# Tag Info

25

Let us define the functions: $$f(a,b)=\frac{\sqrt{a}}{\sqrt{b}}\quad\,\,\text{and}\,\,\quad g(a,b)=\sqrt{\frac{a}{b}}.$$ Then $f$ and $g$ AGREE on the intersection of their domains, but they have different domains: $$\mathrm{Dom}(f)=\{(a,b): a\ge 0,\,\,b>0\},\\ \mathrm{Dom}(g)=\{(a,b): a\ge 0,\,\,b>0\}\cup\{(a,b): a\le 0,\,\,b<0\}.$$ Strictly ...

12

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.

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 $\sqrt[3]{-x} = \sqrt[3]{(-1)x} = \sqrt[3]{-1}\sqrt[3]{x} = -\sqrt[3]{x}$

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 ...

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 ...

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.

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 $\;I-2\cdot II\;$ .

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 ...

4

EDDIITT: pretty good approximation ( for $x>4,$ say) with $$\color{blue}{ h(x) \approx x^{\sqrt 2} + \frac{x^{ \left(\sqrt 2 - 1 \right)}}{\sqrt 2} + (1 - \sqrt 2 ) }$$ It would take me a few days to actually go through the process again, but the following things are true: as long as you just want $C^1$ with no hope of extending to the complex numbers, ...

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 ...

4

There is no such injective function. To solve it we first see that if $x^2 = x$ (which is the case for $x=0$ and $x=1$) then $f(x^2) = f(x)$ and $f(x^2) - f^2(x) \geq \frac{1}{4}$ becomes (for $x=0$ or $x=1$) $$f(x) - f^2(x) - \frac{1}{4} = -\left(f(x)- \frac{1}{2}\right)^2 \geq 0$$ but this is only possible ($-a^2 \geq 0$ implies $a=0$ since a square ...

3

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 ...

3

I like using the triangle inequality here instead: $$|x + 2| = |(x - 2) + (4)| \leq |x - 2| + |4| < 1 + |4| = 5$$

3

Actually, you are right- as functions of two real variables, they have different domains. For example take $a=-1,b=-1$

3

You stumbled upon a classic counterexample to the "theorem" $f\colon \mathbb{R} \to \mathbb{R}$ has an horizontal asymptote if and only if $f'(x) \to 0$ as $x\to \infty$. Now, your function $f(x) = \log(1+x^2)$ is a counterexample to the "$\Leftarrow$" implication, since $\lim_{x\to \infty} f(x) = + \infty$. This can be shown by many means; but let's try ...

3

Suppose that $f$ is continuous, and not the zero function, and put $M={\rm Sup}\{f(x), x\in [0,1]\}$. There exists $x_0$ such that $M=f(x_0)>0$. Taking $x=\sqrt{x_0}$, we get $\displaystyle 4\sqrt{x_0}f(x_0)=f(x)+f(1-x)\leq 2M$, hence $\displaystyle 4\sqrt{x_0}\leq 2$, and $\displaystyle x_0\leq \frac{1}{4}$. By the functional equation with $x=x_0$, we ...

3

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

3

$a_{n+1}-a_n\leq -a_n^2\leq 0$, hence $a_n$ is a decreasing sequence , since $a_n\to 0$ then $0\leq a_n\leq a_0=\frac{2}{3}$. Now $f:[0,\frac{2}{3}]\to \Bbb R$ by $f(x)=x-x^2$ we have $a_{n+1}=f(a_n)$. it is a continuous function on a compact $[0,\frac{2}{3}]$, hence it is uniformly continuous. EDIT: According the remark below we can extend $f$ to an ...

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 ...

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 $x \ne -\sqrt{2}$.

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 $$\frac{x^2-2}{x+\sqrt{2}} \equiv ... 2 You are using the fact that f is increasing to see that if \Delta x > 0, f(x+\Delta x) \ge f(x) \Rightarrow f(x+\Delta x) - f(x) \ge 0 \Rightarrow \frac{f(x+\Delta x) - f(x)}{\Delta x} \ge 0. Analogously for \Delta x < 0 you still have \ge 0. Now using that if g(x) \ge 0 \forall x \in U where U is a neighborhood of x_0, you get that ... 2 Number 1 is simple since f(\frac{p}{q}) = \frac{3p}{3q} = \frac{p}{q} we have f(x) = x for any x \in \mathbb{Q}, which clearly satisfies x=y \Rightarrow f(x)=f(y). Number 2 is a bit more tricky, but really all you have to do is like you wrote your self, prove that if \frac{p}{q} = \frac{p'}{q'} (which is (as you wrote) the same as pq'=p'q), then ... 2 Solving \tan(x)=x^2 is the same as finding the intersection of the two curves y=\tan(x) and y=x^2. If you plot these two curves for x \in [0, 2\pi], you will notice that, beside the trivial root x=0, they intersect just before x=\frac{3\pi}{2}. The function being very stiff, it is then better to search for the solution of \sin(x)=x^2\cos(x) as ... 2 Actually, to prove that f is injective, you have to show that f(p_1,q_1)=f(p_2,q_2) implies (p_1,q_1)=(p_2,q_2). But since f(p,q)=(r,q), then this immediately implies q_1=q_2. The rest of the argument you gave shows that p_1=p_2, thus (p_1,q_1)=(p_2,q_2). For the surjectivity, let (r,q)\in H. Let's find p such that (p,q)\in G and ... 2 I believe that you are right, here's another approach. Set y_n=n^2+1/n and x_n=n^2. Clearly x_n-y_n\to 0 then if f is uniformly continuous we'd have f(y_n)-f(x_n)\to 0 but$$f(y_n)-f(x_n) >n^2(n^2+\frac{1}{n}-n^2)=n\to +\infty$$Contradiction ! then f is not continuous on any interval (a,+\infty). 2 There are several problems with your proof. Firstly, in both directions of your proof, your statements "By hypothesis, ..." seem to assume that all proofs are by contradiction, and that you must refute the negation of the statement. What you are assuming is not a hypothesis. Secondly, your negation of the statement you wish to prove is incorrect. The ... 2 Here’s a technique for finding the first few terms of a formal power series representing the fractional iterate of a given function like f(x)=x+x^2. I repeat that this is a formal solution to the problem, and leaves unaddressed all considerations of convergence of the series answer. I’m going to find the first six terms of f^{\circ1/2}(x), the “half-th” ... 2 I'll expand on my comment to construct distinct functions f,g,h on all \mathbb{R}^2 that satisfy the given constraint. First, define \tilde{f}\equiv f-h and \tilde{g}\equiv g -h. The constraint can be rearranged to yield$$ (e^{\tilde{f}(x,y)} - 1)x + (e^{\tilde{g}(x,y)}-1)y = 0. $$Ignoring division by 0 issues, we can write$$ ...

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?

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