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 |
|---|---|---|---|---|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Leandro Vendramin | 17,845 | <p>For a course on cluster algebras (by S. Fomin): <a href="http://qgm.au.dk/video/mc/cluster/" rel="nofollow">http://qgm.au.dk/video/mc/cluster/</a></p>
<p>EDIT: Some graduate short-courses in FCEyN, UBA, Buenos Aires, Argentina:</p>
<ul>
<li>J. Harris, Intersection Theory</li>
<li>R. Hartshorne, Introduction to Deformation Theory</li>
<li>D. Maclagan, Introduction to Tropical Algebraic Geometry</li>
<li>P. Beelen, Algebraic Geometric Codes</li>
</ul>
<p><a href="http://mate.dm.uba.ar/~visita16/ELGA-2011/version/v1/images-en.shtml" rel="nofollow">Here</a> are the links to the videos of these 4 lectures.</p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Community | -1 | <p>I feel that I have something new to add. I sometimes make mathematical videos, and I should make more. Here is one about a 3D diagram that I made of the happy family, in group theory. There are a lot of other videos on my channel, however, not pertaining to mathematics such as singing and art, but I hope that in the future I will make more mathematical content.</p>
<p><a href="https://www.youtube.com/watch?v=_4IjnIcECoQ&t=11s" rel="nofollow noreferrer">https://www.youtube.com/watch?v=_4IjnIcECoQ&t=11s</a></p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| debapriyay | 9,586 | <p>The video that I am adding here is a new upload in Youtube.
Here is the link: <a href="https://www.youtube.com/watch?v=aUl28Pjz89M" rel="nofollow noreferrer">https://www.youtube.com/watch?v=aUl28Pjz89M</a></p>
<p>I have seen many videos on 21-card trick and also read many blogs on it. A very few of them has attempted to explain why this trick works. The ones give the proof, is pretty involved and requires some inclination towards the subject to understand this. But, this video has taken a totally different approach to explain the trick and why it works in a visual way. The approach is easy to understand for anyone. This is a great way of helping students get attracted towards Mathematics.</p>
|
736,624 | <blockquote>
<p>Heaviside's function $H(x)$ is defined as follows:</p>
<p>$H(x) = 1$ if $x > 0$, </p>
<p>$H(x) = \frac 1 2$ if $x= 0$</p>
<p>$H(x) = 0$ if $x < 0$ </p>
</blockquote>
<p>Define $F(x) := \int^x_0 H(t) \ dt, \ x \in \mathbb R$. </p>
<p>$H(x)$ is continuous except at $x=0$ so the integral exists?</p>
<p><strong>Is $F$ continuous?</strong></p>
<p>I have $F(x) = x$ for $x > 0$ and $F(x) = 0$ for $x \le 0$. So $F$ is indeed continuous on $\mathbb R$.</p>
<p><strong>Is F differentiable?</strong></p>
<p>According to the fundamental theorem of calculus $F^{'}(x_0) = H(x_0)$.</p>
<p>However now to the final question:</p>
<p><strong>Is the result contrary to the Fundamental Theorem of Analysis (Fundamental Theorem of Calculus)?</strong></p>
<p>I don't see why it should be? Have I made an error?</p>
| Siminore | 29,672 | <p>The left derivative of $F$ at zero is zero, the right derivative is one. Hence $F$ cannot be differentiable at zero. This is your mistake. Be careful, since the FTC requires that the function to be integrated be continuous in the whole interval. </p>
|
361,755 | <p>Let $S$ be a multiplicatively closed subset of a commutative noetherian ring $A$. Let $M$ and $N$ be finitely generated $A$-modules. If $M_S$ is isomorphic to $N_S$, show that $M_t$ is isomorphic to $N_t$ for some $t \in S.$</p>
| Community | -1 | <p>Since $M$ is finitely presented, we have that $$\text{Hom}_A(M,N)_S \cong \text{Hom}_{A_S}(M_S, N_S) \text{ (*)}$$</p>
<p>via the map which takes $f/s$ to the product of the constant map $1/s$ and the map from $M_S$ to $N_S$ induced by $f$, by THM 7.11 in Matsumura's "Commutative Ring Theory". Take $f \in \text{Hom}_{A_S}(M_S, N_S)$ to be an isomorphism. Suppose $g/s \in \text{Hom}_R(M,N)_S$ maps to $f$ under isomorphism (*). Then $g$ must induce an isomorphism from $M_S$ to $N_S$. </p>
<p>We have an exact sequence </p>
<p>$$0 \rightarrow \ker g \rightarrow M \rightarrow N \rightarrow \text{coker }g \rightarrow 0.$$ </p>
<p>Notice that $(\ker g)_S=0=(\text{coker } g)_S$. Since $\ker g$ and $\text{coker } g$ are finitely generated, we may choose $a \in \text{Ann}_A(\ker g) \cap S$ and $b \in \text{Ann}_A(\text{coker } g) \cap S$. Now, take $t:=ab$. </p>
|
1,452,425 | <p>From what I have been told, everything in mathematics has a definition and everything is based on the rules of logic. For example, whether or not <a href="https://math.stackexchange.com/a/11155/171192">$0^0$ is $1$ is a simple matter of definition</a>.</p>
<p><strong>My question is what the definition of a set is?</strong> </p>
<p>I have noticed that many other definitions start with a set and then something. A <a href="https://en.wikipedia.org/wiki/Group_%28mathematics%29#Definition" rel="noreferrer">group is a set</a> with an operation, an equivalence relation is a set, a <a href="https://en.wikipedia.org/wiki/Function_%28mathematics%29#Definition" rel="noreferrer">function can be considered a set</a>, even the <a href="https://en.wikipedia.org/wiki/Natural_number#Constructions_based_on_set_theory" rel="noreferrer">natural numbers can be defined as sets</a> of other sets containing the empty set.</p>
<p>I understand that there is a whole area of mathematics (and philosophy?) that deals with <a href="https://en.wikipedia.org/wiki/Set_theory#Axiomatic_set_theory" rel="noreferrer">set theory</a>. I have looked at a book about this and I understand next to nothing.</p>
<p>From what little I can get, it seems a sets are "anything" that satisfies the axioms of set theory. It isn't enough to just say that a set is any collection of elements because of various paradoxes. <strong>So is it, for example, a right definition to say that a set is anything that satisfies the ZFC list of axioms?</strong></p>
| Mauro ALLEGRANZA | 108,274 | <p><em>Long Comment</em> to Asaf's answer, trying to add some more background.</p>
<p>We can compare the issue regarding the "definition" of <em>set</em> with Geometry.</p>
<p>Euclid's <a href="https://en.wikipedia.org/wiki/Euclid%27s_Elements" rel="noreferrer">Elements</a> opens with five <a href="http://aleph0.clarku.edu/%7Edjoyce/java/elements/bookI/bookI.html" rel="noreferrer">definitions</a> :</p>
<blockquote>
<ol>
<li><p>A <em>point</em> is that which has no part.</p>
</li>
<li><p>A <em>line</em> is breadthless length. [...]</p>
</li>
<li><p>A <em>surface</em> is that which has length and breadth only.</p>
</li>
</ol>
</blockquote>
<p>They can be of some help in grasping the basic concepts, but hardly they can be conceived as real definitions at all.</p>
<p>In 1899 David Hilbert's published his groundbraking book on the axiomatization of geometry : <a href="https://en.wikipedia.org/wiki/Hilbert%27s_axioms" rel="noreferrer">Grundlagen der Geometrie</a>, based on previous lectures. These are the <a href="https://archive.org/stream/abr1237.0180.001.umich.edu#page/2/mode/2up" rel="noreferrer">first paragraphs</a> (page 3) :</p>
<blockquote>
<p>Let us consider three distinct systems of things. The things composing the first system, we will call <em>points</em> and designate them by the letters <span class="math-container">$A, B, C,\ldots$</span>; those of the second, we will call <em>straight lines</em> and designate them by the letters <span class="math-container">$a, b, c,\ldots$</span>; and those of the third system, we will call <em>planes</em> and designate them by the Greek letters <span class="math-container">$\alpha, \beta, \gamma, \ldots$</span>. [...]</p>
<p>We think of these points, straight lines, and planes as having certain mutual relations, which we indicate by means of such words as “are situated,” “between,” “parallel,” “congruent,” “continuous,” etc. The complete and exact description of these relations follows as a consequence of the <em>axioms of geometry</em>.</p>
</blockquote>
<p>Hilbert's work on foundations of mathematics and logic has been called <a href="http://plato.stanford.edu/entries/formalism-mathematics/" rel="noreferrer">Formalism</a> and it is still the prevailing philosophical view between "working" mathematicians.</p>
<hr />
<p>For <em>set</em> we can consider Georg Cantor's mature definition of <em>set</em> in "<a href="https://books.google.it/books?id=5v_BAgAAQBAJ&pg=PA85" rel="noreferrer">Beiträge zur Begründung der transfiniten Mengenlehre</a>", <em>Mathematische Annalen</em> (1895-97, Engl.transl.1915 - Dover reprint), §1, page 85 :</p>
<blockquote>
<p>By an "aggregate" (<em>Menge</em>) we are to understand any collection into a whole (<em>Zusammenfassung su einem Ganzen</em>) <span class="math-container">$M$</span> of definite and separate objects <span class="math-container">$m$</span> of our intuition or our thought. These objects are called the "elements" of <span class="math-container">$M$</span>.</p>
</blockquote>
<p>Compare it with a modern textbook on set theory : Nicolas Bourbaki, <a href="https://books.google.it/books?id=IL-SI67hjI4C&pg=PA65" rel="noreferrer">Elements of Mathematics : Theory of sets</a> (1968 - 1st French ed : 1939-57), page 65 :</p>
<blockquote>
<p>From a "naive" point of view, many mathematical entities can be considered as collections or "sets" of objects. We do not seek to formalize this notion, and in the formalistic interpretation of what follows, the word "set" is to be considered as strictly synonymous with "term". In particular, phrases such as "let <span class="math-container">$X$</span> be a set" are, in principle, quite superfluous, since every letter is a term. Such phrases are introduced only to assist the intuitive interpretation of the text.</p>
</blockquote>
<hr />
<p>Thus, from a mathematical perspective, <em>points</em> and <em>lines</em> are "things" satisfying the <em>axioms of geoemtry</em>; in the same way, <em>sets</em> are "objects" satisfying the <em>axioms of set theory</em>.</p>
<p>Of course, also if a definition "inside" set theory of the notion of <em>set</em> is impossible, we can still have attempts to elucidate the notion of set from a philosophical perspective.</p>
<p>See e.g. Paul Benacerraf & Hilary Putnam (editors), <a href="https://books.google.it/books?id=cUSbAAAAQBAJ" rel="noreferrer">Philosophy of Mathematics: Selected Readings</a>, (2nd ed : 1983), <strong>Part IV. The concept of set</strong>.</p>
|
87,538 | <p>My problem is the following:</p>
<p>I have a finite surjective morphism $f: X\rightarrow Y$ of noetherian schemes and know that $Y$ is a regular scheme.
(Indeed, in my situation, the two schemes are topologically the same and the arrow is topologically the identity.)</p>
<p>I don't know if $f$ is étale or smooth. But I know that $f$ has a section.</p>
<p>I want to conclude somehow that $X$ is also a regular scheme. What further minimal conditions (on $f$ or $X$) would imply this?</p>
<p>I would be glad about just some hints of what one could do here.</p>
| Mahdi Majidi-Zolbanin | 16,046 | <p>Here is a possible affirmative answer. At any point $P\in X$, write $f_P^{\#}$ for the induced local homomorphism of stalks $\mathcal{O}_{Y,f(P)}\rightarrow\mathcal{O}_{X,P}$. Assume: </p>
<ol>
<li>$X$ is Cohen-Macaulay;</li>
<li>At any point $P\in X$, we have: $\dim \mathcal{O}_{X,P}=\dim {\mathcal{O}}_{{Y,f(P)}}+\dim (\mathcal{O}_{X,P}/f_P^{\#}(\mathfrak{m}_{f(P)}){\mathcal{O}}_{{X,P}})$.</li>
<li>At any point $P\in X$, the ring $\mathcal{O}_{X,P}/f_P^{\#}(\mathfrak{m}_{f(P)}){\mathcal{O}}_{{X,P}}$ is regular.</li>
</ol>
<p>Then 1) and 2), and the assumption that $Y$ is regular imply that for every $P\in X$, $f_P^{\#}$ is flat (see, e.g., Matsumura, <em>Commutative ring theory</em>, Theorem 23.1, p. 179). This flatness plus condition 3) plus the assumption that $Y$ is regular imply that $X$ is also regular (see, e.g., Matsumura, <em>Commutative ring theory</em>, Theorem 23.7, p. 182).</p>
<p>I don't know if conditions 1)-3) are satisfied in the case you are considering, but maybe you can say more about your special case.</p>
|
1,722,948 | <blockquote>
<p>$$\frac{1}{x}-1>0$$</p>
</blockquote>
<p>$$\therefore \frac{1}{x} > 1$$</p>
<p>$$\therefore 1 > x$$</p>
<p>However, as evident from the graph (as well as common sense), the right answer should be $1>x>0$. Typically, I wouldn't multiple the x on both sides as I don't know its sign, but as I was unable to factories the LHS, I did so. How can I get this result algebraically?</p>
| MathematicsStudent1122 | 238,417 | <p>Your answer would be correct were we assuming $x$ is positive. You must keep in mind, though, that the inequality fails for $x<0$ (recall that the inequality sign switches when we multiply by a negative number). </p>
<p>For $x>0$, we have a solution $1>x$. In other words, $0<x<1$ is a solution. </p>
<p>If we have $$\frac 1x > 1$$ and $x<0$, we get solutions $x>1$. But that's an absurdity, and so there are no solutions if $x<0$. </p>
|
1,893,280 | <p>How to show $\frac{c}{n} \leq \log(1+\frac{c}{n-c})$ for any positive constant $c$ such that $0 < c < n$?</p>
<p>I'm considering the Taylor expansion, but it does not work...</p>
| RRL | 148,510 | <p>Hint: With $x = (c/n)/(1 - c/n)$</p>
<p>$$\log(1 + x) = \int_1^{1+x} \frac{dt}{t} \geqslant \frac{x}{1+x}$$</p>
|
1,893,280 | <p>How to show $\frac{c}{n} \leq \log(1+\frac{c}{n-c})$ for any positive constant $c$ such that $0 < c < n$?</p>
<p>I'm considering the Taylor expansion, but it does not work...</p>
| Hagen von Eitzen | 39,174 | <p>Starting with
$$ e^t\ge 1+t\qquad\text{for all }t\in\Bbb R$$
(possibly the most useful inequality about the exponential)
we find by plugging in $-\frac cn$ for $t$
$$ e^{-c/n}\ge 1-\frac cn= \frac{n-c}{n}=$$
and then after taking reciprocals (both sides are positive!)
$$e^{c/n} \le \frac n{n-c}=1+\frac c{n-c}$$
Finally, apply logarithm to obtain
$$\frac cn\le \ln\left(1+\frac c{n-c}\right) $$</p>
|
2,948,118 | <p>I understand that for a function or a set to be considered a vector space, there are the 10 axioms or rules that it must be able to pass. My problem is that I am unable to discern how exactly we prove these things given that my book lists some weird general examples.</p>
<p>For instance: the set of all third- degree polynomials is not a vector space but the set of all fourth degree or less polynomials is? Is this because I can have <span class="math-container">$$f(x) = x^3$$</span> <span class="math-container">$$g(x) = 1 + x - x^3$$</span> <span class="math-container">$$f(x) + g(x) = 1 + x$$</span>
which isn't in 3rd degree where as the fourth degree or less means I can have it not have to be in 4th degree?</p>
<p>Other curious sets I can't seem to discern or wrap my head around include</p>
<p>The set <span class="math-container">$${(x, y)}$$</span> where <span class="math-container">$$x>=0$$</span> and y is a real number. </p>
<p>A 4x4 matrix with symmetrical ordering except the diagonal is 0, 0, 0, 1 in descending order.</p>
<p>And the set of all 2x2 singular matrices.</p>
<p>The most confusing of all is what I like to call the modified set which changes an operation or two into something like this:</p>
<p>Let V be a set of all positive real numbers; determine whether V is a vector space with the operations shown below</p>
<p><span class="math-container">$$x + y = xy$$</span>
<span class="math-container">$$cx = x^c$$</span></p>
<p>If anyone could help explain why these sets break whatever axiom (because I just feel like these sets fit as a vector space but my book says otherwise) it would really help me out. I just started the vector space unit in my class and I gotta say being this clueless is scary.</p>
| Miles Zhou | 596,572 | <p>With any Riemann sum with maximum partition size <span class="math-container">$\delta \lt \epsilon/2$</span>, we have <span class="math-container">$|U-L|<(f(1^+)-f(1^-))*\delta=2\delta<\epsilon$</span>.</p>
|
1,464,747 | <p>I am trying to solve this question:</p>
<blockquote>
<p>How many ways are there to pack eight identical DVDs into five indistinguishable boxes so that each box contains at least one DVD?</p>
</blockquote>
<p>I am very lost at trying to solve this one. My attempt to start this problem involved drawing 5 boxes, and placing one DVD each, meaning 3 DVDs were left to be dropped, but I am quite stuck at this point.</p>
<p>Any help you can provide would be great. Thank you.</p>
| General Thiha | 344,304 | <p>It is called stirling numbers.
S(8,5)=1050 ways</p>
|
1,464,747 | <p>I am trying to solve this question:</p>
<blockquote>
<p>How many ways are there to pack eight identical DVDs into five indistinguishable boxes so that each box contains at least one DVD?</p>
</blockquote>
<p>I am very lost at trying to solve this one. My attempt to start this problem involved drawing 5 boxes, and placing one DVD each, meaning 3 DVDs were left to be dropped, but I am quite stuck at this point.</p>
<p>Any help you can provide would be great. Thank you.</p>
| Ranendra Bose | 354,326 | <p>Identical objects in indistinguishable(identical) boxes.
Use partition of numbers but $x_1 \geq x_2 \geq x_3 \geq x_4 \geq x_5 \geq 1$ since $(3,0,0,0,0)$ is the same as $(0,3,0,0,0)$ //identical offices.
Now $$x_1+x_2+x_3+x_4+x_5=8$$ but $x_i' = x_i-1$ so all $x_i'\geq 0$.
Hence, $$x_1'+x_2'+x_3'+x_4'+x_5'=3 =8-(5 \cdot 1)$$
So, $$(x_1',x_2',x_3',x_4',x_5') \in \{(3,0,0,0,0),(2,1,0,0,0),(1,1,1,0,0,0)\}$$
Ans) 3 ways.</p>
<p>NOTE: Stirling's Number(first or second kind depends on the problem) is for distinct objects in identical boxes. </p>
|
194,954 | <p>Is there a reason to use <code>Hold*</code> attributes for functional code (e.g. no intention to mutate input)? I'd expect performance gains as in pass by value vs pass by reference. </p>
<p>E.g. </p>
<pre><code>data = RandomReal[1, 10^8];
data // Function[x, x[[1]]] // RepeatedTiming
</code></pre>
<blockquote>
<pre><code>{8.*10^-7, 0.0372378}
</code></pre>
</blockquote>
<pre><code>data // Function[x, x[[1]], HoldAllComplete] // RepeatedTiming
</code></pre>
<blockquote>
<pre><code>{8.0*10^-7, 0.0372378}
</code></pre>
</blockquote>
<p>The question is, why doesn't it matter? Or are there cases where it matters, performance tuning wise.</p>
<p>Due to my limited understanding of internals of WL implementation I am puzzled by this and I often have a case where I need to traverse a big expression by keeping a 'reference' to the expression as a whole. So I can do:</p>
<pre><code>parse[bigExpression]:=parse[bigExpression, #] & /@ bigExpression
</code></pre>
<p>or I can make <code>parese</code> a <code>HoldFirst</code>. Or I could even <code>MapIndexed</code> and keep track of the index I could use to access a global <code>bigExpression</code>.</p>
<h2>Edit:</h2>
<p>As noted in comments, the initial example wasn't the best, here I hope is a better one:</p>
<pre><code>data = RandomReal[1, 10^8];
g1[x_] := x[[1]];
f1[x_] := g1[x];
SetAttributes[{g2, f2}, HoldAll];
g2[x_] := x[[1]];
f2[x_] := g2[x];
f1[data] // RepeatedTiming
f2[data] // RepeatedTiming
%%/%
</code></pre>
<blockquote>
<p>{8.38*10^-7, 0.487565}</p>
<p>{7.840*10^-7, 0.487565}</p>
<p>{1.07, 1.}</p>
</blockquote>
| Carl Woll | 45,431 | <p>I don't understand why you think there will be a difference in timing for your example. In both cases the head is evaluated, the data variable is evaluated, the <a href="http://reference.wolfram.com/language/ref/Function" rel="noreferrer"><code>Function</code></a> is evaluated, and then the <a href="http://reference.wolfram.com/language/ref/Part" rel="noreferrer"><code>Part</code></a> is evaluated. It's just a matter of order. Let's take a smaller example so that we can follow the trace:</p>
<pre><code>data = RandomReal[1, 3];
f1 = Function[x, x[[1]]];
f2 = Function[x, x[[1]], HoldAllComplete];
</code></pre>
<p>The first example trace:</p>
<pre><code>Trace[data //f1]
</code></pre>
<blockquote>
<pre><code>{
{f1,Function[x,x[[1]]]},
{data,{0.717387,0.557898,0.102441}},
Function[x,x[[1]]][{0.717387,0.557898,0.102441}],
{0.717387,0.557898,0.102441}[[1]],
0.717387
}
</code></pre>
</blockquote>
<p><em>(the HoldForm wrappers have been suppressed)</em></p>
<p>Notice how the head is evaluated, data is evaluated, the <a href="http://reference.wolfram.com/language/ref/Function" rel="noreferrer"><code>Function</code></a> is evaluated, and then finally <a href="http://reference.wolfram.com/language/ref/Part" rel="noreferrer"><code>Part</code></a> is evaluated.</p>
<p>The second example:</p>
<pre><code>Trace[data //f2]
</code></pre>
<blockquote>
<pre><code>{
{f2,Function[x,x[[1]],HoldAllComplete]},
Function[x,x[[1]],HoldAllComplete][data],
data[[1]],
{data,{0.717387,0.557898,0.102441}},
{0.717387,0.557898,0.102441}[[1]],
0.717387
}
</code></pre>
</blockquote>
<p>In this case, the head is evaluated, the <a href="http://reference.wolfram.com/language/ref/Function" rel="noreferrer"><code>Function</code></a> is evaluated, data is evaluated, and then <a href="http://reference.wolfram.com/language/ref/Part" rel="noreferrer"><code>Part</code></a> is evaluated. The <a href="http://reference.wolfram.com/language/ref/HoldAllComplete" rel="noreferrer"><code>HoldAllComplete</code></a> attribute only changes the order of evaluation, it doesn't eliminate any evaluations.</p>
|
1,130,302 | <p>I am struggling with following challenge in my free time programming project $-$ how is it possible to make reflection vector that reflects along normal with angle that is not larger than some $\alpha$?</p>
<p><img src="https://i.stack.imgur.com/6bmgv.png" alt="Example of normal and limited angle reflection"></p>
<p>I have already seen classical reflection formula $ r = d - 2 (d \cdot n) n $, which unfortunately does not provide answer to my problem.</p>
<p>I tried to fiddle around with JS atan2() function, as described <a href="https://stackoverflow.com/questions/21483999/using-atan2-to-find-angle-between-two-vectors">here</a>, but it didn't work for all cases and I would appreciate some elegant general solution instead of patching special cases.</p>
<p>Thank you in advance.</p>
| mesel | 106,102 | <p>Let <span class="math-container">$G$</span> act on the left coset of <span class="math-container">$H$</span> by left multiplication. Then we have an homomorphism <span class="math-container">$\phi$</span> from <span class="math-container">$G$</span> to <span class="math-container">$S_3$</span>. As <span class="math-container">$G$</span> has no subgroup of index <span class="math-container">$2$</span>, <span class="math-container">$\phi(G)\leq A_3$</span>.</p>
<p>Notice that <span class="math-container">$\phi$</span> can not be trivial thus, <span class="math-container">$\phi(G)=A_3$</span>. Then clearly <span class="math-container">$\ker \phi=H$</span>.</p>
|
3,399,195 | <p>So I've seen various questions with the limit 'equal' to <span class="math-container">$\infty$</span> or that the limit doesn't exist in a case where the function tends to <span class="math-container">$\infty$</span>.</p>
<p>For example, the limit of <span class="math-container">$\sqrt{x}$</span> as <span class="math-container">$x$</span> tends to <span class="math-container">$\infty$</span>. Is the answer <span class="math-container">$\infty$</span> or that the limit doesn't exist?</p>
<p>Obviously the function tends to <span class="math-container">$\infty$</span> as <span class="math-container">$x$</span> tends to <span class="math-container">$\infty$</span> but I don't know what to give as an answer.</p>
<p>I've seen similar questions where the function tends to <span class="math-container">$\infty$</span> as <span class="math-container">$x$</span> tends to a certain value where the answer has been that the limit doesn't exist. I've also seen where, in a similar situation, the limit has been 'equal' to <span class="math-container">$\infty$</span>.</p>
<p>So which is the one to use? What's the difference? Thanks!</p>
| Allawonder | 145,126 | <p>Depends on the context in which you're working.</p>
<p>If it's with the reals, for example, then such limits simply fail to exist. However, we do sometimes work with the extended reals <span class="math-container">$[-\infty,+\infty],$</span> and clearly in that case we can say something like <span class="math-container">$\lim_{x\to+\infty}x=+\infty.$</span></p>
|
3,009,362 | <p>I need to find
<span class="math-container">$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}$$</span></p>
<p>Looking at the graph, I know the answer should be <span class="math-container">$\frac{20}{17}$</span>, but when I tried solving it, I reached <span class="math-container">$0$</span>.</p>
<p>Here are the two ways I approached this:</p>
<p>WAY I:</p>
<p><span class="math-container">$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35} =
\lim_{x\rightarrow -5} \frac{\require{cancel} \cancel{x^2}(2- \frac{50}{x^2})} {\require{cancel} \cancel{x^2}(2+ \frac{3}{x}-\frac{35}{x^2})}
=\frac{2-2}{\frac {42}{5}}=0
$$</span></p>
<p>WAY II:
<span class="math-container">$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35} =
\lim_{x\rightarrow -5} \frac{\require{cancel} \cancel{2}(x^2- 25)} {\require{cancel} \cancel{2}(x^2+ \frac{3}{2}x-\frac{35}{2})}
=\lim_{x\rightarrow -5} \frac{{\require{cancel} \cancel{(x-5)}}(x+5)}{{\require{cancel} \cancel{(x-5)}}(x+3.5)}= \frac{-5+5}{-5+3.5}=0
$$</span></p>
<p>What am I doing wrong here?</p>
<p>Thanks!</p>
| Siong Thye Goh | 306,553 | <p>For your way <span class="math-container">$1$</span>, check the computation of your denominator, it should give you <span class="math-container">$0$</span> again.</p>
<p>For your way <span class="math-container">$2$</span>, check your factorization in your denominator as well.</p>
<p>Use L'hopital's rule:</p>
<p><span class="math-container">$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}= \lim_{x\rightarrow -5} \frac{4x}{4x+3}=\frac{-20}{-17}=\frac{20}{17}$$</span></p>
|
2,628,149 | <p>I am having trouble finding the general solution of the following second order ODE for $y = y(x)$ without constant coefficients: </p>
<p>$3x^2y'' = 6y$<br>
$x>0$</p>
<p>I realise that it may be possible to simply guess the form of the solution and substitute it back into the the equation but i do not wish to use that approach here. </p>
<p>I would appreciate any help, thanks. </p>
| nonuser | 463,553 | <p>You can try to use this approximation $$f(x+h)\doteq f(x)+hf'(x)$$ if $h$ is suficently small. Take $f(x)=\sqrt[4]{x}$, $x=16$ and $h=-1$. So we get
$$ \sqrt[4]{x+h} \doteq \sqrt[4]{x} +{h\over 4\sqrt[4]{x^3}}$$</p>
<p>So $$ \sqrt[4]{15} \doteq \sqrt[4]{16} -{1\over 4\sqrt[4]{16}^3}= 2-{1\over 32}$$</p>
|
912,217 | <p>Let $X$ be a R.V whose pdf is given by
$$f(x)=p\frac{1}{\sqrt{2\pi\sigma_1^2}}\exp\left(-\frac{(x-\mu_1)^2}{2\sigma_1^2}\right)+
(1-p)\frac{1}{\sqrt{2\pi\sigma_2^2}}\exp\left(-\frac{(x-\mu_2)^2}{2\sigma_2^2}\right)$$</p>
<p>clearly $X\sim pN(\mu_1,\sigma_1^2)+(1-p)N(\mu_2,\sigma_2^2)=N(p\mu_1+(1-p)\mu_2,p^2\sigma_1^2+(1-p)^2\sigma_2^2)$</p>
<p>therefore <strong><em>$\kappa=E(e^X)-1=e^{p\mu_1+(1-p)\mu_2+\frac{1}{2}(p^2\sigma_1^2+(1-p)^2\sigma_2^2)}-1$.</em></strong></p>
<p>However, if I let $Y=e^X$, then
Let $G(y)$ be the cdf of $Y$, then
$G(y)=P(Y<y)=P(e^X<y)=p(X<\ln(y))=F_X(\ln(y))$
there fore $g(y)=f(\ln(y))\frac{1}{y}$
so
the pdf $g(y)$ of $Y$ is given by </p>
<p>$$g(y)=p\frac{1}{y\sqrt{2\pi\sigma_1^2}}\exp\left(-\frac{(\ln(y)-\mu_1)^2}{2\sigma_1^2} \right) + (1-p) \frac{1}{y\sqrt{2\pi\sigma_2^2}} \exp\left(-\frac{(\ln(y)-\mu_2)^2}{2\sigma_2^2}\right)$$</p>
<p>i.e $Y=p\times \mathrm{logNormal}(\mu_1,\sigma_1^2)+(1-p)\times \mathrm{logNormal} (\mu_2,\sigma_1^2)$</p>
<p>where $\mathrm{logNormal}(\mu_1,\sigma_1^2)$ means a log-normal distribution.</p>
<p>hence <strong><em>$\kappa=E(Y)-1=pE(\mathrm{logNormal}(\mu_1,\sigma_1^2))+(1-p) E(\mathrm{logNormal} (\mu_2,\sigma_2^2))=
pe^{\mu_1+\frac{\sigma_1^2}{2}}+(1-p)e^{\mu_2+\frac{\sigma_2^2}{2}}-1$.</em></strong></p>
<p>My question is : why I have different values for $\kappa$? Could some one explain for me where I am wrong ? thank you for your time.</p>
| Alex | 38,873 | <p>Let's assume $\Delta x>0$. Then it immediately follows from the 'increasing' part that $ f(x + \Delta x)- f(x) \geq 0$, and, since it is differentiable, the ratio is also positive. </p>
|
1,335,878 | <p>Let $N\unlhd K$ be a normal subgroup of a given group $K$ and let $$q:K\to K/N$$ be the natural quotient map. Let $A\subseteq K$ be a subset of $K$ and let the <a href="https://en.wikipedia.org/wiki/Conjugate_closure" rel="nofollow">conjugate closure</a> of $A$ in $K$ be denoted by $\langle A^K\rangle$.</p>
<p><strong>Question</strong></p>
<p>Is it true that if $\langle A^K\rangle$ is Abelian, then $\langle(q(A))^{K/N}\rangle$ is also Abelian?</p>
| DanielWainfleet | 254,665 | <p>Let $B$ be a real or complex Banach space, and let $B^*$ be its dual space.That is, $B^*$ is the space of continuous linear functionals $g:B\to S$ (where $S$ is $ R$ for real space $X$ ,and $ S$ is $ C$ for complex space $X$)......For $g\in B^*$ we define $||g||=\sup \{|g(x)|/||x|| :x \ne 0\}$ . Now let $ M_g=\{x\in X: g(x)=0\}$.It is an easy exercise to show that $$(1)... \forall x \in X (|g(x)|=||g||.d(x,M_g))$$ and $$(2)... g\ne 0 \implies \forall x\in X (g(x)\ne 0\implies X=\{s x +y : s\in S\wedge y\in M_g\}.$$Now for $g\neq 0$ we say that $ g$ attains its norm iff $$(3)...\exists x (|g(x)|=||g||.||x||\ne 0).$$ Obviously $g$ attains its norm iff $\exists x (||x||=1\wedge |g(x)|=||g||\ne 0).$ And for $ x\in X$ we say that $x$ attains its distance to $ M_g$ iff $$(4)...\exists y\in M_g (||x-y||=d(x,M_g).$$Now from (1) and (2) we can easily prove : For any $g\in B^*$ with $ g\ne 0 $, $$g\text{ attains its norm }\iff$$ $$\forall x\in X (x \text{ attains its distance to } M_g) \iff$$ $$\exists x\in X \backslash M_g (x \text{ attains its distance to } M_g) .$$ In your Q, we have $g(f)=\int_0^1 f $ .So prove that $||g||=1$. (To show $||g||\geq 1$,consider $f(x)=\min (x,1/n)$ for $ n\in N$.) Then show that ($g(f)\ne 0\implies |g(f)|<||f||$).(Consider that for $f\ne 0$ there exists $d>0$ such that $x\in [0,d]\implies |f(x)|<||f||/2 $ .) Hence $$g \text { does not attain its norm.}$$ Hence by (1) $$(||f||=1\wedge g(f)\ne 0)\implies 1=||g||.||f||>|g(f)=||g||d(f,M_g)=d(f,M_g).$$ Furthermore by (4),we have $$g(f)\ne 0\implies \forall m\in M_g (||f-m||>d(f,M_g).$$ So for an example we may take any $f\in X$ for which $||f||=1$ and $\int_0^1f\ne 0$ . For instance $f(x)=x$ for $x\in [0,1]$.</p>
|
84,034 | <p>Vieta's theorem states that given a polynomial
$$ a_nx^n + \cdots + a_1x+a_0$$
the quantities
$$\begin{align*}s_1&=r_1+r_2+\cdots\\
s_2&=r_1 r_2 +r_1 r_3 + \cdots \end{align*}$$
etc., where $r_1,\dots, r_n$ are the roots of the given polynomial, are given by
$$s_i = (-1)^i \frac{a_{n-i}}{a_n} .$$</p>
<p>So my question is: can we use this to find all the roots of a polynomial?</p>
| B. Decoster | 9,143 | <p>Quick answer: you're not going to find the roots in any quicker way with this method. Remember that in general, for polynomials of degree 5 or more, you cannot find explicit formulas for the roots. You simply cannot. With this method or another.</p>
<p>Now, what is Vieta's theorem? It is in fact just expanding the product
$$ a_n \prod_{i=1}^n (x - r_i) = a_nx^n + \cdots + a_1x+a_0$$</p>
|
84,034 | <p>Vieta's theorem states that given a polynomial
$$ a_nx^n + \cdots + a_1x+a_0$$
the quantities
$$\begin{align*}s_1&=r_1+r_2+\cdots\\
s_2&=r_1 r_2 +r_1 r_3 + \cdots \end{align*}$$
etc., where $r_1,\dots, r_n$ are the roots of the given polynomial, are given by
$$s_i = (-1)^i \frac{a_{n-i}}{a_n} .$$</p>
<p>So my question is: can we use this to find all the roots of a polynomial?</p>
| J. M. ain't a mathematician | 498 | <p><em>Numerically</em>, however, for a <em>monic</em> polynomial $p(x)=x^n+c_{n-1} x^{n-1}+\cdots+c_1 x+c_0$, one can treat the Vieta equations relating the $n$ roots $x_k$ and the $n$ remaining coefficients $c_j$ as a system of simultaneous nonlinear equations, and then apply the multivariate version of Newton-Raphson on them.</p>
<p>To wit, note that the Jacobian of the system (I use the quartic case here to keep things simple)</p>
<p>$$\begin{align*}x_1+x_2+x_3+x_4=&-c_3\\x_1 x_2+x_3 x_2+x_4 x_2+x_1 x_3+x_1 x_4+x_3 x_4=&c_2\\x_1 x_2 x_3+x_1 x_4 x_3+x_2 x_4 x_3+x_1 x_2 x_4=&-c_1\\x_1 x_2 x_3 x_4=&c_0\end{align*}$$</p>
<p>is</p>
<p>$$\begin{split}&\mathbf J(\mathbf x)=\mathbf J(x_1,x_2,x_3,x_4)=\\&\small\begin{pmatrix} 1 & 1 & 1 & 1 \\ x_2+x_3+x_4 & x_1+x_3+x_4 & x_1+x_2+x_4 & x_1+x_2+x_3 \\ x_2 x_3+x_4 x_3+x_2 x_4 & x_1 x_3+x_4 x_3+x_1 x_4 & x_1 x_2+x_4 x_2+x_1 x_4 & x_1 x_2+x_3 x_2+x_1 x_3 \\ x_2 x_3 x_4 & x_1 x_3 x_4 & x_1 x_2 x_4 & x_1 x_2 x_3\end{pmatrix}\end{split}$$</p>
<p>with inverse</p>
<p>$$\begin{split}&\mathbf J^{-1}(\mathbf x)=\mathbf J^{-1}(x_1,x_2,x_3,x_4)=\\&\tiny \begin{pmatrix}\frac{x_1^3}{(x_1-x_2) (x_1-x_3) (x_1-x_4)} &\frac{x_1^2}{(x_1-x_2) (x_1-x_3) (x_4-x_1)} &\frac{x_1}{(x_1-x_2) (x_1-x_3) (x_1-x_4)} & \frac1{(x_2-x_1)(x_1-x_3) (x_1-x_4)} \\ \frac{x_2^3}{(x_2-x_1) (x_2-x_3) (x_2-x_4)} &\frac{x_2^2}{(x_1-x_2) (x_2-x_3) (x_2-x_4)} &\frac{x_2}{(x_2-x_1) (x_2-x_3) (x_2-x_4)} & \frac1{(x_1-x_2) (x_2-x_3) (x_2-x_4)} \\\frac{x_3^3}{(x_1-x_3) (x_2-x_3) (x_3-x_4)} &\frac{x_3^2}{(x_1-x_3) (x_3-x_2) (x_3-x_4)} &\frac{x_3}{(x_1-x_3) (x_2-x_3) (x_3-x_4)} & \frac1{(x_1-x_3)(x_3-x_2) (x_3-x_4)} \\\frac{x_4^3}{(x_4-x_1) (x_4-x_2) (x_4-x_3)} &\frac{x_4^2}{(x_1-x_4) (x_4-x_2) (x_4-x_3)} &\frac{x_4}{(x_4-x_1) (x_4-x_2) (x_4-x_3)} & \frac1{(x_1-x_4)(x_4-x_2) (x_4-x_3)}\end{pmatrix}\end{split}$$</p>
<p>and thus the Newton-Raphson iteration function looks like this:</p>
<p>$$\mathbf x-\mathbf J^{-1}(\mathbf x)\begin{pmatrix}s_1+c_3\\s_2-c_2\\s_3+c_1\\s_4-c_0\end{pmatrix}$$</p>
<p>or explicitly,</p>
<p>$$\begin{align*}x_1-&\frac{c_3 x_1^3+c_2 x_1^2+c_1 x_1+c_0+x_1^4}{(x_1-x_2) (x_1-x_3)(x_1-x_4)}\\x_2-&\frac{c_3 x_2^3+c_2 x_2^2+c_1 x_2+c_0+x_2^4}{(x_2-x_1)(x_2-x_3) (x_2-x_4)}\\x_3-&\frac{c_3 x_3^3+c_2 x_3^2+c_1 x_3+c_0+x_3^4}{(x_3-x_1) (x_3-x_2) (x_3-x_4)}\\x_4-&\frac{c_3 x_4^3+c_2 x_4^2+c_1 x_4+c_0+x_4^4}{(x_4-x_1) (x_4-x_2) (x_4-x_3)}\end{align*}$$</p>
<p>This application of the multivariate Newton-Raphson method to the Vieta formulae is called the <a href="https://en.wikipedia.org/wiki/Durand%E2%80%93Kerner_method" rel="noreferrer"><em>Durand-Kerner</em> algorithm</a> for simultaneously determining the roots of a polynomial:</p>
<p>$$x_i^{(k+1)}=x_i^{(k)}-\frac{p(x_i^{(k)})}{\prod\limits_{j\neq i} (x_i^{(k)}-x_j^{(k)})},\qquad i=1\dots n;\; k=0,1,\dots$$</p>
<p>where the $x_i^{(0)}$ are initial approximations (which, as with any Newton-Raphson method, are required). As with the usual Newton-Raphson method, it is quadratically convergent. It is however remarkable that the method is not as finicky with initial conditions as is usual with Newton-Raphson. (In practice, however, one usually takes equally spaced points around a circle in the complex plane as a starting point for the method, where the radius is determined from bounds for the roots.)</p>
<p>See <a href="http://dx.doi.org/10.1007/BF02162564" rel="noreferrer">Kerner's original paper</a> for more details of the derivation.</p>
<hr>
<p>The algorithm is also sometimes referred to as the Weierstrass-Durand-Kerner method, since Karl Weierstrass used this in his constructive proof of the fundamental theorem of algebra. See <a href="http://quod.lib.umich.edu/u/umhistmath/aan8481.0003.001/262?view=image" rel="noreferrer">Weierstrass's paper</a> for more on this.</p>
|
2,972,957 | <p><strong>Artin's Theorem-</strong> Let <span class="math-container">$E$</span> be a field and <span class="math-container">$G$</span> be a group of automorphisms of <span class="math-container">$E$</span> and <span class="math-container">$k$</span> be the set of elements of <span class="math-container">$E$</span> fixed by <span class="math-container">$G$</span>. Then <span class="math-container">$k$</span> is a subfield of <span class="math-container">$E$</span> and <span class="math-container">$E$</span> has finite degree over <span class="math-container">$k$</span> iff <span class="math-container">$G$</span> is finite. In that case, <span class="math-container">$[E:k]=|G|.$</span></p>
<p><strong>Question-</strong> Use Artin's Theorem to show that for every field <span class="math-container">$E$</span> with <span class="math-container">$n$</span> distinct automorphisms, if <span class="math-container">$k$</span> is the fixed field of this set of automorphisms, then <span class="math-container">$[E:k] \ge n$</span>.</p>
<p><strong>Attempt-</strong> If <span class="math-container">$[E:k]$</span> is infinite we are done. </p>
<p>So we take the case where <span class="math-container">$[E:k]$</span> is finite. Let <span class="math-container">$A= \{\sigma_{1}, \ldots\ ,\sigma_{n}\}$</span> be the set of distinct automorphisms of <span class="math-container">$E$</span>. Then <span class="math-container">$k=\{a\in E\ | \sigma_{i}(a)=a\ , 1\le i \le n \}$</span>. </p>
<p>Now the <em>hint given in class</em> says "consider the group <span class="math-container">$G$</span> generated by <span class="math-container">$\sigma_{i}'s$</span> and <span class="math-container">$F=\{a\in E \ |\ \sigma(a)=a\ \forall\ \sigma \in G\}$</span>. (Here <span class="math-container">$G$</span> is the subgroup of all automorphisms of <span class="math-container">$E$</span>)" </p>
<p>But isn't <span class="math-container">$A$</span> itself a group so <span class="math-container">$G$</span> must be equal to <span class="math-container">$A$</span> and thus <span class="math-container">$F=k.$</span> And thus <span class="math-container">$[E:k]=[E:F]$</span>. What did we achieve this way?</p>
| R.C.Cowsik | 293,582 | <p>The group <span class="math-container">$G$</span> generated by <span class="math-container">$n$</span> automorphisms has atleast <span class="math-container">$n$</span> elements, and the fixed field of <span class="math-container">$G$</span> is the same as the fixed field
of the set of <span class="math-container">$n$</span> automorphims. So by Artin <span class="math-container">$$[ E : F] = |G| \geq n$$</span></p>
|
2,252,579 | <p>$$ \lim_{n\to\infty}\left (\frac n {n+1} \right )^{2n} = \lim_{n\to\infty}\left (\frac{n+1}{n} \right )^{-2n} =\lim_{n\to\infty} \left (1 + \frac 1n \right )^{-2n}= \left (\lim_{n\to\infty}\left (1 + \frac 1n \right )^{n} \right )^{-2} = e^{-2}$$</p>
<p>What I don't understand is why is it a -2 and not +2? Also, is there a better way to solve this?</p>
<p>Thank you</p>
| Mark Viola | 218,419 | <p>Note that $x=\left(\frac{1}{x}\right)^{-1}$. But, perhaps a better way to avoid any confusion is to write</p>
<p>$$\lim_{n\to \infty}\left(\frac{n}{n+1}\right)^{2n}=\lim_{n\to \infty}\left(\frac{1}{\left(1+\frac1n\right)^n}\right)^2=\left(\frac{1}{e}\right)^2=\frac1{e^2}=e^{-2}$$</p>
|
221,017 | <p>If I have the following list:</p>
<pre><code>https://pastebin.com/nqyf4yY5
</code></pre>
<p>How can I find the closest value to "89" in the "T[C]" column and its corresponding value in the "DH,aged-DH,unaged (J/g)" column?.</p>
<p>Thank you in advanced,</p>
| Rohit Namjoshi | 58,370 | <pre><code>data = Import["~/Downloads/data.txt"] // ToExpression // Part[#, 6 ;;] &;
nf = Nearest[data[[All, 2]] -> {"Index", "Element"}];
data[[nf[89][[1, 1]]]]
(* {3.87*10^-6, 89.2592, 5.13099, 0.0107504, 0.0102723, 0.0123268, 0.0000417117} *)
</code></pre>
|
221,017 | <p>If I have the following list:</p>
<pre><code>https://pastebin.com/nqyf4yY5
</code></pre>
<p>How can I find the closest value to "89" in the "T[C]" column and its corresponding value in the "DH,aged-DH,unaged (J/g)" column?.</p>
<p>Thank you in advanced,</p>
| creidhne | 41,569 | <p>With your data, assuming the column headings are in row 5:</p>
<pre><code>data[[5]]
(* {"Time(s)","T[C]","K(T)=k^(1/n)","dx/dT","x(t)","DH,aged-DH,unaged (J/g)","Check dx"} *)
</code></pre>
<p>... find the column numbers for columns "T[C]" and "DH,aged-DH,unaged (J/g)":</p>
<pre><code>{c1, c2} = Flatten@{
Position[data[[5]], "T[C]"],
Position[data[[5]], "DH,aged-DH,unaged (J/g)"]};
</code></pre>
<p>It's easier to search the data without the extra rows at the top. When there are multiple values that match, we find only the first one. Get the nearest value and the matching column:</p>
<pre><code>values = data[[6 ;;]];
v = 89.;
First@Extract[values[[All, {c1, c2}]],
Position[values[[All, c1]], First@Nearest[values[[All, c1]], v]]]
(* {89.2592, 0.0123268} *)
</code></pre>
|
3,314,561 | <p>Consider the triangle <span class="math-container">$PAT$</span>, with angle <span class="math-container">$P = 36$</span> degres, angle <span class="math-container">$A = 56$</span> degrees and <span class="math-container">$PA=10$</span>. The points <span class="math-container">$U$</span> and <span class="math-container">$G$</span> lie on sides TP and TA respectively, such that PU = AG = 1. Let M and N be the midpoints of segments PA and UG. What is the degree measure of the acute angle formed by the lines MN and PA?</p>
<p>It would be very helpful if anyone had a solution using complex numbers to this problem.</p>
| Philip Roe | 430,997 | <p>It is a fact that not all physical problems have smooth solutions. Often this situation arises from a set of conservation laws that are expressed mathematically by applying such laws to a finite control volume to obtain an integral equation. Then we let the size of the control volume go to zero and arrive at some PDEs if the flow is smooth. But then we discover that the PDEs are unable to solve many important problems and have to rethink our strategy.</p>
<p>When this first occurred to me I found it a bit shocking because surely differential calculus was the natural language for describing continua? After a bit I realised that the integral calculus is more fundamental. It can be applied to functions that are more general (Anything can be integrated, but not everything can be differentiated) and it is the form in which much physical knowledge comes to us. </p>
<p>I suspect you felt the same surprise that I did. I thought that I wanted to solve differential equations, so why would I start integrating things? The truth is the reverse. I really want to solve integral equations, and the PDE is a powerful tool, but only if it is valid. That it often is should come as another surprise.</p>
|
122,468 | <p>I know how to find the number of solutions to the equation:</p>
<p><span class="math-container">$$a_1 + a_2 + \dots + a_k = n$$</span></p>
<p>where <span class="math-container">$n$</span> is a given positive integer and <span class="math-container">$a_1$</span>, <span class="math-container">$a_2$</span>, <span class="math-container">$\dots$</span>, <span class="math-container">$a_n$</span> are positive integers. The number of solutions to this equation is:</p>
<p><span class="math-container">$$\binom{n - 1}{k - 1}$$</span></p>
<p>This can be imagined as <span class="math-container">$n$</span> balls arranged on a straight line and selecting <span class="math-container">$k - 1$</span> gaps from a total of <span class="math-container">$n - 1$</span> gaps between them as partition boundaries. The <span class="math-container">$k - 1$</span> partition boundaries divide the <span class="math-container">$n$</span> balls into <span class="math-container">$k$</span> partitions. The number of balls in the <span class="math-container">$i$</span>th partition is <span class="math-container">$a_i$</span>.</p>
<p>Now, I don't know how to find the number of solutions to the same equation when we have an additional constraint: <span class="math-container">$$0 < a_1 \leq a_2 \leq \dots \leq a_k.$$</span> Could you please help me?</p>
| Minh Hien | 821,566 | <p>Number of solutions is <span class="math-container">$S(n, k)$</span>.</p>
<p>If <span class="math-container">$a_1=1$</span>, we need the number of solutions of: <span class="math-container">$a_2+⋯+a_k=n-1$</span>, which <span class="math-container">$S(n-1, k-1)$</span></p>
<p>If <span class="math-container">$a_1>1$</span>, minus all <span class="math-container">$k$</span> numbers by 1: <br/>
<span class="math-container">$$(a_1-1)+(a_2-1)⋯+(a_k-1)=n-k$$</span>
The number of solutions of this is <span class="math-container">$S(n-k, k)$</span>.</p>
<p>So <span class="math-container">$S(n, k)=S(n-1, k-1)+S(n-k, k)$</span></p>
|
2,398,215 | <p>If $f$ is continuous on $\mathbb{R}$ any of the following conditions are satisfied then $f$ must be a constant.</p>
<p>(1).$f(x)=f(mx),\forall x\in \mathbb{R},|m|≠1,m\in \mathbb{R}$</p>
<p>(2).$f(x)=f(2x+1),\forall x\in \mathbb{R}$</p>
<p>(3).$f(x)=f(x^2),\forall x\in \mathbb{R}$.</p>
<p>Suppose $f$ satisfy (1). I assumed, $f$ is not a constant, since f is not defined in the closed and bounded interval I am not able to use the major theorems of continuity.</p>
<p>Similarly I tried to apply the same theorems to (2) and (3). I am not able to proceed. Please help me to solve this question.</p>
| Gabriel Romon | 66,096 | <ol start="2">
<li>Let $y=x+1$, so that $\forall y \in \mathbb R, f(y-1)=f(2y-1)$. Then define $g(y)=f(y-1)$, so that $\forall y, g(y)=g(2y)$. </li>
</ol>
<p>Using the result of 1., $g$ is constant and equal to $g(0)=f(-1)$, so $f$ is also constant and equal to $f(-1)$.</p>
<ol start="3">
<li>Note that $f(-x)=f((-x)^2)=f(x^2)=f(x)$ so $f$ is even.</li>
</ol>
<p>Consider $x>0$. Prove by induction on $n$ that $f(x^{1/n})=f(x)$ and let $n\to \infty$. Since $x^{1/n} \to 1$, then $f(x^{1/n})\to f(1)$, hence $f(x)=f(1)$.</p>
<p>Since $f$ is even, $f(x)=f(1)$ everywhere except $0$, and since $f$ is continuous, $f(0)=f(1)$.</p>
|
809,336 | <p>If G is abelian then factor group G/H is abelian.</p>
<p>How about the converse of this statement? </p>
<p>Is it true?</p>
| Brian Fitzpatrick | 56,960 | <p>If $G$ is your favorite nonabelian group, then $G/G$ is the trivial group!</p>
<p>Also, if $H$ has prime index $p$, then $G/H\simeq\Bbb Z_p$.</p>
|
1,232,532 | <p>First, I'm not looking for an answer here, I'm just looking to understand the problem so that I can prove it. I'm trying to analyzing the worst case running time of an algorithm, and it must has summation notation. What keeping me back is that I don't understand how to express <code>doSomething(n-j)</code> in summation ( I know that <code>doSomething(k)</code> takes <code>c * k</code> operations for some constant <code>c > 0</code> (stated in the problem), so it is not constant in this case). The other two loops have starting points (e.g. <code>i = 1</code> or <code>j = i</code>). Anyway, the pseudo-code is stated blow: </p>
<pre><code>function(n)
for int i from 1 to n
for int j from i to n
doSomething(n - j)
endfor
endfor
endfunction
</code></pre>
<p>I can express the nested for loop in summation as follow:</p>
<p>$\sum_{i=1}^n \sum_{j=i}^n doSomething(n-j)$</p>
<p>I think I need one more summation, it's just that I don't know how to express it, maybe something like:</p>
<p>$\sum_{k=?}^{n-j}$</p>
<p>I could be wrong here.
Could anyone please provide me with some hints in this problem? Thanks a lot.</p>
<p>EDIT: since <code>doSomething(k)</code> takes <code>c * k</code> operations, can I express it as follow:</p>
<p>$\sum_{i=1}^n \sum_{j=i}^n c*k$</p>
| Michael Hoppe | 93,935 | <p>Lhs is defined for $x\geq1/2$; for those $x$ the lhs is non-negative whereas the rhs is negative.</p>
<p>Some explanation: squaring is “if” but not “iff”: if $x=5$ then $ x^2=25$, but if $x^2=25$ then $x$ may be $-5$ as well.</p>
|
680,319 | <p>Let's restate this question in using mathematical notation. Let $n,k \in \mathbb{N}$. Let $f(n)=\left\lfloor{\frac{n}{k}}\right\rfloor$. Is it possible to rewrite this using the addition, multiplication, and exponentiation operators? I know it's possible for the case where $k=1$. Quite simply, note that $f(n)=\left\lfloor{\frac{n}{1}}\right\rfloor=\lfloor{n}\rfloor=n$. I know it's possible when $k=2$. Let $f(n)=\left(\frac{n}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}\right)(-1)^n$. The trick there was to start halfway between $\frac{n}{2}-\frac{1}{2}$ and $\frac{n}{2}$ and using the $(-1)^n$ term to figure out the rest. </p>
<p>I'm wondering if it's possible if $k=3$? I think it's possible with the use of infinite sums and trigonometric functions (though I haven't thought about that yet). However, I'd like to avoid those if necessary. I feel they may not be necessary -- though I may be wrong.</p>
<p>Even a slight discussion about the topic would be great.</p>
<p>EDIT: </p>
<p>So, I like the @user44197's solution. It gives rise to the solution I was looking for -- which was simpler than I thought. Let $f(n)=\left\lfloor{\frac{n}{k}}\right\rfloor$, where $n$ is a variable and $k$ is a constant (for simplicity, let $k$ be some constant in the naturals). Why not treat it like a recursion? Let $R(n) = R(n-k) + 1$. To solve this, we need to solve the characteristic polynomial $\lambda^k-1=0$. Therefore, $R_{Homogeneous}(n)=\sum_{i=0}^{k-1}n^{i}c_{i}(1)^n=\sum_{i=0}^{k-1}n^{i}c_{i}$ where $c_i$ is the coefficient that will solve the recursion. Now you just need to find the non-homogeneous part, solve for the coefficients, and you're done.</p>
| user44197 | 117,158 | <p>Here is something to get started.
Suppose you have the function
$$
f(n,k) = \left\{\begin{array}{ll}1 & \text{if $k | n$}\\\\0 & \text{if $k \not | n$}\end{array}\right.
$$
Then you should be able to cobble together the floor function (havent't thought this through).</p>
<p>Let
$$
\theta = e^{i 2 \pi/k}
$$
Then
$$
f(n,k) = \frac{1}{k}\sum_{m=0}^{k-1} \theta^ { n m}
$$
will give you such a function.</p>
|
2,170,382 | <p>I'm working on a question that asks to:</p>
<p>Find the area in the first quadrant bounded by the curves;
$\ xy = 1, xy=5, y=e^2x, y=e^5x $. </p>
<p>I would very much appreciate help solving this question (including the method of how to find the transformation expressions for $\ u$ and $v$ to use in the Jacobian). Thanks!</p>
| DonAntonio | 31,254 | <p>An idea: </p>
<p>Try to write everything as function of $\;x\;$ to find intersection points. After all we have two hyperbolas and two straight lines:</p>
<p>$$\begin{cases}y=\frac1x\\{}\\y=\frac5x\\{}\\y=e^2x\\{}\\y=e^5x\end{cases}\;\;\implies\begin{cases}e^2x=\frac1x\implies x=\frac1e\\{}\\e^2x=\frac5x\implies x=\frac{\sqrt5}e\end{cases}$$</p>
<p>and do the same with $\;e^5x\;$ . Now, since $\;\frac5x>\frac1x\;$ and $\;e^2x<e^5x\;\;\forall\,x>0$ , find out the limits...and you may need more than one integral as the domain of integration isn't a simple one (a geometric sketch could help a lot)</p>
|
4,488,991 | <p>Let <span class="math-container">$M,M'$</span> be oriented connected compact smooth manifolds of the same dimension, let <span class="math-container">$S$</span> be a smooth manifold,
and let <span class="math-container">$\nu : S\times M\rightarrow M'$</span> be some smooth map.
Let <span class="math-container">$\nu_s : M\rightarrow M'$</span> denote <span class="math-container">$\nu_s(\cdot) = \nu(s,\cdot)$</span> for <span class="math-container">$s\in S$</span>.</p>
<p>If I'm not mistaken, we can formalize the claim that the per-<span class="math-container">$s$</span> degree map <span class="math-container">$s \mapsto \mathrm{degree}(M\xrightarrow{\nu_s} M')$</span>,
as a map <span class="math-container">$S \rightarrow \mathbb{Z}$</span>, should depend continuously on <span class="math-container">$s \in S$</span> (hence be locally constant) by using e.g. de Rham theory.</p>
<p>(Details: Fix some top-degree differential form <span class="math-container">$\omega$</span> on <span class="math-container">$M'$</span> which is non-exact; then <span class="math-container">$\mathrm{deg}(\nu_s) = (\int_M \nu_s^*\omega)/(\int_{M'}\omega)$</span>, which we can probably argue depends continuously on <span class="math-container">$s$</span>.)</p>
<hr />
<p>More generally, if <span class="math-container">$M,M'$</span> are Poincaré duality spaces (with chosen fundamental classes), and <span class="math-container">$S$</span> is a topological space and "smooth" everywhere above is replaced by "continuous", does the same claim hold? How can the proof be formalized?</p>
| poetasis | 546,655 | <p>In support of the answer by @heropup , all triples where
<span class="math-container">$\space B =\dfrac{A^2-1}{2}\space$</span> and
<span class="math-container">$\space C-B=1\space$</span> can be generated by</p>
<p><span class="math-container">\begin{align}
A &=&&2k+1\\
B &= 2 k^2 + &&2 k\\
C &= 2 k^2 + &&2 k + 1
\end{align}</span></p>
<p>A similar set of all-primitive where
<span class="math-container">$\space C-A=2\space$</span> can be generated by
<span class="math-container">$\space A=4n^2-1\qquad B=4n\qquad C=4n^2+1\qquad$</span>
et seq
<span class="math-container">$(3,4,5)\quad (15,8,17)\quad (35,12,37)
\quad (63,16,65)\quad (99,20,101)\space\cdots$</span></p>
<p>It would be interesting to see if the latter triples have properties similar to the former.</p>
|
2,657,301 | <p>On the set of natural numbers$\mathbb { N} $, define the operations $a \oplus b := \max(a,b)$
and $a\otimes b := a+b$ Is $(\mathbb {N},\oplus,\otimes)$ is ring? commutative ring with unity? Field?</p>
<p>My solution :</p>
<p>1- $(\mathbb {N},\oplus) $ is abelian groub because : </p>
<p>a. It is comutative
$a \oplus b = \max(a,b)= \max(b,a) =b \oplus a$</p>
<p>b. It is associative
$(a \oplus b) \oplus c =a \oplus( b\oplus c) =\max(a,b,c)$ </p>
<p>C. The identity of element $a$is $a $
$$a \oplus a= \max(a,a)= a $$
d. The inverse also a
$a \oplus a = \max(a,a)=a $</p>
<p>$(\mathbb {N},\times) $ </p>
<p>1.it is comutative :
$a\otimes b = a+b= b+a= b\otimes a$</p>
<ol start="2">
<li><p>$\otimes $has a identity and it is 0
$a\otimes 0 = a+0=a$</p></li>
<li><p>$\otimes $ has inverse and inverse of a is -a
$a\otimes -a = a-a=0$</p></li>
</ol>
<p>Distributive law:
$a\otimes (b\oplus c) =a\otimes (max (b,c))= a+max( b,c) $</p>
<p>It is true ? If not, why?</p>
<p>Thanks </p>
| Stefan4024 | 67,746 | <p>The identity can't be dependent on $a$. Anyway you want to have $a = a \oplus e = \max(a,e)$. So for each $a$ we must have $a \ge e$, which in natural numbers is $0$ (or $1$ if you consider $\mathbb{N}$ as the set of positive integers). But then there aren't any inverses, as for $a \not = 0$ we have that $a \oplus b = \max(a,b) \ge a > 0$, so there isn't equality. Therefore $(\mathbb{N},\oplus)$ isn't a group.</p>
|
2,657,301 | <p>On the set of natural numbers$\mathbb { N} $, define the operations $a \oplus b := \max(a,b)$
and $a\otimes b := a+b$ Is $(\mathbb {N},\oplus,\otimes)$ is ring? commutative ring with unity? Field?</p>
<p>My solution :</p>
<p>1- $(\mathbb {N},\oplus) $ is abelian groub because : </p>
<p>a. It is comutative
$a \oplus b = \max(a,b)= \max(b,a) =b \oplus a$</p>
<p>b. It is associative
$(a \oplus b) \oplus c =a \oplus( b\oplus c) =\max(a,b,c)$ </p>
<p>C. The identity of element $a$is $a $
$$a \oplus a= \max(a,a)= a $$
d. The inverse also a
$a \oplus a = \max(a,a)=a $</p>
<p>$(\mathbb {N},\times) $ </p>
<p>1.it is comutative :
$a\otimes b = a+b= b+a= b\otimes a$</p>
<ol start="2">
<li><p>$\otimes $has a identity and it is 0
$a\otimes 0 = a+0=a$</p></li>
<li><p>$\otimes $ has inverse and inverse of a is -a
$a\otimes -a = a-a=0$</p></li>
</ol>
<p>Distributive law:
$a\otimes (b\oplus c) =a\otimes (max (b,c))= a+max( b,c) $</p>
<p>It is true ? If not, why?</p>
<p>Thanks </p>
| rschwieb | 29,335 | <p>As soon as you saw $a\oplus a=a$ for every $a$, you should have immediately seen that if this were a ring, $a=a\oplus a\oplus -a=a\oplus -a=0$, a contradiction if there exists $a\neq 0$. So it clearly can't be a ring.</p>
<p>But it is easy to see that it is a sub-semiring of the <a href="https://en.wikipedia.org/wiki/Max-plus_algebra" rel="nofollow noreferrer">max-plus algebra</a>.</p>
|
3,888,766 | <p>I need to prove this identity:</p>
<p><span class="math-container">$$2\cos\left(2\pi ft + \phi\right)\cos(2\pi ft) = \cos(4\pi ft + \phi) + \cos(\phi)$$</span></p>
<p>I know I have to use some identity or property but I can't find any to do it.</p>
| Fourier_T | 575,542 | <p>Multiply the two matrices as shown and you will have following equations:</p>
<ol>
<li><span class="math-container">$$a+3b=1$$</span></li>
<li><span class="math-container">$$4a-b=0$$</span>
From (2), we have: <span class="math-container">$$b=4a$$</span>
Putting in (1), we get: <span class="math-container">$$a+3(4a)=1$$</span>
this implies, <span class="math-container">$$13a=1$$</span>
So <span class="math-container">$$a=\frac{1}{13} $$</span> and <span class="math-container">$$b=\frac{4}{13}$$</span></li>
</ol>
|
19,596 | <p>I am trying to rearrange and manipulate some vector differential equations in <em>Mathematica</em>. As far as I understand you have to tell <em>Mathematica</em> that a variable is a vector by specifying the components of the vector. For example</p>
<pre><code>r = {x, y, z};
</code></pre>
<p>If I want to define vector fields I have to do it in the following way</p>
<pre><code>v = {vx[r, t], vy[r, t], vz[r, t]};
n = {nx[r, t], ny[r, t], nz[r, t]};
</code></pre>
<p>Now I can use the <code>Div</code> operator to express a differential equation, in this case the continuity equation in hydrodynamics:</p>
<pre><code>D[n, t] + Div[n*v, r] == 0
</code></pre>
<p>My problem is the output I get from this. It looks absolutely horrible and I can't do anything with it.</p>
<blockquote>
<p>({0, 0, 0},1) ({0, 0, 1},0) {nx
[{x, y, z}, t] + vz[{x, y, z}, t] nz [{x, y, z}, t] +
({0, 0, 1},0) ({0, 1, 0},0) nz[{x, y, z}, t] vz [{x, y, z}, t] + vy[{x,
y, z}, t] ny [{x, y, z}, t] +
({0, 1, 0},0) ({1, 0, 0},0) ny[{x, y, z}, t] vy [{x, y, z}, t] + vx[{x,
y, z}, t] nx [{x, y, z}, t] +
({1, 0, 0},0) nx[{x, y, z}, t] vx [{x, y, z}, t], ({0, 0, 0},1)<br>
({0, 0, 1},0) ny [{x, y, z}, t] + vz[{x, y, z}, t] nz<br>
[{x, y, z}, t] +
({0, 0, 1},0) ({0, 1, 0},0) nz[{x, y, z}, t] vz [{x, y, z}, t] + vy[{x,
y, z}, t] ny [{x, y, z}, t] +
({0, 1, 0},0) ({1, 0, 0},0) ny[{x, y, z}, t] vy [{x, y, z}, t] + vx[{x,
y, z}, t] nx [{x, y, z}, t] +
({1, 0, 0},0) nx[{x, y, z}, t] vx [{x, y, z}, t], ({0, 0, 0},1)<br>
({0, 0, 1},0) nz [{x, y, z}, t] + vz[{x, y, z}, t] nz<br>
[{x, y, z}, t] +
({0, 0, 1},0) ({0, 1, 0},0) nz[{x, y, z}, t] vz [{x, y, z}, t] + vy[{x,
y, z}, t] ny [{x, y, z}, t] +
({0, 1, 0},0) ({1, 0, 0},0) ny[{x, y, z}, t] vy [{x, y, z}, t] + vx[{x,
y, z}, t] nx [{x, y, z}, t] +
({1, 0, 0},0) nx[{x, y, z}, t] vx [{x, y, z}, t]} == 0</p>
</blockquote>
<p>I would like <em>Mathematica</em> to write the equation in vectorial form so that I can rearrange it and use vector identities to manipulate it.</p>
<p>Is there a way to do this?</p>
| jonsq | 5,506 | <p>Our package VEST (Vector Einstein Summation Tools) is designed to do exactly this. It's described <a href="http://arxiv.org/abs/1309.2561">here</a> and can be downloaded from <a href="https://github.com/jonosquire/VEST">github</a> along with a tutorial. </p>
<p>I realize this question is now very old, but thought this answer might be helpful for others looking for similar capability. </p>
|
66,671 | <p>$$\text{ABC- triangle:} A(4,2); B(-2,1);C(3,-2)$$<br>
Find a D point so this equality is true:</p>
<p>$$5\vec{AD}=2\vec{AB}-3\vec{AC}$$</p>
| Américo Tavares | 752 | <p>$$\text{The given vectors } \overrightarrow{AB}=B-A\text{ and }\overrightarrow{AC}=C-A \text{ and the solution }D=A+\overrightarrow{AD}$$</p>
<p><img src="https://i.stack.imgur.com/Iunwu.jpg" alt="enter image description here"></p>
<p>Let $(x,y)$ be the coordinates of $D$. The equation</p>
<p>$$5\overrightarrow{AD}=2\overrightarrow{AB}-3\overrightarrow{AC}\tag{0}$$</p>
<p>means that</p>
<p>$$5(x-4,y-2)=2(-2-4,1-2)-3(3-4,-2-2),\tag{1}$$</p>
<p>because the vectors $\overrightarrow{AD}=D-A$, $\overrightarrow{AB}=B-A$ and
$\overrightarrow{AC}=C-A$.</p>
<p>The vectors $5\overrightarrow{AD}=5\left( D-A\right) =\left( 5D-5A\right) $,
$2\overrightarrow{AB}=\left( 2B-2A\right) $, etc.</p>
<p>A possible way of solving the equation $(1)$ is as follows.</p>
<p>$$5(x-4,y-2)=2(-2-4,1-2)-3(3-4,-2-2)$$</p>
<p>$$\begin{eqnarray*}
&\Leftrightarrow &(5x-20,5y-10)=2(-6,-1)-3(-1,-4) \\
&\Leftrightarrow &(5x-20,5y-10)=(-12,-2)-(-3,-12) \\
&\Leftrightarrow &(5x-20,5y-10)=(-12,-2)+(3,12) \\
&\Leftrightarrow &(5x-20,5y-10)=(-12+3,-2+12) \\
&\Leftrightarrow &(5x-20,5y-10)=(-9,10) \\
&\Leftrightarrow &\left\{
\begin{array}{c}
5x-20=-9 \\
5y-10=10
\end{array}
\right. \Leftrightarrow \left\{
\begin{array}{c}
x=\frac{11}{5} \\
y=4
\end{array}
\right.
\end{eqnarray*}\tag{2}$$</p>
|
478,517 | <blockquote>
<p>Construct a topological mapping of the open disk $|z|<1$ onto the whole plane.</p>
</blockquote>
<p>I represent $z=re^{i\theta}$. I thought about the bijection from $(0,1)$ to $(0,\infty)$, which is given by $x\rightarrow \dfrac1x-1$. Applying this to the norm, we will get the mapping $re^{i\theta}\rightarrow\left(\dfrac1r-1\right)e^{i\theta}$. The only problem is that the point $0$ has not been mapped to or from yet. If I map $0$ to itself, the map becomes non-continuous.</p>
| miracle173 | 11,206 | <p>Put a sphere on a plane. Draw a line from the center $M$ of the sphere to a point $P$ in the plane. This line intersects the surface of lower hemisphere of the sphere in a point $Q$. so you get a 1-1 mapping from the surface of the lower hemisphere to the plan. Draw Line through $Q$ vertical to the plane. This line intersects the plane in $R$. This gives you a 1-1 mapping from the lower hemisphere to the projection of this hemispere to the plane which is disk. Both give you a 1-1 from the open disk to the plane. A beam through $R$ vertical to the plane is mirrored by the sphere in $Q$ to $P$.
<img src="https://i.stack.imgur.com/TB2Ck.jpg" alt="enter image description here"></p>
|
454,040 | <p>I need to know whether There exists any continuous onto map from $(0,1)\to (0,1]$</p>
<p>could any one give me any hint?</p>
| Mathronaut | 53,265 | <p>$f(x)=|\sin (\pi x)|$ will work</p>
|
3,128,862 | <p>I'm really stuck in this fairly simple example of conditional probability, I don't understand the book reasoning:</p>
<blockquote>
<p>An ordinary deck of 52 playing cards is randomly divided into 4 piles of 13 cards each.
Compute the probability that each pile has exactly 1 ace. </p>
<p><strong>Solution.</strong> Define events <span class="math-container">$E_i, i = 1, 2, 3, 4$</span>, as follows:</p>
<p><span class="math-container">$E_1$</span> = {the ace of spades is in any one of the piles}</p>
<p><span class="math-container">$E_2$</span> = {the ace of spades and the ace of hearts are in different piles}</p>
<p><span class="math-container">$E_3$</span> = {the aces of spades, hearts, and diamonds are all in different piles}</p>
<p><span class="math-container">$E_4$</span> = {all 4 aces are in different piles}</p>
<p>The desired probability is <span class="math-container">$P(E_1E_2E_3E_4)$</span>, and by the multiplication rule, </p>
<p><span class="math-container">$P(E_1E_2E_3E_4) = P(E_1)P(E_2|E_1)P(E_3|E_1E_2)P(E_4|E_1E_2E_3)$</span></p>
<p>Now, <span class="math-container">$P(E_1) = 1$</span> since <span class="math-container">$E_1$</span> is the sample space S. Also, <span class="math-container">$P(E_2|E_1) = \frac{39}{51}$</span> since the pile containing the ace of spades will receive 12 of the remaining 51 cards (...)</p>
</blockquote>
<p>I was lost with <span class="math-container">$P(E_2|E_1)$</span>, I don't understand why it's <span class="math-container">$\frac{39}{51}$</span>. I tried to think like this: </p>
<p>by definition, <span class="math-container">$P(E_2|E_1) = P(E_1E_2)/P(E_1)$</span> and, since <span class="math-container">$P(E_1) = 1$</span>, <span class="math-container">$P(E_2|E_1) = P(E_1E_2)$</span>. But it's obvious that <span class="math-container">$E_2 \subset E_1 \Rightarrow E_1 \cap E_2 = E_2 \Rightarrow (E_2|E_1) = P(E_2)$</span>. </p>
<p>So I tried to calculate <span class="math-container">$P(E_2)$</span> to see if it matched the answer of the book. By definition, <span class="math-container">$E_2$</span> is the event where the ace of spades and the ace of hearts are in different piles. So the sample space is <span class="math-container">$52\choose13,13,13,13$</span>. Now, suppose you take out the ace of hearts and the ace of spades of your deck, now you have 50 cards and there are <span class="math-container">$50\choose12,12,13,13$</span> ways of dividing the deck and <span class="math-container">$4\choose2$</span> ways of deciding which piles receive 12 cards and which receives 13. After making this division, there are <span class="math-container">$2!$</span> ways, for each result, to put back the 2 aces you took off (each one in one of the piles with 12 cards). So</p>
<p><span class="math-container">$P(E_2) =$</span> <span class="math-container">${50}\choose{12,12,13,13}$$ {4}\choose{2}$$2!$$/$${52}\choose{13,13,13,13}$$= \frac{39}{51}$</span></p>
<p>In the end, I found the right answer for <span class="math-container">$P(E_2)$</span>, but I went through a whole line of reasoning that was not trivial. In the other hand, the book apparently deduces it in a trivial way:</p>
<blockquote>
<p>since the pile containing the ace of spades will receive 12 of the remaining 51 cards</p>
</blockquote>
<p>What I'm not getting?</p>
| Floris Claassens | 638,208 | <p>An automorphism is a permutation which is by definition a bijection from <span class="math-container">$\mathcal{P}$</span> to <span class="math-container">$\mathcal{P}$</span>. Let <span class="math-container">$\phi$</span> be an automorphism and <span class="math-container">$L,L'$</span> parallel lines. Suppose <span class="math-container">$\phi(L)$</span> and <span class="math-container">$\phi(L')$</span> intersect in <span class="math-container">$y$</span>, then <span class="math-container">$\phi^{-1}(y)\in L,L'$</span> from which it follows that <span class="math-container">$L$</span> and <span class="math-container">$L'$</span> intersect and thus are not parallel.</p>
|
2,867,042 | <blockquote>
<p>Find the value of
$$\tan\theta \tan(\theta+60^\circ)+\tan\theta \tan(\theta-60^\circ)+\tan(\theta + 60^\circ) \tan(\theta-60^\circ) + 3$$
(The answer is $0$.)</p>
</blockquote>
<p>My try: Let $\theta$ be $A$, $60^\circ -\theta$ be $B$, and $60^\circ + \theta$ be $C$. I simplified the result and got the expression
$$1 + 1/\cos A\cos B\cos C$$ but after that I can't simplify it.</p>
| Ross Millikan | 1,827 | <p>The number of symmetric chains is much higher. You have $n \choose i$ ways to choose $C_i$. Then you choose the $n-2i$ elements that will get added, but you choose them in order, so the total number of chains is $${n \choose i}{n-i \choose n-2i}(n-2i)!$$
For example, if $n=6, i=2$ we have ${6 \choose 2}=15$ ways to choose $C_2$, then $4$ ways to choose the element to add to make $C_3$, and $3$ ways to choose the element to add to make $C_4$, for a total of $15 \cdot 4 \cdot 3=180$. The above formula gives $15 \cdot 6 \cdot 2=180$</p>
|
1,654,545 | <p>My teacher explained this problem to us - "There are $3$ mailboxes. $3$ people put letters in at random. There is no preference for any of the $3$ mailboxes. Compute the probability that each mailbox contains $1$ letter."</p>
<p>I tried this problem on my own and got the wrong answer. I understand the teacher's solution but I don't get what went wrong with my answer. Can someone explain where I went wrong?</p>
<p>My answer: Treat each outcome as a sequence of $3$ numbers denoting the number of letters in each mailbox, e.g $300$ means $3$ in the 1st, $0$ in the second and $0$ in the 3rd mailbox.
Sample space $S = \{300, 030, 003, 111,012,021, 120,102,210,201\}$
$S$ contains $10$ sample points.
Let $A =$ event that each mailbox is chosen once.
$P(A) = 1/10$ </p>
<p>Teacher's solution: Treat each outcome as a sequence of $3$ numbers denoting the person and their corresponding chosen mailbox, e.g $312$ means the first person chose mailbox 3, the second chose mailbox 1, the third chose mailbox 3. So the sample space is $S = \{ 111, 112...\}$ and contains $27$ sample points.
Now let $A =$ event that each mailbox is chosen once.
$A =\{123,132,213,231,312,321\}$
So the answer is $P(A) = 6/27$</p>
| Archis Welankar | 275,884 | <p>Please note that each letter is different so $111$ can have $3!$ ways same for $1,2$ so there are more than $10$ ways . thus this is the flaw of your solution put a,b,c as letters and A,B,C as mailboxes and then count ways but still personally i think teachers solution is fast and easy as always writing sample space isnt useful.</p>
|
1,654,545 | <p>My teacher explained this problem to us - "There are $3$ mailboxes. $3$ people put letters in at random. There is no preference for any of the $3$ mailboxes. Compute the probability that each mailbox contains $1$ letter."</p>
<p>I tried this problem on my own and got the wrong answer. I understand the teacher's solution but I don't get what went wrong with my answer. Can someone explain where I went wrong?</p>
<p>My answer: Treat each outcome as a sequence of $3$ numbers denoting the number of letters in each mailbox, e.g $300$ means $3$ in the 1st, $0$ in the second and $0$ in the 3rd mailbox.
Sample space $S = \{300, 030, 003, 111,012,021, 120,102,210,201\}$
$S$ contains $10$ sample points.
Let $A =$ event that each mailbox is chosen once.
$P(A) = 1/10$ </p>
<p>Teacher's solution: Treat each outcome as a sequence of $3$ numbers denoting the person and their corresponding chosen mailbox, e.g $312$ means the first person chose mailbox 3, the second chose mailbox 1, the third chose mailbox 3. So the sample space is $S = \{ 111, 112...\}$ and contains $27$ sample points.
Now let $A =$ event that each mailbox is chosen once.
$A =\{123,132,213,231,312,321\}$
So the answer is $P(A) = 6/27$</p>
| N. F. Taussig | 173,070 | <p>By counting how many envelopes were placed in each mailbox, you treated the three letters as if they were identical. However, they are distinct. Consequently, the outcomes you listed are not equally likely.</p>
<p>Suppose the letters are $a$, $b$, and $c$ and the mailboxes are $A$, $B$, $C$. Since there are three mailboxes in which each letter can be placed, there are $3^3 = 27$ equally likely assignments of letters to mailboxes. They can be listed as ordered pairs. For instance, $(a, C)$, $(b,A)$, $(c,A)$ is the assignment of the first letter to the third mailbox and the other two letters to the first mailbox.</p>
<p>There is exactly one way to place all three envelopes in the first mailbox, namely by making the assignment $(a, A)$, $(b, A)$, and $(c, A)$. There is also exactly one way to place all three envelopes in the second mailbox and exactly one way to place all three envelopes in the third mailbox.</p>
<p>On the other hand, there are $3! = 6$ ways to place the three letters in three different mailboxes. For such an assignment to occur, once the first person to arrive has selected one of the three mailboxes, the next person must select one of the other two mailboxes, and the final person must choose the mailbox the first two people did not select. By counting how many envelopes were placed in each mailbox, you treated these six outcomes as identical. </p>
<p>Similarly, by counting how many envelopes were placed in each mailbox rather than which envelopes were placed in each mailbox, you treated each of the six cases in which two letters were placed in one mailbox and one was placed in another mailbox as one case. Each such case can occur in $3$ ways since there are $3$ ways of selecting which letter will not be in the mailbox that contains two letters.</p>
<p>If $(k, m, n)$ denotes $k$ letters in mailbox $A$, $m$ letters in mailbox $B$, and $n$ letters in mailbox $C$, then the corresponding number of outcomes is
\begin{array}{c c}
\text{case} & \text{number of outcomes}\\
(3, 0, 0) & 1\\
(2, 1, 0) & 3\\
(2, 0, 1) & 3\\
(1, 2, 0) & 3\\
(1, 1, 1) & 6\\
(1, 0, 2) & 3\\
(0, 3, 0) & 1\\
(0, 2, 1) & 3\\
(0, 1, 2) & 3\\
(0, 0, 3) & 1
\end{array}
As you can see, there are a total of $27$ outcomes. By counting how many envelopes were in each mailbox, you reduced the problem to $10$ cases that are not equally likely.</p>
|
2,114,619 | <p>An intruder has a cluster of 64 machines, each of which can try 10^6 passwords per second. How long does it take him to try all legal passwords if the requirements for the password are as follows:</p>
<ul>
<li>passwords can be 6, 7, or 8 characters long</li>
<li>each character is either a lower-case letter or a digit</li>
<li>at least one character has to be a digit</li>
</ul>
<p>My approach is that, for password of length k, there are 36^k possibilities - 26^k forbidden possibilities (because can't be all letters):
∑(k=6 to 8)(36^k−26^k) = 2,684,483,063,360 so
2,684,483,063,360 passwords × (1 second / 64*10^6 passwords) × (1 minute/ 60 seconds) × (1 hour / 60 minutes) = 11.651 hours</p>
<p>Am I approaching this the right way? Is my solution correct?</p>
| BranchedOut | 364,830 | <p>The number of possible passwords is $\sum_{k=0}^{2}10\cdot(10+26)^{5+k}=806,014,126,080=8.06\cdot10^{11}$.</p>
<p>The rate of password input is $64\cdot10^6$ passwords per second. Thus the time to try all of them is </p>
<p>$$\frac{8.06\cdot10^{11}\,\text{passwords}}{64\cdot10^6\,\text{passwords/second}}=12,593.971\,\text{seconds}\approx4.5\,\text{hours}$$</p>
|
259,308 | <p>The output of <code>ListPointPlot3D</code> is shown below:
<a href="https://i.stack.imgur.com/ypt73.png" rel="noreferrer"><img src="https://i.stack.imgur.com/ypt73.png" alt="enter image description here" /></a>
I only want to connect the dots in such a way that it forms a ring-like mesh. However, when I use <code>ListPlot3D</code>, the output is like this:
<a href="https://i.stack.imgur.com/zmaI1.png" rel="noreferrer"><img src="https://i.stack.imgur.com/zmaI1.png" alt="enter image description here" /></a>
Which is not what I want.</p>
<p>May I know how I should modify the code to achieve the intended result? Thank you.</p>
<p>Here is the data I used (with fewer data points):</p>
<pre><code>{{15.,-4.33154,0.015},{15.4114,-3.89839,0.0154114},{15.7792,-3.46523,0.0157792},{16.1037,-3.03208,0.0161037},{16.3847,-2.59893,0.0163847},{16.6224,-2.16577,0.0166224},{16.8169,-1.73262,0.0168169},{16.968,-1.29946,0.0169681},{17.076,-0.866309,0.017076},{17.1408,-0.433154,0.0171408},{17.1624,-6.01122*10^-16,0.0171624},{17.1408,0.433154,0.0171408},{17.076,0.866309,0.017076},{16.968,1.29946,0.0169681},{16.8169,1.73262,0.0168169},{16.6224,2.16577,0.0166224},{16.3847,2.59893,0.0163847},{16.1037,3.03208,0.0161037},{15.7792,3.46523,0.0157792},{15.4114,3.89839,0.0154114},{14.698,-4.79395,2.99474},{15.1194,-4.31455,3.08061},{15.4564,-3.83516,3.14927},{15.7246,-3.35576,3.2039},{15.9361,-2.87637,3.24701},{16.1009,-2.39697,3.28058},{16.2264,-1.91758,3.30616},{16.3185,-1.43818,3.32492},{16.3815,-0.95879,3.33775},{16.4181,-0.479395,3.34521},{16.4301,-6.65294*10^-16,3.34766},{16.4181,0.479395,3.34521},{16.3815,0.95879,3.33775},{16.3185,1.43818,3.32492},{16.2264,1.91758,3.30616},{16.1009,2.39697,3.28058},{15.9361,2.87637,3.24701},{15.7246,3.35576,3.2039},{15.4564,3.83516,3.14927},{15.1194,4.31455,3.08061},{13.8101,-5.25915,5.85509},{14.2225,-4.73323,6.02996},{14.5189,-4.20732,6.15563},{14.7316,-3.6814,6.24579},{14.8837,-3.15549,6.31026},{14.9917,-2.62957,6.35607},{15.0675,-2.10366,6.3882},{15.1193,-1.57774,6.41015},{15.1527,-1.05183,6.42432},{15.1714,-0.525915,6.43225},{15.1774,-7.29853*10^-16,6.4348},{15.1714,0.525915,6.43225},{15.1527,1.05183,6.42432},{15.1193,1.57774,6.41015},{15.0675,2.10366,6.3882},{14.9917,2.62957,6.35607},{14.8837,3.15549,6.31026},{14.7316,3.6814,6.24579},{14.5189,4.20732,6.15563},{14.2225,4.73323,6.02996},{12.3716,-5.68073,8.48201},{12.7528,-5.11265,8.74341},{13.0014,-4.54458,8.91387},{13.1635,-3.97651,9.02496},{13.2689,-3.40844,9.09726},{13.3374,-2.84036,9.14419},{13.3815,-2.27229,9.17442},{13.4094,-1.70422,9.19355},{13.4263,-1.13615,9.20515},{13.4353,-0.568073,9.21134},{13.4382,-7.88359*10^-16,9.21329},{13.4353,0.568073,9.21134},{13.4263,1.13615,9.20515},{13.4094,1.70422,9.19355},{13.3815,2.27229,9.17442},{13.3374,2.84036,9.14419},{13.2689,3.40844,9.09726},{13.1635,3.97651,9.02496},{13.0014,4.54458,8.91387},{12.7528,5.11265,8.74341},{10.4398,-6.03187,10.7708},{10.7685,-5.42868,11.1099},{10.9656,-4.8255,11.3133},{11.0838,-4.22231,11.4352},{11.1547,-3.61912,11.5083},{11.1971,-3.01594,11.552},{11.2223,-2.41275,11.5781},{11.2372,-1.80956,11.5935},{11.2457,-1.20637,11.6022},{11.25,-0.603187,11.6066},{11.2513,-8.3709*10^-16,11.608},{11.25,0.603187,11.6066},{11.2457,1.20637,11.6022},{11.2372,1.80956,11.5935},{11.2223,2.41275,11.5781},{11.1971,3.01594,11.552},{11.1547,3.61912,11.5083},{11.0838,4.22231,11.4352},{10.9656,4.8255,11.3133},{10.7685,5.42868,11.1099},{8.09191,-6.3005,12.6302},{8.35003,-5.67045,13.033},{8.49481,-5.0404,13.259},{8.57602,-4.41035,13.3858},{8.62155,-3.7803,13.4569},{8.64707,-3.15025,13.4967},{8.66132,-2.5202,13.5189},{8.66922,-1.89015,13.5313},{8.67348,-1.2601,13.5379},{8.67557,-0.63005,13.5412},{8.67619,-8.74371*10^-16,13.5421},{8.67557,0.63005,13.5412},{8.67348,1.2601,13.5379},{8.66922,1.89015,13.5313},{8.66132,2.5202,13.5189},{8.64707,3.15025,13.4967},{8.62155,3.7803,13.4569},{8.57602,4.41035,13.3858},{8.49481,5.0404,13.259},{8.35003,5.67045,13.033},{5.42138,-6.48148,13.986},{5.5955,-5.83333,14.4352},{5.68862,-5.18519,14.6754},{5.73842,-4.53704,14.8039},{5.76505,-3.88889,14.8726},{5.77928,-3.24074,14.9093},{5.78687,-2.59259,14.9289},{5.79089,-1.94444,14.9393},{5.79297,-1.2963,14.9446},{5.79396,-0.648148,14.9472},{5.79425,-8.99486*10^-16,14.9479},{5.79396,0.648148,14.9472},{5.79297,1.2963,14.9446},{5.79089,1.94444,14.9393},{5.78687,2.59259,14.9289},{5.77928,3.24074,14.9093},{5.76505,3.88889,14.8726},{5.73842,4.53704,14.8039},{5.68862,5.18519,14.6754},{5.5955,5.83333,14.4352},{2.53472,-6.57254,14.7843},{2.6163,-5.91529,15.2601},{2.65878,-5.25803,15.5079},{2.6809,-4.60078,15.6369},{2.69241,-3.94352,15.704},{2.69841,-3.28627,15.739},{2.70152,-2.62902,15.7572},{2.70313,-1.97176,15.7666},{2.70395,-1.31451,15.7713},{2.70433,-0.657254,15.7735},{2.70444,-9.12123*10^-16,15.7742},{2.70433,0.657254,15.7735},{2.70395,1.31451,15.7713},{2.70313,1.97176,15.7666},{2.70152,2.62902,15.7572},{2.69841,3.28627,15.739},{2.69241,3.94352,15.704},{2.6809,4.60078,15.6369},{2.65878,5.25803,15.5079},{2.6163,5.91529,15.2601},{-0.452986,-6.5727,14.9932},{-0.467535,-5.91543,15.4747},{-0.475076,-5.25816,15.7243},{-0.478985,-4.60089,15.8537},{-0.481011,-3.94362,15.9207},{-0.48206,-3.28635,15.9555},{-0.482603,-2.62908,15.9734},{-0.482883,-1.97181,15.9827},{-0.483024,-1.31454,15.9873},{-0.483089,-0.65727,15.9895},{-0.483108,-9.12146*10^-16,15.9901},{-0.483089,0.65727,15.9895},{-0.483024,1.31454,15.9873},{-0.482883,1.97181,15.9827},{-0.482603,2.62908,15.9734},{-0.48206,3.28635,15.9555},{-0.481011,3.94362,15.9207},{-0.478985,4.60089,15.8537},{-0.475076,5.25816,15.7243},{-0.467535,5.91543,15.4747},{-3.42264,-6.48177,14.6043},{-3.5319,-5.8336,15.0705},{-3.58953,-5.18542,15.3164},{-3.61993,-4.53724,15.4462},{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</code></pre>
| kglr | 125 | <pre><code>bdr = BoundaryDiscretizeRegion @ DelaunayMesh[pts]
</code></pre>
<p><a href="https://i.stack.imgur.com/dRX8U.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dRX8U.png" alt="enter image description here" /></a></p>
<p>With some manual tweaking of the threshold <code>t</code>, we can filter the polygons that lie on the outer ring based on the <code>y</code> component of the face normal vector:</p>
<pre><code>t = .9;
cellindices = First /@ Select[Abs[#[[2, 2]]] <= t &]@
Transpose[{MeshCellIndex[bdr, 2], Region`Mesh`MeshCellNormals[bdr, 2]}];
Graphics3D[{LightBlue, Opacity[.5], EdgeForm[LightGray],
MeshPrimitives[bdr, cellindices], Black, Point @ pts},
ImageSize -> 600, Boxed -> False]
</code></pre>
<p><a href="https://i.stack.imgur.com/yO19h.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yO19h.png" alt="enter image description here" /></a></p>
|
2,397,564 | <p>Question:</p>
<p>Prove that if $ \ A\cup B \subseteq C \cup D,\ A \cap B =$ ∅ $\land \ C \subseteq A \implies B \subseteq D$.</p>
<p>My attempt:</p>
<p>Let $ \ x\in B \implies x \in A \cup B \implies x \in C \cup D \because A\cup B \subseteq C \cup D$.</p>
<p>Now, $ x \in C \lor x\in D$. If $\ x \in C \implies x \in A \because C \subseteq A$. But that's not possible $\because x \notin A \cap B$, in particular $ x \notin A$. So we must have $ x \in D$.</p>
<p>I found this proof a little challenging. Not quite sure if this is the correct way to prove it. Is my logic correct?</p>
| Eric | 309,041 | <blockquote>
<p><strong>Theorem.</strong> Let $A,~B,~C,~D$ be (any) sets. If $A\cup B\subseteq C\cup D$ and $A\cap B=\emptyset$ and $C\subseteq A$, then $B\subseteq D$.</p>
</blockquote>
<p><em>Proof.</em>(in much detailed, giving you the instruction that what should you do in every step)</p>
<p>Let $A,~B,~C,~D$ be sets. Claim:
$A\cup B\subseteq C\cup D\wedge A\cap B=\emptyset\wedge C\subseteq A\to B\subseteq D$.</p>
<p>Suppose $A\cup B\subseteq C\cup D\wedge A\cap B=\emptyset\wedge C\subseteq A$. Claim: $B\subseteq D$, namely $\forall x,x\in B\to x\in D$.</p>
<p>Let $x$ in the domain of discourse, claim: $x\in B\to x\in D$.</p>
<p>Suppose $x\in B$(claim $x\in D$). Then $x\in A\vee x\in B$. Hence $x\in A\cup B$. Since $A\cup B\subseteq C\cup D$, we have $x\in C\cup D$, namely $x\in C\vee x\in D$.</p>
<ul>
<li>If $x\in D$, then done.</li>
<li>Suppose $x\in C$. by the hypothesis that $C\subseteq A$, we have $x\in A$. Hence $x\in B\wedge x\in A$, namely $x\in A\cap B$. However, by the hypothesis, $A\cap B=\emptyset$. So it is a contradiction(Reductio Ad Absurdum). It is not the case that $x\in C$.</li>
</ul>
<p>(The following are the ending steps, which is usually omitted in mathematical literature.)
Hence we have shown that $x\in B\to x\in D$. Since $x$ is arbitrary, we have shown that $\forall x,x\in B\to x\in D$, which is $B\subseteq D$.</p>
<p>Hence we have shown that $A\cup B\subseteq C\cup D\wedge A\cap B=\emptyset\wedge C\subseteq A\to B\subseteq D$. Since the sets $A,B,C,D$ are all arbitrary, we have proved the theorem.</p>
|
3,715,715 | <p>Let’s say I have a set <span class="math-container">$X$</span> and a set <span class="math-container">$Y$</span>, and <span class="math-container">$X \subseteq Y$</span>. Is it possible to state that <span class="math-container">$|X| \leq |Y|$</span> (<span class="math-container">$|X|$</span> cardinality of <span class="math-container">$X$</span>)? How can I demonstrate that?</p>
| Arturo Magidin | 742 | <p>Not only can you say it, you can prove it!</p>
<p>Recall the definitions: for sets <span class="math-container">$A$</span> and <span class="math-container">$B$</span>, we say that</p>
<ul>
<li><span class="math-container">$|A|\leq |B|$</span> if and only if there exists a one-to-one function <span class="math-container">$f\colon A\to B$</span>;</li>
<li><span class="math-container">$|A|=|B|$</span> if and only if there exists a bijective function <span class="math-container">$f\colon A\to B$</span>;</li>
<li><span class="math-container">$|A|\lt |B|$</span> if and only if <span class="math-container">$|A|\leq |B|$</span> and <span class="math-container">$|A|\neq|B|$</span>; that is, if and only if there is an injection <span class="math-container">$f\colon A\to B$</span> but no bijection <span class="math-container">$f\colon A\to B$</span>.</li>
</ul>
<p>(In the presence of the Axiom of Choice, you get more equivalences; for example, <span class="math-container">$|A|\lt |B|$</span> if and only if there is an injection <span class="math-container">$A\to B$</span> but no surjection <span class="math-container">$A\to B$</span>; and <span class="math-container">$|A|\geq |B|$</span> if and only if there is a surjection <span class="math-container">$A\to B$</span>; and more).</p>
<p>To prove that if <span class="math-container">$A\subseteq B$</span> then <span class="math-container">$|A|\leq |B|$</span>, then simply use the inclusion function <span class="math-container">$\iota\colon A\to B$</span> given by <span class="math-container">$\iota(a)=a$</span> for all <span class="math-container">$a\in A$</span>. This function is injective, as <span class="math-container">$a\neq b\implies \iota(a)\neq\iota(b)$</span>. This shows <span class="math-container">$|A|\leq|B|$</span>. </p>
<p>Careful, that it is possible for <span class="math-container">$A\subset B$</span>, <span class="math-container">$A\neq B$</span>, yet <span class="math-container">$|A|=|B|$</span>; the inclusion function does not prove the equality of cardinality, but other functions might. An example of this occurs with <span class="math-container">$\mathbb{N}\subseteq\mathbb{Z}$</span>. </p>
|
2,539,693 | <p>A number theory textbook asked us to compare $\tan^{-1}(\frac{1}{2})$ and $\sqrt{5}$. In fact, these are rather close:</p>
<p>\begin{eqnarray*}
\tan^{-1} \frac{1}{2} &=& 0.46364 \\ \\
\frac{1}{\sqrt{5}} &=& 0.44721
\end{eqnarray*}</p>
<p>So at least numerically I think we have the answer that the first one is bigger. Momentarily, I thought we had an exact answer: $\tan^{-1}\frac{1}{\sqrt{2}} = \frac{\pi}{4} $, but that's totally different. So we are left with:</p>
<p>$$ \tan^{-1} \frac{1}{2} > \frac{1}{\sqrt{5}} > 0 $$</p>
<p>It's impressive that we could have so many decimal places, and I wonder if I should take the computer on faith for that. And I noticed these two answers are close, so I also wonder if we estimate the difference... I don't have any conjecture in either case yet.</p>
<p>For now I just want proof of the inequality as stated above.</p>
| Robert Z | 299,698 | <p>Hint. Note that for $x\in (0,1)$, if $d$ is an odd positive integer then
$$\arctan(x)> \sum_{k=0}^d\frac{(-1)^kx^{2k+1}}{2k+1}.$$</p>
|
2,539,693 | <p>A number theory textbook asked us to compare $\tan^{-1}(\frac{1}{2})$ and $\sqrt{5}$. In fact, these are rather close:</p>
<p>\begin{eqnarray*}
\tan^{-1} \frac{1}{2} &=& 0.46364 \\ \\
\frac{1}{\sqrt{5}} &=& 0.44721
\end{eqnarray*}</p>
<p>So at least numerically I think we have the answer that the first one is bigger. Momentarily, I thought we had an exact answer: $\tan^{-1}\frac{1}{\sqrt{2}} = \frac{\pi}{4} $, but that's totally different. So we are left with:</p>
<p>$$ \tan^{-1} \frac{1}{2} > \frac{1}{\sqrt{5}} > 0 $$</p>
<p>It's impressive that we could have so many decimal places, and I wonder if I should take the computer on faith for that. And I noticed these two answers are close, so I also wonder if we estimate the difference... I don't have any conjecture in either case yet.</p>
<p>For now I just want proof of the inequality as stated above.</p>
| Ross Millikan | 1,827 | <p>You can use the Taylor series for $\arctan x$
$$\arctan (\frac 12) \gt \frac 12-\frac 1{2^3\cdot 3}\gt 0.45833$$ where we have an alternating series so the error is of the sign of the first neglected term and $$11^2=121 \lt 5\cdot 5^2\\\frac 1{\sqrt 5} \lt \frac 5{11}\lt 0.45455$$</p>
|
1,163,033 | <p>I want to calculate $ 8^{-1} \bmod 77 $ </p>
<p>I can deduce $ 8^{-1} \bmod 77$ to $ 8^{59} \bmod 77 $ using Euler's Theorem.</p>
<p>But how to move further now. Should i calculate $ 8^{59} $ and then divide it by $ 77 $ or is there any other theorem i can use ? </p>
| paw88789 | 147,810 | <p>For small to intermediate size moduli, you can use the method of adding the modulus, and cancelling when possible. (For larger moduli, you may want to use the extended Euclidean Algorithm.)</p>
<p>Here's adding the modulus on your problem:</p>
<p>$8x\equiv 1\equiv 78 \pmod{77}$. So $4x\equiv 39 \pmod{77}$.</p>
<p>Then $4x\equiv 39\equiv 116 \pmod{77}$.
So $x\equiv 29 \pmod{77}$.</p>
<p>$8^{-1}\equiv 29\pmod{77}$</p>
|
1,163,033 | <p>I want to calculate $ 8^{-1} \bmod 77 $ </p>
<p>I can deduce $ 8^{-1} \bmod 77$ to $ 8^{59} \bmod 77 $ using Euler's Theorem.</p>
<p>But how to move further now. Should i calculate $ 8^{59} $ and then divide it by $ 77 $ or is there any other theorem i can use ? </p>
| Bill Dubuque | 242 | <p>Below are a few methods.</p>
<hr>
<p>${\rm mod}\ 77\!:\ \ \dfrac{1}8\equiv \dfrac{-76}8\equiv \dfrac{-19}{2}\equiv \dfrac{58}2\equiv 29\,\ $ by fraction fiddling</p>
<hr>
<p>${\rm mod}\ 77\!:\ \ \dfrac{1}8 \equiv \dfrac{10}{80}\equiv \dfrac{10}{3}\equiv\dfrac{87}3\equiv 29\,\ $ by <a href="https://math.stackexchange.com/a/174687/242">Gauss's algorithm.</a></p>
<hr>
<p>${\rm mod}\ 7\!:\,\quad 8\equiv 1\,\Rightarrow\, 8^{-1}\equiv 1.\ $ </p>
<p>${\rm mod}\ 11\!:\,\ \dfrac{1}8\equiv \dfrac{12}{-3}\equiv -4\equiv \color{#0a0}{7}.\ $ Now we apply CRT.</p>
<p>${\rm mod}\ 7\!:\quad 1\equiv 8^{-1}\equiv \color{#0a0}{7}+11n\equiv 7-3n\iff 3n\equiv 6\iff n\equiv \color{#c00}2$</p>
<p>Thus we find $\,\ 8^{-1}\equiv 7+11(\color{#c00}2+7k)= 29+77k$</p>
<hr>
<p><strong>Beware</strong> $ $ Modular fraction arithmetic is well-defined only for fractions with denominator <em>coprime</em> to the modulus. <a href="https://math.stackexchange.com/a/921093/242">See here</a> for further discussion.</p>
|
623,428 | <blockquote>
<p>Suppose $$
Y = X^TAX,
$$ where $Y$ and $A$ are both known $n\times n$, real, symmetric matrices. The unknown matrix $X$ is restricted to $n\times n$.</p>
</blockquote>
<p>I think there should be at least one real valued solution for $X$. How do I solve for $X$? </p>
| Disintegrating By Parts | 112,478 | <p>Suppose $Y$ is of full rank, but $A$ is not. You can't do what you want then.</p>
|
4,336,659 | <p>For a beta distribution with parameters <span class="math-container">$a$</span> and <span class="math-container">$b$</span>, we can interpret it as the distribution of the probability of heads for a coin we tossed <span class="math-container">$a+b$</span> times and saw <span class="math-container">$a$</span> heads and <span class="math-container">$b$</span> tails. At the same time, if we draw <span class="math-container">$n$</span> uniform random numbers and sort them, the <span class="math-container">$k$</span>-th order statistic is also Beta distributed with parameters <span class="math-container">$a=k$</span> and <span class="math-container">$b=n+1-k$</span>. So, its like we tossed <span class="math-container">$n+1$</span> coins and got <span class="math-container">$k$</span> heads. Is there an intuitive explanation for this? I can see the derivations mechanically but any logical reason the two distributions should be the same?</p>
| Shannon Starr | 1,003,218 | <p>You answered your own question completely and entirely, using the observation by <a href="https://math.stackexchange.com/users/356647/lee-david-chung-lin">lee-david-chung-lin</a>. Actually a Beta(a,b) distribution for <span class="math-container">$a,b \in \{1,2,\dots\}$</span> is the same as the Bayesian posterior for <span class="math-container">$U$</span> when <span class="math-container">$U$</span> is uniform, <em>a priori</em>, and we observe that the outcomes of a Binomial(U,n) random variable is <span class="math-container">$a-1$</span> for <span class="math-container">$n=(a-1)+(b-1)=a+b-2$</span>. That is just because the density of a Beta(a,b) random variable is <span class="math-container">$x^{a-1}(1-x)^{b-1} \mathbf{1}_{[0,1]}(x)/B(a,b)$</span> where <span class="math-container">$B(a,b)$</span> is the Beta integral. But the probability for a binomial(n,p) random variable to equal <span class="math-container">$k$</span> is <span class="math-container">$\binom{n}{k} p^k (1-p)^{n-k}$</span>. So changing <span class="math-container">$p$</span> to <span class="math-container">$x$</span>, using Bayes's rule, and matching <span class="math-container">$a-1$</span> to <span class="math-container">$k$</span> and <span class="math-container">$b-1$</span> to <span class="math-container">$n-k$</span> gives the result. That is what was pointed out by <a href="https://math.stackexchange.com/users/356647/lee-david-chung-lin">lee-david-chung-lin</a>. So then the <span class="math-container">$a$</span> for the <span class="math-container">$k$</span>th order statistic is <span class="math-container">$a=k-1+1$</span> following your explanation <a href="https://math.stackexchange.com/users/155881/rohit-pandey">Rohit Pandey</a> and <span class="math-container">$b=n-k+1$</span>. Those match the desired connection.</p>
<p>There is a trivial discrete analog. Imagine <span class="math-container">$L$</span> boxes in a linearly-ordered row, with <span class="math-container">$n$</span> balls in the boxes, such that there are no more than 1 ball in each of the boxes. The probability for the <span class="math-container">$k$</span>th ball (reading left-to-right) to be in box <span class="math-container">$x$</span> is <span class="math-container">$\binom{x-1}{k-1} \binom{L-x}{n-k}/\binom{L}{n}$</span> for each possible choice of <span class="math-container">$x$</span>. (Note that, summing over <span class="math-container">$x$</span>, this generalizes the ``hockeystick identity'' for binomial coefficients in a trivial way.) This is not quite the same: here the primary interpretation is in terms of the <span class="math-container">$k$</span>th order-statistics.</p>
<p><a href="https://i.stack.imgur.com/22IBb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/22IBb.png" alt="example of balls in boxes" /></a></p>
<p>The binomial(n,p) distribution is replaced by a capture-recapture hypergeometric distribution, which is not as straightforward as the binomial. Imagine a population of <span class="math-container">$L-1$</span> elephants and you capture and tag a uniformly distribution random number <span class="math-container">$\mathsf{Y}$</span> elephants, and set <span class="math-container">$x=y+1$</span> if <span class="math-container">$\mathsf{Y}=y$</span>. (This hypothesis is admittedly a bit odd in this context.) Then you randomly collect a new sample of <span class="math-container">$n-1$</span> of the elephant population. You discover <span class="math-container">$k-1$</span> of your new sample were tagged and <span class="math-container">$n-k$</span> were not. What is your updated guess for <span class="math-container">$y$</span>?</p>
<p>This example is sometimes interesting in classes because it gives an explicitly calculable example for Bayes's rule that is purely discrete, setting it up just as you did <a href="https://math.stackexchange.com/users/155881/rohit-pandey">Rohit Pandey</a> for the continuous case. This could be a good example if one teaches basic probability to students who are concurrently learning multivariable calculus. Then you do not have to delay the Bayes' rule examples to the 2nd half of the semester. (For example, following Durrett's <em>Elementary Probability for Applications</em> [which is available free on his website now] but getting to this combinatorial Bayes' rule example before doing the multivariable calculus stuff in the later chapter on continuous distributions.)</p>
|
679,904 | <p>The question is let $a \in \mathbb{R} $ does not contain 0. Prove that $|a+\frac{1}{a}| \ge 2$. I have no idea how to start this problem and any help on it would be greatly appreciated.</p>
| triwer23 | 129,160 | <p>Note that it is enough to consider the case $a>0$. Also note that for $x,y$ non negative, $x+y\geq 2\sqrt{xy}$. </p>
|
239,136 | <p>I was given this question and I'm not really sure how to approach this...</p>
<p>Assume $(r,s) = 1$. Prove that If $G = \langle x\rangle$ has order $rs$, then $x = yz$, where $y$ has order $r$, $z$ has order $s$, and $y$ and $z$ commute; also prove that the factors $y$ and $z$ are unique.</p>
| Dan Shved | 47,560 | <p>Here is a hint: you can set $y=x^{sn}$ and $z=x^{rm}$. To find the appropriate $n$ and $m$ you can use <a href="http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity" rel="nofollow">Bezout's identity</a>.</p>
|
621,742 | <p>How do you get from$$\int^\infty_0\int^\infty_0e^{-(x+y)^2} dx\ dy$$to
$$\frac{1}{2}\int^\infty_0\int^u_{-u}e^{-u^2} dv\ du?$$ I have tried using a change of variables formula but to no avail.<br>
Edit: Ok as suggested I set $u=x+y$ and $v=x-y$, so I can see this gives $dx dy=\frac{1}{2}dudv$ but I still can't see how to get the new integration limits. Sorry if I'm being slow.</p>
| Yiorgos S. Smyrlis | 57,021 | <p>Hint. Set $u=x+y$, $v=x-y$.</p>
<p>Then
$$
\{(x,y): x,y\ge 0\}=\{(u,v) : u>0, v\in(-u,u)\},
$$
and
$$
dx\,dy=\frac{1}{2}du\,dv,
$$
as
$$
\frac{\partial (x,y)}{\partial(u,v)}=\frac{1}{2}
$$</p>
|
621,742 | <p>How do you get from$$\int^\infty_0\int^\infty_0e^{-(x+y)^2} dx\ dy$$to
$$\frac{1}{2}\int^\infty_0\int^u_{-u}e^{-u^2} dv\ du?$$ I have tried using a change of variables formula but to no avail.<br>
Edit: Ok as suggested I set $u=x+y$ and $v=x-y$, so I can see this gives $dx dy=\frac{1}{2}dudv$ but I still can't see how to get the new integration limits. Sorry if I'm being slow.</p>
| Felix Marin | 85,343 | <p>$\newcommand{\+}{^{\dagger}}%
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
\newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
\newcommand{\dd}{{\rm d}}%
\newcommand{\ds}[1]{\displaystyle{#1}}%
\newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
\newcommand{\fermi}{\,{\rm f}}%
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
\newcommand{\half}{{1 \over 2}}%
\newcommand{\ic}{{\rm i}}%
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}%
\newcommand{\isdiv}{\,\left.\right\vert\,}%
\newcommand{\ket}[1]{\left\vert #1\right\rangle}%
\newcommand{\ol}[1]{\overline{#1}}%
\newcommand{\pars}[1]{\left( #1 \right)}%
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}%
\newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
\newcommand{\sech}{\,{\rm sech}}%
\newcommand{\sgn}{\,{\rm sgn}}%
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}%
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
With $x \equiv \rho\cos\pars{\theta}$, $y \equiv \rho\sin\pars{\theta}$ where $\rho \geq 0$ and $0 \leq \theta < 2\pi$ we'll get
$\ds{{\partial\pars{x,y} \over \partial\pars{\rho,\theta}} = \rho}$ such that</p>
<blockquote>
\begin{align}
&\color{#0000ff}{\large%
\int_{0}^{\infty}\int_{0}^{\infty}\expo{-\pars{x + y}^{2}}\,\dd x\,\dd y}
=
\int_{0}^{\pi/2}\dd\theta\int_{0}^{\infty}\expo{-\rho^{2}\bracks{1 + \sin\pars{2\theta}}}\rho\,\dd\rho
\\[3mm]&=
\int_{0}^{\pi/2}\dd\theta\,\left.%
{-\expo{-\rho^{2}\bracks{1 + \sin\pars{2\theta}}} \over 2\bracks{1 + \sin\pars{2\theta}}}
\right\vert_{\rho = 0}^{\rho \to \infty}
=
\half\int_{0}^{\pi/2}
{\dd\theta \over 1 + \sin\pars{2\theta}}
=
{1 \over 4}\int_{0}^{\pi}
{\dd\theta \over 1 + \sin\pars{\theta}} = \color{#0000ff}{\large\half}
\end{align}
</blockquote>
<p>since
\begin{align}
&\color{#0000ff}{\large{1 \over 4}\int_{0}^{\pi}{\dd\theta \over 1 + \sin\pars{\theta}}}
=\half\int_{0}^{\pi/2}{\dd\theta \over 1 + \sin\pars{\theta}}
=\half\int_{0}^{\pi/2}{1 - \sin\pars{\theta} \over \cos^{2}\pars{\theta}}\,\dd\theta
\\[3mm]&=
\half\,\lim_{\theta \to \pars{\pi/2}^{-}}\bracks{%
{\sin\pars{\theta} - 1 \over \cos\pars{\theta}}} + \half = \color{#0000ff}{\large\half}
\end{align}</p>
|
964,387 | <p>I'm attempting to teach myself topology for graduate school this summer, but I'm having a tough time.
I'm trying to prove that the Euclidean topology on $\mathbb{R}^{m+n}$ is equivalent to the product topology on $\mathbb{R}^m \times \mathbb{R}^n$. I realize to do this that I should make a homeomorphism between them, and the identity function would work for this, but I'm unsure on what to do from there.</p>
<p>Here's what I have so far:
Let $x \in U \subseteq \mathbb{R}^{m+n}$ be some open set in $\mathbb{R}^{m+n}$, then there exists an open ball $B_{\epsilon}(x) \subseteq U$. But I'm not sure where to go from there. </p>
<p>I have read this <a href="https://math.stackexchange.com/questions/755586/product-topology-and-standard-euclidean-topology-over-mathbbrn-are-equival">Product topology and standard euclidean topology over $\mathbb{R}^n$ are equivalent</a> but I do not understand why you are allowed to assume that each of the subsets in $B^1_{\epsilon}(x_i)$ are open in $\mathbb{R}$. thank you</p>
| Pedro | 23,350 | <p><em>Hint</em> Suppose $\lVert \cdot \rVert_1$ and $\lVert \cdot \rVert_2$ are the Euclidean norms in $ \Bbb R^n,\Bbb R^m$ respectively. Then $\lVert (x,y)\rVert = (\lVert x\rVert_1^2+\lVert y\rVert_2^2)^{1/2}$ is the Euclidean norm in $\Bbb R^n\times\Bbb R^m$. </p>
|
964,387 | <p>I'm attempting to teach myself topology for graduate school this summer, but I'm having a tough time.
I'm trying to prove that the Euclidean topology on $\mathbb{R}^{m+n}$ is equivalent to the product topology on $\mathbb{R}^m \times \mathbb{R}^n$. I realize to do this that I should make a homeomorphism between them, and the identity function would work for this, but I'm unsure on what to do from there.</p>
<p>Here's what I have so far:
Let $x \in U \subseteq \mathbb{R}^{m+n}$ be some open set in $\mathbb{R}^{m+n}$, then there exists an open ball $B_{\epsilon}(x) \subseteq U$. But I'm not sure where to go from there. </p>
<p>I have read this <a href="https://math.stackexchange.com/questions/755586/product-topology-and-standard-euclidean-topology-over-mathbbrn-are-equival">Product topology and standard euclidean topology over $\mathbb{R}^n$ are equivalent</a> but I do not understand why you are allowed to assume that each of the subsets in $B^1_{\epsilon}(x_i)$ are open in $\mathbb{R}$. thank you</p>
| Ri-Li | 152,715 | <p>Just use $f:\Bbb R^m \times \Bbb R^n \to \Bbb R^{m+n}$. $((x_1,...,x_m),(x_{m+1},...,x_{m+n}))\to (x_1,...,x_{m+n})$ Now see that it is bijective and continuous. Actually it some kind of identity function. [As you are having problem, some things you need to clarify,
1) Eucledean topology is same as the metric topology, here euclidean norm is the metric,$d(x,y)=||x-y||$ (check!!)
2) For finite products, box topology is same as product topology. I don't know whether you come across the term box topology or not, but advise you to make your conception clear in these cases]</p>
|
582,478 | <p>Please simplify this logic expression for me with helping boolean algebra :</p>
<p>A'C'D + A'BD + BCD + ABC + ACD'</p>
<p>I know that must use consensus theorem .</p>
<p>my solve :</p>
<p>STEP 1 : Terms 1 & 3 ---eliminate---> Term 2</p>
<p>STEP 2 : Terms 3 & 5 ---eliminate---> Term 4</p>
<p>STEP 3 : Terms 2 & 4 ---eliminate---> Term 3</p>
<p>But truth table said step 3 is incorrect . but why ?</p>
<p>please tell me why step 3 is Wrong ?</p>
<p>and tell me What is the simplest form of it ?</p>
| Suraj M S | 85,213 | <p>given
$$ A'C'D+A'BD+BCD+ABC+ACD'$$
use distributive law $xyz+pqr=(xyz+p).(xyz+qr)$
$$\to (A'C'D+A'D).(A'C'D+B) +(BCD+BC).(BCD+A)+ACD'$$
$$ \to A'C'D+B+BCD+A+ACD'$$
$$ \to A'C'D+B(1+CD)+A(1+CD')$$
$$ \to A'C'D+A+B$$
$$ \to (A+A').(A+C'D) +B$$
$$ \to A+B+C'D$$</p>
|
582,478 | <p>Please simplify this logic expression for me with helping boolean algebra :</p>
<p>A'C'D + A'BD + BCD + ABC + ACD'</p>
<p>I know that must use consensus theorem .</p>
<p>my solve :</p>
<p>STEP 1 : Terms 1 & 3 ---eliminate---> Term 2</p>
<p>STEP 2 : Terms 3 & 5 ---eliminate---> Term 4</p>
<p>STEP 3 : Terms 2 & 4 ---eliminate---> Term 3</p>
<p>But truth table said step 3 is incorrect . but why ?</p>
<p>please tell me why step 3 is Wrong ?</p>
<p>and tell me What is the simplest form of it ?</p>
| Donald Splutterwit | 404,247 | <p>Using a truth table Let $X=A'C'D + A'BD + BCD + ABC + ACD'$. (We shall leave entries blank where their value is zero)
\begin{eqnarray*}
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
A & B & C & D & & A'C'D & A'BD & BCD & ABC & ACD' & & X \\ \hline
0&0&0&0& & & & & & & & & \\ \hline
0&0&0&1& &1& & & & & &1& \\ \hline
0&0&1&0& & & & & & & & & \\ \hline
0&0&1&1& & & & & & & & & \\ \hline
0&1&0&0& & & & & & & & & \\ \hline
0&1&0&1& &1&1& & & & &1& \\ \hline
0&1&1&0& & & & & & & & & \\ \hline
0&1&1&1& & &1&1& & & &1& \\ \hline
1&0&0&0& & & & & & & & & \\ \hline
1&0&0&1& & & & & & & & & \\ \hline
1&0&1&0& & & & & &1& &1& \\ \hline
1&0&1&1& & & & & & & & & \\ \hline
1&1&0&0& & & & & & & & & \\ \hline
1&1&0&1& & & & & & & & & \\ \hline
1&1&1&0& & & & &1&1& &1& \\ \hline
1&1&1&1& & & &1&1& & &1& \\ \hline
\end{array}
\end{eqnarray*}</p>
<p>From the table above it is clear that we only need to "or" the first, third & fifth terms in order to construct $X$. So the minimal expression is $X=\color{red}{A'C'D + BCD + ACD'}$.</p>
|
3,428,668 | <p>How to prove this</p>
<p><span class="math-container">$$S = \{(x, y) | Ax + By ≥ c, x ≥ 0, y ≥ 0\}$$</span>
where <span class="math-container">$A$</span> is an <span class="math-container">$m \times n$</span> matrix, <span class="math-container">$B$</span> is a positive semi-definite <span class="math-container">$m \times m$</span> matrix and
<span class="math-container">$c \in \Bbb R^m$</span>. The author explicitly assumed the set <span class="math-container">$S$</span> is compact in <span class="math-container">$\Bbb R^{n+m}$</span>. A
reviewer of the paper pointed out that the only compact set of the above form
is the empty set. Prove the reviewer’s assertion</p>
| user2661923 | 464,411 | <p><span class="math-container">$27 \times 24 = 648 \;\equiv 18 \mod(63).$</span></p>
<p>Therefore, the 2nd equation should be </p>
<p><span class="math-container">$18y + 48 \;\equiv 24 \mod(63).$</span></p>
|
3,498,199 | <p>Suppose if a matrix is given as</p>
<p><span class="math-container">$$ \begin{bmatrix}
4 & 6\\
2 & 9
\end{bmatrix}$$</span></p>
<p>We have to find its eigenvalues and eigenvectors.</p>
<p>Can we first apply elementary row operation . Then find eigenvalues.</p>
<p>Is their any relation on the matrix if it is diagonalized or not.</p>
| Simply Beautiful Art | 272,831 | <p>As noted, you cannot apply elementary row operations to <span class="math-container">$A$</span> and expect the eigenvalues/vectors be preserved. However, you can apply elementary row operations to <span class="math-container">$|A-\lambda I|=0$</span> to solve for <span class="math-container">$\lambda$</span>. In your example, we have:</p>
<p><span class="math-container">\begin{align}|A-\lambda I|&=\left|\begin{matrix}4-\lambda&6\\2&9-\lambda\end{matrix}\right|\\&=\frac1{4-\lambda}\left|\begin{matrix}4-\lambda&6\\2(4-\lambda)&\lambda^2-13\lambda+36\end{matrix}\right|\\&=\frac1{4-\lambda}\left|\begin{matrix}4-\lambda&6\\0&\lambda^2-13\lambda+24\end{matrix}\right|\\&=\lambda^2-13\lambda+24\end{align}</span></p>
<p>if you so desired to do so.</p>
|
221,351 | <p>I asked the following question (<a href="https://math.stackexchange.com/questions/1487961/reference-for-every-finite-subgroup-of-operatornamegl-n-mathbbq-is-con">https://math.stackexchange.com/questions/1487961/reference-for-every-finite-subgroup-of-operatornamegl-n-mathbbq-is-con</a>) on math.stackexchange.com and received no answers, so I thought I would ask it here. I've asked several people in my department who were all stumped by the question.</p>
<p>The question is: why is every finite subgroup of $\operatorname{GL}_n(\mathbb{Q})$ conjugate to a finite subgroup of $\operatorname{GL}_n(\mathbb{Z})$?</p>
<p>Note that at least for $n=2$ the question of isomorphism is much easier, since one can (with some effort) work out exactly which finite groups can be subgroups of $\operatorname{GL}_2(\mathbb{Q})$. Further, there are isomorphic finite subgroups of $\operatorname{GL}_2(\mathbb{Q})$ that are not conjugate to each other. For example, the group generated by $-I_{2 \times 2}$ and $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ are both isomorphic to $C_2$, but they cannot be conjugate to each other because the eigenvalues of the two generators are different.</p>
<p>If there is a relatively simple proof, that would be ideal, but a reference with a potentially long proof is fine as well.</p>
<p>Thanks for any assistance.</p>
| p Groups | 84,772 | <p>This is not exact answer to your question, but long back, I had seen this result in an article in Monthly. I hope it will be useful.</p>
<p><a href="http://www.jstor.org/stable/pdf/2695329.pdf?acceptTC=true" rel="nofollow">http://www.jstor.org/stable/pdf/2695329.pdf?acceptTC=true</a></p>
|
465,255 | <p>Does there exists any form of Algebra where operators can be assumed as variables?</p>
<p>For example:
$$
1+2\times3=7
$$
can be considered as:
$$
1\:(\mathrm{\,X})\:2\:(\mathrm{Y})\:3=7
$$
?</p>
| qaphla | 85,568 | <p>The total number of points scored on the exam would be $200 * 25 + 304 * 25 + 350 * 25 + 250 * 25 = 27,600$, out of a maximum possible $400 * (25 * 4) = 40,000$.</p>
<p>Taking $\frac{27,600}{40,000}$ and simplifying it gives $\frac{276}{400} = \frac{69}{100}$.<br>
As there are $100$ total points on the exam, each student scored, on average, $\frac{69}{100} * 100 = 69$ points.</p>
|
10,427 | <p>I like Mathematica, but it's syntax baffles me.</p>
<p>I am trying to figure out how to minimize the whitespace around a graphic.</p>
<p>For example,</p>
<pre><code>ParametricPlot3D[{r Cos[t], r Sin[t], r^2}, {r, 0, 1}, {t, 0, 2 \[Pi]},
Boxed -> True, Axes -> False]
</code></pre>
<p><img src="https://i.stack.imgur.com/jGpvA.png" alt="3d bounding box on"></p>
<p>Puts the 3d bounding box at the limits of the view. But if I don't show the 3d bounding box,</p>
<p><img src="https://i.stack.imgur.com/jGpvA.png" alt="3d bounding box on"></p>
<pre><code>ParametricPlot3D[{r Cos[t], r Sin[t], r^2}, {r, 0, 1}, {t, 0, 2 \[Pi]},
Boxed -> False, Axes -> False]
</code></pre>
<p><img src="https://i.stack.imgur.com/E8skr.png" alt="3d bounding box off"></p>
<p>there is all this white space around the actual object.</p>
<p>Is there some way (syntax) that can put the view just around the visible objects?</p>
<h1>Edit in response to answers</h1>
<p>Ok, from the below answers, I have two solutions; 1) use ImageCrop, or 2) use <code>Method->{"ShrinkWrap" -> True}</code>. However both of these options do a little something strange to the plot I want (maybe it is just a problem with the plot itself). </p>
<p>So the actual plot I am after is,</p>
<pre><code>Module[{r = 1, \[Theta] = \[Pi]/2, \[CurlyPhi] = \[Pi]/6, \[Psi] = \[Pi]/12},
Framed@Show[
Graphics3D[
{
{Arrowheads[.025],
Arrow[{{0, 0, 0}, {1.1, 0, 0}}], Text["x", {1.2, 0, 0}],
Arrow[{{0, 0, 0}, {0, 1.1, 0}}], Text["y", {0, 1.2, 0}],
Arrow[{{0, 0, 0}, {0, 0, 1.1}}], Text["z", {0, 0, 1.2}],
Arrow[{{0, 0, 0}, r {Cos[\[Theta]] Sin[\[CurlyPhi]],
Sin[\[Theta]] Sin[\[CurlyPhi]], Cos[\[CurlyPhi]]}}]},
{Specularity[White, 50], Opacity[.1], Sphere[{0, 0, 0}, r]}
},
Boxed -> False,
ImageSize -> 600,
PlotRange -> 1.1 {{-r, r}, {-r, r}, {0, r}}
]]
]
</code></pre>
<p><img src="https://i.stack.imgur.com/emm3z.png" alt="enter image description here"></p>
<p>Which has too much whitespace. If I replace <code>Framed@Show[</code> with <code>Framed@ImageCrop@Show[</code> I
get,
<img src="https://i.stack.imgur.com/AUCNY.png" alt="enter image description here"></p>
<p>which actually crops some of the (hemi)sphere. If just use <code>Method -> {"ShrinkWrap" -> True},</code> in the <code>Show</code> options, I get,</p>
<p><img src="https://i.stack.imgur.com/WhX2Y.png" alt="Mathematica graphics"></p>
<p>which looks almost correct, but the <code>x</code> and <code>z</code> textboxes have now not included. Seems like I can't win!</p>
| Dr. belisarius | 193 | <p>Actually, there isn't white space at all:</p>
<pre><code>Show[RegionPlot3D[True, {x, -1, 1}, {y, -1, 1}, {z, 0, 1},
PlotStyle -> Directive[Yellow, Opacity[0.5]], Mesh -> None,
Boxed -> False, Axes -> False, PlotRangePadding -> 0],
ParametricPlot3D[{r Cos[t], r Sin[t], r^2}, {r, 0, 1}, {t, 0, 2 Pi} ]]
</code></pre>
<p><img src="https://i.stack.imgur.com/l9fmH.png" alt="Mathematica graphics"></p>
<p><strong>Edit</strong> </p>
<p>If you want to crop the image in <strong>2D</strong>:</p>
<pre><code>p = ParametricPlot3D[{r Cos[t], r Sin[t], r^2}, {r, 0, 1}, {t, 0, 2 Pi},
Boxed -> False, Axes -> False, PlotRangePadding -> 0];
Framed@ImageCrop@p
</code></pre>
<p><img src="https://i.stack.imgur.com/b5R5s.png" alt="Mathematica graphics"></p>
<p><strong>Edit</strong></p>
<p>For your plot. Use .2 as <code>Opacity</code>. It has been reported elsewhere in this site that lowering the opacity too much makes other functions unable to detect the object.</p>
<pre><code>Module[{r =
1, \[Theta] = \[Pi]/2, \[CurlyPhi] = \[Pi]/6, \[Psi] = \[Pi]/12},
Framed@ImageCrop@Show[
Graphics3D[
{{Specularity[White, 50], Opacity[.2],
Sphere[{0, 0, 0}, r]}, {Arrowheads[.025],
Arrow[{{0, 0, 0}, {1.1, 0, 0}}], Text["x", {1.2, 0, 0}],
Arrow[{{0, 0, 0}, {0, 1.1, 0}}], Text["y", {0, 1.2, 0}],
Arrow[{{0, 0, 0}, {0, 0, 1.1}}], Text["z", {0, 0, 1.2}],
Arrow[{{0, 0, 0},
r {Cos[\[Theta]] Sin[\[CurlyPhi]],
Sin[\[Theta]] Sin[\[CurlyPhi]], Cos[\[CurlyPhi]]}}]}},
Boxed -> False, ImageSize -> 600,
PlotRange -> 1.1 {{-r, r}, {-r, r}, {0, r}}]]]
</code></pre>
<p><img src="https://i.stack.imgur.com/geidR.png" alt="Mathematica graphics"></p>
|
10,427 | <p>I like Mathematica, but it's syntax baffles me.</p>
<p>I am trying to figure out how to minimize the whitespace around a graphic.</p>
<p>For example,</p>
<pre><code>ParametricPlot3D[{r Cos[t], r Sin[t], r^2}, {r, 0, 1}, {t, 0, 2 \[Pi]},
Boxed -> True, Axes -> False]
</code></pre>
<p><img src="https://i.stack.imgur.com/jGpvA.png" alt="3d bounding box on"></p>
<p>Puts the 3d bounding box at the limits of the view. But if I don't show the 3d bounding box,</p>
<p><img src="https://i.stack.imgur.com/jGpvA.png" alt="3d bounding box on"></p>
<pre><code>ParametricPlot3D[{r Cos[t], r Sin[t], r^2}, {r, 0, 1}, {t, 0, 2 \[Pi]},
Boxed -> False, Axes -> False]
</code></pre>
<p><img src="https://i.stack.imgur.com/E8skr.png" alt="3d bounding box off"></p>
<p>there is all this white space around the actual object.</p>
<p>Is there some way (syntax) that can put the view just around the visible objects?</p>
<h1>Edit in response to answers</h1>
<p>Ok, from the below answers, I have two solutions; 1) use ImageCrop, or 2) use <code>Method->{"ShrinkWrap" -> True}</code>. However both of these options do a little something strange to the plot I want (maybe it is just a problem with the plot itself). </p>
<p>So the actual plot I am after is,</p>
<pre><code>Module[{r = 1, \[Theta] = \[Pi]/2, \[CurlyPhi] = \[Pi]/6, \[Psi] = \[Pi]/12},
Framed@Show[
Graphics3D[
{
{Arrowheads[.025],
Arrow[{{0, 0, 0}, {1.1, 0, 0}}], Text["x", {1.2, 0, 0}],
Arrow[{{0, 0, 0}, {0, 1.1, 0}}], Text["y", {0, 1.2, 0}],
Arrow[{{0, 0, 0}, {0, 0, 1.1}}], Text["z", {0, 0, 1.2}],
Arrow[{{0, 0, 0}, r {Cos[\[Theta]] Sin[\[CurlyPhi]],
Sin[\[Theta]] Sin[\[CurlyPhi]], Cos[\[CurlyPhi]]}}]},
{Specularity[White, 50], Opacity[.1], Sphere[{0, 0, 0}, r]}
},
Boxed -> False,
ImageSize -> 600,
PlotRange -> 1.1 {{-r, r}, {-r, r}, {0, r}}
]]
]
</code></pre>
<p><img src="https://i.stack.imgur.com/emm3z.png" alt="enter image description here"></p>
<p>Which has too much whitespace. If I replace <code>Framed@Show[</code> with <code>Framed@ImageCrop@Show[</code> I
get,
<img src="https://i.stack.imgur.com/AUCNY.png" alt="enter image description here"></p>
<p>which actually crops some of the (hemi)sphere. If just use <code>Method -> {"ShrinkWrap" -> True},</code> in the <code>Show</code> options, I get,</p>
<p><img src="https://i.stack.imgur.com/WhX2Y.png" alt="Mathematica graphics"></p>
<p>which looks almost correct, but the <code>x</code> and <code>z</code> textboxes have now not included. Seems like I can't win!</p>
| Vitaliy Kaurov | 13 | <p>I think what you are looking for is <a href="http://reference.wolfram.com/mathematica/ref/ViewAngle.html" rel="noreferrer"><strong>ViewAngle</strong></a> option. The graph below compares default <code>Automatic</code> versus custom setting for <code>ViewAngle</code>. The image are framed intentionally to see clearly the removal of surrounding white space.</p>
<pre><code>Framed[#, FrameMargins -> 0] & /@
(ParametricPlot3D[{r Cos[t], r Sin[t], r^2}, {r, 0, 1}, {t, 0, 2 \[Pi]},
Boxed -> False, Axes -> False, PlotRangePadding -> 0,
ImageSize -> 350 {1, 1}, ViewAngle -> #] & /@ {Automatic, .31})
</code></pre>
<p><img src="https://i.stack.imgur.com/lqH1m.png" alt="enter image description here"></p>
<p>Mathematica has rich set of options for so called "simulated camera" that "views" all 3D objects. To learn in detail about this i suggest downlading notebbok and watching free video of the @Yu-SungChang course <a href="http://www.wolfram.com/events/virtual-conference/2011/presentations/" rel="noreferrer"><strong>Lights, Camera, Graphics!</strong></a> . Here is a illustrious diagram from that course explaining how <code>ViewAngle</code> option (measured in your example in 0.31 radians) sets up the view of a 3D objects.</p>
<p><img src="https://i.stack.imgur.com/iijxH.png" alt="enter image description here"></p>
<p>You actually can change the <code>ViewAngle</code> setting interactively by holding CTRL or ALT key and dragging graphics with mouse - this effectively zooms in or out. See <a href="http://reference.wolfram.com/mathematica/howto/RotateZoomAndPanGraphics.html" rel="noreferrer">this tutorial</a> for details.</p>
|
2,416,424 | <p>It is known that the collection of finite mixtures of Gaussian Distributions over $\mathbb{R}$ is dense in $\mathcal{P}(\mathbb{R})$ (the space of probability distributions) under convergence in distribution metric.</p>
<p>I'm interested to know the following:</p>
<p>Let $P_X$ be a random variable with finite $p$ th moment i.e. $\mathbb{E}_{P_X}[|X|^p]<\infty$, and $P_{X_n}\stackrel{d}{\to} P_X$ where $P_{X_n}$ are mixtures of Gaussian distributions. Then, suppose $X_n \sim P_{X_n}$ and $X\sim P_X$, does it follow that
$$\mathbb{E}[|X_n|^p] \to \mathbb{E}[|X|^p]$$? $\quad (*)$</p>
<p><strong>My attempt:</strong></p>
<p>I've been able to show that
$$\liminf_{n \to \infty} \mathbb{E}[|X_n|^p] \geq \mathbb{E}[|X|^p]$$
In fact this didn't even use the mixtures part. However I'm having difficulty showing the upper bound. Here are some steps:
\begin{eqnarray}
\mathbb{E}[|X_n|^p] &=& \mathbb{E}[|X_n|^p1_{\{|X_n|\leq A\}}] + \mathbb{E}[|X_n|^p1_{\{|X_n|>A\}}] \\
&=& \mathbb{E}[|X_n|^p1_{\{|X_n|\leq A\}}] + A^pPr[|X_n|\geq A] + \mathbb{E}[(|X_n|^p-A^p)1_{\{|X_n|>A\}}]
\end{eqnarray}
Let $f(x)=x1_{0\leq x\le A} + A1_{x\geq A}$ which is a continuous and bounded function. Then the first two terms of RHS are equal to $\mathbb{E}[f(|X_n|^p)]$, which by definition of weak convergence, will converge to $\mathbb{E}[f(|X|^p)]$ as $n \to \infty$. Now we may use MCT as $A\to \infty$ to get $\mathbb{E}[|X|^p]$. Hence in a nutshell, we need to show:
\begin{equation}
\limsup_{A\to\infty} \limsup_{n \to \infty} \mathbb{E}[(|X_n|^p-A^p)1_{\{|X_n|>A\}}] \leq 0
\end{equation}
However I do not know how to proceed from here.</p>
<p>I'd like to point out that I don't know the answer to the question I asked in $(*)$ but my guess is that this is true. The reason is because I read a similar result in Soren Asmussen's book. The result states that for distributions over non-negative reals, we have phase type distributions not only being weak dense but also the moments converging. But in the provided proof, they said "It can be easily shown that the moments converge".</p>
<p><strong>Update:</strong>
Sincerest apologies for I forgot one very essential condition. The mixtures are not any mixtures but specific ones. Namely, I'm looking at distributions $P_X$ such that $P_{X_n}$ has the form $\sum_{k=1}^n \alpha_k \mathcal{N}(\mu_k;P_k)$ ($\mathcal{N}(\mu_k;P_k)$ stands for Gaussian distribution with mean $\mu_k$ and variance $P_k$.) where given $P<\infty$, $\sum_{k=1}^n \alpha_k =1$, $\alpha_k >0$ and $\sum_{k=1}^n\alpha_k P_k = P$. </p>
<p><strong>Update 2:</strong> The reference I mentioned earlier is Soren Asmussen "Applied probability and queues", 2nd edition, page 84.</p>
<p><strong>Update 3:</strong> Looks like I misinterpreted Asmussen. What he wanted to say was that for any distribution $P_X$ with finite $p$th moment, <strong>there exists</strong> a sequence of phase type distributions $P_X^k$ such that $P_X^k\stackrel{d}{\to}P$ and $E_{P_X^K}[|X|^p] \to E_{P_X}[|X|^p]$. This doesn't mean any weak converging mixture will have moment convergence as the answer points out.</p>
| Jaroslaw Matlak | 389,592 | <p>By denoting $t=2^x$ you obtain the following inequality:</p>
<p>$$\frac{16-t}{t-8}>0$$</p>
<p>You can now multiplicate both sides of inequality by $-1$:
$$\frac{t-16}{t-8}<0$$</p>
<p>Ranges of $t$ satisfying this inequality are the same, as ranges of $t$ satisfying this one:
$$(t-16)(t-8)<0$$
with condition $t-8\neq 0$</p>
<p>Resulting range is then $t\in (8,16)$, so $x\in (3,4)$.</p>
|
1,524,349 | <p>This is Problem 45 in Chapter 19 in Michael Spivak's book "Calculus".</p>
<ol start="45">
<li>(a) Suppose that $\frac {f(x)} x$ is integrable on every interval [a, b] for $0$ < a < b, and that $\lim_{x\to0}f(x)=A$ and $\lim_{x\to\infty}f(x)=B$. Prove that for all $\alpha$, $\beta$ > $0$ we have</li>
</ol>
<p>$\int_0^\infty \frac {f(\alpha x) - f(\beta x)}{x}dx = (A-B)log(\frac \beta \alpha)$.</p>
<p>(b) Now suppose instead that $\int_0^\infty\frac{f(x)}xdx$ converges for all $a>0$ and that $\lim_{x\to0}f(x)=A$. Prove that</p>
<p>$\int_0^\infty \frac {f(\alpha x) - f(\beta x)}{x}dx=Alog(\frac \beta \alpha)$.</p>
| John Dawkins | 189,130 | <p>Another approach: Observe that
$$
\int_a^b {f(\alpha x)-f(\beta x)\over x}\,dx =\int_{\alpha a}^{\beta a}{f(t)\over t}\,dt - \int_{\alpha b}^{\beta b}{f(t)\over t}\,dt.
$$
The first integral on the right side of the above display can be written as
$$
\int_{\alpha a}^{\beta a}{f(t)-A\over t}\,dt+A\log(\beta/\alpha),
$$
and the integral in <em>this</em> display converges to $0$ as $a\to 0$. Likewise $\int_{\alpha b}^{\beta b}{f(t)\over t}\,dt=B\log(\beta/\alpha)+o(1)$ as $b\to\infty$. Taken together these show that
$$
\lim_{a\to 0,b\to\infty}\int_a^b{f(\alpha x)-f(\beta x)\over x}\,dx=(A-B)\log(\beta/\alpha).
$$
This is (a). Similar reasoning works for part (b).</p>
|
1,747,696 | <p>First of all: beginner here, sorry if this is trivial.</p>
<p>We know that $ 1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2 $ .</p>
<p>My question is: what if instead of moving by 1, we moved by an arbitrary number, say 3 or 11? $ 11+22+33+44+\ldots+11n = $ ?
The way I've understood the usual formula is that the first number plus the last equals the second number plus second to last, and so on.
In this case, this is also true but I can't seem to find a way to generalize it.</p>
| Christopher Carl Heckman | 261,187 | <p>You get what is called an arithmetic series. More details are at <a href="https://en.wikipedia.org/wiki/Arithmetic_progression" rel="nofollow">https://en.wikipedia.org/wiki/Arithmetic_progression</a> .</p>
|
3,554,646 | <p>Let <span class="math-container">$f:(a,b)\to \mathbb{R}$</span> be injective and continuous. Prove that</p>
<ul>
<li><span class="math-container">$f$</span> is monotonic.</li>
<li>The image of <span class="math-container">$f$</span> is <span class="math-container">$(c,d)$</span> or maybe (<span class="math-container">$-\infty,+\infty$</span>).</li>
</ul>
<p>I proved <span class="math-container">$f$</span> is monotonic but I don't know how to the second proposition begin. </p>
<p>Thankyou so much.</p>
| Ansar | 719,973 | <p>Put <span class="math-container">$c = \mbox{inf Im} f$</span> and <span class="math-container">$d = \mbox{sup Im} f$</span>. If <span class="math-container">$c<y<d$</span>, then <span class="math-container">$c\leq c'<y<d'\leq d$</span> for some <span class="math-container">$c',\ d'\in\mbox{Im} f$</span>. By the intermediate value theorem <span class="math-container">$y\in \mbox{Im} f$</span>. We use the fact that <span class="math-container">$f$</span> is monotonic to prove that <span class="math-container">$c,\ d \not\in \mbox{Im} f$</span>.</p>
|
4,226,455 | <blockquote>
<p>Let <span class="math-container">$X$</span> be a set and <span class="math-container">$Y$</span> a topological space. What is the topology on <span class="math-container">$X$</span> induced by constant maps <span class="math-container">$f:X \to Y$</span>?</p>
</blockquote>
<p>The induced topology is <span class="math-container">$\tau_X = \{f^{-1}[V] : V \in \tau_Y\}$</span> and <span class="math-container">$f^{-1}[V] = \{x \in X : f(x) \in V \}$</span></p>
<p>So if <span class="math-container">$f(x ) = c(x)$</span>, then I initially considered that <span class="math-container">$f^{-1}[V]$</span> would be just the singletons <span class="math-container">$\{x\}$</span>, but I don’t see why there couldn’t be a open set where the there could be multiple <span class="math-container">$f(x) \in V$</span> and thus the preimage wouldn’t be just the singleton?</p>
| Bram28 | 256,001 | <p>You typically can't remove quantifiers and still have a proposition.</p>
<p>For example, take <span class="math-container">$\forall x \ P(x)$</span>. If you take away the quantifier, you are left with <span class="math-container">$P(x)$</span>. But <span class="math-container">$P(x)$</span> is not a proposition. With an unquantified ('free') variable like that, we can't assign a truth-value. Ogf course, if you sound it out, it <em>looks</em> to be a proposition ("<span class="math-container">$x$</span> has property P" certainly sounds like a statement), but since we don;t know what this <span class="math-container">$x$</span> refers to, it is in fact not a claim. Indeed, since it is a variable, we can put either a universal or an existential in front of it, and clearly <span class="math-container">$\forall x \ P(x)$</span> and <span class="math-container">$\exists x \ P(x)$</span> mean different things.</p>
<p>So, a claim like <span class="math-container">$\forall x \ P(x)$</span> is molecular.</p>
<p>I should mention one small caveat: If you have a quantifier in front of something that does not contain the variable that is quantified by that quantifier (e.g. <span class="math-container">$\forall x \exists y P(y)$</span>), then we are dealing with what is called a <em>null</em> quantifier, and as it turns out you can simply remove null quantifiers without changing the meaning of the statement. That is, <span class="math-container">$\forall x \exists y P(y)$</span> is equivalent to <span class="math-container">$\exists y P(y)$</span>. And since the latter is a proposition, one could argue that <span class="math-container">$\forall x \exists y P(y)$</span> is not molecular .... though I think that even in that case some would disagree and say that anytime you have a quantifier, it is molecular.</p>
|
4,226,455 | <blockquote>
<p>Let <span class="math-container">$X$</span> be a set and <span class="math-container">$Y$</span> a topological space. What is the topology on <span class="math-container">$X$</span> induced by constant maps <span class="math-container">$f:X \to Y$</span>?</p>
</blockquote>
<p>The induced topology is <span class="math-container">$\tau_X = \{f^{-1}[V] : V \in \tau_Y\}$</span> and <span class="math-container">$f^{-1}[V] = \{x \in X : f(x) \in V \}$</span></p>
<p>So if <span class="math-container">$f(x ) = c(x)$</span>, then I initially considered that <span class="math-container">$f^{-1}[V]$</span> would be just the singletons <span class="math-container">$\{x\}$</span>, but I don’t see why there couldn’t be a open set where the there could be multiple <span class="math-container">$f(x) \in V$</span> and thus the preimage wouldn’t be just the singleton?</p>
| ryang | 21,813 | <ol>
<li><p>The formula <span class="math-container">$$\forall x{\in} D\:\:P(x)$$</span> is logically equivalent to
<span class="math-container">$$\forall x\,\big(x{\in} D\to P(x)\big).$$</span> These formulae are vacuously true when <span class="math-container">$∀x\;x\notin D,$</span> since a <strong>vacuously true</strong> sentence is a conditional whose antecedent is false.</p>
<p>Your example statement is vacuously true because <span class="math-container">$∀x\:\:x\notin\emptyset=\{x\mid3<x<2\}.$</span></p>
</li>
<li><p>On the other hand, the formula <span class="math-container">$$\exists x{\in} D\:\:P(x)$$</span> is
logically equivalent to <span class="math-container">$$\exists x\,\big(x{\in}
D\:\land\:P(x)\big),$$</span> which can't be converted to a conditional
statement, so can't have an antecedent to speak of as being false.
So, it is never <em>vacuously</em> true.</p>
</li>
<li><p>An atomic formula is one that <em>cannot</em> be broken down into smaller ones;
in contrast, a molecular formula <em>does</em> contain quantifier(s) and/or
logical operator(s).</p>
</li>
<li><p><span class="math-container">$ E=P\to Q\;$</span> is not a well-formed formula; perhaps you mean <span class="math-container">$\;(E\leftrightarrow P)\to Q.$</span></p>
</li>
</ol>
|
2,965,717 | <p>How would you prove that <span class="math-container">$$\displaystyle \prod_{k=1}^\infty \left(1+\dfrac{1}{2^k}\right) \lt e ?$$</span></p>
<p>Wolfram|Alpha shows that the product evaluates to <span class="math-container">$2.384231 \dots$</span> but is there a nice way to write this number? </p>
<p>A hint about solving the problem was given but I don't know how to prove the lemma.</p>
<p>Lemma : Let, <span class="math-container">$a_1,a_2,a_3, \ldots,a_n$</span> be positive numbers and let <span class="math-container">$s=a_1+a_2+a_3+\cdots+a_n$</span> then <span class="math-container">$$(1+a_1)(1+a_2)(1+a_3)\cdots(1+a_n)$$</span> <span class="math-container">$$\le 1+s+\dfrac{s^2}{2!}+\dfrac{s^3}{3!}+\cdots+\dfrac{s^n}{n!}$$</span></p>
| Awe Kumar Jha | 605,905 | <p>Proof of the Lemma by Induction.
Let <span class="math-container">$P(n): (1+a_1)(1+a_2)...(1+a_n) ≤ 1 + s_n + ... + \frac {s_n^n}{n!}$</span>
For<span class="math-container">$n=1$</span> the result is obvious.
Let <span class="math-container">$P(m): (1+a_1)(1+a_2)...(1+a_m) ≤ 1 + s_m + ... + \frac {s_m^m}{m!}$</span> be true. Then,
<span class="math-container">$$1 + s_{m+1} + ... + \frac {s_{m+1}^{m+1}}{(m+1)!}$$</span>
<span class="math-container">$$= 1 + (s_m + a_{m+1}) + ... + \frac {(s_m + a_{m+1})^{m+1}}{(m+1)!}$$</span>
<span class="math-container">$$= 1 + s_m + ... + \frac {s_m^m}{m!} + Q , Q>0$$</span>
Thus we have,
<span class="math-container">$$(1 + s_{m+1} + ... + \frac {s_{m+1}^{m+1}}{(m+1)!}) - Q = 1 + s_m + ... + \frac {s_m^m}{m!} ≥ (1+a_1)(1+a_2)...(1+a_m) < (1+a_1)(1+a_2)...(1+a_m)(1+a_{m+1})$$</span>
Hence <span class="math-container">$P(m+1)$</span> is true whenever <span class="math-container">$P(m)$</span> is true.
Therefore <span class="math-container">$P(n)$</span> is true in general.</p>
|
1,768,700 | <p>According to my knowledge, to prove that $24^{31}$ is congruent to $23^{32}$ mod 19, we must show that both numbers are divisible by 19 i.e. their remainders must be equal with mod 19. Please correct me if I'm wrong.</p>
<p>So, I was able to reduce $23^{32}$ and find its mod 19, which is 17 but I am having a bit of problem with $24^{31}$ since 31 is a prime number and I do not know how to break it down. Please help me with that. </p>
| Umberto P. | 67,536 | <p>After some calculation you have
\begin{align*} 24 &\equiv 5 \bmod 19 \\
24^2 &\equiv 25 \equiv 6 \bmod 19 \\
24^4 &\equiv 36 \equiv -2 \bmod 19 \\
24^8 & \equiv 4 \bmod 19 \\
24^{16} & \equiv 16 \equiv -3 \bmod 19 \end{align*}</p>
<p>Now multiply:
$$24^{31} \equiv 5\cdot 6 \cdot (-2) \cdot 4 \cdot -3 \equiv (30) \cdot (24) \equiv 11 \cdot 5 \equiv 55 \equiv 17 \bmod 19$$</p>
|
4,255,587 | <p>There was a quiz posted on F*cebook by someone. Here's the problem.</p>
<p><a href="https://i.stack.imgur.com/fGu2W.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fGu2W.jpg" alt="enter image description here" /></a></p>
<p>And here's my attempt:</p>
<p><a href="https://i.stack.imgur.com/EMP6g.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EMP6g.jpg" alt="enter image description here" /></a></p>
<p>First, as you can see there, I drew a line from one of the corner of the pink square to the one of the corner of the green square. The degree has been symbolized as alpha (<span class="math-container">$\alpha$</span>).</p>
<p><span class="math-container">$$\begin{align}
\alpha &= \tan^{-1}\left(\frac{16}{23}\right)\\
\end{align}$$</span></p>
<p>That means the degree between the line (almost diagonally to the green square) and the side line of the green square is <span class="math-container">$90-\alpha$</span>. Let's say the side green square is <span class="math-container">$x$</span></p>
<p>We know <span class="math-container">$\cot(x)=\tan(90^{0} - x)$</span>, so</p>
<p><span class="math-container">$$\begin{align}
\tan(90^{0} - \alpha) = \cot(\alpha) &= \frac{16}{x}\\
x &= 16 \tan(\alpha)\\
&= 16 \tan\left(\tan^{-1}\left(\frac{16}{23}\right)\right)\\
&= \frac{16^2}{23} = \frac{256}{23}
\end{align}$$</span></p>
<p>Now, I assume (I can't tell reason, it's just my guesswork) that <span class="math-container">$x$</span> I've found earlier has the same length as the blue side one. I also assume those two blue are right triangles.</p>
<p>Finally, the last calculation is:</p>
<p><span class="math-container">$$\begin{align}
& \left(\frac{1}{2}\cdot \frac{256}{23}\cdot 16\right) + \left(\frac{1}{2}\cdot 7 \left( \frac{256}{23} + 9\right)\right)\\
&= 159.5
\end{align}$$</span></p>
<p>The poster gave me cry emoji and didn't say anything, I conclude my answer is incorrect. Where's the mistake?</p>
| Vasili | 469,083 | <p>Let <span class="math-container">$\alpha$</span> is the base angle. Then the area of one (bigger) blue triangle is <span class="math-container">$\frac{16^2}{2} \tan (2\pi-2\alpha)$</span>, the area of the second blue triangle is <span class="math-container">$\frac{7^2}{2} \tan \alpha$</span>. On the other hand, <span class="math-container">$\tan \alpha=\frac{16\tan (2\pi-2 \alpha)+9}{7}$</span> or (using double-angle formula and considering that <span class="math-container">$\tan (2\pi-2\alpha)=-\tan 2\alpha$</span>) <span class="math-container">$$7\tan \alpha \cdot (1-\tan^2 \alpha)=-32 \tan \alpha-9 \tan^2 \alpha +9 \implies \tan \alpha =3, \tan 2\alpha=-\frac{3}{4}$$</span> Thus, the blue area is <span class="math-container">$8 \cdot 16 \cdot \frac{3}{4}+\frac{7^2 \cdot 3}{2}=169.5$</span><p>
Your assumption is incorrect as the blue side is <span class="math-container">$12 \ne \frac{256}{23}$</span></p>
|
3,068,197 | <p>I'm working through this problem and I haven't been able to make any progress. The textbook provides the answer of <span class="math-container">$ {9 \choose 4}$</span> but I'm not sure as to how they got this result. </p>
| trancelocation | 467,003 | <ul>
<li>If you write the <span class="math-container">$8$</span> <span class="math-container">$1's$</span> horizontally beside each other, the number of possible places to put zeros between them or at the left-most or right-most position is <span class="math-container">$\color{blue}{9}$</span></li>
<li>The number of ways to choose <span class="math-container">$4$</span> of these <span class="math-container">$9$</span> places to put exactly one <span class="math-container">$0$</span> is <span class="math-container">$\color{blue}{\binom{9}{4}}$</span></li>
</ul>
|
3,068,197 | <p>I'm working through this problem and I haven't been able to make any progress. The textbook provides the answer of <span class="math-container">$ {9 \choose 4}$</span> but I'm not sure as to how they got this result. </p>
| robjohn | 13,854 | <p>Consider the two atoms: <span class="math-container">$x\to01$</span> and <span class="math-container">$y\to1$</span>. If we arrange <span class="math-container">$4$</span> <span class="math-container">$x$</span>s and <span class="math-container">$5$</span> <span class="math-container">$y$</span>s, we get all the allowable strings suffixed with a <span class="math-container">$1$</span>. For example:
<span class="math-container">$$xxxyyxyyy\leftrightarrow010101110111\color{#AAA}{1}$$</span>
<span class="math-container">$$yyxyyyxxx\leftrightarrow110111101010\color{#AAA}{1}$$</span>
<span class="math-container">$$yxyxyxyxy\leftrightarrow101101101101\color{#AAA}{1}$$</span></p>
<p>So the number of possible strings is the number of ways to arrange the <span class="math-container">$4$</span> <span class="math-container">$x$</span>s and <span class="math-container">$5$</span> <span class="math-container">$y$</span>s: <span class="math-container">$\binom{9}{4}$</span>.</p>
|
75,791 | <p>When will a probabilistic process obtained by an "abstraction" from a deterministic discrete process satisfy the Markov property?</p>
<p>Example #1) Suppose we have some recurrence, e.g., $a_t=a^2_{t-1}$, $t>0$. It's a deterministic process. However, if we make an "abstraction" by just considering the one particular digit of each $a_t$, we have a probabilistic process. We wonder whether it satisfy a Markov property or not?</p>
<p>Example #2) Suppose we have a finite state automaton. Now we make an "abstraction" by grouping the states into sets of states and obtaining a probabilistic finite state automaton. We consider this automaton in time and we wonder whether it satisfies the Markov property or not.</p>
<p>The particular examples are not important, of interest are some general conditions when a deterministic process becomes a Markov process after an "abstraction" of the kind above (in any context). I'm looking for any references on this matter. </p>
<hr>
<p>Edit: as pointed out in the comments below, the examples #1 and #2 were not well specified. Now there's a distribution on the starting state $a_0$ and $a_t=f(a_{t-1})$ is a deterministic function. Then, $a_t$, $t\geq 0$ is a degenerate Markov chain. Now the question is whether grouping some of the states of such a chain can yield a Markov chain (i.e., pointers to literature where the conditions would be discussed is required). </p>
<p>A more general problem seems to be "Given any Markov chain (i.e., not a degenerate one), group the states into sets: what are the conditions under the resulting process satisfies the Markov assumption?"</p>
| Community | -1 | <p>As @Did points out, you want to know when "lumping" preserves the Markov property. See Lemma 2.5 in <em>Markov Chains and Mixing Times</em> by Levin, Peres, and Wilmer. The book is available online <a href="http://pages.uoregon.edu/dlevin/MARKOV/markovmixing.pdf" rel="nofollow">here</a>.</p>
|
3,449,274 | <p>I have equation <span class="math-container">$2b^2 - 72b - 406=0$</span>. I divided it with 2 and I got <span class="math-container">$b^2 - 36b - 203=0$</span>. My teacher then wrote <span class="math-container">$(b-29)(b-7)=0$</span> but I don’t understand how he got that. When I try to solve that equation I get <span class="math-container">$18 \pm\sqrt{(527)}$</span>. How did he get <span class="math-container">$29$</span> and <span class="math-container">$7$</span> and how did he factorize that?</p>
| Matt Samuel | 187,867 | <p>A horizontal asymptote only tells you that it tends to a certain value as <span class="math-container">$x$</span> goes to infinity in some direction. It can certainly cross the asymptote any number of times without contradicting this, including infinitely many times. For example,
<span class="math-container">$$f(x) =\frac{\sin x} x$$</span>
has a horizontal asymptote at <span class="math-container">$y=0$</span>, but continually crosses the asymptote as <span class="math-container">$x$</span> approaches infinity. </p>
|
2,245,631 | <blockquote>
<p>$x+x\sqrt{(2x+2)}=3$</p>
</blockquote>
<p>I must solve this, but I always get to a point where I don't know what to do. The answer is 1.</p>
<p>Here is what I did: </p>
<p>$$\begin{align}
3&=x(1+\sqrt{2(x+1)}) \\
\frac{3}{x}&=1+\sqrt{2(x+1)} \\
\frac{3}{x}-1&=\sqrt{2(x+1)} \\
\frac{(3-x)^{2}}{x^{2}}&=2(x+1) \\
\frac{9-6x+x^{2}}{2x^{2}}&=x+1 \\
\frac{9-6x+x^{2}-2x^{2}}{2x^{2}}&=x \\
\frac{9-6x+x^{2}-2x^{2}-2x^{3}}{2x^{2}}&=0
\end{align}$$</p>
<p>Then I got:
$-2x^{3}-x^{2}-6x+9=0$ </p>
| zwim | 399,263 | <p>$f(x)=x+x\sqrt{2x+2}$</p>
<p><a href="https://i.stack.imgur.com/m0hnj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/m0hnj.png" alt="enter image description here"></a></p>
<p>We are trying to find $f(x)=3$ but notice that for $-1\le x\le 0$ then $f(x)\le 0$.</p>
<p>$f'(x)=1+\sqrt{2x+2}+\frac {x}{\sqrt{2x+2}}>0$ for $x>0$ </p>
<p>so $f$ is increasing from $f(0)=0$ to $\lim\limits_{x\to+\infty}f(x)=+\infty$</p>
<p>Thus there is an unique $x$ such that $f(x)=3$ and since $f(1)=3$ we are done.</p>
<p>It's not fordidden to have a look at the graph, before trying more complicated stuff.</p>
|
4,215,724 | <p><span class="math-container">$f\colon \mathbb{R}^2\to \mathbb{R}$</span> such that <span class="math-container">$f_x(2,-1)=1$</span> and <span class="math-container">$f_y(2,-1)=1$</span> and <span class="math-container">$g(x,y)=\langle x^2y,x-y\rangle$</span> and <span class="math-container">$h = f\circ g$</span> then find <span class="math-container">$h_y(1,2)$</span>. The options given are <span class="math-container">$-2,2,0,5,-5,-3,10,3$</span></p>
<p><strong>My Attempt</strong></p>
<p><span class="math-container">$$
h_y(1,2)=(f\circ g)_y(x,y)\Big|_{x=1,y=2}=f_y(g(1,2))\cdot g_y(1,2)=f_y(2,-1)\cdot g_y(1,2)\\=1 \cdot g_y(1,2)
$$</span></p>
<p>Is it the right way towards finding the solution ? How do I find <span class="math-container">$g_y(1,2)$</span> ?</p>
| Sebathon | 482,453 | <p>If <span class="math-container">$\langle \cdot,\cdot \rangle$</span> defines a vector, then you did it wrong. Using the <a href="https://en.wikipedia.org/wiki/Chain_rule#General_rule" rel="nofollow noreferrer">chain rule</a> (the case when <span class="math-container">$k=1$</span>) we have</p>
<p><span class="math-container">$$ h_{y}(x,y) =(\nabla f)(g(x,y)) \cdot g_{y}(x,y)=\langle f_{x}(g(x,y)),f_{y}(g(x,y)) \rangle \cdot \langle x^{2},-1 \rangle.$$</span></p>
<p>Thus</p>
<p><span class="math-container">\begin{align*} h_{y}(1,2) &=\langle f_{x}(g(1,2)),f_{y}(g(1,2)) \rangle \cdot \langle 1,-1 \rangle \\ &= \langle f_{x}(2,-1),f_{y}(2,-1) \rangle \cdot \langle 1,-1 \rangle \\ &= \langle 1,1 \rangle \cdot \langle 1,-1 \rangle \\ &=0
\end{align*}</span></p>
|
233,169 | <p>I had to redo the problem because there was a mistake. With the given function from a previous problem, I was solving <a href="https://mathematica.stackexchange.com/questions/231664/adding-a-point-in-a-manipulate-command">link</a>, I found that the parabola created a trajectory on the graph, ie another parabola.</p>
<p>I'm trying to create a plot where I collect the coordinate points of Min. value for 21 values of parameter <strong>a</strong> (from -7 to 7), and find a,b, and c such that the points are on the curve of the quadratic equation <span class="math-container">$ax^2+bx+c=y$</span>. The plot looks something like this</p>
<p><a href="https://i.stack.imgur.com/msUOn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/msUOn.png" alt="enter image description here" /></a></p>
<p>The curve of <span class="math-container">$ax^2 + b+x +c=y$</span> goes through all the minimum points for the curves created</p>
| cvgmt | 72,111 | <p>To get the trajectory equation about the minimum, we can do as below.</p>
<pre><code>f[x_, a_] := x^2 - 2*(a - 2)*x + a - 2;
min = Minimize[f[x, a], x];
Eliminate[{x, y} == ({x, y} /. Last@min /. y -> First@min), a]
</code></pre>
<blockquote>
<p>-y == -x + x^2</p>
</blockquote>
<p>To plot the minimum points, here we also use <code>Mesh</code>.</p>
<pre><code>Clear["`*"];
f[x_, a_] := x^2 - 2*(a - 2)*x + a - 2;
curves = Plot[Table[f[x, a], {a, -7, 7, 14/20}], {x, -20, 20},
PlotStyle -> {Thickness[Small], Cyan}];
min = Minimize[f[x, a], x];
trajectory =
ParametricPlot[{x, y} /. Last@min /. y -> First@min, {a, -9, 9},
MeshFunctions -> Function[{x, y, a}, a],
Mesh -> {Range[-7, 7, 14/20]},
MeshStyle -> {PointSize[Medium], Red}, PlotStyle -> Yellow];
Show[curves, trajectory]
</code></pre>
<p>Another way is <code>Show</code> the minimum points by <code>Graphics</code></p>
<pre><code>points = Graphics[{Red,
Table[Point[{x, y} /. Last@min /. y -> First@min], {a, -7, 7,
14/20}]}];
</code></pre>
<p><a href="https://i.stack.imgur.com/8lcl9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8lcl9.png" alt="enter image description here" /></a></p>
|
108,060 | <p>Suppose:
$$\sum_{n=2}^{\infty} \left( \frac{1}{n(\ln(n))^{k}} \right) =\frac{1}{ 2(\ln(2))^{k} } +\frac{1}{ 3(\ln(3))^{k} }+...,
$$
by which $k$ does it converge?</p>
<p>When I use comparison test I get inconclusive result:</p>
<p>$\lim_{n\rightarrow\infty} \frac{u_{n+1}}{u_{n}}=\frac{n\ln(n)^{k}}{(n+1)\ln(n+1)^{k}}
=\lim_{n\rightarrow\infty} =\frac{n\ln(n)^{k}+\ln(n+1)^{k}-\ln(n+1)^{k}}{(n+1)\ln(n+1)^{k}}\approx
1- \frac{\ln(n+1)^{k}}{(n+1)\ln(n+1)^{k}}=\\1-\lim_{n\rightarrow \infty}\frac{1}{n+1}=1$</p>
<p>Now my conclusion would be when $k\in\mathbb R$ but I feel I am doing something wrong because I am pretty sure I have done this kind of problems earlier where I used some well-known series comparison. WA nothing <a href="http://www.wolframalpha.com/input/?i=Integrate%28x%5E%7Bk%7Dln%28x%29,x,0,%5Cinfty%29" rel="nofollow">here</a>.</p>
<p><strong>[Update] Trying to use Cauchy condensation test</strong></p>
<p>$$\sum_{n=2}^{\infty}\left( n^{1/k} ln(n)\right)^{-k}$$</p>
<p>and now C-test:</p>
<p>$$\sum_{n=2}^{\infty}\left( \left(2^{n} \right) 2^{n/k} ln(2^{n})\right)^{-k}=$$
$$\sum_{n=2}^{\infty} e^{-k \left( n(1+\frac{1}{k})ln(2)+ln(n)+ln(ln(2) \right) }$$</p>
<p>so now as a geometric series, can I conclude something in terms of $k$? Look $k$ is still in one denominator not the just first factor in the exponent. </p>
| Community | -1 | <p>I think Cauchy's Condensation test will do the deal.</p>
<p>$$\sum a_n \text{converges} \iff \sum2^na_{2^n} \text{converges}$$</p>
<p>And, on applying this test, and simplifying the test, you'll need to compare it with the series $\sum n^{-p}$. For what values of $p$ does this series converge?</p>
|
327,860 | <p>Let <span class="math-container">$A$</span> be a symmetric <span class="math-container">$d\times d$</span> matrix with integer entries such that the quadratic form <span class="math-container">$Q(x)=\langle Ax,x\rangle, x\in \mathbb{R}^d$</span>, is non-negative definite. For which <span class="math-container">$d$</span> does it imply that <span class="math-container">$Q$</span> is a sum of finitely many squares of linear forms with integer coefficients
<span class="math-container">$$
Q(x)=\sum_{i=1}^N (\ell_i(x))^2\quad \text{for some}\, N?
$$</span>
For <span class="math-container">$d=2$</span> this is true, I know it from the <a href="https://artofproblemsolving.com/community/c6h219889p1219430" rel="noreferrer">problem</a> proposed by Sweden to IMO in 1995, but probably all this stuff is known for a longer time. </p>
<p>I think, I may prove it for some other small dimensions, although not so elementary (using Minkowski theorem on lattice points in convex bodies: if <span class="math-container">$Q$</span> is positive definite, we may find a linear form <span class="math-container">$\ell(x)$</span> such that <span class="math-container">$Q-\ell^2$</span> is still non-negative definite, this is equivalent to finding an integer point in an ellipsoid), but for large <span class="math-container">$d$</span> this argument fails. </p>
| Zhi-Wei Sun | 124,654 | <p>Such problems were investigated by L. J. Mordell in the 1930s. Two related papers of Mordell are as follows:</p>
<ol>
<li><p>L. J. Mordell, A new Warings problem with squares of linear forms, Quarterly J. (Oxford series) 1 (1930), 276–288. </p></li>
<li><p>L. J. Mordell, On binary quadratic forms expressable as a sum of three linear squares with integer coefficients, J. Reine Angew. Math. 167 (1932), 12–19. </p></li>
</ol>
<p>For an introduction to Mordell's results on sums of squares of linear forms, see Section 2 of D. W. Hoffmann's recent preprint <a href="https://arxiv.org/abs/1902.07109" rel="nofollow noreferrer">Sums of integers and sums of their squares</a> where the author applied Mordell's results to study sums of squares with certain linear restrictions.</p>
|
78,725 | <p>The general theorem is: for all odd, distinct primes $p, q$, the following holds:
$$\left( \frac{p}{q} \right) \left( \frac{q}{p} \right) = (-1)^{\frac{p-1}{2}\frac{q-1}{2}}$$</p>
<p>I've discovered the following proof for the case $q=3$:
Consider the Möbius transformation $f(x) = \frac{1}{1-x}$, defined on $F_{p} \cup {\infty}$. It is a bijection of order 3: $f^{(3)} = Id$.</p>
<p>Now we'll count the number of fixed points of $f$, modulo 3:</p>
<p>1) We can calculate the number of solutions to $f(x) = x$: it is equivalent to $(2x-1)^2 = -3$. Since $p \neq 2,3$, the number of solutions is $\left( \frac{-3}{p} \right) + 1$ (if $-3$ is a non-square, there's no solution. Else, there are 2 distinct solutions, corresponding to 2 distinct roots of $-3$).</p>
<p>2) We know the structure of $f$ as a permutation: only 3-cycles or fixed points. Thus, number of fixed points is just $|F_{p} \cup {\infty}| \mod 3$, or: $p+1 \mod 3$.</p>
<p>Combining the 2 results yields $p = \left( \frac{-3}{p} \right) \mod 3$. Exploiting Euler's criterion gives $\left( \frac{p}{3} \right) = p^{\frac{3-1}{2}} = p \mod 3$, and using $\left( \frac{-1}{p} \right) = (-1)^{\frac{p-1}{2}}$, we get:
$$\left( \frac{3}{p} \right) \left( \frac{p}{3} \right) = (-1)^{\frac{p-1}{2}\frac{3-1}{2}} \mod 3$$
and equality in $\mathbb{Z}$ follows.</p>
<p>My questions:</p>
<ul>
<li>Can this idea be generalized, with other functions $f$?</li>
<li>Is there a list\article of proofs to special cases of the theorem?</li>
</ul>
| Grigory M | 152 | <p>Well, at least it can be extended to a proof of the case $q=5$.
$\def\lf#1#2{\left(\dfrac{#1}{#2}\right)}$</p>
<p><strong>0.</strong> $\lf5p=1\iff\exists\phi\in\mathbb F_p:\phi^2+\phi-1=0$.</p>
<p>// Note that in $\mathbb R$ one can take $\phi=2\cos(2\pi/5)$. So in the next step we'll use something like 'rotation by $2\pi/5$' (unfortunately $\sin(2\pi/5)\notin\mathbb Q(\sqrt5)$ so we'll have to use a slightly different matrix).</p>
<p><strong>1.</strong> $\lf5p=1\implies\lf p5=1$.</p>
<p>$\sigma\colon x\mapsto\dfrac1{\phi-x}$ is a Möbius transformation of $P^1(\mathbb F_p)$ of order 5 (its matrix has two 5th roots of unity as eigenvalues) which has either 0 or 2 fixed points. So $p\pm1$ is divisible by 5, or equivalently $\lf p5=1$.</p>
<p><strong>2.</strong> $\lf p5=1\implies\lf5p=1$.</p>
<p>$\lf p5=1$ means that $p^2+1$ is not divisible by 5, so the action of $\sigma$ on $P^1(\mathbb F_p(\phi)=\mathbb F_{p^2})$ has fixed points. A fixed point of $\sigma$ is a solution of the equation $x^2-\phi x+1=0$, so $\phi=x+x^{-1}\in\mathbb F_p$.</p>
<p>// It, of course, would be nice to have a different, more geometric proof of (2)…</p>
|
62,000 | <p>Let $I,J,K$ be three non-void sets, and let $\gamma$:$I\times J\times K\rightarrow\mathbb{N}$.
Is there some nonempty set $X$, together with some functions {$\{ f_{i}:X\rightarrow X;i\in I\} $},
some subsets {$\{ \Omega_{j}\subset X;j\in J\} $}, and some
points {$\{p_{k}\in X;k\in K} $} s.t. $\mid f_{i}^{-1}\left(p_{k}\right)\cap\Omega_{j}\mid=\gamma\left(i,j,k\right)$
$\left(i\in I,j\in J,k\in K\right)$, and $\mid f_{i}^{-1}\left(p\right)\mid\leq\mid\mathbb{R\mid}$$\left(i\in I,p\in X\right)$
? In other words, is $\gamma$ ''representable'' as the number of
solutions of some ''reasonable'' equations? [An elementary problem,
indeed.] </p>
| Zev Chonoles | 1,916 | <p>Here is a second attempt (see edit history for previous version).</p>
<p>For each $t\in\mathbb{N}$, let
$$P_{i,j,k,t}=\{1_{i,j,k,t},\ldots,n_{i,j,k,t},\ldots,\gamma(i,j,k)_{i,j,k,t}\}$$
(so that for each choice of $i\in I$, $j\in J$, $k\in K$, and $t\in\mathbb{N}$, we have a disjoint set of size $\gamma(i,j,k)$). </p>
<p>For each $t\in\mathbb{N}$, let
$$Q_t=\{a_{k,t}\mid k\in K\}$$
(so for each $t\in\mathbb{N}$, this is just a copy of $K$, up to relabeling). </p>
<p>Let
$$X=\coprod_{t\in\mathbb{N}}\left(Q_t\coprod_{\substack{i\in I,j\in J\\\k\in K}}P_{i,j,k,t}\right).$$
Define
$$\Omega_j=\coprod_{i\in I,k\in K}P_{i,j,k,1}\subset X,$$
and $f_i:X\rightarrow X$ by
$$f_{i_0}(n_{i,j,k,t})=\begin{cases}a_{k,1}\text{ if }i=i_0,t=1\\\ n_{i,j,k,t+1}\text{ otherwise}\end{cases}$$
$$f_i(a_{k,t})=a_{k,t+1}$$</p>
<p>Thus
$$f_{i}^{-1}(n_{i,j,k,t})=\begin{cases}\emptyset\text{ if }t=1,2\\\ \{n_{i,j,k,t-1}\}\text{ if }t>2\end{cases}$$
$$f_i^{-1}(a_{k,t})=\begin{cases}\coprod_{j\in J}P_{i,j,k,1}\text{ if }t=1\\\ \{a_{k,t-1}\}\text{ if }t>1\end{cases}$$
We choose $p_k=a_{k,1}$.</p>
<p>Thus $f_i^{-1}(p_k)\cap \Omega_j=P_{i,j,k,1}$, so $|f_i^{-1}(p_k)\cap\Omega_j|=\gamma(i,j,k)$.</p>
<p>Unfortunately this still doesn't address your size concerns, i.e. the preimage of any element of $X$ being countable, because if $J$ is uncountable then $f_i^{-1}(a_{k,1})$ is uncountable (I added the whole mess with the $t$'s to make the preimages of all the other elements countable). I'll leave this as a community wiki, and if anyone sees a way of fixing it they are welcome to edit this.</p>
|
4,513,678 | <p>Suppose <span class="math-container">$f(x) = ax^3 + bx^2 + cx + d$</span> is a cubic equation with roots <span class="math-container">$\alpha, \beta, \gamma.$</span> Then we have:</p>
<p><span class="math-container">$\alpha + \beta + \gamma= -\frac{b}{a}\quad (1)$</span></p>
<p><span class="math-container">$\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}\quad (2)$</span></p>
<p><span class="math-container">$\alpha\beta\gamma = -\frac{d}{a}\quad (3)$</span></p>
<p>We can find <span class="math-container">$\alpha^2\beta + \beta^2\gamma + \gamma^2\alpha + \alpha^2\gamma + \gamma^2\beta + \beta^2\alpha$</span> in terms of <span class="math-container">$a,b,c,d$</span> with the formula:</p>
<p><span class="math-container">$$ \alpha^2\beta + \beta^2\gamma + \gamma^2\alpha + \alpha^2\gamma + \gamma^2\beta + \beta^2\alpha = (\alpha+\beta+\gamma)(\alpha\beta+\alpha\gamma+\beta\gamma) - 3\alpha\beta\gamma $$</span>
<span class="math-container">$$=\left(\frac{-b}{a}\right) \left(\frac{c}{a}\right) - 3\left(-\frac{d}{a}\right).$$</span></p>
<p>But I was wondering if there was some way to find <span class="math-container">$ \alpha^2\beta + \beta^2\gamma + \gamma^2\alpha\ $</span> and therefore also <span class="math-container">$\ \alpha^2\gamma + \gamma^2\beta + \beta^2\alpha\ $</span> in terms of <span class="math-container">$a,b,c,d,\ $</span> with some algebraic manipulation, i.e. without <a href="https://mathworld.wolfram.com/CubicFormula.html" rel="nofollow noreferrer">finding the roots with a cubic formula</a>?</p>
<p>Notice that there are <em>two</em> possible values of <span class="math-container">$\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha,$</span> namely <span class="math-container">$\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha = \beta^2\gamma+\gamma^2\alpha+\alpha^2\beta = \gamma^2\alpha+\alpha^2\beta+\beta^2\gamma$</span> and <span class="math-container">$\alpha^2\gamma+\gamma^2\beta+\beta^2\alpha = \gamma^2\beta+\beta^2\alpha+\alpha^2\gamma = \beta^2\alpha + \alpha^2\gamma+\gamma^2\beta.$</span></p>
| Sourav Ghosh | 977,780 | <p>A function of two or more variables is said to be symmetric function if <span class="math-container">$f$</span> remains unaltered by an interchange of any two of it's variables.</p>
<p>If <span class="math-container">$\alpha, \beta, \gamma$</span> are roots of a cubic equation and the function <span class="math-container">$f$</span> of <span class="math-container">$\alpha,\beta,\gamma$</span> is invariant under the permutations of <span class="math-container">$\alpha, \beta, \gamma$</span>, then we call <span class="math-container">$f$</span> , a symmetric function of roots.</p>
<p>Let us consider a function</p>
<p><span class="math-container">$f(x, y, z) =x^2y+y^2x+z^2x+x^2z+y^2z+z^2y$</span></p>
<p>Then <span class="math-container">$f$</span> is a symmetric function of the roots <span class="math-container">$\alpha, \beta, \gamma$</span>.</p>
<p>Here comes the nice formula</p>
<p><span class="math-container">$\sum_{\text{sym}} \alpha^2\beta= \sum_{\text{sym}} \alpha
\sum_{\text{sym}} \alpha\beta-3\alpha\beta\gamma$</span></p>
<hr />
<p>Now consider the function</p>
<p><span class="math-container">$f(x, y, z) =x^2y+y^2z+z^2x$</span></p>
<p>Then <span class="math-container">$f$</span> is not a symmetric function of the roots <span class="math-container">$\alpha, \beta, \gamma$</span>.</p>
<p>The values of <span class="math-container">$f$</span> changes along with the permutations of the roots.</p>
<p>Hence <span class="math-container">$\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha$</span> may have different values.</p>
<hr />
<p><span class="math-container">$f(\alpha, \beta, \gamma) =\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha$</span> have two different vaues for distinct values of <span class="math-container">$\alpha,\beta,\gamma$</span>. One for all even permutations and other for all odd permutations.</p>
<p>For even permutations :</p>
<p><span class="math-container">$f(\alpha, \beta, \gamma) =f(\beta, \gamma, \alpha) =f(\gamma, \alpha, \beta) $</span></p>
<p>For odd permutations :</p>
<p><span class="math-container">$f(\alpha, \gamma, \beta) =f(\beta, \alpha, \gamma) =f(\gamma, \beta, \alpha) $</span></p>
<p>But still the sum depends on the choice of the order of the roots <span class="math-container">$\alpha, \beta, \gamma$</span>.</p>
|
2,525,573 | <p>Find the domain and range of $y=\sqrt {x-2}$</p>
<p>My Attempt:
$$y=\sqrt {x-2}$$
For $y$ to be defined,
$$(x-2)\geq 0$$
$$x\geq 2$$
So $dom(f)=[2,\infty)$.</p>
| Michael Rybkin | 350,247 | <p>Here, I'm using the property: $a^b=a^c \Leftrightarrow b=c $ where $a > 0$ and $a\ne1$. Here's how it goes:</p>
<p>$$
e^{3x-1} = 5e^{2x}\implies\\
e^{3x-1} = e^{\ln{5}}\cdot e^{2x}\implies\\
e^{3x-1} = e^{\ln{5}+2x}\implies\\
3x-1 = \ln{5} + 2x\implies\\
3x-2x= \ln{5}+1\implies\\
x= \ln{5}+1\implies\\
x= \ln{5}+\ln{e}\implies\\
x= \ln{(5e)}
$$</p>
<p>Your mistake was that $2x$ is the exponent only on $e$, while according to the rule $\log_{a}{b^c}=c\log_{a}{b}$, the exponent should be on the entire expression of the logarithm before it can be pulled out front. You thought that $\ln 5e^{2x}$ was $\ln (5e)^{2x}$, but that's not a correct interpretation of mathematical symbology. So, you can't really bring that $2x$ out front like you did in your solution. And that's where problems arose.</p>
|
416,940 | <p><span class="math-container">$\DeclareMathOperator\Spec{Spec}\newcommand{\perf}{\mathrm{perf}}\DeclareMathOperator\SHC{SHC}$</span>I have just finished reading the paper "The spectrum of prime ideals in tensor triangulated categories" in which Balmer proposes his notion of spectrum which nowadays is considered central in the understanding and classification of the homotopy categories which we want to study in the concrete mathematical practice (to name a few examples: the <span class="math-container">$G$</span>-equivariant stable homotopy category for <span class="math-container">$G$</span> a compact Lie group, or the derived category of quasi-coherent sheaves on a scheme).</p>
<p>Since I am not familiar with this notion I wanted to ask here various questions about the underlying ideas of such concept.</p>
<p>(1) I noticed that all the examples proposed by Balmer in his paper deal with compact objects, in the sense that the proposed tensor triangulated categories (t.t. categories from now on) can be identified with the full-subcategories of compact objects in a larger t.t. category. And from what I remember every other example which I read in different sources does the same thing: we study the Balmer spectrum of compact objects in a larger t.t. category. Balmer does not explicitly state that this must be the case, indeed his definition does not require the involved objects to be compact a priori.</p>
<p>For this abstract machinery to work we only need the t.t. category to be essentially small. I could think that this is the problem: in general we cannot guarantee that the t.t. category we are interested in is essentially small so we restrict to the subcategory of its compact objects for this property to be more likely.</p>
<p>But I have other reasons to believe that this justification is not completely correct: if we indulge in the intuition suggested by the choice of words, we should think of the support of an object in our t.t. category as an higher categorical analogue of the usual support of a function. Fixing the domain of our functions to be compact spaces ensures that the support will also be compact. So if we consider also non-compact objects the support could be non "topologically small".</p>
<p>Thus I am inclined to believe for the complete t.t. categories either the Balmer spectrum is too big to be computed or its is not the correct notion we want to use to classify their tensor subcategories.</p>
<p>(2) Related to the previous question: if the proposed notion of Balmer spectrum should be applied only to categories of compact objects, what can we deduce about the whole category of possibly non-compact objects? Suppose we consider an essentially small t.t. category <span class="math-container">$\mathcal{T}$</span> and we manage to compute the Balmer spectrum of <span class="math-container">$\mathcal{T}^c$</span>, can we deduce any information regarding the thick tensor ideals or localizing tensor ideals of <span class="math-container">$\mathcal{T}$</span>?</p>
<p>Two classical examples of this are <span class="math-container">$D(R)$</span>, the derived category of a commutative ring <span class="math-container">$R$</span>, and <span class="math-container">$\SHC$</span>, the stable homotopy category. For <span class="math-container">$D^{\perf}(R)$</span> this is homeomorphic to the usual Zariski spectrum <span class="math-container">$\Spec(R)$</span>, while for <span class="math-container">$\SHC^\mathrm{c}$</span> we have the classification provided by the thick subcategory theorem from chromatic homotopy theory. But I have never seen a classification (even partial) of their thick tensor subcategories or thick localizing subcategories.</p>
<p>(3) What information does the Balmer spectrum encode? Balmer proves that there is a bijection between the Thomason subsets of this spectrum and the radical thick tensor ideals of the t.t. category. But other than this? At first I expected that if two t.t. categories had isomorphic spectrum then they would have a sufficiently compatible t.t. structure. Then I found the following interesting example: we have that the Balmer spectrum of the category of compact rational <span class="math-container">$S^1$</span>-equivariant spectra is homeomorphic to <span class="math-container">$\Spec(\mathbb{Z})$</span>. If <span class="math-container">$H \leq S^1$</span> is a closed subgroup then the kernel of <span class="math-container">$\phi^H$</span>, the non-equivariant geometric <span class="math-container">$H$</span>-fixed points, provides a Balmer prime. Then <span class="math-container">$\ker \phi^{S^1}$</span> corresponds to the generic point <span class="math-container">$(0)$</span>, while <span class="math-container">$\ker \phi^{C_n}$</span> can be mapped to <span class="math-container">$(p_n)$</span> where we order the prime numbers <span class="math-container">$\{p_n : n \geq 1 \}$</span>.</p>
<p>Therefore <span class="math-container">$S^1\text{-}\SHC^\mathrm{c}_{\mathbb{Q}}$</span> and <span class="math-container">$D^{\perf}(\mathbb{Z})$</span> have the same Balmer spectrum, but they are very different t.t. categories: for one, the latter has a compact generator given by the tensor unit, while this is not the case in the former category. I would have thought that the t.t. structure would have been more rigid with respect to the Balmer spectrum, but this seems not to be the case.</p>
<p>If you wanted a more precise question: if two t.t. categories have homeomorphic Balmer spectra, can we translate this to any information on the two categories? What if the homeomorphism is induced by a monoidal exact functor? Can we deduce it is fully faithful, essentially surjective or any other property?</p>
<p>I hope that my questions are not too vague or naïve.</p>
| Maxime Ramzi | 102,343 | <p>I am not an expert in tt-geometry, but let me try to answer some of your questions.</p>
<p>(1) You are correct, the Balmer spectrum is typically not well-suited to study the "big" categories - this is because all definitions that appear only use "finitary" things : tensor products, cones/extensions, finite direct sums, retracts. This makes it, as defined, ill-suited for studying big categories where you also have interesting infinitary phenomena.</p>
<p>In a big category, you might be more interested in studying localizing ideals for instance, where you can take arbitrary (homotopy) colimits, but then you run into subtle issues about compact generation and telescope conjectures etc. (which are also studied !)</p>
<p>This is not a hopeless situation, though : a lot of work has been done (is probably being done) about finding suitable notions of support for "big" categories (see e.g. Balmer's paper <em>Homological support of big objects in tensor-triangulated categories</em> - this is far from the only one on the topic, see e.g. <em>Big categories, big spectra</em> by Balchin and Stevenson)</p>
<p>In fact, the place where the Balmer spectrum is somehow the best suited is when the monoidal structure interacts well with finiteness: namely in rigid situations (resp. rigidly compactly generated).</p>
<p>There was not a clear question here, so I hope this answers it.</p>
<p>(2) I think I've answered this partially in my answer to (1). Thick tensor ideals in the small world give rise to localizing ideals in the big world, but in general there is no way to go back, and even when there is the comparison is not perfect (a keyword here is telescope conjecture; but it's not the only thing, and the example of <span class="math-container">$SHC$</span> should be enlightening : say we look at a prime <span class="math-container">$p$</span>, then the kernel of <span class="math-container">$K(n)\otimes -$</span> and <span class="math-container">$E_n \otimes -$</span> are very different, as witnessed by the difference between <span class="math-container">$L_{K(n)}$</span> and <span class="math-container">$L_n$</span>, but their kernels agree in <span class="math-container">$SHC^c$</span>). The Balchin-Stevenson paper I mentioned earlier has a section "Comparison maps". Probably other papers that study this kind of thing raise the same kind of question, so you might want to look at that literature (if someone more knowledgeable wants to edit my answer and add some references about this, they would be most welcome !).</p>
<p>The moral is somehow that "big" things are harder to classify.</p>
<p>(3) An abstract homeomorphism of spectra is unlikely to give you any information, except that the "large scale" structure of the two tt-categories is the same (but that would be tautological : one could define this large scale structure by the Balmer spectrum) . This is the same thing with ordinary commutative rings: an abstract homeomorphism of spectra won't tell you much.</p>
<p>You can say much more if the homeomorphism is induced by a tt-functor between them f course, and somehow the functoriality of the Balmer spectrum is key to Balmer's approach, and to computations (e.g. the computation of the spectrum of the equivariant stable homotopy category relies heavily on leveraging the various geometric fixed points functors that one has). If you think in terms of rings, a morphism of commutative rings <span class="math-container">$R\to S$</span> that induces a homeomorphism of spectra doesn't tell you that they are isomorphic : indeed, there is some nilpotent business happening here. But you can think of it as some <a href="https://en.wikipedia.org/wiki/Going_up_and_going_down" rel="nofollow noreferrer">going up/down theorem</a>.</p>
<p>Balmer has a paper about this kind of question, called <em>On the surjectivity of the map of spectra associated to a tensor-triangular functor</em>. He proves there for instance that if the map of spectra is surjective (on closed points), then the original functor is conservative. He further completely characterizes surjectivity in terms of detection of nilpotence (which is related to the nilpotent problem I mentioned earlier for rings).</p>
<p>Certainly, more things can be said if the map is a homeomorphism, and again someone more aware of the literature on the topic could probably say more than I did (if anyone wants to edit and add some references, it would be great, as before).</p>
<p>As a general rule of thumb, a good place to test ideas/conjectures about this stuff is with commutative rings or more generally schemes : the spectrum of the perfect derived category of a (nice) scheme <span class="math-container">$X$</span> is exactly (the underlying space of) <span class="math-container">$X$</span>, in a way compatible with morphisms of schemes. This is to some extent not the most interesting case, but it's a good way to check intuitions.</p>
|
4,492,566 | <blockquote>
<p>To which degree must I rotate a parabola for it to be no longer the graph of a function?</p>
</blockquote>
<p>I have no problem with narrowing the question down by only concerning the standard parabola: <span class="math-container">$$f(x)=x^2.$$</span></p>
<p>I am looking for a specific angle measure. One such measure must exist as the reflection of <span class="math-container">$f$</span> over the line <span class="math-container">$y=x$</span> is certainly no longer well-defined. I realize that preferentially I should ask the question on this site with a bit of work put into it but, alas, I have no intuition for where to start. I suppose I know immediately that it must be less than <span class="math-container">$45^\circ$</span> as such a rotation will cross the y-axis at <span class="math-container">$(0,0)$</span> and <span class="math-container">$(0,\sqrt{2})$</span>.</p>
<p>Any insight on how to proceed?</p>
| Taladris | 70,123 | <p>Let <span class="math-container">$P_\theta$</span> be the parabola obtained from the parabola <span class="math-container">$P$</span> of equation <span class="math-container">$y=x^2$</span> by a rotation of angle <span class="math-container">$\theta\in(0,2\pi)$</span>.</p>
<p>A rotation of <span class="math-container">$\pi$</span> is simply a symmetry about the origin so <span class="math-container">$P_\pi$</span> is the graph of <span class="math-container">$y=-x^2$</span>, the graph of a function. Moreover, if <span class="math-container">$\theta=\frac{3\pi}{2}$</span>, then <span class="math-container">$P_{\theta}$</span> is the graph of <span class="math-container">$x=y^2$</span>, not the graph of a function. Similarly, if <span class="math-container">$\theta=\frac{\pi}{2}$</span>, <span class="math-container">$P_{\theta}$</span> is not the graph of a function. From now on, we assume <span class="math-container">$\theta\notin\{\frac{\pi}{2},\pi,\frac{3\pi}{2}\}$</span>.</p>
<p><span class="math-container">$P_\theta$</span> is the graph of a function if and only if every vertical line intersects its graph in at most one point.</p>
<p>Assume that <span class="math-container">$P_\theta$</span> is the graph of a function. Then the line <span class="math-container">$L$</span> of equation <span class="math-container">$x=0$</span> intersects <span class="math-container">$P_{\theta}$</span> in one point, the origin. Now, rotate <span class="math-container">$P_\theta$</span> and <span class="math-container">$L$</span> by <span class="math-container">$-\theta$</span>. The image of <span class="math-container">$L$</span> by this rotation is the line <span class="math-container">$L_\theta$</span> through the origin with inclination <span class="math-container">$\frac{\pi}{2}-\theta$</span>. <span class="math-container">$L_\theta$</span> should intersects the parabola <span class="math-container">$P$</span> in exactly one point.</p>
<p>Since <span class="math-container">$\theta\notin\{\frac{\pi}{2},\pi,\frac{3\pi}{2}\}$</span>, <span class="math-container">$L_\theta$</span> is neither horizontal nor vertical. It then intersects <span class="math-container">$P$</span> in two points, a contradiction.</p>
<p>As a conclusion, <span class="math-container">$P_\theta$</span> is the graph of a function if and only if <span class="math-container">$\theta=\pi$</span> (assuming <span class="math-container">$0<\theta<2\pi$</span>).</p>
|
4,492,566 | <blockquote>
<p>To which degree must I rotate a parabola for it to be no longer the graph of a function?</p>
</blockquote>
<p>I have no problem with narrowing the question down by only concerning the standard parabola: <span class="math-container">$$f(x)=x^2.$$</span></p>
<p>I am looking for a specific angle measure. One such measure must exist as the reflection of <span class="math-container">$f$</span> over the line <span class="math-container">$y=x$</span> is certainly no longer well-defined. I realize that preferentially I should ask the question on this site with a bit of work put into it but, alas, I have no intuition for where to start. I suppose I know immediately that it must be less than <span class="math-container">$45^\circ$</span> as such a rotation will cross the y-axis at <span class="math-container">$(0,0)$</span> and <span class="math-container">$(0,\sqrt{2})$</span>.</p>
<p>Any insight on how to proceed?</p>
| Brian Drake | 843,797 | <p>You write (emphasis added):</p>
<blockquote>
<p>I am looking for a specific angle measure. <strong>One</strong> such measure must exist as
the reflection of <span class="math-container">$f$</span> over the line <span class="math-container">$y=x$</span> is certainly no longer
well-defined.</p>
</blockquote>
<p>It sounds like you are looking for <span class="math-container">$\theta_\text{min}$</span>, where a rotation
through an angle <span class="math-container">$\theta \geq \theta_\text{min}$</span> produces a graph that is no
longer the graph of a function, but a rotation through
<span class="math-container">$\theta < \theta_\text{min}$</span> still produces the graph of a function.</p>
<p>But, in this case, the equality needs to be on the other side: a rotation
through <span class="math-container">$\theta > \theta_\text{min}$</span> produces a graph that is no longer the
graph of a function, but a rotation through <span class="math-container">$\theta \leq \theta_\text{min}$</span>
still produces the graph of a function.</p>
<p>In fact, as we will show, <span class="math-container">$\theta_\text{min} = 0$</span>: <em>any</em> rotation (except for a
half-revolution) produces a graph that is no longer the graph of a function.</p>
<p>You then write:</p>
<blockquote>
<p>I suppose I know immediately that it must be less than <span class="math-container">$45^\circ$</span> as such a
rotation will cross the y-axis at <span class="math-container">$(0,0)$</span> and <span class="math-container">$(0,1)$</span>.</p>
</blockquote>
<p>Actually, it crosses the <span class="math-container">$y$</span>-axis at <span class="math-container">$(0,0)$</span> and <span class="math-container">$(0,\sqrt{2})$</span>. But the
important point is that it crosses a vertical line at multiple points.</p>
<p>So now we want to rotate the parabola through some other angle <span class="math-container">$\theta$</span> and
check whether it crosses a vertical line at multiple points.
<a href="https://math.stackexchange.com/a/4492567/843797">5xum’s answer</a> shows how to
do this, and that is great. But for all the talk about it being “elementary”,
it requires transformation matrices and trigonometry. A <em>truly</em> elementary
solution would require no knowledge beyond lines, parabolas and basic algebra:
such a solution is presented below.</p>
<p>Instead of rotating the parabola, then checking if it crosses a vertical line
at multiple points, we can rotate the axes, then check if the parabola crosses
a line that is “vertical” relative to the rotated axes at multiple points. This
is the idea behind
<a href="https://math.stackexchange.com/a/4492951/843797">Taladris’s answer</a>.</p>
<p>As stated in that answer, it always does (again, excluding half-revolutions).
But how can we prove this?</p>
<p>This “rotated vertical” line could (relative to the original axes) be anything
other than a vertical line. That is, it could be a line with gradient <span class="math-container">$m$</span>, for
any value of <span class="math-container">$m$</span>. Now, for any such value, we can find a corresponding line
that crosses the parabola at multiple points. To see this, note that for
<span class="math-container">$m = 0$</span>, we can simply use any line above the origin, and for any other value
of <span class="math-container">$m$</span>, we can use the line through the origin, which crosses the parabola
there and at a second point given by:</p>
<p><span class="math-container">\begin{align}
mx &= x^2\\
x &= m\\
y &= m^2.
\end{align}</span></p>
|
2,880,566 | <p>In an optimization problem I finally get to the point where I have to solve</p>
<p>$$x +\sec(x)(\tan(x)\cos(2x)+\tan(x)-2\sin(2x)) =0$$</p>
<p>which obviously leads to</p>
<p>$$x=-\sec(x)(\tan(x)\cos(2x)+\tan(x)-2\sin(2x))$$</p>
<p>Nevertheless, this couldn't in any case help knowing the optimal size of the angle $x$. Wolfram Alpha can give me the solutions, but is not able to give me a step-by-step explanation of how to reach an approximate value for $x$. </p>
<p>Then, how do we solve manually such an equation?</p>
| Multigrid | 541,516 | <p>Hint: Use $\cos(2x) = 2\cos^2(x) - 1$ and $\sin(2x) = 2\sin(x) \cos(x)$</p>
<p>by the way do you get $x = 0$ and 2 other solutions (symmetric)?</p>
|
340,264 | <p>Given that</p>
<p>$L\{J_0(t)\}=1/(s^2+1)$</p>
<p>where $J_0(t)=\sum\limits^{∞}_{n=0}(−1)n(n!)2(t2)2n$,</p>
<p>find the Laplace transform of $tJ_0(t)$. </p>
<p>$L\{tJ_0(t)\}=$_<strong><em>_</em>__<em>_</em>__<em>_</em>___<em></strong>---</em>___?</p>
| Alex Youcis | 16,497 | <p>Now, I want to say something more than just the run-of-the-mill answer, if that's ok with you.</p>
<p>There is a general principle which says that a polynomial $F(T_1,\ldots,T_n)\in\mathbb{Z}[T_1,\ldots,T_n]$ has a zero in $\mathbb{Z}_p$ if and only if $F(T_1,\ldots,T_n)$ has a zero in $\mathbb{Z}/p^n\mathbb{Z}$ for all $n\geqslant 1$.</p>
<p>How does this apply here? Note that if $\alpha$ is a solution to $x^2=2$, then</p>
<p>$$v_p(\alpha)=\frac{1}{2}v_p(\alpha^2)=\frac{1}{2}v_p(2)\geqslant 0$$</p>
<p>so any solution to $x^2=2$ in $\mathbb{Q}_p$ must actually lie in $\mathbb{Z}_p$.</p>
<p>So, how does one go about proving this principle? For the same of simplicity, let's stick the to the univariate case, so we're dealing with a polynomial $F(T)$. The one direction is trivial, since, as I noted in your last question:</p>
<p>$$\mathbb{Z}_p=\left\{(x_n)\in\prod_n\mathbb{Z}/p^n\mathbb{Z}:\text{for all }n\leqslant m,\,\, x_n\equiv x_m\mod p^{n+1}\right\}$$</p>
<p>Thus, you see that </p>
<p>$$F((x_n))=(F(x_n))$$</p>
<p>and so clearly any solution to $F(x)=0$ in $\mathbb{Z}_p$ begets a solution to $F(x)\equiv 0 \mod p^n$ for all $n\geqslant 1$. </p>
<p>Conversely, if you start with a sequence of solutions $(x_n)$ in $\displaystyle \prod_n \mathbb{Z}/p^n\mathbb{Z}$, where $F(x_n)\equiv 0 \mod p^{n+1}$, one can take this sequence and "build" a solution sequence in $\mathbb{Z}_p$. Basically, the issue with our original sequence is that it may not be consistent. To fix this, note that there have to exist infinitely many $n$ for which all $x_n$ are all equivalent to the same thing modulo $p^{1}$. Try taking this subsequence, and noting that there must exist another subsequence all equivalent to the same thing modulo $p^2$. Continue in this way to create a solution sequence that actually lies in $\mathbb{Z}_p$. </p>
<p>So, how does this all factor back into our problem? Namely, the above would solve everything if the problem stated</p>
<blockquote>
<p>Show that $x^2=2$ has a solution in $\mathbb{Q}_p$ for $p\ne 2$ if and only if $x^2=2\mod p^n$ has a solution for all $n\geqslant 1$</p>
</blockquote>
<p>So, how do we pass from a solution modulo $p$ to a solution modulo higher powers of $p$? This is exactly the content of <i><a href="http://en.wikipedia.org/wiki/Hensel%27s_lemma" rel="nofollow">Hensel's Lemma</a></i> which states that if $f(x)\equiv 0\mod p^r$ has a solution $\alpha$ which satisfies $f'(r)\not\equiv 0\mod p^r$, then we can "lift" the solution $\alpha$ to $f(x)\equiv 0\mod p^r$ to solutions $f(x)\equiv 0\mod p^s$ for $s\geqslant r$. </p>
<p>Putting this all together gives us the problem.</p>
|
3,365,112 | <p>Pretty simple, for <span class="math-container">$a,b \in \mathbb R$</span>, show that <span class="math-container">$|a-b|<\frac{|b|}{2}$</span> implies <span class="math-container">$\frac{|b|}{2}<|a|$</span>. I can see this graphically on the number line, but I can't seem to show it algebraically.</p>
<p>I'm think it involves the triangle inequalities:</p>
<blockquote>
<p><span class="math-container">$$|x+y|\le|x|+|y|$$</span></p>
<p><span class="math-container">$$|x-y|\ge||x|-|y||.$$</span></p>
</blockquote>
| azif00 | 680,927 | <p>First, rewrite <span class="math-container">$|a-b|<\cfrac{|b|}{2}$</span> as <span class="math-container">$|b-a|<\cfrac{|b|}{2}$</span>. </p>
<p>Now, I'll use your second version of the triangle inequality (without the extra bars) and we obtain
<span class="math-container">$$|b|-|a| \leq |b-a|<\frac{|b|}{2}$$</span>
which means
<span class="math-container">$$|b|-|a| < \frac{|b|}{2}$$</span>
Finally, add <span class="math-container">$|a|$</span> and substract <span class="math-container">$|b|/2$</span> to both sides of the latter to get the desired result.</p>
|
3,365,112 | <p>Pretty simple, for <span class="math-container">$a,b \in \mathbb R$</span>, show that <span class="math-container">$|a-b|<\frac{|b|}{2}$</span> implies <span class="math-container">$\frac{|b|}{2}<|a|$</span>. I can see this graphically on the number line, but I can't seem to show it algebraically.</p>
<p>I'm think it involves the triangle inequalities:</p>
<blockquote>
<p><span class="math-container">$$|x+y|\le|x|+|y|$$</span></p>
<p><span class="math-container">$$|x-y|\ge||x|-|y||.$$</span></p>
</blockquote>
| Mohammad Riazi-Kermani | 514,496 | <p><span class="math-container">$$ |a-b|<\frac {|b|}{2} \implies -\frac {|b|}{2} < a-b<\frac {|b|}{2}$$</span></p>
<p><span class="math-container">$$\implies b-\frac {|b|}{2} <a< b+\frac {|b|}{2}$$</span></p>
<p>For <span class="math-container">$b>0$</span> we get <span class="math-container">$$\frac {b}{2} <a<\frac {3b}{2}\implies \frac {b}{2} <a $$</span>
for <span class="math-container">$b<0$</span> we get <span class="math-container">$$\frac {3b}{2} <a<\frac {b}{2}\implies -\frac {b}{2} <- a$$</span>
Thus in any case we have <span class="math-container">$$\frac {|b|}{2}< |a|$$</span></p>
|
2,713,391 | <p>I can find the answers to similar questions online, but what I'm trying to do is develop my own intuition so I can find the answers. I am quite sure I am wrong, so could you look over my reasoning?</p>
<p>If $X = (1,2,3,4,5,6,7,8)$,</p>
<blockquote>
<ol>
<li>How many strings over X of length 5?</li>
</ol>
</blockquote>
<p>Reasoning: Each character of the string is a choice of a selection from the string, so $8^5$.</p>
<blockquote>
<ol start="2">
<li>How many strings over X with length 5 don't contain $1$?</li>
</ol>
</blockquote>
<p>Reasoning: This is equivalent to strings of length 5 of an alphabet not containing $1$, so $7^5$.</p>
<blockquote>
<ol start="3">
<li>How many strings over X with length 5 contain $1$?</li>
</ol>
</blockquote>
<p>Reasoning: First get the strings for length 4 ($8^4$). Then select a position to insert a 1 ($5$). Compounded: $5\times 8^4$.</p>
<p>At this point I saw $8^5 \ne 7^4 + 5 \times 8^4$ and lost motivation. I am more sure 1 and 2 are correct than 3 so perhaps the simplest solution would be just $8^5-7^5$, but my logic for 3 "feels" sound. I am frustrated because I have never had these kinds of difficulties before. I look at solutions and though they too "feel" correct, I'd not have come up with them.</p>
<p>I apologize for the irregular question form, and understand if this isn't the site for this. </p>
| drhab | 75,923 | <p>What you call simplest solution is a correct one (and by far the most simple and elegant one).</p>
<p>If you still insist to go the other direction then your answer must be:$$\binom51\times7^4+\binom52\times7^3+\binom53\times7^2+\binom54\times7^1+\binom55\times7^0=15961$$</p>
<p>Here $\binom5{k}$ is the number of ways to choose $k$ ones among $5$ candidates. </p>
|
69,948 | <p>Has anyone ever created a "pairing function" (possibly non-injective)
with the property to be nondecreasing wrt to product of arguments, integers n>=2, m>=2. (We can also assume that n and m are bounded by an integer K, if useful) :</p>
<p>n m > n' m' => p(n,m) > p(n',m') </p>
<p>If yes what does it look like, does it have a name ? </p>
<p>-Luna</p>
| Joel David Hamkins | 1,946 | <p>Your second property is simply inconsistent with the nature
of a pairing function, since we want $p(n,m)=p(n',m')\iff
n=n'$ and $m=m'$. That is, a pairing function must be
one-to-one on pairs, but multiplication is not, since
$2\cdot 6=3\cdot 4$.</p>
<p>The first property, however, is easy to arrange as follows
(and there will be continuum many different such
functions). Let me work only with positive integers.
First, list all the possible product values in order.
For each such product $r$, observe that there are only
finitely many pairs with $nm=r$; list these pairs in any
desired order. Now, concatenate these lists of pairs, and
let $p(n,m)$ be the place of the pair $\langle n,m\rangle$
in your master list. This is a pairing function, since it
is injective on pairs, and it is monotone with respect to
products, since larger products appear later on the master
list.</p>
<p>Finally, note that it is not possible to achieve the property if you allow $0$, since in this case there are infinitely many pairs with product $0=n\cdot 0$, and they cannot all have pairing value before the the other pairs.</p>
<hr>
<p><b>Edit.</b> In your comments, you drop the one-to-one requirement, which makes this very far from what would ordinarily be called a pairing function. Nevertheless, the problem now admits the following rather silly solution in the positive integers: let $p(n,m)=nm$, which has both your stated properties. The "inverse" function is: let $F(r)=r$ and $G(r)=1$. That is, given the product $r$, we return the pair $\langle r,1\rangle$, which of course also has product $r\cdot 1=r$. In otherwords $p(F(r),G(r))=r$, which would seem to be the inverse requirements.</p>
|
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