New answers tagged

1

You have shown that inequality $2f(a)f(b) \ge af(a) + bf(b)$ holds for any choice of real numbers $a$ and $b$. Since you are free to let $b$ be whatever you like, you can choose $b=f(a)$.


1

The author is applying the definition of convergence in probability multiple times. With $\epsilon = 1/2$, there exists $n_1$ such that $P(|X_{n_1} - X| > 1/2) < 1/2$ (because $P(|X_n - X|>1/2) \overset{n \to \infty}{\longrightarrow} 0$). With $\epsilon = 1/2^2$, there exists $n_2 > n_1$ such that $P(|X_{n_2} - X| > 1/2^2) < 1/2^2$. With $\...


0

Let $\mathbb{K}$ denote the Field, $A:\mathbb{K}^n \to \mathbb{K}^n$ lives in. For $x\in \ker(A)= \{x\in \mathbb{K}^n\, | \, Ax = 0\} $ it holds, that $A^2x = AAx = A0 = 0$. This yields $x\in \ker(A^2)$. Remark that $v\in \text{rank}(A^2)$ if $w\in \mathbb{K}^n$ exists with $A^2w = v$. Its clear that $v = A^2w = AAw$ and therefore $v\in \text{rank}(A)$.


0

If $v\in\text{nullspace}(A)$ then $Av=0$. Therefore $A^2v=A(Av)=A0=0$, so $v\in\text{nullspace}(A^2)$. We conclude $\text{nullspace}(A)\subseteq\text{nullspace}(A^2)$. If $v\in\text{columnspace}(A^2)$ then there is some vector $w$ such that $v=A^2w$. Therefore renaming $Aw$ as $u$ we have that $v=A^2w=A(Aw)=Au$, so there is a vector $u$ such that $v=Au$, ...


4

For the first point, let $$S = \{I^+(g) \mid 0 \leq g \leq f\},$$ and $$T = \{I^+(h) \mid 0 \leq h \leq cf\}.$$ To show that $c \sup S = \sup T,$ I'll show that in fact $c \cdot S = T$ for $c > 0.$ (For $c=0,$ things are easy enough to prove directly...) Imagine you have a term $I^+(g) \in S,$ corresponding to some $g$ with $0 \leq g \leq f.$ Define $h = ...


2

For the first question, the equality is clear when $c = 0$. Suppose $c\neq 0$, then for $0\leq g\leq cf$, $I(g)\in\{cI(g):0\leq g\leq f\}$. So, $I^+(cf)\leq cI^+(f)$. Conversely, when $0\leq g\leq f, cI(g) = I(cg)\leq I^+(cf)$, so $cI^+(f)\leq I^+(cf)$. As you have said, for a general $f, I^+(cf) = cI^+(f)$ when $c\geq 0$ from the first part. When $c<0$, ...


1

The function $f$ is assumed to have value $0$ on $V_0$ and $1$ on $V_1$. An integral curve of an ODE/vector field on a compact manifold with boundary can be extended until it hits the boundary. Since $f(\psi(s))=s$ this will happen precisely at $s=0$ and at $s=1$.


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

We can figure this out by making it more abstract. We have a binary operation $\square:A\rightarrow A$, a function $f:A\rightarrow B$, and binary operation $\bigtriangleup:B\rightarrow B$. We also know (we can prove) that $$ f(a_1)\bigtriangleup f(a_2) = f(a_1\square a_2). $$ If $f$ is surjective then each element of $B$ can be denoted as $f(a)$ for some $a:...


0

Welcome to MSE! There is some fun model theory happening here, but I'll refrain from mentioning it because you aren't familiar with morhpisms yet. If you're interested, I go into some detail in my answer here. The idea is that "identities" are preserved under morphisms. Let's work with commutativity first: Let $x,y \in \mathbb{Z}/n$. Then, by ...


2

$\sqrt[n]{c_n}\leq \sqrt[n]{c_N\beta ^{-N}}\beta$ implies that $\limsup_{n\to\infty}\sqrt[n]{c_n}\leq \limsup_{n\to\infty}\sqrt[n]{c_N\beta ^{-N}}\beta=\beta$, since $\lim_{n\to\infty}\sqrt[n]{a}=1\quad\forall a>0$


2

It does not imply that $(c_n)^{1/n} \leq \beta$. But what is being used here is the fact that $x^{1/n} \to 1$ as $n \to \infty$ for any $x>0$. Note that the previous inequality holds for $n \geq N$. So fix $N$ and let $n \to \infty$ to see that $(c_N\beta^{-N})^{1/n} \to 1$ ad $ n \to \infty$.


0

‘Replace $x$ by $-x$’ just means substitute $-x$ for $x$ in the equation $$\sum_{n=0}^kc(k,n)x^n=x^{(k)}\,.$$ When you do that, you get $$\begin{align*} \sum_{n=0}^kc(k,n)(-x)^n&=(-x)^{(k)}\\ &=(-x)(-x+1)\ldots(-x+k-1)\\ &=(-1)^kx(x-1)\ldots(x-k+1)\\ &=(-1)^k(x)_n\\ &=(-1)^k\sum_{n=0}^ks(k,n)x^n\\ &=\sum_{n=0}^k(-1)^ks(k,n)x^n\,. \end{...


1

Do you know dot product? If you do, regard the formula as a dot product and take $(a,b)$ as a vector. Since $\vec v=(\cos (\theta ),\sin (\theta ))$ is the vector moving along the unit circle, and the dot product can be interpreted as a projection [in this case the $\vec v$ is projected to $(a,b)$]. Also, if you admit that $\cos (\theta )$ is nothing more ...


0

The definition of the author is logically equivalent to the usual one. It's just "stronger" in the sense that it makes applying the induction easier. His definition can be quite useful in some cases e.g. if the proof for $N+1$ requires you to know that the statement is true not just for $N$, but also for some smaller values than $N$ (say for $N-1$ ...


0

I'll give you an example. According to Pythagoras theorem, we have that: Let $ABC$ be a triangle. Then, $AB^2 = BC^2 + AC^2$ if and only if $ABC$ is rectangle in $C$. To prove this theorem from left to right would be to assume "$AB^2 = BC^2 + AC^2$" as a hypothesis and "$ABC$ is rectangle in $C$" as conclusion. So you'd try to develop ...


1

You can also do the other way as you stated. Now, $\cos\theta = \frac{b}{\sqrt{a^2+b^2}}$ and $\sin\theta = \frac{a}{\sqrt{a^2+b^2}}$ $\Rightarrow a= \sqrt{a^2+b^2}\sin\theta$ and $b= \sqrt{a^2+b^2}\cos\theta$ $\Rightarrow f(x) = \sqrt{a^2+b^2}\left[\sin\theta\sin x \pm \cos\theta\cos x\right] = \sqrt{a^2+b^2}\left[\mp\cos(x\pm\theta)\right]$ The range of $\...


0

Let $a_n=\sup_{k\geq n}x_k$. Then the usual definition of limit superior is that $\limsup x_n=\lim a_n$. Suppose $\limsup x_n=\lim a_n=+\infty$. Then for any $m>0$, there exists $N$ such that $a_n>m$ for all $n\geq N$. Choose $m$ systematically in the following way: Starting with $m=1$, there exists $n_1$ such that $a_{n_1}>1$. By definition, this ...


2

Hanoi[n]+1 == 2*Hanoi[n-1]+2 So Hanoi[n]+1 == 2 * (Hanoi[n-1] + 1). Then TowerOf[n] = Hanoi[n]+1 has the two specializations TowerOf[n] = Hanoi[n]+1 TowerOf[n-1] = Hanoi[n-1]+1. Using those in the above, TowerOf[n] == 2 * (Hanoi[n-1] + 1) == 2 * TowerOf[n-1]. Now, what is TowerOf[n] == 2 * TowerOf[n-1] TowerOf[n-1] == 2 * TowerOf[n-2] TowerOf[n-2] ...


2

Since multiplication distributes over addition, 2*Hanoi[n-1]+2 = 2*(Hanoi[n-1]+1). (And Hanoi[n-1]+1=TowerOf[n-1] by definition.)


5

If you understand this one: Hanoi[n]+1 == 2*Hanoi[n-1]+1+1 then you can simply rewrite it as: Hanoi[n]+1 == 2*(Hanoi[n-1]+1) which is equivalent to: TowerOf[n] == 2*(TowerOf[n-1])


1

In my opinion, by taking the countable union, the author shows that the Brownian motion is not monotonic in $[a,b]$ under the assumption that $[a,b]$ can only be decomposed as a countable union of intervals with rational endpoints (rational endpoints because is a countable union). But he has to prove it for an interval without the assumption above. ...


0

Note that: Statements: If $m$ and $n$ are odd, so $m+n$ is even. Now, you have the form $$H_{1} \wedge H_{2} \implies C $$ where: $H_{1}:$ $m$ is odd. $H_{2}:$ $n$ is odd. $C:$ $m+n$ is even. Note also that the contrapositive is of the form $$[\neg C \implies \neg(H_{1} \wedge H_{2})] \equiv [ \neg C \implies \neg H_{1} \vee \neg H_{2}]$$ In words this is: &...


-1

Whenever $a<0$, then the statement is true. It is asking for a value $a$ that exists to prove this statement right for the very first part. since $-b^2\leq0$, with $a<0$, we can always get a value that is lower than $0$. For example let $a=-0.1$ and $b=0$, then for the first statement is true. There exists an $a$, for all $b$s. As for the second ...


0

This is a proof of the uniform continuity of the function $f(x) = \sqrt{x}$ at the point $x = 1$. Notice that since $\sqrt{x}$ is continuous on the closed interval $[0, 1]$, which is compact, it is also uniformly continuous on this interval, so it satisfies this hypothesis for all $x \in [0, 1]$. Thus we can assume that $x > 1$ (we ignore negative values ...


1

[It seems you are] asking about the relative Künneth Theorem applied to two copies of $(X,\ast)$. -- Tyrone.


3

Each number $n\in\mathscr{N}$ is of the form $2^a3^b5^c7^d\cdots q^z$ for some non-negative integers $a,b,c,d,\ldots,z$, where $q$ denotes the largest prime less than or equal to $y$. It follows that $$\sum_{n\in\mathscr{N}}{1\over n}=\sum_{a=0}^\infty\sum_{b=0}^\infty\cdots\sum_{z=0}^\infty{1\over2^a3^b\cdots q^z}=\left(\sum_{a=0}^\infty{1\over2^a} \right)\...


0

Consider $\sqrt{n} < m \le n$ and suppose further that no integer $j: 1< j \le\sqrt {n}$ divides $m$. Now suppose $m$ is not prime. Then $m$ has a factor not equal to $1$ or to $m$. Call that factor $d$. Now we just said we can't have $1< d \le \sqrt {n}$ so $ \sqrt{n}< d < m \le n$. But then $1 < \frac md < \sqrt n$. (Just algebra ...


2

You can argue by contradiction. To make things clear, when I talk about a number's 'prime factors' I'm going to count multiple instances of the same prime distinctly; for instance, $36=2^2\cdot3^2$ has four prime factors: $\{2, 2, 3, 3\}$. (This is sometimes referred to as a multiset of prime factors, but that's an aside...) Suppose there were a non-prime ...


0

Hint: You can use the fact that if $f(x)$ is a monic polynomial with integer coefficients, then any rational root of $f(x)$ is necessarily an integer. Now, applying this to the polynomial $f(x) = x^{2} − 3$ you can conclude that $\sqrt{3}$ is irrational number.


0

$\sqrt3$ is a root of $X^2-3$ which is irreducible over $\mathbb{Q}$ (Eisenstein)


2

For $t>v$, $$ |g_n(t,\omega)-g_n(v,\omega)|\le M\int_v^t \psi_n(s-t)\, ds\le M'(t-v). $$ because $\psi_n$ is bounded (it is continuous on $[-1/n,0]$). Therefore, $g_n(\cdot,\omega)$ is continuous, i.e., for each $\epsilon>0$, $|g(t,\omega)-g(v,\omega)|\le \epsilon$ whenever $|t-v|\le \epsilon/M'$. (In fact, it is uniformly continuous.) Also $g_n$ is ...


2

I believe the issue here is that the theorem is claiming that, w.p. 1, the Brownian path is non-monotonic on every interval, simultaneously. What the first part shows is that, if you are given any arbitrary interval (but only one), the path is non-monotonic on it w.p. 1. Since there are uncountably many possible intervals, some additional arguments are ...


0

As mentioned, we should use the cross multiplication approach. $$\frac{e}{f}<\frac{g}{h}$$ Therfore $$eh<gf \tag{1}$$ Now lets deal with the first part of the inequality, namely \begin{align*} \frac{e}{f}&<\frac{e+g}{f+h}\\ e(f+h)&<f(e+g)\\ ef+eh&<fe+fg \tag{substracting $fe$}\\ eh&<fg . \end{align*} Which is actually true ...


4

From $\frac{e}f<\frac{g}h$ we have $eh<fg$. $\frac{e}f<\frac{e+g}{f+h}\iff ef+eh<ef+fg\iff eh<fg$ $\frac{e+g}{f+h}<\frac{g}h\iff eh+gh<fg+gh\iff eh<fg$ Here cross multiplication is valid since $e, f, g, h$ are positive.


0

hint Since for $ n\ge 4$, we have $$n!= n.(n-1)!$$ You just need prove that $$(\forall n\ge 4)\;\; (n-1)! > n$$ or, in an equivalent way $$(\forall n\ge 3)\;\; n!> n+1$$ which is much easier.


1

They noted that $2e$ is the sum of the degrees of the vertices in the graph. If each vertex has degree at least $4$, then the sum of degrees is at least $4v$. So $2e \ge 4v$.


2

If all vertices have degree at least $4$, then the sum of the degrees of the vertices is at least $4v$. We know from the handshaking lemma that the sum of the degrees of the vertices is $2e$, so $2e\ge 4v$ if each vertex has degree $4$ or more. But from $(*)$ we know that $2e\le 4v-8<4v$, so we have a contradiction, and therefore at least one vertex must ...


0

Since $\cal C$ was a chain, every finite subset of $A$ is a subset of some $X\in\cal C$, but since $X\in\frak F$ by assumption, so must be all of its finite subsets. Therefore every finite subset of $A$ is a finite subset of some $X\in\frak F$, and therefore an element of $\frak F$. The point here is to take the usual argument using Zorn's lemma in almost ...


0

You know that it is a proof by contradiction. The assumption is that $\sqrt 2$ is not irrational, i.e. $\sqrt 2$ is rational. This means $\sqrt 2 = p/q$ for some integers $p,q$. There are many such representations as a fraction, but we may cancel the gcd of $p, q$ and get a representation in which $p, q$ do not have a common factor. The proof shows then ...


-1

if p and q have common factors divide by their hcf and you have a new fraction with p and q as coprime. hence your assumption was wrong that p and q are coprime and √2 is irrational.


1

To me, it seems all it says is that we can't assume that p,q have no common factors. That would mean that you could reduce the fraction. But you can not reduce a fraction infinitely many times. At some point, $p$ and $q$ have to coprime, what means that they have no common factor. That means, you can assume that you can write $\sqrt 2$ as a fully reduced ...


0

$\forall r\in \Bbb R\setminus \{0\}\,\exists! s\in \Bbb R \,(rs=sr=1).$


2

Yes, your reasoning is correct. For a fixed $\omega$, $g(t,\omega)$ is a continuous function of $t$. This means it is uniformly continuous (in $t$) on the compact interval $[S,T]$. Hence, for any given $\epsilon > 0$, we can find $\delta >0$ such that $$ |g(s, \omega) - g(t,\omega)| < \epsilon \quad \forall s, t \in [S,T]: |s-t| < \delta. $$ ...


0

I think the wording of the proof you have attached above is what you are having difficulty with. Let me show a much simpler worded translation of the same proof below, which is also a well known application of Lagrange multipliers. Let $\mathbf{a} = (a_1,.....,a_n) \in \mathbb{R}^n,$ and $\mathbf{b} = (b_1,....., b_n) \in \mathbb{R}^n.$ We need to show that $...


3

For some integers $z_i$ and $a_i$, we have $$x_i \equiv z_i \pmod{2^{k_i}} \implies x_i = 2^{k_i}a_i + z_i \tag{1}\label{eq1A}$$ Next, there are $k_i$ odd integer results in a row after repeated applications of the $T$ function starting with $x_i$. This gives for the first one, $$\begin{equation}\begin{aligned} T^{1}(x_i) & = \frac{3x_i + 1}{2} \\ & =...


2

Suppose that $x\in Z$; then $x\in X_n\setminus Y_n$ for some $n\ge 0$, and $f(x)\in f[X_n\setminus Y_n]$. Clearly $f[X_n\setminus Y_n]\subseteq f[X_n]=X_{n+1}$. Moreover, $f$ is injective, $x\notin Y_n$, and $Y_{n+1}=f[Y_n]$, so $f(x)\notin Y_{n+1}$, and therefore $f(x)\in X_{n+1}\setminus Y_{n+1}\subseteq Z$. Thus, $f[Z]\subseteq Z$, and hence $f[Z]$ and $X\...


0

Quick question - why can we use algebra here when we don't know which metric we are supposed to be using? Surely we can only use the 3 properties of metrics, and not assume we are using the Euclidian distance?


0

keep in mind exactly one and only one of the following must be true. $n=0\pmod 3; n=1\pmod 3; n\pmod 3$. So $[n\not\equiv 1\pmod 3$ and $n\not\equiv 0\pmod 3]\iff n\equiv 2\pmod 3$. So the converse is $n\not \equiv 0\pmod 3$ and $n\not \equiv 1\pmod 3 \implies n^2 \not \equiv n\pmod 3$. And to prove it. $n\not\equiv 1\pmod 3$ and $n\not\equiv 0\pmod 3$...


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