New answers tagged

0

You just need to find the absolute maxima or minima of $f(x)$ in its domain $[-3,3]$. Write $f(x)=√g(x)$, where $g(x)=9-x^2$. Now, $g(x)$ has an extrema if $g'(x)=0$ which gives $x=0$ as the point of maxima as $g''(0)<0$. Hence $f(0)=√g(0)=3$, $f(-3)=f(3)=0$. Clearly, $Range(f)=[0,3]$.


1

$$9-x^2\ge 0\implies D_f=[-3,3]$$ set $x=3\sin\theta$ $$y=\sqrt{9-9\sin^2\theta}=3\,|\,\cos\theta\,|\implies R_f=[0,3]$$


1

The usual convention is that the principal square root function $f(x) = \sqrt x$ returns only the non-negative root. So in this case, the range corresponds to the semicircular upper half of the circle defined by $x^2 + y^2 = 9$. Which gives the range as $[0,3]$.


1

You correctly found the domain, although you meant that the expression that's under the square root, which is called the radicand, must be non-negative. Observe that $y = \sqrt{9 - x^2} \implies y \geq 0$. Moreover, if we square both sides of the equation, we obtain $y^2 = 9 - x^2$, which is equivalent to $$x^2 + y^2 = 9$$ This is the equation of a circle ...


1

Hint: Since a square is non-negative, $0\le 9-x^2\le9\;$ on $\;[-3,3]$, and $\;\sqrt x\;$ is a continuous increasing function.


4

Let: $$f(x)={x+\tan{x}\over A +B(x+\tan{x})^{2n}}$$ Set $x:=-u$ \begin{align} f(-u)&={-u+\tan{-u}\over A +B(-u+\tan{-u})^{2n}} \\ &={-u-\tan{u}\over A +B(-1)^{2n}(u+\tan{u})^{2n}} \\ &=(-1){u+\tan{u}\over A +B(u+\tan{u})^{2n}} \\ &=-f(u) \end{align} Since x and u are dummy variables, $f(-x)=-f(x)$. Thus $f(x)$ is an odd function. Your ...


1

Let us call the integrand as $$f(x)={x+\tan{x}\over A +B(x+\tan{x})^{2n}}$$ Then it is quite evident that it is an odd function of $x$ $$f(-x)=-f(x)$$ and hence you can easily conclude $$\int_{-n}^{n}f(x)dx=0$$


0

HINT: Use $$I=\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$$ $$\implies I+I=\int_a^b[f(x)+f(a+b-x)]\ dx$$ Here $a=?,b=?$ and $\tan(-x)=-\tan x$


1

I see what you're getting at based on our comment discussion now. The conditions also change. For example, say we have $f(x) = x^2$ if $x > 10$. Then $f(2x) = (2x)^2$ if $2x> 10$. (Other pieces are irrelevant for this discussion and the same thing happens to them, so it's sufficient to consider one piece.) Similar reasoning for horizontal ...


1

The composition of $g$ and $f$ is defined. Let's examine the definition of a composite function. Let $u$ and $v$ be any two functions. We define a new function $u \circ v$, the composition of $u$ and $v$, by $$(u \circ v)(x) = u(v(x))$$ The domain of $u \circ v$ is the set of $x$ in the domain of $v$ such that $v(x)$ is in the domain of $u$. Since $f: (...


2

Sorry, I got it! For anyone liking future help I figured it out with my textbooks: $$f(g(x))\\ g(x) = x + 3\\ f(x + 3) = f(g(x)) = (x + 3)^2 -3$$ $$g(f(x)) = g(x^2 - 3) = x^2 - 3 + 3 = x^2$$


1

The minimum of $|f(z)|$ on $D$ is not $1$ but $1/e$, which is achieved at $z=-1$ on the boundary.


4

$\sin x=0$ is rejected because when $\sin x=0$, the value $\csc x$ does not exist, so $(\sin^2 x)(1+\csc x)$ does not exist, so it can't be equal to $0$ or to anything else either.


2

Let's say $(\sin x)^2=0$. Then, take square root of both sides to get $\sin x=0$. Now, we have either $x=0$ to $x=\pi$. Then, substitute back in: $$x=0 \implies (\sin 0)^2(\csc 0+1)=\text{undefined because} \csc 0 \text{ is undefined}$$ $$x=\pi \implies (\sin \pi)^2(\csc \pi+1)=\text{undefined because} \csc \pi \text{ is undefined}$$ Always remember to ...


0

It can be simply defined as an infinite product $$f(x)=\prod_{n=1}^{\infty}\cos^{2n}(\frac{\tau x}{2})$$ Where $\tau=2\pi$


1

There exists an infinite family of analytic solutions. But most are not clear until you have the machinery of Q-calculus! We start by "standardizing" your invariance equation into a form that is ripe for analysis using "q" series. So $$ f(x) = a f(bx) \rightarrow \frac{f(bx) - f(x)}{(b-1)x} = \frac{(1-a)}{(b-1)x} f(x)$$ You can observe that ...


1

Your question doesn't make sense as stated. The notation $f:\mathbb{R}^2/\lbrace(0,0)\rbrace \to \mathbb{R}$ means that $f$ is a function with domain $\mathbb{R}^2/\lbrace(0,0)\rbrace$ and codomain $\mathbb R.$ You surely didn't mean to ask "is this function a function?" Now the question "Is the expression $\arctan (x/y)$ well defined for all $(x,y) \ne (0,...


2

$$x^{ log_{ 2 }x }+\frac { 16 }{ x^{ log_{ 2 }x } } =17\\ { x }^{ 2\log _{ 2 }{ x } }-17x^{ log_{ 2 }x }+16=0\\ \left( x^{ log_{ 2 }x }-16 \right) \left( x^{ log_{ 2 }x }-1 \right) =0\\ x^{ log_{ 2 }x }=16\Rightarrow \log _{ 2 }{ x^{ log_{ 2 }x } } =\log _{ 2 }{ 16 } \Rightarrow { \left( \log _{ 2 }{ x } \right) }^{ 2 }=4\Rightarrow \log _{ 2 }{ x } =\pm ...


2

let $$y=x^{log_2x}$$ your equation becomes, $$y+\frac{16}y=17$$ solve it, you get two solutions: 1 and 16. Now it becomes less horrible, $$x^{log_2x}=1 ~or~ 16 $$ This leads to solution $x =1$, $2^2$ and $2^{-2}$


0

Adding to @Adriano's answer, if you want to reflect $y=f(x)$ across $y=mx+b$ you have to reflect the function by the x-axis and then rotate it by $\theta=\pi+2\tan^{-1}(m)$ radians. So if we have $y=f(-x)$, then by using the rotation matrix in linear algebra, we can set $x=x'\cos(\theta)+y'\sin(\theta)$ and $y=y'\cos(\theta)-x'\sin(\theta)$. So the ...


1

The expression $\arctan(\frac{x}{y})$ is not defined for $y=0$, so with that as the definition your $f$ will only be defined on $\mathbb R\times(\mathbb R\setminus\{0\})$. But on that set it assign exactly one value to each pair $(x,y)$ and that makes it a function. If you want a function that has the values you mention in a comment, you need to define it ...


2

Well, first notice that $\frac{x}{y}$ is not defined for $y=0$, so you should not consider in the domain the set $A = \{(x, y) : y = 0\}$. Now, you should verify when the function $f(z) = arctan(z)$ is well-defined.


1

Not sure what to say for minimum length. http://www.wolframalpha.com/input/?i=arctan%28x%2Fy%29


-1

for graph Y = x^2 + Kx -2 to cut x - axis y coordinate must be zero thus, x^2 + Kx -2 = 0 Now for this equation for solution to be finitly 2 , b^2 - 4ac > 0. here b = K , a = 1 and c = -2. for every value of K , b^2 - 4ac will be greater than zero. for example k = 0 => (0)^2 - 4(1)(-2) = 8 for k = -1 => (-1)^2 - 4(1)(-2) = 9 hence for all value of K ...


0

We are basically attempting to show that $$ y = x^2 +kx - 2$$ Equals 0, for exactly two, distinct x, for any choice of real number k. A trick here is to basically factor the expression, and show that the factors give rise to distinct roots. So we want to find an $a,b$ such that $$ (x-a)(x-b) = x^2 + kx - 2$$ And show that regardless of the $k$, $a \...


2

Use the quadratic function, with a= 1, b= k, and c= -2. There's three different outcomes for a quadratic: ${b^2-4ac}\gt 0 \rightarrow$ we get two solutions, which means we have two x-intercepts. ${b^2-4ac}=0 \rightarrow$ we get one solution, which means we have one x-intercept ${b^2-4ac}\lt 0 \rightarrow$ no real solutions, which means no x-intercepts. ...


2

We are looking for $n$ such that $3\pi$ is a period, but not necessarily the smallest period. We have $2\cos nx\sin\frac{5x}{n}=\sin(\frac{5x}{n}+nx)+\sin(\frac{5x}{n}-nx)$. So we require $3(n\pm\frac{5}{n})$ to be an even integer. Clearly that is impossible for $|n|>15$ or for $n$ not a factor of 15. But equally clearly any factor of 15 (positive or ...


2

You cannot ignore the exponent or the multiplication. In this case, set temporarily $t=\log_3x$ so the equation becomes $$ t^2-3t+2=0 $$ which has roots $t=1$ and $t=2$. Thus you get the two equations $$ \log_3x=1 $$ and $$ \log_3x=2 $$ Can you finish them? Your method would be sound, too, provided you did the decomposition right.


5

should be $$\left( \log _{ 3 }{ x } -1 \right) \left( \log _{ 3 }{ x } -2 \right) =0$$ $$\log _{ 3 }{ x } =1\Rightarrow \quad x=3\\ \log _{ 3 }{ x } =2\Rightarrow x=9$$


2

Graph it using Desmos. $$\tan\left(\frac{\pi}{2}\text{floor}(x)\right)$$ Now, it's easy to see that this is $0$ when $\text{floor}(x)$ is even and undefined when $\text{floor}(x)$ is undefined, so we have a period of $2$.


1

You must be talking about bounded linear operators, not limited. In normed vector spaces it holds that a linear operator is bounded if and only if it is continuous. This even is given as a definition sometimes. Now, if your definition of bounded linear operator is that it maps bounded sets to bounded sets then: From your expression for $||A||$ you get that $...


0

Let $f(n)$ and $g(n)$ be two function,then if $\lim \limits_{x \to \infty} \frac{f(n)}{g(n)} =0$ then $f(n)=O(g(n))$ and $g(n)=\Omega(f(n))$ if $\lim \limits_{x \to \infty} \frac{f(n)}{g(n)} =\infty$ then $f(n)=\Omega(g(n))$ and $g(n)=O(f(n))$ if $\lim \limits_{x \to \infty} \frac{f(n)}{g(n)} =k$(constant),then $f(n)=\theta(g(n))$ and $g(n)=\theta(f(n))$ ...


1

Your approach is lacking. You say Now,above equation could hold only if $c \lt 0$ ,which contradicts our assumption Which is the correct idea, but it is not entirely true. The above equation, for example, holds if $c=1$ and $n=10$. Now, I assume you know that my counterexample is not good, but it works because you forgot to say that the equation ...


1

The idea being discussed doesn't need a rigorous definition of wavelet, or even to use wavelets at all; the notion of a wavelet is, I think, mainly a clever way of generating a convenient basis in a systematic fashion starting from a basic shape. (the particular basis you listed could probably stand to be scaled so as to be orthonormal basis rather than just ...


0

a homogeneous function is a polynomial function which all the terms have the same degree. then in your example (in this case of one dimension) $v \rightarrow av + z$ is not a homogeneous polynomial since $z$ is a vector constant, not a variable.


0

as you go solving for $f^{-1}$ if $y = f^{-1}$ then: $X = \frac {y}{1-y^2}\\ xy^2 + x - y = 0$ Now here you divided through by $x$ in order to apply the quadratic formula. But in so doing, you have obliterated $x = 0$ and must make a note to yourself to return to this before you are complete. $y = \dfrac{-1+\sqrt{1+4x^2}}{2x} \text { if } x\ne 0$, or $x ...


0

$f^{-1}$ is actually defined at $0$, and $f^{-1}(0) = f^{-1}(f(0)) = 0$. The problem is that you arrived at the expression of $f^{-1}$ under the (implicit) assumption that $y \neq 0$. In fact, $$y = f(x) = \frac{x}{1 - x^2} \iff yx^2 - y+ x = 0$$ This is a quadratic equation in $x$ iff $y \neq 0$, so the obtained expression is valid only when $y \neq 0$. If ...


0

let us assume that the following equation is discontinuous at point c... f(x) + f(2x) + f(4x) where f is a continuous function. If that is the case, then the limit as x goes to c should not exist. However, the limit operation is linear. This means we have the following three limits that should exist: limit x -> c : f(x) = f(limit x -> c : x) = f(x) It ...


0

Well you shouldn't care actually. As $n$ gets bigger, all terms $n^i$ with $i<4$ are negligible when compared to $n^4$ (This is because $\lim_{n \rightarrow \infty} \frac{n^i}{n^4} = 0$ when $i<4$). Hence, you know that starting from a certain $n_0$ (you don't have to know its value yet), $n⁴ + 100n² + 50 < c n^4$ (for any $c>1$ you want). And ...


1

The biggest term in this polynomial is $n^4$, so you have to choose a term bigger than $n^4$. Since $2n^4 > n^4$, you can choose $2n^4$ to find some constant $n_0$ where $n > n_0 \implies f(n) \leq 2n^4$. In this case, $n_0=11$. However, you didn't have to pick $2n^4$. For example, you could've picked $3n^4$ since $3n^4 > n^4$. In this case, we ...


0

Well, you can basically take a guess for $c$. Notice if you were to take a larger $n$, you could make $c$ smaller. But $2$ seems right, so let's try that. You could then graph $2n^4$ and $n^4 + 100n^2 +50$ and see that our desired inequality holds for all $n\geq 11$.


1

If the indicated homeomorphisms are $h_A$ and $h_D$, just take $$g = h_B\circ f\circ h_A^{-1}$$ It's continuous and onto because each of the elements of the composition are.


4

This one is pretty straightforward. Intuitively, all we have to do is pick a function that maps set $A$ to the image of your initial fuction, and a function that "extends" the image of $f$ to set $B$, namely the identity function. Here, the functions are named "$map$" and "$ext$", respectively and are defined as follows: $map:A\rightarrow Imf\subseteq B, ...


0

Rather a comment than an answer. The only thing missing in Georgy's answer is an integral: $$ \int_0^\infty n^2xe^{-nx} dx = -\frac{1}{n} \int_0^\infty n^2x\, d e^{-nx} = \left[ - \frac{n^2x e^{-nx}}{n} \right]_0^\infty + \int_0^\infty n e^{-nx} dx = 0-\left[e^{-nx}\right]_0^\infty = 1 $$ And a picture, where the grey area is independent of $\,n$ , i.e. the ...


1

The first equality is true, because $f^2(x)$ is usually defined to be $(f \circ f)(x) = f\big(f(x)\big)$, and $(f \circ g)(x)$ is defined as $f\big(g(x)\big)$. Because of these definitions, the second equality does not hold. However, if the notation $f^2$ or $\circ$ is differently defined, any one or possibly both equalities may be true.


1

It is nowhere near as neat. $$\begin{align}\mathsf M_{X/\, Y}(t) =&~ \mathsf E(e^{tX/\, Y}) \\[1ex]=&~\mathsf E(\mathsf E(e^{tX /\, Y}\mid Y)) \\[1ex]=&~ \mathsf E(\mathsf M_{X\mid Y}(t/Y))\end{align}$$ Now, if $X$ and $Y$ are independent (as required for the summation formula you gave) then this becomes: $$\begin{align}\mathsf M_{X/\, Y}(t) =&...


1

Just to better asses the solution given by Kitter Catter, premised that $$ \eqalign{ & x = \left\lfloor x \right\rfloor + \left\{ x \right\} \cr & \left\lfloor { - x} \right\rfloor = - \left\lceil x \right\rceil \cr & \left\lceil x \right\rceil = \left\lfloor x \right\rfloor + \left\lceil {\left\{ x \right\}} \right\rceil \cr} $$...


2

Starting from $n=0$, even-$n =2k$ terms are $(4^k - 1)/3$ and the subsequent odd-$n = 2k+1$ terms are $2\times (4^k- 1)/3$. So overall: \begin{align} \sum_{k=0}^{49}\left[(1+2)\frac{4^{k} - 1}{3}\right] + \frac{4^{50}-1}{3} &= \sum_{k=0}^{49}(4^{k} - 1) + \frac{4^{50}-1}{3} \\ & = \sum_{k=0}^{49}(4^{k}) + \frac{4^{50}-1}{3}-50 \\ &= \frac{4^{50}-...


5

The sequence of $[2^n/3]$ for $n\in \mathbb{N}$ and $n\geq 2$ would be like this: $$1,\ 2,\ 5,\ 10,\ 21,\ 42,\ 85,\ \cdots $$ So, comparing the numbers in the sequence successively, you may guess that it is a recursive sequence like $a_1=1$, and for $n>1$, we have $a_n=\begin{cases} 2a_{n-1} & \text{if } n\ \text{ is even, }\\ 2a_{n-1}+1 & \...


9

I think I have an answer. Look at $\frac{1}{3}$ in binary format and you will notice that it is $0.\overline{01}$. This tells us something, namely that $[2^{2i}/3]+[2^{2i+1}/3] = 4^i-1$. This immediately tells me that your sum reduces to: $$\sum \limits_{i=0}^{50} [2^{2i}/3]+\sum \limits_{i=0}^{49} [2^{2i+1}/3]=\sum \limits_{i=0}^{49} [2^{2i}/3]+\sum \...



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