# Tag Info

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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