=1}^n \frac{1}{a_k+1} \right) \geq \frac{n}{\sqrt[n]{(1+a_1)(1+a_2)...(1+a_n)}}$$ If ...


2

Such a function exists. It is defined by: $$f(A, B, ..., N) = \begin{cases} A & \text{if } A \ge B, A \ge C, ... , A \ge N\\ B & \text{if } B \ge A, B \ge C, ... , B \ge N\\ \vdots\\ N & \text{if } N \ge A, N \ge B, ... , N \ge M\end{cases}$$ Like cosine and sine, this function even has its own special three-letter symbol: $\max$. Sarcasm ...


2

Define $a_n = f(x_n)$, with $x_n$ just as you defined it. Now let $L = sup_{x\in[a,b]} f(x)$. We have: $|f(x_n)| \leq \frac{L}{2^n}$. As $n \rightarrow \infty$ we have $|f(c)| \leq0$ .


2

The answer is quite easy if you think of what is going on with the transformation. We are taking a vector of $\mathbb{R}^2$ and assigning to it the unique unit vector which has the same direction and orientation. So the inverse of each unit vector $u$ is the ray which goes from 0 to u (without including the 0). Note however that there is not an inverse ...


2

If $f(0)\neq 0$, then $\frac{f(x)}{x^2}$ diverges, since $$\lim_{x\to 0}\frac{c}{x^2}=\pm\infty$$


2

Because multiplying by a negative quantity reverse the inequality, so you have to separe the two case: if the denominator is $>0$ or $<0$ ( and obviously exclude the case that it is $=0$).



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