qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
272,193 | <p>It is well known that the Petersen Graph is not Hamiltonian. I can show it by case distinction, which is not too long - but it is not very elegant either.</p>
<p>Is there a simple (short) argument that the Petersen Graph does not contain a Hamiltonian cycle? </p>
| Chris Godsil | 16,143 | <p>If you can use the symmetry (as Jernej suggests), the case argument has a lot going for it. </p>
<p>There is a proof using interlacing. Observe that if $P$ has a Hamilton cycle then its line graph $L(P)$ contains an induced copy of $C_{10}$.
Eigenvalue interlacing then implies that $\theta_r(C_{10}) \le \theta_r(L(... |
717,084 | <p>I'm slightly confused as to how </p>
<p>$$\{\emptyset,\{\emptyset,\emptyset\}\} = \{\{\emptyset\},\emptyset,\{\emptyset\}\}$$</p>
<p>are equivalent. I thought two sets were equivalent if and only if "$A$" and "$B$" have exactly the same elements. In this case, we have one element which is in both sets but then two... | Emanuele Paolini | 59,304 | <p>Two sets are equal if every element of one set is also an element of the other and vice-versa.</p>
<p>So you must convince yourself that $\{\emptyset,\emptyset\}=\{\emptyset\}$ and the exercise becomes very simple...</p>
<p>By the way it happens that natural numbers are usually defined by $0=\emptyset$, $n+1=\{n\}... |
1,011,560 | <p>The problem is as stated in the title. With this problem and I am restricted to modus tollens (MT), modus ponens (MP), repetition (R), and double negation (DN). I'm just getting used to logic derivation, so I've really no idea where to proceed. The issue I'm having is just not knowing how to properly start this. Thi... | Forever Mozart | 21,137 | <p>Suppose $P\to (Q\to R)$. (1)</p>
<p>We want to show $Q\to (P\to R)$.</p>
<p>Well, suppose $Q$. (2)</p>
<p>We want to show $P\to R$.</p>
<p>Suppose $P$. (3)</p>
<p>Then $Q\to R$ by 1 and 3. (4)</p>
<p>Then $R$ by 4 and 2.</p>
|
1,693,078 | <p>I have read this theorem in my book but i do not know how to prove it.<br><br>
If a family of straight lines can be represented by an equation $\lambda^2 P+\lambda Q+R=0$ where $\lambda $ is a parameter and $P,Q,R$ are linear functions of $x$ and $y$ then the family of lines will be tangent to the curve $Q^2=4PR.$<b... | Matthew Leingang | 2,785 | <p>This is a pretty brute-force method but it works.</p>
<h2>Test</h2>
<p>To figure out what was going on, I tried a simple case: $P=x$, $Q=y$, $R=1$. The family of lines is $$\lambda^2x +\lambda y + 1 = 0$$ and the quadratic is $$y^2 = 4x$$ To say that the family of lines is tangent to the curve means that whereve... |
1,693,078 | <p>I have read this theorem in my book but i do not know how to prove it.<br><br>
If a family of straight lines can be represented by an equation $\lambda^2 P+\lambda Q+R=0$ where $\lambda $ is a parameter and $P,Q,R$ are linear functions of $x$ and $y$ then the family of lines will be tangent to the curve $Q^2=4PR.$<b... | Blue | 409 | <p>You are seeking the <a href="https://en.wikipedia.org/wiki/Envelope_(mathematics)#Envelope_of_a_family_of_curves" rel="nofollow">envelope</a> of the family of lines. As described in the Wikipedia entry, there is a standard approach to this task, using Calculus.</p>
<p>Write $$F(\lambda) \;=\; \lambda^2 P + \lambda ... |
1,693,078 | <p>I have read this theorem in my book but i do not know how to prove it.<br><br>
If a family of straight lines can be represented by an equation $\lambda^2 P+\lambda Q+R=0$ where $\lambda $ is a parameter and $P,Q,R$ are linear functions of $x$ and $y$ then the family of lines will be tangent to the curve $Q^2=4PR.$<b... | Narasimham | 95,860 | <p>To find envelope by C-discriminant method, we eliminate $\lambda$ between</p>
<p>$$f(\lambda) \;=\; \lambda^2 P + \lambda Q + R =0, \; f^\prime(\lambda) \;=\; 2\lambda P + Q =0 \; $$</p>
<p>( the latter is the <em>characteristic</em> for the parameter $\lambda$ )</p>
<p>yielding:</p>
<p>$$ Q^2 = 4 P R $$</p>
<p... |
897,815 | <p>how do I go forward with sketching the graphs of the following two functions?</p>
<p>i) $y(t)=|2+t^3|$</p>
<p>ii) $f(x)=4x+|4x-1|$</p>
<p>thanks in advance!</p>
| whacka | 169,605 | <p>The transitive $G$-sets are precisely those isomorphic to the coset space $G/H$ for some subgroup $H\le G$. Any of our $G$-sets decomposes as a disjoint union of orbits. So we should be able to decompose $G/H\times G/K$ into orbits. Indeed we have the decomposition</p>
<p>$$G/H\times G/K\cong\bigsqcup_{HxK}\frac{G}... |
2,793,380 | <blockquote>
<p>How many ways are there to seat n people in 4 benches so that no bench
is left empty with order?</p>
</blockquote>
<p>Hints from the teacher</p>
<ul>
<li><p>Each bench should have at least 1 person</p></li>
<li><p>This question is similar to distributing different object among n children</p></li>
... | Mr Pro Pop | 513,695 | <p>Let k = number of benches = 4</p>
<p>It is like we are distributing people to benches. If there were no restrictions, then it would have been $${n+4-1\choose 4-1}$$</p>
<p>Since we have a rule that each bench should have at least one person, then $$ x1+....+xn = 4$$</p>
<p>For any pair of positive integers n and ... |
3,276,699 | <p>If in a game of chance you have a certain probability <span class="math-container">$p$</span> to receive <span class="math-container">$x$</span> dollars else you receive <span class="math-container">$y$</span> dollars. How would you calculate the average money you would make per game? To clarify, for each game you e... | David | 651,991 | <p>You are right indeed. It is easy to reach that conclusion if you think of it in a "frequentist" way. Let's say you get a million tries:</p>
<p>On average, you get <span class="math-container">$x$</span> <span class="math-container">$1.000.000 \cdot p$</span> times</p>
<p>You get <span class="math-container">$y$</s... |
284,053 | <p>Some people have taken the view that morphisms in category theory ought to simply be called functions.</p>
<p>However, not all morphisms look like functions at first sight. For example, in categories of partial orders, such as the category in which the objects are the natural numbers and a morphism exists between t... | Uday Reddy | 28,521 | <p>[It is better to put "function" in quotes, because we don't mean here what a mathematician normally means by a function, but some more abstract, intuitive idea.]</p>
<p>Whenever we build an algebraic theory of something, we often end up with degenerate examples that fit the theory but don't look anything like what ... |
2,785,160 | <p>Suppose the inequality $\frac {1}{2}-\frac{x^2}{24}<\frac{1-\cos(x)}{x^2}<\frac {1}{2}$ then $\lim_{x\to 0}\frac{1-\cos(x)}{x^2}$ is? I already solved this question by taking limit on the inequalities the answer is $\frac{1}{2}<\lim_{x\to0}\frac{1-\cos(x)}{x^2}<\frac{1}{2}$ but I have a small doubt, as... | Hans Lundmark | 1,242 | <p>Limits don't necessarily preserve <em>strict</em> inequalities. If you know that $f(x) < g(x)$ and $f(x)\to A$ and $g(x) \to B$, then you can only conclude $A \le B$ (not $A < B$).</p>
|
2,869,753 | <p>I get that $∅$ is subset of every set thus $∅ ⊆ \{\{∅\}\}$.
However, I'm not sure if $∅ ⊂ \{\{∅\}\}$.
From definition of proper subset, the relation between two sets require the larger set to have at least one element not in the other one.
What I'm confused is, does $\{\{∅\}\}$ have an element that $∅$ doesn't have?... | Ashwani Bhat | 580,298 | <p>Let A={{∅}} be a set, then {∅}∈A.
But always remember, ∅ ≠ {∅}, the former is an empty set and the latter is the element of a set. </p>
<p>Hence set A contains one element ({∅}). And as per your definition,
<code>From definition of proper subset, the relation between two sets require the larger set to have at leas... |
438,925 | <p>There are many statements in abstract algebra, often asked by beginners, which are just <em>too good to be true</em>. For example, if <span class="math-container">$N$</span> is a normal subgroup of a group <span class="math-container">$G$</span>, is <span class="math-container">$G/N$</span> isomorphic to a subgroup ... | Exit path | 101,861 | <p>An example might be that the category of abelian groups is hereditary. That is, every complex of abelian groups is quasi-isomorphic to the (graded) direct sum of its cohomologies. Although this is very straightforward, I know of at least one person (me) who was surprised by this statement when learning homological ... |
438,925 | <p>There are many statements in abstract algebra, often asked by beginners, which are just <em>too good to be true</em>. For example, if <span class="math-container">$N$</span> is a normal subgroup of a group <span class="math-container">$G$</span>, is <span class="math-container">$G/N$</span> isomorphic to a subgroup ... | darij grinberg | 2,530 | <p>Quite a few things in the Hopf algebra world are surprising:</p>
<ul>
<li><p>Takeuchi's theorem: Every connected graded bialgebra is a Hopf algebra. (No finiteness assumptions!) Takeuchi was actually more general: If <span class="math-container">$H$</span> is an <span class="math-container">$\mathbb N$</span>-graded... |
438,925 | <p>There are many statements in abstract algebra, often asked by beginners, which are just <em>too good to be true</em>. For example, if <span class="math-container">$N$</span> is a normal subgroup of a group <span class="math-container">$G$</span>, is <span class="math-container">$G/N$</span> isomorphic to a subgroup ... | rschwieb | 19,965 | <ol>
<li><p>There is some theory about <em>maximal valuation rings</em> (a special type of ring with linearly ordered ideals, not necessarily a domain) and then there are <em>almost-maximal valuation rings</em> which is of course a relaxation. But if you look for an almost-maximal valuation ring which is not maximal, ... |
438,925 | <p>There are many statements in abstract algebra, often asked by beginners, which are just <em>too good to be true</em>. For example, if <span class="math-container">$N$</span> is a normal subgroup of a group <span class="math-container">$G$</span>, is <span class="math-container">$G/N$</span> isomorphic to a subgroup ... | Gerry Myerson | 3,684 | <p>As fields, the algebraic closures of the fields <span class="math-container">${\bf Q}_p$</span> are isomorphic, and are isomorphic to the complex numbers.</p>
|
438,925 | <p>There are many statements in abstract algebra, often asked by beginners, which are just <em>too good to be true</em>. For example, if <span class="math-container">$N$</span> is a normal subgroup of a group <span class="math-container">$G$</span>, is <span class="math-container">$G/N$</span> isomorphic to a subgroup ... | Wojowu | 30,186 | <p>A subring of a Noetherian ring need not be Noetherian.</p>
<p>Given all the stability properties that Noetherian rings enjoy, this may sound surprising at first, but it becomes much more obvious if you think about the fact that any domain can be embedded in a Noetherian ring - its field of fractions.</p>
<p>Perhaps ... |
438,925 | <p>There are many statements in abstract algebra, often asked by beginners, which are just <em>too good to be true</em>. For example, if <span class="math-container">$N$</span> is a normal subgroup of a group <span class="math-container">$G$</span>, is <span class="math-container">$G/N$</span> isomorphic to a subgroup ... | Dave Benson | 460,592 | <p>In ZFC, the complex number field has <span class="math-container">$2^{2^{\aleph_0}}$</span> automorphisms, whereas the real number field has just one, the identity.</p>
|
3,171,530 | <p>can you help me solving this problem, I got stucked answering it in the beginning and hoping that you can help me with this. Thanks! </p>
<p>~Q ∧ (P→Q) ∧ (R ∨ ~Q) </p>
| Bableen Kaur | 486,471 | <p>If we go by truth table as follows:</p>
<p><span class="math-container">\begin{array} {|r|r|r|r|r|r|r|r|r|}
\hline
P &Q &R &\neg Q &P\rightarrow Q &R\vee \neg Q & \neg Q \wedge (P\rightarrow Q) \wedge (R\vee \neg Q) \\
\hline
T &T &T &F &T &T &F \\
\hline
T &T &am... |
1,902,731 | <p>Imagine a group of 8 people, initially arranged in two rows like so:</p>
<p>1 2 3 4</p>
<p>8 7 6 5</p>
<p>How do I proceed to make sure that each person is able to have conversation with the person immediately across from them (1-8,2-7, etc) for a fixed period of time and then rotate the arrangement in such a way... | Doug Spoonwood | 11,300 | <p>Suppose we rewrite the definition of well-formed formula of M in Polish notation. Then, it follows that the correlates of the above axiom schema can get written as the following:</p>
<pre><code> level
Ax1 CaCba 0
Ax2 CCaCbcCCabCac 0
Ax3 CCNbNaCba 0
</code></pre>
<p>Formal mo... |
1,902,731 | <p>Imagine a group of 8 people, initially arranged in two rows like so:</p>
<p>1 2 3 4</p>
<p>8 7 6 5</p>
<p>How do I proceed to make sure that each person is able to have conversation with the person immediately across from them (1-8,2-7, etc) for a fixed period of time and then rotate the arrangement in such a way... | DanielV | 97,045 | <blockquote>
<p>$$P \implies Q \implies R \vdash_M Q \implies P \implies R \tag{1}$$</p>
</blockquote>
<p>A proof of this can be constructed using typed lambda calculus.</p>
<p>Assuming a lambda expression $A$ exists of type $P \to Q \to R$. Then the following lambda expression:</p>
<p>$$\lambda ~ b~c~.~A~c~b$$</... |
1,876,086 | <p>The definition of limit says that Let $f(x)$ be a function defined on some open interval that contains the number $a$, except possibly at $a$ itself. Then we say that the limit of $f(x)$ as $x$ approaches $a$ is $L$ If....{the rest of definition is left to make the question easier}. </p>
<p>What does the phrase "ex... | Michael Hardy | 11,667 | <p>An open interval that contains $a$ also contains all points sufficiently close to $a$. That is why an open interval is used in the definition.</p>
<p>"except possibly at $a$ itself" was included because the limit of the function at $a$ is determined by the behavior of the function near $a$ but not at $a$.</p>
|
836,841 | <p>Calculation of $\displaystyle \lim_{x\rightarrow 1}\frac{(1-x)\cdot(1-x^2)\cdot(1-x^3)\cdots (1-x^{2n})}{\{(1-x)\cdot(1-x^2)\cdot (1-x^3)\cdots(1-x^n)\}^2} = $</p>
<p><b>My Trial</b> After simplification, we get $$\displaystyle \lim_{x\rightarrow 1}\frac{(1-x^{n+1})\cdot(1-x^{n+2})\cdot(1-x^{n+3})\cdots(1-x^{2n})}{... | Ivo Terek | 118,056 | <p>We can use the identity: $$1 - x^m = (1-x)(1 + x + \cdots+ x^{m-1})$$
in every single factor, there. All of those $1-x$ will cancel, and we'll be left with $$\lim_{x \to 1} \frac{(1+x+\cdots+ x^n)(1+x+\cdots+ x^{n+1})\cdots (1+x+\cdots +x^{2n - 1})}{(1+x)(1+x+x^2)\cdots(1+x+x^2+\cdots +x^{n-1})} = \frac{(n+1)(n+2)\c... |
208,485 | <p>Two years ago, I made a conjecture <a href="https://math.stackexchange.com/questions/334448/1-a21-b21-c2-8abc-a-b-c-in-mathbbq-has-infinitely-many-sol/336803#336803">on stackexchange</a>:</p>
<p>Today, I tried to find all solutions in integers $a,b,c$ to
$$(1-a^2)(1-b^2)(1-c^2)=8abc,\quad a,b,c\in \mathbb{Q}^{+}.$$... | Joe Silverman | 11,926 | <p>Call (the projective completion of) your surface $S$. It admits three double covers of $\mathbb P^2$, namely $$\pi_1(a,b,c)=(a,b), \quad\pi_2(a,b,c)=(a,c),\quad\pi_3(a,b,c)=(b,c).$$ Each double cover induces an involution, so we get
three involutions $$\sigma_1,\sigma_2,\sigma_3:S\to S.$$ These involutations don't c... |
208,485 | <p>Two years ago, I made a conjecture <a href="https://math.stackexchange.com/questions/334448/1-a21-b21-c2-8abc-a-b-c-in-mathbbq-has-infinitely-many-sol/336803#336803">on stackexchange</a>:</p>
<p>Today, I tried to find all solutions in integers $a,b,c$ to
$$(1-a^2)(1-b^2)(1-c^2)=8abc,\quad a,b,c\in \mathbb{Q}^{+}.$$... | Allan MacLeod | 56,617 | <p>The original proposer asks for "simple methods". Simplicity, like beauty, is in the eye of the beholder. I am sure
that Noam Elkies and Joe Silverman feel their answers are extremely simple. The following discussion is, in my humble opinion,
simpler.</p>
<p>We can express the underlying equation as a quadratic in $... |
2,810,410 | <p>Define $B: \mathbb{R^3} × \mathbb{R^3} \to \mathbb{R}$ with $B((x_1,x_2,x_3),(y_1,y_2,y_3)) := -2x_1y_1-x_2y_3-x_3y_2$.</p>
<p>How to check if vectors $v \in \mathbb{R^3}, v \neq 0$ exist such that $B(v,v) = 0$?</p>
<p>I don't know how to start here.</p>
<p>Also, how to find a basis of $\mathbb{R^3}$ such that th... | robjohn | 13,854 | <p>If $[0\le x\le1]$ is the pdf for $x$, then the cdf for $x$ is $x\,[0\le x\le1]$. Thus, the cdf for $y=\log(x)$ is $e^y\,[y\le0]$, and therefore the pdf for $y$ is $e^y\,[y\le0]$. The pdf for the sum of $n$ values of $y$ is the $n$-fold convolution of the pdf $e^y\,[y\le0]$ with itself. The Fourier Transform of this ... |
28,321 | <p>Let $K$ be a nonempty compact convex subset of $\mathbb R^n$ and let $f$ be the function that maps $x \in \mathbb R^n$ to the unique closest point $y \in K$ with respect to the $\ell_2$ norm. I want to prove that $f$ is continuous, but I can't seem to figure out how.</p>
<p>My thoughts: Suppose $x_n \to x$ in $\mat... | joriki | 6,622 | <p>Consider two points $a$ and $b$, and for simplicity and without loss of generality, assume $f(a)=0$. Then the line through $f(a)$ and $f(b)$ can be described by $\lambda f(b)$, and all points with $0\le\lambda\le1$ are in $K$. The point nearest to some point $c$ on this line is determined by</p>
<p>$$(\lambda_c f(b... |
1,511,477 | <p>So I tried and found that $$7^{100} \equiv 1 \pmod{100}$$ but I got stuck with $8^{100}$. Help me out please. </p>
| Mikael Jensen | 71,727 | <p>Following David Cipras advice let me give this answer:
The conjugate rule applied twice gives us
$7^{100}-8^{100}= (7^{25}-8^{25})(7^{25}+8^{25})(7^{50}+8^{50})$
The first factor is 1-1 mod (25) = 0 mod (25) so the last numbers are 00,-25 or -75 mod(100) </p>
<p>And considering mod 3 we have:</p>
<p>$7^{100}-8^{... |
107,882 | <p>Can someone recommend a good basic book on Geometry? Let me be more specific on what I am looking for. I'd like a book that starts with Euclid's definitions and postulates and goes on from there to prove thereoms about triangles, circles and other plane shapes. I'm not interested (at this time) in a book that tie... | Community | -1 | <p>"Problems and Solutions in Euclidean Geometry" by M. N. Aref and W. Wernick. <a href="http://www.farzanegan2t.ir/Repositary/RadEditor/SentDoc/amozesh/ryazi/ketab%20hendese-%20tamrinat%20bishtar.pdf" rel="nofollow">Here</a> is a free e-book. </p>
<p>It features 200 problems of increasing complexity with worked-out ... |
3,090,052 | <p>My question concerns the answer to exercise 1.3:</p>
<blockquote>
<p>Given a partition <span class="math-container">$P$</span> on a set <span class="math-container">$S$</span>, show how to define a relation <span class="math-container">$\sim$</span> on <span class="math-container">$S$</span> such that <span class... | Mark Viola | 218,419 | <p><strong>HINT:</strong></p>
<p>Note that</p>
<p><span class="math-container">$$\begin{align}
\frac{n}{(2n+1)!!}&=\frac12\left( \frac{2n+1-1}{(2n+1)!!}\right)\\\\&=\frac12\left(\frac{1}{(2n-1)!!}-\frac1{(2n+1)!!}\right)
\end{align}$$</span></p>
<p>Now, telescope.</p>
|
849,619 | <p>When we draw a Venn diagram, we use circle to represent a Set. We can use any closed plane figure but most of the time it is a circle. Why? are there any specialty about that?</p>
| Squirtle | 29,507 | <p>Look at this noncircle Venn Diagram:</p>
<p><img src="https://i.stack.imgur.com/NG4wf.png" alt="Venn Diagram"></p>
<p>Source : <a href="http://www.learnnc.org/lp/media/authors/walbert/venn/venn-edwards-food.png" rel="nofollow noreferrer">http://www.learnnc.org/lp/media/authors/walbert/venn/venn-edwards-food.png</a... |
287,154 | <p>Note: I asked the question below last week on MathSE but received no answer. </p>
<p>Background:</p>
<p>I have read the claim that perverse sheaves behave more like sheaves than like complexes of sheaves. This refers to the fact that they can be glued. </p>
<p>For instance, suppose that $X$ is a complex analyti... | Geordie Williamson | 919 | <p>One can ask whether the derived category forms a stack (of triangulated categories). The answer is no:</p>
<p>Let $S^2$ denote the two-sphere (or $\mathbb{P}^1$ if you prefer) and let $k$ denote a ring of coefficients (e.g. $\mathbb{Z}$). Let $k_{S^2}$ denote the constant sheaf on $S^2$. Consider a map $\alpha: k_{... |
125,709 | <p>Suppose $\lambda = (\lambda_1,\lambda_2,.....,\lambda_k)$ is a partition of $2n$ where $n \in \mathbb N$ satisfying the following conditions:</p>
<p>(1) $\lambda_{k} = 1$.</p>
<p>(2) $\lambda_{i} - \lambda_{i+1} \leq 1$ for every $i\leq k-1$.</p>
<p>(3) In the partition $\lambda$, the number of odd parts in odd p... | Matt Fayers | 6,771 | <p>For $j\geqslant0$ let $c_j$ denote the $2$-core partition $(j,j-1,\dots,1)$.</p>
<p>Your conditions on partitions of $2n$ can be re-phrased as asking for $2$-restricted partitions of $2$-weight $n$ and $2$-core $c_0$. Now for any $j$, there is a standard way to biject between $2$-restricted partitions of weight $n$... |
125,709 | <p>Suppose $\lambda = (\lambda_1,\lambda_2,.....,\lambda_k)$ is a partition of $2n$ where $n \in \mathbb N$ satisfying the following conditions:</p>
<p>(1) $\lambda_{k} = 1$.</p>
<p>(2) $\lambda_{i} - \lambda_{i+1} \leq 1$ for every $i\leq k-1$.</p>
<p>(3) In the partition $\lambda$, the number of odd parts in odd p... | Marc van Leeuwen | 19,077 | <p>This is the bijection indicated in the answer by Matt Fayers, described more explicitly. Note that I needed to correct the wrong definition of "conormal nodes" that I had found in the document mentioned in my comment to that answer. </p>
<p>Represent partitions by their Young diagram, laid out on a checkerboard pat... |
386,899 | <p>Show that $$\sum_{k=0}^{n}(-1)^k\binom{n}{k}(n-2k)^{n+2}=\frac{2^{n}n(n+2)!}{6}.$$</p>
| akotronis | 34,294 | <p>The only thing I have to add on @robjohn's answer is the derivation of the identities used:</p>
<p>We have that</p>
<p>$\displaystyle\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}k^j=\frac{d^{j}}{d x^j}\left[(e^x-1)^n\right]\Big|_{x=0}=\frac{d^{j}}{d x^j}\left[x^n+\frac{n}{2}x^{n+1}+\frac{n(3n+1)}{24}x^{n+2}+\mathcal O(x^{n... |
1,280,882 | <p>This may be a silly question, but one that I am confused about nonetheless.</p>
<p>With regards to the compound trig identities such as $\cos(A+B)=\cos A\cos B - \sin A\sin B$ etc., I'd like to know why they are used. What's the purpose? Surely, one would ask themselves that if we can just add $A$ and $B$ together ... | Daniel Rudy | 62,636 | <p>The purpose of these equations are to enable a student, mathematician, engineer, etc... to solve equations. You will see it alot in the following classes: Integral Calculus, Multivariable and Vector Calculus, Differential Equations, etc....</p>
<p>A really good example that is important in signal processing is the... |
3,870,392 | <p>Taking the real numbers to be a complete ordered field, why do we believe that they model distances along a line? How do we know (or why do we believe) that any length that can be drawn is a real number multiple of some unit length?</p>
| Steven Alexis Gregory | 75,410 | <p>I think what you are talking about is called <a href="http://www.math.brown.edu/%7Ebanchoff/STG/ma8/papers/kadams/geometry_extended.html#:%7E:text=Thm%202.1%3A%20The%20Ruler%20Placement,for%20PQ." rel="nofollow noreferrer">The Ruler Postulate</a>. The existence of <a href="https://en.wikipedia.org/wiki/Nonstandard_a... |
3,870,392 | <p>Taking the real numbers to be a complete ordered field, why do we believe that they model distances along a line? How do we know (or why do we believe) that any length that can be drawn is a real number multiple of some unit length?</p>
| user21820 | 21,820 | <p>Firstly, we have some geometrical intuition about what an <strong>ideal line</strong> should be like, which is based on our experience of the physical world. This ideal line satisfies:</p>
<ol>
<li><p>The points (i.e. possible positions) on the ideal line are closed under translation and scaling, and addition and mu... |
2,233,369 | <p>If there exists a homeomorphism $f$ between the closed unit interval and some cartesian product $A\times B$, either $A$ is a singleton or $B$ is a singleton.</p>
<p>The proof I have argues as follows:</p>
<p>Since $[0,1]$ is connected, $A\times B$ is, so $A$ and $B$ are as well. If $a_1,a_2 \in A$ and $b_1,b_2\in ... | Angina Seng | 436,618 | <p>The sets $f^{-1}(\{a_1\}\times B)$ etc., are connected closed subsets of $[0,1]$. The connected subsets of $[0,1]$ are intervals.</p>
<p>I would attack the question a different way: the set $[0,1]$ can be disconnected by removing one point. Prove that $A\times B$ cannot be disconnected by removing one point wheneve... |
2,132,994 | <p>I need to prove that absolute value of any real number is greater than or equal to that real number, where $|a| = a ; a\ge0 , |a| = -a ; a<0 $</p>
<p>I came across this on real analysis.
I need this proven Filed and Order Axioms and basic definitions.</p>
| Anna SdTC | 410,766 | <p>What is $|a|$? It is either $a$ itself, when $a\geq0$, or $-a$, when $a\leq0$. Just separate in cases, $a\leq0$ and $a\geq0$, and substitute the absolute value by these.</p>
|
1,467,946 | <p><strong>10 fair coins are tossed. How many outcomes have 3 Heads?</strong>
I'm supposed to solve it with combination C(10, 3). But...</p>
<p>How do you know it's a combination that will solve it? I'm <strong>not</strong> interested in what makes it a combination, instead of a permutation. I know the answer is (some... | Graham Kemp | 135,106 | <p>The $50:50$ odds on each toss means that all distinct outcomes are equally weighted, and so we don't worry about it any further. Here an outcome is an <em>arrangement</em> of heads and tails in ten coin tosses. Hence we are counting permutations.</p>
<p>So to count the favoured outcomes, which are arr... |
1,556,609 | <p>Let $C$ be the curve of intersection of the two surfaces $x+y=2 , x^2+y^2+z^2=2(x+y)$ . The curve is to be traversed in clockwise direction as viewed from the origin . The what is the value of $\int_Cydx+zdy+xdz$ ? I am not even able to parametrize the curve of intersection . Please help . Thanks in advance </p>
| Community | -1 | <p><strong>Hint</strong>:</p>
<p>In spherical coordinates,</p>
<p>$$r\cos(\phi)\sin(\theta)+r\cos(\phi)\cos(\theta)=2,\\
r^2=2(r\cos(\phi)\sin(\theta)+r\cos(\phi)\cos(\theta))=4.$$</p>
<p>Then from the first equation,</p>
<p>$$\cos(\phi)=\frac1{\sin(\theta)+\cos(\theta)}$$and$$\sin(\phi)=\sqrt{1-\cos^2(\phi)}.$$</p... |
353,374 | <p>If we throw three dice at the same time and see a sum of numbers. What number will have the greatest probability. I think that that number is 11 but I am not positive. Any help will be appreciated. </p>
| Max | 53,326 | <p>Possible Sums with three dice are 3 ... 18</p>
<p>3 -> 1+1+1 Possible Combinations: 1</p>
<p>4 -> 1+2+1 Possible Combinations:3</p>
<p>5 -> (1+3+1),(1,2,2) Possible Combinations:6</p>
<p>6 -> (1,4,1), (1,3,2),(2,2,2) Possible Combinations:10</p>
<p>7 -> (1,4,2), (1,3,3),(5,1,1),(3,2,2) Possible Combinations:15<... |
2,870,910 | <blockquote>
<p>I wish to show that if $z$ is real, then
$$\left|\frac{e^{iz}}{z^2+1}\right|\leq\frac{1}{|z|^2+1}$$</p>
</blockquote>
<p>I have shown this result, although my inequality is the wrong way around.</p>
<p>I considered
\begin{align}
|z^2+1|&\leq |z^2|+|1| \ \ \ \ \ \ \ \text{(triangle inequality)... | Kavi Rama Murthy | 142,385 | <p>$\geq $ and $\leq$ don't contradict each other! When $z$ is real $|z^{2}+1|=|z|^{2}+1$. This proves that the stated inequality is actually an equlity for real $z$. To see that the inequality may not hold for complex $z$ take $z=e^{-in}$ where $n$ is a large positive integer.</p>
|
793,930 | <p>I was asked to evaluate the integral </p>
<p>$$\int_{-1}^{1} \frac{\sin{x}}{1+x^2}dx$$</p>
<p>if it exists.</p>
<p>This is a problem from Calculus and the student has been taught how to use trigonometric substitution. My intuition was to do trig sub with $$x=\tan{\theta}$$ and eliminating $$\frac{dx}{1+x^2}$$ b... | Brad | 108,464 | <p>We will begin by evaluating $$I(x) = \int\!\dfrac{\sin(x)}{1+x^2}\mathrm{d}x$$</p>
<p>$$I = \dfrac{1}{2i} \int\!\dfrac{\sin(x)}{x-i}\mathrm{d}x - \dfrac{1}{2i}\int\!\dfrac{\sin(x)}{x+i}\mathrm{d}x$$</p>
<p>$$I = \dfrac{1}{2i} \int\!\dfrac{\sin(x)}{x-i}\mathrm{d}x - \dfrac{1}{2i}\int\!\dfrac{\sin(x)}{x+i}\mathr... |
763,499 | <p>This seems trivial, and yet after a bit of thinking, I couldn't supply a simple proof.</p>
<p>Is the following true?</p>
<p>The series $$\underset{n=1}{\overset{\infty}\sum}\cos(nx)$$ is divergent for almost every $x\in[-\pi,\pi]$ (or at least for a positive measure subset, though I believe a.e., or likely everywh... | André Nicolas | 6,312 | <p>The series does not converge for any real $x$. We show that in the "calculus" way you suggested, by showing there is no $x$ such that $\lim_{n\to\infty} \cos(nx)=0$. </p>
<p>Suppose to the contrary that $\lim_{n\to\infty} \cos(nx)=0$. Then $\lim_{n\to\infty} \cos(2nx)=0$. This is impossible, since $\cos(2nx)=2\cos... |
3,865,636 | <p>I was wondering if one can explicitly solve the following equation
<span class="math-container">$$
x^2 y'' + x y' + k^2 x^2 (x^\beta+1) y = a^2 y
$$</span>
for real constants <span class="math-container">$k,a,\beta$</span>?
If <span class="math-container">$\beta = 0$</span> then the solution is a linear combination ... | 2'5 9'2 | 11,123 | <p>The trace of the matrix would have to be <span class="math-container">$-1$</span>, since <span class="math-container">$1$</span> is the coefficient of your linear term. So it's at least <span class="math-container">$$\begin{bmatrix}a&*\\*&-1-a\end{bmatrix}$$</span></p>
<p>And the determinant would have to be... |
13,649 | <p>I've been exposed to various problems involving infinite circuits but never seen an extensive treatment on the subject. The main problem I am referring to is</p>
<blockquote>
<p>Given a lattice L, we turn it into a circuit by placing a unit resistance in each edge. We would like to calculate the effective resista... | Benoît Kloeckner | 4,961 | <p>The book by Peres and Lyons, freely available here <a href="http://php.indiana.edu/~rdlyons/prbtree/prbtree.html" rel="nofollow">http://php.indiana.edu/~rdlyons/prbtree/prbtree.html</a>, should give you much information at least for the probability part of the question.</p>
|
682,107 | <p>I think I have to use the fact that $( 1 + \frac{1}{x})^{x^2}$ tends to $e$ as $x$ tends to minus infinity. But I'm not sure how to apply it . . . Maybe I can compare it to something; like, if it's smaller than something that tends to $0$ then will it tend to zero?</p>
<p>Thanks.</p>
| Community | -1 | <p>If you know the calculation of the canonical bundle of the usual Grassmannian (nice writeup <a href="http://concretenonsense.wordpress.com/2009/08/17/tangent-bundle-of-the-grassmannian/" rel="nofollow">here</a>), you can adapt it directly to the relative situation. The result is that $K_G=\mathcal O_{\mathbf P}(-r)_... |
3,713,900 | <p>I have the function:</p>
<p><span class="math-container">$$f : \mathbb{R} \rightarrow \mathbb{R} \hspace{2cm} f(x) = e^x + x^3 -x^2 + x$$</span></p>
<p>and I have to find the limit:</p>
<p><span class="math-container">$$\lim\limits_{x \to \infty} \frac{f^{-1}(x)}{\ln x}$$</span></p>
<p>(In the first part of the ... | Paramanand Singh | 72,031 | <p>Expanding my comment into an answer.</p>
<hr>
<p>The derivative <span class="math-container">$$f'(x) =e^x+\text{ a polynomial in } x$$</span> clearly tends to <span class="math-container">$\infty$</span> as <span class="math-container">$x\to\infty $</span> and hence is positive in some neighborhood of type <span c... |
3,713,900 | <p>I have the function:</p>
<p><span class="math-container">$$f : \mathbb{R} \rightarrow \mathbb{R} \hspace{2cm} f(x) = e^x + x^3 -x^2 + x$$</span></p>
<p>and I have to find the limit:</p>
<p><span class="math-container">$$\lim\limits_{x \to \infty} \frac{f^{-1}(x)}{\ln x}$$</span></p>
<p>(In the first part of the ... | hunter | 108,129 | <p>Since <span class="math-container">$f$</span> is monotone increasing for <span class="math-container">$x$</span> large (for all <span class="math-container">$x$</span> even), we can substitute <span class="math-container">$f(y)$</span> for <span class="math-container">$x$</span> in the limit as follows.
<span class=... |
259,129 | <p>In dimension 3 we have that for $T=\int_{[0,\infty)}1_{B_{t}\in B(0,1)}dt$ has the Laplace transform
$$E[e^{-\lambda T}]=sech(\sqrt{2\lambda}).$$</p>
<p>And in dimension 1 we have the same for $\tau=\min\{t: |B(t)|=1\}$:</p>
<p>$$E[e^{-\lambda \tau}]=sech(\sqrt{2\lambda}).$$</p>
<p>Inverting this in Mathematica d... | Carlo Beenakker | 11,260 | <p><strong>Q2</strong> $\qquad$ <em>"Is there a closed formula for the inverse Laplace transform of ${\rm sech}(\sqrt{2\lambda})$"</em><br>
might still benefit from an explicit answer in terms of a special function.</p>
<p>The inverse Laplace transform of ${\rm sech}(\sqrt{2\lambda})=1/\cosh(\sqrt{2\lambda})$ follows ... |
3,054,362 | <p>Any idea on how to solve the following definite integral?</p>
<blockquote>
<p><span class="math-container">$$\int_0^1\frac{\ln{(x^2+1)}}{x+1}dx$$</span></p>
</blockquote>
<p>I have tried to parameterize the integral like <span class="math-container">$\ln{(a^2x^2+1)}$</span> or <span class="math-container">$\ln{(... | Yuriy S | 269,624 | <p><strong>Not a full answer (yet)</strong></p>
<p>I'll try to provide a solution which doesn't use any special functions, only the well known "classic" series.</p>
<p><span class="math-container">$$I=\int_0^1 \frac{\ln{(x^2+1)}}{x+1}dx=\sum_{k,n=0}^\infty \frac{(-1)^{k+n}}{k+1} \int_0^1 x^{2(k+1)+n}dx=\sum_{k,n=0}^\... |
3,014,104 | <p>Let <span class="math-container">$A$</span> be the set <span class="math-container">$A = \{1,2,3,...,20\}$</span>.
<span class="math-container">$R$</span> is the relation over <span class="math-container">$A$</span> such that <span class="math-container">$xRy$</span> iff <span class="math-container">$y/x = 2^i$</spa... | Giuseppe Negro | 8,157 | <p>This is maybe more of a trick than of a proof, but I find this approach much easier to memorize. </p>
<p>Let <span class="math-container">$x_1, x_2, x_3$</span> be the Cartesian coordinates of the space. It is clear that the reflection around the plane <span class="math-container">$x_1x_2$</span> is the map
<span c... |
526,107 | <p>I have to give a presentation on vector analysis. One of many important things I want to emphasize is that a division by a vector does not make sense.</p>
<p>How do you explain to your students, for example, that division by a vector does not make sense?</p>
<p>Bonus question: Also how do you explain that integrat... | Emanuele Paolini | 59,304 | <p>Division should be the inverse of multiplication. What is multiplication for vectors? You could multiply a vector by a scalar obtaining a vector, or we could multiply two vectors to obtain a scalar.</p>
<p>In the first case on could possibly define division... but it would be defined only in the very restrictive ca... |
3,491,595 | <blockquote>
<p>Evaluate the sum <span class="math-container">$$\frac{1}{3} + \frac{1}{3^{1+\frac{1}{2}}}+\frac{1}{3^{1+\frac{1}{2}+\frac{1}{3}}}+\cdots$$</span></p>
</blockquote>
<p>It seems that <span class="math-container">$1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{n}$</span> approaches <span class="ma... | heropup | 118,193 | <p>We can certainly impose some bounds on the value of the sum, via the asymptotic expansion <span class="math-container">$$H_n \sim \log n + \gamma + \frac{1}{2n} - \frac{1}{12n^2} + \cdots. \tag{1}$$</span> The crudest bound is to note for <span class="math-container">$0 < z < 1$</span> the sum <span class="ma... |
754,575 | <p>I am slightly stuck on this seemlingly simple problem that I encoutered as part of a problem to show that the orthogonality condition of $M_{2\times2}$ matrices given by $\sum_i a_{ij}a_{jk} = \delta_{jk}$, where the $a_{ij}$ are the matrix entries, implies that $\det(M) = \pm 1$.</p>
<p>If $b(1-a^2) = a(1-b^2)$, s... | Adriano | 76,987 | <p><strong>Hint:</strong> By rearranging and factoring, notice that:
\begin{align*}
(ba^2 - ab^2) - (b - a) &= 0 \\
ab(a - b) + (a - b) &= 0 \\
(ab + 1)(a - b) &= 0 \\
\end{align*}</p>
|
754,575 | <p>I am slightly stuck on this seemlingly simple problem that I encoutered as part of a problem to show that the orthogonality condition of $M_{2\times2}$ matrices given by $\sum_i a_{ij}a_{jk} = \delta_{jk}$, where the $a_{ij}$ are the matrix entries, implies that $\det(M) = \pm 1$.</p>
<p>If $b(1-a^2) = a(1-b^2)$, s... | Umberto | 92,940 | <p>Technically speaking your equation has also the solution $a=-1/b$ with $a,b \neq 0$...</p>
|
1,690,210 | <p>What is $$\int \frac{4t}{1-t^4}dt$$ is there some kind of substitution which might help .Note that here $t=\tan(\theta)$</p>
| Dr. Sonnhard Graubner | 175,066 | <p>HINT: we get $$\frac{4t}{1-t^4}=- \left( t-1 \right) ^{-1}- \left( t+1 \right) ^{-1}+2\,{\frac {t}{{t}
^{2}+1}}
$$</p>
|
1,690,210 | <p>What is $$\int \frac{4t}{1-t^4}dt$$ is there some kind of substitution which might help .Note that here $t=\tan(\theta)$</p>
| ex.nihil | 312,791 | <p>Observe that
$$1-t^4=1-(t^2)^2$$</p>
<p>Let
$$u=t^2$$
with
$$du = 2t\,dt$$</p>
<p>The integral becomes</p>
<p>$$I=\int \frac{2}{1-u^2}du$$</p>
<p>Which is equal to</p>
<p>$$I = \ln{\frac{|u+1|}{|u-1|}}+C=\ln{\frac{|t^2+1|}{|t^2-1|}}+C$$</p>
<p>The above answer has a formula, but can be most conveniently reach... |
23,454 | <p>In an exercise asking to mark true or false, it shows:</p>
<p>$$\frac{1}{a/x-b/x}=\frac{1}{a-b}$$</p>
<p>It really look like <strong>false</strong> to me. But the answer is <strong>true</strong>! How can it be?</p>
| cobbal | 37 | <p>$\frac{1}{\frac{a}{x}-\frac{b}{x}} = \frac{1}{\frac{a-b}{x}} = \frac{x}{a-b} \not\equiv \frac{1}{a-b}$</p>
<p>So this is <em>not</em> true in the general case.</p>
|
174,938 | <p>Adam and Bob are running on a circular track of circumference 1500 m. They start simultaneously from point A in the same direction. Ratio of their speeds is 5:3 respectively. If they keep running , then at how many different points can they meet?</p>
<p>a)Two b)One c)Three d)Data Insufficient</p>
| lab bhattacharjee | 33,337 | <p>$\dfrac{|z|^2+|\alpha|^2}{1+|\alpha z|^2}<1$</p>
<p>iff $|z|^2+|\alpha|^2 < 1+|\alpha z|^2$ as both expressions are > 0</p>
<p>iff $(1-|z|^2)(1-|\alpha|^2)<0$</p>
<p>$(1-|z|^2)<0\ if\ |z|^2,1\ if\ |z|<1 $</p>
|
174,938 | <p>Adam and Bob are running on a circular track of circumference 1500 m. They start simultaneously from point A in the same direction. Ratio of their speeds is 5:3 respectively. If they keep running , then at how many different points can they meet?</p>
<p>a)Two b)One c)Three d)Data Insufficient</p>
| Zev Chonoles | 264 | <p>Because $0\leq |z|<1$ and $0\leq |\alpha|<1$, we have that $|z|^2<1$ and $|\alpha|^2<1$. Thus
$$|z|^2-1<0\qquad\text{ and }\qquad |\alpha|^2-1<0,$$
so that
$$0<(|\alpha|^2-1)(|z|^2-1).$$
Multiplying out the right side,
$$0<|\alpha|^2|z|^2-|\alpha|^2-|z|^2+1$$
and of course $|\alpha|^2|z|^2=... |
2,282,063 | <p>This is Exercise 2.6.11 of Howie's <em>"Fundamentals of Semigroup Theory"</em>.</p>
<h2>The Details:</h2>
<p>Let <span class="math-container">$S$</span> be a semigroup.</p>
<blockquote>
<p><strong>Definition 1:</strong> We say <span class="math-container">$S$</span> is <em>regular</em> if for all <span cla... | ALife | 559,415 | <p>A "direct" proof for (b) <span class="math-container">$\rightarrow$</span> (c) (using quasigroup + semigroup <span class="math-container">$\leftrightarrow$</span> group):</p>
<p>This is to show that <span class="math-container">$\forall a, b\in S (\exists! x$</span> s.t. <span class="math-container">$ax=b)... |
740,154 | <p>I'm trying to prove this sequence: $a_n = \sqrt{n}-\sqrt{n^2-1}$ to be divergent. How would I do this? I'm thinking of proving that it's not bounded below, but I'm not sure how to do that with induction, as I've only done that to prove it's bounded.</p>
| David H | 55,051 | <p>Hint: compare the sequence $a_n=\sqrt{n}-\sqrt{n^2-1}$ to the sequence $b_n=-\sqrt{n}$, which clearly diverges to $-\infty$.</p>
|
1,683,375 | <blockquote>
<p>If Salvatore has achieved the test and Carmela has achieved the
test,then Benedetto also has achieved the test.But Salvatore didn't
achieve the test.So:</p>
<p>A)Benedetto didn't achieve the test.</p>
<p>B)Benedetto could have achieved the test.</p>
<p>C)Benedetto or Carmela didn't ... | Τίμων | 319,592 | <p><em>Ex falso sequitur quodlibet</em>. That is, if your assumption is false (and your assumption $S \land C $ is assumed false), than anything follows.</p>
|
3,506,899 | <p>I need to find the solutions of the following equations:</p>
<p><span class="math-container">$$\ddot{x}=-g+\alpha\dot{x}$$</span></p>
<p>and
<span class="math-container">$$\ddot{x}=-g+\alpha\dot{x}^2$$</span></p>
<p>Considering <span class="math-container">$g$</span> and <span class="math-container">$\alpha$</sp... | Daniel | 727,769 | <p>I presume the first isn't the case, since if we had an integral like <span class="math-container">$$ \int xe^x$$</span>
we would need to use the chain rule and we can't simply take out <span class="math-container">$$ e^x $$</span>
in front of the integral. But is there any operation we can do to factor out a deriva... |
3,506,899 | <p>I need to find the solutions of the following equations:</p>
<p><span class="math-container">$$\ddot{x}=-g+\alpha\dot{x}$$</span></p>
<p>and
<span class="math-container">$$\ddot{x}=-g+\alpha\dot{x}^2$$</span></p>
<p>Considering <span class="math-container">$g$</span> and <span class="math-container">$\alpha$</sp... | Community | -1 | <p>No. What you do have, with integration by parts, is: <span class="math-container">$\int uv'=uv-\int u'v$</span>.</p>
<p>So for a counterexample, we just need <span class="math-container">$v\int u\ne uv-\int u'v$</span>. For instance, take <span class="math-container">$u=1,v=x$</span>.</p>
|
3,480,050 | <p>I've been struggling to solve the following exercise:</p>
<p>For <span class="math-container">$x\in\mathbb{R}$</span>, find the radius of convergence of the series <span class="math-container">$\sum_{n=1}^{\infty}\frac{x^n}{n+\sqrt{n}}$</span>.</p>
<p>My approach so far: Compute <span class="math-container">$\lims... | DanielWainfleet | 254,665 | <p>Method 1. For <span class="math-container">$n\in \Bbb Z^+$</span> we have <span class="math-container">$$1<(n+\sqrt n)^{1/n}\le (n+n)^{1/n}=2^{1/n}\cdot n^{1/n}. $$</span> For <span class="math-container">$2\le n\in \Bbb Z^+$</span> let <span class="math-container">$n^{1/n}=1+x_n.$</span> Then <span class="math-c... |
2,469,841 | <p>I wonder why this is true</p>
<p>$$ \sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 58 n^2} = - \frac{\pi \ln( 27 + 5 \sqrt {29})}{\sqrt {58}} $$</p>
<p>Where the sum omits the case $n = m = 0$ ofcourse.</p>
| Tito Piezas III | 4,781 | <p>(<em>Too long for a comment</em>.)</p>
<p>We have,</p>
<p>$$ \sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 10n^2} = - \frac{2\pi \ln( \sqrt2\; U_{5})}{\sqrt {10}} $$</p>
<p>$$ \sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 58 n^2} = - \frac{2\pi \ln( \sqrt2\; U_{29})}{\sqrt {58}} $$</p>
<p>with <em><a h... |
1,372,558 | <p>$y=\sqrt{x^x}$</p>
<p>How do I convert this into a form that is workable and what indicates that I should do so? </p>
<p>Anyway, I tried this method of logging both sides of the equation but I don't know if I am right.</p>
<p>$\ln\ y=\sqrt{x} \ln\ x$</p>
<p>$\frac{dy}{dx}\cdot \frac{1}{y}=\sqrt{x}\ \frac{1}{x} +... | Rory Daulton | 161,807 | <p>That way could work though you made some mistakes, but an easier way shifts the square root to a fractional exponent.</p>
<p>$$\begin{align}
y&=\sqrt{x^x} \\[2ex]
&= \left(x^x\right)^{1/2} \\[2ex]
&= x^{x/2} \\[2ex]
\ln y&= \ln x^{x/2} \\[2ex]
&= \frac x2\ln x \\[2ex]
\frac{dy}{dx}\frac 1y &a... |
2,732,220 | <p>A friend of mine asked me to help him evaluate the series</p>
<p>$$\mathcal{S} = \sum_{n=1}^{\infty} \frac{\sin (n \pi y) \sin \left ( n \pi x \right )}{n^2 \pi^2} \quad , \quad x , y \in (0, 1)$$</p>
<p>It does not ring any bells as to what it could be behind. The only thing I see is Fourier series and probably a... | Felix Marin | 85,343 | <p>$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\... |
3,720,083 | <p><strong>My attempt</strong>
<span class="math-container">$$1-\frac{1}{3\cdot 3}+\frac{1}{5\cdot 3^2}-\frac{1}{7\cdot 3^3}+\cdots=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)3^{n-1}}$$</span></p>
<p>By Leibniz alternative test for convergence. It is a convergent alternative series. How do I evaluate this limit?</p>
| Felix Marin | 85,343 | <p><span class="math-container">$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\exp... |
2,167,265 | <p>Zeno, a follower of Parmenides, reasoned that any unit of space or time is infinitely divisible or not. If they be infinitely divisible, then how does an infinite plurality of parts combine into a finite whole? And if these units are not infinitely divisible, then calculus wouldn't work because $n$ couldn't tend t... | Bram28 | 256,001 | <p>Zeno's paradox is called a paradox exactly because there is a mismatch between a seemingly logical argument that concludes that motion is impossible, and our experience in dealing with reality, which says that there is motion. </p>
<p>To resolve the paradox, then, you need to figure out <em>where</em> the argument... |
2,167,265 | <p>Zeno, a follower of Parmenides, reasoned that any unit of space or time is infinitely divisible or not. If they be infinitely divisible, then how does an infinite plurality of parts combine into a finite whole? And if these units are not infinitely divisible, then calculus wouldn't work because $n$ couldn't tend t... | user21820 | 21,820 | <p>Your response is mathematically wrong, but intuitively not that far off. The key is to note that the error in the 'paradox' is:</p>
<blockquote>
<p>an infinite many tasks must be performed [CORRECT] ... an infinite many tasks to perform can never be completed [WRONG]</p>
</blockquote>
<p>To get a better understandin... |
457,347 | <blockquote>
<p>Proposition: The field of fractions of an integral domain is a field</p>
</blockquote>
<p>I founded the above proposition from Wikipedia.
"In abstract algebra, the field of fractions or field of quotients of an integral domain is the smallest field in which it can be embedded" (<a href="http://en.wi... | Jyrki Lahtonen | 11,619 | <p>What the proposition precisely means is the following. Assume that $D$ is an integral domain. Assume that you are given an injective ring homomorphism
$f\colon D\to F$, where $F$ is some field. Let $i\colon D\to Q(D)$ be the embedding of $D$
into its field of fractions $Q(D)$ given by the mapping $i(d)=d/1$ for all ... |
76,299 | <p>This is a bit embarrassing, but I can't seem to solve for $x$ in
$$2x=\frac{x}{y}-\frac{1}{1-y}.$$
Could someone please give me a hand!</p>
| Blue | 409 | <p>Here's a slightly different approach, which also uses coordinates, and which fails to enlighten as to <em>why</em> the phenomenon works.</p>
<p>Let $C = C(\theta)$ and $r = r(\theta)$ be the center and radius of the Focal Chord Circle (FCC) for which the focal chord makes angle $\theta$ with the $x$ axis. Let $K$ a... |
1,025,499 | <p>The unit circle is defined to be <span class="math-container">$x^2 + y^2 = 1$</span>. Makes sense. Its an equation of a circle. Now from here if we think about cosine as the <span class="math-container">$x$</span> value and sine as the <span class="math-container">$y$</span> value then we get a trig identity most of... | curious_mind | 91,374 | <p>Please see this reference :
in this please read section 4.14.. This is a very good source of basic trigonometry.
<a href="http://gujarat-education.gov.in/textbook/Images/11sem1/maths11-eng/chap4.pdf" rel="nofollow">http://gujarat-education.gov.in/textbook/Images/11sem1/maths11-eng/chap4.pdf</a></p>
|
171,521 | <p>I have nested associations like this:</p>
<pre><code><|{"Country1","YEAR1"} -> {<|{"VARIABLE1", "MEASURE1"} ->
"Value1"|>, <|{"VARIABLE2", "MEASURE2"} -> "Value2"|>}, {"Country1",
"YEAR2"} -> {<|{"VARIABLE1", "MEASURE1"} ->
"Value1"|>, <|{"VARIABLE2", "MEASURE2"} -> "Va... | José Antonio Díaz Navas | 1,309 | <p>Quite simple. As $Y=1$, then $X$ simplifies to:</p>
<pre><code>$Version
</code></pre>
<blockquote>
<p>11.3.0 for Mac OS X x86 (64-bit) (March 7, 2018)</p>
</blockquote>
<pre><code>X=2*Pi*Y1*Integrate[1/((1 + s)^2*Sqrt[Y1^2 + s]), {s, 0, Infinity}]
</code></pre>
<blockquote>
<p>$$\text{ConditionalExpression}\... |
3,285,697 | <p>I have a matrix <span class="math-container">$B$</span> which it's dimension is <span class="math-container">$nm$</span> (with <span class="math-container">$n>m$</span>). During an iterative process I'll change it to get a desired state of matrix <span class="math-container">$B$</span>, but in each step, I should... | John Hughes | 114,036 | <p>In general, you cannot hope to find <span class="math-container">$B$</span> by knowing <span class="math-container">$B^t B$</span>. For instance, if <span class="math-container">$B$</span> is <em>any</em> rotation matrix, then <span class="math-container">$B^t B = I$</span>. So if you're told that <span class="math-... |
52,249 | <p>Is it true that all zeros of the Riemann Zeta Function are of order 1?</p>
<p>Let <span class="math-container">$h(z) = \frac{\zeta'(z)}{\zeta(z)}\frac{x^z}{z}$</span>, where <span class="math-container">$x$</span> is a positive real number (<span class="math-container">$x > 1$</span>, probably?) , and <span class... | draks ... | 19,341 | <p>You might be interested in this recent arXiv paper: <a href="http://xxx.lanl.gov/abs/1302.5018" rel="nofollow">On simple zeros of the Riemann zeta-function</a></p>
<blockquote>
<p>We show that <strong>at least 19/27 of the zeros of the Riemann zeta-function are simple</strong>, assuming the Riemann Hypothesis (RH... |
3,335,615 | <p>Let <span class="math-container">$f,g$</span> be Riemann integrable on <span class="math-container">$[0,1]$</span> such that <span class="math-container">$\int_0^1 f=\int_0^1 g=1$</span>. Show that there exists <span class="math-container">$0\leq a<b\leq 1$</span> such that <span class="math-container">$\int_a^b ... | quarague | 169,704 | <p>This is a partial solution assuming that <span class="math-container">$f, g > 0$</span>. There is probably some clever trick to generalize it but I don't see it right now.</p>
<p>Let <span class="math-container">$s \in (0,1)$</span> be the unique point where <span class="math-container">$\int_0^s f(x)dx=\frac{1}... |
2,121,863 | <blockquote>
<p>Show that a subring of a division ring must be a domain.</p>
</blockquote>
<p>Let $S$ be a subring of $R$ and let $R$ be a division ring.
Can we just say that since every element has an inverse in $R$, then every element also has an inverse in $S$, then we can deduce that since for all elements in $S... | Bill Dubuque | 242 | <p><strong>Hint</strong> $\ $ If $\,ax=0\,$ has unique root $\,x=0\,$ in $R\,$ then the same holds true in every subring.</p>
|
2,121,863 | <blockquote>
<p>Show that a subring of a division ring must be a domain.</p>
</blockquote>
<p>Let $S$ be a subring of $R$ and let $R$ be a division ring.
Can we just say that since every element has an inverse in $R$, then every element also has an inverse in $S$, then we can deduce that since for all elements in $S... | Nguyen Dang Son | 439,718 | <p>A subring of a division ring may not be a division ring. But a finite subring of a division ring is a division ring because a finite domain is a division ring and subring of a division ring is a domain. Fix <span class="math-container">$s \in S$</span>, we can consider left translation <span class="math-container">$... |
287,203 | <blockquote>
<p>Prove the statement:</p>
<p><span class="math-container">$\forall n \in \mathbb{N}$</span>,<span class="math-container">$\forall m \in \{2, 3,...,floor(\sqrt{n})\}$</span>, <span class="math-container">$m$</span> does not divide <span class="math-container">$n \implies n$</span> is prime</p>
</blockquot... | Andreas Caranti | 58,401 | <p>If you write the natural number $n$ as the product of two natural number $a$ and $b$, then $a$ and $b$ cannot be both $> \sqrt{n}$.</p>
|
419,807 | <p>Given a homomorphism of rings <span class="math-container">$S \rightarrow R$</span>, for a pair of <span class="math-container">$R$</span>-modules <span class="math-container">$M, N$</span> the machinery of relative homological algebra defines relative <span class="math-container">$Ext$</span>-groups</p>
<p><span cl... | Tyler Lawson | 360 | <p>Let <span class="math-container">$K(S)$</span> be the category of chain complexes of <span class="math-container">$S$</span>-modules; this category has Hom-complexes <span class="math-container">$hom_S(-,-)$</span> making it a dg-category, and thus produces a quasicategory via the construction in Higher Algebra sect... |
10,730 | <p>While the question is stated with reference to the iPhone, my actual question is about phones in general. Just as there was much talk about the use of Computers in the classroom over the past fifty years (or so), how much talk is there about the use of cellphones in the classroom?</p>
<p>What benefits are there to ... | Brendan W. Sullivan | 80 | <p>I teach undergraduate students (ages ~17-22), with class sizes ranging from 10 to 30. I have come to realize that students will have their phones with them no matter what, so it would be unnecessarily punitive of me to try to limit their use. Instead, I have accepted the fact that they're in the classroom and have s... |
298,179 | <p>Are these sets conformally equivalent, That is, is there a conformal bijection between them. $S_1=\{z\in\mathbb C\mid0<|z|<1\}$ and $S_2=\{z\in \mathbb C\mid1<|z|<2\}$?<br/><br/>The only thing that I could find is a theorem of (F.H. Schottky, 1877) see <a href="https://www.google.ca/search?q=F.H.+Schottk... | James S. Cook | 36,530 | <p>I believe the mapping which sends $|z|$ to $|z|+1$ and maintains the angle will do. You're just blowing-up the punctured disk into an annulus. I haven't worked out the details, but it seems plausible.</p>
<p><strong>Added 2-9-13</strong> Follow-up on why my conjecture fails:</p>
<p>$$ f(z) = (|z|+1)\frac{z}{|z|} $... |
3,455,482 | <p>I'm stuck trying to solve an inequality: Let <span class="math-container">$X \in \mathcal{L}^{2}$</span>, then </p>
<p><span class="math-container">$$\mathbb{E}[\vert X-\mathbb{E}[X\mid \mathcal{G}]\vert^{2}] \leq \mathbb{E}[\vert X - \mathbb{E}[X]\vert^{2}].$$</span></p>
<p>My attempts:</p>
<p>First I (mistakenl... | Olórin | 187,521 | <p>You are not mistaken about <span class="math-container">$\mathbf{Var}[X] = \mathbf{E}\left[\left(X - \mathbf{E}\left[X\right]\right)^2\right]$</span>, just develop the square etc to find that it is the same as <span class="math-container">$\mathbf{E}\left[X^2\right] - \mathbf{E}\left[X\right]^2$</span>.</p>
<p>The ... |
144,567 | <p>Fix a homomorphism $f:A\rightarrow B$.
Choose $\{b_1,\dots,b_n\}$, $\{b'_1,\dots,b'_m\}$ subsets of elements in $B$. Suppose that $B$ is algebraic over $f(A)[b_1,\dots,b_n]$ and $\{b_1,\dots,b_n\}$ are algebraically independent over $f(A)$. Suppose that $\{b'_1,\dots,b'_m\}$ satisfies the same condition. Does it imp... | Sasha Anan'in | 40,352 | <p>The problem is actually the following. Let $A\subset B$ be a subring and suppose that the subrings $C:=A[b_1,\dots,b_n]$ and $C':=A[b'_1,\dots,b'_m]$ of $B$ are the rings of polynomials in the indicated variables such that $B$ is algebraic over both. Does it follow $m=n$ ?</p>
<p>If "$B$ is algebraic over $C$" mean... |
1,026,975 | <p>Given a topology $(X,T)$, $A\subset X$, $x \in X$ is a limit point of A if $\forall$ open $U$ that contains $x$, $(U\cap A)$\ {$x$} $\neq \emptyset$. $x \in X$ is in $cl(A)$ if $\forall$ open $U$ that contains $x$, $U\cap A$ $\neq \emptyset$. Is there any example that a point in the the closure of $A$ is not a limit... | NickC | 189,951 | <p>consider a finite set A of singletons in $\mathbb{R}$, there are no limit points, but they are al in $cl(A)$</p>
|
266,738 | <p>Say I have an expression that has multiple subvalues how do I return the definition that would be applied to it.</p>
<p>Say I define for example:</p>
<pre><code>fun[y_][x_] := {x, y};
fun[3][x_] := 2;
</code></pre>
<p>Then I would like <code>findSubValue[fun[3][5]]</code> to return <code>HoldPattern[fun[3][x_]] :>... | lericr | 84,894 | <p>How about something like this:</p>
<pre><code>SetAttributes[FindMatchingSubValue, HoldAll];
FindMatchingSubValue[sym_, expr_] :=
SelectFirst[SubValues[sym], MatchQ[Unevaluated[expr], First@#] &]
</code></pre>
<p>Testing:</p>
<pre><code>FindMatchingSubValue[fun, fun[3][5]]
(*HoldPattern[fun[3][x_]] :> 2*)
... |
20,634 | <p>From the <a href="https://en.wikipedia.org/wiki/Peter%E2%80%93Weyl_theorem" rel="nofollow noreferrer">Peter–Weyl theorem in Wikipedia</a>, this theorem applies for compact group. I wonder whether there is a non-compact version for this theorem.</p>
<p>I suspect it because the proof of the Peter–Weyl theorem heavily ... | Marty | 3,545 | <p>I don't know anything about quantum groups, but the Peter-Weyl theorem for compact groups generalizes nicely to Type I second-countable locally compact topological groups, a result of Segal and Mautner. For such groups, there is a decomposition of the unitary biregular representation:
$$L^2(G) \cong \int_{\pi \in \... |
2,276,404 | <p>I stumbled upon the following inequality in a scientific paper which estimates a lower bound for $\frac{k!}{k^k}$ for $k \in \mathbb{N}$:
$$\frac{k!}{k^k} > e^{-k}$$
They did not explain why this holds true, and I could not find any answer by myself yet.</p>
| B. Goddard | 362,009 | <p>Use the Taylor series:</p>
<p>$$e^k = 1+k+\frac{k^2}{2!} +\cdots+\frac{k^k}{k!} +\cdots.$$</p>
<p>Because all terms on the right are positive, we have</p>
<p>$$e^k > \frac{k^k}{k!},$$</p>
<p>then just take reciprocals of both sides.</p>
|
3,561,242 | <blockquote>
<p>Determine if the following integral will converge.
<span class="math-container">$$\int_0^1\frac{e^{\sqrt x}-1}{x}dx$$</span></p>
</blockquote>
<p>My approach was something like this. I made the assumption that <span class="math-container">$e^{\sqrt{x}} − 1 ≈ x$</span> and then followed like this:</... | Michael Hardy | 11,667 | <p>Rather than <span class="math-container">$\dfrac{e^\sqrt{x}}{\left( 2\sqrt{x}\over1 \right)}$</span> you should have <span class="math-container">$\dfrac{\left( \dfrac{e^{\sqrt x}}{2\sqrt x} \right)}{1}$</span> in your application of L'Hopital's rule.</p>
<p>Once you've got that, you can do this:
<span class="math-... |
3,561,242 | <blockquote>
<p>Determine if the following integral will converge.
<span class="math-container">$$\int_0^1\frac{e^{\sqrt x}-1}{x}dx$$</span></p>
</blockquote>
<p>My approach was something like this. I made the assumption that <span class="math-container">$e^{\sqrt{x}} − 1 ≈ x$</span> and then followed like this:</... | mjw | 655,367 | <p>Let <span class="math-container">$y=\sqrt{x}$</span> so that <span class="math-container">$x=y^2$</span> and <span class="math-container">$dx=2y\,dy.$</span></p>
<p><span class="math-container">$$\int_0^1 \frac{e^{\sqrt{x}}-1}{x} dx = 2\int_0^1 \frac{e^y-1}{y}\,dy.$$</span></p>
<p>Because the denominator goes to z... |
2,982,942 | <p>Our professor gave us definitions for closed and open intervals. </p>
<p>A set <span class="math-container">$U$</span> is open if <span class="math-container">$\forall x \in U$</span>, <span class="math-container">$\exists \epsilon \gt 0$</span> such that <span class="math-container">$(x- \epsilon,x+ \epsilon)\subs... | José Carlos Santos | 446,262 | <p>Yes, the limit is <span class="math-container">$0$</span>. You can also prove that using the fact that<span class="math-container">$$\frac{xy\sin y}{3x^2+y^2}=\frac{xy^2}{3x^2+y^2}\times\frac{\sin y}y,$$</span>that<span class="math-container">$$\lim_{(x,y)\to(0,0)}\frac{xy^2}{3x^2+y^2}=0$$</span> and that <span clas... |
1,892,669 | <p>Let $F=\left \langle a,b \right \rangle$ be the rank 2 free group. Then, a map $f:\left \langle a \right \rangle \rightarrow \left \langle b \right \rangle$ with $f(a):=b$ is an isomorphism with $\left \langle x,f(x) \right \rangle$ is a rank 2 free group for any $x\in\left\langle a\right\rangle - \{ 1\}.$</p>
<p>C... | Olin | 344,157 | <p>The answer is positive; this is exactly the statement of Proposition 0.63 from the book <a href="http://www.math.hawaii.edu/~lee/book/" rel="nofollow noreferrer">Finite Rank Torsion Free Modules Over Dedekind Domains</a> by E. L. Lady. (<a href="http://www.math.hawaii.edu/~lee/book/zero.pdf" rel="nofollow noreferrer... |
2,571 | <p>Square of an irrational number can be a rational number e.g. $\sqrt{2}$ is irrational but its square is 2 which is rational.</p>
<p>But is there a irrational number square root of which is a rational number?</p>
<p>Is it safe to assume, in general, that $n^{th}$-root of irrational will always give irrational numbe... | Bill Dubuque | 242 | <p>It's true precisely because the rationals $\mathbb Q$ comprise a multiplicative <em>subsemigroup</em> of the reals $\mathbb R$,<br>
i.e. the subset of rationals is closed under the multiplication operation of $\mathbb R$. Your statement arises by taking the contrapositive of this statement - which transfers it into ... |
2,357,091 | <p>Let us consider $\text{trace}(f(AA^\top))$ where $f$ is some smooth function and $A \in \mathbb{R}^{n \times m}$. Here, function $f$ is a function of matrices (c.f., Higham's books). If $f(x) = x$, then $\text{trace}(f(AA^\top)) = \|A\|_F^2$ (Frobenius norm) and if $f(x) = x^{p/2}$, then $\text{trace}(f(AA^\top))$ i... | pitchounet | 61,409 | <p>Let $\mathrm{Mat}(n,\mathbb{R})$ be the vector space of real $n \times n$ matrices. With $f \, : \, \mathrm{Mat}(n,\mathbb{R}) \, \rightarrow \, \mathrm{Mat}(n,\mathbb{R})$ a function which is differentiable on $\mathrm{Mat}(n,\mathbb{R})$, let $g \, : \, X \in \mathrm{Mat}(n,\mathbb{R}) \, \mapsto \, \mathrm{tr}\bi... |
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