New answers tagged

1

I think your examples suffer because they depend on notational conventions. AFter all, why do we denote a single number by both $\frac12$ and $\frac24$? Because rationals are defined by equivalence classes, but that's hidden in our familiarity with the notation. The same goes for modular arithmetic. I suggest the following. Let's break the integers into ...


0

$f(x,y) = 0 \implies (x^2+y^2)(ye^{|x|}-1)=0 \implies (y=x=0) \vee \left(y = e^{-|x|}\right)$ So, when $x=0$ then we could have $y=0$ or $y=1$. I don't see how $\eta$ could be uniquely defined here.


0

Suppose $K$ is in quadrant III or IV and $C$ is in quadrant I or II. Let $P$ be a point on the $x$ axis with coordinates $(x,0).$ Given the coordinates of $K$, find an expression for the distance $KP$ in terms of $x.$ Find an expression for $t_{KP},$ the time to travel in a straight line from $K$ to $P$, in terms of $x.$ Given the coordinates of $C$, find an ...


0

For all $a \in A$, we define $[a] \in A/R$ by $[a] = \{x \in A: xRa\}$. Since $f$ is compatible with $R$. Then $xRa$ implies $f(x) = f(a)$. So, for all$[x],[a] \in A/R$, $[x] = [a]$ implies $xRa$ implies $f(x)=f(a)$. It follow that the function $h:A/R \to B$ defined by $h([a]) = f(a)$ is a well-defined function.


0

Here is a simpler polynomial with fractional exponent instead of a trig function. Function is taken in form $ y={x^2(1-x^m)}$ and normalized. All curves pass through $(0,0),(1,0)$ on x-axis: $$y= A\dfrac{x^2(1-x^m)}{(\dfrac{2}{2+m})^{\frac{2}{m}} (\dfrac{m}{m+2})}$$ $y_{max}=A,$ the constant height occurs at a variable $x$ $$x_m =(\dfrac{2}{2+m})^{\frac{2}{m}...


0

I do not think that you are interpreting the phrase "loan balance" the way that a bank will, with respect to paying premiums for loan insurance. Let $k \in \{0, 1, 2, \cdots, 9\}.$ After the $k$-th payment, which will occur $k$ years in the future, the loan balance at that time will be interpreted as $$(10 - k) \times 1000.$$ The idea behind this ...


1

So, let say you have a function $f : \mathbb R^2 \to \mathbb R$. If you want to rigorously prove that you can use your theorem, you should proceed as follow. Let $y\in\mathbb R$. Then we can define $g : \mathbb R\to \mathbb R$ by $$ g(x) = f(x,y) $$ You can now apply your theorem on the function $g$. This implies for example that for any $x\in\mathbb R$, $...


0

Let $f(x,y)=x+y$. If you want to prove some property of the function $x+3$, you can fix $y$ and prove for $f(x,3)$. That's what it means.


1

Typically, to show $f$ is measurable you need to show that $$ \{x: f(x)< c\} $$ is measurable for all $c$. (Equivalently, one can change $<$ to $\leq$, $\geq$, or $>$). It should be pretty easy to break down $c$ into cases and show that these sets amount to intervals, which are measurable.


0

The condition should be $$ \forall_{a\in D_f}\,\forall_{b\in D_f}(a\ne b\to f(a)\ne f(b)) $$ which is very different from the one stated in your question. You need to express the condition “if $a\ne b$, then $f(a)\ne f(b)$”. It's a common error to state this improperly as “$a\ne b$ and $f(a)\ne f(b)$”, but as you see from the negation it is nonsense. By the ...


2

$\forall_{a \in D_f} \forall_{b \in D_f} \left( a \neq b \land f(a) \neq f(b) \right)$ implies that $\forall_{a \in D_f} \forall_{b \in D_f} \left( a \neq b\right),$ i.e. if you take any $a,b\in D_f$ then these are different. But when you take $a$ you don't remove it from $D_f$; it's still there, so $b$ can be the same element. Then $a \neq b$ is absurd.


1

Actually, the first definition should be$$f\text{ injective}\iff(\forall a\in D_f)(\exists b\in D_f):a\ne b\implies f(a)\ne f(b).$$Its negation is$$f\text{ not injective}\iff(\exists a\in D_f)(\exists b\in D_f):a\ne b\wedge f(a)=f(b)$$and that makes sense.


3

Let $$ f(x)=\begin{cases} (1-\text{frac}(\log_k|x|))^{-1} & x\ne 0\\ 0 & x=0 \end{cases} $$ Here, $\text{frac}(x):=x-\lfloor x\rfloor$ denotes the fractional part of $x$. Then $f(x)$ is unbounded. P.S. Proof that $f(x)$ is unbounded. Let $M>1$ be arbitrary, and let $x=k^{1-1/M}$. We then have $f(x)=(1-\text{frac}(1-\frac1M))^{-1}=M$.


3

Suppose such $f$ exists. I found that $f(0) = 0, f(1) = 0, f(2) = 8$. With $x=1, y=1$, $$ f(3) = f(2) + f(1) + 4 = 12$$ With $x=1, y=2$, $$f(4) = f(2) + f(2) + 8 = 24 $$ With $x=2, y=0$, $$ f(4) = f(3) + f(0) + 0 = f(3) = 12$$ Contradiction.


4

(Suppose $k=2$) Assuming continuity: $\sup_{x\in\mathbb R} \left| f(x)\right|=\sup_{x\in[-1,1] }\left| f(x)\right|<\infty$ Without continuity: Let $f((2n+1)2^m)=2n+1$ for $n,m\in \mathbb Z$ and $f(x)=0$ for others.


3

In other words, we never run out of $0$s or $1$s. There are only countably many functions where we do, so there must be uncountably many where we don't. To see the finitely-many-$1$s are countable, define$$\Phi(f)=\prod_{f(k)=1}p_k,$$where $p_k$ is the $k$th prime number. This product is finite since $f$ has only finitely many ones. On the other hand, this ...


0

I presume it's supposed to be $f(x)=f(y)$ instead of $f(x)=f(x)$? If so, an injection $\Phi:\{0,1\}^{\Bbb N}\hookrightarrow S$ can be devised, for instance, by setting $$\Phi(f)(n)=\begin{cases}f(n/3)&\text{if }n\equiv 0\pmod 3\\ 1&\text{if }n\equiv 1\pmod 3\\ 0&\text{if }n\equiv 2\pmod 3\end{cases}$$ Therefore $\lvert S\rvert=\left\lvert\{0,1\}^{...


0

Using you attempt $$\frac {y_{n+1} - y_{n}}{y_{n} - y_{n-1}} = c\implies y_{n+1}-(c+1) y_n+c y_{n-1}=0$$ The characteristic equation $$r^2-(c+1)r +c = 0$$ has two solutions $r_1=1$ and $r_2=c$. So, the solution is $$y_n= k_1+k_2 c^n$$ But the condition imposes $k_1=0$.


0

I'll take $Domf = D$ & $Rngf = \mathit{R}$. $f$ is bounded Clearly $(1,0) \in D$ thus $f((1,0))=2 \in \mathit{R}$ ,Now construct a ball $B(f((1,0)),1000)$,clearly $\mathit{R} \subset B(f((1,0)),1000)$,Thus $\mathit{R}$ is bounded. $f$ is closed Let $b$ be a limit point of $\mathit{R}$,then $\exists x_n \in D \ \forall n$ for which $f(x_n)\to b$. Now ...


0

Alternately you can set $f(x)=\ln(y(x))$ and solve $f(x+1)-f(x)=c$. For instance all $f(x)=cx+T_1(x)$ where $T_1$ is of period $1$ fits in.


2

It’s not the graph of another function you’d recognize. But here is an award-winning, absolutely terrific article about that function and so much more: http://eretrandre.org/rb/files/Knoebel1981_158.pdf


2

How about trig. functions like $y=cos(2\pi x)$? Here the ratio is always $=1$.


2

OK here is my try. Let's set $y(0)=a$. $a$ is just a constant. Now $y(x+1)/y(x)$ is constant. I will call this constant $c$. So $y(1)/y(0)=c$. This means that $y(1)=ac$. $y(2)/y(1)=c$ and so $y(2)=ac^{2}$. $y(3)=ac^{3}$. In general $y(n)=ac^{n}$. Technically, the proof of this is by induction but I hope it is clear. You know what? I'm going to do it. To ...


0

Your mestake is : $(-3)^{-\frac{1}{3}} = (-\frac{1}{3})^{\frac{1}{3}}$ not $ - (\frac{1}{3})^{\frac{1}{3}} $ And $(-\frac{1}{3})^{\frac{1}{3}}=0.34668063717532 + 0.600468477588 i$ So for $x<0 f(x) $no defined at $\mathbb{R} $


2

As a real function, $x^y$ is defined as $$x^y\overset{\text{def}}{=}\mathrm e^{y\ln x}, \enspace\text{ so }\quad x^{\tfrac 1x}= \mathrm e^{\tfrac{\ln x}x},$$ which requires $x>0$ to be defined.


1

$f(x) =x^{\frac{1}{x}} =e^{\frac{1}{x} \ln(x)} $ So : The Domain is $ ] 0,+\infty [$ Because the domain of $ \ln(x) $ is $ ] 0,+\infty [$ And the domain of $\frac{1}{x}$ is $\mathbb{R^*} $ And the domain of $e $ is $\mathbb{R} $


-1

Its $$(-3)^{1/(-3)}$$ not $$-(3^{1/(-3)})$$


1

Here's one approach. Let $A(z)=\sum_{n\ge 0} a_n z^n$ be the ordinary generating function for $a_n$. Then the recurrence relation implies that \begin{align} &A(z) - a_0 - a_1 z - a_2 z^2 \\ &= \sum_{n\ge 0}\left(5a_{n+2}-7a_{n+1}+3a_n+16+24n^2+36\cdot3^n\right)z^{n+3} \\ &= 5z\sum_{n\ge 0} a_{n+2} z^{n+2} - 7z^2 \sum_{n\ge 0} a_{n+1} z^{n+1} + ...


1

Note that each piece of the function is continuous, so you only need to check if the different pieces are continuous at each junction. You can do this by checking if the limits approached from the right and left are equal. If you are still stuck, here is the solution:


0

Use middle definition to get values $f(2)=8$ and $f(3)=13$. Then get a pair of equations from continuity: $2c+4d=8$ and $9d+\frac{6}{c}+1=13$. Solve for $c$ and $d$.


0

This is essentially covered by one of my answers to Open sets on the unit circle $S^1$ . Let us give a short proof following these lines. As you know, it suffices to show that for any $(a,b) \subset (0,1)$ the set $f((a,b))$ is open in $C$ (because $f((a,b)) \subset C \setminus \{0\}$). Consider the map $$ F : [0,1] \to C, F(t) = (\cos(2\pi t),\sin(2\pi t)) ....


0

$$g(g(x))=g(2-4x)=2-4(2-4x)=16x-6$$ and $$(f+g)(-2)=f(-2)+g(-2)=(-2)^2-1+2-4(-2)=13$$


0

Here are some results that I found. First, let $u_n=f(n)$ and $w_n = u_n +1$ we have $$ u_{n+1}+1=f(u_n+1) $$ so $$ w_{n+1}=f(w_n)=\cdots=f^{n+1}(w_0) $$ then $$ f(n)+1=f^n(f(0)+1). $$ If $c=f(0)+1 \ge 2$, we have $$ f(n)=f^n(c)-1. $$ Let's see if we can find some solutions: if $f(n)=n^k$ ($k \ge 2$) we have $f^m(n)=n^{k^m}$ so $n^k=c^{k^n}=\exp(k^n\ln(c))$ ...


0

Edit: as noted in the comments, the property only holds for natural numbers, not all real numbers. Assuming that it does hold for real numbers, the following can help. Let us look at the function with this same property, but on $\mathbb{R}$. Then: $$\Big[ f(x+1)+1 \Big]' = f'(x) = f'(x) \cdot f'(f(x)+1)$$ by the chain rule. Therefore: $$f'(f(x)+1) = 1 \ \ \ \...


1

You can go with $f:\Bbb N\to \mathcal P\left(\bigcup_{n\in\Bbb N}\Bbb R^n\right)$ and then specify the additional condition that $f(n)\subseteq\Bbb R^n$ for all $n$, or you can be arguably a little more obscure and say that $f\in\prod\limits_{n\in\Bbb N}\mathcal P(\Bbb R^n)$. Personally, I don't know which one to recommend, but probably in general I'd choose ...


0

You can put it in WA e.g. the recurrence See the Recurrence equation solution section there. Then just use the values for $a_0, a_1, a_2$ to find the constants $c_i$. Probably there's good amount of theory behind that and that's what WA implemented. And here is the solution with the constants found. complete solution All in all, I don't think this is a ...


0

An inflection point at (0,18) gives two equations: $f''(0) = 0$ and $f(0)=18$. The critical point gives rise to the equation $f'(2)=0$ and you have $f(2)=2$. Then you have four linear equations in four unknowns, which you can solve by substitution.


2

Little lemma: $\forall x \in \mathbb{R}:$ $$ 0<\frac{2}{1+e^x} < 2$$ Proof: We know that $\forall x \in \mathbb{R}:$ $$ 0< e^x <\infty$$ Using this, we observe $$ \frac{2}{1+e^x} < \frac{2}{1 + 0} = 2 \\ \frac{2}{1+e^x} > \frac{2}{1+\infty}=0$$ Hence, $$ 0<\frac{2}{1+e^x} < 2 \ \ $$ as desired. Transforming the form of the original ...


1

You'll have hard time to build such a solution based on elementary methods. See this answer for more details.


0

If we remove the condition that $f$ is differentiable, then a counterexample can be found (instead of defining nondecreasing via the gradient, we would need to define it by requiring that $f$ is nondecreasing in every variable $x_i$). I believe even if $f$ is differentiable you will encounter similar problems. Let $d=2$, $f(x)=\min(x_1,x_2)$, $\bar x=(1,1)$. ...


3

It's basically fine* to think of the object part of a functor $C\to D$ as a function between the set of objects of $C$ and the set of objects of $D$. However, it's important to note that that's not all the information of the functor: it also involves a function from the set of morphisms of $C$ to the set of morphisms of $D$, together with some requirements ...


1

Since $\mathbf{Set}$ is large (i.e., it has more objects than can be accounted for with a set), and since function application takes all values from one set to values in another set, a functor from $\mathbf{Set}$ can't technically be a function.


0

Define $f\colon[0,2\pi)\to [0,\pi)\setminus \Bbb Q$ as $$ f(x)=\begin{cases}(x+7)\bmod 2\pi&\text{if }x\equiv a+7b\pmod{2\pi}\text{ for some }a\in\Bbb Q\cap[0,2\pi), b\in\Bbb N_0\\x&\text{otherwise}\end{cases}$$ Note that if $x$ is of the form mentioned, then so is $f(x)$, and only if it is of this from with $b=0$ (and so necessarily $k=0$) it is ...


5

After realizing the flaw in my previous, affirmative, answer: No such function exists. If $f$ has a nonempty domain at every point, then it grows arbitrarily large, but since it is continuous, after sufficiently many iterations of log its range will always be $(-\infty,\infty)$. WLOG, take $f$ to be such a function (if any valid $f$ exists, all of its ...


0

An interesting example is one defined by Hellmuth Kneser, as an analytic solution $E$ to Abel's equation which I see has been mentionned several times already: $\forall x>c,E(x+1)=\exp(E(x))$, where $c$ is a real constant. The function is also strictly increasing and all of its derivatives are positive on (distinct) intervals $(b,+\infty)$, $b\geq c$. ...


1

First, consider sums instead of integrals. Let's look at what happens in the discrete case, with successive differences. Suppose you have a polynomial like $f(x)=x^2$. Look at what happens with the differences between neighboring terms, and the differences between those differences: $$\begin{array}{cccccccc}f(n)=0&&1&&4&&9&&...


3

$f(x)$ has $x$ in the denominator. So it includes both positive and negative values of $x$. $g(x)$ has $\sqrt{x^2} = |x|$ (correction to be made here). This involves only positive values. So, $g(x)$ is identical to $f(x)$ for $x>0$ and is the reflection of $f(x)$ about $x$ axis for $x<0$ as, $|x| = \begin{cases}x &\text{for } x\ge0 \\ -x &\text{...


0

It's $$\frac{|\sin{x}|}{x}-\frac{|\sin{x}|}{|x|},$$ which is $0$ for $x>0$ and $\frac{2|\sin{x}|}{x}$ for $x<0$.


2

Hint: For real $y,$ $$\sqrt{y^2}=|y|=\begin{cases} y &\mbox{if } y\ge0 \\ -y & \mbox{if } y<0 \end{cases} $$


3

They will have different signs at $x<0$. Watch out their denominators. Note that $\sqrt{x^2}=\left|x\right|$ instead of merely $x$.


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