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0

The relation will be transitive if $xRy$ and $yRz$ imply $xRz$ for all $x,y,z \in A$. In the language of your relation, the implication you would need to prove is $y=3x$ and $z=3y \overset{?}{\implies} z=3x$ (If you wanted to prove this, you could try substituting equations.) Now, to prove that it is not transitive, you just need to find some $x,y,z$ in ...


0

hint...The focal chord property is $pq=-1$, whereupon you can calculate the gradient of $RU$...


0

For (1), yes, that suffices since every choice gives the same result. This assumes points are (time, distance). But note you have made a slight error in the units, which should be m/sec, not just m. For (2), the point on the other end of the segment appears to be $(15,0)$, doesn't it? The starting point is $(x_1,y_1) =(0,1200)$ and the ending point is ...


0

Since, the locus of the parachute is linear in nature so its slope is given as that of a straight line passing through $(0, 1200)$ & $(15, 0)$ as follows $$\text{slope of straight line }=\frac{1200-0}{0-15}=-80\ m/min$$ Negative sign indicates that the parachute is descending. If the acute angle of linear trajectory of parachute with horizontal is ...


2

max{x,3x} + min{x,3x} = 4x max{x,3x} - min{x,3x} = |2x| ---------------------------- 2max{x,3x} = 4x + 2|x| max{x,3x} = 2x + |x| As a bonus, min{x,3x} = 2x - |x|


1

Notice, $$\frac{x+1}{x^3-x}$$$$=\frac{x+1}{x(x^2-1)}$$ Apply, $a^2-b^2=(a-b)(a+b)$ $$=\frac{x+1}{x(x-1)(x+1)}$$ $$=\frac{1}{x(x-1)}$$ $$=\color{blue}{\frac{1}{x^2-x}}$$ Hence, option (d) is correct.


2

Without imagination... $$ \frac{x+1}{x^3-x} = \frac{x+1}{x(x^2-1)} = \frac{x+1}{x(x-1)(x+1)} = \frac{1}{x(x-1)} $$


1

There are 16 teams that can be chosen as the first team in a game, and for each team, 15 teams they can play. But this approach overcounts by a factor of 2, because you have counted team A (first player) playing team B (second player) as well as team B (first player) playing team A (second player). Thus the answer is $16 \times 15/2 = 120$.


2

$$\begin{align} f(x+k.346)&=f(x+(k-1).346+346)\\ &=\frac{1+f(x+(k-1).346)}{1-f(x+(k-1).346)}\\ &=\frac{1+\frac{1+f(x+(k-2).346)}{1-f(x+(k-2).346)}}{1-\frac{1+f(x+(k-2).346)}{1-f(x+(k-2).346)}}\\ &=\frac{-1}{f(x+(k-2).346)}\\ &=-\frac{1}{\frac{1+f(x+(k-3).346)}{1-f(x+(k-3).346)}}\\ &=\frac{f(x+(k-3).346)-1}{f(x+(k-3).346)+1}\\ ...


2

Let $a = 346$. $$f(x +2a) = f\left( (x + a) + a \right) = \frac{1 + f(x+a)}{1 - f(x+a)} = \frac{1 + \frac{1 + f(x)}{1 - f(x)}}{1 - \frac{1 + f(x)}{1 - f(x)}} = \ldots = -\frac1{f(x)}$$ Now, $f(x + 4a) = ?$


6

I will probably put more emphasis on those transformation properties of the $\max(\cdot)$ function which is reasonably "obvious" and easy to remember/use. For example, $\max(a+b,a+c) = a + \max(b,c)$. $\max(ab,ac) = a \max(b,c)$ when $a > 0$. $\max(a,-a) = |a|$. This may help a student to get familiar with such "tools" for attacking similar problems. ...


4

Hint: consider $f(x+k\cdot 346)$ for the first few values of $k$.


1

Christian's answer nicely addresses how to solve the second part, but let me see if I can give some additional insight on why we should interpret the first part as he did. (A.G.'s comment alludes to the problem, too.) If we are supposed to assume only that $(1)$ holds for all (appropriate) $n$, then it is not possible to verify that ...


0

You need to do some analysis.I recommend take the following points. From basic function f and g: See when are f and g zero Find the max and min value of the f and g (example : for sin(x) +1 and -1) Plot the envelopes of the shape in the enlarged size to get an idea of the graph. OR Calculate the roots of function(sum) if possible. Analyse the value at ...


1

Since $h(x)=(f+g)(x):=f(x)+g(x)$ for every $x$ in the domain, the graph is the one that you obtain summing the two functions pointwise. That is, at $x=x_0$ will correspond the point $h(x_0)=f(x_0)+g(x_0)$. Edited after seeing the comment about discontinuities: if one of the functions $f$ and $g$ has a discontinuity, remember that the domain of $f+g$ is ...


0

You can think about the graph of h(x) pointwise, adding the heights of the two graphs f(x) and g(x) at each point x. For example, if f(1) = 2 and g(2) = 3. h(1) = f(1) + g(1) = 2 + 3 = 5.


-2

Those are all idempotent functions. More examples are here.


1

Assuming that you are talking only about $\Bbb R\to\Bbb R$ functions, for a such function, this equation holds: $$f(x)=x\;\forall x\in f(\Bbb R)$$ So, for any nonempty $A\subset \Bbb R$ define $f(x)=x$ for $x\in A$, and $f(x)$ to be any $y\in A$ otherwise. Any such function will hold $f\circ f=f$. The sets $A$ correspondig to your examples are, ...


1

Assuming we are using the usual topology of $\mathbb{R}$. a. $A_1 = \left\{ \frac{1}{n\pi} \,|\, n \in \mathbb{Z} -\{0\}\right\}$. So clearly $A_1$ is NOT open , since it has isolate points (actually all of its points are isolated points). It is also clear that $A_1$ is NOT closed, since $0$ is in its closure and $0\notin A_1$. In other words, there is a ...


21

You can use the fact that, $\max\{x,y\}=\frac{x+y+|x-y|}{2}$. Therefore, $\max\{x,3x\}=\frac{x+3x+|x-3x|}{2} = \frac{4x+|-2x|}{2} =2x+|x|$.


1

Note that $f(-x) = -ax + b|x|+ c$, thus breaking into symmetrical and anti symmetrical combinations one obtains $$ f(x) - f(-x) = 2ax\\ f(x) + f(-x) = 2b|x| + 2c $$ Thus $$ a = \frac{f(1) - f(-1)}{2} = \frac{3 + 1}{2} = 2\\ c = f(0) = 0\\ b = \frac{f(1) + f(-1)}{2} - c = \frac{3 - 1}{2} = 1 $$ Second variation of the same idea: $$ f(-x) = \max(-x, -3x) = ...


2

I like the question! It is not immediately obvious to me, however, that it suffices to match your form to the given function at three points. Perhaps that is the case, but if so it requires a separate argument. Alternatively, having discovered the final form you could verify it directly. I would address the original problem this way: Your function, $f$, ...


4

One has $$\cos(0x)=1,\quad \cos(1x)=\cos x,\qquad\cos(2x)=2\cos^2x-1\ .$$ This shows that for $0\leq n\leq 2$ the functions $x\mapsto \cos(nx)$ can be written "as polynomials in $\cos x\>$". Furthermore, it follows from the addition theorem for $\cos$ that $$\cos\bigl((n+1)x\bigr)+\cos\bigl((n-1)x\bigr)=2\cos(nx)\cos x\ ,$$ since the $\sin$ terms cancel. ...


1

The continuity of $F$ is proved by zhw. As you probably know, the key point is to show that for any compact set $K\subseteq \mathbb H$, one can find an integrable function $g_K$ such that $\vert f(x,y,t)\vert\leq g_K (t)$ when $(x,y)\in K$. Then you can apply the "standrad" theorem on continuity for integrals depending on a parameter; or reprove this theorem ...


1

Hope you've solved it by now, but here is a complete proof for others who may find themselves looking for answers Showing the first inclusion is straight forward def. $f(A) = \{f(x) : x \in A \}$ def. $f^{-1}(B) = \{x : f(x) \in B \}$ $x \in A \subset X$ $f(x) \in f(A)$ $f^{-1}(f(A)) = \{x : f(x) \in f(A) \}$ $x \in f^{-1}(f(A))$ $A \subset ...


3

For any function $f : \mathbb R \rightarrow \mathbb R$, the set $f^{-1}( a, \infty)$ always exists, and it is defined by $$ \{ x: \in \mathbb R : f(x) > a \}, $$ although it is not always measurable.


10

$f^{-1}$ here does not mean the inverse of $f$. It is a horrifically bad overuse of notation and can be very misleading. $f^{-1}(A)$ where $A$ is a set is the pre-image, i.e. $$f^{-1}(A) = \{x\in X: f(x) \in A\}.$$ The pre-image always exists. If $f$ were invertible, then this would coincide with what you think; however the pre-image takes care of the ...


19

$f^{-1}(A)$ is the preimage of the set $A$, it exists even if $f$ is not invertible. For $f: X\to Y$, it is defined as $$f^{-1}(A) = \{x\in X: f(x)\in A\}.$$ The notation is slightly confusing, I would say, but one gets used to it. It isn't really that bad of a notation, since, if $f$ is actually invertible and its inverse is $g$, then $g(A) = f^{-1}(A)$ ...


1

As the discussion in the comments is getting involved, I'll summarize the conclusions here. EDIT: I have worked out the count in the final case and have included that here. Total number: The total number of functions $A\rightarrow B$ is $m^n$ because each of n slots can be filled with any of m things. Case I: $n = m$ In that case surjections are the ...


0

I read this from the book called "The Malliavin Calculus and related Topics", p.27. In fact, it is an expectation question. $(E[|F|^{p}]+\sum_{i}E[||D^{i}F||_{H^{i}}^{p}])^{1/p} \leq (E[|F|^{q}]+\sum_{i}E[||D^{i}F||_{H^{i}}^{q}])^{1/q}$ After simplification, it suffices to prove $[\int \left(|f(x)|^{p} + \sum_{i=1}^n |f^{(i)}(x)|^{p} ...


0

Given any irrational number r, and rational number k, since k=r+(k−r), the sum of two irrational numbers, for any rational k, f(x)=f(x+k). As this property applies to all real numbers, rational or not, the function must be equal for all real numbers, and henceforth, it is constant.


1

Hint: It is enough to show that you have $f(x) = f(x+r)$ for every real number $r$. You have it for all irrational $r$, so you just need to prove it for rational $r$. You can write any rational number as a sum of two irrational numbers.


1

$a): D = \{x: x > -1\}, \text{ asymptote } x = -1$ $b):$ $x-$intercept: Put $y = 0 \to \log(x+1) = 0 \to x+1 = 1\to x = 0$. This turns out to be the $y-$intercept as well. $c)$: $y' = \dfrac{2}{(x+1)\ln 10} > 0 $ since $x > -1$. Thus $g$ is always increasing.


1

Let $E_R =[-R,R]\times [1/R,\infty).$ Claim: There exists a constant $C_R>0$ such that $$f(x,y,t)\le C_R, \text {for}\,(x,y) \in E_R, t\in \mathbb {R}.$$ Suppose the claim is true. Fix $(a,b)$ in the upper half plane. Then there is $R> 0$ such that $D((a,b),b/2)\subset E_R.$ Let $(x_n,y_n) \to (a,b).$ Then for large $n, (x_n,y_n) \in E_R.$ For such ...


1

The question is: Given the relationship $$f_{n+1}(x)=2xf_n(x)-f_{n-1}(x)$$ demonstrate that $\cos(n\theta)=f_n(\cos\theta)$. To show this, induction may be useful. The second part can probably be done through the substitution $x = \cos \theta$ and the relationship from above.


3

HINT: Don't work hard, work smart. Recall, or prove, that ${}^C({}^BA)\sim{}^{C\times B}A$. Now use the facts that $\mathcal P(X)\sim{}^X2$ and that $\Bbb{Z^+}\sim\Bbb{Z^+\times Z^+}$.


1

$$f(x)=\sin^2 (x)$$ $$f(0)=0$$ $$f'(x)=\cos(x)2\sin(x)=\sin(2x)$$ $$f'(0)=0$$


1

f=0. Now it turns out a reply must be at least thirty characters. Hmm. Humdee hum...


3

For the first highlighted step, note that \begin{align*} \| x - a \| &\le \max(\|x - a\|, \|y - b\|) = \|(x,y) - (a,b)\| \\ \| y - b \| &\le \max(\|x - a\|, \|y - b\|) = \|(x,y) - (a,b)\| \\ \| a \| &\le \max(\|a\|, \|b\|) = \|(a,b)\| \\ \| b \| &\le \max(\|a\|, \|b\|) = \|(a,b)\|. \end{align*} Therefore, \begin{align*} \| B \| \| x - a \| \| ...


1

Maybe you can try using sequential continuity, equivalent to continuity in metric (gen. 1st countable) : Let $x_n \rightarrow x , y_n \rightarrow y$ , we want to show $x_ny_n \rightarrow xy$ : We have that $(x_n-x)(y_n-y) \rightarrow 0$ as $n \rightarrow \infty$ , since each of the terms on the left can be made small -enough. Now: $x_ny_n -xy= x_ny_n ...


2

The elements in $a$ must be non-negative. (Take $a=-1$ for a start of a counterexample) If $a\ge 0$ then $$ a^Tb_i = \sum_{j=1}^d a_j b_{i,j} \le \sum_{j=1}^d a_j \max_i b_{i,j}= a^T (\max b_i). $$ Now take the maximum on the left-hand side.


1

You're on the right track, but you're having trouble with the endpoints. The points $(-3, -2)$ and $(4, -2)$ are on the graph in a), which means that your domain should actually be $x \in [-3, 4]$ in interval notation to indicate that $x = -3$ and $x = 4$ are included. This would be reflected in the set-builder notation by using inequalities with "or equal ...


1

Presumably you want a continuous (or better) such function, because otherwise the problem's trivial. :) Also, I assume $L, C, A, a, b$ are all constants. Recall the situation in one variable (that is, if I'm given the values of a one-variable function at two points): if I want a function $g$ such that $g(a)=u$ and $g(b)=v$, one easy way to get such a $g$ is ...


0

I know that, $\sin\theta=\frac{y}{r}$, and, $\cos\theta=\frac{x}{r}$. Probably you mean the "trigonometric functions as quantities computed from a right triangle" definition. To make it clearer that everything is well defined it is better to use the more modern "point on a fixed circle of radius $r$" definition, but this answer will use the presumed ...


1

In general we have $$f_{\omega a+b}(c) \approx \underbrace{3 \rightarrow 3 \cdots 3 \rightarrow 3}_{a+1} \rightarrow c \rightarrow (b+1)$$ Let's evaluate some small numbers. $f(1)=2$, $g(1)=1$, $g(2)=4$ $$g(3)<f(2)=f_{\omega2}(2)=f_{\omega+2}(2)=f_{\omega+1}(f_{\omega+1}(2))=f_{\omega+1}(f_{\omega+1}(2))=f_{\omega+1}(f_{\omega}(8))\approx3 \rightarrow 3 ...


0

Set $f(x)=2^x$ $$\sum_{i=0}^{n-1} f(i)=\sum_{i=0}^{n-1} 2^i=2^n-1<2^n=f(n)$$


3

$\int_{-\infty}^xe^{2t}\,dt=e^{2x}/2.$


1

By contradiction: assume $y > \frac{1}{2}$ $$\frac{x}{x^2 +1} > \frac{1}{2} \Rightarrow 0 > (x-1)^2$$ and $y<-\frac{1}{2}\Rightarrow 0 > (x+1)^2 $. But we also see that $y(-1) = -\frac{1}{2}$ and $y(1) = \frac{1}{2}$.


1

by $AM-GM$ we have $\frac{x^2+1}{2}\geq |x|$, from here we obtain $\frac{|x|}{x^2+1}\le \frac{1}{2}$



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