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16

Hint: what is $f(a,1)$? $\,\,\,\,$


11

Since this question is about pronunciation of $\sinh$, or the hyperbolic sine, I thought it might be useful to see what the authoritative Oxford English Dictionary had to say: This reveals that there are actually three pronunciations often encountered. From left to right: sh-eye-n $\qquad$ (i.e., shine) s-i-n-t-sh $\qquad$ (i.e., cinch) s-eye-n-ay-tch ...


7

The easiest way probably to prove this is by proving that $f$ is strictly increasing. You can do that quite easily using induction.


5

Suppose $f(n)=f(m)$. Then $(n+1)!-1=(m+1)!-1$. Adding $1$ to each side we conclude that $(n+1)!=(m+1)!$. We assume without loss of generality, by swapping the names of $m,n$ if necessary, that $n\ge m$. We now divide both sides by $(m+1)!$ to conclude $$1=\frac{(n+1)!}{(m+1)!}=\frac{(m+1)!(m+2)(m+3)\cdots(n+1)}{(m+1)!}=(m+2)(m+3)\cdots (n+1)$$ If there ...


5

for $x > 0,$ the function equals $$ \frac{1}{1+1/x} $$ $x$ is increasing so $1/x$ is decreasing so $1+1/x$ is decreasing so $$ \frac{1}{1+1/x} $$ is increasing. For $x<0,$ a similar argument. Joining the two cases together is easy.


5

If $x,y\not\in\mathbb{Z}$ and $x+y\in\mathbb{Z}$, then $\left\lfloor x\right\rfloor+\left\lfloor y\right\rfloor=x+y-1$. Therefore, $$ \left\lfloor j\frac ab\right\rfloor+\left\lfloor(b-j)\frac ab\right\rfloor=a-1\tag{1} $$ If $(a,b)=1$, then for $0\lt j\lt b$, we have $j\frac ab\not\in\mathbb{Z}$. Summing $(1)$, we get $$ \sum_{j=1}^{b-1}\left\lfloor j\frac ...


3

My guess is that $f(x)$ is a more complicated function, but that, as $x \to \infty$, $f(x)$ behaves like $\sqrt{\frac{(x^2+x)^3}{\pi}} $. That can mean either $\dfrac{f(x)}{\sqrt{\frac{(x^2+x)^3}{\pi}}} \to 1 $ or $f(x)-\sqrt{\frac{(x^2+x)^3}{\pi}} \to 0 $. I do not know what the "$\pm$" means, since the term following is monotonically increasing and ...


3

Proof: Let $\eta$ be an arbitrary element in $\mathbb{N}$. Then $$ f(\eta,1) = \eta^1 = \eta\in\mathbb{N}. \blacksquare $$ This answer gives you the proof that the mapping $f\colon\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ defined by $f(a,b)=a^b$ is onto, and it makes use of Cameron's hint. Can you see why the above constitutes a proof?


3

Hint: First, prove that $f(1)=0$ (writing $1=\frac11$). Then, prove that $f(-1)=0$ (writing $-1=\frac{-1}1$ and $-1=\frac1{-1}$). Then, using this prove that $f(-x)=f(x)$ (writing $x-=\frac{x}{-1}$) for any $x\neq0$.


3

You can consider a function to be a bit string of length three, i.e. $000$, $111$, $101$, etc. This would define all possible assignments of $\{a,b,c,\}$ to $\{0, 1\}$ and there are indeed 8 possible functions: $f(x) = 0 \rightarrow f(a) = f(b) = f(c) = 0$ $f(a) = 0$, $f(b) = 0$, $f(c) = 1$ $f(a) = 0$, $f(b) = 1$, $f(c) = 0$ $f(a) = 0$, $f(b) = 1$, $f(c) ...


3

Compose with the strictly increasing function $x=2\tan(z/2)$, $z\in(0,\pi)$. Such composition doesn't change the monotonicity. We get $$f(2\tan(z/2))=\frac{(1+\tan^2(z/2))z/2}{2\tan(z/2)}=\frac{z}{2\sin(z)}$$ And now $g(z)=\frac{\sin(z)}{z}$ is easy(er) to study in $(0,\pi)$. For this we can proceed similar to what you did, by looking at ...


3

Hint: $e^x$ is convex, hence stays above its tangent at $x=1$.


3

(Due to the nature of this question I've created a tongue-in-cheek response) I never short cut the pronunciation of my trig functions. Sine becomes Sin, Sine is most certainly not sinful. Cos becomes Cos, Tangent becomes tan, secant becomes sec Cosecant becomes csc!!! (that should be the real question here) To answer your question, your typical ...


3

If $f^8(x)$ means the application of the function $f$ to $x$ 8 times, i.e. $f(f(f(f(f(f(f(f(x))))))))$, then the first is not the same as the second. If however $f^8(x)$ means the $f(x)$ multiplied by itself 8 times, i.e. $f(x)\times f(x)\times ... \times f(x)$ (eight times), then yes they are equal.


3

You're pretty much right on with seeing the connection between the first line and the continuity condition. Looking at this problem, we see that the property we're looking for is something local - that is, we're only interested in what happens near the point $x^*$. Continuity is also a local property, so we're on the right track. Furthermore, all we know ...


3

$$\lim_{x\to -1} \frac{x-y}{x^3-y}=\frac{-1-y}{-1-y}=1$$ Whereas for your second limit, $$\lim_{y\to -1} \frac{x-y}{x^3-y}=\frac{x+1}{x^3+1}$$ Using the identity $a^3+b^3=(a+b)(a^2+b^2-ab)$, $$\frac{x+1}{x^3+1}=\frac{1}{x^2-x+1}$$


2

Let $\epsilon > 0$. (*) There is a $\delta_1 > 0$ such that $|y - y_0|<\delta_1$ implies $|g(y) - l| < \epsilon$. Using $\delta_1$ in the definition of limit for $f(x) \to y_0$, there is a $\delta > 0$ such that $|x - x_0| < \delta$ implies $|f(x) - y_0| < \delta_1$; for such $x$, by (*) with $y = f(x)$, we have $$ |g(f(x)) - l| < ...


2

if $y=\sqrt{x} : \mathbb{R}^+ \to \mathbb{R}$, then it certainly can not be $onto$, since $\sqrt{x} \geq 0$ and then by looking at the graph you will see that negative values are not taken by the function. Also by looking the graph you see that any horizontal line intersects at most one point of the graph, then the function must be 1 to 1.


2

No, the square root function is not surjective as a function $\sqrt\cdot:\Bbb R^+\to\Bbb R$, because the square root of a positive real number is positive real; hence $\sqrt x=-1$ has no solution. It is "into", because the square root always gives real numbers for positive real input. Generally, a function $f:A\to\Bbb R$ is onto if it touches every ...


2

No It is not In terms of elementary functions, But we can write in Infinite Series form. Using $$\displaystyle \sin x= x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+..........$$ So $$\displaystyle \frac{\sin x}{x} = 1-\frac{x^2}{3!}+\frac{x^4}{4!}-\frac{x^6}{6!}.........$$ So $$\displaystyle \int \frac{\sin x}{x}dx = \int ...


2

Your problem can be rephrased as $$\arg\max_y\{\sum_{r_1\in \Omega_1}p(r_1,s_2)u(y,r_1,s_2)\}\ge \arg\max_x\{\sum_{r_1\in \Omega_1}p(r_1,t_2)u(x,r_1,t_2)\}\text{ if }s_2>t_2.$$ Note that if $s_2>t_2$ and $x>y$, we will have $$u(x,r_1,s_2)-u(y,r_1,s_2)>u(x,r_1,t_2)-u(y,r_1,t_2).$$ Denote the left part to be $a_{r_1}$ and right part be $b_{r_1}$, ...


2

Thank you for this problem. I was not familiar with the notation $h^2(x)=h(h(x))$, but this is clearly what must be meant in this problem since interpreting $h^2(x)=(h(x))^2$, yields no real solutions. If $h(x)=2-\frac{a}{x}$, then we can solve for $h^{-1}(x)$ in the following manner. $x=2-\frac{a}{h^{-1}(x)} \Rightarrow x-2=-\frac{a}{h^{-1}(x)} ...


2

When $G$ is map from $\mathbb R$ to $\mathbb R^2$, we write $G\colon \mathbb R\to\mathbb R^2$. This means that $G$ takes any real number $x\in\mathbb R$ and depending on $x$ returns an element of $\mathbb R^2$, i.e. a pair of real numbers $(y,z)$. For example \begin{align*} G\colon \mathbb R &\longrightarrow \mathbb R^2 \\ x &\longmapsto (2x, x+1) ...


2

The formula you propose is only valid for asymptotes of the form $$y = mx + q,$$ where $m \in \mathbb R \backslash \{0\}$ and $q \in \mathbb R$. Since $m \in \mathbb R$, these lines cannot be vertical. On the other hand, the asymptotes you seek are of the form $x = q$. You can find them by researching all the points $\alpha$ for which $$\left|\lim_{x \to ...


2

Use the inverse functions of $\sin$, and $\cos$ $$\arcsin : [-1, 1] \rightarrow [\frac{-\pi}{2}, \frac{\pi}{2}]$$ $$\arccos : [-1, 1] \rightarrow [0, \pi]$$ $f^{-1}(x, y) = \arcsin x$, if $\arcsin x = \arccos y$ $f^{-1}(x, y) = 2\pi -\arccos x $, if $\arcsin x < 0$ and $\arccos y > \frac{\pi}{2}$ $f^{-1}(x, y) = \pi -\arcsin x$, if $\arcsin x \geq ...


2

A fixed point of $f$ is a root of $g(x)=f(x)-x$ and vice versa.


2

Let $A$ and $B$ be two non empty sets with $m\ \text{and}\ n$ elements respectively. Then the number of functions from $A$ to $B$ is $n^m.$ here $A=\{a,b,c\}$ and $B=\{0,1\}.$ Then the number of functions from $A$ to $B$ is $2^3(=8).$ For, $f:A \rightarrow B$ to be a function, Every element in $A$ should have a unique image. $a$ has 2 ways to be mapped, ...


2

Subtract your equations to obtain $f(x/y)-f(-x/y)=f(x)-f(-x)=f(-x)-f(x)$



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