# Tag Info

11

Consider $g(x)=xf(x)$. We have $g(x^2)=x^2f(x^2)=xf(x)=g(x)$. By induction we have $g(x)=g(\sqrt[2^n]{x})$ for any $x > 0$. By the well known $\sqrt[2^n]{x} \to 1$, we obtain $g(x)=g(1)$ by continuity. Hence $g$ is constant, $f$ is easy to determine now.

7

Hint: Let $g(x) = xf(x)$. Then $g(x) = g(x^2)$ for all $x>0$. Use the continuity of $g$ to show that $g$ is constant.

7

Suppose , $gcd(a,b)=1$ , $x=\frac{a}{b}$ and $x^2=2^x$ We have $$a^2=b^2\times 2^{a/b}$$ implying $$a^{2b}=b^{2b}\times 2^a$$ This is impossible, if $gcd(a,b)=1$ and $b>1$. The negative solution is obviously not an integer. If $x$ is irrational algebraic, then $2^x$ is transcendental, but $x^2$ is not. So, $x$ , the negative solution, must be a ...

7

Assume that $u=\frac{x}{x^2+x+1}=\frac{1}{1+x+\frac{1}{x}}$. We have $\frac{1}{u}-1=x+\frac{1}{x}$, and by squaring: $$\frac{1}{u^2}-\frac{2}{u} = x^2+\frac{1}{x^2}+1,$$ hence: $$\frac{u^2}{1-2u} = \frac{1}{1+x^2+\frac{1}{x^2}} = \frac{x^2}{1+x^2+x^4}$$ and: $$f(u) = \color{red}{\frac{u^2}{2-4u}}.$$

6

When $x$ gets close to zero, $\sin x \approx x-\frac{x^3}{6}$. So $$\sin(x^2)\approx x^2-\frac{x^6}{6}\\x\sin(x)\approx x^2-\frac{x^4}{6}$$ Now, when $x$ is small, $x^4$ and $x^6$ are "very small." So the functions are dominated by $x^2$ near $x=0$. Indeed, if you graphed $y=x^2$ alongside, you'd see that both of your functions are close to but smaller ...

5

What happens if $X=Z=\{1\},$ $Y=\{1,2\}$ and $f:X\to Y$ is defined by $f(1)=1$ and $g:Y\to Z$ is defined by $g(1)=g(2)=1$ ?

5

$f\circ g$, being an automorphism, has an inverse $h$. Then $(h\circ f)\circ g$ is the identity, so $g$ has a left inverse and hence is a split monomorphism. Similarly, $g\circ f$ has an inverse $h'$, so $g\circ (f\circ h')$ is the identity, hence $g$ has a right inverse, hence is a split epimorphism. Thus $g$ is an isomorphism. Obviously this is a symmetric ...

5

Note that, by induction, $f(r_nx)=f(r_0x)$ for all $n\in\Bbb N$ and $x\in\Bbb R$. Let $a\in\Bbb R$ and $x_n=a/r_n$ ($n\in \Bbb N$). Then $$f(a)=\lim_{n\to\infty}f(a)=\lim_{n\to\infty}f(r_nx_n)=\lim_{n\to\infty}f(r_0a/r_n)=f(0),$$because $r_n\to\infty$ and by the continuity of $f$. Since $a$ is arbitrary, $f$ is constant.

4

Left-hand derivative of $f(x)$=$Lf'(x)$ $=\lim_\limits{h\to 0^-}\frac{f(1+h)-f(1)}{h}$ $=\lim_\limits{h\to 0^+}\frac{f(1-h)-f(1)}{-h}$ $=\lim_\limits{h\to 0^+}\frac{(1-h)^2+2-3}{-h}$ $=\lim_\limits{h\to 0^+}\frac{h^2-2h}{-h}$ $=\lim_\limits{h\to 0^+}(2-h)=2$ AND Right-hand derivative of $f(x)$=$Rf'(x)$ $=\lim_\limits{h\to 0^+}\frac{f(1+h)-f(1)}{h}$ ...

4

This is definitely not true. The easiest example is a constant function. $f(x,y)=c$ for all $(x,y)\in\Bbb R^2$. For a less trivial example, consider $f(x,y)=x^2+y^2$. Here $f(\Bbb R^2)=[0,\infty)$.

4

That is false in general; consider the map $f: (x,y) \mapsto \sin x: \Bbb{R}^{2} \to \Bbb{R}$.

4

Here is a sketch of the proof that no such function exists. Say that such a function does exist. Take $n$ be such that $f(n)=m$ is the minimum value that $f$ takes. If $n\neq1$, then then $m$ is expressed as the sum of two outputs of the function $f$ by the given relation. Since the outputs of $f$ are positive integers, this is impossible. If the minimum ...

4

One systematic way to do this is to find such a function satisfying: $$f\left(x+\frac{2}5\right)=f(x)+1$$ Once you've defined it over the interval $[0,\frac{2}5]$ the rest of the values can be found from this relation. The only constraints on $f$ in that interval will be that $f(\frac{1}5)=f(1)-2=f(0)-2$ and that $f(\frac{2}5)=f(0)=1$. Any continuous $f$ ...

4

It will be hard to do that by modifying the argument of the sine. Much easier to modify the result of the sine. More precisely, you want $$f(\sin(2x))$$ where $f$ is some function that is odd (such that your modified curve is still nicely glide symmetric), has a given slope at $0$, satisfies $f(1)=1$, and is strictly increasing from $-1$ to $1$. There ...

4

This is a common misconception that is fostered by blanket use of the term "cancelling." It is certainly true that $$f(x)=\frac{(x+1)(x-1)}{x-1}=(x+1)\cdot\frac{x-1}{x-1}.$$ Moreover, it is true that, for $x\ne 1,$ we have $x-1\ne 0,$ so that $\frac{x-1}{x-1}=1,$ whence $f(x)=x+1.$ However, $$f(1)=2\cdot\frac00.$$ Since $\frac00$ is undefined, then so is ...

4

Alternate way to do this is to find an $x$ such that $g(x)=\frac{1}{2}$. Since $$g(\frac{1}{\sqrt{2}}) = \frac{1}{2}$$ We have $$f(\frac{1}{2})=f(g(\frac{1}{\sqrt{2}}))=\frac{(\frac{1}{\sqrt{2}})^4+(\frac{1}{\sqrt{2}})^2}{1+(\frac{1}{\sqrt{2}})^2} = \frac{\frac{1}{4}+\frac{1}{2}}{1+\frac{1}{2}} = \frac{\frac{3}{4}}{\frac{3}{2}} = \frac{1}{2}$$

4

I think that the matrices represent permutations on the set $\{1,2,3\}$. So they are the same as saying $$f(1)=2,\quad f(2)=3,\quad f(3)=1\\ g(1)=2,\quad g(2)=1,\quad g(3)=3$$ Now, $f\circ g(1)=f(g(1))=f(2)=3$ and so on. Can you do this for all elements, both for $f\circ g$ and $g\circ f$? What can you see, then?

3

A polynomial is usually not considered as a function, which is a key distinction, though you can use a polynomial to define a function. When we have a polynomial in a variable $x$, $x$ is frequently called an indeterminate. This means that it is a symbol, not a number. The way we get a function from a polynomial is called evaluation; it is the act of ...

3

If two functions have the same output and the same domain, they are the same function. You know that $g(x)=kx+n$, and that $g^{-1}(x) = g(x)=kx+n$. Now, using the properties of inverse functions, you also know that for each $x$, you have $x = \mathrm{id}(x) = (g\circ g^{-1})(x) = (g\circ g)(x)=g(g(x))$, and this equation should give you a lot of ...

3

So we have $x^2 = 2^x$. Taking the square root of both sides and assume the solution is negative gives $x=-\sqrt{2}^x$. We can then establish a recursive sequence, $x_n = -\sqrt2^{x_n-1}$. Assuming that this converges gives us the answer, $x=-\sqrt2 ^{-\sqrt2 ^ {-\sqrt2 ^\cdots}}$. After five iterations, we get $$x\approx-0.76961847524.$$ Substituting the ...

3

The claim is true, but you have definitely not proven it. Case 1: you wrote $f(z-z) = f(z) - f(z)$. How do you know that? A priori all you know is $f(x+y) = f(x) + f(y)$, so here it becomes $f(z-z) = f(z+(-z)) = f(z) + f(-z)$. You still need to show that $f(-z) = -f(z)$, which is not part of what you assumed at the beginning. To do case 1 properly, write ...

3

You may write $$\frac{1-x^2}{1+x^2}=y$$ giving $$x^2=\frac{1-y}{1+y},\qquad y\neq-1.$$ Then you have to ensure that $$\frac{1-y}{1+y}\geq0$$ in order to obtain $$x=\sqrt{\frac{1-y}{1+y}}.$$ Finally the domain you are looking for is $$y \in (-1,1].$$

3

I believe you probably saw this, not the one you said $$\ln n\leq \sum_{i=2}^{n}\frac{1}{i-1}$$ Your inequality doesn't hold for $n\geq3$. Check with your calculator that $$\ln 3 > \frac{1}{2} + \frac{1}{3} + \frac{1}{4}$$ For your reference to show $\ln n\leq \sum_{i=2}^{n}\frac{1}{i-1}$: ...

3

By definition of derivative, the problem requires one to show that the following limit exists: $$\lim_{x\to 1}\frac{f(x)-f(1)}{x - 1}$$ The limit exists if and only if the following one-sided limits exist and are of equal value: $$\lim_{x\to 1^-}\frac{f(x)-f(1)}{x - 1}$$ and $$\lim_{x\to 1+}\frac{f(x)-f(1)}{x - 1}$$ Based on the definition of limit, the ...

3

The square root of a negative number is defined when discussing complex numbers, but when only real numbers are discussed (as is usual with most of calculus) there is no advantage to using the complex numbers as square roots.

3

Not an answer, but something heavy I want to get of my chest here: Some of you make statements as "The square root of a negative number is complex" That is from a mathematical standpoint a very poor way of "introducing" complex numbers. It falls in the same category as stating $i=\sqrt{-1}$ whereas $i^2=-1$ is a far better concept. A Squareroot of negative ...

2

Speaking geometrically, you can see some shape of the graph: it is evident that $f$ is continuous at $x=1$ but it can't be differentiable because there are infinitely many tangents to the graph at the corresponding point $(1,3)$ so the conclusion.On the other hand, --analytically now-, right-hand derivative gives $1$ and left-hand derivatives gives $2$. Your ...

2

Assuming $\mathbb{N}$ contains $0$ (if not, just shift everything by $1$): Let $$f(i,j) = \left\{\begin{matrix} \frac{i(i+1)}{2} + j &\text{ if }i\le j \\ f(j,i) &\text{ if } i> j\end{matrix}\right.$$ The idea is to let $f(0,0) = 0$, $f(1,0) = 1$, $f(1,1) = 2$, $f(2,0) = 3$, $f(2,1) = 4$, $f(2,2) = 5$, $f(3,0)=6$, and so on.

2

You can do this by induction over the size of $X$ Or rather prove something stronger: If $f:Y\rightarrow Z$ is a function between sets of equal size, then $f$ is injective if and only if $f$ is surjective Base: $|Y|=|Z|= 1$ (or possibly $0$ but that is completely trivial), then every function is both surjective and injective. Induction assumption: If ...

2

By Descartes' Rule of signs, there is only one change in sign of coefficients of $f(x)$ and no change in sign of coefficients of $f(-x)$. So the equation $f(x)=0$ has exactly $1$ positive real root, no negative real root and $2$ complex conjugate roots. Now use $f(a)\cdot f(b)<0$ in the interval $[3.1,3.2]$ since the root does not lie either at $x=3.1$ ...

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