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H: Proving all roots of a sequence of polynomials are real Let the sequence of polyominoes $R_n(z)$ be defined as follows for $n\geqslant1$: $$R_n(z)\;= \;\sum_{r=0}^{\lfloor\frac{n-1}{2}\rfloor} \tbinom{n}{2r+1}(4z)^r.$$ I would like to prove that all the roots of $R_n(z)$ are real (and thus negative since the coefficients are all positive). Perhaps there's some way of getting the result by using Sturm’s theorem, but that seems too computationally intensive to be practical. Also, Kurtz’s condition on the coefficients only holds for small $n$. Any ideas? In case it helps, $R_n(z)$ satisfies the following recurrence: \begin{align} R_1(z)&\;=\;1 \\ R_2(z)&\;=\;2 \\ R_n(z)&\;=\;2R_{n-1}(z)+(4z-1)R_{n-2}(z)\qquad \text{for } n>2, \end{align} and also has the following closed form: $$ R_n(z)\;=\;\frac{(1+2\sqrt{z})^n-(1-2\sqrt{z})^n}{4\sqrt{z}}. $$ Thanks. AI: Due to your closed form, every root of $R_n$ must satisfy $(1+2\sqrt z)^n=(1-2\sqrt z)^n$. But that requires that $\|1+2\sqrt z\|=\|1-2\sqrt z\|$ which can't be true if $\sqrt z$ has nonzero real part. So $\sqrt z$ is pure imaginary, which is only possible if $z$ itself is a negative real.
H: Usual topology and Heine Borel theorem Heine-Borel theorem say: A subset in $\mathbb{R}^n$ is compact if and only if it is closed and bounded. Is this theorem independent of the topology in $\mathbb{R}^n$? If the answer is no, which is a counterexample? I have seen a demonstration but uses the usual topology for $\mathbb{R}^n$. Thank you for your help. AI: It is very dependent of the usual topology in $\,\Bbb R^n\,$ , of course. For example, if you take the discrete topology on this same space, then a subset is compact iff it is finite (meaning: it contains a finite number of elements.) Can you prove this?
H: Rotate a 3D vector on a plane I have a 3D line vector with end points x0 and x1, which lies along the x-axis of a subsection of the plane, P. However P has been translated, rotated and translated back from the global coordinate system by theta degrees along the global x-axis. The following image should illustrate my point. I need to rotate my 3D line vector by a known angle theta to find the line between x0 and x2. Can this be done by a set of transformation matrices (i.e. translate, x-rotate to global xy-plane, z-rotate by theta then x-rotate and translate back)? If so, how do I do this? I know the 3D vector equation is as below, but I'm not sure how to integrate that into the transformation. UPDATE: Thanks for the answers guys. I apologise for not making myself clear - my head's got in a bit of a muddle with all this! So I have a world system as in the 1st image, where the Z-axis faces vertically up. $P$, the red plane section, is defined by it's normal (giving orientation) and it's distance, $d$, from the origin using the equation $$\textbf{n}\cdot\textbf{x} = d$$ $\textbf{n}$ is formed using the rotation of the plane in terms of the global coordinate system. That is, assume that $P$ was flat along the x-y plane, as with the black one. It was translated to the origin, rotated by (e.g.) 25 degrees, then translated back. The actual problem I'm trying to solve is to simulate an object moving from the first (black) plane onto the 2nd (red). I know the direction and speed of the object's movement on the black plane, but I need to know it on the red. I know $x_0$, this is the objects position at the intersection of the two planes. I can work out $x_1$ by simply taking the unit direction vector from one end of P to the other (call this $l$) and multiplying it by the object's speed: $$x_1 = x_0 + s*l$$ However, my object may not be moving directly up the plane as assumed for $l$. We know the angle it's moving at relative to $l$, call this $\theta$. I need to work out the new vector for that direction, so I can add some multiple of it to $x_0$. AI: Is this what you want? $$\left(\begin{matrix} x \\ y \\ z \end{matrix}\right) = \left(\begin{matrix} 0 \\ a \\ 0 \end{matrix}\right) + \left(\begin{matrix} 0 \\ a\,(\cos(\theta)-1) \\ a\,\sin(\theta) \end{matrix}\right)\;t $$ where $X_2$ lies on the $y$-axis a distance $a$ from $X_0$. How? A point on the line is defined by the origin $$\vec{X}_2 = \left(\begin{matrix} 0 \\ a \\ 0 \end{matrix}\right) $$ and point $$\vec{X}_1 = {\rm Rx}(\theta) \vec{X}_2 $$ to give the equation: $$\vec{r} = \vec{X}_2 + \left( {\rm Rx}(\theta) \vec{X}_2-\vec{X}_2 \right)\; t $$ with $$ {\rm Rx}(\theta) = \left(\begin{matrix} 1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \end{matrix}\right) $$
H: About $\|\operatorname{Sym}(\Omega)\|$ when $\Omega$ is an infinite set. Here is a problem: Show that if $\Omega$ is an infinite set, then $\|\operatorname{Sym}(\Omega)\|=2^{\|\Omega\|}$. I have worked on a problem related to a group that is $S=\bigcup_{n=1}^{\infty } S_n$. Does it make sense we speak about the relation between $S$ and $\operatorname{Sym}(\Omega)$ when $\Omega$ is an infinite set. Moreover, I know that $\|\operatorname{Sym}(\Omega)\|=\|\Omega\|^{\|\Omega\|}$. How we can reach from $\|\Omega\|^{\|\Omega\|}$ to $2^{\|\Omega\|}$. Thanks. AI: You have $\|\Omega\|^{\|\Omega\|} \geq 2^{\|\Omega\|}$, since $\Omega$ is infinite. Then, $\|\Omega\| < 2^{\|\Omega\|}$ (Cantor's theorem) so $\|\Omega\|^{\|\Omega\|} \leq (2^{\|\Omega\|})^{\|\Omega\|}=2^{\|\Omega\|.\|\Omega\|}=2^{\|\Omega\|}$, since $\Omega$ is infinite.
H: Please explain this notation equation I am confused by this equation as I rarely use math in my job but need this for a program that I am working on. What exactly does the full expression mean? Note that $m^_i{_j}$ refers to a matrix whose values have already been obtained. Define the transition matrix $M = ${$m_i{_j}$} as follows: for $i\not=j$ set $m_i{_j}$ to $m^_i{_j}/\|U\|$ and let $m_i{_i} = 1-\Sigma_{j\not=i} m_i{_j}$ AI: To obtain the transition matrix $M$ from the matrix $M^=(m^_{ij})$, the rule gives us two steps. First, for all off-diagonal terms $m^_{ij}$ where $i\neq j$ we simply divide the existing entry by $\lvert U\rvert$ (in this case $\lvert U\rvert =24$), and we temporarily replace the diagonal entries $m^_{ii}$ by $0.$ Second, to get the $i^{\rm th}$ diagonal entry $m_{ii}$ of $M$ we sum up all entries in the $i^{\rm th}$ row of this intermediate matrix and subtract the resulting sum from $1,$ giving $1-\sum_{j\neq i}m_{ij}.$
H: Max of Brownian motion with drift is finite almost surely For $B_t$ Brownian Motion with drift $\mu<0$, I need to prove that the max value, $X = \max_{0<t<\infty}B_t$ is finite almost surely, ie $P(X<\infty)=1$. Now, I know that because the mean is negative, it will go more and more negative, and it is also a supermartingale. But I don't know how to prove almost surely... Appreciate any hints. AI: Hint: Try the strong law of large numbers. What does it say about $\lim_{t \to \infty} B_t/t$? What does this say about the sign of $B_t$ for large $t$?
H: An ignorant question about the incompleteness theorem Let me preface this by saying that I have essentially no background in logic, an I apologize in advance if this question is unintelligent. Perhaps the correct answer to my question is "go look it up in a textbook"; the reasons I haven't done so are that I wouldn't know which textbook to look in and I wouldn't know what I'm looking at even if I did. Anyway, here's the setup. According to my understanding (i.e. Wikipedia), Godel's first incompleteness theorem says that no formal theory whose axioms form a recursively enumerable set and which contain the axioms for the natural numbers can be both complete and consistent. Let $T$ be such a theory, and assume $T$ is consistent. Then there is a "Godel statement" $G$ in $T$ which is true but cannot be proven in $T$. Form a new theory $T'$ obtained from $T$ by adjoining $G$ as an axiom. Though I don't know how to prove anything it seems reasonably likely to me that $T'$ is still consistent, has recursively enumerable axioms, and contains the axioms for the natural numbers. Thus applying the incompleteness theorem again one deduces that there is a Godel statement $G'$ in $T'$. My question is: can we necessarily take $G'$ to be a statement in $T$? Posed differently, could there be a consistent formal theory with recursively enumerable axioms which contains arithmetic and which can prove every true arithmetic statement, even though it can't prove all of its own true statements? If this is theoretically possible, are there any known examples or candidates? Thanks in advance! AI: You are right that $T'=T+G$ must be consistent. If it were not, then we could prove a contradiction from $T$ together with $G$, which would amount to a proof by contradiction of $\neg G$ in $T$. But that contradicts the fact that $G$ is independent of $T$! On the other hand, when you ask Posed differently, could there be a consistent formal theory with recursively enumerable axioms which contains arithmetic and which can prove every true arithmetic statement, even though it can't prove all of its own true statements? that cannot be, because the independent statement $G$ produced by Gödel's construction is always an arithmetic statement. ($G$ is often popularly explained as stating something about provability or not, which is correct so far as it goes -- but the major achievement of Gödel's work is to show how such claims can be expressed as arithmetic statements). Indeed, if the original theory $T$ was something like Peano Arithmetic, the language of the theory does not allow one to even state any sentence that is not arithmetic -- and no amount of added axioms can change that.
H: Further reading on the $p$-adic metric and related theory. In his book Introduction to Topology, Bert Mendelson asks to prove that $$(\Bbb Z,d_p)$$ is a metric space, where $p$ is a fixed prime and $$d_p(m,n)=\begin{cases} 0 \;,\text{ if }m=n \cr {p^{-t}}\;,\text{ if } m\neq n\end{cases}$$ where $t$ is the multiplicty with which $p$ divides $m-n$. Now, it is almost trivial to check the first three properties, namely, that $$d(m,n) \geq 0$$ $$d(m,n) =0 \iff m=n$$ $$d(m,n)=d(n,m)$$ and the only laborious was to check the last property (the triangle inequality). I proceeded as follows: Let $a,b,c$ be integers, and let $$a-b=p^s \cdot k$$ $$b-c=p^r \cdot l$$ where $l,k$ aren't divisible by $p$. Then $$a-c=(a-b)+(b-c)=p^s \cdot k+p^r \cdot l$$ Now we have three cases, $s>r$, $r>s$ and $r=s$. We have respectively: $$a-c=(a-b)+(b-c)=p^r \cdot(p^{s-r} \cdot k+ l)=p^r \cdot Q$$ $$a-c=(a-b)+(b-c)=p^{s} \cdot( k+p^{r-s} \cdot l)=p^s \cdot R$$ $$a-c=(a-b)+(b-c)=p^s \cdot (k+l)=p^s \cdot T$$ In any case, $$d\left( {a,c} \right) \leqslant d\left( {a,b} \right) + d\left( {b,c} \right)$$ since $$\eqalign{ & \frac{1}{{{p^r}}} \leqslant \frac{1}{{{p^s}}} + \frac{1}{{{p^r}}} \cr & \frac{1}{{{p^s}}} \leqslant \frac{1}{{{p^s}}} + \frac{1}{{{p^r}}} \cr & \frac{1}{{{p^s}}} \leqslant \frac{1}{{{p^s}}} + \frac{1}{{{p^s}}} \cr} $$ It might also be the case $k+l=p^u$ for some $u$ so that the last inequality is $$\frac{1}{{{p^{s + u}}}} \leqslant \frac{1}{{{p^s}}} + \frac{1}{{{p^s}}}$$ $(1)$ Am I missing something in the above? The author asks to prove that in fact, if $t=t_p(m,n)$ is the exponent of $p$, that $$t\left( {a,c} \right) \geqslant \min \left\{ {t\left( {a,b} \right),t\left( {b,c} \right)} \right\}$$ That seems to follow from the above arguement, since if $s \neq r$ then $$t\left( {a,c} \right) = t\left( {a,b} \right){\text{ or }}t\left( {a,c} \right) = t\left( {b,c} \right)$$ and if $s=r$ then $$t\left( {a,c} \right) \geqslant t\left( {a,b} \right){\text{ or }}t\left( {a,c} \right) \geqslant t\left( {b,c} \right)$$ $(2)$ Is there any further reading you can suggest on $p$-adicity? AI: I am currently trying to learn about $p$-adic numbers and analysis too, so I too would be really interested to hear the opinions of people who know more than I do about this. I am currently using the following three texts, but don't intend to work through them fully; just enough to get something useful for a better understanding of how they can be used in number theory: Koblitz - "$p$-adic Numbers, $p$-adic Analysis, and Zeta-functions" - The first chapter I have found very interesting, and pretty well-written, with lots of easy exercises to get used to the concepts, as well as some harder ones to test deeper understanding. Robert - "A Course in $p$-adic Analysis" - covers much more material at a more advanced level than Koblitz, but isn't (quite) as off-putting as it seems at first, and it possible to pick out quite a few bits from the first two chapters which are illuminating. Borevich and Shafarevich - "Number Theory" - This one has some relatively understandable stuff on $p$-adic numbers in the first chapter which I have found really useful as it gives a different feel to the topic in a rather more old-fashioned approach.
H: Minimal diameter I have a closed bounded convex shape in $\mathbb{R}^3$. I want to calculate what I would call the minimal diameter: find the plane (or, generally, the space of codimension 1) which minimizes the maximum distance from points in the shape to the plane, then take that maximum distance. I suppose one should take the supremum if the set is not closed. How can I find this efficiently? And is there a standard name for it? It seems too obvious to be without a name. I'm particularly interested in the case that the shape is a conic section or otherwise defined by low-degree polynomial inequalities. AI: Up to the factor of 2, this is known as the minimum width, and sometimes simply as the width of a convex body. Another, and perhaps clearer definition of it is: the length of the shortest projection onto a line. Unfortunately I do not know anything about the algorithmic side of the issue.
H: Turn fractions into $\mathbb Z_7$ elements I had to perform a division between two polinomials $2x^2+3x+4$ and $3x+4$, my book suggests to do this operation without worrying about the modulo. So my result is $(3x+4)(\frac{2}{3}x+\frac{1}{9})+\frac{32}{9}$. Unfortunately my book fails to explain how should I perform the conversion from fractions to $\mathbb Z_7$ elements, like it's kind of obvious, it only reports $\frac{1}{3}=5$, $\frac{2}{3} = 10 = 3$, $\frac{32}{9}=2$. Will you please break it down for me? AI: Remember that $\frac{1}{x}$ means "the element which, when multiplied by $x$, gives $1$." Alternatively, $\frac{a}{b}$ is just shorthand for "the solution to the equation $bx=a$." So, in $\mathbb{Z}_7$, the fraction $\frac{1}{3}$ is shorthand for "the solution to the equation $3x=1$", which, when interpreted in $\mathbb{Z}_7$, is just "the solution to the integer congruence $3x\equiv 1\pmod{7}$." Similarly, $\frac{2}{3}$ means "the solution to the integral congruences $3x\equiv 2\pmod{7}$", and $\frac{32}{9}$ means "the solution to the integral congruence $9x\equiv 32\pmod{7}$." Since $\gcd(3,7)=\gcd(9,7)=1$, each of these congruences has a unique solution modulo $7$. $3(5)\equiv 1\pmod{7}$, so $5 = \frac{1}{3}$. Hence, $\frac{2}{3} = 2\frac{1}{3}=2(5) = 10 \equiv 3\pmod{7}$, so $\frac{2}{3}$ is $3$. And since $32\equiv 4\pmod{7}$, for $\frac{32}{9}$ you have $4(5)(5) = 100\equiv 2\pmod{7}$.
H: Strong markov property on max of brownian motion For $B_t$ Brownian Motion with drift $\mu<0$, I have the max value, $X = \max_{0<t<\infty}B_t$ . I need to prove with the Strong Markov Property that, $P(X>c+d)=P(X>c)P(X>d)$ a. It seems weird to me since one of the right hand terms is redundant... b. I'll appreciate any hints, especially on how to use the markov property on X. I know I have it on the BM process itself, but not sure how they relate. Thanks. AI: Let $T=\inf\{t\geqslant0\mid B_t\geqslant c\}$. Then $[T\lt\infty]=[X\geqslant c]$ and, conditionally on the event $[T\lt\infty]$, $(B_{t+T})_{t\geqslant0}$ is distributed like $(c+B_t)_{t\geqslant0}$ and is independent on $(B_t)_{t\leqslant T}$. Hence, conditionally on the event $[T\lt\infty]$, the maximum of $(B_{t+T})_{t\geqslant0}$ is distributed like $c+X$. Thus, $\mathrm P(X\geqslant c+d\mid T\lt\infty)=\mathrm P(c+X\geqslant c+d)=\mathrm P(X\geqslant d)$.
H: $\mathcal{K}(\mathcal{H})$ is separable? I am reading Arveson's Notes On Extensions of $C^*$-algebras, in proving the Corollary to Thm2, he seems to assume that $\mathcal{K}(\mathcal{H})$, the space of compact operators on a separable Hilbert space, is separable as a topological space, i.e., it contains a countable dense set. This is a little bit surprising to me. If we just take $\mathcal{H}=\ell^2$ and $F$ defined by:\begin{equation} F(\sum\alpha_n e_n)=(\sum\alpha_n f_n)e_1, \end{equation} where $\{e_n\}$ is the canonical basis for $\ell^2$. That is, $F$ maps everything to the first coordinate, $F e_n=f_n e_1$. Since $(f_n)$ can be be arbitrary sequences in $\ell^{\infty}$, which is a non-separable space. My argument seems to imply that even the space of rank-one operators is not separable. Where did I make a mistake? Thanks! AI: The space of rank-one operators is separable, being a continuous image of $\mathcal H\oplus \mathcal H$ under the map $(u,v)\mapsto T_{u,v}$ where $T_{u,v}x=\langle x,u\rangle v$. Hence the space of finite-rank operators is separable (they are finite sums of rank-1 operators), and its closure in $B(\mathcal H)$ is $\mathcal K(\mathcal H)$.
H: Existence of a prime ideal in an integral domain of finite type over a field without Axiom of Choice Let $A$ be an integral domain which is finitely generated over a field $k$. Let $f \neq 0$ be a non-invertible element of $A$. Can one prove that there exists a prime ideal of $A$ containing $f$ without Axiom of Choice? I came up with this question to solve this problem. This is a related question. AI: This answer builds on Qiaochu's and uses the same definition as Qiaochu, to wit: A ring $R$ is noetherian if, for any nonempty collection of ideals $\mathcal{I}$, there is some $I \in \mathcal{I}$ which is not properly contained in any $J \in \mathcal{I}$. Theorem: If $R$ is noetherian, then $R[x]$ is noetherian. This proof is basically taking the standard proof and rephrasing it to use Qiaochu's definition and be careful about choice. I'm going to try to systematically use the following conventions: Ideals in $R[x]$ get capital letters; ideal in $R$ get overlined capital letters. Sets of ideals in $R[x]$ get calligraphic letters. I found that I could manage to write this without ever assigning a name to a set of ideals in $R$. For any ideal $I \subseteq R[x]$, and any integer $j \geq 0$, define $$s_j(I) := \{ r \in R : \mbox{there is an element of $I$ of the form } r x^j + r_{j-1} x^{j-1} + \cdots +r_0 \}$$ Observe that $s_j(I)$ is an ideal and $s_j(I) \subseteq s_{j+1}(I)$. Lemma: If $R$ is noetherian, and $I \subseteq R[x]$ is an ideal, then there is an index $j$ such that $s_k(I) = s_j(I)$ for $k \geq j$. Proof: Let $s_j(I)$ be a maximal element in $\{ s_j(I) \}_{j \geq 0}$. Let $k \geq j$. Then we observed above that $s_k(I) \supseteq s_j(I)$. But, by the definition of a maximal element, we do not have $s_k(I) \supsetneq s_j(I)$, so $s_k(I) = s_j(I)$. $\square$. We will denote the ideal $s_j(I)$ defined in the above lemma as $s_{\infty}(I)$. For $\mathcal{I}$ a collection of ideals in $R[x]$, we will write $s_j(\mathcal{I})$ or $s_{\infty}(\mathcal{I})$ for the result of applying $s_j$ or $s_{\infty}$ to each element of $\mathcal{I}$. So $s_j(\mathcal{I})$ is a set of ideals in $R$. Let $\mathcal{I}$ be a collection of ideals in $R[x]$. Since $R$ is noetherian, there is a maximal element $\bar{J}$ in $s_{\infty}(\mathcal{I})$. Let $\mathcal{J}$ be the set of all $I \in \mathcal{I}$ with $s_{\infty}(I) = \bar{J}$. Note that no element of $\mathcal{I} \setminus \mathcal{J}$ contains an element of $\mathcal{J}$, by the maximality of $\bar{J}$, so it is enough to show that $\mathcal{J}$ has a maximal element. Choose an ideal $K \in \mathcal{J}$. (Making one choice does not use AC.) Let $m$ be an index such that $s_m(K) = \bar{J}$. Let $\mathcal{K}$ be the collection of ideals $I \in \mathcal{J}$ for which $s_m(I) = \bar{J}$. Note that no element of $\mathcal{J} \setminus \mathcal{K}$ can properly contain an element of $\mathcal{K}$, so it is enough to show that $\mathcal{K}$ has a maximal element. We now make finitely many dependent choices. Choose a maximal element $\bar{J}^{m-1}$ in $s_{m-1}(\mathcal{K})$; let $\mathcal{K}_{m-1}$ be the set of $I \in \mathcal{K}$ with $s_{m-1}(I)=\bar{J}^{m-1}$; it is enough to show that $\mathcal{K}_{m-1}$ has a maximal element. Choose a maximal element $\bar{J}^{m-2}$ in $s_{m-2}(\mathcal{K}_{m-1})$; let $\mathcal{K}_{m-2}$ be the set of $I \in \mathcal{K}^{m-1}$ with $s_{m-2}(I)=\bar{J}^{m-2}$; it is enough to show that $\mathcal{K}_{m-2}$ has a maximal element. Continue in this manner to construct $\mathcal{K}_{m-3}$, $\mathcal{K}_{m-4}$, ..., $\mathcal{K}_0$. Since we are only making finitely many choices, we don't need AC; see my answer here. At the end, we have a nonempty collection $\mathcal{K}_0$ of ideals such that, for any $I$ and $J \in \mathcal{K}_0$, and any $j \geq 0$, we have $s_j(I)= s_j(J)$. I claim that any element of $\mathcal{K}_0$ is maximal. Let $I$ and $J \in \mathcal{K}_0$ and suppose that $I \supseteq J$. I will prove that $I=J$. This shows that every element of $\mathcal{K}_0$ is maximal. Let $I_{\leq d}$ be the set of polynomials in $I$ of degree $\leq d$. I will show by induction on $d$ that $I_{\leq d} = J_{\leq d}$. The base case is $d=-1$, where both sides are $\{ 0 \}$. Since $I \supseteq J$, I just need to show that $I_{\leq d} \subseteq J_{\leq d}$. Let $f \in I_{\leq d}$ and let the leading term of $f$ be $r x^d$. Then $r \in s_d(I) = s_d(J)$ so there is some $g \in J_{\leq d}$ with leading term $r$. Since $I \supseteq J$, we have $g \in I$ and hence $f-g \in I$. Since $\deg(f-g) < d$, by the induction hypothesis, we have $f-g \in J$. So $f = (f-g)+g \in J$. QED
H: Tracing a point on a rotating circle I have a large circle and a small circle as shown in the image. The distance between the centers of those circles. The Larger circle is rotating about it's center while traveling in direction D such that it travels its circumference in 1 revolution. There is a point on the smaller circle called X. I need a function that can calculate the path of x (for graphing purposes). Is there a mathematical formula for this? If not then what group of math formulas could be used to solve this problem? AI: See here for more information (especially look in the Related Curves section). In particular, this appears to be a curtate cycloid, based on my reading of your question.
H: When is $(6a + b)(a + 6b)$ a power of two? Find all positive integers $a$ and $b$ for which the product $(6a + b)(a + 6b)$ is a power of $2$. I havnt been able to get this one yet, found it online, not homework! any help is appreciated thanks! AI: Hint: $(6a+b)(a+6b)$ is a power of $2$ iff $(a+6b)$ and $(6a+b)$ are powers of $2$ individually.
H: Inequality's root Given $\sqrt x + \sqrt y < x+y$, prove that $x+y>1$. Havnt been able to try this yet, found it online, any help is appreciated thanks! AI: If $\sqrt{x}+\sqrt{y}<x+y$, then after squaring both sides we have $x+2\sqrt{xy}+y<x^2+2xy+y^2$. Rearrange to isolate the square root: $$2\sqrt{xy}<x^2+2xy+y^2-x-y=(x+y)^2-(x+y)=(x+y)(x+y-1)\;.$$ Can you see why this implies that $x+y>1$?
H: What means the boundary of a space By definition topological spaces are clopen and then their boundaries are empty, but for example, is said that the boundary of the closed unit interval is it's two endpoints a so on. whath is the meanining of "boundary" in this context? AI: There are two notions of boundary in mathematics. One is the boundary of a subset of a topological space, and it is a relative notion: it takes as input two objects, a topological space and a subset of it. As a subset of $\mathbb{R}$, the boundary of $[0, 1]$ is its endpoints. (However, as a subset of $\mathbb{R}^2$, for example, the boundary of $[0, 1]$ is $[0, 1]$. Again, relative.) There is also the boundary of a manifold with boundary, which takes as input one object, a manifold with boundary. As a manifold with boundary, the boundary of $[0, 1]$ is also its endpoints.
H: Let $A\subseteq\mathbb{R}$ be open, with $A\cup (0,1)$ connected Let $A\subseteq\mathbb{R}$ be open. If $A\cup (0,1)$ is connected then A must be connected. A must have one or two component. $A\setminus(0,1)$ has at most two component. $A$ must be a cantor set. Take $A=(0,1/2)\cup(1/2,1)$. Then $A\cup (0,1)=(0,1)$ is connected but $A$ is not, so $1$ is false. $2$ is also false as I can take 3 or more components. I am not sure about 3 and 4, thank you for help. AI: HINT Can you characterize the connected subsets of $\mathbb{R}$? EDIT If you just want to answer the question, you can easily rule out $1$, $2$ and $4$, by taking $$A = (-1/2,1/4) \cup (1/2,3/4) \cup (8/9,2)$$
H: the derivative for a Lipschitz function For the solution of the ODE $dy/dt=f(t,y)$, $f$ has to be Lipschitz for all $t$. So, if $f$ is a function that is not differentiable with respect to $y$ but Lipschitz, what can I say about $f_y$? Can I estimate some norm of it? I can't say $\lVert f_y\rVert_C$ because $f$ is not $C^1$, however, I can say that $f_y$ is defined in a weak sense and then this weak derivative is bounded (by the Lipschitz property) and I can estimate $\lVert f_y\rVert_{L^{\infty}}$ and have it bounded. Is that correct? AI: Yes, every Lipschitz function (on an open subset $\Omega\subset \mathbb R^n$) has a weak derivative which belongs to $L^\infty(\Omega)$. The $L^\infty$ norm of the derivative is bounded by the Lipschitz constant of the function. Under appropriate assumptions on the geometry of $\Omega$ one can reverse this implication and obtain the global Lipschitz condition from the boundedness of the derivative.
H: How complex exponential converges and "sum of exponents" rule holds How is it the complex exponential converges for any value of $z$ in the complex plane? $$e^{z} = 1 + \frac{z}{1!} + \frac{z^2}{2!} \cdots\cdots$$ How is it the "sum of exponents" rule holds for complex exponential, that is $e^{w}e^{z} = e^{w+z} $? ...using only the definition of $e^z$? AI: $$e^z=\sum_{n=0}^\infty\frac{z^n}{n!}$$ Putting $$a_n:=\frac{1}{n!}\Longrightarrow R:=\frac{1}{\lim_{n\to\infty}\sqrt[n]{\|a_n\|}}=\lim_{n\to\infty}\sqrt[n]{n!}=\infty$$ So the series has infinite convergence radius and is thus absolutely convergent for any $\,z\in\Bbb C\,$ , and from here it follows that $$e^{w+z}=\sum_{n=0}^\infty\frac{(w+z)^n}{n!}=\sum_{n=0}^\infty\sum_{k=0}^n\binom{n}{k}\frac{w^kz^{n-k}}{n!}=\sum_{n=0}^\infty\sum_{k=0}^n\frac{w^kz^{n-k}}{k!(n-k)!}=$$ $$=\sum_{k=0}^\infty\frac{w^k}{k!}\sum_{n=0}^\infty\frac{z^n}{n!}=:e^we^z$$
H: Some questions on Laplace equation . While i am revising for exam : I am facing some problems to understand things clearly . Here are my doubts : a) If $u$ solves $\Delta u =0 , x\in \Omega; u=g , x\in \partial \Omega$ for non constant boundary data $g$ with $g\ge0$ and $g(x_0)\> 0$ for some $x_0 \in \partial \Omega $ then why is it true that $u(x)>0$ for all $x\in \Omega$ ? b) It's about Harnack inequality . Let $V$ be open and connected and $V\subset\subset \Omega$ ie $V$ is compactly contained in $\Omega$ then there existz a constant $C<\infty$ such that $sup_V u \le C inf_V u$ . My question is why only upto $V$ and not upto $\bar V $ ie why not upto closure ? AI: a) You are missing an important assumption that $g$ is continuous. Under this assumption $g(y)$ agrees with the limit of $u(x)$ as $x\to y\in\partial\Omega$. By the maximum principle $u\ge 0$ in $\Omega$, and the stronger form of the same principle says that either $u\equiv 0$ or $u>0$ everywhere. The first case is impossible, for then we would have $g(x_0)=\lim_{x\to x_0}u(x)=0 $. b) You have my permission to replace $V$ with $\overline{V}$ in Harnack's inequality. It does not matter.
H: Probability of a number being a multiple of another number For a random whole number, n, between 0 and 4000 billion (don't know what that's called), is the probability that n is a multiple of 4096, 1/4096? AI: If one end (either $0$ or $4000$ billion) is included, the probability is exactly $\frac 1{4096}$ as $4096=2^{12}$ and 4000 billion has at least $14$ factors of $2$ (if billion = $10^9$). If both ends are excluded, there is an error of $1$ in $4000$ billion. If both ends are included, there is an error of about $1$ in $976562500$ Added: as both ends are included, the probability is $\frac {976562501}{4000000000001}$ which is not exactly $\frac 1{4096}$
H: Notation for a function from all members of a tuple minus one. Is there any compact, mainstream notation for a function from all members of a tuple minus one? What I have in mind is $f\left(a_{1},\ldots,a_{n}\right)$ (except for $a_{i}$) $=a_{i}$ AI: If you mean an ordered tuple, then a common notation for $$(a_1,a_2,\ldots,a_{i-1},a_{i+1},\ldots,a_n)$$ is $$(a_1,a_2,\ldots,\widehat{a_i},\ldots,a_n).$$ But it is usually specified in text what it means the first time it is used.
H: Should one imagine diagrams/figures when working? I'm working through Baby Rudin and find it exceedingly difficult to understand what's happening without drawing a small figure. For instance when proving properties of compactness, I would often draw figures like : (Black is the set, red are the finite sub-covers) My question is: Am I handicapping myself by continuously drawing figures (limited to $\mathbb{R}^2$)? I am tuned to imagine things which align neatly. For instance, if someone says imagine a triangle, I imagine an equilateral one and this usually prevents me from understanding subtle points. AI: I can't do mathematics at all without drawing and I have practically never given a course or a talk without making coloured doodles on the blackboard. There are a few exceptions, as when I had to teach the Sylow theorems, but then I had the feeling that I didn't understand what I was talking about, although the proofs were (I hope!) correct in the logical sense. The same goes for hard analysis, say linear partial differential equations: I tried to read Hörmander but gave up, because I couldn't get a gut feeling for the inequalities there (even though geometry is definitely present in that book). I am in a field (algebraic geometry) where it is easy to make drawings and I started to understand scheme theory only when I saw Mumford's drawings of $Spec (\mathbb Z)$, $Spec (\mathbb Z[T])$ and his cartoonesque rendition of the spectrum $Spec(\mathcal O_{X,O})$ of the local ring at the origin of the plane $X=Spec(\mathbb A^2_k)$ over the field $k$, where the closed points of curves have disappeared and only their generic point is left behind , exactly like the grin of the Cheshire cat in Alice in Wonderland, the masterpiece of that wonderful mathematician who also loved drawings. [If you have never seen a live scheme in its natural habitat, look at pages 72-75 of Mumford 's Red Book where the pictures I evoke above are to be found. Or here pages 111-112] In conclusion, if you feel drawings help you, by all means go ahead: I find your version of a covering of a compact space truly ingenious an illuminating. And to finish on a lighter note, there is this story of an engineer asking a mathematician how he could visualize 4-space: "Very easy, I imagine $\mathbb R^n$ and I specialize to $n=4$"
H: Condition(s) that satisfy this equality I am having difficulty understanding how my book came up with this answer. Define $a \star b =ab+2b$, and suppose $x \star y = y \star x$. Then which of the following must be true? A. $x+y=1$. B. $y=0$. C. $x=y$. D. $x=-2$. E. $xy=0$. How did the text conclude that $x=y$ or C is the answer ? AI: Hint: So you are told that $xy+2y=yx+2x$. Cancel.
H: A question related to Novikov's condition The well-known 'Novikov condition' says: Let $ L = (L_t)_{t \geq 0} $ be a continuous local martingale null at 0 and $ Z = \exp(L - \frac{1}{2} \langle L \rangle) $ its stochastic exponential. If $ E[\exp(\frac{1}{2} \langle L \rangle_\infty)] \ < + \infty $, then $ Z $ is a (uniformly integrable) martingale on $ [0, +\infty] $. Now to my question: Is it also true that in this case $ Z_\infty > 0, P-a.s. $? I'm interested in this question, because $ Z_\infty > 0 $ would ensure that the measure $ Q $ defined by $ Q[A] := E_P [Z_\infty 1_A] $ is equivalent to $ P$. Thanks for your help! Regards, Si AI: Yes because $\mathbb E [L_{\infty},L_{\infty}] < \infty$ will make L an $\mathbb L^2$ bounded martingale, so $L_{\infty}$ is finite (and the limit of $L_t$.)
H: Integral of $e^{(x-x^3)/3n}$ from $0$ to $\infty$ How can you compute the following integral assuming $n>0$? $$\int_{x=0}^{\infty}e^{\frac{x -x^3}{3n}}dx $$ Mathematica etc. fail to produce anything useful. EDIT: I would be happy with an asymptotic result in $n$ if it is too hard to compute exactly. I don't know if it helps but $$\int_{x=0}^{\infty}e^{\frac{-x^3}{3n}}dx \approx 1.29 n^{1/3}.$$ AI: The asymptotic behavior ($n\gg 1$) is given by (the $x$ contribution disappears and the integral becomes solvable) : $$\int_0^\infty e^{\frac{x -x^3}{3n}}dx \sim (3n)^{1/3}\Gamma\left(\frac 43\right)$$ To get the next terms of the expansion, and give you some confidence in this result, let's expand the $e^{x/(3n)}$ factor in series and use (since this integral may be rewritten as a gamma integral) : $$\int_0^\infty \left(\frac x{3n}\right)^k e^{\frac{-x^3}{3n}}dx=\frac {(3n)^{(1-2k)/3}}{k+1}\Gamma\left(\frac{k+4}3\right)$$ Observe that we will have to divide by $(3n)^{2/3}$ at each step $k\to k+1$ (as you may see in GEdgar's answer) getting with the expansion $e^r=1+r+\frac {r^2}2+\cdots$ $$\int_0^\infty e^{\frac{x -x^3}{3n}}dx \sim (3n)^{1/3}\left(\Gamma\left(\frac 43\right)+\frac {\Gamma\left(\frac 53\right)}{2(3n)^{2/3}}+\frac 1{2\cdot 3 (3n)^{4/3}}+\rm{O}\left(n^{-2}\right)\right)$$
H: Question about the socle of a finite-dimensional algebra Let $n\in \mathbb{N}$ and $k$ be an arbitrary field. Is the socle of the algebra $k[x,y]/\langle x^2,y^{n+2}\rangle$ isomorphic to $k$? Is $k[x,y]/\langle x^2,y^{n+2}\rangle$ a symmetric algebra or a Frobenius algebra or a self-injective algebra? I would be very grateful for an answer. AI: Your algebra $A$ has the set $\mathcal B=\{x^iy^j:0\leq i<2, 0\leq j<n+2\}$ as a basis. An element $a$ of $A$ is in the socle iff $xa=ya=0$, because $x$ and $y$ generate the radical of $A$ (This last statement has to be checked of course: the ideal $I$ generated by $x$ and $y$ is nilpotent, because $x$ and $y$ are, and the quotient $A/I$ is a field, so $I$ is the radical by a well-known characterization of the radical) You can easily find which linear combinations of the elements of $\mathcal B$ satisfy these two conditions.
H: Using a single scalar equation to describe a line in space A line in three-dimensional space may be described as an intesection of two planes, for example: $$\begin{align}x+y+z=0\tag{1}\\3x+7y=1\tag{2}\end{align}$$ This can be understood as two separate scalar equations or as a single matrix equation. (One may also describe a line parametrically.) This poses a question: is it possible to express the same line using a single scalar equation? It turns out that it is. The equation $$x^2+y^2=0\tag{3}$$ can be understood as describing the set of all points $(x,y,z)\in\Bbb R^3$ for which $x=0$ and $y=0$. In other words, it describes the $z$-axis. So, we have described a line in $\Bbb R^3$ using a single scalar equation. But this means any line in $\Bbb R^3$ (or $\Bbb R^n$ for that matter) can be described by a single scalar equation, simply by using an appropriate affine transformation on the equation $(3)$. As pointed out by Pantelis Damianou in the comment below, this gives us the equation $$(x+y+z)^2+(3x+7y-1)^2=0$$ in the case described above. Note that this tells us exactly the same thing as the equations $(1)$ and $(2)$, since $z^2+w^2=0$ is just another way to say that $z=0$ and $w=0$. My question is: Is this point of view ever useful? Does it have any striking applications? Is there an area of mathematics that uses such equations in a fruitful way? Thanks. AI: More generally, we can "pack up" any finite system of equations (over the reals) as a single equation involving a sum of squares. If you want an approximate solution of the system of equations, you can try to minimize that sum of squares: this is the starting point for Least Squares Approximation, which is very useful in data analysis.
H: $G=\mathbb{Z}_2\times\mathbb{Z}_3$ is isomorphic to? $G=\mathbb{Z}_2\times\mathbb{Z}_3$ is isomorphic to $S_3$ A subgroup of $S_4$. A proper subgroup of $S_5$ $G$ is not isomorphic to a subgroup of $S_n$ for all $n\ge 3$ What I know is Any finite group is isomorphic to a subgroup of $S_n$ for some suitable $n$(Caley's Theorem), 1 is not true as $G$ is abelian but $S_3$ is not,$4$ violates Caleys Theorem, for 3 I saw that there is an order $6$ element in $G$ namely $(1,1)$ but No element of the subgroup of $S_4$ have order $6$ right? Not sure about 3, Thank you for the help. AI: You are correct that $G$ is not isomorphic to $S_3$ but that it is isomorphic t some subgroup of $S_n$ for some $n$ (in fact, any $n\geq 6$ clearly works). Note that $G\cong \mathbb Z_6$, the cyclic group of order $6$. Thus the existence of a subgroup isomorphic to $G$ is equivalent to the existence of an element of order $6$. Any element of $S_4$ is either a $4$-cycle, which has order $4$, a $3$-cycle, which has order $3$, or a product of disjoint $2$-cycles, which have order $2$. Thus $G$ is not isomorphic to a subgroup of $S_4$. Can you think of an element of $S_5$ with order $6$?
H: Lipschitz constant on a functional Let $C$ be the space of continuous and nondecreasing functions defined on $[0,1]$ and endowed with the sup norm. Let $T:C\rightarrow C$ be a continuous mapping, and consider the following expression: $$ U(Tz(x);z)=\int_{0}^{Tz(x)}\left[\int_{0}^{1}F(z(\xi))f(\xi)d\xi+\int_{s\in \Gamma(t,z)}\left\{F(t)-F(z(s))\right\}f(s)ds\right]^{n-1}dt $$ where $\Gamma(t,z)=\{s:[0,1]\|s\geq z(t)\}$, and $F:[0,1]\rightarrow [0,1]$ is continuously differentiable and increasing; $F'=f$, $f(s)>0, s\in[0,1]$ and $f(s)=0,\,s\notin [0,1]$, $x\in[0,1]$, and $n>2$. Suppose that I've managed to show that there exists a constant $K$ such that $$ \|\|U(Tz(x);z)-U(Ty(x);y)\|\|\leq K\|\|z-y\|\| \qquad z,y\in C $$ holds true. What I want is to make a claim about $T$. Particularly, I'd like to show whether is true or not that the following hold: $$ \|\|Tz(x)-Ty(x)\|\|\leq C\|\|z-y\|\| \quad z,y\in C $$ where $C$ is a constant independent of $z$ and $y$. Does anyone has any idea how I can prove/disprove this? I have no clue whatsoever on how to proceed, and I do need some help to get through it. Can someone help me or point me the right direction on how to tackle this problem? Any help/suggestion/insight/reference is greatly appreciated it! AI: In the earlier question we saw that $\|U(Ty(x);z)-U(Ty(x);y)\|\le M\\|z-y\\|$. By the triangle inequality, $\|U(Tz(x);z)-U(Ty(x);z)\|\le (M+K)\\|z-y\\|$. On the other hand, $$\|U(Tz(x);z)-U(Ty(x);z)\|\ge \left\|\int_{Ty(x)}^{Tz(x)} \left[1-\int_{s}^{1}F(z(\xi))f(\xi)d\xi\right]^{n-1}ds \right\| \\ \ge \|Tz(x)-Ty(x)\|\min_{[Ty(x),Tz(x)]} \left[1-\int_{s}^{1}F(z(\xi))f(\xi)d\xi\right]^{n-1}$$ So if you can estimate the minimum from below by a positive number, you win. In other words, you need an upper bound on $\int_{s}^{1}F(z(\xi))f(\xi)d\xi$ which is less than $1$.
H: Irrational distances, rational area triangles Given any positive integer $n\ge3$ how to show that there are $n$ distinct points in the plane such that 1- the distance between any two points is irrational number and 2- each set of three points determines a non-degenerate triangle whose area is a rational number AI: Look at the points $(x,x^2)$ where $x$ is a non-negative integer. These are lattice points, so the area of any triangle they determine is rational, indeed of the form $k/2$, where $k$ is an integer. Now calculate the distance between $(a,a^2)$ and $(b,b^2)$, where $b\gt a$. The square of the distance is $$(b-a)^2+(b^2-a^2)^2,$$ which is $(b-a)^2(1+(b+a)^2)$. This cannot be a perfect square, since $1+(b+a)^2$ cannot be: $x^2+1$ is a perfect square only when $x=0$.
H: Pythagorean Theorem for imaginary numbers If we let one leg be real-valued and the other leg equal $bi$ then the Pythagorean Theorem changes to $a^2-b^2=c^2$ which results in some kooky numbers. For what reason does this not make sense? Does the Theorem only work on real numbers? Why not imaginary? AI: The Pythagorean Theorem is specific to right triangles in Euclidean space, which have nonnegative real lengths. It is properly generalized by considering inner product spaces, which are vector spaces $V$ equipped with a function $\langle \cdot,\cdot \rangle:V\to \mathbb R$ (or $\mathbb C$) satisfying certain axioms. This allows us to define a right triangle as a triple of points $(\vec a,\vec b,\vec c)$ such that $\langle \vec b-\vec a,\vec c-\vec a\rangle=0$, which can be seen as saying the sides $\vec b-\vec a$ and $\vec c-\vec a$ are orthogonal. It also gives us a notion of length, with the length of a vector $\vec v$ being $\sqrt{\langle \vec v,\vec v\rangle}$. The generalization then states that $$\langle \vec b-\vec a,\vec b-\vec a\rangle+\langle \vec c-\vec a,\vec c-\vec a\rangle=\langle \vec b-\vec c,\vec b-\vec c\rangle$$ assuming $\vec b-\vec c$ is the longest side, i.e. that the length of the side $\vec b-\vec a$ squared plus the length of the side $\vec c-\vec a$ square equals the length of the side $\vec b-\vec c$ squared. This is very different from the generalization you gave.
H: limit question on Lebesgue functions Let $f\in L^1(\mathbb{R})$. Compute $\lim_{\|h\|\rightarrow\infty}\int_{-\infty}^\infty \|f(x+h)+f(x)\|dx$ If $f\in C_c(\mathbb{R})$ I got the limit to be $\int_{-\infty}^\infty \|f(x)\|dx$. I am not sure if this is right. AI: Let $f$ a continuous function with compact support, say contained in $-[R,R]$. For $h\geq 2R$, the supports of $\tau_hf$ and $f$ are disjoint (they are respectively $[-R-h,R-h]$ and $[-R,R]$ hence \begin{align} \int_{\Bbb R}\|f(x+h)+f(x)\|dx&=\int_{[-R,R]}\|f(x+h)+f(x)\|+\int_{[-R-h,R-h]}\|f(x+h)+f(x)\|\\ &=\int_{[-R,R]}\|f(x)\|+\int_{[-R-h,R-h]}\|f(x+h)\|\\ &=2\int_{\Bbb R}\|f(x)\|dx. \end{align} If $f\in L^1$, let $\{f_n\}$ a sequence of continuous functions with compact support which converges to $f$ in $L^1$, for example $\lVert f-f_n\rVert_{L^1}\leq n^{—1}$. Let $L(f,h):=\int_{\Bbb R}\|f(x+h)+f(x)\|dx$. We have \begin{align} \left\|L(f,h)-L(f_n,h)\right\|&\leq \int_{\Bbb R}\|f(x+h)-f_n(x+h)+f(x)-f_n(x)\|dx\\ &\leq \int_{\Bbb R}(\|f(x+h)-f_n(x+h)\|+\|f(x)-f_n(x)\|)dx\\ &\leq 2n^{-1}, \end{align} and we deduce that $$\|L(f,h)-2\lVert f\rVert_{L^1}\|\leq 4n^{-1}+\|L(f_n,h)-2\lVert f_n\rVert_{L^1}\|.$$ We have for each integer $n$, $$\limsup_{h\to +\infty}\|L(f,h)-2\lVert f\rVert_{L^1}\|\leq 4n^{—1}.$$ This gives the wanted result.
H: Numerical methods book I'm looking for an introductory book on numerical methods. I'm beginning to learn to program (in Haskell, a functional language, if that would affect the recommendations). The reason I want such a book is to practice my programming skills by implementing easy math-related algorithms. For example, calculations of transcendental or special functions (perhaps the Gamma function and Bessel functions) and interesting constants, solving differential equations (by Runge-Kutta and series solutions, for example), implementing things like Newton-Raphson approximation, primality testing (Miller-Rabin, for example) and so on. I would prefer something with a wide sampling of algorithms. I'm looking for breadth, not depth. My intention is to use this as sort of a programming exercise book. I care about having a lot of different, fun, and hopefully somewhat simple algorithms to implement. I would say I have a fairly good mathematical background, so don't shy away from recommending something that is proof based or theoretical at times. If the book includes a lot of mathematics, great! I'm always up for learning more math. I don't have formal computer science training, but I would be willing to do a small amount of reading on the basics of algorithms and data structures if I had to. If the book included this, that would be great. I should mention that I haven't taken more than an introductory course in ODEs and have no experience with PDEs other than solving the basic heat and wave equations, so I would not be able to understand or appreciate advanced material in these areas (I get the impression that solving PDEs numerically is a big, important field). However, I do know some harmonic analysis, so seeing fast Fourier transforms might be fun. AI: I really enjoyed reading Forman Acton's Numerical Methods that Usually Work. He writes with a lot of personality and presents interesting problems. I don't know that this is the best book to learn the field from because (1) it was written in 1970 and last revised in 1990 and (2) I am not an expert, so I simply don't know how to evaluate it. But it sounds like these might not be major obstacles from your perspective.
H: Which of the following metric spaces are complete? [NBHM_2006_PhD Screening test_Topology] Which of the following metric spaces are complete? $X_1=(0,1), d(x,y)=\|\tan x-\tan y\|$ $X_2=[0,1], d(x,y)=\frac{\|x-y\|}{1+\|x-y\|}$ $X_3=\mathbb{Q}, d(x,y)=1\forall x\neq y$ $X_4=\mathbb{R}, d(x,y)=\|e^x-e^y\|$ $2$ is complete as closed subset of a complete metric space is complete and the metric is also equivalent to our usual metric. $3$ is also complete as every Cauchy sequence is constant ultimately hence convergent. $4$ is not complete I am sure but not able to find out a counter example, not sure about 1.thank you for help. AI: For (1), consider the sequence $\left\langle\frac1{2^n}:n\in\Bbb N\right\rangle$. Is it $d$-Cauchy? Does it converge to anything in $X_1$? For (4), what about $\langle -n:n\in\Bbb N\rangle$?
H: How to construct a Bernstein set and what are their applications? Bernstein Set: A subset of the real line that meets every uncountable closed subset of the real line but that contains none of them. It's from wiki. My question is this: How to construct a Bernstein set? And what's its application in mathematics? Thanks ahead for any help:) AI: One uses the axiom of choice in some form to construct a Bernstein set. The most straightforward way is by transfinite recursion. There are $2^\omega$ uncountable closed subsets of $\Bbb R$, so we can list them as $\{F_\xi:\xi<2^\omega\}$, and it can be proved that each of them has cardinality $2^\omega$. Now suppose that $\eta<2^\omega$, and for each $\xi<\eta$ you’ve chosen points $x_\xi,y_\xi\in F_\xi$ so that all of these points are distinct. Let $X_\eta=\{x_\xi:\xi<\eta\}$ and $Y_\eta=\{y_\xi:\xi<\eta\}$. Then $\|X_\eta\cup Y_\eta\|<2^\omega$, so $F_\eta\setminus(X_\eta\cup Y_\eta)$ is infinite, and we can choose distinct $x_\eta,y_\eta\in F_\eta\setminus(X_\eta\cup Y_\eta)$ to continue the construction. Now let $X=\bigcup_{\xi<2^\omega}X_\xi$ and $Y=\bigcup_{\xi<2^\omega}Y_\xi$; by construction $X$ and $Y$ are disjoint sets meeting each uncountable closed subset of $\Bbb R$, so both are Bernstein sets. Added: In my experiences they are most useful as a tool for constructing (counter)examples. This post in Dan Ma’s Topology Blog is a good example of such use.
H: The inhomogenous Wave Equation I've been trying to come up with a good way to get rid of an inhomogeneity in this PDE, but I have two different solutions. I am not sure if this is a question for the Math community. If not I'll ask the physics community about this Consider this PDE which models the Wave Equation $$u_{tt} = u_{xx} + c\sin(2\pi x) $$ With these conditions $$u(0,t) = u(1,t) = u(x,0) = u_t(x,0) = 0$$ So I thought about making a substitution where $u_1 = \sum_{n\in \mathbb{N}} T(t)X(x)$ and to get rid of that sine function, I let $X(x) = \sin(n\pi x)$ and all the other $n$s to be $0$ for convenience and let $u(x,t) = v(x,t) + u_1$ where $u_1$ is my particular solution and $v(x,t)$ is my homogenous solution. After a while, I managed to solved that $v(x,t) = 0$ and there only exists $u_1$. So I thought about making another substitution. Using the theory of the heat equation where I let $$u(x,t) = w(x,t) + z(x)$$ where $z(x)$ is the steady state solution to the Wave Equation (if I can even use such a terminology) and $w(x,t)$ is the transient state (this is my "particular" solution). I reapplied the techniques as I did for the heat equation where I first set all the time derivatives to $0$ and solved for the steady state and then the transient state. Neither the transient state nor steady state were $0$ and after computing, I got my solution to be the same for both PDE So my question is, is it a coincidence they both worked? Does the Wave Equation have those steady/transient states solution? Thank you AI: Your equation is not "the Wave Equation", but it is an inhomogeneous wave equation. Inhomogeneous wave equations of the form $u_{tt} = u_{xx} + g(x)$, where the inhomogeneous term doesn't depend on $t$, do have particular solutions that don't depend on $t$, namely solutions of the ODE $u_{xx} + g(x) = 0$. If $v(x)$ is such a solution, by linearity $u(x,t)$ satisfies the (homogeneous) Wave equation if and only if $u(x,t) + v(x)$ satisfies this inhomogeneous wave equation.
H: How many points are necessary to find a parallel ellipse, and how to do it? So, I understand that to find an ellipse for sure you need at least five points. Why? The ellipse equation has only four variables ($x_0, y_0, a,\text{ and }b$). That's not actually my true question, just a curiosity. My question is, what if I know that the ellipse is parallel to the y-axis---how many points then do I need to find it, and how to do it? I have three, but I can get to four if necessary. The problem is I just can't see, on the ellipse formula, how to indicate that it is parallel, and I keep with 3 variables and 4 equations, because I can't 'insert' this information! By using 'find and ellipse' I mean finding it's center and axes length (that is, $x_0, y_0, a\text{ and }b$). By points I mean points that the ellipse contains (that is, x and y in the formula). AI: I assume that you mean major or minor axis of ellipse is parallel to y-axis. The equation of an ellipse whose major and minor axes coincide with the Cartesian axes is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ For your case you only need to shift the origin to some other point say $(x_0,y_0)$ and equation of ellipse transforms as $$\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1$$ Evidently you need four variables to solve it for $x_0,y_0,a,b$. On a side note for a general ellipse you need to rotate the co-ordinate axes and also shift the origin to some other point. suppose you rotate by angle $\theta$ anticlockwise and shift origin to $(x_0,y_0)$ then $x\mapsto((x-x_0)\cos\theta+(y-y_0)\sin\theta)$ and $y\mapsto(-(x-x_0)\sin\theta+(y-y_0)\cos\theta)$ then equation of ellipse transforms as $$\frac{((x-x_0)\cos\theta+(y-y_0)\sin\theta)^2}{a^2}+\frac{(-(x-x_0)\sin\theta+(y-y_0)\cos\theta)^2}{b^2}=1$$ Above has five variable so you would need five points.
H: Every ideal has an approximate identity? Averson's 1970 paper on extensions of $C^$-algebras seems to assume that every ideal has an approximate identity. However, I am a little bit suspicious here, since he does not assume the closeness of these ideals-at certain steps, he proves something for the ideal of all finite rank operators, which is not closed. Since closed ideals of a $C^$-algebra are themselves $C^*$-algebras, we know that closed ideals have approximate identities. However, if we leave out the condition of closeness, how can we show that an ideal still has an approximate identity? Or, in other words, when you remove the 'skin' of an closed ideal, how can you be sure that enough elements in the approximate identity remains there? Thanks! AI: You can find the proof of this theorem at page 7 in this lecture notes.
H: Two proofs I'm having difficulty with I've been given an assignment. Almost done except the last two are tripping me up. They are as follows: 1) if $2x^2-x=2y^2-y$ then $x=y$ 2) if $x^3+x=y^3+y$ then $x=y$ I imagine they use a similar tactic as they both involve powers, but I've tried factoring,completing the square, difference of squares and difference of cubes and nothing seems to help. Any hints would be appreciated. AI: Question $1$: The equation is equivalent to $2x^2-2y^2=x-y$. The left-hand side factors as $2(x+y)(x-y)$. So our equation can be rewritten as $$2(x+y)(x-y)=x-y.$$ Thus any pair $(x,y)$ such that $x\ne y$ and $2(x+y)=1$ is a counterexample to the assertion that $x$ must be equal to $y$. This was pointed out by ncmathsadist. Edit: It turns out that one is supposed to show that the only integer solutions have $x=y$. This follows from the above calculation, since $2(x+y)=1$ has no integer solutions. Question $2$: We look at the question as amended. We can factor and rewrite the equation as $(x-y)(x^2+xy+y^2)=-(x-y)$. If $x\ne y$, we can cancel $x-y$, and obtain $x^2+xy+y^2=-1$. But the equation $x^2+xy+y^2=-1$ has no real solutions, since $x^2+xy+y^2\ge 0$ for all real $x$ and $y$. One way to see this is to complete the square, getting $$x^2+xy+y^2=\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2.$$ The right-hand side is clearly never negative for real $x$ and $y$. So it is true that the only solutions of the original equation have $x=y$.
H: Trying to prove $\frac{2}{n+\frac{1}{2}} \leq \int_{1/(n+1)}^{1/n}\sqrt{1+(\sin(\frac{\pi}{t}) -\frac{\pi}{t}\cos(\frac{\pi}{t}))^2}dt$ I posted this incorrectly several hours ago and now I'm back! So this time it's correct. Im trying to show that for $n\geq 1$: $$\frac{2}{n+\frac{1}{2}} \leq \int_{1/(n+1)}^{1/n}\sqrt{1+\left(\sin\left(\frac{\pi}{t}\right) -\frac{\pi}{t}\cos\left(\frac{\pi}{t}\right)\right)^2}dt$$ I checked this numerically for several values of $n$ up through $n=500$ and the bounds are extremely tight. I've been banging my head against this integral for a while now and I really can see no way to simplify it as is or to shave off a tiny amount to make it more palatable. Hopefully someone can help me. Thanks. AI: It is not hard to show (for example, in the problem following the problem you posted, exercise 10 in Section 1-3 of Differential Geometry of Curves and Surfaces by Do Carmo) that the arc length of and arc with endpoints $x$ and $y$ is at least the length of the straight line segment connecting them. In any case, the problem only asks for a "geometrical" proof. That integral is the arc length of the curve $f(t) =(t,\sin (\pi/t))$ between the points $t=1/(n+1)$ and $t=1/n$. These points are $(1/(n+1), \sin((n+1)\pi)/(n+1))$ and $(1/n, \sin(n\pi)/n)$ (so the $y$ coordinates are $0$). Call them $A$ and $B$, respectively. The arc passes through the point $(1/(n+1/2),\sin(n\pi/2)/(n+(1/2))=(1/(n+1/2),\pm 1/(n+(1/2))$. Call this $C$. We see the arc length is at least the sum of the length of the segments $AC$ and $CB$. These each have length at least $\frac{1}{n+\frac{1}{2}}$ (draw the picture. This is the length of the perpendicular to the $x$-axis).
H: Convergent fraction for constant $e$? I've just learned about e. I am very much the novice and my problem is that while trying to calculate the convergent fractions for e. For instance: $${2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{2}{3+\cfrac{3}{4}}}}}$$ I end up with 144/53? I was wondering are there specific steps that I'm missing? For me I've been starting at the end of the continued fraction and working my way left. For instance: $\frac{3}{1} + \frac{3}{4}$ And get 15/4 and then: $\frac{2}{1} / \frac{15}{4}$ Until I finish with 144/53, which I'm not seeing this anywhere as one of the first few convergents of e. AI: You’re using a generalized continued fraction; the convergents that you normally see listed are those for the standard continued fraction expansion of $e$, i.e., the one with $1$ for each numerator: $$e=[2;1,2,1,1,4,1,1,6,1,1,8,\dots]\;.$$ This can also be written $$[1;0,1,1,2,1,1,4,1,1,6,1,1,8,\dots]$$ to emphasize the pattern even more strongly.
H: Combinatorics: how many unique albums... I want to record a set of music albums so that each one is unique. Each album has ten tracks, and I've recorded 4 versions of each track. How many unique albums can I compile so that no two albums has the exact same set of tracks while still having only one version of track 1 one of track 2 etc...my initial assumption was $4\times10!$ which is $$815915283247897734345611269596115894272000000000$$ but this seems a bit too much (although certainly enough for double platinum;) AI: Each album must have 10 different songs. For each song, you can choose one of the four versions. A choice out of 4, done 10 times, gives you $4^{10}$ options.
H: Overlaying Latin squares of order 4 Here are two latin squares overlayed upon each other to make one latin square, if you will. One "sub-latin" square is labeled with $1,2,3,4$ while the other is represented with $a,b,c,d$. There must be an $a$ corresponding to each of $1,2,3,4$ as you will see(and so on) and there must be $1$ corresponding to each $a,b,c,d$ as well and so on. (1,a) (2,d) (3,b) (4,c) (2,c) (1,b) (4,d) (3,a) (4,b) (3,c) (2,a) (1,d) (3,d) (4,a) (1,c) (2,b) Can we find three latin squares of order four such that any two of them will overlie on each other perfectly? How many different such Latin squares can we have? EDIT: Is there a method to creating mutually orthogonal squares to any extent? AI: You’re asking about mutually orthogonal Latin squares. The answer to the specific question is that yes, it is possible to find three mutually orthogonal Latin squares of order $4$, and that is the maximum. You’ll find a little more information and a few references at OEIS A001438.
H: A question of a Buzyakova's 2005 paper I came across a question of a Buzyakova's 2005 paper. This is a paragraphy of his paper, which is the last example in his paper. (I know it is a little complex to ask the complete question without the paper. ) I want to know in the last line, why $N-\cup_{\beta <\alpha}N_\beta$ is not empty. Thanks for any help and everyone who comes across this question, please vote me:) AI: Let $N$ be an infinite subset of $Q$ that is closed in $\bigcup_{\beta<\alpha}X_\beta$; then every point of $\left(\bigcup_{\beta<\alpha}X_\beta\right)\setminus N$ has a nbhd disjoint from $N$. In particular, for each $\beta<\alpha$ there is a $B_\beta\in\mathcal{B}_{x_\beta}$ such that $N\cap B_\beta=\varnothing$. If you go on to read the construction of the local base $\mathcal{B}_{x_\alpha}$, you’ll see that each $\mathcal{B}_{x_\beta}$ is countable and nested, so that if $N\cap B_\beta=\varnothing$, then $N$ intersects at most finitely many members of $\mathcal{B}_{x_\beta}$. By definition, therefore, $N\notin\mathcal{N}_\beta$ for any $\beta<\alpha$, i.e., $N\in\mathcal{N}\setminus\bigcup_{\beta<\alpha}\mathcal{N}_\beta$. (By the way, Raushan Buzyakova is a woman.)
H: Power series identity Possible Duplicate: Summation of $\sum\limits_{n=1}^{\infty} \frac{x(x+1) \cdots (x+n-1)}{y(y+1) \cdots (y+n-1)}$ Through a numerical computation, I stumbled across the following identity. It takes place in the ring $(\mathbb{Z}[x])[[t]]$, which is complete with respect to the $t$-adic valuation. The apparent identity is $$\sum_{n=0}^\infty\frac{x(x+1)(x+2)\cdots(x+n-1)}{(1+t)(1+2t)\cdots(1+nt)}t^n=\frac1{1-xt}$$ I have numerically verified that this holds $\bmod t^{50}$. Does anyone have any ideas about how to prove this identity? AI: Your identity is a special case of Gauss's hypergeometric identity (see here for a proof), $$\begin{align} \sum_{n=0}^\infty\frac{\prod_{j=0}^{n-1} (x+j)}{\prod_{j=0}^{n-1} (1+(j+1)t)}t^n&=\sum_{n=0}^\infty\frac{\prod_{j=0}^{n-1} (x+j)}{\prod_{j=0}^{n-1} \left(1+\frac1{t}+j\right)}\\ &=\sum_{n=0}^\infty \frac{(x)_n}{\left(1+\frac1{t}\right)_n}=\sum_{n=0}^\infty \frac{(x)_n (1)_n}{\left(1+\frac1{t}\right)_n}\frac1{n!}\\ &={}_2 F_1\left({{1,x}\atop{1+\frac1{t}}}\mid 1\right)=\frac{\Gamma\left(1+\frac1{t}\right)\Gamma\left(\frac1{t}-x\right)}{\Gamma\left(\frac1{t}\right)\Gamma \left(1+\frac1{t}-x\right)} \end{align}$$ where ${}_2 F_1\left({{a,b}\atop{c}}\mid z\right)$ is the Gaussian hypergeometric function, and since $\Gamma(1+z)=z\Gamma(z)$, $$\require{cancel} {}_2 F_1\left({{1,x}\atop{1+\frac1{t}}}\mid 1\right)=\frac{\cancel{\Gamma\left(\frac1{t}\right)}\cancel{\Gamma\left(\frac1{t}-x\right)}}{t\left(\frac1{t}-x\right)\cancel{\Gamma\left(\frac1{t}\right)}\cancel{\Gamma \left(\frac1{t}-x\right)}}=\frac1{t\left(\frac1{t}-x\right)}=\frac1{1-xt}$$
H: A misbehaved Power Series (contradiction with ln(2)) Possible Duplicate: Explain why calculating this series could cause paradox? Using the power series expansion $\ln(x+1)=\sum _{k=1}^\infty \frac{(-1)^{k+1}}{k}x^k$ we have $\ln(2)=(1-\frac{1}{2})+(\frac{1}{3}-\frac{1}{4})+\cdots+(\frac{1}{2k-1}-\frac{1}{2k})+\cdots=\sum _{k=1}^\infty (\frac{1}{2k-1}-\frac{1}{2k})$. Let's rearrange the terms of this sequence so that two negative terms follow each positive term: $\ln(2)=(1-\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{6}-\frac{1}{8})+\cdots+(\frac{1}{2k-1}-\frac{1}{4k-2}-\frac{1}{4k})+\cdots=\sum _{k=1}^\infty (\frac{1}{2k-1}-\frac{1}{4k-2}-\frac{1}{4k})$. However, $\frac{1}{2k-1}-\frac{1}{4k-2}-\frac{1}{4k}=\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k})$; therefore, $\sum _{k=1}^\infty (\frac{1}{2k-1}-\frac{1}{4k-2}-\frac{1}{4k})=\frac{1}{2}\sum _{k=1}^\infty (\frac{1}{2k-1}-\frac{1}{2k})$, but how can the latter inequality hold if $\sum _{k=1}^\infty (\frac{1}{2k-1}-\frac{1}{2k})$ and $\sum _{k=1}^\infty (\frac{1}{2k-1}-\frac{1}{4k-2}-\frac{1}{4k})$ are equal to $\ln(2)$? Feel free to come up with a better title... AI: Nice argument! You have produced a special case of the Riemann Rearrangement Theorem. If we have a conditionally convergent series, the terms can be rearranged so that the sum is anything we like, or so that the sum does not exist. You have rearranged a conditionally convergent series for $\log 2$ so that it has sum $\frac{1}{2}\log 2$. Remark: Results that are clear for finite sums do not necessarily extend to infinite "sums." And continuous functions do not necessarily behave like the smooth curves of our imagination. That is part of what makes analysis necessary.
H: $(a+b+c)^p-(a^p+b^p+c^p)$ is always divisible by...? $(a + b + c)^p - (a^p + b^p + c^p)$ is always divisible by (a) $p - 1\quad$ (b) $a + b + c\quad$ ( c ) $p\quad$ ( d ) $p^2 - 1$ $p$ is prime I am able to solve this by substituting values and by euler theorem by assuming $( a + b + c )$ are co prime with $p$. But I am unable to solve it by expansion nothing is working AI: $(a+b+c)^p-(a^p+b^p+c^p)=(a+b+c)^p-(a+b+c)-(a^p-a)-(b^p-b)-(c^p-c)$ Since $x^p-x$ is 0 mod $p$, the above is always divisible by $p$ whether or not $a+b+c$ and $p$ are co-prime! So option (c) is true (b) and (d) are ruled out by setting $a=1,b=c=2,p=3$
H: Description of a matrix in first-order logic Assume a 9 by 9 matrix with variable elements that are natural numbers ranging from 1 to 9 (like a Sudoku puzzle). I want to describe the entire matrix in first-order logic, but I'm having trouble thinking of a way to do so that isn't too horribly verbose. Anyone have ideas? In general, I guess I don't really understand well how to quantify objects in first-order logic... AI: To do this you need some way to describe the relations between the numbers in the matrix. A standard one is using set theory, with the membership relation $\in$ as our only primitive relation. Using that, we first define some convenience notation: \begin{align} x = \left\{ a, b \right\} :=& \forall{y} (y \in x \leftrightarrow y = a \vee y = b) \\\\ x \in \left\{ a, b \right\} :=& (x = a \vee x = b). \end{align} This lets us define the notation for singletons as $\left\{ a \right\} := \left\{ a, a \right\}$. Now we can define an ordered pair using the usual Kuratowski definition: $$\langle a, b \rangle := \left\{ \left\{ a \right\}, \left\{ a, b \right\} \right\}.$$ From ordered pairs we move to finite sequences $\langle a_0, a_1, \dotsc, a_n \rangle$ of length $n$. These we can define by recursion using ordered pairs. Let a sequence of $n$ elements be defined as follows \begin{align} \langle a_1, a_2, \dotsc, a_n \rangle :=& \langle \langle a_1, a_2, \dotsc, a_{n - 1} \rangle, a_n \rangle \end{align} Then we can define an $n$ by $m$ matrix as a sequence of length $m$ of sequences of length $n$ of numbers. To get back a representation of a particular matrix in the language of set theory, $\mathcal{L}_\in$, just start with a matrix and replace recursively with the right-hand side of these definitions. Eventually you'll get a very long formula $\varphi$ in $\mathcal{L}_\in$. There is a reason we don't do this very often!
H: Partial Latin squares of even order Can we show that $P$ is a partial latin square that is $n \times n$ where $n$ is even, where the upper quadrant $\frac{n}{2} \times \frac{n}{2}$ is filled and the rest is blank, then $P$ can be completed to a Latin square? AI: Yes it can always be completed. This is a consequence of an important result in design theory/combinatorics, proved by Ryser in: Ryser, H. J., A combinatorial theorem with an application to latin rectangles. Proc. Amer. Math. Soc. 2, (1951). 550–552. More generally: an $r \times s$ matrix containing symbols from $\{1,2,\ldots,n\}$ can be completed to an $n \times n$ Latin square if and only if the number of copies of the symbol $i$ is at least $r+s-n$, for all $i \in \{1,2,\ldots,n\}$. In the specific case of $r=s=n/2$, Ryser's condition is trivially satisfied.
H: Is the empty set partially ordered ? Also, is it totally ordered? I am not sure on how to go about this. Please provide clear explanations. AI: The question as phrased isn’t really meaningful, since you didn’t specify a relation on the empty set. However, there is only one, so I’ll assume that it’s the one that you meant. A relation on a set $A$ is a subset of $A\times A$. Since $\varnothing\times\varnothing=\varnothing$, the only subset of $\varnothing\times\varnothing$ is $\varnothing$. Thus, the question can be interpreted as: Is $\varnothing$ a partial order on $\varnothing$? Is it a total order? Let’s recall the definitions: A relation $R$ on a set $A$ is a partial order if it is reflexive, antisymmetric, and transitive. Reflexive: For each $a\in A$, $a\,R\,a$. Antisymmetric: For all $a,b\in A$, if $a\,R\,b$ and $b\,R\,a$, then $a=b$. Transitive: For all $a,b,c\in A$, if $a\,R\,b$ and $b\,R\,c$, then $a\,R\,c$. If in addition it’s true that for all $a,b\in A$, $a\,R\,b$ or $b\,R\,a$, then $R$ is a total order on $A$. The thing to notice here is that each of these is a universally quantified statement: something must be true of every element, pair of elements, or trio of elements of $A$. Thus, in order to show that $R$ is not reflexive, you must find an $a\in A$ such that $a\,\not R\,a$; in order to show that $R$ is not antisymmetic, you must find elements $a,b\in A$ such that $a\,R\,b$ and $b\,R\,a$, but $a\ne b$; in order to show that $R$ is not transitive, you must find elements $a,b,c\in A$ such that $a\,R\,b,b\,R\,c$, and $a\,\not R\,c$; and in order to show that $R$ is not total, you must find elements $a,b\in A$ such that $a\,\not R\,b$ and $b\,\not R\,a$. The crucial point is that in each case you must find elements of $A$ that actually have certain properties. If such elements don’t exist, then $R$ is reflexive (or antisymmetric, transitive, or total). In your question $A$ is $\varnothing$, the empty set. Is it possible to find elements of $\varnothing$ that have certain properties? Is it possible to find elements of $\varnothing$ at all?
H: Write down the sum of sum of sum of digits of $4444^{4444}$ Let $A = 4444^{4444}$; Then sum of digits of $A = B$; Then sum of digits of $B = C$; Then sum of digits of $C = D$; Find $D$. What should be the approach here? AI: The approach is to use the fact that $4444 \equiv 7 \pmod 9,$ so that $4444^3 \equiv 1 \pmod 9,$ and then get $4444^{4444} \equiv 7 \pmod 9$. Then use the fact that for any integer $N$, the sum of the digits of N is equivalent to $N \pmod 9$. Finally use logs to base 10 to get a limit on the size of $A$, hence $B$ etc. The answer is 7, if I remember correctly.
H: Continuous extensions of continuous functions on dense subspaces I thought that if I have a function $f: \mathbb Q \to \mathbb R$ that is continuous then I can (uniquely) extend it to a continuous function $F: \mathbb R \to \mathbb R$ as follows: for $r \in \mathbb R \setminus \mathbb Q$ pick a sequence $q_n$ converging to $r$ and then define $F(r) = \lim_{n \to \infty} f(q_n)$. So I thought there must be a theorem saying that given a continuous function $f: D \to Y$ where $D$ is a dense subset of a metric space $X$ one can uniquely extend it to $F: X \to Y$. Instead I found a theorem stating this but with the additional requirement that $f$ has to be uniformly continuous. Now I'm confused: is my example above wrong? Where does uniform continuity come in here? Thanks. AI: Uniform continuity ensures that the Cauchy sequence $(q_n)$ in $\mathbb Q$ is mapped to a Cauchy (and hence convergent) sequence $\bigl(f(q_n)\bigr)$ in $\mathbb R$. If $f$ is just continuous, $\bigl(f(q_n)\bigr)$ needn't converge (remember: $f$ is just continuous on $\mathbb Q$, so you can't argue, that convergent sequences $q_n \to r \not\in\mathbb Q$ are mapped onto convergent sequences by $f$). For example $x\mapsto \frac 1{x - \sqrt 2}$ is continuous (not uniformly!) on $\mathbb Q$ and can't be extendend. In fact, you only need the property, that Cauchy sequences are mapped to Cauchy sequences, which is a little weaker than uniform continuity, and a little stronger than continuity (for some properties of such functions, see here for start).
H: Proving a theorem from topology Theorem: Suppose $Y \subset X$. A subset $E \subset Y$ is open relative to $Y$ if and only if $E = Y \cap G$ for some open subset G of X. I don't understand what's happening neither can I follow the proof. AI: For metric spaces you can argue as follows: First we note, that balls in $Y$ (which I will denote by $B^Y_\epsilon(y) = \{z \in Y \mid d(z,y) < \epsilon\}$ are intersections of balls in $X$ with $Y$, more concretely $B^X_\epsilon(y) \cap Y = B^Y_\epsilon(y)$: To prove this, first let $z \in B^Y_\epsilon(y)$, then $z \in Y$ and $d(z,y) < \epsilon$, hence $z \in B^X_\epsilon(y) \cap Y$. The other way round: If $z \in Y \cap B^X_\epsilon(y)$, then $z \in Y$ and $d(y,z) < \epsilon$, that is $z \in B^Y_\epsilon(y)$ by defition of a ball in $Y$. (If you want to draw a picture to see what is happening, you should imagine $y$ lying on the "boundary" of $Y$). No to the proof of your theorem: First suppose that $E \subseteq Y$ is relatively open, that is for each $y \in E$ there is an $\epsilon_y > 0$ such that $B_{\epsilon_y}^Y(y) \subseteq E$. We want an open set in $X$ and only know something about balls, so we let $G = \bigcup_{y\in E} B^X_{\epsilon_y}(y)$. As a union of open balls in $X$, $G$ is clearly open in $X$ and (by our above lemma) \[ G \cap Y = \bigcup_{y \in E} B^X_{\epsilon_y}(y) \cap Y = \bigcup_{y\in E} B^Y_{\epsilon_y}(y) = E. \] Otherwise if $E = G \cap Y$ for some open $G \subseteq X$, given $y \in E$ we have an $\epsilon > 0$ with $B_\epsilon^X(y) \subseteq G$, hence $B^Y_\epsilon(y) = Y \cap B^X_\epsilon(y) \subseteq G \cap Y = E$. So $E$ is relatively open.
H: Solving an exponential distribution In a simulation, I am trying to find the value of $d_i$ where: $\displaystyle d_i \sim \frac{\epsilon_i}{\lambda_i}$ where $\epsilon_i$ is i.i.d. exponentially distributed with parameter = 1 and $i=1...n$. Conditional on $\lambda_i$ the $d_i$ have an exponential distribution of $\lambda_i$. I know the value of $\lambda_i$ but I don't know how to find the value of $d_i$. What are the steps should I undertake to find $d_i$? How relevant is the $\epsilon_i$? AI: All you need to do is simulate $\epsilon_i$ and then divide by $\lambda_i$. So for example if $U_i$ is uniformly distributed on $[0,1)$ then you can take $\epsilon_i = -\log_e (1-U_i)$ and $d_i = \dfrac{-\log_e (1-U_i)}{\lambda_i}$.
H: Roots equal constant \[ \sqrt{a-b} + \sqrt{b-c} + \sqrt{c-d} + \sqrt{d-a} = K \] for some real constant $K$ and some real numbers $a$, $b$, $c$ and $d$. Find $K$. Again i apologize for the syntax and appreciate anyhelp thanks! AI: Here $a, b,c,d,K$ are real numbers. $\sqrt{a-b} , \sqrt{b-c} , \sqrt{c-d} , \sqrt{d-a}$ are real numbers because K is real number. So,$\sqrt{a-b}$ real numbers means $ a \geq b$ and rest of also same as thing as. That is, $ a \geq b, b \geq c, c \geq d, d \geq a$ This implies $a=b=c=d$, The value of K should be zero.
H: Determine if $(\mathbb N, \Sigma)$ is a poset, look for $\min(\mathbb N, \Sigma)$ and $\max(\mathbb N, \Sigma)$ Let $D(x) = \{y \in \mathbb N : y\text{ is a divisor of } x\}$ and let the relation $\Sigma$ be defined as follows: $$\begin{aligned}x \Sigma y \Leftrightarrow D(x) \subseteq D(y) \end{aligned}$$ check $\Sigma$ is a partial order, if it is total and determine $\min(\mathbb N, \Sigma)$ and $\max(\mathbb N, \Sigma)$. In order to prove $\Sigma$ to be a partial order I need to gain proof of its reflexivity, anti-simmetry and transitivity, so: Reflexivity $x \Sigma x \Leftrightarrow D(x) \subseteq D(x)$ $\{x\in \mathbb N : x \text{ is a divisor of } x\} \subseteq \{x\in \mathbb N : x \text{ is a divisor of } x\}$ Anti-symmetry $x \Sigma y \land y\Sigma x \Rightarrow x= y$ $\{y \in \mathbb N : y \text{ is a divisor of } x\} \land \{x \in \mathbb N : x \text{ is a divisor of } y\}$ for the fundamental theorem of arithmetic $x = y$ Transitivity $x \Sigma y \land y\Sigma z \Rightarrow x \Sigma z$ $\{y\in \mathbb N : y \text{ is a divisor of } x\} \land \{z \in \mathbb N : z \text{ is a divisor of } y\} \Rightarrow \{z \in \mathbb N : z \text{ is a divisor of } x\}$. I should prove if $\Sigma$ is a total order, but I have no idea how I could do that, can it involve the fact that element $0 \in \mathbb N$ doesn't belong to this relation because $x\Sigma 0$ would mean $\{0 \text{ is a divisor of } x\}$ which is not possible? And my gut feeling tells me that $\max(\mathbb N, \Sigma) = 1$ and $\nexists\min(\mathbb N, \Sigma)$ but I don't know if I am right. Will you please give me a nudge in the right direction? AI: For $\Sigma$ to be a total order every two elements need to be comparable. However $D(2)=\{1,2\}$ and $D(3)=\{1,3\}$ so neither is a subset of the other. Therefore $\lnot(2\Sigma3\lor 3\Sigma2)$ and so this is not a total order. For $\min$ and $\max$, as you noted every number is a divisor of $0$; while no number except $1$ is a divisor of $1$, so $D(1)=\{1\}$ and $D(0)=\mathbb N$. It remains to show that $1\Sigma x$ for all $x$, but $1$ divides every number so it holds.
H: Reccurence relation: Lucas sequence I need to solve the given recurrence relation:$$L_n = L_{n-1} + L_{n-2},$$ $n\geq3$ and $ L_1 = 1, L_2 =3$ I'm confused as to what $n\geq3$ is doing there, since $L_1$ and $L_2$ are given I got $t = \frac{1\pm\sqrt 5}{2}$ Which got me the general solution, $ L_n = a $(golden ratio)$^n$ + $b$(silver ratio)$^n$ But when I try to plug in numbers and solve from there, the answers don't come out evenly...I feel I'm doing something wrong. Any help would be appreciated! AI: $L_n=a(\frac{1+\sqrt 5}{2})^n+b(\frac{1-\sqrt 5}{2})^n$. Putting $n=1$ gives, $$ L_1=\frac{a+b}{2}+\frac{(a-b)\sqrt 5}{2}=1$$ and putting $n=2$ gives $$L_2=3\frac{(a+b)}{2}+\frac{(a-b)\sqrt 5}{2}=3$$ $$\implies a+b=2,a=b\implies a=b=1$$ which gives $$L_n=(\frac{1+\sqrt 5}{2})^n+(\frac{1-\sqrt 5}{2})^n$$ Condition $n\geq 3$ is there as the definition of the sequence(that this recurrence is defined for $n\geq 3$).
H: Prove by mathematical induction that $2n ≤ 2^n$, for all integer $n≥1$? I need to prove $2n \leq 2^n$, for all integer $n≥1$ by mathematical induction? This is how I prove this: Prove:$2n ≤ 2^n$, for all integer $n≥1$ Proof: $2+4+6+...+2n=2^n$ $i.)$ Let $P(n)=1 P(1): 2(1)=2^1\implies 2=2$. Hence, $P(1)$ is true. $ii.)$ Assume that $P(n)$ is true for $n=k$, i.e, $2+4+6+...+2k=2^k$, and prove that $P(n)$ is also true for $n=k+1$, i.e, $2+4+6...+2(k+1)=2^{(k+1)}$ from the assumption add $2(k+1)$ on both sides so we have $2+4+6...2k+2(k+1)=2^k+2(k+1)$ I'm confused with $2^k+2(k+1)$, I don't know how to make $2^k$ be equivalent to $2^{k+1}$. I feel i'm doing something wrong. Any help would be appreciated! AI: Begin with the basis case: $$P(1): 2(1)\leq2^{1}\implies P(1) \text{ is true}$$ Next let's look at the inductive step: $$P(n)\implies P(n+1): 2(n+1)=2n+2, 2^{n+1}=2\cdot2^{n}$$ Through our inductive hypothesis we assume that $2n\leq2^{n}$, and as $\forall n\in\mathbb{N}$, $2^{n}>1$, doubling this must be greater than or equal to adding 2 to the other side. Therefore we can see that if $P(n)$ is true, then it implies $P(n+1)$ must also be true. To conclude, as we have shown that $P(1)$ is true, and that if $P(n)$ is true, then $P(n+1)$ is also true. Therefore, $P(n)$ must hold for all $n\in\mathbb{N}$. Q.E.D.
H: Solve $k_1 a = k_2 b + c$ Find all $k_1, k_2$ that satisfy $k_1 a = k_2 b + c$ where everything are integers. It feels like there should be some easy way to describe this in terms of congruence and gcd. AI: Let $d=\gcd(a,b)$. If $d$ does not divide $c$, there is no solution. So assume from now on that $d$ divides $c$. Suppose that we have found one particular solution $(x_0,y_0)$ of the equation $ax=by+c$. Then all solutions $(x,y)$ are given by $$x=x_0 +\frac{b}{d}t, \qquad y=y_0+\frac{a}{d}t,\tag{$1$}$$ where $t$ ranges over the integers, positive, negative, and $0$. So now look for a particular solution $(x_0,y_0)$. In "small" cases, a particular solution can be found by experimentation. In other cases, use the Extended Euclidean Algorithm to find integers $s$ and $t$ such that $as=bt+d$. Then a particular solution $(x_0,y_0)$ of our original equation is given by $$x_0=\frac{c}{d}s,\qquad y_0=\frac{c}{d}t.$$ Now using $(1)$ we can generate all the solutions.
H: Prove the statement : $\log(k + 1) -\log k>\frac{ 3}{10k}$ Prove the statement : $\log(k + 1) - \log k > \frac{3}{10k}$ Approach : $$\log(k+1)-\log{k} > \frac{3}{10k}$$ Clearly, $k\in\mathbb{Z}^{+}$ $$\log(k+1)-\log{k}=\log\bigg(1+\frac{1}{k}\bigg)$$ given base is $10$, so $$\log\left(1+\frac{1}{k}\right) > \log\left(\frac{1}{k}\right) \implies \log\left(1+\frac{1}{k}\right) > \frac{1}{k}$$ Since, $0.3 < 1$ $$ \log\left(1+\frac{1}{k}\right) > \frac{3}{10k}$$ QED AI: Solution №1. Consider function $f(x)=\log(x)$ and fix $k\in\mathbb{Z}_+$. By mean value theorem there exist $c\in[k,k+1]$ such that $$ \log(k+1)-\log(k)=(\log x)'\|_{x=c}((k+1)-k)=\frac{1}{c} $$ Since $c>k+1$ then $$ \log(k+1)-\log(k)=\frac{1}{c}>\frac{1}{k+1} $$ Since $k\in\mathbb{Z}_+$, then $k+1<10/3k$ and we obtain $$ \log(k+1)-\log(k)>\frac{1}{k+1}>\frac{3}{10 k} $$ Solution №2. It is enough to show that $\log(1+x)>0.3x$ for all $x\in (0,1)$. Then you can take $x=1/k$ for each $k\in\mathbb{Z}_+$ and prove your inequality. In order to prove inequality $\log(1+x)>0.3x$ for all $x\in (0,1)$, consider function $$ f(x)=\log(1+x)-0.3x $$ You can check, that $f(0)=0$ $f'(x)=\frac{0.7-0.3x}{x+1}>0$ for $x\in (0,1)$. Hence $f$ is non-negative on $(0,1)$, which is equivalent to $$ \log(1+x)>0.3x\quad\text{ for }\quad x\in(0,1) $$ The rest is clear.
H: $f:X\rightarrow Y$ is a continuous surjection, $X$ and $Y$ are topological space $f:X\rightarrow Y$ is a continuous surjection Then If $V$ is open, does this imply $f(V)$ is open? If $F$ is closed, does this imply $f(F)$ is closed? If $A$ is an infinite subset, does this imply that $f(A)$ is so in $Y$? Thank you for help. AI: None of these is true. Let $\Bbb S$ be the Sierpiński space, i.e. $\Bbb S=\{0,1\}$ where the open sets are $\emptyset,\{1\}$ and $\{0,1\}$. Define $f:\Bbb R\to\Bbb S$ as follows: $$f(x)=\begin{cases}0;&x\leq 0\\1;&x>0\end{cases}$$ This is a continuous surjection, since $f^{-1}(\{1\})$ is open, but none of the properties holds: $f((-\infty,0))=\{0\}$ is not open $f([1,\infty))=\{1\}$ is not closed the image of $\Bbb R$ is finite.
H: How to do this summation: $\sum_{n=1}^\infty \log_{2^\frac{n}{2^n}}256$ How to do this summation? $$\sum_{n=1}^\infty \log_{2^\frac{n}{2^n}}256=?$$ All I'm getting is $8(2 + 2 + \frac83 + \frac{16}{4} + \cdots )$ which is a diverging series. AI: A single term of series is $$\log_{2^{\frac{n}{2^n}}} 256=\frac{\log_2 256}{\log_2 2^{\frac{n}{2^n}}}=\frac{2^n8}{n}$$ First condition for series to be convergent is that it's terms should converge to $0$, but here $\frac{2^n8}{n}\to \infty$ as $n\to \infty \implies $ this series is not convergent.
H: Properties of $xy^2/(x^2+y^4)$ near the origin $f:\mathbb{R}^2\rightarrow \mathbb{R}$ Defined by $$f(x,y)= \frac{xy^2}{x^2+y^4}$$ if $x\neq 0,y\in\mathbb{R}$ and $$f(x,y)=0$$ if $x=0,y\in\mathbb{R}$ Then it is continuous but not differentiable at origin differentiable at origin has all first order partial derivative at origin. does not have all first order derivatives at origin. consider the limit $(x,y)\rightarrow (0,0)$ along the curve $y=m\sqrt{x}$ we get lim$$(x,y)\rightarrow(0,0)\frac{x^2m^2}{x^2+m^4x^2}=\frac{m^2}{1+m^4}$$ which is different for different values for $m$ hence $f$ is not continuous at origin, so 1 is false, and 2 is clearly false. I have checked that $f_x$ and $f_y$ exists at $(0,0)$ so only $3$ is correct and all others are false. could any one confirm me am I right? Thank you. AI: $\displaystyle \lim_{x\rightarrow0}\frac{f(x,0)-f(0,0)}x$is found to be 0. Similarly for $y$. So 3 is true and not 4
H: Fibonacci sequence, strings without 00, and binomial coefficient sums Refer to the sequence $S$ where $S_n$ denotes the number of n-bit strings that do not contain the pattern 00. By considering the number of n-bit strings with exactly i 0's, show that $\displaystyle f_{n+2} = \sum_{i=0}^{\frac{n+1}{2}} \binom{n+1 - i}{i}$ for $n=1,2,\dots$ where $f$ denotes the Fibonacci sequence. So I know the Fibonacci sequence consists of $0, 1, 1, 2, 3,\dots$ And $S_n=f_{n+2}$. But what else? Any help is appreciated. Thanks! AI: How many strings of length $n$ which do not contain "00" and contains precisely $i$ zeroes are there? Notice that any zero must be followed by 1 or be at the and. For the sake of simplicity, let us consider string of length $n+1$ which is obtained by adding 1 to the original string. So no we are counting the strings of length $n+1$ which are obtained from $i$ strings of the form "01" with 1's on the remaining position. If we regard the whole string "01" as one character, then there are only $n+1-i$ positions and we have to place this "combined character" on $i$ of them. So we have $\binom{n+1-i}i$ positions. BTW this can be regarded as a more detailed solution based on the hint from this answer.
H: What is The 3rd side length of Isosceles Triangle I've a isosceles triangle which length is $10\;\mathrm{cm}$ , $10\;\mathrm{cm}$ and $x$. If I want to make this triangle $120^\circ$ degree then what should be the $x$? AI: Angles opposite to equal sides are equal, so one angle is $120^0$ while others are $30^0$ each (angles opposite to equal sides can't be $120^0$,otherwise sum of angles of triangle would be greater than $180^0$). Draw perpendicular from vertex (intersection of two equal sides) to the opposite side, it divides the opposite side into two equal halves. Let one half of that be $x$,then $$\frac{x}{10}=\cos30^0\implies x=5\sqrt 3$$ Thus the side length=$2x=10\sqrt 3$
H: A continuous map from $D$ unit disk, to $S^1$ $f:D\rightarrow S^1$ is a continuous then $\exists x\in S^1$ such that $f(x)=x$? $f:S^1\rightarrow S^1$ then same as 1 holdd? $f:E\rightarrow E$ then same as 1 hold? $E=\{(x,y):2x^2+3y^2\le 1\}$ by Fixed point Theorem I know 2,3 are correct, what about 1? AI: In #1, you can think of $f: D\rightarrow D$. The fixed-point theorem says that there exists $x\in D$ so that $f(x) = x$. In #2, any rotation that is not a complete rotation has no fixed point. In #3, $E$ is homeomorphic to the unit disk. It must have a fixed point.
H: Finding a well-defined solution to a matrix equation I have the following problem: Given two 2D real positive-definite symmetric matrices $M_1$ and $M_2$, find a matrix $T$ such that $$ M_2=TM_1T^t$$ Clearly, the solution is not unique, but I don't care too much about that. All I need is a well-defined solution - something like a principal branch. To be concrete, let's say I have some $T_{input}$, and I generate $M_1$ randomly and calculate $M_2=TM_1T^t$. I want a function $T_{output}=f(M_1,M_2)$ that will give me the same result (which is not necessarily $T_{input}$) regardless of my choice of $M_1$. My thoughts: For a symmetric positive definite matrix $A$, the principal square root of the matrix is uniquely defined, ans is also symmetric. One solution of of the problem is thus $$T=\sqrt{M_2}\left(\sqrt{M_1}\right)^{-1}\ .$$ It is easily seen that this is a solution. Since $\sqrt{M_1}\sqrt{M_1}=M_1$, we have $$\left(\sqrt{M_1}\right)^{-1}M_1\left(\sqrt{M_1}\right)^{-1}=I\ ,$$ and thus $$ \begin{align} TM_1T^t&=\sqrt{M_2}\left(\sqrt{M_1}\right)^{-1}\ M_1\ \left(\sqrt{M_1}\right)^{-1} \sqrt{M_2}\\ &=\sqrt{M_2}\ I \ \sqrt{M_2}=M_2\\ \end{align} $$ But this solution depends on the choice of $M_1$. Any suggestions will be greatly appreciated. AI: Let me restate the question, because I'm not entirely sure I understand it correctly. As I understand it, you want a function that takes $M$ and $TMT^\top$ and returns a matrix $S$ that depends only on $T$ and not on $M$, such that $SMS^\top=TMT^\top$. This is impossible. Given $T$, since $SMS^\top=TMT^\top$ for all positive-definite symmetric $M$, we have $$T^{-1}SMS^\top T^\top{}^{-1}=(T^{-1}S)M(T^{-1}S)^\top=M$$ for all positive-definite symmetric $M$. In particular, for $M=I$, we have $(T^{-1}S)(T^{-1}S)^\top=I$, so $T^{-1}S$ is orthogonal. Thus we have $$(T^{-1}S)M=M(T^{-1}S)$$ for all positive-definite symmetric $M$. But the only matrices that commute with all positive-definite symmetric matrices are multiples of the identity. Thus $S=\pm T$. So your function would have to return $T$ (up to a sign), which it can't, since different $T$s can lead to the same inputs.
H: how to solve system of linear equations of XOR operation? how can i solve this set of equations ? to get values of $x,y,z,w$ ? $$\begin{aligned} 1=x \oplus y \oplus z \end{aligned}$$ $$\begin{aligned}1=x \oplus y \oplus w \end{aligned}$$ $$\begin{aligned}0=x \oplus w \oplus z \end{aligned}$$ $$\begin{aligned}1=w \oplus y \oplus z \end{aligned}$$ this is not a real example, the variables don't have to make sense, i just want to know the method. AI: As I wrote in my comment, you can just use any method you know for solving linear systems, I will use Gauss: $$ \begin{array}{cccc\|c\|\|l} \hline x & y & z & w &\ & \\ \hline\hline 1 & 1 & 1 & 0 & 1 & \\ 1 & 1 & 0 & 1 & 1 & \text{$+$ I}\\ 1 & 0 & 1 & 1 & 0 & \text{$+$ I}\\ 0 & 1 & 1 & 1 & 1 & \\ \hline 1 & 1 & 1 & 0 & 1 & \\ 0 & 0 & 1 & 1 & 0 & \text{III}\\ 0 & 1 & 0 & 1 & 1 & \text{II}\\ 0 & 1 & 1 & 1 & 1 & \text{$+$ III}\\ \hline 1 & 1 & 1 & 0 & 1 & \\ 0 & 1 & 0 & 1 & 1 & \\ 0 & 0 & 1 & 1 & 0 & \\ 0 & 0 & 1 & 0 & 0 & \text{$+$ III}\\ \hline 1 & 1 & 1 & 0 & 1 & \\ 0 & 1 & 0 & 1 & 1 & \\ 0 & 0 & 1 & 1 & 0 & \\ 0 & 0 & 0 & 1 & 0 & \\\hline \end{array} $$ Now we can conclude $w = 0$ from line 4, which gives $z = 0$ from 3 and $y = 1$ from 2, and finally $x = 0$. So $(x,y,z,w) = (0,1,0,0)$ is the only solution.
H: Proving that a linear isometry on $\mathbb{R}^{n}$ is an orthogonal matrix I wish to prove that if $T:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is defined by $T(v)=Av$ (where $A\in M_{n}(\mathbb{R})$) is an isometry then $A$ is an orthogonal matrix. I am familiar with many equivalent definition for $A\in M_{n}(\mathbb{R})$ to be orthogonal, and it doesn't matter to me which one to show. What I tried to do is the following: $\|\|x-y\|\|=\|\|Ax-Ay\|\|\implies\langle x-y,x-y\rangle=\langle Ax-Ay,Ax-Ay\rangle\implies\langle x-y,x-y\rangle=\langle x-y,A^{t}A(x-y)\rangle$, from here I thought that I will be able to deduce $A^{t}A=I$ and complete the proof, but I was unable to do so. How can I complete the proof, or prove this in another fashion ? Help is appreciated! AI: For every $x, y \in \mathbb{R}^n$ we have $$ \langle Ax,Ay\rangle=\frac{1}{4}\left[\|A(x+y)\|^2-\|A(x-y)\|^2\right]=\frac{1}{4}\left[\|x+y\|^2-\|x-y\|^2\right]=\langle x,y\rangle. $$ Hence, for every $x,y \in \mathbb{R}^n$ we have $$ \langle A^TAx,y\rangle=\langle x,y\rangle, $$ i.e. $A^TA=I_n$.
H: If $H\unlhd G$ with $(\|H\|,[G:H])=1$ then $H$ is the unique such subgroup in $G$. Here is a problem from "An introduction to the Theory of Groups" by J.J.Rotman: Let $G$ be a finite group, and let $H$ be a normal subgroup with $(\|H\|,[G:H])=1$. Prove that $H$ is the unique such subgroup in $G$. I assumed there was another normal subgroup like $H$, say $K$, such that $(\|K\|,[G:K])=1$. My aim was to show that $[K: K\cap H]=1 $ that was not held if I didn’t suppose $\|H\|=\|K\|$ . My question is if my last assumption about two subgroups is right? If it isn’t, please guide me. Thanks. AI: Alternatively, you might note that if $K$ is any other subgroup of order $\|H\|$, whether or not $K$ is normal (but assuming $H$ is normal), then $HK$ is a subgroup of $G$, so its order must divide $\|G\|$.
H: $GL_n(k)$ (General linear group over a algebraically closed field) as a affine variety? In the context of linar algebraic groups, I read in my notes from the lecture that's already some while ago that $GL_n(k)$ is an algebraic variety because $GL_n=D(\det)$, $ \det \in k [ (X_{ij})_{i,j} ]$. Now, $k$ is an algebraically closed field, $\det$ is the determinant and $k [ (X_{ij})_{i,j} ]$ are the polynomials in unkwnowns $X_{ij}$. But I cannot find what $D$ meant, maybe it's also a typo or uncommon notation -.- How can I interpret $GL_n (k)$ as a variety? (Or what does this $D$ stand for?) AI: I think $D$ becomes a more common notation when working with schemes, but here $D(f)$ for $f \in k[X_1, \ldots, X_n]$ should mean $\{a \in k^n : f(a) \neq 0\}$. So it's the open set where $f$ does not vanish. The important thing is that $D(f)$ is isomorphic to the algebraic set defined by $fX_{n + 1} - 1$ in $k^{n + 1}$ via the map $(a_1, \ldots, a_n) \mapsto (a_1, \ldots, a_n, 1/f(a_1, \ldots, a_n))$.
H: Determine monic and degree 3 polynomial in $\mathbb Z_p$ I stumbled upon this kind of problem and I really can't get the hang of it. Will anyone please outline the way to solve it? Determine for which of the first $p > 0$ values the polynomial $f = 42x^4+21x^3-x+1 \in \mathbb Z_p$ is monic and has degree 3. Then factor it as product of irreducibile polynomials in the polynomial rings found. AI: Hint $\ $ If $\rm\:f\ mod\ p\:$ is cubic then $\rm\:p\:\|\:42,\ p\nmid 21\:$ hence $\rm\:p = \ldots$ Further, mod this $\rm\:p\:$ we see $\rm\:f\:$ has no roots, so $\rm\:f\:$ is an irreducible cubic. The point of this is that it implies that if $\rm\:f\:$ factors over $\Bbb Q\,$ then it must split as a linear times a cubic. Thus to show $\rm\:f\:$ is irreducible over $\Bbb Q\,$ it suffices to show it has no root, e.g. by using the Rational Root Test. Edit $\ $ The OP later clarified that factorization over $\Bbb Q$ is not needed, so the second half of my answer is not needed. But I'll leave it since it may still prove of interest. In fact this is the way some polynomial factorization algorithms work: by deducing constraints on the degrees of possible factors from factorizations mod $\rm p$ for various primes $\rm p.$
H: I have n flavors of icecream. I choose k scoops, where k can be larger or smaller than n. How to generate all possible sequences? Say I have 4 flavors of icecream, a, b, c, and d, and I want to get 3 scoops. So basically I want to generate all possible 4-tuples where the sum of all elements in each tuple adds up to 3, like: (3, 0, 0, 0) (0, 3, 0, 0) (0, 0, 3, 0) (0, 0, 0, 3) (2, 1, 0, 0) (2, 0, 1, 0) ... and so on, so that to generate all possible 3-scoop outcomes that can exist in menu of 4 flavors. So for example the 5th 4-tuple I typed above would translate into an item that has 2 scoops of flavor a, 1 scoop of flavor b, no scoops of flavor c and no scoops of flavor d. Is there a nice way of counting these things and generating them, for all cases where flavors=scoops, flavors>scoops and scoops>flavors...? Thank you all in advance! AI: Let $x_1,x_2,x_3,x_4$ denotes the number of scoops of different flavors,then since the total number of scoops $=3$$$\implies x_1+x_2+x_3+x_4=3$$ where $x_1,x_2,x_3,x_4\geq 0$. The number of solutions of this equation $={3+4-1\choose 4-1}={6\choose 3}=20$
H: Proving that for every real $x$ there exists $y$ with $x+y^2\in\mathbb{Q}$ I'm having trouble with proving this question on my assignment: For all real numbers $x$, there exists a real number $y$ so that $x + y^2$ is rational. I'm not sure exactly how to prove or disprove this. I proved earlier that for all real numbers $x$, there exists some $y$ such that $x+y$ is rational by cases and i'm assuming this is similar but I'm having trouble starting. AI: Why isn't this trivial? For take $z$ to be any positive rational you like greater than $x$, put $y = \sqrt{(z-x)}$ ...
H: show that integral converges even if it has a singularity i am currently reading through a book on generalized functions, and there it is said that: ... $\int_{\|x\|\le r} \|x\|^{-t} dx$ converges for $t < n$ (in $n$ dimensions) and diverges for $t \ge n$. Why does it converge for $t < n$, its area/volume still goes to infinity near zero? AI: Use spherical coördinates. The $c_n$ shown below is the surface area measure of $S^{n-1}$. $$\int_{B_r(0)} \|x\|^{-t}\, dx = \int_{S^{n-1}}\int_0^r \rho^{-t} \rho^{n-1}\,d\rho\,d\sigma = c_n \int_0^r \rho^{n-1-t}d\rho$$ The last integral converges iff $n-1-t > -1$, or if $n > t$.
H: Showing $\mathbb{Q}(\sqrt[4]{2},i)=\mathbb{Q}(\sqrt[4]{2}+i)$ using the Galois orbit of $\sqrt[4]{2} + i$ The following is a problem from I. Martin Isaac's Algebra. Let $E=\mathbb{Q}(\sqrt[4]{2}+i)$. I am trying to show $\mathbb{Q}(\sqrt[4]{2},i)=E$ with the following hint: Find at least five different elements in the orbit of $i+\sqrt[4]{2}$ under $\text{Gal}(E/\mathbb{Q})$. I have solved the problem in the usual manner i.e. by showing $\mathbb{Q}(\sqrt[4]{2}+i)\subseteq \mathbb{Q}(\sqrt[4]{2},i)$ and $\mathbb{Q}(\sqrt[4]{2},i)\subseteq \mathbb{Q}(\sqrt[4]{2}+i)$ with a few calculations. My question is the following: What is the theoretical framework behind Isaacs' hint? AI: Let $K/k$ be a Galois extension with group $G$. Let $\alpha$ be an element of $K$ with minimal polynomial $f$ over $k$. Then $G$ acts transitively on the set of roots of $f$ in $K$, which all have multiplicity $1$ because our extension is separable. Hence $$ [k(\alpha) : k] = \deg f = \#(G\alpha) \quad \text{divides} \quad \#(G) = [K : k]. $$ Is clear to you how one might find five conjugates in this case, by the way?
H: Changing the order of $\lim$ and $\sup$ Suppose that $f_n:X\to [0,1]$ where $X$ is some arbitrary set. Suppose that $$ f_n(x)\geq f_{n+1}(x) $$ for all $x\in X$ and all $n = 0,1,2,\dots$ so there exists $\lim_n f_n(x)$ point-wise, let's call it $f(x)$. Define $f^_n:=\sup\limits_{x\in X}f_n(x)$, $f^:=\sup\limits_{x\in X}f(x)$ and $\hat f:= \lim\limits_n f^_n$. I wonder when $f^ = \hat f$, i.e. $$ \lim\limits_n \sup\limits_{x\in X}f_n(x) = \sup\limits_{x\in X}\lim\limits_n f_n(x). $$ I was googling the topic, but strangely have not found any information, strangely because I expected it to be available as for changing the order of limits or of integration. Some simple facts: $\hat f\geq f^$ and the reverse is true at least when $f_n$ converges uniformly to $f$. This does not hold in general, e.g. when $f_n = 1_{[n,\infty)}$. I would appreciate any other ideas that you can advise me. Also related to this. A similar question was asked here. AI: Given $\epsilon>0$, let $E_n=\{x\in X: f_n(x)\ge \hat f-\epsilon\}$ and $E^=\{x\in X: f(x)\ge \hat f-\epsilon\}$. We know that the sets $E_n$ are nonempty and nested: $E_{n+1}\subset E_n$. We would like to show that $E^$ is nonempty. Since $f_n$ decreases to $f^$ pointwise, it follows that $E^*=\bigcap E_n$. So the problem becomes: how do we show that a nested sequence of nonempty sets has nonempty intersection? I know three ways: $E_1$ has finite measure and the measures of $E_n$ are bounded from below by $c>0$. each $E_n$ is compact $X$ is complete, each $E_n$ is closed, and $\mathrm{diam}\, E_n\to 0$.
H: Question about proof of Arzelà-Ascoli (Arzelà-Ascoli, $\Longleftarrow$) Let $K$ be a compact metric space. Let $S \subset (C(K), \\|\cdot\\|_\infty)$ be closed, bounded and equicontinuous. Then $S$ is compact, that is, for a sequence $f_n$ in $S$ we can find a convergent subsequence (conv. in $\\|\cdot\\|_\infty)$. Proof: Note that a compact metric space is separable. Let $D$ be a dense, countable subset of $K$. By assumption, $S$ is bounded, hence there exists $M$, such that $\\|f\\|_\infty \leq M$ for all $f$ in $S$, in particular, for $f_n$. Now consider the space of all functions from $D$ to $[-M,M]$. By Tychonoff, $[-M,M]^D$ is compact. Define $g_n = f_n\mid_D$. Then $g_n$ is a sequence in $[-M,M]^D$, hence has a convergent subsequence $g_{n_k}$. This is what it says in my notes. Now I've been thinking about what exactly "convergent" means in the last sentence. I'm quite sure it means pointwise. But theoretically, I can endow $[-M,M]^D$ with a norm or metric (that induces the product topology). So "convergent" could mean convergent with respect to that metric, no? Or does that not make sense? AI: You even need a metric to conclude a convergent subsequence. And the metric d for the product topology is obvious for e.g. $D=\{ x_n\| n\in N\}$, i.e. \begin{align} d(f,g)=\sum_{n=1}^\infty 2^{-n}\|f(x_n)−g(x_n)\|. \end{align}
H: Birthday probability problem There are four people in a room, namely P, Q, R and S. Q's birthday is different from everyone else. What is the probability that P and R share the same birthday? I'm getting $1/364$ as answer. $(3653641364)/(365364^3) = 1/364$ AI: Two cases: If $P$ and $R$ share same birthday, the number of choices $=364$, otherwise, the number of choices $=2{364\choose 2}=364363$. Thus, the probability that $P$ and $R$ same birthday $=\frac{364}{364+364363}=\frac{364}{364^2}=\frac{1}{364}$
H: Determine $a$ values allowing $x^2+ax+2$ to be divided by $x-3$ in $\mathbb Z_5$ Determine for which $a$ values $f = x^2+ax+2$ can be divided by $g= x-3$ in $\mathbb Z_5$. I don't know if there are more effective (and certainly right) ways to solve this problem, I assume there definitely are, but as I am not aware of them, I thought I could proceed like this: I have divided $f$ by $g$, pretending $a$ to be a constant in $\mathbb Q$, the resulting quotient is $x+(a+3)$, the reminder is $2+3(a+3)$. In order to have an exact division it needs to happen: $$\begin{aligned} 2+3(a+3) = 0 \end{aligned}$$ $$\begin{aligned} 2+3a + 4 = 0 \end{aligned}$$ $$\begin{aligned} 3a + 1 = 0 \Rightarrow 3a=-1 \Rightarrow 3a = 4 \Rightarrow a= \frac{4}{3}=3 \end{aligned}$$ now I would expect $x^2+3x+2 = (x+1)(x-3)$, but it isn't the case because $(x+1)(x-3) = x^2-2x-3$. Is my way to solve this exercise totally wrong and it would be better if I'd set my notebook on fire (and in this case please feel free to jump in) or I am just doing some calculation wrong? AI: $x-a$ is the factor of polynimial $f(x)$ iff $f(a)=0\implies (x-3)$ divides $f(x)=x^2+ax+2$ iff $f(3)=11+3a\equiv 0\pmod 5\implies 3a\equiv-11\pmod 5\implies 9\pmod 5\implies a\equiv3\pmod 5$. In your solution you are getting $(x+1)(x-3)=x^2-2x-3\equiv x^2+3x+2\pmod 5$ as $-2\equiv 3\pmod 5$ and $-3\equiv 2\pmod 5$. so your solution is fine.
H: Prove that the dihedral group $D_4$ can not be written as a direct product of two groups I like to know why the dihedral group $D_4$ can't be written as a direct product of two groups. It is a school assignment that I've been trying to solve all day and now I'm more confused then ever, even thinking that the teacher might have missed writing out that he means normal subgroups. On another thread it was stated (as the answer to this question) that the direct product of two abelian groups is again abelian. If we consider the direct product of abelian subgroups $H$,$K\in G$ where $HK=G$ (for all $g \in G$, $g=hk$ $h \in H$, $k \in K$.) I can't understand why this would imply $g=kh$? It is not stated anywhere that $H$,$K$ has to be normal! But if this implication is correct I do understand why $D_4$ (that is non-abelian) can't be written as a direct product of two groups. But if it's not, as I suspect, I need some help! We know that all groups of order 4 and 2 are abelian, (since $4=p^2$), but only 4 of the subgroups of $D_4$ are normal: Therefor I can easily show that $D_4$ can't be a direct product of normal subgroups: The only normal subgroups of $D_4$ is the three subgroups of order 4, (index 2 theorem): $\{e,a^2,b,a^2b\}$, $\langle a\rangle$, $\{e,a^2,ab,a^3b\}$ and the center of $D_4=\{e,a^2\}$ We can see that these are not disjoint. So $D_4$ can't be a direct product of normal subgroups. The reason for this being that the center is non-trivial. But why can't $D_4$ be a direct product of any two groups? If we write the elements of $D_4$ as generated by $a$ and $b$, $a^4=e$, $b^2=e$, $ba=a^3b$ why isn't $D_4=\langle b\rangle\langle a\rangle$ ? I calculated the products of the elements of these two groups according to the rule given above and ended up with $D_4$, and also $\langle a\rangle$, $\langle b\rangle$ is disjoint...? Why is this wrong? Very thankful for an answer! AI: For $G$ to be the direct product of $H$ and $K$, by definition we must meet the following conditions: $H$ and $K$ are normal subgroups of $G$; $G=HK$; and $H\cap K=\{e\}$. Note that you do not take all three conditions in your first paragraph explicitly. The normality of $H$ and $K$ is implicit when you say that it is a direct product. This is what you are missing: the definition of "direct product" requires normality of the two subgroups. In general, if $H$ and $K$ are normal and $H\cap K=\{e\}$, then $hk=kh$ for all $h\in H$ and $k\in K$: indeed, consider the element $hkh^{-1}k^{-1}$. Writing it as $(hkh^{-1})k$, it is a product of two elements of $K$ (by normality of $K$), so it lies in $K$. But writing it as $h(kh^{-1}k)$, then, by normality of $H$, it is a product of two elements of $H$, so it lies in $H$. Hence, $hkh^{-1}k^{-1}$ lies in $H\cap K=\{e\}$. So $hkh^{-1}k^{-1}=e$. Multiplying by $kh$ on the right, we get $hk=kh$. In particular, if $H$ and $K$ are abelian, then $G$ is abelian: given $hk$ and $h'k'$ in $G$ (using the fact that $G=HK$), then $$(hk)(h'k') = h(kh')k' = h(h'k)k' = (hh')(kk') = (h'h)(k'k) = h'(hk')k = h'(k'h)k = (h'k')(hk)$$ so $G$ is abelian. You are correct, however, that if $G$ is a product (not a direct product, but just a product) of two abelian subgroups $H$ and $K$ (which only requires that $HK=G$), then one cannot conclude that $G$ itself is abelian. For example, consider the nonabelian group of order $27$ and exponent $3$, presented by $$\Bigl\langle a,b,c \;\Bigm\|\; a^3=b^3=c^3=e,\ ba=abc,\ ac=ca,\ ab=ba\Bigr\rangle.$$ Let $H= \langle a,c\rangle$ and $K=\langle b\rangle$. Then $H$ and $K$ are each abelian, and $G=HK$, but $G$ is not abelian. So, if you are asking whether $D_4$ can be written as a direct product of two proper subgroups, you agree that it cannot, because "direct product" necessarily requires $H$ and $K$ to be normal in $G$, and that would necessarily make $G$ abelian, which it is not. Now, a separate question is: can we write $G$ as a product, not necessarily direct, of two subgroups? This only requires $G=HK$, and it does not even require $H\cap K=\{e\}$. In this case, the answer is "yes": you can. You can even do it with $H\cap K=\{e\}$. For example, writing $$D_4 = \Bigl\langle r,s\;\Bigm\|\; r^4 = s^2 = e,\ sr=r^3s\Bigr\rangle$$ then we can take $H=\langle e, r, r^2, r^3\rangle$, and $K=\{e,s\}$. Then $HK$ has: $$\|HK\| = \frac{\|H\|\,\|K\|}{\|H\cap K\|} = \frac{4\times 2}{1} = 8$$ elements, hence $HK=D_4$. In fact, $D_4$ is a semidirect product of $C_4$ by $C_2$, which is what I exhibit above; in order for $G$ to be an (internal) semidirect product of $H$ and $K$, we require $H$ and $K$ to be subgroups such that: $H\triangleleft G$; $G=HK$; and $H\cap K=\{e\}$. In particular, every expression of $G$ as a direct product of two subgroups is also an expression as a semidirect product, but not conversely; and every semidirect product is also an expression as a product, but not conversely.
H: Value of a scaled Bessel function for negative argument Is the function $\hat{i}_0(x) = e^{-\|x\|} \sqrt{\frac{\pi}{2x}} I_{\frac{1}{2}}(x)$ positive or negative for negative $x$? $I_{\alpha}(x)$ above is a modified Bessel function. Here are my arguments. Considering that $I_{\frac{1}{2}}(x) = \sqrt{\frac{2}{\pi x}} \sinh(x)$, $\hat{i}_0(x)$ can be represented as follows: $$ \hat{i}_0(x) = e^{-\|x\|} \sqrt{\frac{\pi}{2x}} \sqrt{\frac{2}{\pi x}} \sinh(x)$$ $$ \hat{i}_0(x) = \frac{e^{-\|x\|} \sinh(x)}{(\sqrt{x})^2}$$ Using the convention that $\sqrt{x} = i\sqrt{-x}$ for negative $x$, we get $$ \hat{i}_0(x) = \frac{e^{-\|x\|} \sinh(x)}{x}$$ which is positive for negative $x$: However, using the original formula, Wolfram Alpha says that the function is negative for negative $x$: Am I missing something? AI: What you're missing is that $\sqrt{1/x}$ is not the same as $1/\sqrt{x}$ when $x$ is negative. Indeed, $\sqrt{1/(-1)} = \sqrt{-1} = i$ but $1/\sqrt{-1} = 1/i = -i$. You might try asking Wolfram Alpha for $e^{-\|x\|} \dfrac{\sqrt{\pi}}{\sqrt{2x}} I_{1/2}(x)$.
H: The trace norm cannot be increased by composing with a unitary operator $\newcommand{\tr}{\operatorname{tr}}$ I was reading a proof for the statement $\|\tr(US)\|\leq \|\tr(S)\|$, for every endomorphism $S$ on a complex vector space $H$ and every unitary operator $U$ on the same space. Though the proof is short it uses the polar decomposition and the cauchy swartz inequality. I came out with the very simple "proof": \begin{eqnarray} \left\|\tr(US)\right\| &=& \left\|\tr\left(\sum_iu_{ii}\|i\rangle\langle i\|\sum_{jk}s_{jk}\|j\rangle\langle k\|\right)\right\|\\ &=& \left\|\tr\left(\sum_i\sum_k u_{ii} s_{ik} \|i\rangle\langle k\|\right)\right\| \\ &=& \left\|\sum_{i}u_{ii}s_{ii}\right\|\\ &\leq& \left\|\sum_i s_{ii}\right\| \\ &=& \left\|\tr(S)\right\| \end{eqnarray} where I just rely in the fact that unitary operators are normal and admit the spectral decomposition theorem (that is $U=\sum_i u_{ii}\|i\rangle\langle i\|$ for some orthonormal basis $\|i\rangle$) I would like to know if I am missing something very obvious or the proof is indeed correct. AI: Your statement of the "spectral decomposition theorem" is incorrect. Unitary operators don't necessarily have point spectrum. Also the result is incorrect. Try $$ S = U = \pmatrix{1 & 0\cr 0 & -1\cr}$$
H: Memory of a neural network: is it forever? As far as I remember a neural network cannot forget anything. Does this mean that no matters how evolved the network is, it's always going to throw me back the right output if I feed it an input? And when I say "right output" I mean "precise output" without getting a bit wrong AI: neural networks only have a finite capacity to "remember" (called capacity). Assuming that you don't keep trying to update it with new information, then it would theoretically 'last forever'. But if you keep adjusting the weights to learn new and different things, eventually it will forget old patterns.
H: Two notions of uniformizer Let $X$ be a projective algebraic curve and consider a `uniformizing' map $h:X \rightarrow \mathbb{P}^1$. Is there any connection between this notion of uniformizer and a uniformizer of the maximal ideal of the local ring at a point $P \in X$? AI: There is no connection because, as far as I know, there is no such thing as a "uniformizing map $X\to \mathbb P^1$". Assume $X$ is a smooth irreducible projective curve over an algebrically closed field $k$ of arbitrary characteristic. Then a morphism $h:X\to \mathbb P^1$ (which is not the constant map with value $\infty$) is exactly the same as a rational function $f\in Rat(X)$. Each point $P\in X$ has an associated local ring $k\subsetneq \mathcal O_P\subsetneq k(T)$, which is is a discrete valuation ring by smoothness of $X$. The function $h$ is said to be a uniformizer of $X$ at $P$ if $h$ is a generator of the maximal ideal $\mathfrak m_P$ of $\mathcal O_P$. This is equivalent to saying that $h$ is non-ramified (or in alternative terminology étale) at $P$. The interpretation of "uniformizing map" in your question might be that $h$ be a uniformizer at each $P\in X$. However such a situation only occurs if $h:X \xrightarrow {\simeq} \mathbb P^1$ is an isomorphism: this result follows from Riemann-Hurwitz and is the algebraic geometer's way of saying that $\mathbb P^1_k$ is simply connected.
H: Realization of graded algebras with Poincaré duality Question: Given a finite dimensional positively graded algebra $A$ over some ring $R$ that satisfies Poincaré duality in some dimension $n$, is there necessarily a topological space $X$ such that $H^*(X;R) \cong A$? I recognise this is some sort of realization question but I don't know much algebraic topology. The case I am most interested in is when $R$ is a field. As vague motivation, I'm interested in whether, given such an $A$ over $\mathbb{Q}$, there is an elliptic Sullivan algebra $(\Lambda V, d)$ such that $H(\Lambda V, d) \cong A$. The converse appears in the textbook Rational Homotopy Theory by Felix et. al.: Theorem: If $(\Lambda V,d)$ is an elliptic Sullivan algebra (i.e. $V$ and $H(\Lambda V, d)$ are finite dimensional vector spaces) over a field of characteristic 0, then $H(\Lambda V, d)$ satisfies Poincaré duality. There is at least some Sullivan algebras $(\Lambda V, d)$ quasi-isomorphic to $A$ (since $A^0 \cong R$) but whether any of them are elliptic is the question. I may make this another post later. AI: The answer is, in general, no. For example, the following is corollary 4L.10 of Hatcher's Algebraic Topology book (freely available): If $H^\ast(X;\mathbb{Z})$ is a polynomial algebra $ \mathbb{Z}[\alpha]$, possibly truncated by $\alpha^m = 0$ for $m > 3$, then $\|\alpha\| = 2$ or $4$. Here, $\|\alpha\|$ denotes the degree of $\alpha$, meaning $\alpha \in H^{\|\alpha\|}(X;\mathbb{Z})$.
H: Fourier transform eigenvalues Since I've studied the Fourier transform extension to the Hilbert space $L^2$, I wondered if there is a complete study relative to its eigenvalues. I know that its adjoint operator is the inverse transform, which means that I can't use the theory of self-adjoint operators to state something about the eigenvalues. Could someone of you tell me something about them? Are they countable? Is there any "algorithm" to determine them? (just like the one working for compact self-adjoint operators). If you have references to books, they are welcome! Thank you! (sorry for a possibly repeated or stupid question) AI: The eigenvalues of the Fourier transform are $\pm 1$ and $\pm i$. Note that if $\cal F$ is the Fourier transform operator, $({\cal F})^4 = I$.
H: Counting marbles In how many ways can you distribute 9 marbles in 3 piles? The marbles are all identical. I know its not a very deep question but I don't know how to solve it. There is a difference between (pile a getting 4 pile b getting 4 and pile c getting 1) and (pile a getting 1 pile b getting 4 and pile c getting 4). I first thought it was 9^3 but then i realized you needed to eliminate various combinations that gave the same result. Is there a special name for the combination I'm asking for? AI: You are asking for "combinations with repetitions" (you want to choose the 3 piles, 9 times, allowing the same pile to be chosen more than once, but you don't care the order in which you select the piles); it is sometimes also called (because of a particular method of proving the formulas) stars and bars: Imagine all $9$ marbles on a line, indicated by stars: $$\star\quad\star\quad\star\quad\star\quad \star\quad\star\quad\star\quad \star\quad\star$$ Now, insert two vertical bar to indicate where you will finish the piles. For example, $$\star\quad\star\quad\star\quad\star\quad\Bigm\|\quad \star\quad\star\quad\Bigm\|\quad\star\quad \star\quad\star$$ means that you have 4 marbles in the first pile, two on the second pile, and three in the last pile; or $$\Bigm\|\quad\star\quad\star\quad\star\quad\star\quad\Bigm\|\quad\star\quad\star\quad\star\quad \star\quad\star$$ means no marbles in the first pile, four on the second pile, and five in the third pile; and $$\star\quad\star\quad\star\quad\star\quad\Bigm\|\quad\Bigm\|\quad \star\quad\star\quad\star\quad \star\quad\star$$ means four marbles in the first pile, no marbles on the second pile, and five on the third pile. So you need to choose where to put two vertical bars. How many ways are there to do so?
H: Lipschitz continuity:$f_2 - f_1$ , $f_2$ constant $L_2$ and $f_1$ constant $L_1$ Suppose you have this two Lipschitz continuous functions: $f_1$ ,with constant $L_1$ and $f_2$ with constant $L_2$. I have to prove that $f_2 - f_1$ is Lipschitz continuous with constant $L_1+L_2$. I did like this: $\|(f_2-f_1)(y)-(f_2-f_1)(x)\|=\|f_2(y)-f_2(x)-f_1(y)+f_1(x)\|=$ $=\|(f_2(y)-f_2(x))+(-f_1(y)+f_1(x))\|$ By the triangle inequality: $\leq\|f_2(y)-f_2(x)\|+\|-f_1(y)+f_1(x)\|$ $\leq L_2\|y-x\|+\|-y+x\|$ But I´m having problems in this last line,above...I should have $\|y-x\|$ in both,isn´t it?What´s wrong?how to proceed?Thanks? AI: One of the properties of a norm tells you that $\|\lambda u\|=\|\lambda\|\|u\|$ for every scalar $\lambda$ and every vector $u$. Therefore $\|-f_1(y)+f_1(x)\|=\|f_1(y)-f_1(x)\|\le L_1\|y-x\|$ .
H: Proof of the lemma used in proving that a finite-dimensional normed space is complete I'm trying to understand the proof for the lemma: $$\\|\alpha _1 e_1 + \alpha _2 e_2 + \cdots + \alpha_n e_n\\| \geq c (\|\alpha_1\|+\|\alpha_2\|+\cdots+\|\alpha_n\|)$$ where $c>0$ and the $e_i$s are finite and linearly independent. The proof I'm referring to is found in many places, like Kreyszig’s book and on page 16 of this PDF file; and in other locations. The proof is based on proving that no sequence of the form $(\alpha_1 e_1 + \alpha _2 e_2 + \cdots + \alpha _n e_n)$ could converge to a sequence with zero norm. What I don't quite get is the following: Why the proof is not simply stated as: If the lemma is not true, then $\\|\alpha _1 e_1 + \cdots + \alpha_n e_n\\| = 0$ which is not possible since the $e_i$s are independent. I mean, why do they involve the sequence convergence business here? Even when they assume that a sequence has to converge to a zero normed sequence, why does it have to be with the constraint that $\|\alpha_1^{(m)}\|+\|\alpha_2^{(m)}\|+\cdots+\|\alpha_n^{(m)}\| = 1$? Shouldn’t we consider that any possible sequence converges to zero normed sequence without the restriction that the sum should always be one? Thanks a lot! AI: 1- The lemma they are trying to prove is that there exists some $c>0$ such that, for any $\alpha_1,\ldots,\alpha_n$, we have $\\|\alpha_1e_1+\cdots+\alpha_ne_n\\|\geq c(\|\alpha_1\|+\cdots+\|\alpha_n\|)$. The key point here is that $c$ does not depend on $\alpha_1,\ldots,\alpha_n$. What you propose (showing that the norm is not $0$) only allows us to say we can pick some $c>0$ dependent on $\alpha_1,\ldots,\alpha_n$, which is a weaker statement. 2- It suffices to consider the case $\|\alpha_1\|+\cdots+\|\alpha_n\|=1$ to prove that if $\|\alpha_1\|+\cdots+\|\alpha_n\|=1$, then $\\|\alpha_1e_1+\cdots+\alpha_ne_n\\|\geq c$. The full lemma can then be proved by dividing both sides by $\|\alpha_1\|+\cdots+\|\alpha_n\|$.
H: Permutation problem scheduling games There are: 14 teams $t$ numbered from 0 to 13 13 different games $g$ numbered from 0 to 12 13 rounds $r$ numbered from 0 to 12 I want to make a planning such that: each team plays against each other team each team plays all 13 games 2 times the same game during one round is not allowed Is there an efficient way to construct such a planning? Is there a way to define a "simple" function $f(t,r)$ that returns the game $g $ played by team t in round r. I am more interested in how to approach the problem than in the actual solution. AI: Yes, the bridge players call this a Howell movement. You can see this particular example at this site or this site. Your games are the pairs of boards. Basically 13 of the pairs (your teams) move one direction around the room (the other is stationary) and the boards move the other direction around the room. It gets more complicated if the number of teams is 4n or 4n-1. You can think of seating the teams at a long table. The team at one end is stationary and the rest rotate. Each team plays the team across the table from it. The games rotate around the matches, in the opposite direction from the teams, with a pile sitting out not being played (only 7 are played each round). Note your games and rounds should be numbered 0 to 12 or 1 to 13.
H: which of the following metric spaces are separable? which of the following metric spaces are separable? $C[0,1]$ with usual 'sup norm' metric. the space $l_1$ of all absolutely convergent real sequences, with the metric $$d_1(a_i,b_i)=\sum_{1}^{\infty}\|a_i-b_i\|$$ The space $l_{\infty}$ of all bounded real sequences with the metric $$d_{\infty}(a_i,b_i)=\sup\|a_i-b_i\|$$ Well, 1 is separable as polynomials are dense in $C[0,1]$ so I can construct a set of polynomial with rational coefficients that is going to be a countable dense set for $C[0,1]$ I have no idea about 2,3 . Well, along with this question I just want to ask The closed unit ball is compact with respect to $l_1$ metric? I guess no, because Sequence $e_1=(1,0,\dots),\dots e_n=(0,0,\dots,1(nth place),0,0\dots)$ this seqquence has no convergent subsequence so not sequentially compact. Am I right? AI: HINTS: For (2), notice that if $x\in\ell_1$, then the terms of $x$ converge to $0$. Consider sequences that have rational terms that are eventually $0$. For (3), let $D=\{x_n:n\in\Bbb N\}$ be a countable subset of $\ell_\infty$. For each $n\in\Bbb N$ choose $y(n)\in[0,1]$ so that $\|y(n)-x_n(n)\|\ge\frac12$. Let $y=\langle y(n):n\in\Bbb N\rangle$; is $y$ in the closure of $D$?
H: Can any Polynomial be factored into the product of Linear expressions? Specifically I am wondering if... Given a Polynomial of n degree in one variable with coefficients from the Reals. Will every Polynomial of this form be able to be factored into a product of n linear (first degree) Polynomials, with the coefficients of these factors not being constrained to $\mathbb{R}$ but to $\mathbb{C}$ instead. By a product of linear polynomials I mean something of the form: $$(Ax+a)(Bx+b)(Cx+c)...$$ I am also interested in the general behaviour of polynomials if we also play around with the parameters of my question. Such as factoring polynomials with complex coefficients, or coefficients from any set for that matter. As well Polynomials in any (Positive Integer?) number of variables. AI: The Fundamental Theorem of Algebra states precisely that: Fundamental Theorem of Algebra. Every nonzero polynomial $p(x)$ with coefficients in $\mathbb{C}$ can be factored, in essentially a unique way, as a product of a constant and linear terms, in the form $$p(x) = a(x-r_1)\cdots(x-r_n)$$ where $a$ is the leading coefficient of $p(x)$ and $n$ is the degree of $p(x)$. This result does not hold in general. In some cases, you cannot factor them (you cannot in $\mathbb{R}$, for instance, where $x^2+1$ cannot be written as a product of linear terms; over $\mathbb{Q}$, there are polynomials of arbitrarily high degree that cannot be factored at all, let alone into a product of linear terms). In some cases, the factorization is not unique: in the quaternions, the polnomial $x^2+1$ can be factored in infinitely many essentially distinct ways as a product of linear terms, e.g., $x^2+1 = (x+i)(x-i) = (x+j)(x-j) = (x+k)(x-j) = \cdots$. A field $F$ is said to be algebraically closed if and only if every nonconstant polynomial $p(x)$ with coefficients in $F$ has at least one root. It is then easy to verify that $F$ is algebraically closed if and only if every nonzero polynomial can be factored into a product of linear terms, as above. Subject to some technical assumptions (The Axiom of Choice), every field is contained in an algebraically closed field, and for every field there is a "smallest" algebraically closed field that contains it, so there is a "smallest" larger field $K$ where you can guarantee that every polynomial factors. This does not hold for arbitrary sets of coefficients (e.g., the quaternions, because they are non-commutative; or rings with zero divisors; and others). Once you go beyond one variable, it is no longer true that a polynomial can always be factored into terms of degree one, even in $\mathbb{C}$. For example, the polynomial $xy-1$ in $\mathbb{C}[x,y]$ cannot be written as a product of polynomials of degree $1$, $(ax+by+c)(rx+sy+d)$. If you could, then $ar=bs=0$, and we cannot have $a=b=0$ or $r=s=0$, so without loss of generality we would have $b=0$ and $r=0$, so we would have $(ax+c)(sy+d) = xy-1$. Then $ad=0$, so $d=0$ (since $a\neq 0)$, but then we cannot have $cd=-1$. So no such factorization is possible.
H: Poisson random variable with mean going to infinity I was looking for some facts on the probability theory and I found this exercise on Billingsley's "Probability and Measure" book (exercise 27.3, page 379). It doesn't look like a hard exercise but I'm having a hard time trying to prove the general case. Here it goes: Exercise: Let $Y_\lambda$ be a Poisson random variable with mean $\lambda$ on a probability space $(\Omega, \Sigma,P)$. Show that $$P\left( \dfrac{Y_\lambda - \lambda}{\sqrt{\lambda}} \leq t\right) \rightarrow \dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^t e^{-\frac{x^2}{2}} dx$$ as $\lambda\rightarrow\infty$. I have a partial solution in the sense that if we consider $\lambda\in\mathbb{N}$ and make $P_i=Y_1, \ i\in\mathbb{N}$, ie, $P_i$ a Poisson random variable with mean 1. Now observe that $$\dfrac{Y_\lambda - \lambda}{\sqrt{\lambda}} = \dfrac{\sum_{i=1}^{\lambda} P_i - \lambda \cdot 1}{\sqrt{\lambda\cdot 1}} $$ The result on this special case follows using the CLT for $\{P_i=Y_1\}$. Do you guys have any idea of how to solve it for any $\lambda$? I'm pretty sure that the general solution is realated to this case but I can't see how. Thank you guys in advance!! AI: You could use characteristic functions and the Lévy continuity theorem.
H: Dense subspace of $\ell^2$ Is the set \begin{align} A=\left\{a=(a_1,a_2,\dots)\in\ell^2 \ \ \lvert \ \ \sum_{k=1}^\infty \frac{a_n}{n}=0 \right\}\subset\ell^2 \end{align} dense in $\ell^2$ Is the following argument correct? Let $x=(1,0,0,\dots)$ and assume $\forall \epsilon>0\ \ \exists a\in A$ such that $\lVert x-a\lVert_2<\epsilon $. This implies $\lvert\sum_{n\geq1}\frac{x_n-a_n}{n}\lvert\leq \frac{\pi}{\sqrt{6}}\epsilon$, i.e. $\lvert\sum_{n\geq 1}\frac{a_n}{n}\lvert\geq1-\frac{\pi}{\sqrt{6}}\epsilon$. Choosing $\epsilon$ sufficiently small therefore implies $a\not\in A$ AI: Actually, $A$ is closed. Indeed, let $L\colon \ell^2\to \Bbb C$, $La=\sum_{k=1}^{+\infty}\frac{a_k}k$. It's a linear map, and by Cauchy-Schwarz inequality it's continuous. We have $A=\ker L$. Since $A$ is strictly contained in $\ell^2$ ($(1,1,0,\dots)\in\ell^2\setminus A$), it cannot be dense.
H: Is $M(x)=O(x^σ)$ possible with $σ≤1$ even if the Riemann hypothesis is false? The wiki page on Mertens conjecture and the Connection to the Riemann hypothesis says Using the Mellin inversion theorem we now can express $M$ in terms of 1/ζ as $$ M(x) = \frac{1}{2 \pi i} \int_{\sigma-i\infty}^{\sigma+i\infty} \frac{x^s}{s \zeta(s)}\, ds $$ which is valid for $\color{blue}{1} < σ < 2$, and valid for $\color{red}{1/2} < σ < 2$ on the Riemann hypothesis. ... From this it follows that $$ M(x) = O(x^{\color{red}{1/2}+\epsilon}) $$ for all positive ε is equivalent to the Riemann hypothesis, ... The $\color{red}{\text{red}}$ color indicates the question My question changed, due to anon's comment's, to: If Riemann was false, would this imply a bound of $M(x)=O(x^{\color{blue}{1}+\epsilon})\! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \!\! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \!\! \! \! \! \! \! \! \! \! \!\! \! \! \! \! \! \! \! \! \! ---------\;\;$? $ \phantom{------------------------------------------------}$ Is $M(x)=O(x^σ)$ possible with $σ≤1$ even if RH is false? A look at Mertens function, makes me think that it should be easy to prove this. But I still don't have a clue... AI: There is a trivial bound of $\|M(x)\| \le x$ for all $x\ge 0$, because the Möbius function is bounded by $1$. So we already have $M(x) = O(x^1)$ regardless of whether RH is true or false. This turns out to be rather close to the best known unconditional bound on $M(x)$, which looks like $$M(x) = O(x \exp( -c\, \log^{0.6} x \log\log^{-0.2} x))$$ (for instance see this paper of Nathan Ng). In other words, we do not even have a proven upper bound of the form $O(x^{0.999})$. Because the Dirichlet series for $\mu(x)$ is just $1/\zeta(s)$, bounds on $M(x)$ can be obtained by Perron's formula using knowledge of the poles of $1/\zeta(s)$, in other words the zeros of $\zeta(s)$. (It is worth noting that Granville and Soundararajan have recently developed a new approach to many such problems without intimate knowledge of $\zeta(s)$ in the critical strip.) The fact that we are still rather ignorant about the zeros of $\zeta(s)$ is the reason we don't know a significantly better bound for $M(x)$ than the trivial one. At the same time, RH being false is not a very strong statement: it just means there is some zero of $\zeta(s)$ with $\Re(s) > 1/2$. While this does preclude a bound of $M(x) = O_\epsilon(x^{1/2+\epsilon})$, it does not rule out $M(x) = O(x^{3/4+\epsilon})$, if all the zeros of $\zeta(s)$ happen to lie to the left of $\Re(s) = 3/4$. One last comment: it is somewhat naive to use the observed differences as evidence of how easy it is to prove a bound. For a more striking example, try graphing the prime gaps function $d(n) = p_{n+1} - p_n$. You will find that $d(n)$ appears to be $O(\log^2 n)$ (with a pretty small constant), but even assuming RH we don't know how to prove $d(n) = O(\sqrt{n}).$
H: How to see that the shift $x \mapsto (x-c)$ is an automorphism of $R[x]$? In the process of studying irreducibility of polynomials, I encountered the criterion that $p(x)$ is irreducible if and only if $p(x-c)$ is irreducible. When trying to determine what properties of the ring were preserved under this map $x \mapsto (x-c)$, which appears sometimes to be called the shift isomorphism, I read that it was an isomorphism of the polynomial ring $R[x]$, but my attempts to prove that fact only led me through some difficult calculations, at which I generally fail. So how does one prove that the map is an isomorphism of $R[x]$? Is it an isomorphism for all rings $R$ and for any number of variables? Is this just a specific case of a more general phenomenon? AI: As Lubin noted in a comment above, the first thing to check is that the map which sends $p(x)$ to $p(x-c)$ is indeed a ring homomorphism. To do this, we have the following tool: Theorem: Let $R$ be a commutative ring. The polynomial ring $R[X]$ satisfies the following universal property: given a commutative ring $A$ containing $R$ and an element $a$ of $A$, there is a unique ring homomorphism $$\phi:R[X] \to A$$ such that $\phi(X)=a$. (Note that there exists a similar version for the polynomial ring in finitely many variables.) Hence, by this universal property there exists a unique ring homomorphism that sends $X$ to $X-c$, for any $c\in R$. And as azarel noted above, this is in fact a ring isomorphism, since you have an inverse (namely, $X\mapsto X+c$).
H: Kernel of $T$ is closed iff $T$ is continuous I know that for a Banach space $X$ and a linear functional $T:X\rightarrow\mathbb{R}$ in its dual $X'$ the following holds: \begin{align}T \text{ is continuous } \iff \text{Ker }T \text{ is closed}\end{align} which probably holds for general operators $T:X\rightarrow Y$ with finite-dimensional Banach space $Y$. I think the argument doesn't work for infinite-dimensional Banach spaces $Y$. Is the statement still correct? I.e. continuity of course still implies the closedness of the kernel for general Banach spaces $X,Y$ but is the converse still true? AI: The result is false if $Y$ is infinite dimensional. Consider $X=\ell^2$ and $Y=\ell^1$ they are not isomorphic as Banach spaces (the dual of $\ell^1$ is not separable). However they both have a Hamel basis of size continuum therefore they are isomorphic as vector spaces. The kernel of the vector space isomorphism is closed (since is the zero vector) but it can not be continuous.

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