New answers tagged

3

First, we want to make sure that the term inside the logarithm is always positive, as $\log{x}$ is defined only for $x > 0$. $$x-x^2 = x(1-x) > 0 \implies 0 < x < 1$$ Next, we need the term inside the square root to be positive or equal to 0, so we get: $$\begin{align*} \log_{0.4}(x-x^2) &\geq 0\iff\\\frac{\ln(x-x^2)}{\ln{0.4}} &\geq ...


0

You have here two conditions on the domain: $$ x-x^2 >0 \tag{1}$$ because $\log_{0.4}(t)$ is defined only for $t>0$, and $$\log_{0.4}(x-x^2)\geq 0 \tag{2}$$ because $\sqrt{t}$ is defined only for $t\geq 0$. From $(1)$ we want that $x>x^2$ and so $x\in (0,1)$ because $x^2 \geq 0$ and $x^2\geq x$ for $x\geq 1$. For condition $(2)$, we know that ...


-1

First, I have to rewrite the question. If $X$ and $Y$ are two non-empty sets where $f:X\to Y$ is a function. A function $F:P(X)\to P(Y)$ is defined such that $$F(C) = \{f(x):x\in C\}\text{ for }C \subseteq X$$ and $$F^{-1}(D) = \{x:f(x)\in D\}\text{ for }D \subseteq Y$$ for any $A \subseteq X$ and $B \subseteq Y$ then: $F^{-1}(F(A))=A$ ...


1

It seems you are having more conceptual difficulties. The problem is fairly straightforward in this case, the solution (as mentioned by others) is quite easy to come up with. I just wanted to outline how one might approach this problem, since you did not know how to do just that. First we are asked to find a bijection between the sets $\{1,2,3,...\}$ and ...


0

Try the function $n \mapsto -n$


1

The question seems garbled, but I'll try to clear it up. We're given a function $f\colon x\mapsto x^2 - 3, \colon A\to \Bbb R$. defined on $A=\{x\mid x\ge 0\}$, the nonnegative reals. So $A = \Bbb R_{\ge 0}$. On $\Bbb R_{\ge 0}$, $f$ sends $0$ to $-3$, its strictly increasing, unbounded and continuous, and more. It's an injection, with range $B=\{x\mid ...


0

I have proved it as follows. Taking the expression for $f'(x)$, multiplying through by $(1+x^{2} \sin^{4} \frac{1}{x})$, we need to prove the following inequality, call it (a): $$\left(1+x^{2} \sin^{4} \frac{1}{x}\right) \arctan(x \sin^{2} \frac{1}{x}) + x\sin^{2}\frac{1}{x} > 2\sin\frac{1}{x} \cos\frac{1}{x} \qquad \mbox{(a)}$$ Observe that since ...


3

The domain is correct; the solution to $f(x)=1$ is correct too: indeed $$ \frac{2x-1}{x^2}=1 $$ becomes $$ (x-1)^2=0 $$ Now for the inequality it is essentially the same. You want to see for what value of $x$ you have $f(x)\le1$, that is, $$ \frac{2x-1}{x^2}\le1 $$ This translates into $$ (x-1)^2\ge0 $$ so every $x\in D_f$ satisfies the inequality. Or you ...


0

We can first calculate the 1st derivative. $$\frac{df}{dx} = \frac{d(2x-1)}{dx} \cdot x^{-2} + \frac{d(x^{-2})}{dx} \cdot (2x-1) = \frac{2}{x^2} - \frac{2(2x-1)}{x^3}$$ Set $\frac{df}{dx} = 0$, we have $x =1$. So the turning point is $x =1$. By calculating the second derivative, we can find out that this point is the maximum of the function. $f(x=1) = 1$. ...


0

Simple, use derivative, not approximations. You get: $(x\arctan(x\sin^2(\frac{1}{x})))'=\arctan(x\sin^2(\frac{1}{x}) + \frac{x}{1+x^2\sin^4(\frac{1}{x})}(sin^2(\frac{1}{x}) + \frac{x}{1+x^2\sin^4(\frac{1}{x})}(-2x\sin\frac{1}{x}\cos\frac{1}{x}(-\frac{1}{x^2}))$ so all are positive for $x \geq 1$.


3

$f(x)= \frac{x+a}{x^2-1}$ $a \in {1,2,3,.....,100}$ When you take $a=1$, then $f(x)$ will not be able to achieve $0$ as denominator also becomes $0$. In other words $x= \pm 1$ are not in domain of function. For all other values of $a$, $f(x)=0$ for $x=-a$ Now set $\frac{x+a}{x^2-1}=1$ and obtain quadratic in $x$ Try to use the fact that $Ax^2+Bx+C=0$ ...


1

Here's an intuitive way to think of it. Consider a more general case:$$y=\sin(\frac{1}{x})$$ When $x$ gets closer and closer to $0$, there will be an infinite amount of times it crosses the x-axis. This is because $\frac{1}{x}$ approaches $-\infty$ from the left and $\infty$ from the right. When $\frac{1}{x}$ gets very close to the y-axis it will grow ...


1

Since $\sin x$ is zero at $\pi,2\pi,\ldots$, $(\sin x)(\sin x^{-2}) $ is zero at $\dfrac{1}{\sqrt \pi},\dfrac{1}{\sqrt {2\pi}},\ldots $, so the answer to your question is "yes". $\sin\left(\dfrac1x\right)$ is a simpler example with the same property.


0

First, this function $f$ can be defined at $0$ by continuity, with value $0$. Now take $x_k = \frac{1}{\sqrt{2\pi k}}$, $k >0$. All $x_k$ are in $[0,1]$, and $f(x_k)=0$, an infinity of times in the interval, and increasingly denser near $0$. A lot of them are quite easy to build, and motivate of lot of exercices for students to study continuity, ...


1

Yes. Def'n : A linear order $<_S$ on a set $S$ is order-dense iff $\forall x,y\in S\;(x<_S y\implies$ $ \exists z\;(x<_Sz<_Sy)).$ Theorem.(Cantor). If $S$ is countably infinite and $<_S$ is an order-dense linear order on $S$ with no end-points (no $<_S$ max or min) then there is an order-isomorphism from $S$ to $Q .$ So let $S$ be the ...


0

Have you learnt what the Fourier Transform does? it takes a function $g(t)$ in the time domain, and gives a new function $\mathcal{F}\{g\} = G(f)$, which for different values of $f$ (which are frequencies), provides the amplitude of the given frequency in the original $g(t)$. You can take the signal you have, apply the fourier transform on it, take as many ...


1

$$p(2,4)= p((1,2)+(1,2))=p(1,2)+p(1,2)=1+1=2\neq3,$$ so $p$ is non linear.


3

Two of the most frequently used step functions are the "floor" and "ceiling" functions. The floor function takes a real number and returns the largest integer(whole) number less then or equal the real number. For a positive real number this is equivalent to cutting off all the numbers after the decimal place. The ceiling function returns the smallest ...


1

Your question is the same as asking: "are there scalars $a$ and $b$ so that $a \cdot 1+ b \cdot 2 = 1$ and $a \cdot 2 + b \cdot 4 = 3$." You should be able to show easily that this is impossible, so the function is not linear.


1

You want $x+2y=1$ and $2x+4y=3\implies x+2y=\frac{3}{2}$ This is impossible.


-5

Step functions are regularly used to describe cumulative densities for discrete random variables. For example, consider a random variable, $X$, to be the throw of a single die. The possible values that $X$ can take are $\{1, 2, \dots, 6\}$. The cumulative density function $P(X\leq x)={F_X}(x)$ is easily drawn. At each score of the die the function jumps, ...


2

Presumably you're allowed to assume that $f$ is differentiable. The Idea: If you draw a picture you convince yourself that if, say, $y<z<x$ then $$\frac{f(y)-f(z)}{y-z}\le\frac{f(y)-f(x)}{y-x}.$$Now let $z\to y$, the left side of the inequality tends to $f'(y)$ and you're done. Now to make that idea into an actual proof from what we're given we write ...


0

Consider any plane simple differentiable loop. We say that this is convex if one can draw a straight segment connecting any two points, without leaving the "inside" of the loop. Convexity is just the name of this property, a way -if you like- to spend less time conveying the meaning. Now, by Dini's theorem the support (or graph, the actual line in the ...


0

This is what I tell my students. We know what a convex set is, and we need a name for functions satisfying the condition above. By (verbal) analogy we call them convex, too. But in that case the curve is the bottom (down) part of a convex region, so we can say that convex means convex down. But then concave up should equal convex down, i.e., the curve is the ...


0

Here is an example of modulo function using recursion - var modulo = function(x, y) { var result = 0; if (x === 0 && y === 0) { return NaN; } else if (x < y) { return x; } else { result = modulo(x - y, y); } return result; }; // example: modulo(23, 4); //3


2

$$f(x)=\begin{cases} 2x-4,&x\gt0\\ -3x+1,&\text{otherwise} \end{cases},$$ means that $f(x)=2x-4$ when $x$ satisfies the condition “$x\gt0$”, and $f(x)=-3x+1$ when $x$ doesn't satisfy the condition “$x\gt0$”, i.e. when $x\leqslant0$. So to determine the value of $f$ for a certain $x$ we would first have to know whether this number satisfies the first ...


2

First, you can define $\phi$ however you want on $F\setminus Im(f)$, it does not matter for your problem. Then for any $y\in Im(f)$, write $y=f(x)$ and define $\phi(y) = g(x)$. To see that this is well-defined, you have to check that $g(x)$ does not depend on the choice of $x$ but only on $y$. But if $y=f(x')$ then by hypothesis $g(x')=g(x)$. So $\phi$ is ...


0

Well, $erf(x)\le1$; that's probably not the bound you want. But if you're talking about large $x$ that's the best bound on $erf(x)$ that you're going to get. Seems to me what matters is how small $1-erf(x)$ is for large $x$. For that you want to bound the "complementary error function": If $x>0$ then $$\int_x^\infty ...


0

Corrected Question The question was incorrect to begin with. It should ask: $\def\rr{\mathbb{R}}$ $\def\less{\smallsetminus}$ Given a function $f : \rr\less\{\frac12\} \to \rr$ such that $f(\frac{x-2}{2x}) = 2x+5$ for every $x \in \rr\less\{0\}$, find $f^{-1}(3)$. Correct Solution The solution you gave is also logically incorrect. Given $y \in ...


0

This can me asked as a graph theory problem: how many graphs with vertices $\{a,b,c,d\}$ exist such that each vertex lies in exactly one cycle of length at most $2$? Case 1: All cycles are of length 1. This is equivilant to the identity function, which we know is unique. However from a graph theoretic point of view this is equal to $\binom{4}{4}$ because ...


0

The graph is made up of two parts: When $x$ is odd, it is the graph of $f(x) = 3x + 1$. When $x$ is odd, it is the graph of $f(x) = x/2$. That's why you get two lines: the first with slope $3$ on top and the second with slope $1/2$ on the bottom. Note that the lines between the dots should not be there. The domain is only on the integers.


7

The number of such functions is the number of ways of dividing our set, in this case $\{1,2,3,4\}$ into $1$ or $2$ element subsets. For given such a subdivision, we can define $f(x)$ to be $x$ if $x$ is a singleton in the subdivision, and by $f(x)=y$, $f(y)=x$ if in the subdivision $x$ and $y$ are a "couple." Conversely, a function $f$ such that $f(f(x))=x$ ...


2

Note that a function does not necessarily have a closed expression which defines it. that is you to not necessarily need to express it $f(x)=5-x$ or $f(x)=x$, but rather we may just state where each element go. So one function is if $f(1)=4, f(4)=1, f(2)=3, f(3)=2$. Each function satisfying the above condition must be of the form $f(x)=y$ if and only if ...


2

I just wish to contribute a "quicker" development of Ragib Zaman's derivation. Just as he had shown, $$ \cos^k \theta = \left( \frac{ e^{i\theta} + e^{-i\theta} }{2} \right)^k = \frac{1}{2^k} \sum_{n=0}^k \binom{k}{n} (e^{i\theta} )^n (e^{-i\theta})^{k-n} = \frac{1}{2^k } \sum_{n=0}^k \binom{k}{n} e^{i(2n-k)\theta} $$ Now, assuming that $\theta$ is real, we ...


0

Assume by contradiction that one of the roots $z_0$ is not simple. This means that $$f(z)=(z-z_0)^2g(z)$$ for some $g$ which is analytic at $z_0$. Then $$f(z_0)=0 \\ f'(z_0)=2(z_0-z_0)g(z_0)+(z_0-z_0)^2g'(z_0)=0$$ This gives $15z_0^4-3z_0^5=0$. Therefore $z_0=0$, which is not possible, or $z_0=5$ which doesn't lie in $D(0,1)$.


0

I am not familiar with the notation $D(0;1)$. I am going to interpret it as $z_0$ is one of the root with multiplicity 2, which should exists if the solutions are not distinct. Hence $z_0$ should satisfy both $f(z_0)=0$ and $f'(z_0)=0$. $$f(z_0)=e^{z_0}-3z_0^5=0 $$ $$f'(z_0)=e^{z_0}-15z_0^4=0$$ subtracting the equations gives us $$15z_0^4-3z_0^5=0$$ ...


1

A quadratic approximation will usually get worse the further out that you go, and so the $R$ term will only get bigger. This is because $R = f(x,y) - \hat{f}(x,y)$, where $\hat{f}$ is the quadratic approximation (this is not standard notation). Therefore if $|x-x_0|,|y-y_0| \leq \epsilon$ then $|R| \leq |f(\epsilon + x_0,\epsilon + y_0)-\hat{f}(\epsilon + ...


1

The first equality is true: \begin{align*} f^{-1}(Y\setminus F)&=\{x\in X:f(x)\in Y\setminus F\}\\ &=\{x\in X:f(x)\in Y\text{ and }f(x)\not\in F\}\\ &=\{x\in X:f(x)\in Y\}\cap\{x\in X:f(x)\not\in F\}\\ &=\{x\in X:f(x)\in Y\}\cap\left(X\setminus\{x\in X:f(x)\in F\}\right)\\ &=f^{-1}(Y)\cap\left(X\setminus f^{-1}(F)\right)\\ ...


0

Yes to both. To see why the first is true, observe that $x\in f^{-1}(Y\setminus F)$ if and only if $f(x)\in Y\setminus F$ if and only if $x\notin f^{-1}(F)$.


0

Solve the inequality $$ 2x>\frac{x^3}{3} $$ that can be rewritten as $$ x(x-\sqrt{6})(x+\sqrt{6})<0 $$ The expression on the left changes sign only at $-\sqrt{6}$, $0$ and $\sqrt{6}$ (where it vanishes). Since its value at $1$ is $-5$ you have the diagram $$ \begin{array}{ccccccc} \_\!\_\!\_\!\_\!\_\!\_\!\_ & -\sqrt{6} & ...


2

Noting that $\int_0^a x(a-x) \; dx = a^3/6$, let $$ f_n(x) = \cases{ 9 n^3 x (1/n - x) & for $ 0 \le x \le 1/n$\cr 0 & otherwise\cr} $$


1

$$ f_n(x) = \cases{ 3nx & for $ 0<x<\frac{1}{2n}$\cr 3-3nx & for $\frac{1}{2n} \leq x < \frac{1}{n}$ \cr 0 & otherwise\cr} $$


0

You forgot the solution $x=0$. By the Intermediate Value theorem the relative positions of the curves can change only at intersection points. Now, for $x\gg 1$, $f(x)<g(x)$, hence we conclude if $x>\sqrt 6$, $f(x)<g(x)$, if $0<x<\sqrt 6$, $f(x)>g(x)$, if $-\sqrt 6<x< $, $f(x)<g(x)$, if $x<-\sqrt 6$, $f(x)>g(x)$.


0

The roots, $x = 0, -\sqrt{6}, +\sqrt{6}$ show merely the crossover points. You must test the four regions defined by those three points: $-\infty < x \leq -\sqrt{6}$ $-\sqrt{6} \leq x \leq 0$ $0 \leq x \leq +\sqrt{6}$ $\sqrt{6} \leq x < \infty$


1

Well, you're looking for period $p\in\mathbb R$ such that $\forall x\in \mathbb R$ we have $$ f(x+p)=f(x) $$ In the case of $$ f(x)=\sin(x)+\cos(x) $$ we have $$ f(x+p)=\sin(x+p)+\cos(x+p) $$ but since both $\sin(x)$ and $\cos(x)$ are periodic to $p=2\pi$ you are finished here and $f$ is periodic to $p=2\pi$ as well. Remark: As mentioned in the comments by ...


0

I think "composed" has a consistent usage ("$f$ composed with $g$" = $f \circ g$) but "precomposed" doesn't. I'm happy with people using "precompose", but the "pre" here can seemingly refer to either: "before it in the order we write it": so "$f$ precomposed with $g$" is the same as $f \circ g$. "before it in the order we apply functions": so "$f$ ...


3

Step functions can be used in the process of defining the Riemann integral, and in general for approximating continuous functions.


1

I'll assume $f$ is defined in a neighborhood of $a$. Note that you cannot use a particular $f$ for proving the result. For the case $\gamma>0$ (in particular $\gamma=1$), the squeeze theorem suffices: from $\lim_{x\to a}|x-a|^\gamma=0$ it follows that $$ \lim_{x\to0}|f(x)-f(a)|=0 $$ so also $\lim_{x\to0}(f(x)-f(a))=0$ and, finally, $$ ...


0

$$\lim_{x \to a}\frac{f(x) - f(a)}{x - a} = 0$$ means that $f$ is differentiable at $a$ and $f'(a) = 0$ (by definition). If $f'(a) = 0$, then $f$ doesn't necessarily have a (global or local) extremum at $x = a$. For instance, take $f(x) = x^3$ and $a = 0$.



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