Tag Info

New answers tagged

0

According to http://en.wikipedia.org/wiki/Transcendental_number, in 1932 Kurt Mahler gave a further classification of transcendental numbers.


0

I would have to check my homework, but I strongly believe that if you have a polynomial of odd degree $\geq 3$ where all coefficients are real algebraic except for one coefficient that is real transcendental, then at least one of the real roots of the polynomial must be transcendental, if they exist (which in your case, a real root does exist). I'm not sure ...


1

It looks like the quantile function. It's domain is usually $(0, 1)$ and here are some simple properties: $F^{- 1} (x) \leqslant t$ iff $x \leqslant F (t)$. $F^{- 1}$ is increasing and left-continuous. If $F$ is continuous, then $F \circ F^{- 1} = \text{id}$.


2

Any real number is either algebraic or transcendental by definition. The definition of a transcendental number is a real number which is not algebraic.


0

Once you make a choice of ordered bases {$v_1,v_2,..,v_n$} and {$w_1,w_2,..,w_m$}for two vector spaces $V,W$ respectively, over the same field $K$( though I think the same can be done for modules over the same ring ), then, given a linear map $L : V \rightarrow W$, the map taking $L(v_i)=w =c_{i1}w_1+..+c_{ik}w_k..+c_{im}w_m$ , gives you an isomorphism from ...


1

A first observation is that $|x+y|\geqslant|x-y|$ if and only if $xy\geqslant 0$, so defining $$f(x,y):=|x+y|-|x-y|,$$ the positivity of $f$ is linked to that of $xy$. We would like to find a more tractable expression for $f$. Assume that $x\gt 0$ and $ y\gt 0$. Then $f(x,y)=x+y-|x-y|=2\min\{x,y\} =\min\{|x|,|y|\}$. Since $f(-x,-y)=f(x,y)$ we get ...


0

The operation of multiplication on the left by an $n\times n$ matrix $A$, written as $L_A$ is a linear operator on the vector space $\mathbb{R}^n$ or $\mathbb{C}^n$. There are several ways to realize this. One way is to choose a basis set for your vector space. The usual choice is $e_1=(1,0,...,0)^T$, ..., $e_n = (0,...,0,1)^T$. Now if $A$ has a column ...


0

Every $n \times m$ matrix with real entries act as a linear map wen acts over a vector in $\mathbb{R}^n$ by matrix multipication. Using rows-columns multiplication we have: $$ A\vec v= \begin{bmatrix} a_{11}&a_{12}\cdots a_{1n}\\ \cdots\\ a_{m1}&a_{m2}\cdots a_{mn} \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ \cdots\\ v_n \end{bmatrix}= ...


0

Intuitive picture: $v_j$ alternates between $a$ and $b$ increasingly often, so that the two values blend together, proportionally to the measures of the sets $\{v_i=a\}$ and $\{v_i=b\}$. To explain this blending, consider the integral $\int_0^1 v_j\phi$ with continuous $\phi$. Let $v$ denote the constant function $\theta a+(1-\theta)b$. When $j$ is large ...


2

Yes, a linear transformation is not required to be invertible. All that is required is that it is a mapping from one vector space $V$ to another $W$ that satisfies $$L(\alpha_1 v_1 + \alpha_2 v_2) = \alpha_1 L(v_1) + \alpha_2 L(v_2)$$ where $\alpha_i$ are elements of the field over which the V is defined and $v_i$ are elements of $V$. A matrix is such a ...


4

By the AM-GM inequality, $$\frac{a_n}{n} + \frac{n}{a_n^2} + \frac{a_n}{n^3} \ge 3\sqrt[3]{\frac{a_n}{n}\cdot \frac{n}{a_n^2}\cdot \frac{a_n}{n^3}} = \frac{3}{n}.$$ Since $\sum\limits_{n = 1}^\infty \frac{3}{n}$ diverges, so does the series $\sum\limits_{n = 1}^\infty (\frac{a_n}{n} + \frac{n}{a_n^2} + \frac{a_n}{n^3})$ by comparison.


1

I'm not sure if this qualifies as a comparison test, but the first two terms within your parantheses, i.e. $a_n/n + n/a_n^2$ is enough to establish divergence. Note that since all your terms are positive, you can split up the series into two series. if $\sum_n n/a_n^2$ converges then $a_n$ is bounded below by a positive constant (otherwise $n / a_n^2$ would ...


0

For $x > 0$, the equation can be rewritten as $$\frac{1}{x} = \sum_{i=1}^n \frac{x}{x+a_i} = \sum_{i=1}^n\left(1 - \frac{a_i}{x+a_i}\right)\quad\iff\quad \frac{1}{x} + \sum_{i=1}^n \frac{a_i}{x+a_i} = n$$ Since the LHS of the expression on the right is a strictly decreasing in $x$, the equation has at most one positive $x$ solution. In fact, since that ...


3

This is the same as the equation $$1=\sum_{k=1}^{n}\frac{x^2}{x+a_k} = f(x).$$ Note $f$ is a nice $C^1$ function on $[0,\infty).$ Since $$f'(x) =\sum_{k=1}^{n}\frac{x^2+2xa_k}{(x+a_k)^2}>0, x> 0,$$ $f$ is strictly increasing on $[0,\infty).$ We have $f(0) = 0,$ and $f\to \infty$ at $\infty,$ so the intermediate value theorem shows $f$ takes on ...


0

I am not sure this qualifies for an alternative proof anyway, define $$A_j:=\left \{ \liminf_p X_p^{ 1/p}\geqslant \liminf_p\left(\mathbb E[X_p]\right)^{1/p} +\frac 1j\right\}.$$ We have to prove that for each $j$, $\mathbb P(A_j)=0$. If not, then by Markov's inequality and Fatou's lemma, $$\liminf_p\left(\mathbb E[X_p]\right)^{1/p} +\frac 1j\leqslant ...


0

Bessel's inequality still holds for uncountable orthonormal families, $$\sum_{v\in M} \lvert \langle x,v\rangle\rvert^2 \leqslant \lVert x\rVert^2,\tag{1}$$ where the sum of an uncountable number of non-negative terms is defined as the supremum of the sums of all finite subfamilies: $$\sum_{v\in M} \lvert\langle x,v\rangle\rvert^2 := \sup ...


2

No, consider $p(re^{it}) = (\sin r )e^{it}.$ Then $p$ maps $K = \overline {D(0,3\pi /4)}$ onto $\overline {D(0,1)}$ but $p(\partial K)$ doesn't even intersect $\partial (p(K)).$


0

One approach is to note that if $x_k \in \mathbb{R}^n$, then $x_n \to x$ iff $\phi(x_k) \to \phi(x)$ for any linear functional $\phi$ (this works in $\mathbb{R}^n$ since strong and weak convergence coincide). In particular, given some $f_n, f, g$ with $f_n(x) \to f(x)$ and $\|f_n\| \le g$, then for a linear functional $\phi$, we have $|\phi(f_n)| \le ...


0

The dominated convergence theorem applies to measurable functions with values in a Banach space, with the dominating function still being non-negative and integrable with $\|f_n(x)\|\leq g(x)$ The assumption of convergence almost everywhere can be weakened to require only convergence in measure.


1

$$\sum_{k=1}^n\frac{k^3}{\sqrt{n^8+k^4n^4}}=\frac1{n^4}\sum_{k=1}^n\frac{k^3}{\sqrt{1+\left(\frac kn\right)^4}}=$$ $$=\frac1n\sum_{k=1}^n\frac{\left(\frac kn\right)^3}{\sqrt{1+\left(\frac kn\right)^n}}\xrightarrow[n\to\infty]{}\int_0^1\frac{x^3}{\sqrt{1+x^4}}dx$$


0

Suppose $\mathscr{P}$ is a partition such that $s\in \mathscr{P}$. Let $s=x_{i-1}$. Then $U(\mathscr{P},f,\alpha)=\sum_{i=1}^n M_i \Delta\alpha_i=M_i$ which comes from the definition of the unit step function. (Do you see how?) $L(\mathscr{P},f,\alpha)=\sum_{i=1}^n m_i \Delta\alpha_i=m_i$ Therefore, $U(\mathscr{P},f,\alpha)-L(\mathscr{P},f,\alpha)=M_i-m_i\ ...


0

If $A_n=A$ for each $n\in\mathbb N$ then evidently $\limsup A_n=A$. For any function $f:\mathbb N\rightarrow\mathbb N$ we have $A_{f(n)}=A$ for each $n\in\mathbb N$ so that also $\limsup A_{f(n)}=A$. Then $$\limsup A_{f(n)}=A\subseteq A=\limsup A_n$$ but no conditions like $f(n)\rightarrow\infty$ on function $f$ are necessary.


1

Partial answer (The estimate below are a bit hard to make precise. Feel free to edit/criticize/downvote): Let $O = \frac{1}{5}(A_n+B_n+C_n+D_n+E_n)$ be the center of mass of the five points $A_n, B_n, C_n , D_n, E_n$. Note that $O$ is really independent of $n$ as the next five points are midpoints of the previous five. By a translation of the initial data, ...


0

The continiuity of $f^{-1}$ follows from the equality $(f^{-1} )' (y) =\frac{1}{f'(f^{-1} (y))}$ which shows that $f^{-1}$ is differentiable and hence continuous.


3

we have to show that the image of every open set is open. Take $f'>0$ everywhere WLOG. (work with -f otherwise) Consider an open set $U$ and $x \in U.$ Take $y < x< z$ all in $U.$ Then $$ f(y) < f(x) < f(z) $$ since $f'$ is never zero. So the image set contains $(f(y),f(z))$ that is an open interval containing $f(x)$ so the image of $U$ is ...


1

Part a OK Part b What is your definition of open? You have to prove that $B_\epsilon(A)$ is a neighborhood of all its points. Part c To prove an equality between two sets $R = S$, you have to prove that $R \subset S$ and $S \subset R$. One direction is pretty clear from part b. For the other one consider a point $x \notin A$


1

1) If $A$ is Hermitian positive definite, then $Ax=b$ is not ill-posed, since the matrix $A$ is invertible. It could be ill-conditioned if the condition of $A$, i.e., the quotient of its largest and smallest eigenvalue, is large. 2) No. For Hermitian matrices, eigenvectors are orthogonal, which is kind of an ideal situation. Ill-posedness is connected to ...


3

Depends on which topology you are talking about. If it is the topology on the real number system $\mathbb{R}$, then $\mathbb{R}$ is the closure of itself. If it is the topology on the extended real number system $\bar{\mathbb{R}}$, then $\bar{\mathbb{R}}$ is the closure of $\mathbb{R}$.


1

It matters with respected to which sense of convergence you are speaking of. If a point is in the space because it is the limit of a cauchy sequence, then yes R is it's own closure.


1

It looks right. In terms of proof-writing, I would say the following: First, make sure you always state $\epsilon > 0$ rather than just $\epsilon$. It is better to use more words and less symbols. It drastically improves the readability of the proof.


2

Let $y = [x] \Rightarrow x-1 <y \leq x\Rightarrow x < y+1 \leq x+1\rightarrow y+1 = [x+1]\Rightarrow [x]+1=[x+1]$


0

I have partial, but positive results. There is a fundamental book “Differential and Integral Calculus” by Grigorii Fichtenholz. This is a famous book for our students and it has many translations (but except English). I found in Appendix of the vol. I the following Theorem (I-II), with a long and complex proof. A simple curve without special points ...


0

You can think of it this way: taking advantage of the fact that $f$ is strictly positive continuous function, we have $g=1/f$ is also a positive continuous function on $[0,1]$. $g$ being continuous on $[0,1]$ attains a maximum, say, $a>0$. That is $g(x)=|g(x)|\leq a$ on $[0,1]$. Equivalently, $f(x)=|f(x)| \geq a$ on $[0,1]$. Just for fun, with the same ...


1

Note that for $x > 0$, $|x| = x$. Since we're taking $x$ to infinity, $x$ must pass $0$ at some point, so we can claim $$\lim_{x\to +\infty} \frac{x^3}{|x|} = \lim_{x \to +\infty} \frac{x^3}{x}.$$ From here you can make two arguments: either we discount the removable singularity at $x=0$, because we're onwards to $+\infty$ anyways, or we can use ...


3

For positive $x$, $|x|=x$, so $$\lim\limits_{x\to\infty} \frac{x^3}{|x|}=\lim\limits_{x\to\infty} \frac{x^3}{x}=\lim\limits_{x\to\infty} x^2=\infty$$


1

A slight refinement: $$ 0 < f(k) - e^{-k} = \sum_{k=2}^{\infty} e^{-kn^2} < \sum_{k=2}^{\infty} e^{-kn} = \frac{e^{-2k}}{1-e^{-k}}, $$ which is bounded by, say, $2e^{-2k}$ for $1-e^{-k}>1/2$, i.e. $k>\log{2}$. Hence, in the Poincaré-type asymptotics, $$ f(k) = e^{-k} + O(e^{-2k}). $$ Aside: The $(\pi/k)^{1/2}$ bound is useful when $k$ is ...


1

$A$ is the union of all rational numbers in $(0,1)$ and isolation points of {$2,3$}. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, any real number in $[0,1]$ is the limit point of $A$. Also note that no isolation point is the limit point. Denote $A^{'}$ as the limit points set of $A$. Then $A^{'}=[0,1]$ $A^o=\emptyset$ (for $\mathbb{Q}$ has no open set ...


0

You're very close. In particular, there is an open $n$-ball centered at $a$ contained entirely in $f^{-1}(B),$ having radius $\delta$ for some $\delta>0.$ You should probably then use the definition of preimage to finish the proof.


1

You need to find those points that are outside of $A$ but still are infinitely close. By that I mean, there exists an infinite set of points in $A$ that get arbitrarily close to that point. e.g. 0 is a limit point. This is because $1/2^n$ is in $A$ for all $n$ and they get arbitrarily close. You could make the same argument regarding irrationals in ...


1

If $f_{n}$ converged uniformly on $[0,1]$, the limit function would necessarily be continuous, but as you have shown, it is not. Therefore, the sequence does not converge uniformly.


1

It converges pointwise but not uniformly. Take the sequence $x_n=1-\frac{1}{n}$. The we get that the $\sup f_n(x)\geq 1-\frac{1}{1+e}$. The sup is not neccesarilly equal to 1/2. However, it is positive.


1

To “prove” this, you need convince yourself, that (i) every closed subset of a complete metric space is automatically complete with respect to the metric; (ii) the reals under the eukl. norm $(\mathbf{R},|\cdot|)$ is a complete metric space; and (iii) $[a,b]\subseteq\mathbf{R}$ is closed in this topology. Statement (i) is an easy exercise (take a ...


0

Hint: It is a well known result that given a Banach space $X$, and a (topological) subspace $F \subset X$, then $F$ is closed if and only if $F$ is complete. Then you should be able to conclude now, since $\Bbb R$ is complete. Is $[-1,1]$ closed in $\Bbb R$?


1

By definition $\varnothing,X\in\tau$, so $\varnothing = \varnothing\cap A\in\tau_A$ and $A=X\cap A\in\tau_A$. If $U\cap A, V\cap A\in\tau_A$, then $U\cap V\in\tau$ so $$(U\cap A)\cap(V\cap A) = (U\cap V)\cap A\in\tau_A. $$ If $U_\alpha\cap A\in\tau$ then $\bigcup_\alpha U_\alpha\in\tau$ so $$\bigcup_\alpha (U_\alpha\cap A) = \left(\bigcup_\alpha ...


1

For $k\ge 1,$ $$\sum_{n=1}^{\infty}e^{-kn^2} \le \sum_{n=1}^{\infty}e^{-kn} = \frac{e^{-k}}{1-e^{-k}}\le \frac{1}{1-e^{-1}}\cdot e^{-k}.$$


1

Your idea is fairly sound but you would get little or no credit on an exam for your answer. If the $f(x)$ is continuous at some point $x=x_0$ then (by definition) for any given $\epsilon > 0$ you can find a $\delta > 0$ such that whenever $x-\delta < x < x+\delta$, $|f(x)-f(x_0) < \epsilon$. In particular, at any irrational point $x_0$ (so ...


1

Since $\mathbb{Q}$ is dense in $\mathbb{R}$, for every $x \in \mathbb{R}/\mathbb{Q}$ there exists a sequence $\{x_n\} \subset \mathbb{Q}$ of rational numbers such that $ x_n\rightarrow x$ and you have: $$ \lim_{n \rightarrow \infty}f(x_n)= 1 $$ but $f(x)=0$


5

Pick a point $x \in \mathbb{R}$. Either $x$ is rational or irrational. Assume it is the former, so $f(x) = 0$. Let $\epsilon = \frac{1}{2}$. Regardless of $\delta$, there will exist a irrational number $z$ such that $$|x - z| < \delta \ \ \text{but} \ \ \|f(x) - f(z)| > \frac{1}{2}$$ where we have used the density of the reals and that $f(z) = 1$. So, ...


1

Let $R$ be so large that $\Omega\subset B(x,R)$ (where $B(x,R)$ is the ball centered at $x$ and with radius $R$). Then, since the function is positive, you have, by domain-monotonicity, $$ \int_\Omega|x-y|^{1-n}\lambda^n(y)\leq \int_{B(x,R)}|x-y|^{1-n}\lambda^n(y). $$ Now, let $u=x-y$, and you get that the integral to the right equals $$ ...


2

There is no way to express in elementary functions. In fact, go to http://mathworld.wolfram.com/JacobiThetaFunctions.html and you will find $$ f(k)=\frac12(-1+\vartheta_3(0,e^{-k})). $$



Top 50 recent answers are included