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H: Prove a language is NP-Complete $A$ is NP-complete. $B$ is P. $A \cap B = \emptyset $ $A \cup B \neq \sum^{*}$ Prove that $A \cup B $ is NP-complete. How can I prove this ? I think if anything can be P-reducible to A then it can also be reducible to $A \cup B$, but don't know how to formally say this. Thanks. AI: Hint Just use the usual argument. Take a procedure that decides if something is in $A \cup B$ and turn this, with a polynomial-time transformation, into a procedure that decides if something is in $A$. For this you need to use another procedure that decides, in polynomial-time, if something is in $B$. Concretely Let's say you have a procedure $P_{AB}$ that decides if something is in $A \cup B$, that is $P_{AB}(\omega) = \text{true}$ if $\omega \in A \cup B$ and $P_{AB}(\omega) = \text{false}$ if $\omega \not\in A \cup B$. Similarly, let's say $P_B$ is a polynomial-time procedure that decides if something is in $B$, that is $P_B(\omega) = \text{true}$ if $\omega \in B$ and $P_B(\omega) = \text{false}$ if $\omega \not\in B$. Now we have to build a procedure $P_A(\omega)$ that decides if something is in $A$. Ofcourse, first it will call $P_{AB}(\omega)$ and $P_B(\omega)$. Now, when is $\omega \in A$ and when is $\omega \not\in A$. Recall at this point that $A \cap B = \emptyset$.
H: Dividing rational expressions How come I am able to divide the following: $$\frac{2}{2} = 1$$ Yet I am not allowed to divide the $2x$'s in the following: $$x^2 + \frac{2x}{2x} = x^2 + 1$$ Why can't I divide the $2x$ in numerator with $2x$ in denominator to get $1$? They are both equal parts, and just like in the first example should be allowed to be divided. AI: The functions $f(x)=x^2+\frac{2x}{2x}$ and $g(x)=x^2+1$ are in fact equal wherever they are both defined, but they’re not defined on the same subsets of $\Bbb R$: $g$ is defined on all of $\Bbb R$, while $f$ is not defined at $x=0$. Thus, they are not the same function, despite agreeing at every point except $0$.
H: A first order linear inhomogenous ODE Consider this ODE without any initial conditions. $$ax'(t) + bx(t) + c = 0$$ where $a,b,c$ is a nonzero constant. What method could be used to solve this ODE? I am seeking a technique that does not require differentiating both sides and turning it into a second-order ODE. AI: Hint: The equation is separable, so you can separate and integrate. You can also use integrating factor or the Laplace Transform.
H: Integrate $f(x)=x^2\ln\left(2\sqrt{\frac{a^2-x^2}{a^2+4x^2}}+\sqrt{\frac{5a^2}{a^2+4x^2}}\right)$ I am trying to find the integral of $$f(x)=x^2\ln\left(2\sqrt{\frac{a^2-x^2}{a^2+4x^2}}+\sqrt{\frac{5a^2}{a^2+4x^2}}\right)$$ And I am having no luck with it. Does anyone have any ideas? Is it even possible? I also need the integral of $$g(x)=\ln\left(2\sqrt{\frac{a^2-x^2}{a^2+4x^2}}+\sqrt{\frac{5a^2}{a^2+4x^2}}\right)$$ which I also am having trouble with. I'm not sure if these two integrals are solved in a similar way or have vastly different methods of solution. AI: Give common factor $\frac1{\sqrt{a^2+4x^2}}$ inside the logarithm, and then use the fact that $\ln ab=\ln a+\ln b$ , and that $\ln\frac1{\sqrt t}=-\frac12\ln t$ , in order to arrive at $$I=-\frac12\int[x^2]\ln(a^2+4x^2)dx+\int[x^2]\ln\left(2\sqrt{a^2-x^2}+\|a\|\sqrt5\right)dx$$ after which you will proceed to integrate by parts (at least in the case where the term $x^2$ is missing from inside the integral sign), then exploiting the derivates of inverse trigonometric functions such as arcsine and arctangent, so as to arrive at results similar to $$\int\ln(1+x^2)dx=2\arctan x+x\Big[\ln(1+x^2)-2\Big]$$ $$\int\ln\left(1+\sqrt{1-x^2}\right)dx=\arcsin x+x\left[\ln\left(1+\sqrt{1-x^2}\right)-1\right]$$ Use Wolfram Alpha to verify the result, by copy-pasting the same code you used inside your post, and then waiting patiently for a few seconds for the result to finally appear.
H: Proof that $\lim{a_n}=L$ when $n$ goes to infinity, then $\{a_n\}$ its a Cauchy Sequence Proof that $\lim{a_n}=L$ when $n$ goes to infinity, then $\{a_n\}$ its a Cauchy Sequence I start with this hypothesis $\displaystyle\lim_{n \to\infty}{a_n}=L \Leftrightarrow{\forall{\epsilon}>0}$ $\exists{N}\in{\mathbb{N}}$ such that $\forall{n}>N \left \|{a_n-L}\right \|< \epsilon$ The same its true fot any $m>N,\left \|{a_m-L}\right \|=\left \|{L-a_m}\right \|< \epsilon$ I will sum the inequalities but I havent any idea how do it... Help me please!!! AI: $$ \textbf{Solution} $$ Let $\epsilon > 0 $ be given. Suppose $ \lim a_n = L $. Therefore, we can take $N_1, N_2 $ such that $$ \|a_n - L \| < \epsilon/2 \; \; \; \text{for all} \;\;n \geq N_1 $$ $$ \|a_m - L \| < \epsilon/2 \; \; \; \text{for all} \;\;m \geq N_2 $$ We can find such $N_1$ and $N_2$ by the definition of limit. Now, Take $N = \max\{N_1, N_2 \} $ $$ \therefore \|a_n - a_m\| = \|a_n - L + L - a_m\| \leq \|a_n - L\| + \|L - a_m\| < \epsilon/2 + \epsilon/2 = \epsilon $$ For every $m, n \geq N$. Note the triangle trick have been used in the above estimate. Therefore, the sequence must be Cauchy and the problem is solved
H: Congruence question with divisibility I have this question and I have proved that a/d is congruent to b/d mod(m/d) However, I don't know how to go forward to prove a/k is congruent to b/k mod(m/d) Can anyone help me out? THX AI: We have $a\equiv b\pmod m\iff a=b+c\cdot m$ where $c$ is some integer Let $\displaystyle \frac aA=\frac bB=k\implies (A,B)=1$ $\implies k(A-B)=c\cdot m$ Let $(k,m)=D$ and $\displaystyle \frac kK=\frac mM=D\implies (K,M)=1$ $\displaystyle\implies K\cdot D(A-B)=c\cdot M\cdot D\iff K(A-B)=c\cdot M\implies A-B=\frac{c\cdot M}K$ As $(K,M)=1$ and $A-B$ is an integer, $K$ must divide $c,$ $\displaystyle\implies A\equiv B\pmod M\iff \frac ak\equiv \frac bk\pmod {\frac m{(k,m)}}$ as $\displaystyle M=\frac mD=\frac m{(k,m)}$
H: regular expression for a number It is a regular expression for a number. I have several questions about it. (0U1U2U3U4U5U6U7U8U9)* Does it means a set containing a number from 0 to 9 and then concatenate itself n times, or a set containing all of those 10 numbers and then concatenate itself n times. Is the first (0U1U2U3U4U5U6U7U8U9) necessary if (0U1U2U3U4U5U6U7U8U9)* means a set containing all of those 10 numbers and then concatenate itself n times. Is this set represent the natural number or only part of the natural number. Thank you. AI: The regular expression $$0\cup 1\cup 2\cup 3\cup 4\cup 5\cup 6\cup 7\cup 8\cup 9$$ represents any single digit, including $0$. Call this expression $\rho$. Then $\rho^$ represents any finite sequence of things matching $\rho$, which in this case means any finite string of digits. Note that finite includes empty, the string consisting of no digits at all. Thus, $\rho\rho^$, which is what you have, represents any string consisting of a single digit ($\rho$) followed by any finite (possibly empty) string of digits ($\rho^$). In other words, $\rho\rho^$ represents any non-empty finite string of digits; the purpose of the initial $\rho$ is to ensure that there is at least one digit. In words this definition of number amounts to saying that a number is any non-empty finite string of digits, where the digits are $0,1,2,3,4,5,6,7,8$, and $9$. Note that this definition allows leading zeroes: it allows $00123$ as a number, for instance. It does, however, include all natural numbers, including $0$; indeed, it includes each of them infinitely many times, since it allows any number of leading zeroes.
H: Cardinality of a vector space versus the cardinality of its basis Let $V$ an infinite dimensional vector space. How to show that the cardinality of $V$ is the same of a basis of $V$? I saw this argument here link in the main answer (MathOverFlow). AI: You need the additional assumption that $\dim_F V \ge \|F\|$ (where $F$ is the base field). Let $V$ be a vector space over a field $F$ and let $B \subseteq V$ be a basis. Suppose $B$ is infinite. There is a surjection $(F \times B)^{<\omega} \to V$ given by $$(( a_i, v_i) : i < n) \mapsto \sum_{i<n} a_iv_i$$ So we have $$\|V\| \le \|(F \times B)^{<\omega}\| = \sum_{n<\omega} \|F \times B\|^n \le \aleph_0 \cdot \|F\| \cdot \|B\| = \max \{ \aleph_0, \|F\|, \|B\| \}$$ So if $B$ is infinite and $\|B\| \ge \|F\|$ then we get $\|V\| \le \|B\| = \dim_F V$. (And obviously $\|V\| \ge \dim_F V$ so we have equality.)
H: Tensor Projection I'm currently reading "Vector and Tensor Analysis with Applications" by A.I. Borisenko and I.E. Tarapov, and I'm having trouble following a particular mathematical step in where the author projects the moment of inertia tensor onto a set of axes, K. This occurs on page 68 in section 2.4.3. Below is an excerpt: $$ L = \sum_{j=1}^{n}m_{j}[\mathbf{r}_{j} \times ( \boldsymbol{\omega} \times \mathbf{r}_{j} )] = \sum_{j=1}^{n}m_{j}[\boldsymbol{\omega}(\mathbf{r}_{j}\cdot \mathbf{r}_{j})-\mathbf{r}_{j}(\boldsymbol{\omega} \cdot \mathbf{r}_{j} )] $$ Where L is angular momentum in a system composed of n particles, where the j'th particle has mass $$ m_{j} $$ and $$ \omega $$ is the instantaneous angular velocity of the system. The they say the project L onto the axes of K to obtain: $$ L_{i} = \sum_{j=1}^{n}m_{j}(\omega_{i}x_{l}^{(j)}x_{l}^{(j)}-x_{i}^{(j)}\omega_{k}x_{k}^{(j)})\; \; \; (summation \, over \, k \, and \, l) $$ My question then pertains to the indices k and l. Why do we introduce them, What values do they span, and why? ( Assuming R^3 ) I've been stumped on this one, and can't make sense out of it enough do a calculation with it to clarify further. I feel like there's something intrinsic about projection that I'm missing, and it's making the rest of what follows nearly incomprehensible for me unfortunately. AI: They're just writing a couple dot products in index notation. There's no real meaning or content in doing so, except that some people feel that if you're already using indices elsewhere, everything should be written in terms of indices, even something as simple as a dot product. I assume you're familiar with the idea that $$a \cdot b = \sum_i a_i b_i $$ The expression given suppresses the summation symbols and just uses a different letter index than $i$.
H: If $G \cong H/K$, does it follow that $H \cong G \times K$? It is very tempting to perform this step, but I feel like it is not true. I couldn't come up with a counterexample though. AI: No: $H = \mathbb{Z}_4$, $K = \langle 2 \rangle$. More generally: $H = $ your favorite cyclic group, $K = $ your favorite non-trivial subgroup of $H$. As Dan Shved points out below, this doesn't always work - but it does give infinitely many counterexamples.
H: Finite abelian groups Find all finite abelian groups (up to isomorphism) of order 320. So I found the prime factorization to be $2^6 \times 5$. I found the 11 groups to be $\mathbb{Z}_{64} \times \mathbb{Z}_5$, $\mathbb{Z}_{32} \times \mathbb{Z}_2 \times \mathbb{Z}_5$, $\mathbb{Z}_{16} \times \mathbb{Z}_4 \times \mathbb{Z}_5$, $\mathbb{Z}_{16} \times \mathbb{Z}_2 \times\mathbb{Z}_2 \times\mathbb{Z}_5$, $\mathbb{Z}_8 \times \mathbb{Z}_8 \times\mathbb{Z}_5$, $\mathbb{Z}_8 \times \mathbb{Z}_4 \times\mathbb{Z}_2 \times\mathbb{Z}_5$, $\mathbb{Z}_8 \times \mathbb{Z}_2 \times\mathbb{Z}_2 \times\mathbb{Z}_2 \times\mathbb{Z}_5$, $\mathbb{Z}_4 \times \mathbb{Z}_4 \times\mathbb{Z}_4 \times\mathbb{Z}_5$, $\mathbb{Z}_4 \times \mathbb{Z}_4 \times\mathbb{Z}_2 \times\mathbb{Z}_2 \times\mathbb{Z}_5$, $\mathbb{Z}_4 \times \mathbb{Z}_2 \times\mathbb{Z}_2 \times\mathbb{Z}_2 \times\mathbb{Z}_2 \times \mathbb{Z}_5$, $\mathbb{Z}_2 \times \mathbb{Z}_2 \times\mathbb{Z}_2 \times\mathbb{Z}_2\times \mathbb{Z}_2 \times\mathbb{Z}_2 \times\mathbb{Z}_5$. This is all I have to show right? I listed all the groups. AI: You have found the prime factorization to be $2^6 \times 5^1$, so the number of groups would be $P(6) \times P(1)$ where $P(i)$ is the partition function of i i.e the number of ways of expressing natural number $i$ in a distinct manner. 6=6 6=5+1 6=4+2 6=4+1+1 6=3+3 6=3+2+1 6=3+1+1+1 6=2+2+2 6=2+2+1+1 6=2+1+1+1+1 6=1+1+1+1+1+1 Infact if the prime factorization of your order $n$ is $p_1^{a_1} \times p_2^{a_2} \times \dots p_k^{a_k}$ then the total number of abelian groups possible upto isomorphism of order $n$ is $P(a_1) \times P(a_2) \times \dots \times P(a_k)$.
H: If integral is zero and function is continuous and non negative, then what about the function? If $f$ is continuous on $[a,b]$, $f(x)≥0$ on $[a,b]$ and $$\int_{a}^{b} f(x) =0$$ then prove that $f(x)=0$ for all $x \in [a,b]$. I tried with Riemann's definite integral definition but couldn't proceed AI: Hint: Suppose that $f$ were not identically zero, and choose $x_0 \in (a, b)$ for which $f(x_0) > 0$. Then there exists an $\epsilon > 0$ such that $$\|x - x_0\| < \epsilon \implies f(x) > \frac{1}{2} f(x_0)$$ Try considering a partition of the interval containing the interval $[x_0 - \epsilon, x_0 + \epsilon]$, and compute a lower sum.
H: A single transferrable vote question I'm preparing for the Oxford TSA, and am using past papers as a way to practice. One of the questions, I thought I had right, but turned out incorrect. Would love it if you were to analyze my reasoning here --> Question: In the elections for the mayor of Bitton, the single transferable vote system is used. In this, each voter places a '1' beside the candidate they most want and a '2' beside their second choice. In the first round of votes, all the '1' vote of candidates are counted. If no candidate has over 50% of the votes, the bottom candidate in the poll drops out, and the '2' votes for the candidate are added to the appropriate other candidates. This is repeated until one person has over 50% of the votes. The votes for the first round of counting were as follows: MG 87 SJ 63 PG 45 IM 36 WD 18 RM 17 Total votes cast: 254 How many candidates can still win? a) 1 b) 2 c) 3 d) 4 e) 5 My reasoning may have been off. I tried subtracting the voters at the bottom, until at least the highest voter could win. This got me to subtract the bottom 3, leaving 3 more people. Hence, I said 3 more people could win. I didn't take into consideration though if other people would be able to win the the added votes. However, even if I were, my answer would be wrong, because, the correct answer is 4 people. How can 4 people have a chance of winning here? AI: Suppose that IM is the first or second choice of every voter. When RM is dropped, IM picks up $17$ votes, and the table is now: $$\begin{array}{cc} \text{MG}&\text{SJ}&\text{IM}&\text{PG}&\text{WD}\\ 87&63&53&45&18 \end{array}$$ Then WD is dropped, and IM picks up another $18$ votes: $$\begin{array}{cc} \text{MG}&\text{IM}&\text{SJ}&\text{PG}\\ 87&71&63&45 \end{array}$$ Then PG is dropped, giving IM another $45$ votes: $$\begin{array}{cc} \text{IM}&\text{MG}&\text{SJ}\\ 116&87&63 \end{array}$$ And now when SJ is dropped, IM gets $63$ more votes and wins. Clearly if IM can win, so can anyone above IM in the original table. It’s also clear that RM cannot win. Neither can WD, who would still be at the bottom of the table in round two even after receiving all of RM’s second choice votes. (By the way, the actual total is $266$, not $254$.)
H: Propositional Logic - Is my answer correct? I have a question relating to Propositional Logic. Any help will be greatly appreciated. Without changing the meaning of the following formulæ, which rely on operator precedence to be interpreted correctly, introduce brackets in each so that no precedence information is required. (a) $\lnot p \land q \implies r \land p \land q \land \lnot r \iff F$ Ans $((¬p ∧ q) ⇒ (r ∧p )∧(q ∧¬r ))⇔ F$ (b) $¬p ∧ q ∧ r ⇔ ¬p ∨ ¬q ∧¬r$ Ans $(¬p ∧ (q ∧ r)) ⇔ ¬p ∨ (¬q ∧¬r)$ Please can somebody advise me if my answer is correct or incorrect. Thank you all geniuses so much. AI: You're almost there. The missing parens are inserted & highlighted: (a) $((\lnot p \land q) \rightarrow \color{red}{\bf(}(r \land p) \land (q \land \lnot r)\color{red}{\bf)}) \leftrightarrow F$ (b) $(\lnot p \land (q \land r)) \leftrightarrow \color{red}{\bf(}\lnot p \lor (\lnot q \land \lnot r)\color{red}{\bf)}$ But since they want no precedence information assumed, we have to make the formulas uglier: (a) $((\color{red}{\bf(}\lnot p\color{red}{\bf)} \land q) \rightarrow ((r \land p) \land (q \land \color{red}{\bf(}\lnot r\color{red}{\bf)}))) \leftrightarrow F$ (b) $(\color{red}{\bf(}\lnot p\color{red}{\bf)} \land (q \land r)) \leftrightarrow (\color{red}{\bf(}\lnot p\color{red}{\bf)} \lor (\color{red}{\bf(}\lnot q\color{red}{\bf)} \land \color{red}{\bf(}\lnot r\color{red}{\bf)}))$
H: Fast algebraic expansion Is there an algebraic trick to expand the following expression without multiplying each term with another, expanding the standard way it gives $6+6+6=18$ terms and then cancelling the same terms with opposite signs to get the final result, is there any shortcut? $(b+c-a)(y+z)+(c+a-b)(z+x)+(a+b-c)(x+y)$ AI: There is a cyclic symmetry in the expression. The second term is obtained from the first by replacing $a \to b, b \to c, c \to a, x \to y, y \to z, z \to x$. The third term is obtained from the second in the same way. And if you apply the symmetry to the third term, you get the first term again. There are at most 9 terms in the expanded form: each of a,b,c multiplied by each of x,y,z. But the symmetry means that the coefficient of $ax$ will be the same as the coefficients of $by$ and $cz$, and similarly for all the other terms. Now take any particular term that occurs in the expansion of the first term, say $by$. You notice that $by$ also occurs in the third term as a positive term, but not in the second term at all. So $by$ will have a coefficient 2 in the final answer. By applying the cyclic symmetry, the terms $ax$ and $cz$ also have coefficient 2. Now take one of the other terms in the expansion of the first term, say $bz$. The second term has a $-bz$ and the third term, no $bz$. So there is no $bz$ and by symmetry no $cx$ nor $ay$. Similarly, check $bx$. It cancels out of the second and third term. Therefore, by symmetry, so does $cy$ and $az$. So final answer: $2(ax+by+cz)$.
H: Reduced nonintegral domain contains at least two minimal primes Let R be a ring such that R is reduced and R is not an integral domain. Show that R contains at least 2 minimal prime ideals. AI: This is immediate once you know that the nilradical of $R$ is the intersection of its minimal primes.
H: Minimum number of hemispheres covering a sphere Here is a question which seems easy but seems to have many pitfalls. If I give you an arbitrary covering of the sphere by $N$ closed hemispheres. You can pick any of the hemispheres to keep. What is the minimum number you can keep while still covering the sphere? We suspect the answer is $4$ but we can't seem to prove it. AI: Hint: Look up Helly's theorem. EDIT: I should have been more specific, since Helly had more than one theorem. The one I'm talking about is the one found at http://en.wikipedia.org/wiki/Helly%27s_theorem If your closed hemispheres are $H_1, \ldots, H_n$ (in the sphere $S$), let $X_i$ be the convex hull of $S \backslash H_i$. If no $4$ of your hemispheres cover the whole sphere, that says every $4$ of the $X_i$ have nonempty intersection, and then (since we're in ${\mathbb R}^3$) Helly says the intersection of all the $X_i$ is nonempty, and that implies that $H_i$ don't cover the whole sphere.
H: Relations and Functions - Is my answer correct? Could someone please advise if my answer is correct or incorrect? Any help will be greatly appreciated. Given the sets $A = \{1, 2, 3\}$, $B = \{−1, 0, 1, 2\}$ and $C = \{3, 4, 5, 6\}$, indicate the members (pairs) in the following relations. Let $\Bbb N$ denote the natural numbers. (Note that in these definitions a comma ‘,’ is used as a less intrusive way of indicating predicate conjunction ‘$\land$’.) 1) $\{(x, n) \mid x \in A, n \in\Bbb N, n = 2 \cdot x\}$ Ans: $\{(1, 2), (2, 4), (3, 6)\}$ 2) $\{(x, z) \mid x \in A, z \in C, x = z − 1\}$ Ans: $\{(2,3),(3,4)\}$ 3) $\{(x, y, z) \mid x \in A, y \in B, z \in C, z = x + y\}$ Ans: $\{(1,2,3),(2,1,3),(2,2,4),(3,0,3),(3,1,4),(3,2,5)\}$ 4) $\{(x, y, z) \mid x \in A, y \in B, z \in C, z = x \cdot y\}$ Ans: $\{(2,2,4),(3,1,3),(3,2,6)\}$ 5) $\{(x, y, z) \mid x \in A, y \in B, z \in C, z = 100xy\}$ **Ans: NULL EDIT: I have edited this post and put in the answers I believe are correct. Thank you so much. AI: Your answer to (1) is correct. Your answer to (2) is not: $4$ and $5$ are not elements of $A$, so $(4,5)$ and $(5,6)$ don’t meet the requirement that the first component of the ordered pair belong to $A$. Your answer to (3) is wrong in two ways. First, $(1,-1,0)$ and $(2,0,2)$ should not be there, since $0$ and $2$ are not in $C$: the definition requires that the third component be in $C$. Secondly, $(3,2,5)$ should be there: $3\in A$, $2\in B$, $5\in C$, and $3+2=5$, so it satisfies all of the conditions. $(2,2,4)$ should also be there. (I did not make an exhaustive search for missing triples; there might be more.) In (4) you also have both kinds of error: $(1,-1,-1)$ should not be present, since $-1\notin C$, and $(3,2,6)$ should be. There is at least one more error of each kind, but I’ll leave them for you to find. (5) is also wrong: no member of $C$ is a multiple of $100$, so there are no triples $(x,y,z)$ such that $x\in A$, $y\in B$, $z\in C$, and $z=100xy$. This is $\varnothing$, the empty set.
H: $\mathbb{E}[e^{ t \sum_{i=1}^n X_i^2 }] = \Pi_{i=1}^n \mathbb{E}[e^{tX_i^2}]$ $$\mathbb{E}[e^{ t \sum_{i=1}^n X_i^2 }] = \Pi_{i=1}^n \mathbb{E}[e^{tX_i^2}]$$ How do I get this? It seems like in here, $\mathbb{E}[y_1+y_2+...] = \mathbb{E}[y_1]\mathbb{E}[y_2]...$ AI: Note that $$e^{t\sum\limits_{i=1}^n X_i^2}=e^{t X_1^2}\cdot\ldots\cdot e^{t X_n^2}=\prod\limits_{i=1}^n e^{t X_i^2} $$ For any tuple $Y_1,\ldots,Y_n$ of independent random variables we have $$ \mathbb{E}\left[\prod\limits_{i=1}^n Y_i\right] =\mathbb{E}[Y_1\cdot\ldots\cdot Y_n] =\mathbb{E}[Y_1]\cdot\ldots\cdot\mathbb{E}[Y_n] =\prod\limits_{i=1}^n\mathbb{E}[Y_i] $$ It is remains to set $Y_i=e^{t X_i^2}$ to get $$ \mathbb{E}\left[e^{t\sum\limits_{i=1}^n X_i^2}\right]=\prod\limits_{i=1}^n \mathbb{E}[e^{t X_i^2}] $$ Since $X_i$ are identically distributed, then $\mathbb{E}[e^{t X_i^2}]=\mathbb{E}[e^{t X_1^2}]$, so we get $$ \mathbb{E}\left[e^{t\sum\limits_{i=1}^n X_i^2}\right]=\mathbb{E}[e^{t X_1^2}]^n $$
H: What is the limit of $\sum^{i=n}_{i=0} \left(-\frac{2}{3}\right)^i$? What is the limit of $$\sum^{i=n}_{i=0} \left(-\frac{2}{3}\right)^i$$ ? I am currently taking a introductory course in real analysis. Hints will be appreciated! As a soft question, can I ask what should I search for in the search bar to find my answer? I am pretty sure there are similar sort of questions asked here before, and I want to try avoiding duplicating questions as much as possible. AI: You have a geometric progression. What the sum of a^i for i going from 0 to n ? Are you able to continue with this ?
H: What is the domain of $x^x$ as a real valued function? Consider the function $f(x) = x^x$. Wolfram alpha tells me that this function's domain is $x : x>0$, $x \in \mathbb{R}$. I can't see why it cannot be defined for a number like $(-2)$. I mean $(-2)^{-2}=0.25$, the same Wolfram Alpha told me. I realize that fractional powers for negative numbers may cause problems, but it could be defined for integers. Thanks for any help. AI: For most purposes (including the hotly debated question of what to make of $0^0$) you can consider any instances of exponentiation $x^y$ (where $x,y$ can stand for expressions) to stand for on of two quite disparate definitions that happen to coincide on the intersection of their domains: If $y$ designates an integer, then $x^y$ is defined algebraically; recursively by $x^0=1$ and $x^{n+1}=xx^n$ for the case $y\geq0$, and provided $x$ is invertible by $x^{-n}=(x^{-1})^n$ for $y<0$. In other cases one must assume that $x$ is real and positive, and $x^y$ stands for $\exp(y\ln x)$ (note that I did not write $e^{y\ln x}$, which would result in a circular definition). Here $\exp$ is a perfectly defined function $\Bbb C\to\Bbb C$, so one can allow $y$ to be any complex number (or one could even amuse oneself by taking square matrices for $y$), but $x$ must be restricted to avoid ambiguity of $\ln x$. One can of course extend the definition by making choices for $\ln x$, but using a bare $x^y$ for such cases would be confusing, and also one must be aware that many properties of exponentiation will start to fail. Given this, the real function $x\to x^x$ can only be defined using the second variant, which justifies taking the domain to be the (strictly) positive reals. One could extend to domain to contain the non-positive integers as well (using the first definition), but mixing the two definitions of $x^y$ in a single usage is generally not a very fruitful idea.
H: Evaluation of $\lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \binom{2n}{n}$ Evaluate $$\lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \binom{2n}{n}.$$ $\underline{\bf{My\;\;Try}}::$ Let $\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \binom{2n}{n} = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \left(\frac{(n+1)\cdot (n+2)\cdot (n+3)\cdots (n+n)}{(1)\cdot (2)\cdot (3)\cdots (n)}\right)$ $\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \left\{\ln \left(\frac{n+1}{1}\right)+\ln \left(\frac{n+2}{2}\right)+\ln \left(\frac{n+3}{3}\right)+\cdots+\ln \left(\frac{n+n}{n}\right)\right\}$ $\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \sum_{r=1}^{n}\ln \left(\frac{n+r}{r}\right) = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \sum_{r=1}^{n}\ln \left(\frac{1+\frac{r}{n}}{\frac{r}{n}}\right)$ Now Using Reinman Sum $\displaystyle y = \frac{1}{2}\int_{0}^{1}\ln \left(\frac{x+1}{x}\right)dx = \frac{1}{2}\int_{0}^{1}\ln (x+1)dx-\frac{1}{2}\int_{0}^{1}\ln (x)dx = \ln (2)$ My Question is , Is there is any method other then that like Striling Approximation OR Stolz–Cesàro theorem OR Ratio Test If yes then please explain here Thanks AI: Notice $\displaystyle \sum_{k=0}^{2n}\binom{2n}{k} = 2^{2n}$ and $\displaystyle \binom{2n}{n} \ge \binom{2n}{k}$ for all $0 \le k \le 2n$, we have $$ \frac{2^{2n}}{2n+1}\le \binom{2n}{n} \le 2^{2n} \quad\implies\quad \log 2 - \frac{\log{(2n+1)}}{2n} \le \frac{1}{2n} \log \binom{2n}{n} \le \log 2$$ Since $\quad\displaystyle \lim_{n\to\infty} \frac{\log(2n+1)}{2n} = 0\quad$, we get $\quad\displaystyle \lim_{n\to\infty} \frac{1}{2n}\log\binom{2n}{n} = \log 2$.
H: Geometry of Complex Numbers Write down in the form ${Z}\rightarrow{AZ+B}$ the following transformations of the complex plane: (a) translation in the direction $(2,-3)$ (b) rotation about (0,1) through $\pi/4$ I know from my student answer key that the answer for part (b) is ${Z}\rightarrow{AZ-iA+i}$ where $A=\frac{(1+i)}{\sqrt2}$ can someone explain the steps for part (b)? Part (a) is straightforward. AI: If you find part (a) easy, I guess you understand the geometry of complex addition and how complex numbers can be thought of as vectors. If you can't do part (b), maybe you don't understand complex multiplication. If you have complex numbers $z$ and $z'$, then $zz'$ is $z$ with its length multiplied by the length of $z'$, and rotated by the angle $z'$ makes with the real axis. So for instance, if you have a complex number whose vector is of length $5$ and makes an angle of $15°$ with the horizontal, then multiplying $z$ by that complex number will make $z$ five times longer and rotate it $15°$ in the positive direction.
H: Hexadecimal to Octal and Vice Versa Convert Hexadecimal number to Octal - $(FD56.52A)_{16}$ to octal My answer - $(176526.2452)_8$ Convert Octal to Hexadecimal $(37.27)_8$ My answer - $(1F.5C)_{16}$. Correct or incorrect? Please help. AI: First of all, use $ {inline formula} $ and $$ {paragraph formula} $$ to typeset mathematics, to get proper subscripts. One strategy is to just convert to decimal and back, but a much better strategy is to remember that in hexadecimal each digit corresponds to four binary digits ($2^4=16$) and in octal each digit is three bits ($2^3=8$). $$ ({\rm FD56.52A})_{16} = (1111'1101'0101'0110\,.0101'0010'1010)_2 $$ Then we regroup to three digits and pad with zeros: $$ (001'111'110'101'010'110\,.010'100'101'010)_2 = (176526.2452)_8 $$ You can use this method the other around way too.
H: If a functor preserves finite limits, does it preserve subobjects? I have just started learning category theory. My question really appears on the title. In other words, can a subobject be seen as some kind of limit? AI: Let $f : X \to Y$ be a morphism. Then $f : X \to Y$ is a monomorphism if and only if the diagram below is a pullback square, $$\begin{array}{ccc} X & \rightarrow & X \\ \downarrow & & \downarrow \\ X & \rightarrow & Y \end{array}$$ where $X \to X$ is the identity. Thus, any functor that preserves pullback squares preserves monomorphisms.
H: Existence of an unbounded positive sequence $\{x_n\}$ with $f'(x_{n}) Let $a>1$ and let $f:(0,+\infty)\longrightarrow (0,+\infty)$ be differentiable. Show that there exists a positive sequence $\{x_{n}\}$ with $\lim_{n\to\infty}x_{n}=+\infty$, such that $$f'(x_{n})<f(ax_{n}),\ \forall n\in \Bbb N^{+}.$$ My try: I note that $$(f(x)e^{-x})'=f'(x)e^{-x}-f(x)e^{-x}=e^{-x}[f'(x)-f(x)],$$ but I can't move forward. AI: By reduction to absurdity, suppose that the conclusion is false, i.e. there exists $b>0$, such that $$f'(x)\ge f(ax),\quad\forall x\ge b.$$ Since $f>0$, it follows that $f$ is strictly increasing on $[b,+\infty)$, so by mean value theorem, for every $x\ge b$, there exists $\xi\in(x, ax)$, such that $$f(ax)>f(ax)-f(x)=f'(\xi)(ax-x)\ge f(a\xi)(ax-x)> f(ax)(ax-x).$$ However, the inequality above cannot hold when $x\ge \frac{1}{a-1}$, which completes the proof.
H: Why not $f(z)=z^2$ conformal at $z=0$? $$f(z)=z^2$$ is not conformal at $z=0$ Why? Conformal definition: $f$ is conformal at z if f preserves angles there. AI: The angle between the rays $t$ and $it$ with $t\in[0,\infty)$ is $90^\circ$. The angle between thier images $t^2$ and $-t^2$ is $180^\circ$
H: $B(R,R)$ is not closed in the topology of compact convergence I'm doing this exercise in Munkres book, and got no clue to solve this problem. Help someone can help me. Let $B(R,R)$ be the set of bounded functions $f: R \rightarrow R$. Prove that $B(R,R)$ is not closed in the topology of compact convergence Thanks AI: HINT: Let $f(x)=x^2$, say. For $n\in\Bbb Z^+$ let $f_n(x)=\min\{x^2,n\}$.
H: $E[e^{X_1^2 t}] = \frac{1}{\sqrt{1-2t}}$ then $M(t) = \frac{1}{\sqrt[\color{red}n]{1-2t}}$ I have $E[e^{X_1^2 t}] = \frac{1}{\sqrt{1-2t}}$. How do I get to $M(t) = \frac{1}{\sqrt[\color{red}n]{1-2t}}$. Where did the $n$-th root come from? See last line of image Proposition $\bf 2.4.38.\;$ Let $X_1,\ldots,X_n$ be $\color{darkorange}{\textit{independent and identically distributed (i.i.d.)}}$ standard normal raandom variables ($i.e\,\; X_i\overset{i.i.d.}\sim{\cal N}(0,1)$). Let $$Z=X_1^2+\ldots+X_n^2.$$ Then $Z\sim \mathcal X_n^2.$ Proof. We will use moment generating functions. We have $$\begin{align} M(t)\;&=\;\mathbb{E}\big[e^{Zt}\big] \\ &=\;\mathbb{E}\big[e^{t\sum_{i=1}^{n}X_i^2}\big] \\ &=\;\prod_{i=1}^n \mathbb{E}\big[e^{tX_i^2}\big] \\ &=\;\mathbb{E}\big[e^{X_1^2t}\big]^n.\\ \end{align}$$ Here we have used the independent property on the third line and the fact that the random variables are identically distributed on the last line. Now $$\eqalign{\Bbb E\left[e^{X_1^2t}\right] &=\dfrac1{\sqrt{2\pi}}\int_{-\infty}^\infty\exp\left\{-\dfrac12(1-2t)x_1^2\right\}\mathrm dx_1\quad\dfrac12\gt t \\ &=\dfrac1{\sqrt{2\pi}(1-2t)^{1/2}}\int_{-\infty}^\infty \exp\left\{-\dfrac12u^2\right\}\mathrm du \\ &=\dfrac1{(1-2t)^{1/2}}. }$$ Hence $$M(t)=\dfrac1{(1-2t)^{n/2}}\quad\dfrac12\gt t.$$ Also I know $M(t) = E[e^{Xt}]$ but is $M(t) = E[e^{X^{\color{red}{n}}t}]$ note $X^n$ AI: First off, you did not obtain the $n-$th root: $$\sqrt[n]{1-2t}\ne\left(1-2t\right)^{n/2}=\left(\sqrt{1-2t}\right)^{n}$$ The random variable $Z$ is a sum of independent, identically (normally) distributed variables $X_{i}$, squared. Hence, you have that: $$M(t)=\mathbb{E}e^{t(X_{1}^{2}+...+X_{n}^{2})}=\mathbb{E}e^{tX_{1}^{2}}\cdot...\cdot\mathbb{E}e^{tX_{n}^{2}}=\left[\mathbb{E}e^{X_{1}^{2}t}\right]^{n}.$$ From the formula: $$\mathbb{E}[g(X)]=\int^{\infty}_{-\infty}g(x)f(x)\,\mathrm{d}x$$ where $f$ is the pdf of $X$, you obtained: $$\mathbb{E}e^{X_{1}^{2}t}=\frac{1}{(1-2t)^{1/2}}$$ Hence: $$M(t)=\left(\frac{1}{(1-2t)^{1/2}}\right)^{n}=\frac{1}{(1-2t)^{n/2}}$$
H: Epi-Mono factorization in presentable categories If $\mathcal{C}$ is a locally presentable category, then it seems to be well-known that (Strong Epi, Mono) is a factorization system on $\mathcal{C}$. Where can I find a proof of this fact? Actually I only would like to see a proof that every morphism can be factored as an epimorphism followed by a monomorphism. AI: This is Proposition 1.61 in [Adámek and Rosický, Locally presentable and accessible categories]. The proof given merely observes that $\mathcal{C}$ is cocomplete, well copowered, and has pullbacks – so it has a (strong epi, mono) factorisation system. Dually, $\mathcal{C}$ has an (epi, strong mono) factorisation system, because it is complete, well powered, and has pushouts. This in turn is Proposition 4.4.3 in [Borceux, Handbook of categorical algebra, Vol. I].
H: Question about $a \equiv b \pmod{mn} \Leftrightarrow a \equiv b \pmod{m} \wedge a \equiv b \pmod{n}$ So Knuth's 'Discrete Mathematics' states that: $a \equiv b \pmod{mn} \Leftrightarrow a \equiv b \pmod{m} \wedge a \equiv b \pmod{n}$ if $m$ and $n$ are relatively prime. But being a curious human being that I am a tried to prove that and here's my attempt: $x \equiv y \pmod{mz} \Leftrightarrow \exists_{k \in \mathbb{Z}} kmz = y - x \Leftrightarrow \exists_{l \in \mathbb{Z}} lm = y - x \Leftrightarrow x \equiv y \pmod{m}$, because i put $l=km$. I can do the same for $z$ and I get that $x \equiv y \text{ (mod z) }$. Everything came up roses yet I've never used the fact, that $m$ and $z$ are relatively prime. Is my proof incorrect or isn't it necessary for $m$ and $z$ to be relatively prime? AI: The second equivalence in your proof is not correct; that is, $(\exists k \in \mathbf{Z}) (kmz = y -x)$ is not equivalent to $(\exists l \in \mathbf{Z})(lm = y-x)$. You only have $$(\exists k \in \mathbf{Z}) (kmz = y -x) \implies (\exists l \in \mathbf{Z})(lm = y-x).$$. Which "Discrete Mathematics" book by Knuth is this? Do you mean "Concrete Mathematics" with Graham and Patashnik?
H: Finding a basis of eigenvectors For a linear operator $T$ on $V$ find the eigenvalues of $T$ and an ordered basis $\beta$ for $V$ such that $[T]_\beta$ is a diagonal matrix: $V$=$R^3$, $T(a,b,c)$= $(7a-4b+10c,4a-3b+8c,-2a+b-2c)$. I solved this question, and got that, the eigenvalues are $-1,1,2$ and the basis $\beta$ = {$(1,2,0),(1,4,1),(-2,0,1)$}. But, my book gives a different answer for $\beta$, i.e. $\beta$ = {$(1,2,0),(1,-1,-1),(2,0,-1)$}. Is my answer also correct ? What i want to know is, whether this basis $\beta$ for $V$ can be uniquely determined ? I think that there could be many choices for $\beta$, as even the question says "$an$" ordered basis and not "$the$" ordered basis. Am i correct ? AI: There is no canonical choice for a basis of eigenvectors. For instance, if $(1,1,1)$ is an eigenvector, then also $(a,a,a)$ (for $a\ne0$) is, and there's no rule that makes $(1,1,1)$ preferable to $(2,2,2)$. Your matrix is $$ \begin{bmatrix} 7 & -4 & 10 \\ 4 & -3 & 8 \\ -2 & 1 & -2 \end{bmatrix} $$ It's readily checked that $(1,2,0)$ is an eigenvector for the eigenvalue $-1$; $(1,-1,-1)$ is an eigenvector for the eigenvalue $1$; $(2,0,-1)$ is an eigenvector for the eigenvalue $2$. There's no canonical choice, so using $(-2,0,1)$ is as good as using $(4,0,-2)$ or $(2,0,-1)$. However, having made the checks, your vector $(1,4,1)$ cannot be an eigenvector: if it were, it would be a scalar multiple of one of the preceding vectors, which it isn't. Indeed $$ \begin{bmatrix} 7 & -4 & 10 \\ 4 & -3 & 8 \\ -2 & 1 & -2 \end{bmatrix}\, \begin{bmatrix} 1\\4\\1 \end{bmatrix}= \begin{bmatrix} 1\\0\\0\end{bmatrix} $$ If I had to grade your test, I'd consider this a serious mistake, because you have a way to check your computations, namely that the vectors you found are indeed eigenvectors.
H: How does a left group action on the fiber of a principal bundle induce a right action on the total space? Suppose I define a "principal $G$-bundle" as follows: A principal $G$-bundle is a fiber bundle $F \to P \overset{\pi}{\to} X$ with a left group action of $G$ on $F$ that is free and transitive, together with a trivializing cover whose transition maps are $G$-valued. However, it seems that many references define "principal $G$-bundle" via a right action of $G$ on $P$ (not $F$). How does my definition induce a natural right action of $G$ on $P$? Can this be done without saying the phrase "identify $F$ with $G$"? The reason I would like to avoid this identification is two-fold. First, I would like to keep the fiber $F$ and the group $G$ separate in my head -- at least for now -- in part because not all $G$-bundles are principal. Second, and more importantly, I am concerned that any identification of $F$ with $G$ will involve an arbitrary choice of base-point of $F$, and I would rather not make such unnecessary choices if possible. Ultimately, I would like to say that the specified trivializations in my definition of "principal $G$-bundle" are $G$-equivariant with respect to the actions on $P$ and $F$. I would like to deduce this as a consequence of the definition of the $G$-action on $P$, rather than taking this equivariance as the definition of the action. Aside: As usual, this question is a refinement of a previous, less focused question of mine. AI: A clean way to do it (without making an identification you do not like), is to assume that we have two commuting actions of $G$ on $F$ (left and right actions). Below are the details. Let $g_{\alpha\beta}: U_\alpha\cap U_\beta\to G$ be the transition maps satisfying the cocycle condition. Let $F$ be any left $G$-space (the action at this point need not be transitive). Then one forms the bundle $E\to X$ (with the fibers diffeomorphic to $F$ in the smooth setting) by taking the quotient of the disjoint union $$ \tilde E=\sqcup_{\alpha} U_\alpha\times F $$ by the natural equivalence relation: $(x,v)\in U_\alpha \times F \sim (x,w)\in U_\beta\times F$ whenever $g_{\alpha\beta}v=w$ (here we are using the left action). Now, suppose in addition that we have a second action, a right action of $G$ on $F$, this time transitive, commuting with the left action used earlier. For instance, if $F$ is identified with $G$ itself (I know that you do not want to make this identification explicit, so this is only an illustration of the construction), then we can use the action of $G$ on itself via right multiplication for the right action and via left multiplication for the left action. Using this right action, the group $G$ also acts naturally (on the right) on $\tilde E$. This action commutes with the left action used to make the identification and, hence, projects to a (right) $G$-action on $E$. Since we assumed that the right action of $G$ on $F$ is transitive, then the resulting right action of $G$ on $E$ is transitive on the fibers, as required by the usual definition of a principal bundle. Hope it helps to clear the confusion.
H: Lebesgue measure vs. Borel measure Wikipedia states that the Lebesgue measure \lambda is an extension of "the" Borel measure which possesses the crucial property that it is a complete measure (unlike the Borel measure). However I have read that for every Lebesgue-measurable set a subset can be found, which is not measurable (some kind of Vitali set inside a measurable set, if I'm not mistaken). So I could take a null set, e.g. the Cantor set, and with help of the axiom of choice I can create some sort of Vitali set, which is a subset of the Cantor set. This, however would contradict Wikipedia's claim that the Lebesgue measure is a complete measure. Pretty new to the whole measure theory stuff, so I guess my mistake is a pretty obvious one. Any help is greatly appreciated :) AI: You have the wrong definition of complete. A measure is complete if whenever two sets are almost equal (i.e. differ by a set of measure zero), and one of them is measurable, then the other is measurable. Another way to say this is that your $\sigma$-algebra has all the available sets of measure zero.
H: Conformal map example $ f(z)=e^z$ I an studying the example-1. I understand $f(z)=e^z$ has a nonzero derivative at all points, hence it is everywhere conformal and locally $1-1$. But I dont understand th part I underlined with yellow pencil. Please explain it. Thank you so much:) AI: $z=x+iy$, $f(z)=e^z$, so $e^{c+iy}=e^c(\cos y+i\sin y)$ assuming $x=c$ constant is a circle with radius $e^c$ and $e^{x+ic}=e^x(\cos c+i\sin c)$ The Modulus of $f(z) = e^x,$ argument $f(z) = y.$ Horizontal lines in the cartesian plane have a constant $y$ value so the argument of the map of the horizontal line is also constant while the norm changes with $e^x$ so horizontal lines are mapped to rays with argument $y.$ vertical lines have a constant $x$ component so the map of vertical lines have a constant modulus so vertical lines get mapped to circles centered at the origin with radius $e^x$
H: Problem with a set without accumulation points Let $S$ be a nonempty subset in $\mathbb R^m$ without accumulation points in $\mathbb R^m$. Is then $$ \inf \{ \\|x-y\\|: x,y \in S, x\neq y \} >0 \textrm{ ? } $$ AI: Counterexample : $A=\lbrace k; k+\frac{1}{k} \| k\geq 1\rbrace$.
H: why 64 is equal to 65 here? how is this possible? I know there is some trick, should someone please explain?! AI: This is a very well known optical illusion. Count the number of squares in each triangle (or at least in each non-vertical or non-horizontal line) and you'll see that they don't have the same slope. Therefore the triangles cannot magically ''fit'' as they seem to do so. The slope of green and red is 3/8 (0.375), where as the slope of blue and orange is 2/5 (0.4). These numbers are quite close so it's easy to hide one square unit. But the slopes cannot fit the way they look like they do. Hope that helps,
H: Does my logic statement make sense? I'm trying to convert this sentence to logic notation. "there is an integer less than or equal to all other integers greater than 0". "An integer exists that is less than or equal to all other integers greater than 0". So far I have - L(x,y): x is less than or equal to y ∃y ∈ Z.(∀x ∈ Z.L(x,0)) Does this translate correctly? Any help would be appreciated. AI: Your sentence says that there is an integer $y$ such that all integers are less than or equal to $0$, which is clearly not what you wanted. Let’s try to sneak up on the correct expression. How can we say ‘$x$ is an integer less than or equal to each integers that is greater than $0$’? First, $y>0$ has to be translated. One way is $L(0,y)\land y\ne 0$, and another is $\neg L(y,0)$; you can use either. Now we want to say that if $y$ is an integer greater than $0$, then $x\le y$: $$\forall y\in\Bbb Z\Big(\big(L(0,y)\land y\ne 0\big)\to L(x,y)\Big)\;,$$ or $$\forall y\in\Bbb Z\big(\neg L(y,0)\to L(x,y)\big)\;.$$ Finally, we have to say that there is such an integer $x$: $$\exists x\in\Bbb Z\,\forall y\in\Bbb Z\Big(\big(L(0,y)\land y\ne 0\big)\to L(x,y)\Big)\;,$$ or $$\exists x\in\Bbb Z\,\forall y\in\Bbb Z\big(\neg L(y,0)\to L(x,y)\big)\;.$$ But that’s still not quite right, because the other in all other integers greater than $0$ implies that $x$ is supposed to be greater than $0$, and we have to say so. For convenience I’m going to introduce an abbreviation: I’ll write $L'(x,y)$ for $L(x,y)\land x\ne y$ or $\neg L(y,x)$. Then what we want is $$\exists x\in\Bbb Z\,\forall y\in\Bbb Z\left(L'(0,x)\land\Big(\big(x\ne y\land L'(0,y)\big)\to L(x,y)\Big)\right)\;.$$ You can get rid of the new predicate $L'$ simply by expanding each instance using the definition of $L'$.
H: Computing $\sum\limits_{n=0}^{\infty}{\frac{(-1)^n}{\alpha n +1}}$ where $\alpha>0$ I'm trying to calculate $$\sum_{n=0}^{\infty}{\frac{(-1)^n}{\alpha n +1}};\;\; \alpha>0$$ using the fact that $$\frac{1}{1-a}=\sum_{k=0}^{n-1}{a^k}+\frac{a^n}{1-a}$$ So taking $a=-t^\alpha$ we get $$\int_0^1\frac{1}{1+t^\alpha}dt=\sum_{n=0}^{n-1}{\frac{(-1)^n}{\alpha n +1}}+(-1)^n\int_0^1{\frac{t^{\alpha n}}{1+t^\alpha}dt}$$ It remains to calculate $(-1)^n\int_0^1{\frac{t^{\alpha n}}{1+t^\alpha}dt}$ and show that it goes to $0$ as $n$ goes to $\infty$ but i can't go further, thank you for your help to continue this or if there is another way to do it. AI: More directly, for every $n\geqslant0$, $$ \frac1{an+1}=\int_0^1t^{an}\mathrm dt, $$ hence, the series being alternated (meaning, alternating signs and decreasing absolute values), $$ \sum_{n\geqslant0}\frac{(-1)^n}{an+1}=\int_0^1\sum_{n\geqslant0}(-1)^nt^{an}\mathrm dt=\int_0^1\frac{\mathrm dt}{1+t^a}. $$ Nota: About the mention that: It remains to calculate $(-1)^n\int_0^1{\frac{t^{\alpha n}}{1+t^\alpha}dt}$ and show that it goes to $0$ as $n$ goes to $\infty$. Well, $\displaystyle\int_0^1\frac{t^{\alpha n}}{1+t^\alpha}\mathrm dt\leqslant\int_0^1t^{\alpha n}\mathrm dt=\frac1{\alpha n+1}$ hence this seems clear.
H: the power series converges in compact convergence topology Consider the sequence of functions $f_{n}: (-1,1) \rightarrow R$ defined by:$$f_{n}(x) = \sum_{k=1}^{n}{kx^{k}}$$ a) Prove that $(f_{n})$ converges in the topology of compact convergence, conclude that the limit function is continuous b)Show that $(f_{n})$ does not converge in the uniform topology. I'm learning function space in topology and find it hard to imagine, especially the compact convergence topology. I got stuck in this problem As I know, we don't even have the clear formula for the function that this sequence converges to, so how can we prove that sequence converges at all. One more thing, is there other ways to prove a sequence of functions converges in compact convergence topology except proving it converges uniformly in every compact subspace, cause in some spaces, we don't know all their compact subspaces Thanks. AI: In this case you can actually determine what the limit has to be, if it exists. The series $\sum_{k\ge 1}kx^k$ is absolutely convergent for $\|x\|<1$. We know that $$\sum_{k\ge 0}x^k=\frac1{1-x}$$ for $\|x\|<1$; taking the derivative and multiplying by $x$ yields $$\sum_{k\ge 1}kx^k=\frac{x}{(1-x)^2}\;,$$ which is the only reasonable candidate for the limit of the sequence $\langle f_n:n\in\Bbb Z^+\rangle$. Note that $$\sum_{k=1}^nkx^k=\sum_{k\ge 1}kx^k-\sum_{k\ge n+1}kx^k\;,$$ and you can use the same trick to get a closed form for $\sum_{k\ge n+1}kx^k$, so you can actually get an explicit formula for each $f_n$. Finally, every compact subset of $(-1,1)$ is contained in a closed interval $[-a,a]$ with $0<a<1$, so you need only look at these intervals.
H: Evaluating the limit $\lim_{n \rightarrow +\infty} \frac{e^n+e^{-n}}{e^{n+1}+e^{-n-1}}$ How would you solve the following limit? It's $\frac \infty \infty$ and L'Hospital doesn't seem to help: $$\lim_{n \rightarrow +\infty} \frac{e^n+e^{-n}}{e^{n+1}+e^{-n-1}}$$ AI: Hint: Multiply the numerator and denominator by $e^{-n}$.
H: sequence of elements in $\ell_2$ Consider the sequence in $\displaystyle{ \ell _2 (\mathbb N ) }$ $$ (x^k)_k = \sum_{j=1}^{k} \frac{1}{i} e_i $$ where $\displaystyle{ e_i \in \ell_2 }$ is the $i-\text{th}$ unit vector in $\ell_2$ ,i.e. $\displaystyle{ (e_i)_j = \delta_{ij} }$ Does this sequence converge in $\ell_2$ ? And if it converges can we find its limit ? Edit: I correct the defition of $(x^k$. I used superscripts, instead of subscripts, since each $x^k$ is itself sequence. AI: Since $$\lVert x^{m+n}-x^m\rVert^2_{\ell_2}=\sum_{j=m+1}^{m+n}\frac 1{j^2}\leqslant \sum_{j\geqslant m+1}\frac 1{j^2}\to 0, \mbox{ as }m\to \infty,$$ the sequence $(x^{n},n\geqslant 1)$ is Cauchy in the complete normed space $\ell_2$. Therefore, the sequence is convergent to some $x\in\ell_2$. We have if $i$ is an integer that $\langle x,e_i\rangle=\lim_{n\to \infty}\langle x^{n},e_i\rangle=\frac 1i$, which gives the expression of the limit.
H: Probability distributing ice cream satisfying the taste of each person. Distributing randomly 5 vanilla ice-creams and 5 chocolate ice-creams to 10 people among which 3 prefer vanilla, 2 prefer chocolate and the others do not have preference, what is the probability that everyone has an ice-cream he likes? AI: Fix the 3 vanilla and 2 chocolate ice-cream to each person who prefer them. We now have 2 vanilla and 3 chocolate ice-creams left to distribute among 5 persons who are indifferent. We can do this in $5 \choose 2$ or $5 \choose 3$ ways, depending on distributing the 2 vanilla or 3 chocolate ice-creams left among the 5 persons, leading to only one way to giving the other flavor. The sample space would be $10 \choose 5$ ways in distributing the 10 ice-creams to these 5 persons. So the answer would be: $$P = \frac{5 \choose 2}{10 \choose 5} $$ .
H: Findin the most general harmonic polynomial of the form $ax^2 + bxy + cy^2$ The question says to find the most general harmonic form of $ax^2 + bxy + cy^2$. And I've seen one or two answered questions here on this topic but I couldn't understand $why$ certain steps were took and didn't see how this was applicable to a 2-degree function. I'd like some help, if you may. AI: Let $f(x) = ax^2 + bxy + cy^2$. We know that a harmonic function satisfies the equation $$ \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0. $$ We have $$ \frac{\partial f}{\partial x} = 2ax + by, \ \frac{\partial^2 f}{\partial x^2} = 2a. $$ And $$ \frac{\partial f}{\partial y} = bx + 2cy, \ \frac{\partial^2 f}{\partial y^2} = 2c. $$ We evaluate the partial derivative with respect to $x$ by considering $y$ to be constant (and vice versa). Now, we want $\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0$ for all $x$ and $y$. This gives $2a + 2c = 0$, or $c = -a$. It follows that the general harmonic polynomial of degree $2$ has the form $$ ax^2 + bxy -ay^2. $$
H: Matrix topology Let $X$ be the space of all real $n \times n$ matrices. We define the function $f \colon X \rightarrow \mathbb{R}$ as the rank of a matrix. We induce the standard Euclidean topology on $X$. Show that for all $x$ in $X$ and for all $a$ in $\mathbb{R}$ the set $\{x \in X; f(x) \leq a \}$ is closed in $X$. My book says that I should prove that a small change in the components of each vector of a matrix doesn't change the rank of a matrix. I don't understand this because I thought that I have to prove that the boundary of the set is in the set. We view a matrix as a vector in $\mathbb{R}^{n^{2}}$. There has to exist a matrix such that for every open ball with center in the matrix there exists an element of the ball that has the same rank as the matrix and there exists an element of the ball which has a different rank. Intuitively this means that small changes both preserve and change the rank of a matrix which is in contradiction with the hint in my book. Any help would be appreciated. AI: You want to show that a small change can only increase the rank. For example, let $$ A=\begin{pmatrix} 1&0\\0&0\end{pmatrix}. $$ Then changing $A_{2,2}$ by a little will change the rank, but only increase it. To show that small changes can only increase the rank, note that a column of a matrix represents the vector to which that matrix maps the basis vector that corresponds to that column. After you show the above fact, then take a convergent sequence of matrics $A_1,A_2,\ldots$ all with rank less than or equal to $n$. Show that the limit point of this sequence has rank less than or equal to $n$.
H: How to Prove the $ {L}_{\infty} $ Ball Is Convex? I want to prove that a ball for infinity norm is convex: $$ B_\infty=\{x\in\mathbb R^n : \\|x\\|_\infty\le1\} $$ I came up with this proof and appreciate it if someone can help to verify if this is correct: \begin{align} \\|x\\|_\infty&=\\|(1-\lambda)x+\lambda y\\|_\infty\\&=\max_j\|(1-\lambda)x\|+\max_j\|\lambda y\|\\&=(1-\lambda)\max_j\|x\|+\lambda\max_j\|y\|\\&=(1-\lambda)+\lambda=1 \end{align} P.S. This is a homework. Thanks! AI: Suppose $x, y \in B_{\infty}$. Let $x_i$ and $y_i$ denote components of $x$ and $y$. Then for any $\lambda \in [0, 1]$, \begin{align} \\|(1 - \lambda)x + \lambda y\\|_\infty & = \max_i \|(1 - \lambda)x_i + \lambda y_i\| & & \text{by definition of $\\|\cdot\\|_\infty$}\\ & \le \max_i (1 - \lambda)\|x_i\| + \max_i \lambda\|y_i\| & & \text{by $\|a + b\| \le \|a\| + \|b\|$} \\ & = (1 - \lambda)\\|x\\|_\infty + \lambda\\|y\\|_\infty & & \text{by definition of $\\|\cdot\\|_\infty$}\\ & \le (1 - \lambda) + \lambda & & \text{because $x \in B_\infty$ and $y \in B_\infty$}\\ & \le 1 \\ \therefore (1 - \lambda)x + \lambda y & \in B_{\infty} \end{align} Therefore, $B_{\infty}$ is convex.
H: Transformations Difference On what basis we depend when we choose the tool to transform to frequency domain, I can't distinguish in what case we use one of these transformations ( trig. fourier series, complex fourier series, fourier transform, laplace transform, z-Transform) AI: These transformations are generally applicable to different types of time-domain signals: The Fourier series (real or complex, here the difference is of presentation and not substance) exists to periodic signals (which are integrable over a period). Note that in the general case periodic signals aren't integrable over the entire line (time domain). The Fourier transofrm can be applied to integrable (or finite-energy, if you develop the theory carefully enough) signals on the entire line. These signals aren't periodic. The Laplace transform is a method of generalizing the Fourier transform to signals which aren't integrable (or have bounded energy), but whose growth can be exponentially dampened. Note that the imaginary axis is in the region of convergence of the Laplace transform iff the signal has a Fourier transform, in which case its restriction to that axis is the function's Fourier transform. The $Z$-transform is applicable to signals in discrete time-domain, unlike former transforms which are applicable to signals of continuous time-domain.
H: How come when $2^{k} \| (x-1)(x+1)$ one of the terms is divisible by $2$ and not by $4$ when $k \in \mathbb{N} $ and $3 \leq k$ So I'm reading Knuth's 'Discrete Mathematics' at the moment and there's a paragraph detailing how many solutions are there for $x^{2} \equiv 1 \pmod{p}$. So other cases (when $p$ is an odd prime or when p is equal to $2$ or $4$) are clear to me, but I'm having trouble wrapping my head around the case when $p=2^{k}$ with $3 \leq k$. Here's what's troubling me If $2^{k}\|(x - 1)(x + 1)$ then either $(x - 1)$ or $(x + 1)$ is divisible by $2$ but not by $4$, so the other one must be divisible by $2^{k-1}$. How come? I don't see that at all... AI: In general, there are cases in which neither $x+1$ nor $x-1$ is divisible by $2$ (take $x$ to be an even number). It is given that $2^k \| (x+1)(x-1) $. Suppose neither $x+1$ nor $x-1$ is divided by $2$, then $2$ does not divide $(x+1)(x-1)$. So, there is no point of considering $2^k \| (x+1)(x-1)$. So, without loss of generality, suppose $2$ divides $x+1$ suppose both $(x+1)$ and $(x-1)$ are divisible by $4$ then, we should have $2^{2m} \|(x+1)(x-1) $ which need not be the case always... (it is not specified that power of $2$ is even) So, we can suppose $x+1$ is divisible by $2$ but not by $4$. So, we should have $2^{k-1} \| (x+1)$
H: Sum of a set normalize by total items in set This might be a be simple but I just want to make sure I didn't use the wrong notation. If I have a set of weighted terms, ${w_1, w_2, \dots, w_n}$ and the score is the sum of $w_1$ to $w_n$ normalize by count of $w$. Count of $w$ is just the total number of terms, i.e. $n$. so, can I write: $score = \frac{\sum{w_i}}{\left \| w \right \|}$ ? p/s: $W = \{w_1, w_2, \dots, w_n \}$ but I don't really want to introduce $W$ for the sake of showing the count using $\|W\|$. Thanks. AI: If $w$ is the set of weighted terms, you are right.
H: Can I approximate a complex number by its imaginary part, if real part is small compared to imaginary part? I have the following doubt. How do you explain this? Here $j$ means $\sqrt{-1}$. AI: Assuming $\;j=i:=\sqrt{-1}\;$ : $$ix+\frac{x^2}{a+ix}=\frac{-x^2+iax+x^2}{a+ix}=\frac{ax}{a+ix}i\xrightarrow[x\to\infty]{}a$$ The wrong part of your argument is that both summands in your original expression diverge when $\;x\to \infty\;$ , so you can't taka the limits separatedly.
H: Lebesgue Measure as a Countable Sum of Probability Measures Show that Lebesgue measure can be expressed as a countable sum of probability measures. I'm trying to do something with the countable additivity property in order to show this, but so far nothing is working. I don't think this is supposed to be difficult, but I'm not seeing it, so any help you could give would be most appreciated! AI: Hint: For any integer $n,$ note that $0\le\mu\bigl([n,n+1)\cap A)\le 1$ for any measurable set $A\subseteq R.$
H: Find points on perpendicular line I have $P_1=(x_1,y_1)$ and $P_2=(x_2,y_2)$ points and they are specified. Also I have $h$ distance from $P_1$ point that is also known. And I want to find $P_3$ and $P_4$ that located on perpendicular line and $h$ is distance to them. How can i found those $P_4=?$ $P_3=?$ AI: Hints: From the drawing you want to find $\;P_3,P_4\;$ s.t. $\;P_3P_4\perp P_1P_2\;$ and also s.t that $\;P_1\;$ belongs to the line determined by $\;P_3,P_4\;$, so: $$\begin{align}\bullet&\;\;\text{Find the slope of the line segment $\;P_1P_2\;$}\;:\;m_1:=\frac{y_2-y_1}{x_2-x_1}\;,\;\;x_1\neq x_2\\ \bullet&\;\;\text{Find the equation of the line through $\;P_1\;$ with slope}\;\;-\frac1{m_1}\;:\;y-y_1=-\frac1{m_1}\left(x-x_1\right)\\ \bullet&\;\;\text{Find the two points on the above line at a distance of $\,h\,$ from $\,P_1\,$ by solving the quadratic:}\\ &h^2=(x_1-x)^2+(y_1-y)^2\;,\;\;\text{with}\;\;y=-\frac x{m_1}+\frac{x_1}{m_1}+y_1\end{align}$$
H: Is $(m \Leftrightarrow m) \Leftrightarrow (m \Rightarrow m)$ a tautology, contradiction or contingent? Is this a Tautology, contradiction or contingent? $(m \Leftrightarrow m) \Leftrightarrow (m \Rightarrow m)$ My answer is that It is a tautology. But what is yours? Can someone please explain with a truth table? Thank you so much!!!!! AI: m \| m <--> m \| m --> m \| (m <-> m) <--> (m --> m) T \| T \| T \| T F \| T \| T \| T Note that each of $m \rightarrow m$ and $m \leftrightarrow m$ is a tautology (always true, regardless of the truth value of $m$), and hence, $$(m \leftrightarrow m) \leftrightarrow (m \rightarrow m)$$ is necessarily a tautology, as well, which means the following equivalence necessarily holds: $$(m \leftrightarrow m) \leftrightarrow (m \rightarrow m) \equiv T$$
H: Leaky integrator in novice terms I have read few article on the leaky integrator including the Wikipedia. They all give the same equation and the graph and say it is applicable in areas such as neuroscience etc. But I still cannot understand how this works in practice. So I would like to know an example scenario how a leaky integrator works in real world. I do not want equations just a plain scenario would do to make it clear in my mind. AI: An example might be a resistor-capacitor filter in electronics. The charge on the capacitor will be the integral of the current you pour into it (and the voltage across it proportional to the charge). However if the capacitor is in parallel with a resistor, it will discharge ("leak") through that resistor. Thus it's equation of motion will be $$ I = C{dV\over dt} + V/R $$ which you can re-arrange to get the ODE shown on the Wikipedia.
H: An inverse question of uniformly convergence {Edit: since I made some mistake on the pointwise limit and the uniformly continuous.} A classical results in elementary analysis state that if a sequence of continuous function $f_n(x)$ on $[0,1]$ is uniformly convergence to $f$, then $f$ is continuous on $[0,1]$ too. I am wondering that if we know that $f$ is continuous on $[0,1]$, and it is a pointwise limit of a sequence of continuous function $f_n$ on $[0,1]$, can we conclude that $f_n$ convergence to $f$ uniformly on $[0,1]$? I have noted that if $f$ is not continuous, then there has a counter example, $f_n=x^n$, $f(1)=1$ and $f=0$ otherwise. But how about add the continuous to $f$? AI: No. For example, consider the functions $f_n$ on $[0,1]$, where the graph of $f_n$ consists of the straight line segments connecting the points $(0,0)$, $(1/(n+1),1)$, $(1/n,0)$, and $(1,0)$. The sequence of continuous functions $(f_n)$ converges pointwise to the zero function on $[0,1]$; but, the convergence is not uniform. More interesting, is that a pointwise convergent sequence of continuous functions to a continuous limit need not converge uniformly on any open subset of $[0,1]$. See this post for counterexamples.
H: Has the property $\int\Phi =1$ (of the fundamental solution $\Phi$ of the heat equation) a physical interpretation? For each $t>0$ the fundamental solution of the heat equation is given by $$\Phi(x,t)=\frac{1}{(4\pi t)^{n/2}}\exp\left(-\frac{\|x\|^2}{4t} \right )$$ and satisfies $$\int_{\mathbb{R}^n}\Phi(x,t)\,dx=1.$$ Is there any physical interpretation for this property? Thanks. AI: Yes, since $\Phi(t,x)\mathrm dx$ represents the amount of heat contained in the volume $\mathrm dx$, this is saying that the total amount of heat contained in the whole $\mathbb R^n$ does not change with time.
H: Reflection of the origin in the x,y,z plane? The equation of the plane is x-2y-2z=27. It can be written in the form where d gives the distance of the origin from the plane, which I worked out to be 9 (can someone verify?). How do I work out the point which is the reflection of the origin? The answer is (6, -12, -12), but how do I get there? AI: Hint-First find foot of orgin on that plane using equation of staright line passing through orgin and parelel to normal vecter.After that we have orgin and that point and knowing the relation of foot,point and reflection,you can find reflection
H: injectivity of the function $f: \quad \mathbb{R}^2_+ \longrightarrow \mathbb{R}^2 \quad : (x,y)\quad \longmapsto \ (x^2-y^2, 2xy)$ Consider the function: $$f: \quad \mathbb{R}^2_+ \ \longrightarrow \ \mathbb{R}^2 \quad : \quad (x,y) \ \longmapsto \ (x^2-y^2, 2xy)$$ I had to show that this function is injective. According to the exercise, I had to prove the identity: $$ (\tilde{x}^2 + y^2)(\tilde{y}^2-y^2) = 0 \qquad \ \ \ \text{if} \ \ \ \ \ f(x,y) \ = \ f(\tilde{x},\tilde{y})$$ Maybe this this very easy, but I didn't know how to show this, and I didn't know how to establish injectivity by this either. I haven't produced anything fruitful. Could you please help me? AI: If $f(x,y)=f(\tilde x, \tilde y)$ then $$(\tilde x^2 + y^2)(\tilde y^2 - y^2) = \tilde x^2\tilde y^2 - y^4 + y^2(\tilde y^2 - \tilde x^2) = x^2y^2 - y^4 + y^2(y^2 - x^2) = 0$$ Also if this is zero then one of the brackets is zero. Then either $\tilde x = y = 0$ or $\|\tilde y \| = \|y\|$ and in either case you can work out the rest.
H: What is the remainder when $3^{1264}$ is divided by 549? Please explain in detail. I tried a lot by applying normal remainder theorem but I am not able to get anywhere. AI: Here is something to get you started. $549=9\cdot61$, and $9$ and $61$ are mutually prime. So you can compute $3^{1264}\bmod 9$ and $3^{1264}\bmod 61$, then combine results using the Chinese Remainder Theorem (CRT). The first of these is easy, for $3^2\equiv0\bmod9$. And the second one you can do by Fermat's little theorem, noting that $61$ is prime, so $3^{60}\equiv1\bmod 61$. Note that $1264=60\cdot21+4$ … so that $3^{1264}=3^{60\cdot21+4}=(3^{60})^{21}\cdot3^4\equiv3^4=81\bmod 61$. It seems more natural to reduce the last bit mod $61$ and get $20$, but by accident, $81$ is divisible by $9$, so we are saved from doing the last bit involving the CRT.
H: Question regarding compound Poisson random variable I have a question regarding compound Poisson random variable. Let $X:=\sum_{i=1}^{N}{\alpha_i}$ to be a compound Poisson random variable, where $N$ is a Poisson random variable with mean $1$, and $(\alpha_i)_{i \in \mathbb{N}}$ is a family of i.i.d random variables with distribution $\mu$, independent of $N$. Prove that for any $p>0$, $E[\|X\|^p]<\infty$ if and only if $E[\|\alpha_1\|^p]<\infty$. Kindly help as I am new to Probability. Thank you very much. AI: In one direction: If $p\geqslant1$, then $\|X\|^p\leqslant N^{p-1}\sum\limits_{k=1}^N\|\alpha_i\|^p$ hence $E[\|X\|^p\mid N]\leqslant N^{p}E[\|\alpha_1\|^p]$. If $p\lt1$, then $\|X\|^p\leqslant\sum\limits_{k=1}^N\|\alpha_i\|^p$ hence $E[\|X\|^p\mid N]\leqslant NE[\|\alpha_1\|^p]$. In both cases, $E[\|X\|^p]\leqslant E[N^q]E[\|\alpha_1\|^p]$ with $q=\max\{p,1\}$. Since $E[N^q]$ is finite for every positive $q$, if $\alpha_1$ is in $L^p$ then $X$ is in $L^p$. In the other direction: If $N=1$ then $X=\alpha_1$ hence $\|X\|^p\geqslant\|\alpha_1\|^p\mathbf 1_{N=1}$ and $E[\|X\|^p]\geqslant E[\|\alpha_1\|^p]\,P[N=1]$. Since $P[N=1]$ is positive, if $X$ is in $L^p$ then $\alpha_1$ is in $L^p$.
H: Recurrence relation, generating function I am trying to solve this recurrence relation using generating functions $$x_{n+2}+x_{n+1}+x_n=0$$ $$x_0 = x_1=1$$ I have got this generating function $f_a(x)=\frac{2x+1}{x^2+x+1}$. Since the denominator has complex roots I can't factor it. Is there a way to get the series without using complex numbers? AI: Note that $$ f(x)=\frac{(1+2x)(1-x)}{(1+x+x^2)(1-x)}=\frac{1+x-2x^2}{1-x^3}=(1+x-2x^2)\sum_{n\geqslant0}x^{3n}, $$ hence $$ f(x)=\sum_{n\geqslant0}a_nx^{n},\qquad a_{3k}=a_{3k+1}=1,\quad a_{3k+2}=-2, $$ that is, the sequence $(a_n)_{n\geqslant0}$ is $$ 1,\,1,\,-2,\,1,\,1,\,-2,\,1,\,1,\,-2,\,1,\,1,\,-2,\,1,\,\ldots$$
H: In how many ways can $7^{13}$ be represented as product of $3$ natural numbers? How i solved it: all possible non-distinct groups $(a,b,c)$ are, $a = 0 \Rightarrow (b,c) = (0,13)(1,12)(2,11)(3,10)(4,9)(5,8)(6,7)$ $a = 1 \Rightarrow (b,c) = (1,11)(2,10)(3,9)(4,8)(5,7)(6,6)$ $a = 2 \Rightarrow (b,c) = (2,9)(3,8)(4,7)(5,6)$ $a = 3 \Rightarrow (b,c) = (3,7)(4,6)(5,5)$ $a = 4 \Rightarrow (b,c) = (4,5)$ Thus, $7^{13}$ can be written as product of $3$ natural numbers in $7+6+4+3+1 = 21$ ways. Though this gives the required solution, it takes time and is a lengthy way. Is there an alternate method to tackle such problems? (May be by using the "bars and stars" method with some adjustments? I tried but failed to get the correct answer that way.) Please help me out and share your method! Thanks a lot! AI: If you consider different orders equivalent, so that $7^87^37^2$ is considered the same as $7^27^37^8$, you are asking about the number of partitions of $13$ into $3$ parts, while the stars-and-bars answers give you the number of compositions of $13$ into $3$ parts. In the partition page, it states that the number of partitions of $n$ into $1,2,3$ parts is the nearest integer to $\frac {(n+3)^2}{12}$, here $21$. As you area are allowing $0$, which means you include $1$ or $2$ parts, that is your answer.
H: find value (-2)^-(2)^(-2) Find the value of $(-2)^{-(2)^{(-2)}}$. Is it 16/8/-8/none? My attempt: $a^{-x}=\frac1{a^x}$, so, $(-2)^{-(2)^{(-2)}}=(-2)^{\frac{-1}{2^2}}=\frac{1}{(-2)^{\frac14}}$. That is, I would pick 'none of the given options' as my answer. But it is given that answer is 16. And the comments below also suggest that the correct answer is indeed 16. So I will appreciate if one can tell the mistake in my solution. EDIT: I've cross-checked. The statement that I have posed is same as was asked in the exam. No parsing error by me. Maybe by the question-setter, don't know. AI: It appears that you may have confused/misread the problem statement: $$(-2)^{-(2)^{(-2)}} = \frac{1}{\sqrt[4]{(-2)}}\quad \neq \quad\left((-2)^{-2}\right)^{-2} = (-2)^{-2\cdot -2} = (-2)^4 = 16$$ That is, we need to be careful to note where the negation of an exponent occurs $(-2)^{a} \neq -(2^{a})$, and we need to be careful to disambiguate $(a^b)^c$ from $a^{(b^c)}.$ EDIT: If the problem you posted is precisely what appears on the practice exam, then you've correctly argued that none of the given answers are correct. In that case, indeed, there must have been a parsing error in the source, if indeed the answer were to be $16$.
H: What is the kernel of the tensor product of two maps? Assume that $f_1\colon V_1\to W_1, f_2\colon V_2\to W_2$ are $k$-linear maps between $k$-vector spaces (over the same field $k$, but the dimension may be infinity). Then the tensor product $f_1\otimes f_2\colon V_1\otimes V_2\to W_1\otimes W_2$ is defined, and it's obvious that $\ker f_1\otimes V_2+ V_1\otimes \ker f_2 \subseteq \ker (f_1\otimes f_2)$. My question is whether the relation $\subseteq$ is in fact $=$. If this does not hold, how about assuming all these vector spaces are commutative associated $k$-algebras with identity and that all the maps are $k$-algebra homomorphisms? Or can you give a "right" form of the kernel $\ker (f_1\otimes f_2)$? AI: Yes, that's true. Let $f_i : V_i \to W_i$ be two linear maps. Since $\mathrm{im}(f_1) \otimes \mathrm{im}(f_2)$ embeds into $W_1 \otimes W_2$, we may assume that $f_1,f_2$ are surjective. But then they are split, so that we can assume that $V_i = W_i \oplus U_i$ and that $f_i$ equals the projection $V_i \to W_i$, with kernel $U_i$. Then $V_1 \otimes V_2 = W_1 \otimes W_2 \oplus W_1 \otimes U_2 \oplus U_1 \otimes W_2 \oplus U_1 \otimes U_2$ and $f_1 \otimes f_2$ equals the projection of $V_1 \otimes V_2$ onto $W_1 \otimes W_2$. Hence the kernel is $W_1 \otimes U_2 \oplus U_1 \otimes W_2 \oplus U_1 \otimes U_2 = U_1 \otimes V_2 + V_1 \otimes U_2$. This shows even more: The kernel is the pushout $(\ker(f_1) \otimes V_2) \cup_{\ker(f_1) \otimes \ker(f_2)} (V_1 \otimes \ker(f_2))$. By the way, this argument is purely formal and works in every semisimple abelian $\otimes$-category. What happens when we drop semisimplicity, for example when we consider modules over some commutative ring $R$? Then we only need some flatness assumptions: Let $f_1 : V_1 \to W_1$ and $f_2 : V_2 \to W_2$ be two morphisms in an abelian $\otimes$-category (for example $R$-linear maps between $R$-modules). If $f_1,f_2$ are epimorphisms, then we have exact sequences $\ker(f_1) \to V_1 \to W_1 \to 0$ and $\ker(f_2) \to V_2 \to W_2 \to 0$. Applying the right exactness of the tensor product twice(!), we get that then also $\ker(f_1) \otimes V_2 \oplus V_1 \otimes \ker(f_2) \to V_1 \otimes V_2 \to W_1 \otimes W_2 \to 0$ is exact. If $f_1,f_2$ are not epi, we can still apply the above to their images and get the exactness of $\ker(f_1) \otimes V_2 \oplus V_1 \otimes \ker(f_2) \to V_1 \otimes V_2 \to \mathrm{im}(f_1) \otimes \mathrm{im}(f_2) \to 0.$ Now assume that $\mathrm{im}(f_1)$ and $W_2$ are flat. Then $\mathrm{im}(f_1) \otimes \mathrm{im}(f_2)$ embeds into $\mathrm{im}(f_1) \otimes W_2$ which embeds into $W_1 \otimes W_2$. Hence, we have still that the sequence $\ker(f_1) \otimes V_2 \oplus V_1 \otimes \ker(f_2) \to V_1 \otimes V_2 \to W_1 \otimes W_2$ is exact. In other words, we have a sum decomposition $$\ker(f_1 \otimes f_2) = \alpha(\ker(f_1) \otimes V_2) + \beta(V_1 \otimes \ker(f_2)),$$ where $\alpha : \ker(f_1) \otimes V_2 \to V_1 \otimes V_2$ and $\beta : V_1 \otimes \ker(f_2) \to V_1 \otimes V_2$ are the canonical morphisms. In general, these are not monic! However, this is the case, by definition, when $V_1$ and $V_2$ are flat. So in this case we can safely treat $\alpha$ and $\beta$ as inclusions and write $$\ker(f_1 \otimes f_2) = V_1 \otimes \ker(f_2) + \ker(f_1) \otimes V_2.$$
H: Limit with e number, explain me this... Please explain me how somebody got this, step by step what's done here... AI: I think that there is a mistake: it is $\displaystyle \frac{\log(1+x)}{x} = \log(1+x)^{\frac{1}{x}}$. So, remembering that $\displaystyle \lim_{x \to +\infty} \left(1 + \frac{1}{x} \right)^{x}=e$ and putting $1/t = x$...
H: Minimal polynomial of the inverse Given $\mathbf{J}_t(\lambda)=\begin{pmatrix}\lambda&1&&\\&\lambda&1&{\LARGE\mathbf{0}}\\&{\LARGE\mathbf{0}}&\ddots&\ddots\\&&&\lambda\end{pmatrix}\in\mathcal{M}_{t\times t}(\mathbb{F})$, where $\lambda\in\mathbb{F}$. Suppose $\lambda\ne0$, show that $m(x)=\left(x-\lambda^{-1}\right)^t$ is the minimal polynomial of $\mathbf{J}^{-1}$. It is clear that $m(x)$ is monic. I need help in showing that $m\left(\mathbf{J}^{-1}\right)=\mathbf{0}$. Any help would be appreciated. ADDED Note that $m_\mathbf{J}(\mathbf{J})=\mathbf{0}$, i.e. $\left(\mathbf{J}-\lambda\mathbf{I}\right)^t=\mathbf{0}$. Then \begin{align} m\left(\mathbf{J}^{-1}\right)&=\left(\mathbf{J}^{-1}-\lambda^{-1}\mathbf{I}\right)^t\\&=(-1)^t\left(\lambda^{-1}\mathbf{I}-\mathbf{J}^{-1}\right)^t\\&=(-1)^t\left(\lambda^{-1}\mathbf{JJ}^{-1}-\mathbf{J}^{-1}\right)^t\\&=(-1)^t\left(\mathbf{J}-\lambda\mathbf{I}\right)^t\left(\lambda^{-1}\mathbf{J}^{-1}\right)^t\\&=\mathbf{0} \end{align} If I want to show that if $p(x)$ is a nonzero polynomial over $\mathbb{F}$ such that $p\left(\mathbf{J}^{-1}\right)=\mathbf{0}$, then the degree of $m(x)$ $\le$ the degree of $p(x)$, is my following argument correct? Suppose there exists $p(x)=a_0+a_1x+\cdots+a_sx^s$ such that $p\left(\mathbf{J}^{-1}\right)=\mathbf{0}$, then$$a_0+a_1\mathbf{J}^{-1}+\cdots+a_s\left(\mathbf{J}^{-1}\right)^s=\mathbf{0}$$Multiplying both sides by $\mathbf{J}^s$, we have$$a_s+a_{s-1}\mathbf{J}+\cdots+a_0\mathbf{J}^s=\mathbf{0}$$Let $q(x)=a_s+a_{s-1}x+\cdots+a_0x^s$. Since $t$ $=$ degree of $m_\mathbf{J}(x)$ $\le$ degree of $q(x)$ $=$ $s$, we have shown that $t$ = degree of $m(x)$ $\le$ degree of $p(x)$.$\qquad\mathbf{Q.E.D}$ AI: Hint: The minimal polynomial of $\mathbf {J}_t(\lambda)$ is $(x-\lambda)^t$, thus $(\mathbf J-\lambda \mathbf I_{t\times t})^t=\mathbf 0_{t\times t}$. Multiply this equality by $(\lambda \mathbf J)^{-t}$.
H: The MLE of a $N(\theta, 1)$ distribution I am trying to find the Maximum Likelihood Estimator of an i.i.d. sample $X_1, \ldots, X_n$ arising from the model $N(\theta, 1)$, where $\theta \in [0,\infty)$. I have done this problem previously where the mean was not restricted to be non-negative, and found the MLE to be equal to the sample mean (as you would expect). Please could you explain why this situation is different and how it should be approached, as this is confusing me quite a bit! Many thanks. AI: The maximum likelihood estimate is that value in the parameter space such that the likelihood is maximized. In the unrestricted case the MLE should be the sample mean. But in the restricted case the MLE cannot be negative as it will not belong to the parameter space then. Also the likelihood will be maximized at the sample mean. However if the sample mean is negative then the value in $[0,\infty)$ where the likelihood is maximized is $0$. hence the MLE is the sample mean if the sample mean is positive, otherwise the MLE is $0$.
H: Is every theorem of PA true in the standard model of number theory $N$? My understanding is that every theorem $\phi$ of $PA$ is true in $N$ because $N$ is a model for $PA$, $N\models PA$. By completeness of first order logic, "$PA\vdash\phi$" implies that "if $N\models PA$ then $N\models \phi$". Hence $\phi$ is true in $N$, $N\models \phi$. But I was confused upon reading the following from Godel's Theorem: An Incomplete Guide to Its Use and Abuse, p31: We know that there are consistent theories extending PA that prove false mathematical statement ... So we have no mathematical basis for concluding that (say) the twin prime conjecture is true from the two premises "PA is consistent" and "PA proves the twin prime conjecture". According to my understanding above, "PA is consistent" and "PA proves the twin prime conjecture" are enough for concluding truth of the conjecture (in the standard model). Theories extending PA may prove false theorem (relative to the standard model), but surely this is not the case for PA? AI: First a minor remark: I personally prefer to refer to the implication that you use in (2) as soundness and to reserve the term completeness for the other direction. It takes no Gödel to prove soundness ;) You are correct in the mathematics before the quotation. I think the point that the author tries to make in that quotation is merely that the truth of theorems of PA is not a consequence of the consistency of PA. He does not question the truth of theorems of PA. Note that you also never invoke the consistency of PA in your argument, but the (stronger, because of soundness) statement that the natural numbers are a model. Indeed, as Franzén points out, if you have a statement $\phi$ that is true in $N$ but not provable in $PA$ then $PA+\neg\phi$ is a consistent theory that proves $\neg\phi$, but this does not make $\neg\phi$ true. Truth is not a consequence of consistency. And by the way, I think that this is an excellent book.
H: $ \lim_{x \to 0^{-}} 1 + e^{1/x}$ (Definition) Using the definition of limit, I need to show that $\lim_{x \to 0^-} 1 + e^{1/x} =1$. Using the definition, I have that $x > \frac{1}{\ln \epsilon}(\|e^{1/x}\|=e^{1/x} < \epsilon)$, but this is not helping since I need a positive delta. Also, I coudn't deal with the case $\epsilon = 1$ Can you help me with it? Thanks! AI: Hint: $e^{1/x} < \varepsilon \iff \dfrac1x < \ln \varepsilon \iff \dfrac1{\ln\varepsilon} < x < 0$.
H: Existence of a prime between $ap$ and $(a+1)p$ - generalization of Bertrand's postulate Conjecture: There exists at least one prime number $p_{m}$ : $ap_{n} < p_{m} < (a+1)p_{n}$, $\forall$ $a \in \mathbb{N}$ and $\forall$ $p_{n}$ $\in \mathbb{P} $ if $(a+1)p_{n} < p_{n+1}^2$ . Is there a name for this conjecture, and has it been proven or disproven? AI: Bertrand's postulate says that there is always a prime between $n$ and $2n$ for any $n$. So for $a=1$, your conjecture is a special case of this (and you don't need that $(a+1)p_n<p_{n+1}^2$ hypothesis). Note that at the end of that article, it says it has only recently been proven that there is always a prime between $2n$ and $3n$, and between $3n$ and $4n$, so your conjecture is true for $a=2$ and $a=3$ as well (and again, without needing that extra hypothesis). I don't know about the general case (or if there is a name for it), but you might want to start by seeing if the techniques developed for those latter cases can be adapted to prove what you want.
H: Solving a problem with L'Hospital's rule $$\lim _{x \to 2 \pi} (\cos x) ^{1/\sin ^22x} $$ I know that I should get it into $\frac{f(x)}{g(x)}$ form. How should I do it? AI: Hint: $$\ln \left((\cos x)^{1/\sin^2 2x}\right) = \dfrac{\ln \cos x}{\sin^2 2x}.$$
H: Algebraic representation of an absolute value. I tried several ways, but i could not come up with any way to have an equation as such: \|n\| = ... without using the absolute value signs on the right side of the equation. I do not know if there is any way... I tried using some form of n * i^(an + b) but those efforts were futile. And I couldn't think of any other way to approach this. AI: One way is to note that the square root has "branches": $$(\sqrt n)^2=n\tag 1$$ $$\sqrt{n^2}=\|n\|\tag 2$$ Note that $(1)$ would dip into the imaginary numbers for a moment if $n\lt 0$. $(2)$ is true because the square root of a number is one of two solutions to the equation involving the square. $$x^2=4,x=2,-2$$ $$\sqrt 4=2,-\sqrt 4=-2$$
H: How to prove this just by using Natural Deduction? I need your help to prove this by using Natural Deduction: $$(\exists x)(p(x) \implies q) \dashv\vdash (\forall x)(p(x) \implies q).$$ I want to show the proof for both sides. It is a bit easy for me to get the first side from the second side. How can I get the right side from the left side by just using Natural Deduction. Any kind of help will be appreciated. AI: The right hand side does not follow from the left hand side. If $p(x_1)$ is false, $p(x_2)$ is true, and $q$ is false, then $p(x_1) \implies q$ is true but $p(x_2) \implies q$ is false.
H: Multiply a matrix by its transpose I'm using some material found on the internet for learn how to use R. One exercise is asking to return the result of a multiplication of a $15\times 3$ matrix by its transpose. Being the transpose a $3\times 15$ matrix, I would assume the result is a matrix of dimension $15\times 15$. Am I wrong? The code suggested on the answers return a matrix of dimension $3\times 3$. I think I found the right code but before email the professor I would like to check with you if I'm right. Thank you in advance. AI: You are not wrong. If you work through matrix multiplication, you will see that the result always will have the same number of rows as the left factor and the same number of columns as the right factor. So if $A$ is an $m\times n$ matrix, $AA^T$ will have $m$ rows and $m$ columns, while $A^T\!A$ will have $n$ rows and $n$ columns.
H: First Order Differential Equations I am having trouble isolating the $x$ and $y$ to separate side in the differential equations below. Could someone give me a hint as to how to to this. Equation 1: $$\frac{dy}{dx} - \frac{x}{y} = \frac{1}{x}$$ Equation 2: $$xy\frac{dy}{dx} = y^2$$ AI: The first is not separable. For the second, we have: $$xy\frac{dy}{dx} = y^2$$ Dividing, we have: $$\dfrac{y~dy}{y^2} = \dfrac{dx}{x}$$ This is separable and we can now integrate each side as: $$\int \dfrac{1}{y}~ dy = \int \dfrac{1}{x}~ dx$$ I think you can take it from here.
H: How to find angle $v$ in a rectangular diagonal I am unable to find a way to find the angle $v$ (in degrees) in a rectangular diagonal. Here's what I have: Opposite: $15.1$ m Hypotenuse: $23.5$ m Adjacent: $x$ $v$ = measure (in degrees) of the angle opposite to the 'Opposite' side Thanks a lot in advance! AI: HINT: Remember that $\sin(v) = \frac{O}{H}$.
H: There are only two types of groups of order $6.$ There are only two types of groups of order $6.$ Could anyone advise on how to prove a/m claim? Here is my attempt but I'm stuck: If $\exists g\in G$ such that $o(g) =6,$ then $G = \left \langle {g}\right \rangle.$ If not, let $G = \{g_1,g_2,g_3,g_4,g_5,e\},$ where $o(g) \in \{2,3\} , \forall g\in G-\{e\}$ Also, $\exists i \in \{1,2,3,4,5\}$ such that $o(g_i)=2.$ AI: There's an element $a$ of order two and an element $b$ of order three (Cauchy). If they commute, then $ab$ is of order $6$ and $G$ s cyclic. Otherwise, the elements $1,a,b,b^2,ab,ba$ are pairwise distinct. One of them must be $ab^2$ and $ba$ is the only candidate for that. This determines $G$ completely.
H: Why $p \leftrightarrow q$ is equivalent to $(p \wedge q) \vee (\neg p \wedge \neg q)$? Without using the truth table I want to know why $p \leftrightarrow q$ is equivalent to $(p \wedge q) \vee (\neg p \wedge \neg q)$? Without using the truth table. Thanks all AI: Just think about the statement. $p \leftrightarrow q$: This says that $p$ occurs only if $q$ occurs and that $q$ happens only if $p$ does. Meaning, that either they both happen or nothing happens at all. But look at my last sentence. They BOTH happen OR NEITHER happens. They both happen is $p \wedge q$. They both DON'T happen is $\neg p \wedge \neg q$. So either they both happen or they both don't is $$(p \wedge q) \vee (\neg p \wedge \neg q)$$
H: Express lattice axioms using implication and universal quantification I'd like to ask for some help with homework. My task is to express lattice axioms in signature $(\leq, =, \sup, \inf)$ using only implication and universal quantification. Here are these axioms in $(\leq, =)$ signature: $\forall x (x \leq x)$ - reflexivity $\forall x \forall y \forall z (x \leq y \land y \leq z \to x \leq z)$ - transitivity $\forall x \forall y (x \leq y \land y \leq x \to x = y)$ - antisymmetry $\forall x \forall y \exists z (x \leq z \land y \leq z \land (\forall w (x \leq w \land y \leq w \to z \leq w)))$ - supremum $\forall x \forall y \exists z (z \leq x \land z \leq y \land (\forall w (w \leq x \land w \leq y \to w \leq z)))$ - infimum I believe I've done the first 3 axioms right: reflexivity is the same $\forall x \forall y \forall z ((\sup(x, y) \leq \sup(y, z) \to y \leq z) \to x \leq z)$ - transitivity $\forall x \forall y (\sup(x, y) = \inf(x, y) \to x = y)$ - antisymmetry I've stuck with supremum and infimum. I've thought about something like that: $\forall x \forall y \forall z (\sup(x, y) = z \to \forall w (x \leq w \land y \leq w \to z \leq w))$ But then again I should somehow get rid of conjunction. How can I do it? Thank you. AI: It is easy to see that for any three formulas $\phi, \psi$ and $\theta$ the formula $\phi \to \psi \to \theta$ is equivalent to $\phi \land \psi \to \theta$ (where $\phi \to \psi \to \theta$ means $\phi \to (\psi \to \theta)$, i.e. $\to$ associates to the right). Using this we can rewrite, for example, antisymmetry as $$ \forall x \forall y (x \leq y \to y \leq x \to x =y) $$ and transitivity to $$ \forall x \forall y \forall z(x \leq y \to y \leq z \to x \leq z) $$ The supremum axiom would thus be, following your start attempt, written as $$ \forall x\forall y \forall z(\sup(x,y)=z \to \forall w(x \leq w \to y \leq w \to z \leq w)) $$ but you could simplify this by getting rid of the $z$ in the quantification if $\sup$ is a function symbol, i.e. you could state $$ \forall x\forall y \forall w(x \leq w \to y \leq w \to \sup(x,y) \leq w)) $$ but note that these to are not enough to characterize suprema on their own, you also need, for instance $$ \forall x\forall y (x \leq \sup(x,y)) $$ and $$ \forall x\forall y (y \leq \sup(x,y)). $$
H: Proving the cotangent function is uniformly bounded on the complex plane I'm trying to prove that the function $\cot\left(z\right)=i\frac{e^{iz}+e^{-iz}}{e^{iz}-e^{-iz}}$ is uniformly bounded in the complex plane outside $\varepsilon$ neighborhoods of the poles (with the bound depending on $\varepsilon$). The suggested method in my text is to first show that if $z=x+iy$ and $y>0$ then: $$\frac{e^{-2y}}{1+e^{-2y}}<\left\|\cot\left(x+iy\right)+i\right\|<\frac{e^{-2y}}{1-e^{-2y}}$$ And if $y<0$ then:$$\frac{e^{2y}}{1+e^{-2y}}<\left\|\cot\left(x+iy\right)+i\right\|<\frac{e^{2y}}{1-e^{-2y}}$$ The calculations are a bit tedious and they aren't coming out right for me for some reason: \begin{align} \left\|\cot\left(x+i\cdot y\right)+i\right\| &=\left\|i\cdot\frac{e^{iz}+e^{-iz}}{e^{iz}-e^{-iz}}+i\right\| \\ &=\left\|i\cdot\frac{e^{2iz}+1}{e^{2iz}-1}+i\right\| \\ &=\left\|\frac{i\cdot\left(e^{2iz}+1\right)+i\left(e^{2iz}-1\right)}{e^{2iz}-1}\right\| \\ &=\left\|\frac{2ie^{2i\left(x+i\cdot y\right)}}{e^{2i\left(x+i\cdot y\right)}-1}\right\| \\ &=\left\|2i\right\|\cdot\left\|e^{2iz}\right\|\left\|\frac{1}{e^{2iz}-1}\right\|=2\cdot e^{-2y}\cdot\left\|\frac{1}{e^{2iz}-1}\right\| \\ &=2\cdot e^{-2y}\cdot\left\|\frac{e^{2y}}{e^{2ix}-e^{2y}}\right\| \\ &=2\left\|\frac{1}{e^{2ix}-e^{2y}}\right\| =2\frac{1}{\left\|e^{2ix}-e^{2y}\right\|} \\ &=2\cdot\frac{1}{\left\|e^{ix}-e^{y}\right\|\left\|e^{ix}+e^{y}\right\|} \\ &=\frac{2}{\sqrt{\left(e^{y}+\cos\left(x\right)\right)^{2}+\sin^{2}\left(x\right)}\cdot\sqrt{\left(e^{y}-\cos\left(x\right)\right)^{2}+\sin^{2}\left(x\right)}} \\ &=\frac{2}{\sqrt{e^{2y}+2e^{y}\cos\left(x\right)+1}\cdot\sqrt{e^{2y}-2e^{y}\cos\left(x\right)+1}} \\ &=\frac{2}{\sqrt{e^{4y}+2e^{2y}-4e^{2y}\cos^{2}\left(x\right)+1}} \\ &=\frac{2}{\sqrt{\left(1+e^{2y}\right)^{2}-4e^{2y}\cos^{2}\left(x\right)}} \end{align} The denominator is maximal when $\cos^{2}\left(x\right)=1$ and minimal when $\cos^{2}\left(x\right)=0$ and thus: $$\frac{2e^{-2y}}{1+e^{-2y}}=\frac{2}{1+e^{2y}}\leq\left\|\cot\left(x+i\cdot y\right)+i\right\|\leq\frac{2}{\sqrt{\left(e^{2y}-1\right)^{2}}}=\frac{2}{\left\|1-e^{2y}\right\|}=\frac{2e^{-2y}}{1-e^{-2y}}.$$ I can't figure out whether I made an error in the calculations or whether there was an error in the suggested bound. I'm also not sure how to use these bounds in order to reach the required conclusion. Regardless of this method I'm also curious whether someone has an alternative and perhaps less technical method of proving the claim. Thanks in advance! AI: $\cot(z) = \cos(z)/\sin(z)$. The poles are at multiples of $\pi$, so you want to prove that for any $\epsilon > 0$ there is $M$ such that $\|\cos(z)\| \le M \|\sin(z)\|$ whenever $\|z - n \pi\| \ge \epsilon$ for all integers $n$. By periodicity, it suffices to look at the strip $S = \{z: 0 \le \text{Re}(z) \le \pi\}$. After dealing with a compact set (on which a continuous function is bounded), it suffices to look at $\{z \in S: \|\text{Im}(z)\| \ge 1\}$. But if $y \ge 1$, $\|e^{i(x+iy)}\| = e^{-y} \le e^{y} e^{-2} = e^{-2} \|e^{-i(x+iy)}\|$, and then...
H: Does statement 1 imply statement 2? 1) (For some $t, P(t).) \implies Q$. 2) For all $t, (P(t) \implies Q).$ I think so, and my reasoning is this: for Q to be true, we just need P to be true for some t. Therefore, over the range of all possible t's, once P is true, Q is immediately true. AI: Yes. (1) "If (there exists a $t$ such that $P(t)$ holds) then Q." IMPLIES (2) "For any t, (if $P(t)$ holds then also $Q$ holds)." This is because we are ensured in the first proposition that, from the existence of any $t$ such that $P(t)$, it follows that $Q$. And this amounts to saying, $\forall t( P(t) \rightarrow Q)$.
H: What am I doing wrong? I am trying to prove the integral test for series, but got a strange result. Assume that $f$ is decreasing and positive. Because the series can be imagined as the area-sum of $1$-wide rectangles of height $f_n$ and each of those rectangles can be expressed as a constant integral on a $1$-wide interval where $f_n=f(n)$ is the constant when integrating $dx$: $$\sum_{n=1}^\infty f_n = \sum_{n=1}^\infty \int_n^{n+1}f(n)\,dx$$ Instead of $f(n)$ we can use the function value $f(x)$ from either end of the 1-wide interval so either floor or ceiling of $x.$ If we use the floor then the height is the left of the interval in this case always $n.$ If we use the ceiling then the height is the right of the interval in this case always $n+1$ so we should start from $n=0$ not $n=1$ to get the same sum: $$\sum_{n=0}^\infty\int_n^{n+1}f(\lceil x\rceil)\,dx = \sum_{n=1}^\infty f_n = \sum_{n=1}^\infty\int_n^{n+1} f(\lfloor x\rfloor)\, dx$$ Obviously, $$a_1 + \sum_{n=1}^\infty a_{n+1} = \sum_{n=0}^\infty a_{n+1}.$$ Therefore, we have $$f(1) + \sum_{n=1}^\infty \int_n^{n+1} f(\lceil x\rceil)\, dx = \sum_{n=1}^\infty f_n = \sum_{n=1}^\infty \int_n^{n+1} f(\lfloor x\rfloor)\, dx.$$ Since we have a function of $x$, we can concatenate the sum of integrals into a single integral: $$f(1) + \int_1^\infty f(\lceil x\rceil)\, dx = \sum_{n=1}^\infty f_n = \int_1^\infty f(\lfloor x\rfloor)\, dx$$ Because $f$ is decreasing we know that $f(\lceil x\rceil)\leq f(x)\leq f(\lfloor x\rfloor),$ therefore $$f(1) + \int_1^\infty f(x)\, dx \leq \sum_{n=1}^\infty f_n \leq \int_1^\infty f(x)\, dx$$ $$f(1) + \int_1^\infty f(x)\, dx \leq \int_1^\infty f(x)\, dx$$ $$f(1) + A \leq A$$ What happened? $f$ was positive! AI: I'm afraid you went wrong when you introduced the inequality at the very end. Indeed, $f(\lceil x\rceil)\le f(x)$ since $f$ is decreasing, so since $$f(1) + \int_1^\infty f(\lceil x\rceil)\, dx = \sum_{n=1}^\infty f_n$$ then $$f(1) + \int_1^\infty f(x)\, dx\geq f(1) + \int_1^\infty f(\lceil x\rceil)\, dx = \sum_{n=1}^\infty f_n.$$ Also, since $f(x)\le f(\lfloor x\rfloor)$ and $$\sum_{n=1}^\infty f_n = \int_1^\infty f(\lfloor x\rfloor)\, dx,$$ then $$\sum_{n=1}^\infty f_n = \int_1^\infty f(\lfloor x\rfloor)\, dx\ge\int_1^\infty f(x)\,dx.$$ You absolutely had the right idea. You just misapplied the inequalities.
H: How do I calculate the area of a rectangular diagonal I need to find a way to calculate the Area of a rectangular diagonal, this is what I have so far: Opposite: 15.1 m Hypotenuse: 23.5 m Adjacent: x EDIT: The picture was wrong. AI: Given your link to the picture of a "rectangular diagonal," it looks like you are given the dimensions of one of the right triangles formed by the diagonal of the rectangle. To solve for $x$, we use the Pythagorean Theorem which pertains to right-triangles: and note that $$\text{Hypotenuse Length}^2 = \underbrace{\text{Adjacent Length}^2}_{\large x^2} + \text{Opposite Length}^2$$ Then once you've found $x$, you can multiply $(x \times \;\text{Opposite Length})$ to determine the area of the rectangle.
H: Arranging Prime Factors to form Integer Solutions I have a problem as such: How many solutions in positive integers are there to the equation $x_1 \cdot x_2 \cdot x_3 \cdot x_4 = 2^{20} \cdot 13^{13}$? Let $x_1,\ldots,x_4$ all be distinguishable, so $x_1=a,x_2=b,x_3=c,x_4=d$ is distinct from $x_1=d,x_2=c,x_3=b,x_4=a$. I realized that $2^{20} \cdot 13^{13}$ has a unique prime factorization that is, of course, $2^{20} \cdot 13^{13}$. So all positive integer solutions must be products of these terms. However, I'm pretty stumped as to how to 'arrange' all the $33$ integers to form all possible factors $x_1,x_2,x_3,x_4$. Can anyone help me out? AI: You are looking for pairs of solutions in nonnegative integers to the following system of equations: $$ \alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 = 20$$ $$ \beta_1 + \beta_2 + \beta_3 + \beta_4 = 13$$ Then you have $x_i=2^{\alpha_i}13^{\beta_i}$. More details, as requested. Here's one example: $\alpha_1=18, \alpha_2=0, \alpha_3=0,\alpha_4=2, \beta_1=1, \beta_2=9, \beta_3=2,\beta_4=1$. This corresponds to $x_1=2^{18}13^1, x_2=2^013^{9}, x_3=2^{0}13^{2}, x_4=2^213^1$. When we multiply $x_1x_2x_3x_4=2^{18+0+0+2}13^{1+9+2+1}=2^{20}13^{13}$. These are called weak compositions, so your final answer is $${23\choose 3}{16\choose 3}=991760$$
H: Free abelian group has a subgroup of index n A nonzero free abelian group has a subgroup of index n for every positive integer n. Proof: Consider the free abelian group F(S), where S is the set of generators. Let x be some element of S, and let xn be some element of S'. Then F(S') is a subgroup of F(S), and F(S)/F(S') is isomorphic to Z/nZ. So showing this answers my question right. By saying its isomorphic we see that the abelian group has a subgroup of index n? AI: The way is the good one, but your answer is not correct. But since you're really near to the solution I gonna give you some step to complete the proof. Try to write down explicitly what are $S$ and $S'$; motivate formally why $F(S)/F(S')$ should be $\mathbb Z/n \mathbb Z$. If you complete all the step above you should be able to build up an good proof of your claim. Anyway I would suggest to avoid using the notation $F(S')$ to denote the subgroup of $F(S)$ generated by some subset $S'$ of $F(S)$. It would be better if you denote that subgroup with $\langle S' \rangle$, or even $\langle x_1,x_2,x_2,\dots,x_n\rangle$ (when $S'=\{x_1,\dots,x-n\}$). Feel free to ask if you have any other problem.
H: How to derive this second derivative using the quotient rule? If a given first derivative is: $\ {dy \over dx} = {-48x \over (x^2+12)^2} $ What are the steps using the quotient rule to derive the second derivative: $\ {d^2y \over dx^2} = {-144(4-x^2) \over (x^2+12)^3} $ My Steps: \begin{align} {d^2y \over dx^2} &= {-48(x^2 +12)^2 - 2(x^2+12)(2x)(-48x)\over (x^2+12)^4} \\ &=-48{(x^2 +12)^2 - 4x^2(x^2+12)\over (x^2+12)^4} \\ &= -48{(x^2 +12)( - 4x^2 +(x^2+12))\over (x^2+12)^4} \\ &= -48{(x^2 +12)( - 4x^2)\over (x^2+12)^3} \\ &= {???} \end{align} AI: We can factor out the constant term to make life easier (and just multiply by that $-48$ at the end of our calculation) as: $$-48 \dfrac{x}{(x^2+12)^2}$$ This makes it easier to use the quotient and chain rule (I will assume you know these). The derivative of $\ {dy \over dx} = {-48x \over (x^2+12)^2} $, using the quotient and chain rule is: $\dfrac{d^2y}{dx^2} = -48 \dfrac{(1)(x^2+12)^2 - 2(x^2+12)(2x)x}{(x^2+12)^4} = -48 \dfrac{(x^2+12) - 2(2x)x}{(x^2+12)^3} = \dfrac{-144(4-x^2)}{(x^2+12)^3}$
H: Prove Heine-Borel Theorem Prove Heine-Borel Theorem: "A subset $S$ of $\mathbb{R}$ is compact if and only if every open cover for $S$ has a finite subcover." Suggestions: Let $S \subset \mathbb{R}$. If every open cover for $S$ has a finite subcover, then $S$ must be compact. Why? Now, assume that $S$ is compact and let $\mathcal O$ be any open cover for $S$. Let $\epsilon > 0$ so that for all x in $S$, there is come $E \in \mathcal O$ such that $D(x,\epsilon) \subset E$. Assume that S is totally bounded. AI: This is too long for a comment, so I'll make it an answer. I'm not sure if I understand your question. There seems to be a typo. As you know, it is not the case that every open cover of $\mathbb{R}$ has a finite subcover. If $S$ is unbounded, it is easy to find an open cover (that covers $\mathbb{R}$, not just $S$) with no finite subcollcection covering $S$. If $S$ is not closed, $S$ has a limit point $x$ that doesn't belong to $S$, and it is easy to construct an open cover of $\mathbb{R} \setminus \{x\}$ (which contains $S$) with no finite ssubcollection covering $S$. The existence of $\epsilon$ that you are using in your argument is far from obvious (since $\epsilon$ does not depend on $x$), though such an $\epsilon$ does exist. Are you allowed to assume this $\epsilon$ exists, or do you need to prove it? You only need it if you are going to use the idea of sequential compactness, which is not necessary for what you seem to be trying to prove. Proving the other direction, that any closed and bounded subset of $\mathbb{R}$ is compact, is more difficult, but not terribly difficult. You can probably Google it and find a complete, excellent proof faster than you will get one here. You will also find proofs of the easier direction, as I describe above. It would also not surprise me if this question and been asked and answered in this forum before.
H: Find $a\in\mathbb{N}$ such that $n^4+a$ is not prime $\forall n\in\mathbb{N}$ How would I go about finding such an $a$? I've been thinking it is something to do with modular arithmetic, but don't know what base to consider. AI: $n^4 + b^4 = (n^2-\sqrt{2}nb+b^2)(n^2+\sqrt{2}nb+b^2)$. Set $b = 2\sqrt{2}$. Then, $n^4 + 64= (n^2-4n+8)(n^2+4n+8)$. (... and it can be verified that both factors are greater than $1$ for every $n\geq 1$)
H: Random variable of twice another random variable If Y is a random variable with a mean $\mu$ and a standard deviation, $\sigma$, how do I calculate W if W = 2Y? Is it just $2\mu$ and $2\sigma$? AI: E(X) = mu E(W) = cE(X) = cmu VAR(W) = VAR(cX) = c^2VAR(X) SD(W) = cSD(X)
H: How can I get the limit of $(-1)^{2n} $ when $n$ goes to infinity How can I get the limit of $(-1)^{2n}$ when $n$ goes to infinity? I checked it with wolframalpha but the result is $e^{2i}$, $0$ to $\pi$, why $i$??? AI: Note that $(-1)^{2n}=((-1)^2)^n=(1)^n$ As $n$ tends toward infinity, what happens now?
H: Using Weyl sequences to prove relation between quadratic form and spectral radius I know that the formula $$\lVert A\lVert=\sup_{\lVert x\lVert=1} \langle x,Ax\rangle$$ holds true for self adjoint operators. While reading Teschl's book I saw a comment that on can prove this formula for normal operators using Weyl sequences. It's clear that $\lVert A\lVert=\sup_{\lambda\in\sigma(A)}\|\lambda\|$ for a normal operator. Since one of the inequalities is trivial it would remain to show that for a given $\lambda\in\sigma(A)$ one has $$\|\lambda\|\leq\sup_{\\|x\\|=1}\langle x,Ax\rangle.$$ By Weyl's criterion I get a sequence $x_n$ with $\\|x_n\\|=1$ and $\\|(A-\lambda)x_n\\|\to0$. Are those $x_n$'s sufficient for the inequality above? Any ideas about how to use this sequence to prove above's inequality? I tried to expand the norm but then didn't no what to do about the $\\|A x_n\\|^2$ term. AI: Note that $\langle x_n,x_n\rangle=1$. So $$ \|\langle Ax_n,x_n\rangle-\lambda\|=\|\langle (A-\lambda)x_n,x_n\rangle\|\leq\\|(A-\lambda)x_n\\|\to 0 $$
H: Isomorphic Gaussian Integers Can we show that the ring of Gaussian integers $$\mathbb{Z}[\sqrt{17}]:=\{a+b\sqrt{17}:a,b\in\mathbb{Z}\}$$ $$\mathbb{Z}[\sqrt{11}]:=\{a+b\sqrt{11}:a,b\in\mathbb{Z}\}$$ equipped with standard addition and multiplication are not isomorphic? AI: An isomorphism of rings would map $1$ to $1$. Hence it would fix all integers. In $\mathbb Z[\sqrt{17}]$ we have $(\sqrt{17})^2 = 17$. It is easy to check that no element $x$ of $\mathbb Z[\sqrt{11}]$ satisfies $x^2 = 17$.
H: Find number of solutions of the equation $ x_{1}+x_{2}+x_{3} = 41$, where $x_{1}, x_{2}\ \text{and}\ x_{3}$ are odd and non negative integers There are two constraints to this problem: $x_{1}, x_{2}\ \text{and}\ x_{3}$ are non negative integers $x_{1}, x_{2}\ \text{and}\ x_{3}$ are odd If there had been just the first constraint (non negative integer), i would have simply used "bars and stars" method and found the answer as: $$^{43}C_2 [41 + 2\ \text{bars}] = 903$$ But since there is another requirement that all the non-negative integral solutions must be odd as well, even after trying different approaches, am unable to solve further. Please help me out! Thanks a lot! AI: Look, according to your second constraint, $x_1$, $x_2$ and $x_3$ have the following forms: $x_1=2y_1 + 1$, $x_2=2y_2 + 1$ and $x_3=2y_3 + 1$ with $y_1$, $y_2$, $y_3$ non-negative. So, substituting them in the equation and simplifying we have, $y_1 + y_2 + y_3 = 19.$ Then we can apply the “bars and stars” method and get the solution ${21 \choose 2} = 210$.
H: Where am I going wrong on this second derivative? If a given first derivative is: $\ {dy \over dx} = {-48x \over (x^2+12)^2} $ What are the steps using the quotient rule to derive the second derivative: $\ {d^2y \over dx^2} = {-144(4-x^2) \over (x^2+12)^3} $ My Steps: $$ {{d^2y \over dx^2} = {-48(x^2 +12)^2 - 2(x^2+12)(2x)(-48x)\over (x^2+12)^4} =-48{(x^2 +12)^2 - 4x^2(x^2+12)\over (x^2+12)^4} = -48{(x^2 +12)( - 4x^2 +(x^2+12))\over (x^2+12)^4} = -48{(x^2 +12)( - 4x^2)\over (x^2+12)^3} = ???} $$ Once I get to this point I am unsure how to derive the second derivative shown. AI: $$y'=-\frac{48x}{(x^2+12)^2}\Rightarrow y''=\left(-\frac{48x}{(x^2+12)^2}\right)'=-48\left(\frac{x}{(x^2+12)^2}\right)'$$$$=-48\frac{x'\cdot(x^2+12)^2-x((x^2+12)^2)'}{(x^2+12)^4)}=-48\frac{(x^2+12)^2-4x^2(x^2+12)}{(x^2+12)^4}$$ $$=-48\frac{(x^2+12)(x^2+12-4x^2)}{(x^2+12)^4}=-48\frac{12-3x^2}{(x^2+12)^3}=-48\cdot 3\frac{4-x^2}{(x^2+12)^3}=-144\frac{4-x^2}{(x^2+12)^3}$$
H: Maximum N that will hold this true Find the largest positive integer $N$ such that $$\sqrt{64 + 32^{403} + 4^{N+3}}$$ is an integer Is $N = 1003$? AI: Note that with $N=2008$ we have $ (2^{N+3}+8)^2=4^{N+3}+2\cdot 8\cdot 2^{N+3}+64=4^{N+3}+2^{2015}+64=64+32^{403}+4^N,$ so we conjecture that the maximal value is $2008$. If $2^{2015}+2^6+2^{2N+6}$ is a perfect square then also $\frac1{64}$ of it, i.e. $2^{2009}+1+2^{2N}=m^2$ for some $m\in\mathbb N$. But if $N> 2008$, we have $$(2^N)^2=2^{2N}<2^{2009}+1+2^{2N}<1+2^{N+1}+2^{2N}=(2^N+1)^2$$ so that $2^N<m<2^N+1$, which is absurd.
H: Prime gaps with respect to the squared primes Conjecture If we have two consecutive prime numbers $p_{a}$ and $p_{a+1}$, and two other consecutive primes $p_n$ and $p_{n+1}$, so that $p_{a} < p_{a+1} < p^2_{n+1}$, then $p_{a+1} - p_{a} < 2p_{n} $. Are there any known counter examples and are there any known similar conjectures? AI: No known counterexamples. A slightly stronger conjecture, true as far as anyone has been able to check (up to $4 \cdot 10^{18}$) is that $$ \unicode{x2E2E} \unicode{x2E2E} \unicode{191} \unicode{191} \; p_{a+1} < p_a + 2 \; \sqrt {p_a} \; \unicode{63} ? $$ This is currently unprovable. What people actually suspect is that, $$ \unicode{191} ¿ \; \mbox{if} \; \; p_a \geq 11, \; \; \mbox{then} \; \; p_{a+1} < p_a + \log^2 {p_a} \; ? $$ Really, really beyond proof. See Prime pair points slope approaches 1 Note: as far as using the extensive tables of prime gaps, there is a detail involving the fact that $\log^2 x > 2 \sqrt x$ for an interval, roughly $19.6 < x < 187.8.$ So, it was necessary to make a separate confirmation of my version of the conjecture for $p_a < 188.$
H: If $n^2+10$ is odd then $n$ is odd. I have been having a little trouble with proofs, and would like to ask for a hint for this statement. I've already started it, however I'm unsure what I should be doing next. Prove that if $n^2 + 10$ is odd then $n$ is odd. My answer so far: Suppose that $n$ is an odd integer, and we want to prove $n^2 + 10$ is odd. There exists a $k$ that $n=2k+1$. By substituting for $n$, we get... $(2k+1)^2 +10$ = $n^2 = 10$ $4k^2 + 4k + 11$ = $n^2 +10$ I am unsure what to do next. EDIT: Thank you all for your hints and answers! They were all brilliant and my understanding for all kinds of proofs is a heck lot better! So much appreciated! AI: Unfortunately, you're working backward. You've (very nearly) proved that if $n$ is odd, then $n^2+10$ is odd. Now, if you wanted a proof by contrapositive (the simplest way, really), then suppose that $n$ is even, and prove that $n^2+10$ is also even. For a direct proof, assume that $n^2+10$ is odd, and prove that $n^2$ is then odd, and so $n$ is odd (why?). For a proof by contradiction, assume that $n^2+10$ is odd, but $n$ is even, and derive an absurdity.
H: Counting ways to partition a set into fixed number of subsets Suppose we have a finite set $S$ of cardinality $n$. In how many ways can we partition it into $k$-many non empty subsets? Example: There is precisely one way to partition such a set into $n$-many subsets. and there is one way to partition into a single (sub)set. AI: What you want is the Stirling numbers of the second kind. A Stirling number of the second kind counts the number of ways to partition a set of $n$ objects into $k$ non-empty subsets. It is usually denoted by $\left\{ n \atop k \right\}$. See the Wikipedia article for methods to compute such numbers.
H: Let $f(x) \in F[x]$ and assume that $f(x)\|g(x)$ for every nonconstant $g(x) \in F[x]$. Show that $f(x)$ is constant because $f(x)\|g(x)$ $f(x)$ must share at least one root with $g(x)$ and because $g(x)$ could be any degree polynomial that don't share the same roots (ex: $x+1$, $x-2$) $f(x)$ must be a constant value. Do I have the right idea with this one? AI: Your idea is not sufficient as $F$ need not be algebraically closed. For example $X^2+1\in\mathbb R[X]$ or $X^2-2\in\mathbb Q[X]$ have no roots but are not constant. But if we know that $f\|X$ and $f\|X+1$, then also $f\|(X+1)-X=1$, which implies $\deg f\le\deg 1=0$.
H: In the proof that in a PID, every non-zero non-unit is the product of irreducibles... In proving that all non-zero non-units of a PID are a product of irreducibles, theres: "We now show that $a$ is a product of irreducibles. If $a$ is irreducible, we are done. Otherwise let $p_1$ be an irreducible such that $p_1\|a$. Then $a=p_1c_1$. If $c_1$ is a unit, we are done. " How did they get that they're done? I've got $aR = p_1 c_1 R = p_1 R$ if $c_1$ is a unit. Thanks. AI: If $c_1$ is a unit, then $a=p_1c_1$ is in fact irreducible. Since we assumed that was not the case, then $c_1$ is not a unit. I suspect that what they meant to say is "If $c_1$ is irreducible, we are done."

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