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0

Theorem For each $a\in (0,1)$ there are infinitely many $f\in C([0,1])$ so that $f(0)=0$, $f(1) = 1$ and $$(*)\ \ f (g(x) ) = af(x)\ \ \ \ \forall x\in [0,1].$$ Remark: We set $f(1) = 1$ just as a normalization as $(*)$ is linear, as observed in another answer. Proof of theorem: Let $$C = \{x\in [0,1]: \ g(x) = x\}.$$ $C$ is nonempty as $0\in C$. Let ...


4

Hint: What is $\frac{\mathrm{d}}{\mathrm{d}x}Ce^{-x}$, where $C$ is a real constant?


0

My answer... Now in order for z to be a function of x and y points must uniquely satsify the equation for example in function $x^4+y^4+z^4=16$ $$f(0,0,2)=f(0,0,-2)$$ Meaning more than one point satisfy the equation. Where as for the given function we can solve for itself uniquely given that we know (x,y). $$6x-4y+2z=10$$ $$z=5-3x+2y$$


3

Think of a critical point as a "good candidate" for a point at which a local extremum could occur. To come up with a sensible way of formalizing this, think about common places where local extrema occur: for, say, $f(x)=x^2$, it's where $f'(x)=0$, but for $g(x)=|x|$, it's where $g'(x)$ is undefined (i.e. at $x=0$). We also want a critical point of a function ...


2

Where does the function $|x|$ attain it's minimum? Where does $\sqrt[3]{x}$ change concavity?


0

A period of $\sin(x)+\{x\}$ would be a common period of $\sin(x)$ and $\{x\}$. That would require finding $k,n\ne0\in\mathbb{Z}$ so that $2\pi n=k$. That can't be done since $\pi\not\in\mathbb{Q}$.


1

If the function is periodic, then integer multiples of the periods would have to match, i.e. $2\pi n = 1m$ for some integer $n,m$. However this says that $2\pi = \frac{m}{n}$. This contradicts the irrationality of $\pi$.


0

"Behave nicely" must mean, among other things, that $f$ has a gradient $\displaystyle(\nabla f)(x_1,\ldots,x_n)$. This is a vector in $\mathbb R^n$ assigned to each point $(x_1,\ldots,x_n)$ in the domain of $f$. There is a chain rule that says $$ \frac{df}{da} = ((\nabla f)(x_1,\ldots, x_n))\cdot \frac d {da} (x_1,\ldots,x_n) $$ where the product is the ...


1

for $y = \alpha + (x + \beta)^3$ take the graph of $x^3$, shift it $\beta$ unit to the left\right, and then $\alpha$ units upwards\downward it depends on the sign of $\alpha,\beta$.


7

Hint: take the graph of $x^3$, shift it one unit to the left, and then two units upwards.


1

Clearly we need $0<\sin x,\cos x<1$ All Sin Tan Cos Rule says: we need $x$ to be in the first Quadrant i.e., $2m\pi<x<2m\pi+\dfrac\pi2$ where $m$ is any integer


1

you are defining a linear transformation h on C[0,1] defined by h(f)=f(g) and you are trying to find eigen vectors and values. (a) which you are considering is an eigen value and f is an eigen function. Why not consider analytic functions and then we see that g(x)=cx+dx^2+... while f(x)=A+Bx+Cx^2+.... Then we see that A=0. It seems the problem is solvable ...


4

Its reduced discriminant is $$\Delta'=4a^2-(12a^2-8a+8)=-8(a^2-a+1)<0$$ as $\,a^2-a+1$ has no real roots (its roots are the complex cubic roots of $-1$). Alternatively, completing the square you can show $\,a^2-a+1\ge \dfrac34$.


7

if you prove no real roots, in Fact we have $$(x+2a)^2+8a^2-8a+8=(x+2a)^2+8\left(a-\dfrac{1}{2}\right)^2+6>0$$ proof 2: $$\Delta =16a^2-4(12a^2-8a+8)=-32a^2+32a-32=-32\left(a-\dfrac{1}{2}\right)^2-24<0$$


2

Hint: Take $x,y$ such that: $$(f\circ f \circ f)(x) = (f\circ f \circ f)(y).$$ Then $\color{blue}{f}$ injective gives: $$\color{blue}{f}(f(f(x))) = \color{blue}{f}(f(f(y)) \implies f(f(x)) = f(f(y)). $$ Repeat until you conclude that $x=y$.


2

As you might have noticed, we are integrating with respect to $g$. So you need to express $f(x)$ as a function of $g(x)$, i.e. $f(g(x)) = $ something to use regular integration methods. So because $f(x)$ is known, try expressing it as function of $g(x)$ and apply regular methods.


0

From Hadamard's theorem about products versus the summability of powers of zeros, from the functional equation (and Phragmen-Lindelof) we know that $\sum {1\over |\rho|^\sigma}$ is (absolutely) convergent for $\sigma>1$.


1

$F^{-1}(r) = \{z \in \mathbb{C} : |z|=r\}$ for $r \in \mathbb{R}$ and $F^{-1}(z) = \emptyset$ for $z \in \mathbb{C} \setminus\mathbb{R}$.


1

Hint: Start by showing that if $f$ and $g$ are both injective, then $f\circ g$ is injective. (Once you've proven that, let $g=f\circ f$.) To show that $f\circ g$ is injective, suppose it is not and ask why? If it is not, then there are two values $x_1\neq x_2$ so that $f(g(x_1))=f(g(x_2))$. Is $g(x_1)=g(x_2)$? Why? Let $x_3=g(x_1)$ and $x_4=g(x_2)$. If ...


0

Note that $f$ is not only locally injective, it is also a local homeomorphism: Each point $x\in X$ has a neighborhood $U$ on which $f$ is injective. By local compactness, we can find a compact neighborhood $V\subseteq U$, and $f|_V:V\to f(V)$ is then a homeomorphism. You correctly noted that $f^{-1}(y)$ must be finite since it is compact and fibers are ...


0

In the real numbers the domain of the logarithm function ehich is defined as: $a^{x}=b$ implies, $x=\log_{a}(b)$. If we are talking about the real numbers, the domain of the the logarithm function is: $x\in[-\infty,\infty]$ $b\in[1,\infty]$ $a\in[2,\infty$] In the complex numbers the only thing that changes is the domain of $b$ it extends to $b<0$ ...


10

The easy argument is that the two equations are incorrect because they violate the definition. The logarithm function permits a base that is strictly positive and not equal to one, and a domain that is strictly positive. Another way to see that your two examples can lead us to trouble is that you lose two very important properties of the logarithm. Namely, ...


8

They are correct in the sense that $(-3)^{2}=9$ and $(-2)^{3}=-8$. The difficulty arises because you want to think of $\log_{a}(x)$ as the inverse function of $a^{x}$. If $a>0$ (and $a\neq 1$) then $a^{x}$ is a continuous, real-valued (whenever $x$ is real), injective function (so $\log_{a}(x)$ makes sense). If $a=1$ then $a^{x}$ is not injective and ...


0

Absolutely not. What would then be $\log _{-3} 7$?


2

Any function of the form $$ y = ( a_1x-b_1)^2(a_2 x -b_2)^2 ( a_3 x - b_3)^2 $$ will have your triple well form with centres $b_i/a_i$, this fixes 3 free variables. To determine the heights, you want the derivative at the peaks to be zero. Solve the resulting equations and you'll have your conditions. In your example we have $$ y = x^2 (x-3)^2(x+3)^2 $$ You ...


2

It is unbounded. Take the sequence $(\frac{1} {n \pi} ) $.


1

I assume that by "domain" we mean an open, connected subset of $\mathbb{R}^n$. Note, in particular, that if $D$ is a domain and $K\subseteq D$, then $K\cap \partial D=\varnothing$. Let's solve this using the contrapositives. First suppose that $f$ is not proper. Then there exists a compact set $K\subseteq D_2$ such that $f^{-1}(K)$ is not compact. But since ...


0

By definition $\; f:X\to X\;$ means that $\;\forall a\in X\;( f(a)\in X)$ Now, $\;f: X\to X\;$ is injective iff $\;\forall a\in X\;\forall b \in X \;\Big( f(a)=f(b) \to a=b\Big)\;$. So first show that: $\;\forall n\in\Bbb Z^+\;\forall a\in X\;\forall b\in X\; \Big(f\mathop{\circ}^{n} (a)=f\mathop{\circ}^n (b) \to ...


3

The issue I seewith your proof is that we haven't really established any form of contradiction. You can do it more directly by simply proving that composition of injective functions is injective. Which is done here


2

You can also prove it directly. For simplicity let $Y^k$ be $Y$ nested $k$ times. We'll use the fact that function composition is associative. Suppose $Y^3(x)=Y^3(y)$ for $x,y\in X$. We can also write this as $Y(Y^2(x))=Y(Y^2(y))$. By injectivity of $Y$, it follows that $Y^2(x)=Y^2(y)$. In other words, $Y(Y(x))=Y(Y(y))$. Again by injectivity $Y$, this ...


1

Hint: $$x^2-a^2\geq 0\Longleftrightarrow x^2\geq a^2\Longleftrightarrow x\geq a\cup x\leq -a$$


2

Hint: This is one of the many functions from $S = \{ s_1, s_2, \ldots, s_6 \}$ to $\{ \alpha, \beta \}$: \begin{array}{c|c|c|c|c|c|c} s_i \in S & s_1 & s_2 & s_3 & s_4 & s_5 & s_6 \\ \hline f(s_i) & \beta & \beta & \alpha & \beta & \alpha & \alpha \end{array} This is a subset $A$ of $S$: $A = \{ s_1, s_2, ...


9

You need every $x$ in the domain to have a $y$ in the codomain because it's a function, and that's the definition of a function. It had nothing to do with being injective or surjective.


2

Hint: $\{\alpha, \beta\}$ is in bijection with the set $\{\text{yes, no}\}$. Given a set $A \subseteq S$ of stars, can you think of any way to encode information about $A$ into a function from $S$ to the set $\{\text{yes, no}\}$?


3

$$\frac{\sin^2 (\pi/2-x)}{\sqrt {\pi/2 -x}} = (\sin(\pi/2 -x))^{3/2}\cdot \left ( \frac{\sin (\pi/2-x)}{\pi/2 -x}\right )^{1/2} \to 0\cdot 1^{1/2} = 0.$$


3

It may help to let $\frac{\pi}{2}-x=t$. Then you will be on more familiar ground. For $t$ positive and not too large, we have $0\lt \sin t\lt t$. So our expression is positive and less than $t^{3/2}/\sqrt{2}$, which is less than $t$ if $t\lt 1$.


1

$$ \lim\limits_{x \to \pi/2-} \frac{sin^2(\pi/2-x)}{\sqrt{\pi-2x}} $$ Take $t=\pi/2-x$, $$ \lim\limits_{t \to 0^+} \frac{sin^2(t)}{\sqrt{2t}} $$ $$ \lim\limits_{t \to 0^+} \frac{1-cos(2t)}{2\sqrt{2t}} $$ Using L'Hospital's Rule, $$ \lim\limits_{t \to 0^+} \frac{2sin(2t)}{-\frac{\sqrt{2}}{\sqrt{t}}} $$ $$ =0 $$


2

A few illustrative examples: $f(f(x))=x^2+1$: It suffices to find $g\colon [0,\infty)\to[0,\infty)$ with $g(g(x))=x^2+1$ and let $f(x)=g(|x|)$. Note that $g\circ g$ is strictly increasing and has no fixed point. Therefore the sequence $x_0=0, x_1=\frac12$, $x_n=x_{n-2}^2+1$ is strictly increasing and diverges to $\infty$. Assume we have defined continuous ...


2

For injectivity: Assume $f(x) = f(y)$. Either $x < y$, $x = y$ or $x > y$. If $x < y$ then $f(x) < f(y)$ by monotonicity, which is a contradiction. If $x > y$ then $f(x) > f(y)$ by monotonicity, which is a contradiction. So $x = y$. For surjectivity: Let $i \in [f(a),f(b)]$. By the intermediate value theorem, there is an $x \in [a,b]$ such ...


2

For the surjectivity it's not right. Let $y\in f([a,b])$. Then there is an $x\in[a,b]$ such that $y=f(x)$. By the fact that $g$ is increasing, $y\in[f(a),f(b)]\subset f([a,b])$. The fact that $[f(a),f(b)])$ is obvious by the intermediate value theorem.


3

This is a fairly simple question, and questions like these are often asked by professors at the beginning of one's mathematical path, so that's what I'll assume. By that assumption, I will also presume that the level of strictness in your proof is similar to the typical level. For injectivity, you will need to be more specific. Did you prove a theorem ...


1

$\newcommand{\Reals}{\mathbf{R}}$The cross product of functions is defined pointwise: If $U$ is a non-empty open set in $\Reals^{n}$ for some positive integer $n$, and if $f = (f_{1}, f_{2}, f_{3})$ and $g = (g_{1}, g_{2}, g_{3})$ are functions from $U$ to $\Reals^{3}$, then $f \times g:U \to \Reals^{3}$ is defined by $$ f \times g = (f_{2}g_{3} - ...


1

Solution $\chi_E$ is continuous iff the inverse image of any open subset $U \subset \mathbb{R}$ under $\chi_E$ is open. If $0 \notin U$ and $1 \notin U$ then $\chi_E^{-1}(U) = \emptyset$ which is open. If $0 \in U$ and $1 \notin U$ then $\chi_E^{-1}(U) = \mathbb{R} \setminus E$ has to be open. So $E$ has to be closed. If $0 \notin U$ and $1 \in U$ then ...


1

In (1) the answer is negative even for a compact case. Let $X=[0,1]^2$, $Y=[0,1]$, and $f:X\to Y$ be the projection onto first coordinate. Put $$D=\{(1/n,1/n):n\in\Bbb N\}\cup\{(0,1)\}.$$ Then the set $D$ is discrete, but its image $f(D)$ is a convergent sequence $$\{1/n:n\in\Bbb N\}\cup\{0\}.$$


1

You have many questions, I'll try to adress them all. A binary relation, as you read is just some set $R$ which is a subset of the cartesian product of two sets $A$ and $B$, that is $R \subseteq A\times B$. An example may ilustrate this: Let $A=\{\dots,-4,-2,0,2,4,\dots \}$ (the set of even numbers), $B=\{1,3,5\}$. Then a relation $R_1$ could be ...


2

$$f(x)=[(2x-5)(3x+1)^{-1}]^{1 \over 2}$$ So, $$\begin{align} f'(x) & =\frac12\left(\frac{2x-5}{3x+1}\right)^{-1/2}\frac{(3x+1)2-(2x-5)3}{(3x+1)^2},\quad\text{chain and quotient rules}\\ & = \frac12\left(\frac{2x-5}{3x+1}\right)^{-1/2}\frac{17}{(3x+1)^2}. \end{align}$$ This simplifies to Lucas' answer.


5

this is better done with logarithmic differentiation. here is how it works. we have $$y =\sqrt{\frac{2x-5}{3x+1}}\to \ln y = \frac12\left(\ln(2x-5) - \ln(3x+1)\right) $$ differencing the las equation we get $$\frac{2}{y}\frac{dy}{dx} = \frac{2}{2x-5} - \frac{3}{3x+1 } = \frac{17}{(2x-5)(3x+1)}$$ you can simplify more if it is needed.


3

You have right, from here you have to use this formula: $(u^n)'=n\cdot u^{n-1}\cdot u'$, where $u=(2x-5)(3x+1)^{-1}\Rightarrow u'=\frac{17}{(3x+1)^2}$ So you'll obtain: $\frac{1}{2} \cdot u^{-\frac{1}{2}} \cdot u'=\frac{1}{2}\cdot (2x-5)^{-\frac{1}{2}}(3x+1)^{\frac{1}{2}}\cdot\frac{17}{(3x+1)^2}=\frac{17}{2\sqrt{2x-5}(3x+1)^{\frac{3}{2}}}$


1

this comes from the first derivative of $h(x)=\frac{ax+b}{cx+d}$ this is $$h'(x)=\frac{ad-bc}{(cx+d)^2}$$


1

It's too complicated and some aguments seem to be not justifief: e.g. Why do ypu try to prove $x> 0$ for all $x$? $x$ is the variable, it can take any value. As you noticed, we can't have $f(x=0, because this is the same as $\lvertf(x)\rvert=0$, which would imply that if $x' From there, as $f$ is continuous, $f(x)$ has always the same sign (else it ...



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