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0

Firstly, let $g(x)$ be equal to $x^2+x$. Now we can say that $g(x)=g(y)$ is equivalent to $x=y$ or $x+y=-1$. So $g(g(x))=g(g(y))$ means that $g(x)=g(y)$ or $g(x)+g(y)=-1$. But $g(x)=\left(x+\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}$, so the second case can't take place. Thus, $g^n(x)=g^n(y)$ iff $g(x)=g(y)$ for every positive integer $n$. Also $f$ ...


2

There is no such $f:\mathbb C\to\mathbb C$. See this paper: When is $f(f(z)) = az^2 + bz + c$ ? by R. E. Rice, B. Schweizer and A. Sklar The American Mathematical Monthly, vol. 87, no. 4 (Apr., 1980), pp. 252–263 More generally, they prove that a quadratic polynomial has no iterative roots of any order.


0

This touches $y=x$ at $x=0$. Try a Taylor series $f(x)=x+a_2x^2+a_3x^3+a_4x^4+...$ Feed it into the formula $f(f(x))=x+x^2$, and just keep $x$ and $x^2$ terms. I think you get $a_2=1/2$. $$f(f(x))=f(x)+a_2f(x)^2+...\\=x+a_2x^2+...+a_2(x^2+...)=x+2a_2x^2+...$$ Feed the formula back into $f(f(x))=x+x^2$, and this time keep $x^3$ terms as well. Perhaps you ...


2

Actually, to prove that $f$ is injective, you have to show that $f(p_1,q_1)=f(p_2,q_2)$ implies $(p_1,q_1)=(p_2,q_2)$. But since $f(p,q)=(r,q)$, then this immediately implies $q_1=q_2$. The rest of the argument you gave shows that $p_1=p_2$, thus $(p_1,q_1)=(p_2,q_2)$. For the surjectivity, let $(r,q)\in H$. Let's find $p$ such that $(p,q)\in G$ and ...


1

In a one-phase, one-component thermodynamic system there are two independent state variables. Any state function (e.g., entropy, pressure, density, etc.) can be formulated as a function of any two (take-your-pick) variables. In your example, you start with density and temperature as the independent variables but it could have been pressure and entropy. ...


1

It's a pretty interesting question, think of it in the followin way. You're working in two different scenarios! In the first, $T,\rho$ are both independent of each other. In the second, you're defining $T$ as something that depends on $\rho$; it's now a function of $\rho$. It's no wonder you're getting different answers - you're working in different ...


2

When you differentiate $p(\rho,T)$ with respect to $\rho$, you vary $\rho$ while you hold $T$ fixed, which you can do because $T$ is not a function of $\rho,$ and as a result, $p$ varies. On the other hand, when you differentiate $T(\rho,p)$ with respect to $\rho,$ now $p$ is not allowed to vary at all, and instead $T$ varies. In other words, clamp $p$ to a ...


0

I am going to replace some definitions by obviously equivalent ones. The benefit will be that the proofs are easier to understand. I am writing down very detailed proofs that you would not normally find in a college-level math textbook. Definition: For $n \in \mathbb{N}$ define $[n] = \{k \in \mathbb{N} \mid k < n\}$. Definition: The image of a map $f : ...


1

Let me address quickly the first part of the proof, $\Rightarrow$. You don't need to do it by contradiction. You can show that if $f\colon X\to A$ is a bijection, where $A$ is a proper subset of $X$, then there is an injection from $\Bbb N$ into $X$, and therefore $X$ is infinite. To your second question, perhaps it would have been better to say "recursive ...


1

In the case $\Rightarrow$ you are actually arguing by contradiction. Note that $A\subset X, A\ne X\implies |A|< |X|$ if $X$ is finite. In other case you may have $|A|=|X|.$ For example, $\mathbb{N}\subset \mathbb{Z}.$ Now consider the case $\Leftarrow.$ The denumerable subset $A$ is constructed by induction. You start consider any element in $X,$ say ...


4

An inverse of an odd function is odd: $\ f(-x)\, =\, f(-f^{-1}\!fx) \overset{f^{-1}\rm\ odd} = f f^{-1}(-fx)\, =\, -fx$ Remark $\ $ Group-theoretically this may be viewed as a special case of the following observation: $\ $ if $\ g^{-1} = g\ $ then $\,g\,$ commutes with $f^{-1}\!$ $\iff$ $\,g$ commutes with $\,f.\,$ Indeed, with $\,g(x) = -x,\,$ and the ...


8

A function $f$ is odd if for all $x$ in the domain of the function, we have $f(-x)=-f(x)$. In the context of your problem, note that $\sqrt[3]{-x} = \sqrt[3]{(-1)x} = \sqrt[3]{-1}\sqrt[3]{x} = -\sqrt[3]{x}$


2

You are completely correct if you send the second blue dot in $B$ to some element in $C$. Another possible example: let $f$ be the identity on some set $A$ with at least two elements, and let $g$ send all elements of $A$ to a single point $\{c\}$. Then $f \circ g$ is not injective, hence not bijective.


2

I'll try to answer with some comments. This approach to structures origins with Roland Fraïssé. In Heinz-Dieter Ebbinghaus & Jörg Flum & Wolfgang Thomas, Mathematical logic (2nd ed, 1984), we have : XI.1.1 Definition [page 180] : partial isomorphism, XI.1.3 Definition [page 182] : two structures $\mathfrak A$ and $\mathfrak B$ are ...


1

Suppose $(x,y)$ is an intersection point of $f$ and $f^{-1}$. Then $f(x)=y$ and $f^{-1}(x)=y$, thus $f(x)=y$ and $f(y)=x$. If $x < y$ this would mean that $f$ is decreasing on $[x,y]$, whereas $x>y$ would mean that $f$ is decreasing on $[y,x]$. Hence $x=y$, implying that $(x,y)$ lies on the line $y=x$.


0

Suppose $g(x_1, ..., x_n) = g(x_1 + a, ..., x_n + a)$ for all $a \in \mathbb{R}$ (1). Then define $f(y_1, ..., y_{n-1}) := g(y_1 + ... + y_{n-1}, y_2 + ... + y_{n-1}, ..., y_{n-1}, 0)$ Then $f(x_1 - x_2, ..., x_{n-1} - x_n) = g(x_1 - x_n, x_2 - x_n, ..., x_{n-1} - x_n, x_n - x_n) = g(x_1, x_2, .., x_{n-1}, x_n)$ (the last equality taking $a = x_n$ in (1)) ...


0

We may suppose that $a\not =0$. The line $x+y=|a|$ passes through two points $(0,|a|),(|a|,0)$. And the line $ax-y=1\iff y=ax-1$ passes through the point $(0,-1)$ with the slope $a$. Since the two lines intersect in the first quadrant, the slope $a$ of the second line has to be larger than $a_0$ such that $y=a_0x-1$ passes through the point $(|a_0|,0)$. ...


6

Should mean that $f$ is continuous on $\mathbb R$. This is also often written as $C^0(\mathbb R)$ where $C^n(\mathbb R)$ means that the functions in this set are $n$ times continuously differentiable.


1

Here is a hint: estimate $\vert f(1/2)-f(0)\vert$ and $\vert f(1)-f(1/2)\vert$ just like you presumably did (using Taylor's formula), and then use the triangle inequality.


0

If C and c are constant: $$cf'(cx)=Cf'(x)$$ $$f'(cx)=\frac Ccf'(x)=\frac{C^2}{c^2}f'(\frac xc)=\cdots=\frac{C^n}{c^n}|_{n\to\infty}f'(0)$$ If $c>1$: $$f(cx)=\frac{C^n}{c^n}|_{n\to\infty}f'(0)x+C$$ $$f(x)=\left[\lim_{n\to\infty}\left(\frac Cc\right)^n\right]f'(0)\frac xc+\mathcal{Constant}$$ If $C>c$: If $f'(0)=0$ , $f(x)=\mathcal{Constant}$ If ...


2

Number 1 is simple since $f(\frac{p}{q}) = \frac{3p}{3q} = \frac{p}{q}$ we have $f(x) = x$ for any $x \in \mathbb{Q}$, which clearly satisfies $x=y \Rightarrow f(x)=f(y)$. Number 2 is a bit more tricky, but really all you have to do is like you wrote your self, prove that if $\frac{p}{q} = \frac{p'}{q'}$ (which is (as you wrote) the same as $pq'=p'q$), then ...


0

Let $g(x)=g(y)$Now $f$ onto and note that $x,y \in B$ hence $ \exists a,b \in A : f(a)=x,f(b)=y$ So $g(f(a))=g(f(b))\Rightarrow gof(a)=gof(b)\Rightarrow a=b$(Since $gof$ one one and $a,b\in A$) Then $a=b\Rightarrow f(a)=f(b)\Rightarrow x=y\Rightarrow g$ one one


0

You could try proof by contradiction. Assume g is not one-to-one, then $\exists y_1 \neq y_2$ whereas $g(y_1)=g(y_2)$. Since f is onto, $\exists x_1,x_2$ s.t. $f(x_1)=y_1, f(x_2) = y_2$. Of course $x_1 \neq x_2$, but $g(f(x_1)) = g(f(x_2))$. So this contradicts with the condition that g(f(x)) is one-to-one.


0

$g(f(x))$ is one to one implies that if $g(f(a_1))=g(f(a_2))$, then $a_1=a_2$ $f:A\to B$ is onto implies that for every element $b\in B$, there is some element $a\in A$ for which $f(a)=b$. Now you want to show that $g$ is one to one. You can start off by saying "Suppose $g(b_1)=g(b_2)$", then try to show that $b_1=b_2$ using the two observations I listed ...


0

The first answer: functions are never undefined. It's kind of a loaded word: don't talk about something undefined, because, then what are we talking about? What is undefined is an expression in $x$'s and $y$'s that tell you about a function; an expression that doesn't know how to define a function everywhere. So we need to talk about how to go from an ...


0

Let $x_n=-1+\frac{1}{n}$ and $y_n=-1+\frac{1}{n+1}$ for $n=1,2,3,\ldots$. Notice $\{x_n\}_{n \in \mathbb{N}},\{y_n\}_{n \in \mathbb{N}} \subset (-1, \, \infty)$. Now given $\delta>0$ there is $n \in \mathbb{N}$ so that $|x_n-y_n|=\frac{1}{n(n+1)}<\delta$, but $|f(x_n)-f(y_n)|=1$.


1

You are almost correct. $|x|$ can never be less than $0$, by definition. Now, for $x=0$, what is $f_n(0)?$ Also, if $|x|>0$, then there always exists $N$ such that for all $n>N$, $|x|\geq 1/n$ (This follows from the fact that the limit of $1/n$ is 0). So, $f_n(x)=1$ for all $n>N$, which means $\cdots$?


3

If I were to draw this graph, I would get a straight line with a "gap" at $x = -b$. But if I were to approximate b in the following manner $$\frac{x^2 - b^2 }{ x + \bar{b}}$$ where $b$ is equal to some sort of approximation of $b$ (so imagine $b$ is $\pi$ and $\bar b$ is 3.1415).... now, if I were to draw this graph, the result would be a ...


2

It is important to realise what you are doing when we say that a common factor "cancels" from the numerator and denominator. We are actually dividing both the numerator and denominator by that common factor. As you have already pointed out: you can't divide by zero. It is true that $$\frac{x^2-2}{x+\sqrt{2}} \equiv ...


0

For relations $R\subseteq A\times B$ and $S\subseteq B\times C$ one define $R;S=\{(a,c)\in A\times C|\exists b\in B: (a,b)\in R\wedge (b,c)\in S\}$. If $A=B=C=\{0,1\}$ and $R=S$, then $R\subseteq \{(0,0),(0,1),(1,0),(1,1)\}$. Suppose $R=\{(0,1),(1,0)\}$, then $R;R=\{(0,0),(1,1)\}\nsubseteq\{(0,1),(1,0)\}=R$. Counter example thus.


3

When you cancel a factor from the numerator with one in the denominator, you are saying that the ratio of the two is the same as the number one. This is true unless the two factors you happen to be cancelling are both zero. In this case the ratio is not one, it is undefined. So technically the ratio you obtain is $x-\sqrt{2}$ if $x \ne -\sqrt{2}$.


0

Hint As $t\sim 0$ we have $$\tan(t) \sim t $$ Added: $$ \frac{x^2+y^2}{\tan(xy)} \sim_{t\sim 0} \frac{x^2+y^2}{(xy)}. $$ Now polar coordinates tells you that the limit does not exist since $$ \frac{x^2+y^2}{xy} = \frac{1}{\cos\theta\,\sin\theta} $$ depends on $\theta$. See here.


1

In an attempt to answer before its closed we want to show that the limit of the expression as $x,y \rightarrow 0$ is $0$. You can write it as $$\frac{xy}{\sin(xy)}\frac{x^2+y^2}{xy}\cos (xy)$$ Now $\frac{xy}{\sin(xy)} \rightarrow 1$ and $\cos (xy)\rightarrow 1$ so the problem it to investigate the limit of $\frac{x^2+y^2}{xy}$ but this does not exist so ...


0

Let $\displaystyle\dfrac{1+\cos\theta}{2+\sin\theta}=R$ As $\displaystyle0\le\sin\theta,\cos\theta\le1, R\ge0$ Using Weierstrass Substitution, setting $\displaystyle c=\cot\frac\theta2$ We have $$R=\frac{1+\dfrac{c^2-1}{c^2+1}}{2+\dfrac{2c}{c^2+1}}$$ $$\iff c^2=R(c^2+c+1)\iff c^2(R-1)+Rc +R=0$$ which is a Quadratic Equation in $c$ As $\theta$ is real, ...


0

Let $\displaystyle\dfrac{1+\cos\theta}{2+\sin\theta}=R$ As $\displaystyle0\le\sin\theta,\cos\theta\le1, R\ge0$ $\displaystyle\iff1+\cos\theta=2R+R\sin\theta\iff\sqrt{1+R^2}\cos\left(\theta+\arctan R\right)=2R-1$ Now, $\displaystyle-1\le\cos\left(\theta+\arctan R\right)\le1\iff -1\le\frac{2R-1}{\sqrt{1+R^2}}\le1$ ...


4

Another way of tackling this would be to make use of the half angle formulas for cos and sin:$$\sin(\theta)=\frac{2t}{1+t^2}$$$$\cos(\theta)=\frac{1-t^2}{1+t^2}$$this leads to:$$\frac{1+\cos(\theta)}{2+\sin(\theta)}=\frac{1}{t^2+t+1}=\frac{1}{(t+\frac{1}{2})^2+\frac{3}{4}}$$and from this one can observe that the maximum occurs when $(t+\frac{1}{2})$ is zero ...


0

Let $$ f(\theta) = \frac{1+\cos \theta}{2+\cos \theta}. $$ You are asked to prove that $\min f=0$ and $\max f=4/3$. You can find maximum and minimum values by studying the sign of the derivative.


6

$$\frac{1-(-\cos\theta)}{2-(-\sin\theta)}$$ represents the slope of a line on which $(2,1)$ and $(-\sin\theta,-\cos\theta)$ exist. Since $x=-\sin\theta,y=-\cos\theta$ satisfies $x^2+y^2=1$, we know that a point $(-\sin\theta,-\cos\theta)$ is on the circle whose center is the origin with the radius $1$. Now, you can draw both the circle and a point ...


7

Since $\sin(\theta) , \cos(\theta) \in [-1,1]$ LHS part is obviously true. for RHS part we have to prove $$ \frac{1+\cos(\theta}{2+\sin(\theta)} - \frac{4}{3} \le 0 $$ which is equivalent to, $$ \frac{3\cos(\theta)-4\sin(\theta)-5}{2+\sin(\theta)} \le 0$$ So it suffice to show that $$ 3\cos(\theta)-4\sin(\theta)-5 \le 0$$ since $2+\sin(\theta)$ will ...


0

The truthtable of your problem looks as follows: Only one LED is on for every given input combination. The LEDs indicate how many inputs are 1: black: 0 inputs are 1 blue: 1 input is 1 green: 2 inputs are 1 red: 3 or 4 inputs are 1 The terms in your expressions are already minimal in the sense that they cannot be reduced to fewer terms via ...


3

$x \longmapsto -\tan \left[\frac{\pi}{2} \left(\frac{2 x}{R}-1\right)\right]$


5

$$\begin{align*}x=5:&\;\;\;\;I\;\;\;\;\;\;f(5)+2f\left(\frac15\right)=5\\ x=\frac15:&\;\;\;\;II\;\;\;f\left(\frac15\right)+2f(5)=\frac15\end{align*}$$ Now try $\;I-2\cdot II\;$ .


1

No, that is not standard and is not likely to be understood as intended. As Sam points out in a comment, $C(M)$ is usually understood as $C(M,\mathbb R)$. So for your purpose a better notation should be $C(C(\mathbb R),C(\mathbb R))$. Furthermore, echoing PVAL's comment, it is not obvious what topology you have in mind on $C(\mathbb R)$, and this ...


2

Other properties for which this true include "odd", "unbounded" and "one-to-one".


2

The claims are false for "analytic" and "smooth". If $g(x)=x^3$ and $f(x)=x^{1/3}$, then $f\circ g$ is the identity.


0

For $m=1$ and any $n \geq 1$, we can consider the function $f \cdot g: \mathbb R^n \to \mathbb R$ defined by $$x \mapsto f(x)g(x).$$ For $n=1$ and $m>1$, there isn't as obvious a way to define multiplication of functions. Trying to mimic the case for $n=1$, we would want to define $f\cdot g: \mathbb R \mapsto \mathbb R^n$ by "multiplying the resulting ...


2

$x\pm 3$ means $x+3$ or $x-3$ not $x+3-3$, so the $\pm$ doesn't cancel. Are you thinking about simplifying the quadratic formula perhaps?


0

Generating function $$ \sum_{k=0}^\infty A_k(x)\frac{t^k}{k!} = e^{g(x+t)-g(x)} $$ obtained from Taylor's theorem. Take the rest of the Wikipedia page Hermite Polynomials and see what else you can imitate.


0

You can use Faà di Bruno's formula: $$ {d^n \over dx^n} f(g(x)) =\sum \frac{n!}{m_1!\,m_2!\,\cdots\,m_n!} f^{(m_1+\cdots+m_n)}(g(x)) \prod_{j=1}^n\left(\frac{g^{(j)}(x)}{j!}\right)^{m_j} $$ where the sum is over all $(m_1,m_2,\dots,m_n)$ such that $$ 1\cdot m_1+2\cdot m_2+3\cdot m_3+\cdots+n\cdot m_n=n $$ With $f=\exp$, this becomes $$ {d^n \over dx^n} ...


2

Hint: $[-1,1]$ is closed and bounded (compact), and $f',f''$ is bounded on $[-1,1]$.



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