New answers tagged

0

$\displaystyle y = x^2 + bx + c$ or, $\displaystyle y = \left( x + \frac{b}{2} \right)^2 + \frac{4c - b^2}{4}$ Hence $\displaystyle x_v = -\frac{b}{2}$ and $\displaystyle y_v =\frac{4c - b^2}{4}$ Our only task is to compute $b$ in terms of $x_1, x_2, y_1, y_2$ and $c$ Since $\displaystyle y_1 = x_1^2 + bx_1 + c$ and $\displaystyle y_2 = x_2^2 +...


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Since you only have one unknown, $b$, then you can use one of the two points you know the parabola passes through (say $p_1$) to solve the equation $y_1 = x_1^2 + bx_1 + c$ where the unknown is just $b$. Once you have that you know that the coordinates of the vertex are $(x_v,y_v)$ for: $$ y= (x−x_v)^2+y_v\,, $$ obtained from your $y= x^2 + bx + c$ by ...


0

$0=f'(0)=a(a-1)(a+2)\implies a=0,1$ or $-2$. Meanwhile, $0\le f''(0)=2a\implies a\ge0$. Hence $a=0,1$.


0

Maybe a more intuitive way of thinking about this: Consider $g(x):=f(x)-f(-x)$, then $g(-x)=f(-x)-f(x)=-g(-x)$, then: $$\int_{-\infty}^{\infty} f(x)-f(-x) dx=\int_{-\infty}^{\infty} g(x)dx=0$$since $g(x)$ is odd. Hence $\int_{-\infty}^{\infty} f(x)=\int_{-\infty}^{\infty} f(-x)$ (P.S. I did not intend to argue in a very rigorous way)


1

We have that $$\int_{-a}^{a}f(x)\mathrm{d}x=\int_{a}^{-a}f(-x)\mathrm{d}(-x)=\int_{-a}^{a}f(-x)\mathrm{d}x$$ And take the limits. It will imply that the integral of $f(-x)$ is the same as the integral of $f(x)$, if it exists. Can you prove that it exists?


0

Explicit Function: A function in which the dependent variable, and independent variable are separated on opposite sides of the equality are known as explicit function. i.e., If a function is written (or can be written) as $~y=f(x)~$, then it is explicit. e.g., Take $~x^2+y^2-1=0~$, solving for $~y~$ gives an explicit solution: $~y=\pm \sqrt{1-x^2}~$. ...


1

$$\mathbb P(Z\leqslant t)=\mathbb P(\max \{X_1,X_2\}\leqslant t)=\mathbb P(X_1\leqslant t , X_2\leqslant t)=\mathbb P(X_1\leqslant t)\mathbb P(X_2\leqslant t)=\frac t4\times\frac t4$$ So, $$\mathbb P(Z= t)=\frac{t^2}{16}-\frac{(t-1)^2}{16}=\frac{2t-1}{16}.$$


2

Assuming the die is fair, there are $4^2 = 16$ equally likely outcomes. The maximum of a set, if it exists, is the largest element of the set. \begin{array}{c c} \text{outcome} & \text{maximum}\\ \hline (1, 1) & 1\\ (1, 2) & 2\\ (1, 3) & 3\\ (1, 4) & 4\\ (2, 1) & 2\\ (2, 2) & 2\\ (2, 3) & 3\\ (2, 4) & 4\\ (3, 1) & ...


3

The traditional thing is this: take three distinct real numbers $u,v,w$ so the pairwise differences are nonzero. To get a quadratic $q(x)$ that gives $q_u(u) =1$ while $q_u(v)=q_u(w) = 0.$ $$ q_u(x) = \frac{(x-v)(x-w)}{(u-v)(u-w)} $$ Back to your $\alpha, \beta, \gamma$ do the same and make $$ \beta q_\alpha + \gamma q_\beta + \alpha q_\gamma $$ I'm ...


1

By "invertible" it specifically means there is a smooth inverse. ("A $C^\infty$ map $f$... is locally invertible... if $f$ has a $C^\infty$ inverse...") The cubing map $x\mapsto x^3$ from $\mathbb{R}^1$ to itself is topologically invertible with zero Jacobian at $x=0$, but the inverse is not smooth, specifically it is not differentiable at 0. If there is ...


2

The equation at $x=t=0$ gives us: $f (0)=2*f (0) + 2×f (0) $. So $0 <f (0)=0$. Thus, there isn't any $f $ satisfying this.


6

We have $$f(x+t)=\left[\sqrt {f(x)}+\sqrt {f(t)}\right]^2$$which leads to$$\sqrt {f(x+t)}=\sqrt {f(x)}+\sqrt {f(t)}$$since $f(x)\ge 0$. By defining $g(x)=\sqrt {f(x)}\ge0$ we obtain $$g(x+t)=g(x)+g(t)$$Therefore the final answer becomes: $f(x)=g^2(x)$ is the general solution, provided $g(x)$ is a non-negative additive function.


1

From the point of view of set theory, a function $f$ from $X$ to $Y$ is a subset of the Cartesian product $X\times Y$ which satisfies the property that $$ (x,y) \in f \ \text{and}\ (x',y) \in f \implies x=x'. $$ In essence, a function is a subset of the Cartesian product which satisfies the "vertical line property". However, we typically don't like to think ...


1

$$f(x):=e^{px}$$ is a solution when $$1+a\,e^{p\alpha}=b\,e^{p\beta}.$$ This equation has an infinity of solutions, and $p$ complex is possible. And by linearity, $$\sum_{i=1}^n f_i\,e^{p_ix}$$ is also a solution. Note that you can assume $a=1$ WLOG, because with $f(x):=e^{-(x\log a)/\alpha}g(x)$, the equation takes the form $$e^{-(x\log a)/\alpha}g(x)...


3

Interesting question. So when you found that $\sin x$ is such a function, and you mentioned argument shift, means that $\cos x$ is also a solution. Moreover, you can also have $\sin Cx$ and so on. This lead me to think about Fourier transforms. So let's multiply the equation by $e^{ikx}$ and integrate over $x$ from $-\infty$ to $\infty$ $$\int_{-\infty}^\...


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That means if $$0<x_1<x_2$$ then $$f(x_1)\ge f(x_2)$$


2

The statement seems poorly stated. Perhaps it means that there is an $\epsilon>0$ such that $f$ is non-decreasing on the interval $(0,\epsilon)$.


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We have to use the Implicit Function Theorem. We consider the curve defined by $f(x, y)=0$. Now, pick two points $x_0$ and $y_0$ such that $f(x_0,y_0)=0$. If it holds $$ \left. \frac{\partial f}{\partial y} \right|_{(x_0, y_0)} \neq 0, $$ then there exists a differentiable function $y(x)$ such that, for $x$ close to $x_0$, $$ f(x, y(x)) = 0. $$ In ...


0

As long as $f(x)$ is continuous in the Lipschitz sense, we have $$ 2\lim_{t\to 0}\left(\frac{f(x+t)-f(x-t)}{2t}\right) = 2f'(x) = 4x\Rightarrow f(x) = x^2+C $$


1

For arbitrary reals $a$ and $b$ plug $x=\frac{a+b}{2}$ and $t=\frac{a-b}{2}$. Then, we obtain that $$ f(a)-f(b)=4\cdot\frac{(a+b)(a-b)}{4} $$ or $$ f(a)-f(b)=a^2-b^2. $$ Hence, $f(x)-x^2\equiv c$ for some constant $c$. It's easy to see that function $f(x)=x^2+c$ is satysfying the condition.


2

Set $x=t$ and we get $$f(2x)-f(0)= 4x^2$$ so $f(x) =x^2+a$ where $a= f(0)$. Check: If we now put this in to starting equation we get: $$ (x+t)^2+a-(x-t)^2-a = 4xt$$ which is always true. So $\boxed{f(x) =x^2+a}$ for all real $a$.


1

Let $f(q) = q^2$ + const., that solves it directly. Is your question for a solution or for all solutions?


1

For convexity of a function $f(x)$ you like to have the graph of your function on an interval [a,b] falls below or on the graph of a straight line segment connecting $f(a)$ and $f(b)$. You can check arbitrary points or in case the second derivative exist you want your second derivative be non negative. That makes the slopes to increase and the curve ...


1

For particular functions, there are indeed easier ways of checking. For example, for any function $g(x)$ which is twice differentiable in an interval $(a,b)$ $$( \forall x\in(a,b) (\frac{d^2g}dx^2 \leq 0) ) \implies g(x) \mbox{ is convex in } (a,b)$$ That is, a function with non-positive second derivative in an interval is convex in that interval. ...


4

What you gave is the standard definition of a convex function. If $f$ is supposed to be continuous, it is enough to check that $$f(\frac{x+y}{2}) \le \frac{f(x)+f(y)}{2}$$ for all $x,y$. If $f$ is twice differentiable, it is enough to check that the second derivative is non negative.


2

In fact, a function really isn't defined until you define the domain. It is a common exercise to ask, say: What is the domain of: $$f(x)=\left(\sqrt{x}\right)^2\tag{1}$$ Now, $f(x)=x$ where $f$ is defined. And if we take $\sqrt{\cdot}$ to mean a complex branch of the square root function, then $f(x)$ could have a domain equal to any subset of $\mathbb ...


0

This is because you are doing a composition of functions: $dom_{f\circ g} = \{x\in dom_g ~\colon~ g(x)\in~dom_f \} $` and by operating the inside of the function, you use properties without considering the original domain. So and the end, it yields $2x$ but in the original domain.


2

There’s a problem with definitions here. To identify whether a function $f:X\to Y$ is surjective, you must specify the codomain $Y$. Ordinarily, when an expression is given without mentioning the codomain, one assumes that the codomain of the associated function is all of $\Bbb R$. Under this assumption, your function is not surjective, as you yourself said....


3

A function $f:X\to Y$ is surjective if for all $y\in Y$ there exists at least one element $x\in X$ such that $f(x)=y$. For $f(x)=\ln\left(\frac{x+1}{x-1}\right)$, the domain is $X=\{x\in\mathbb R:x<-1 ~\text{o}r~ x>1\}$ and the range is $Y=\mathbb R\setminus \{0\}$. Sketch of proof: Let $y\in\mathbb R\setminus \{0\}$ be arbitrary. Then, we can ...


0

For any $y \neq 0$ $x = \frac{e^y+1}{e^y-1}$


2

...I was wondering how I would go about augmenting this function such that the curve between $0$ and $1$ is more harsh (e.g. the curve goes from $y = 0$ to $y = 1$ jumps up more abruptly) or more gradual (e.g. the curve resembles something that looks like $y = x$) it seems to me that what you want is something like what computer graphics practitioners call ...


2

$f(x)=2x$ whenever $f$ is continuous at $x$. Since $f$ is increasing it is continuous except at a countable number of points Hence it is continuous on a dense set. Given any $x$ choose an increasing sequence $(x_n)$ and a decreasing sequence $(y_n)$ both tending to $x$ such that $f$ is continuous at each $x_n$ and at each $y_n$. Then $2x_n=f(x_n)\leq f(x) \...


1

This answer fleshes out the hint of Edward H. in the comments under the question. Hint Since $\log$ and $\exp$ are strictly increasing (and since $\exp$ is onto the domain of $f$), if we can find the maximum value $m$ of $$\log f(e^t) = t e^{-t} ,$$ then the maximum value of $f(x)$ is $e^m$. The binomial expansion gives $$e^{t - 1} = \lim_{n \to \infty} \...


2

The following function is a good choice to satisfy your requirement: $$ y(x)=x^{t}(2-x)^{t}$$ You may vary the parameter $t$ to regulate steep or slow rise of the curve from 0 to 1.


4

By Lebesgue differentiation theorem, we have $F' = f$ almost everywhere. So we have $f(x) = 2x$ almost everywhere. In particular, for at least two different points $x_1 < x_2$, we have $f(x_1) = 2 x_1$ and $f(x_2) = 2 x_2$, so $f$ can't be non-increasing. Now, if $f(x_0) \neq 2 x_0$ for some $x_0$, then either $f(x_0) < 2 x_0 - \varepsilon$ or $f(x_0) ...


0

In fact, the limit equalities that you wrote are valid if $f$ is supposed to be continuous. The function $$f(x)=\begin{cases} 2x & x \notin I \\ 0 & x \in I \end{cases}$$ where $I=\{1/n \ ; \ n \in \mathbb N\}$satisfies the requested equality but is not increasing. The function $f(x)=2x$ for all $x \in \mathbb R$ is the only increasing one ...


2

Your question seems to be mostly about getting the notation straight. Below let me introduce regression problem and the relevant notations. I think it should clarify the notation and so answer your questions. Problem introduction: In general (for linear regression) your assumption is that $y$ (dependent) and $x$ (independent) are related by (plus some ...


3

A series of circular arcs will give you a family that includes the "S"- and "C"-shaped curves. ("U" doesn't fit the pattern, so I'll ignore it here.) Specifically, consider these arcs of measure $2\alpha$ (in radians), with a fixed opening of $2a$ at the "bottom" or "top". The radius of the arcs is $a\csc\alpha$, and the displacement of their centers from ...


1

They are not the same function. We say that two functions $f$ and $g$ are equal if $f(x)=g(x)$ for every $x$. Since the expressions you wrote are not equal at $x=2$ (because one of them is undefined), they do not define the same function. However, it is true that both functions have the same limit as $x$ approaches $2$. This is because when we evaluate the ...


2

A function is given by a mapping and its domain. In your question, the domains are not the same. The domain of the first function is $\mathbb R\setminus \{2\}$. The domain of the second one is $\mathbb R$. What is true as you refer to calculus is that $$\lim\limits_{x \to 2} f(x) = -1$$ hence you can extend $f$ by continuity at $2$.


2

Wikipedia calls this action lifting an operation pointwise, thereby forming a pointwise operation. From the Wikipedia article "Pointwise": A binary operation $o: Y × Y → Y$ on a set $Y$ can be lifted pointwise to an operation $O: (X→Y) × (X→Y) → (X→Y)$ on the set $X→Y$ of all functions from $X$ to $Y$ as follows: Given two functions $f_1: X → Y$ and $f_2:...


1

I do not think that it has its own name (at least one that is widely accepted). I would just "multiply" the functions and specify which "multiplication" is being used. That is mostly based on the fact that one often uses the word "multiplication" for the binary operation of a semigroup/monoid/group etc. So I guess that is just the way to go.


1

Hint: The given function can be decomposed in the form of a system of real equations $$u=P(x,y),\\v=Q(x,y)$$ where $P,Q$ are bivariate quadratic polynomials. When you vary $u$ and $v$, you obtain pencils of conics. Hence the solutions in $x,y$ are formed by the intersections of two conics, and this leads to a quartic equation, having up to four distinct ...


2

For the statement "$f$ injective implies $f$ surjective", induct on the cardinality of $A$ and $B$. You can check $n=0$. For $n=k+1$, let $f:A\to B$ be injective. $A$ has at least one element $a$. Define $f^-:A\setminus\{a\}\to B\setminus\{f(a)\}$, gotten by forgetting about $a$ and its value $f(a)$. $f^-$ is injective as restricting injections preserves ...


3

If the sets have cardinality $n$ and $f$ is injective, then the image of $f$ must be an $n$ element subset of $B$ and so equal to $B.$ If $f$ is surjective, then the preimage of each element of $B$ contains at least one element, and the preimages are disjoint. So the union of the preimages of the elements of $B$ has at least $n$ elements. Since $A$ has only ...


1

From your question, I take it that codomain and range are not synonyms to you. Note that some authors/texts use range as a synonym for codomain and then image for the set of all $y$-values. This may sound trivial but are all injective functions bijective from its domain to codomain? Obviously, not all injective functions are bijective from the domain to ...


0

If $T_1,T_2,T_3,...$ are the periods of individually added functions and if the ratios of these periods are rational the period of the the whole function is given by the $LCM(T_1,T_2,T_3...)$ If $T$ is period of $g(x)$, the period of $g(ax)$ is $T/a$. So the period of the whole given function is $$T=LCM(2\pi,\pi,2\pi, 4\pi, 8\pi, 16 \pi)=16 \pi$$


2

It is easy to check that $f(x + 16\pi) = f(x)$ for all $x \in \mathbb{R}$. Since $f$ is continuous on its domain (or even continuous as function $\mathbb{R} \to \mathbb{R}\cup\{\infty\}$) and non-constant, it follows that the fundamental period of $f$ is of the form $16\pi/n$ for some positive integer $n$. Now if $n$ is a positive integer for which $16\pi/...


2

$|g(z)| \leq 12$ whenever $|z|=1$ so MMP implies that $|g(0)| \leq 12$. This means $|f(0)|^{2} \leq 12$ or $|f(0)| \leq \sqrt {12}$.


0

In the given function 16π is the fundamental period. The reason is if "T " is one of period then $ \frac{T}{n} $ where n$ {\in N } $ can be a fundamental period if $\\ f(x\,+\,T)\,=f(x) $ . Generally it happens in function which transform from one to the other given e.g |Sin((π/2)+x)| = | Cos x | and vise versa there fundamental period get refused to π/2.


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