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 |
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3,321,098 | <p>I'm wondering if there is also another easy way of solving question C.
In the book, they use <span class="math-container">$A = PDP^{-1}$</span> but the only method I know is by drawing the transformation with vectors <span class="math-container">$ e_1$</span> and <span class="math-container">$e_2$</span> and then look at the result and that would be the standard matrix. I tried this method and it did not work for me. I took into account that the vectors only turned so I kept the total length of the vector equal by using unit vectors of the transformation.</p>
<p><a href="https://i.stack.imgur.com/uAMs6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uAMs6.png" alt="enter image description here"></a></p>
| Anti | 523,511 | <p>I think I found an answer to my own question yesterday before I fell asleep ...</p>
<p>This is how the two urns look after 1 exchange of balls:</p>
<p><a href="https://i.stack.imgur.com/86akC.png" rel="nofollow noreferrer">The 2 urns after 1 exchange of balls</a></p>
<p>Now let's first define</p>
<ul>
<li><span class="math-container">$\alpha:=m_{12} \times n_1^{t_0}$</span></li>
<li><span class="math-container">$\beta:=m_{21} \times n_2^{t_0}$</span></li>
<li><span class="math-container">$\gamma:= \big( 1 - m_{12} \big) \times n_1^{t_0}$</span></li>
<li><span class="math-container">$\delta:= \big( 1 - m_{21} \big) \times n_2^{t_0}$</span></li>
</ul>
<p>If we draw 2 balls randomly from both urns at <span class="math-container">$t_1$</span> we would get balls from</p>
<ul>
<li><span class="math-container">$U_1^{t_0}$</span> and <span class="math-container">$U_2^{t_0}$</span> with a probability of <span class="math-container">$\frac {\gamma}{\beta + \gamma} \times \frac {\delta}{\alpha + \delta}$</span> which represents the measure <span class="math-container">$P_3^{t_0}$</span>.</li>
<li><span class="math-container">$U_2^{t_0}$</span> and <span class="math-container">$U_2^{t_0}$</span> with a probability of <span class="math-container">$\frac {\beta}{\beta + \gamma} \times \frac {\delta}{\alpha + \delta}$</span> which represents the measure <span class="math-container">$P_2^{t_0}$</span>.</li>
<li><span class="math-container">$U_1^{t_0}$</span> and <span class="math-container">$U_1^{t_0}$</span> with a probability of <span class="math-container">$\frac {\gamma}{\beta + \gamma} \times \frac {\alpha}{\alpha + \delta}$</span> which represents the measure <span class="math-container">$P_1^{t_0}$</span>.</li>
<li><span class="math-container">$U_2^{t_0}$</span> and <span class="math-container">$U_1^{t_0}$</span> with a probability of <span class="math-container">$\frac {\beta}{\beta + \gamma} \times \frac {\alpha}{\alpha + \delta}$</span> which represents the measure <span class="math-container">$P_3^{t_0}$</span>.</li>
</ul>
<p>Thus:</p>
<p><span class="math-container">$$P_3^{t_1} = \frac {\gamma}{\beta + \gamma} \times \frac {\delta}{\alpha + \delta} \times P_3^{t_0} + \\
+ \frac {\beta}{\beta + \gamma} \times \frac {\delta}{\alpha + \delta} \times P_2^{t_0} + \\
+ \frac {\gamma}{\beta + \gamma} \times \frac {\alpha}{\alpha + \delta} \times P_1^{t_0} + \\
+ \frac {\beta}{\beta + \gamma} \times \frac {\alpha}{\alpha + \delta} \times P_3^{t_0}$$</span></p>
<p>Does it sound logical?</p>
|
72,794 | <p><img src="https://i.stack.imgur.com/jabWi.png" alt="enter image description here"></p>
<p>I am looking for something like this:</p>
<pre><code>SetOptions[EvaluationNotebook[], "PaperSize" -> "A3"]
</code></pre>
<p>The options are listed, but where do I find the arguments? Say for A3 paper, do I use <code>A3</code> or "A3" or "a3paper"?</p>
<p>and it looks like it takes two inputs? How?</p>
<p>I can change them everytime from the drop down menu, but it's not very efficient.</p>
| Jinxed | 24,763 | <p>This should get you started for A3:</p>
<pre><code>SetOptions[EvaluationNotebook[],
PrintingOptions -> {"PaperSize" ->
Round[72*
QuantityMagnitude@
UnitConvert[Quantity[#, "Millimeters"] & /@ {2*210, 297},
"Inches"]]}]
</code></pre>
<p><strong>Background</strong></p>
<p><em>Mathematica</em> assumes 72pt per inch for the <code>PaperSize</code>-option. The example above just uses the <code>Quantity</code> functionality for conversion from the DIN-dimensions (I used the equivalent of two A4 sheets for reference).</p>
<p>Be aware, though, that the <code>PrintingOptions</code> feature might change in the future (see the online reference for that warning).</p>
|
3,849,649 | <p>Given <span class="math-container">$A^2=A$</span>, <span class="math-container">$2A−B−AB=I$</span>, prove that <span class="math-container">$A−B$</span> is invertible.</p>
<hr />
<p>I have got <span class="math-container">$(I+A)(A-B)$</span>, what is the next step?</p>
<p>Thanks a lot.</p>
| egreg | 62,967 | <p>It's an important fact that</p>
<blockquote>
<p>a square matrix that has a one-sided inverse is invertible</p>
</blockquote>
<p>So if you arrive at <span class="math-container">$(A+I)(A-B)=I$</span>, you're done.</p>
<p>How did you get there? From <span class="math-container">$2A-B-AB=I$</span>, you can write
<span class="math-container">$$
A^2+A-B-AB=I
$$</span>
and therefore <span class="math-container">$(A+I)A-(A+I)B=I$</span>, which yields
<span class="math-container">$$
(A+I)(A-B)=I
$$</span>
With the previously mentioned fact you're done.</p>
<p>Why is the mentioned fact true? If you consider an <span class="math-container">$n\times n$</span> matrix <span class="math-container">$M$</span> to define a linear map <span class="math-container">$\mathbb{R}^n\to\mathbb{R}^n$</span> (or whichever base field you are using), then the linear map induced by <span class="math-container">$A-B$</span> has a left inverse, which implies it is injective. But as the space is finite dimensional, the map is also surjective and it has a unique two-sided inverse.</p>
|
246,827 | <p>Let $A$ and $B$ be two positive definite $n \times n$ matrices. It is, of course, not true that $AB+BA$ is necessarily positive definite. </p>
<p>Consider, though, the results of the following numerical experiment. I generated $A$ by letting its eigenvalues be random in $[0,1]$, and selecting its eigenvectors by generating a random matrix of standard Gaussians and applying Gram-Schmidt to it. The matrix $B$ is generated in the same way.</p>
<p>I did this 1000 times and checked what proportion of times the matrix $AB+BA$ has at least one negative eigenvalue [1]. Here are the results for different dimensions $n$:</p>
<ul>
<li>$n=2, ~~~~94.8 \%$</li>
<li>$n=3, ~~~~89.4 \%$. </li>
<li>$n=4, ~~~~78 \%$. </li>
<li>$n=5, ~~~~72.7 \%$.</li>
<li>$n=10, ~~~40.3 \%$.</li>
<li>$n=15, ~~~20.1 \%$. </li>
<li>$n=20, ~~~11.4 \%$. </li>
<li>$n=50, ~~~0.3\%$. </li>
<li>$n=100, ~~0 \%$. </li>
</ul>
<p>This suggests that, as a function of $n$, examples with $AB+BA$ not psd tend to get rarer and rarer. Is it possible to give a proof of this? </p>
<p>It may be more natural to consider a different random model of randomly generated psd matrices; I only generated them in the way I described above because it seemed easiest. </p>
<p>[1] Actually, I checked if there is an eigenvalue less then $-1 \cdot 10^{-5}$ in MATLAB to account for rounding error. </p>
| Guillaume Aubrun | 908 | <p>It is in theory possible to use free probability to describe the limit eigenvalue distribution (as the dimension tends to <span class="math-container">$+\infty$</span>) of the anticommutator <span class="math-container">$AB+BA$</span> when <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are independent random matrices which are unitarily invariant (which is the case for the description used in the OP). The same is true actually for any self-adjoint polynomial in <span class="math-container">$A$</span>, <span class="math-container">$B$</span>.</p>
<p>The drawback of this method it that the limit distribution will be obtained in an indirect way. If one does this for the uniform measure on <span class="math-container">$[0,1]$</span> as asked in the OP, it may be hard to decide from this approch whether the limit distribution is supported on <span class="math-container">$\mathbf{R}^+$</span> or not.</p>
<p>However, such computations are possible in simple cases. Propositon 6.11 from
<em>Fevrier, Maxime; Mastnak, Mitja; Nica, Alexandru; Szpojankowski, Kamil</em>, <a href="http://dx.doi.org/10.1090/tran/8122" rel="nofollow noreferrer"><strong>Using Boolean cumulants to study multiplication and anti-commutators of free random variables</strong></a> says the following: if <span class="math-container">$E$</span> and <span class="math-container">$F$</span> are subspaces of dimension <span class="math-container">$n/2$</span> which are chosen independently according to the <span class="math-container">$O(n)$</span>-invariant measure on the Grassmann manifold, and <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are the corresponding orthogonal projections, then as <span class="math-container">$n \to \infty$</span> the eigenvalue distribution of <span class="math-container">$AB+BA$</span> is described by an explicit density which is supported on <span class="math-container">$[-1/4,2]$</span> (see also Figure 2 in the same paper). In particular, the probability that <span class="math-container">$AB+BA$</span> is positive semidefinite tends to <span class="math-container">$0$</span> as <span class="math-container">$n \to \infty$</span>. For this toy model, this gives a rigorous proof of the fact that the anticommutator of positive matrices is typically not positive. (This is probably overkill, ane can presumably analyze everything in terms of principal angles between the subspaces <span class="math-container">$E$</span> and <span class="math-container">$F$</span>.)</p>
|
97,579 | <p>Is there some simple upper bound on $||(B^{-1}+A^{-1})^{-1}||$, where $A,B$ are $n \times n$ symmetric matrices?</p>
| Suvrit | 8,430 | <p>To expand on my first comment, if $A, B > 0$ are symmetric positive definite matrices. Then, it is known that</p>
<p>$$\left(\frac{A^{-1}+B^{-1}}{2}\right)^{-1} \le A\sharp B \le \frac{A+B}{2},$$
where the inequalities are in the Löwner partial order, and $A\sharp B := A^{1/2}(A^{1/2}BA^{1/2})^{-1/2}A^{1/2}$ denotes the matrix geometric mean. </p>
<p>These operator inequalities are of course, <strong>stronger</strong> than corresponding norm inequalities (based on unitarily invariant norms).</p>
<p>For the case where you don't have positive matrices, I think the conjecture mentioned in my second argument can be expanded into a proof --- maybe if I get time, I'll try to expand that.</p>
|
3,639,122 | <blockquote>
<p>Let <span class="math-container">$X:\varOmega \to \mathbb{R}$</span> be a random variable. Random variable <span class="math-container">$X$</span> is <em>degenerate</em> if
for some <span class="math-container">$c\in \mathbb{R}$</span> we have <span class="math-container">$\mathrm{P}(X=c)=1.$</span> </p>
<p>True or false? <span class="math-container">$X$</span> is a non degenerate random variable iff for some <span class="math-container">$a\in \mathbb{R}$</span> we have <span class="math-container">$\mathrm{P}(X<a)\in (0,1).$</span></p>
</blockquote>
<p><strong>Attempt</strong>.
<em>Converse</em>: true. Suppose that <span class="math-container">$X$</span> is a.e. equal to a constant <span class="math-container">$c$</span>. If <span class="math-container">$c<a$</span> then <span class="math-container">$1=\mathrm{P}(X=c)\leqslant \mathrm{P}(X<a)<1$</span>, contradiction and if <span class="math-container">$c\geqslant a$</span> then <span class="math-container">$0=\mathrm{P}(X\neq c)\geqslant \mathrm{P}(X<a)>0$</span>, contradiction.</p>
<p>Regarding the other direction I believe the answer is yes (non degenerate: for every <span class="math-container">$c\in \mathbb{R}$</span> we have <span class="math-container">$\mathrm{P}(X=c)<1$</span>), but I haven't been able to reach an <span class="math-container">$a$</span>, as wanted.</p>
<p>Thanks for the help. </p>
| drhab | 75,923 | <p>It might be more handsome to prove the equivalence of negations:<span class="math-container">$$X\text{ is degenerated}\iff P(X<a)\in\{0,1\}\text{ for every }a\in\mathbb R$$</span></p>
<hr>
<p><span class="math-container">$\implies$</span> </p>
<p>Let it be that <span class="math-container">$P(X=c)=1$</span>. Then evidently <span class="math-container">$P(X<a)=1$</span> if <span class="math-container">$c<a$</span> and <span class="math-container">$P(X<a)=0$</span> otherwise.</p>
<p><span class="math-container">$\impliedby$</span> </p>
<p>Let it be that <span class="math-container">$P(X<a)\in\{0,1\}$</span> for every <span class="math-container">$a\in\mathbb R$</span>. </p>
<p>It cannot be that <span class="math-container">$P(X<a)=0$</span> for every <span class="math-container">$a\in\mathbb R$</span> because this leads to <span class="math-container">$P(X\in\mathbb R)=0$</span> which is absurd. So we do have <span class="math-container">$P(X<a)=1$</span> for some <span class="math-container">$a$</span>. </p>
<p>It cannot be that <span class="math-container">$P(X<a)=1$</span> for every <span class="math-container">$a\in\mathbb R$</span> because this leads to <span class="math-container">$P(X\in\varnothing)=1$</span> which is absurd. So we do have <span class="math-container">$P(X<b)=0$</span> for some <span class="math-container">$b$</span>. </p>
<p>Then the set <span class="math-container">$\{a\in\mathbb R\mid P(X<a)=1\}$</span> is not empty and has a lower bound. Consequently it has an infinum <span class="math-container">$c$</span>. It can be proved now that <span class="math-container">$P(X=c)=1$</span>.</p>
<p>Proving that is not difficult.</p>
|
3,891,168 | <p>Let f be the function on <span class="math-container">$[-\pi,\pi]$</span> given by <span class="math-container">$f(0)=9$</span> and <span class="math-container">$f\left( x \right) = \frac{{\sin \left( {\frac{{9x}}{2}} \right)}}{{\sin \left( {\frac{x}{2}} \right)}}$</span> for <span class="math-container">$x\ne 0$</span>. Find the value of <span class="math-container">$\frac{2}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)dx} $</span></p>
<p>My approach is as follow. The function is continuous at <span class="math-container">$x=0$</span>. Hence it is contionuos for <span class="math-container">$[-\pi,\pi]$</span></p>
<p><span class="math-container">$f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\frac{{9x}}{2}} \right)}}{{\left( {\frac{{9x}}{2}} \right)}} \times 9 \times \frac{{\left( {\frac{x}{2}} \right)}}{{\sin \left( {\frac{x}{2}} \right)}} = 9$</span></p>
<p><span class="math-container">$\frac{x}{2}=\theta$</span></p>
<p><span class="math-container">$\frac{2}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)dx} $</span></p>
<p><span class="math-container">$\int\limits_{ - \pi }^\pi {\frac{{\sin \left( {\frac{{9x}}{2}} \right)}}{{\left( {\frac{{9x}}{2}} \right)}}dx} \Rightarrow 2\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{\sin \left( {9\theta } \right)}}{{\sin \left( \theta \right)}}d\theta } \Rightarrow 4\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin \left( {9\theta } \right)}}{{\sin \left( \theta \right)}}d\theta } $</span></p>
<p><span class="math-container">$ \Rightarrow 4\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin \left( {8\theta + \theta } \right)}}{{\sin \left( \theta \right)}}d\theta } \Rightarrow 4\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin \left( {8\theta } \right)\cos \left( \theta \right) + \cos \left( {8\theta } \right)\sin \left( \theta \right)}}{{\sin \left( \theta \right)}}d\theta } $</span></p>
<p><span class="math-container">$ \Rightarrow 4\int\limits_0^{\frac{\pi }{2}} {\left( {\frac{{\sin \left( {8\theta } \right)\cos \left( \theta \right)}}{{\sin \left( \theta \right)}} + \cos \left( {8\theta } \right)} \right)d\theta } $</span></p>
<p><span class="math-container">$\int\limits_0^{\frac{\pi }{2}} {\cos \left( {8\theta } \right)d\theta } = 0$</span></p>
<p>How so I proceed form here as expansion of <span class="math-container">$sin(8\theta)$</span> is cumbersome.</p>
<p>Is their any formula of converting <span class="math-container">$sin(n\theta)$</span> in terms of <span class="math-container">$sin(\theta)$</span></p>
| Christian Blatter | 1,303 | <p>Via <span class="math-container">$$\cos t={e^{it}+e^{-it}\over2},\qquad\sin t={e^{it}-e^{-it}\over2i}$$</span>
you can easily prove that
<span class="math-container">$${\sin(9t)\over\sin t}=1+2\bigl(\cos(2t)+\cos(4t)+\cos(6t)+\cos(8t)\bigr)\ .$$</span></p>
|
649,073 | <p>Show that there exists a continuous function $f: [-1, 1] \rightarrow \mathbb{R}$ such</p>
<p>$f(0) = 1$ and</p>
<p>$f(x) = \frac{2-x^2}{2} \cdot f(\frac{x^2}{2-x^2})$ </p>
<p>$\forall x \in [-1, 1]$</p>
<p>I tried putting in $x = 1$ and $x = -1$ in the second condition to find that $f(1) = f(-1) = 0$.
I also took the derivative of the second equation to find that:</p>
<p>$f'(x) = x (f'(\frac{x^2}{2-x^2})\frac{2}{2-x^2}-f(\frac{x^2}{2-x^2}))$</p>
<p>This gives me</p>
<p>$f'(0) = f'(1) = f'(-1) = 0$ but now I'm stuck. Anybody see a way?</p>
| Andrew Senger | 49,323 | <p>I recall solving this problem before... was it on a Putnam? Sadly, I don't remember what leap of intuition led me to the answer.</p>
<p>One may take $f(x) = \sqrt{1-x^2}$. It is clear that $f$ is continuous on $[-1,1]$ and that $f(0) = 1$, and one sees that</p>
<p>$$\frac{2-x^2}{2} \cdot \sqrt{1-\frac{x^4}{(2-x^2)^2}} = \frac{2-x^2}{2} \cdot \sqrt{\frac{x^4-4x^2+4-x^4}{x^4-4x^2+4}} \\
= \frac{2-x^2}{2} \cdot \sqrt{\frac{4-4x^2}{(2-x^2)^2}} = \sqrt{1-x^2},$$</p>
<p>so that $f(x) = \frac{2-x^2}{2} f(\frac{x^2}{2-x^2})$.</p>
|
10,992 | <p>Alright, I'm trying to figure out how to calculate a critical value using t-distribution in Microsoft Excel... ex. a one-tailed area of 0.05 with 39 degrees of freedom: t=1.685</p>
<p>I know the answer, but how do I get this? I've tried TDIST() TINV() and TTEST() but they all give me different answers. This web calculator: <a href="http://www.danielsoper.com/statcalc/calc10.aspx" rel="nofollow">http://www.danielsoper.com/statcalc/calc10.aspx</a> always gives me what I'm looking for but I cannot manage to get Excel to do the same.</p>
<p>Any help would be greatly appreciated!</p>
| user130182 | 130,182 | <p>As Mike suggested either you can input your alpha level (nominal alpha) as .10 (2*.05) to get the one tailed p value or you can divide the given probability (actual alpha) by two. Both approaches do the same thing. Remember that taking all other things constant, the probability of recting a true null is easier for a one tailed test compared to the two tailed test. One tailed test yields a smaller p value compared to the two tailed test. That is why you should dived the calculated p (actual alpha) by 2. Same goes for SPSS. Hope it helps :) Gulshah</p>
|
2,293,934 | <p>I'm sorry for such a vague question, but I couldn't make it more plain. The other day i was working with a solution to a problem and out of curiosity i decomposed the number into its prime factors and noticed an interesting pattern. The number $216$ distribute in $2,3,2,3,2,3$ and when i grouped it, i had three pairs of 6's which then i could express as $6^3$. What other information could i gather by decomposing a number into its factors or prime factors? What does it say about components of the number?</p>
<p>P.S I'm interested in further readings and external links, too.</p>
| scott | 330,966 | <p>To me, it's more interesting when a number has no prime factors, i.e., it is a prime number. However, being able to break a number into its prime factors can assist with mental calculations. To use your example, </p>
<blockquote>
<p>$72\times3=(9\cdot8)\cdot3=2^3\cdot3^2\cdot3=2^3\cdot3^3=6^3=216$</p>
</blockquote>
<p>Instead of actually multiplying 72 and 3, I broke 72 into its prime factors and found the result will be a perfect cube. Assuming you know the perfect cubes up to $10^3$, this would be quicker than actually carrying out the multiplication.</p>
|
11,651 | <p>Mathematics can come across as a sterile, dead subject - a catalogue of techniques long-ago decided, and forever relearned by each successive generation of students.</p>
<p>This is <em>approximately</em> true for elementary and secondary mathematics, and for the standard progression of undergraduate courses (eg, Calculus 1,2,3, discrete math / combinatorics, ODE + Vector Calc, Analysis and Algebra).</p>
<p>Of course, the subject is alive and kicking, with many thousands of active researchers learning, creating, refining, and publishing every day. But the vast majority of fresh research requires considerable expertise to understand, and are therefore inaccessible to younger students of the subject.</p>
<p><strong>What, then, are some recent results that are interesting and accessible to students at (say) a secondary school level, which might exemplify that the subject remains active?</strong></p>
<p>A couple of examples that come to mind (which could be fleshed out as answers) are the recent progress against the Twin Prime Conjecture, and the surprising observation that primes ending in $X$ 'favor' being followed by a prime ending in $Y$, for various $(X,Y)$ pairs.</p>
<p>Where it's appropriate, please include links to any media treatment of the result.</p>
<p>Let's roughly define 'recent' as being within the lifetime of some collection of students.</p>
| Joseph O'Rourke | 511 | <p>Perhaps: The discovery <strike>a year ago</strike>
in 2015 of a new tiling of the plane by
a convex polygonal tile, found by
Mann, McLoud, and Von Derau (the latter of whom was an undergraduate at
the time of the discovery):</p>
<hr />
<img src="https://i.stack.imgur.com/eMead.png" />
<hr />
<p>Here is a nice article on the discovery in
<a href="https://www.theguardian.com/science/alexs-adventures-in-numberland/2015/aug/10/attack-on-the-pentagon-results-in-discovery-of-new-mathematical-tile" rel="nofollow noreferrer">The Guardian</a>,
by Alex Bellos.
As Alex says, the problem has been studied for <span class="math-container">$100$</span> years now,
since Reinhardt in 1918.
It remains unknown if the current list of <span class="math-container">$15$</span> such tilings is complete. (But see update below.)</p>
<p>Not only is there no algorithm to determine if a candidate tile can indeed
tile the plane, <a href="https://cstheory.stackexchange.com/q/32538/337">it is not even known</a> that the problem is decidable.</p>
<blockquote>
<p>Mann, Casey, Jennifer McLoud-Mann, and David Von Derau. "Convex pentagons that admit <span class="math-container">$ i $</span>-block transitive tilings."
<a href="https://arxiv.org/abs/1510.01186" rel="nofollow noreferrer">arXiv abstract</a> (2015).</p>
</blockquote>
<p><strong>Update</strong>.
Rao, Michaël (2017), Exhaustive search of convex pentagons which tile the plane, <a href="https://arxiv.org/abs/1708.00274" rel="nofollow noreferrer">arXiv:1708.00274</a>.</p>
<blockquote>
<p>We present an exhaustive search of all families of convex pentagons
which tile the plane. This research shows that there are no more than
the already 15 known families. In particular, this implies that there is
no convex polygon which allows only non-periodic tilings.</p>
</blockquote>
|
11,651 | <p>Mathematics can come across as a sterile, dead subject - a catalogue of techniques long-ago decided, and forever relearned by each successive generation of students.</p>
<p>This is <em>approximately</em> true for elementary and secondary mathematics, and for the standard progression of undergraduate courses (eg, Calculus 1,2,3, discrete math / combinatorics, ODE + Vector Calc, Analysis and Algebra).</p>
<p>Of course, the subject is alive and kicking, with many thousands of active researchers learning, creating, refining, and publishing every day. But the vast majority of fresh research requires considerable expertise to understand, and are therefore inaccessible to younger students of the subject.</p>
<p><strong>What, then, are some recent results that are interesting and accessible to students at (say) a secondary school level, which might exemplify that the subject remains active?</strong></p>
<p>A couple of examples that come to mind (which could be fleshed out as answers) are the recent progress against the Twin Prime Conjecture, and the surprising observation that primes ending in $X$ 'favor' being followed by a prime ending in $Y$, for various $(X,Y)$ pairs.</p>
<p>Where it's appropriate, please include links to any media treatment of the result.</p>
<p>Let's roughly define 'recent' as being within the lifetime of some collection of students.</p>
| Sebastian Schoennenbeck | 7,448 | <p>Not a single answer but rather a resource:</p>
<p>The snapshots of modern mathematics from the mathematical research institute Oberwolfach (<a href="http://www.mfo.de/math-in-public/snapshots/" rel="noreferrer">http://www.mfo.de/math-in-public/snapshots/</a>) aim to provide pretty much exactly what you ask for. Research mathematicians present some "recent developments" in their field in a way that is understandable for high school students/ math teachers/ undergraduates (the actual level varies quite a bit between the various snapshots).</p>
<p>The snapshots are freely available on the homepage of the institute and cover quite a lot of topics.</p>
|
2,879,041 | <blockquote>
<p>Let $f: [a,b] \to \mathbb R$ be continuous on $[a,b]$ and differentiable on $(a,b)$. If $f(a) = 0$ and $|f'(x)|\le k|f(x)|$ for some $k$, then $f(x)$ is zero on $[a,b]$. </p>
</blockquote>
<p>I tried proving it using Legrange's Mean Value Theorem but couldn't get it.</p>
<p>$f(x)$ is differentiable on $(a,b)$. $f(x)$ is continuous on $[a,b]$. Let $x$ belong to $[a,b]$ s.t $a < x$. Consider the interval $[a,x]$. By Legrange's Mean Value Theorem, </p>
<p>$$\frac{f(x)-f(a)}{x-a} = f'(t) ; \ \ a \le t \le x.$$</p>
<p>Since $f(a)=0$, </p>
<p>$$f(x)=(x-a)f'(t)$$</p>
<p>$$|f(x)| \le (x-a)k|f(t)|.$$</p>
<p>After this, I was thinking of proceeding with the inequality by putting $f(a)$ in the place of $f(t)$.</p>
| xbh | 514,490 | <p>As I commented, we first prove the conclusion for a smaller interval. </p>
<p><strong>Proof</strong>. $\blacktriangleleft$ We may assume $b-a > 1/(2k)$ [otherwise we proceed on $[a,b]$, and the following argument is still applicable with slight modification] First consider the interval $[a, a + 1/(2k)] \subset [a,b]$. By Lagrange’s MVT,
$$
\forall x \in \left[a, a + \frac {b-a}{2k}\right],\;|f(x)| = |f(x) - f(a)| = |f’(c_1)||x-a| \leqslant k |f(c_1)| \cdot \frac {1}{2k} = \frac 12 |f(c_1)| \quad [c_1\in (a, x)].
$$
Now for $c_1$, the Lagrange’s MVT gives that
$$
|f(c_1)| = |f(c_1) - f(a)| = |f’(c_2)| |c_1 - a| \leqslant k |f(c_2)| \cdot \frac {1} {2k} = \frac 12 |f(c_2)| \quad [c_2 \in (a,c_1)].
$$
Repeatedly we have a sequence $(c_n)_1^\infty$ s.t. $0 < c_n < c_{n-1} < x$ and
$$
|f(c_{n-1})| \leqslant \frac 12 |f(c_{n})|\quad [\forall n \in \mathbb N^*],
$$
hence
$$
\forall n, |f(x)| \leqslant \frac 12 |f(c_1)| \leqslant \frac 14 |f(c_2)| \leqslant \cdots \leqslant \frac 1{2^n} |f(c_n)|.
$$
Since $f$ is continuous on $[a, b]$, $|f|$ is also continuous and attain its maximum on $[a, b]$. Let the maximum be $M$, then
$$
\forall n \in \mathbb N^*, |f(x)| \leqslant \frac M {2^n}.
$$
Let $n \to \infty$, then $|f| \equiv 0$ on $[a, a+1/(2k)]$. </p>
<p>Now for the whole interval $[a,b]$, divide it into $m = \lceil (b-a)/2k \rceil$ subintervals $[a + (j-1)/(2k), a+j/(2k)] [j \leqslant m-1]$ and $[a + (m-1)/2k, b]$, then by induction we could successively show that $f$ vanishes on each subinterval [because $f$ vanishes at the left endpoint of each subinterval]. Hence $f \equiv 0$ on $[a,b]$. $\blacktriangleright$</p>
|
670,345 | <p>Here's what I came up with:</p>
<p>For this problem, I'm required to use a comparison test to determine if $\Sigma1/ln(n)^n$ converges or diverges. By intuition, I am thinking that $\Sigma1/ln(n)^n$ converges. </p>
<p>To prove that it converges by the Direct Comparison Test, I would have to find a convergent series of a sequence that is greater than $1/ln(n)^n$ for $n > \epsilon$. I find that the sequences that I know do not satisfy this condition.</p>
<p>$x^n$, $1/n^n$, $1/n^p$, $1/n!$ each converge, but are not greater than $1/ln(n)^n$.</p>
<p>So then I move on to prove convergence by the Limit Comparison Test. Each of the infinite series of the sequences that I know converge all fail the conditions required by the Limit Comparison Test, since the limit of the ratio of the two ends up being $0$ or $infinity$, and now I am stuck.</p>
<p>What sequence I should compare $1/ln(n)^n$ to?</p>
| Mark Fantini | 88,052 | <p>Since you restricted your question, I'm changing my answer.</p>
<p>No. Consider $a = (1,0,\ldots,0), b = (0,1,\ldots,0)$ and $c=(0,-1,\ldots,0)$. Then $a \cdot b = a \cdot c = 0$ but $b \neq c$.</p>
<p>This is for <strong>given</strong> $a$. If this holds <strong>for all</strong> $a$ then it is true.</p>
|
1,700,457 | <p>Assume $H \leq G$ and $K \leq G$ and assume $aH=bK$ for some $a,b \in G$. Prove that $H=K$</p>
<p>would someone give me a hint. I know that definition of being a subgroup. I know that the element look like this $ah_1=bk_1$ for all $h_1 \in H$ and for all $k_1 \in K$. I do not have an idea from where should I get started. </p>
| eyedropper | 188,031 | <p>Multiply $aH=bK$ by $b^{-1}$ to get</p>
<p>$$b^{-1}aH=K$$</p>
<p>Since there is a $g\in G$ such that $g=b^{-1}a$ we get $gH=K$. This means that for some $h\in H$ and some $k\in K$ we get the equality $gh=k$. $H$ is a subgroup which means that $e\in H$. Set $h=e$ to get $g=k$ and therefore $g\in K$. $K$ is a subgroup which means that also $g^{-1}\in K$. Multiply $gH=K$ by $g^{-1}$ to get
$$H=g^{-1}K=K$$</p>
|
2,285,752 | <p>Does $\int_{-\infty}^\infty \sin(t) \,dt $ converge or diverge? How would I prove it? </p>
<p>Should I use 'principle value' to do:
$$\lim_{a \to \infty} \int_{-a}^a \sin(t)\,dt$$</p>
| Eff | 112,061 | <blockquote>
<p>Does $\int_{-\infty}^\infty \sin (t)\,\mathrm{d}t$ converge?</p>
</blockquote>
<p>No, for the limit to exist with limit $L$, then for every pair of sequences $a_n\to-\infty$ and $b_n\to\infty$, we would have that</p>
<p>$$\lim\limits_{n\to\infty} \int_{a_n}^{b_n} \sin (t)\,\mathrm{d}t = L$$
which is obviously not the case.</p>
<p>The <a href="https://en.wikipedia.org/wiki/Cauchy_principal_value" rel="noreferrer">Cauchy Principal Value</a> is different to usual convergence and this value <em>does</em> exist, it is
$$\lim\limits_{a\to\infty}\int_{-a}^a\sin (t) \,\mathrm{d} t = 0.$$</p>
|
2,371,668 | <p>For a finite group $G$ and a subgroup $H,$ Lagrange's theorem says that $|G|=|G:H||H|,$ where $|G:H|$ is the number of cosets of $H$ in $G.$<br>
My question is for any subgroup $H$ can we find another subgroup with order $|G:H|$?</p>
| DeepSea | 101,504 | <p>Alternatively using AM-GM inequality: $\sqrt{a+b} = \sqrt{1\cdot (a+b)} \le \dfrac{1+(a+b)}{2}\implies \sqrt{a+b} - \dfrac{a+b}{2} \le \dfrac{1}{2}$. Repeat this application of the AM-GM inequality for the $2$ pairs $(b,c)$, and $(c,a)$ and add up the sought inequality follows.</p>
|
640,680 | <p>Could somebody help me to solve these two unrelated questions?
I have to prove or disprove them. The first one is which I have to answer. The second one is just for me, to understand the topic better.</p>
<p>Prove or disprove the following statements:</p>
<ol>
<li><p>If the sequence $(a_n)_{n\in \mathbb{N}}$ is bounded/restricted then the sequence $(a_n)_{n\in \mathbb{N}}$ convergent</p></li>
<li><p>If the sequence $(a_n)_{n\in \mathbb{N}}$ is convergent then the sequence $(a_n)_{n\in \mathbb{N}}$ bounded/restricted</p></li>
</ol>
| mathemagician | 49,176 | <ol>
<li><p>is false. Counter example $a_n=(-1)^n$ the sequence alternates between 1 and -1 so is bounded, but it is not convergent. </p></li>
<li><p>is true. If the sequence converges, say to a point $L\in\mathbb{R}^d$ you have that the infinitely many terms of the tail of the sequence are in an epsilon-ball around $L$. Since the first finite terms of the sequence cannot consist of an unbounded set, they are bounded by some ball of radius $M_1$ centered at the origin. You also have that there exists some ball of radius $M_2$ centered at the origin that contains the epsilon-ball containing the tail of the sequence. Hence whichever radius is bigger ($M_1$ or $M_2$) will be a ball that contains all your sequence, therefore sequence is bounded.</p></li>
</ol>
|
2,392,411 | <p>I am in Adv. Algebra 2 and I have a question. Firstly, would like to say I haven't taken algebra in a year due to geometry (stupid order they do but oh well) and I have a question understanding this: $(x+5)^{0}$. That would be $x^{0} + 5^{0}$ which then, wouldn't that be $1 + 1$ since anything that has a power of $0 = 1$? Maybe I misunderstood but that's what I got.</p>
| md2perpe | 168,433 | <p>Given $\epsilon>0$ you want to find $\delta>0$ such that $|(1-\frac2x)-(-1)|<\epsilon$ when $|x-1|<\delta$.</p>
<p>Now, $|(1-\frac2x)-(-1)| = |2 - \frac2x| = 2|1-\frac1x| = \frac{2}{|x|} |x-1| < \frac{2}{|x|} \delta$ when $|x-1|<\delta.$ But we need to limit $\frac{2}{|x|}.$ To do this we make sure that $|x-1|<\frac12$ because then $x>\frac12$ so $\frac{2}{|x|} < \frac{2}{1/2} = 4.$</p>
<p>Thus, if we take $\delta < \min(\frac12, \frac14\epsilon)$ we get
$|(1-\frac2x)-(-1)| < 4 \cdot \frac14\epsilon = \epsilon.$</p>
|
1,829,086 | <p>So far, I've tried out to reformulate: $$\int{\frac{1}{\cos(x)}}dx$$
to: $$\int{\frac{\sin(x)}{\cos(x)\sin(x)}}dx$$</p>
<p>which is basically: $$\int{\frac{\tan(x)}{\sin(x)}}dx$$
But I'm not sure if this is the right way to go, or if I try something else.</p>
<p>Any tips or methods would be very helpful.</p>
| egreg | 62,967 | <p>You can certainly do it with the substitution $t=\tan(x/2)$: you have
$$
\int\frac{1}{\cos x}\,dx=
\int\frac{1+t^2}{1-t^2}\frac{2}{1+t^2}\,dt=
\int\left(\frac{1}{1+t}+\frac{1}{1-t}\right)\,dt=
\log\left|\frac{1+t}{1-t}\right|+c
$$
The back substitution gives
$$
\log\left|
\frac{\cos\frac{x}{2}+\sin\frac{x}{2}}
{\cos\frac{x}{2}-\sin\frac{x}{2}}
\right|+c=\log\left|\frac{1+\sin x}{\cos x}\right|+c
$$</p>
|
1,829,086 | <p>So far, I've tried out to reformulate: $$\int{\frac{1}{\cos(x)}}dx$$
to: $$\int{\frac{\sin(x)}{\cos(x)\sin(x)}}dx$$</p>
<p>which is basically: $$\int{\frac{\tan(x)}{\sin(x)}}dx$$
But I'm not sure if this is the right way to go, or if I try something else.</p>
<p>Any tips or methods would be very helpful.</p>
| Community | -1 | <p>When I learned how to do this, I used a much different substitution method. Sure, I'm subbing <em>something</em>, but not <span class="math-container">$tan(\frac x2)$</span>.</p>
<p><span class="math-container">$$\int\frac{1}{\cos(x)}\ dx=\int \sec(x)\ dx = \int \sec(x)\ \frac{\sec(x)+\tan(x)}{\sec(x)+\tan(x)}\ dx$$</span></p>
<p>Let <span class="math-container">$u = \sec(x)+\tan(x)$</span>, and so <span class="math-container">$du = \sec(x)\ \tan(x) + {\sec}^2(x)\ dx$</span>.</p>
<p>Substituting this back into the equation yields</p>
<p><span class="math-container">$$\int \frac{{\sec}^2(x) + \sec (x)\ \tan (x)\ dx}{\sec(x) + \tan (x)} = \int \frac{du}u = \ln\ |\ u\ | + C$$</span></p>
<p>Substituting back for u yields <span class="math-container">$ln\ |\ \sec (x) + \tan (x)\ | + C$</span>.</p>
|
233,397 | <p>could anyone please clarify me the meaning of the term 'hypothesis'? </p>
<p>with relation to terms 'reasoning' and 'assumption' ?</p>
<p>Many thanks</p>
| msc zoology | 436,336 | <blockquote>
<p>Hypothesis is a kind of speculation, supposition or assumption that we make on the basis of some argument”. </p>
</blockquote>
<p>Or </p>
<blockquote>
<p>A supposition or proposed explanation made on the basis of limited evidence as a starting point for further investigation. </p>
</blockquote>
<p>Or</p>
<blockquote>
<p>A hypothesis is an educated guess or proposition that attempts to explain a set of facts or natural phenomenon. </p>
</blockquote>
<p>It is the first stage of formulating a law. It plays a very important role in our daily life and in scientific investigation.</p>
<p>As mentioned previously, a hypothesis is a tool of quantitative studies. It is a tentative and formal prediction about the relationship between two or more variables in the population being studied, and the hypothesis translates the research question into a prediction of expected outcomes.</p>
|
898,002 | <p>Let's say I have a continuous piecewise function of a single variable, so that $y = f(x)$ if $x < c$ and $y = g(x)$ if $x>=c$. Is it right to say that the derivative of the function at $x=c$ exists iff $f'(c-)=g'(c+)$, where $f'$ and $g'$ are obtained using derivative rules?</p>
<p>This would seem reasonable to me, and I fail to find an example where this does not hold. However, my calculus professors have always taught me that the only way to evaluate a derivative of such a point is using the limit definition of the derivative.</p>
| Tobias Kienzler | 163 | <p>As the other answers already mention, this won't work. Hoewever, if you are content with distributions such that $y(x)\stackrel!=y(x_0) + \int_{x_0}^x y'(\chi)\,d\chi$, then you <em>could</em> state</p>
<p>$$y'(x) = h'(c) + (g(x)-f(x))\delta(x-c)$$</p>
<p>where $h'(x)$ is $f'(x)$ or $g'(x)$ depending on the region and $\delta(x)$ is the Dirac delta distribution. This still requires that both $\lim\limits_{x\to c}f(x)$ and $\lim\limits_{x\to c}g(x)$ exist, though.</p>
|
146,973 | <p>I want to find the expected value of $\text{max}\{X,Y\}$ where $X$ ist $\text{exp}(\lambda)$-distributed and $Y$ ist $\text{exp}(\eta)$-distributed. X and Y are independent.
I figured out how to do this for the minimum of $n$ variables, but i struggle with doing it for 2 with the maximum.</p>
<p>(The context in which this was given is waiting for the later of two trains, with their arrival times being exp-distributed).</p>
<p>Thanks!</p>
| Davide Giraudo | 9,849 | <p>The sample $(X,Y)$ have a density given by $f_X(x)f_Y(y)$ since $X$ and $Y$ are independent. You have to compute
$$\iint_{\Bbb R^2}\max\{x,y\}f_X(x)f_Y(y)dxdy.$$
Cut this integral in two parts. </p>
|
4,306,776 | <p>I know a lot of similar questions have been asked here, but sadly I can't seem to wrap my head around this topic. The topic is combinatorics.</p>
<p>I am trying to get into math during high school to be more ready for college and this is so hard.</p>
<p>I have a problem.</p>
<p>Making permutation from word <strong>PAPARAZZI</strong>.</p>
<ol>
<li>I need to find <strong>all possible permutations</strong> of this word which was <span class="math-container">$\frac{N!}{N_1!N_2!\ldots}$</span> where <span class="math-container">$N$</span> is the number of characters and <span class="math-container">$N_1$</span> etc. are groups of duplicates. So for this example it was <span class="math-container">$\frac{9!}{3!2!2!}$</span>.</li>
</ol>
<p>Trouble comes with the second and third problems.</p>
<ul>
<li><em><strong>How many permutations can I make if there always has to be exactly <span class="math-container">$2$</span> characters between my Zs?</strong></em></li>
<li><em><strong>How many permutations can I make if P and I cannot be next to each other?</strong></em></li>
</ul>
<p>I thought that for the "two chars problem" I could take the Zs as single block and just do <span class="math-container">$\frac{8!}{3!2!}$</span> and for the second one Reduce total number of chars to <span class="math-container">$6$</span> (<span class="math-container">$9$</span> - both Ps and I) and continue with <span class="math-container">$\frac{6!}{3!2!}$</span>, but sadly both answers were wrong and I haven't got any feedback because of distance education, so I wanted to ask what I did wrong or what I do not understand?</p>
<p>Thanks in advance.</p>
| Peter Smith | 35,151 | <p>But that initial claim [before it was edited] isn't true. You know from the upward Lowenheim-Skolem theorem that any theory with a countable model also has models of all larger cardinalities (which therefore won't be isomorphic with the countable model).</p>
<p>What <em>is</em> true is that all <em>countable</em> models of the theory of dense linear orders without endpoints look the same. It is still false that there is exactly one model: there are still many models. But at least these, the countable models, are all isomorphic.</p>
<p>That special result is proved in any model theory text! Or googling we find e.g. <a href="https://www.math.ucsd.edu/%7Esbuss/CourseWeb/Math260_2012WS/Mar05Ricketts.pdf" rel="nofollow noreferrer">https://www.math.ucsd.edu/~sbuss/CourseWeb/Math260_2012WS/Mar05Ricketts.pdf</a></p>
|
4,306,776 | <p>I know a lot of similar questions have been asked here, but sadly I can't seem to wrap my head around this topic. The topic is combinatorics.</p>
<p>I am trying to get into math during high school to be more ready for college and this is so hard.</p>
<p>I have a problem.</p>
<p>Making permutation from word <strong>PAPARAZZI</strong>.</p>
<ol>
<li>I need to find <strong>all possible permutations</strong> of this word which was <span class="math-container">$\frac{N!}{N_1!N_2!\ldots}$</span> where <span class="math-container">$N$</span> is the number of characters and <span class="math-container">$N_1$</span> etc. are groups of duplicates. So for this example it was <span class="math-container">$\frac{9!}{3!2!2!}$</span>.</li>
</ol>
<p>Trouble comes with the second and third problems.</p>
<ul>
<li><em><strong>How many permutations can I make if there always has to be exactly <span class="math-container">$2$</span> characters between my Zs?</strong></em></li>
<li><em><strong>How many permutations can I make if P and I cannot be next to each other?</strong></em></li>
</ul>
<p>I thought that for the "two chars problem" I could take the Zs as single block and just do <span class="math-container">$\frac{8!}{3!2!}$</span> and for the second one Reduce total number of chars to <span class="math-container">$6$</span> (<span class="math-container">$9$</span> - both Ps and I) and continue with <span class="math-container">$\frac{6!}{3!2!}$</span>, but sadly both answers were wrong and I haven't got any feedback because of distance education, so I wanted to ask what I did wrong or what I do not understand?</p>
<p>Thanks in advance.</p>
| Sahiba Arora | 266,110 | <p>The statement is proved in Theorem 2.5 <a href="https://www.math.uni-hamburg.de/home/geschke/teaching/ModelTheory.pdf" rel="nofollow noreferrer">here</a>.</p>
|
350,211 | <p>I'm looking at the following in Jech's <em>The Axiom of Choice</em> on page 20:</p>
<blockquote>
<p><strong>2.4.1.</strong> <em>Example: The Countable Axiom of Choice implies that every infinite set has a countable subset.</em></p>
<p>Proof. Let $S$ be an infinite set. Consider all finite one-to-one finite sequences $$\langle a_0 , a_1 , \ldots , a_k \rangle$$ of elements of $S$. The Countable Axiom of Choice picks out one $k$-sequence for each natural number; more exactly: $$\mathscr{F} = \{ A_k : k \in \omega \},$$ where $$A_k = \{ \langle a_0 , \ldots , a_k \rangle : a_0 , \ldots , a_k\text{ distinct elements of }S \},$$ and $\mathscr{F}$ has a choice function: $f ( A_k ) \in A_k$ for all $k$. The union of all the chosen finite sequences is obviously countable.</p>
</blockquote>
<hr>
<p>And I'm wondering if I can instead prove it as follows: </p>
<p>Let $S$ be an infinite set, that is, $|S| \ge |\omega|$. By the definition of $|S| \ge |\omega|$ there is an injection $f: \omega \hookrightarrow S$. Then $f(\omega) \subseteq S$ yields the desired result. </p>
<p>I think yes but I might be missing something. Thanks for your help.</p>
| Cameron Buie | 28,900 | <p>We say that $S$ is infinite iff $|S|\not<|\omega|.$ Without some choice principle, we can't conclude that all infinite sets admit a countably infinite subset, as it's entirely possible that $S$ and $\omega$ are of incomparable cardinality.</p>
<p>The least choice principle sufficient to the task is: "$\omega$ is of comparable cardinality with all sets," a strictly weaker principle than $\text{AC}_\omega$. That is, any stronger principle will do the job, but weaker principles will fail to be able to prove that infinite sets have countably infinite subsets. On the other hand, if every infinite set has a countably infinite subset, then $\omega$ is of comparable cardinality with all sets. Thus, the aforementioned choice principle is actually logically equivalent to the result that Jech proves with $\text{AC}_\omega$.</p>
<p>We say that sets with a countably infinite subset are "Dedekind-infinite" or just "D-infinite." D-infinite sets are always infinite, but without a choice principle at least as strong as the one mentioned above, there may be infinite, D-finite sets.</p>
<p>For some equivalent statements to the choice principle I mentioned above, see <a href="https://math.stackexchange.com/questions/144008/equivalences-to-d-finite-finite">here</a>.</p>
|
4,455,201 | <p><em>Question:</em><br />
Let <span class="math-container">$\Omega=\{a,b,c\}$</span>. Give an example for <span class="math-container">$X, F_1, F_2$</span> in which
<span class="math-container">$E(E(X|F_1)|F_2) \neq E(E(X|F_2)|F_1)$</span></p>
<p><em>My answer:</em><br />
I am not at all sure of my answer. If you have any shorter and nicer answer i will be happy to read it.</p>
<p>-Let define:<br />
(a) <span class="math-container">$F=B(\Omega ), \; F_1=\left \{ \Omega;\left \{ \emptyset \right \} ;\left \{ a \right \};\left \{ b;c \right \}\right \}, \; F_2=\left \{ \Omega;\left \{ \emptyset \right \} ;\left \{ b \right \};\left \{ a;c \right \}\right \} $</span>.
By def: <span class="math-container">$Z_{12}=E(X|F_1)$</span> is a rv <span class="math-container">$F_1$</span> measurable, <span class="math-container">$Z_{21}=E(X|F_2)$</span> is a rv <span class="math-container">$F_2$</span> measurable.<br />
(b) X a bijective measurable function from <span class="math-container">$(\Omega ; B(\Omega )) \rightarrow (\left \{ 1;2;3 \right \}; B(\left \{ 1;2;3 \right \})) $</span></p>
<p>-Proof: <span class="math-container">$Z_{12} \neq Z_{21} \; a.s$</span><br />
By absurd, we assume that: <span class="math-container">$Z_{12} = Z_{21} \; a.s \; \Rightarrow E(Z_{12}) = E(Z_{21})$</span>.<br />
(i) But on the other side we have: <span class="math-container">$E(Z_{12}|F_1) = Z_{12}$</span> because is <span class="math-container">$F_1$</span> measurable.<br />
(ii) And by the absurd assumption: <span class="math-container">$E(Z_{21}|F_1) = E(Z_{21}) = E(Z_{12}) $</span><br />
So we get from (i)+(ii): <span class="math-container">$Z_{12}= E(Z_{21})$</span> Wich is not necessarly always true.<br />
And of course <span class="math-container">$Z_{12} \neq Z_{21} \; a.s \; \Rightarrow E(Z_{12}) \neq E(Z_{21})$</span></p>
<p>-Now from what we just wrotte above:<br />
<span class="math-container">$E(E(X|F_1)|F_2)=E(Z_{12}|F_2)=E(Z_{12}) \neq E(Z_{21})=E(Z_{21}|F_1)=E(E(X|F_2)|F_1)$</span></p>
<p>-Q.E.D</p>
| X0-user-0X | 1,059,745 | <p>Here an other answer. I hope it is true.</p>
<p>So let <span class="math-container">$\Omega=\left \{ a,b,c \right \}$</span> be our probabilisable space and a r.v. <span class="math-container">$X(\omega)=1_{\omega=c}$</span>.
So obviously <span class="math-container">$\sigma(X)=\left \{ (a,b) ,(c), \Omega, \varnothing \right \}$</span>.</p>
<p>1)Now let define:<br />
<span class="math-container">$F_1=\left \{\varnothing, \Omega, (a), (b,c) \right \}$</span><br />
<span class="math-container">$F_2=\left \{\varnothing, \Omega, (b), (a,c) \right \}$</span><br />
Obiously none <span class="math-container">$F_1$</span> or <span class="math-container">$F_2$</span> is include into the other.</p>
<p>2)We beguin by computing:<br />
2.1)<br />
<span class="math-container">$Z_1=E(X|F_1)$</span>, so <span class="math-container">$Z_1$</span> can be written as follow: <span class="math-container">$Z_1=\alpha1_{\omega=a}+\beta1_{\omega=b \cup c}$</span> and by definition <span class="math-container">$Z_1$</span> must verify: <span class="math-container">$\forall A \in F_1,E(Z_1.1_A)=E(X.1_A)$</span> If it is true <span class="math-container">$\forall A$</span> so it is true too in particular for <span class="math-container">$A=a$</span> pe <span class="math-container">$A=b \cup c$</span><br />
-For <span class="math-container">$A=a$</span> we get:<br />
<span class="math-container">$E(Z_1.1_A)=\int_{\omega \in \Omega}^{}Z_1.1_ad\mathbb{P}=\alpha.\mathbb{P}(\left \{ \omega \in \Omega:Z_1(\omega)=\alpha \right \})=0=\int_{\omega \in \Omega}^{}1_c.1_ad\mathbb{P}=E(X.1_A)$</span><br />
-For <span class="math-container">$A=(b,c)$</span>
<span class="math-container">$E(Z_1.1_A)=\int_{\omega \in \Omega}^{}Z_1.1_{b,c}d\mathbb{P}=\beta.\mathbb{P}(\left \{ \omega \in \Omega:Z_1(\omega)=\beta \right \})=\mathbb{P}(\left \{ \omega \in \Omega:X(\omega)=1 \right \})=\int_{\omega \in \Omega}^{}1_c.1_{b,c}d\mathbb{P}=E(X.1_A)$</span><br />
So <span class="math-container">$\alpha=0, \beta=P(c)$</span> and <span class="math-container">$Z_1(\omega)=1_{(b,c)}P(c)/P(b,c)$</span><br />
2.2)<br />
By doing exactly the same thing for <span class="math-container">$Z_2=E(X|F_2)$</span> we get: <span class="math-container">$Z_2=1_{(a,c)}P(c)/P(a,c)$</span></p>
<p>3)Similar as before but now we are looking for:<br />
3.1)</p>
<p><span class="math-container">$Z_{12}=E(E(X|F_1)|F_2)=E(p(c)/p(b,c)1_{(b,c)}|F_1)=p(c)/p(b,c)E(1_{(b,c)}|F_2)$</span>. So we want to know: <span class="math-container">$E(1_{(b,c)}|F_2)$</span> so by proceeding as before and according to the definition we are looking for <span class="math-container">$Z_{12}$</span> that looks like this:<br />
<span class="math-container">$Z_{12}=\alpha.1_{(b)}+\beta.1_{(a,c)}$</span><br />
<span class="math-container">$E(Z_{12}.1_{(b)})=\alpha P(b)=P(b)=E(1_{b})=E(1_{(b,c)}.1_{(b)})$</span> So <span class="math-container">$\alpha=1$</span><br />
<span class="math-container">$E(Z_{12}.1_(a,c))=\beta P(a,c)=P(c)=E(1_{c})=E(1_{(b,c)}.1_{(a,c)})$</span> So <span class="math-container">$\beta=P(c)/P(a,c)$</span><br />
3.2)</p>
<p><span class="math-container">$Z_{21}=E(E(X|F_2)|F_1)=E(p(c)/p(a,c)1_{(a,c)}|F_1)=p(c)/p(a,c)E(1_{(a,c)}|F_1)$</span>. So we want to know: <span class="math-container">$E(1_{(a,c)}|F_1)$</span>. By proceeding as before and according to the definition we are looking for <span class="math-container">$Z_{21}$</span> that looks like this:<br />
<span class="math-container">$Z_{21}=\alpha.1_{(a)}+\beta.1_{(b,c)}$</span><br />
<span class="math-container">$E(Z_{21}.1_{(a)})=\alpha P(a)=P(a)=E(1_{a})=E(1_{(a,c)}.1_{(a)})$</span> So <span class="math-container">$\alpha=1$</span><br />
<span class="math-container">$E(Z_{21}.1_(b,c))=\beta P(b,c)=P(c)=E(1_{c})=E(1_{(b,c)}.1_{(a,c)})$</span> So <span class="math-container">$\beta=P(c)/P(b,c)$</span></p>
<p>4)In conclusion we get:<br />
<span class="math-container">$Z_{12}=\frac{P(c)}{P(b,c)}(1_{(b)}+1_{(a,c)}\frac{P(c)}{P(a,c)})$</span> wich is different from <span class="math-container">$Z_{21}=\frac{P(c)}{P(a,c)}(1_{(a)}+1_{(b,c)}\frac{P(c)}{P(b,c)})$</span></p>
<p>Rem: For exemple: <span class="math-container">$P(a,c)=P(a \cup c)$</span> and <span class="math-container">$1_{(a)}$</span> is the indicator function.</p>
|
162,613 | <p>How can I apply a <code>LabelingFunction</code> with an if else condition?
Suppose you have the following plot:</p>
<pre><code>BarChart[{4, 3, 2, 1}, BarOrigin -> Left, LabelingFunction ->(Placed[#,Left] &)]
</code></pre>
<p><a href="https://i.stack.imgur.com/2o9wE.png" rel="noreferrer"><img src="https://i.stack.imgur.com/2o9wE.png" alt="enter image description here"></a></p>
<p>I want to plot it with the first two labels justified as they currently are (left), and the third and fourth label in bold and justified to the right.</p>
| Szabolcs | 12 | <p>It is important to always read the "Details" section of a symbol's documentation: <a href="http://reference.wolfram.com/language/ref/LabelingFunction.html" rel="nofollow noreferrer"><code>LabelingFunction</code></a>. There is it explained how to compute the label based on the element's index rather than its value:</p>
<blockquote>
<p><code>LabelingFunction->f</code> specifies that each chart element should have a label given by <code>f[value,index,lbls]</code>, where <code>value</code> is the data value associated with the element, <code>index</code> is its position in the nested list of datasets, and <code>lbls</code> is the list of relevant labels.</p>
</blockquote>
<p>To understand better what is being passed to the function, simply display all values in the label:</p>
<pre><code>BarChart[data, BarOrigin -> Left,
LabelingFunction -> (Placed[{##}, Left] &)]
</code></pre>
<p><a href="https://i.stack.imgur.com/VSYPe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VSYPe.png" alt="enter image description here"></a></p>
<p>Now that we understand the structure of indices better, we can write our labelling function appropriately:</p>
<pre><code>BarChart[data, BarOrigin -> Left,
LabelingFunction ->
Function[{value, index, label},
If[Last[index] > 2,
Placed[Style[value, Bold], Right],
Placed[value, Left]]]]
</code></pre>
<p><a href="https://i.stack.imgur.com/t7g3F.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/t7g3F.png" alt="enter image description here"></a></p>
|
1,751,825 | <p>Suppose there are two norms on a Banach Space: (X, $||•||_1$) and (X, $||•||_2$) where, for any sequence {$x_n$} in X, $||x_n||_1$->0 implies $||x_n||_2$->0. Show that these norms are equivalent. </p>
<p>I'm fairly certain that making an identity between (X, $||•||_1$) and (X, $||•||_2$) is a reasonable start to this problem. From there, the closed graph theorem and bounded inverse theorem are likely used (since they are introduced in the same section as this exercise) although I'm not really sure as to how each are used. </p>
<p>Let the identity mapping mentioned above be T, then I know that with X being complete, we have that T is closed (unless my knowledge of the closed graph theorem is mistaken). Also, the bounded inverse theorem would imply that if T is bounded linear and bijective then it's inverse is also bounded. The bounded was would likely be helpful with showing equivalency.</p>
<p>The details I'm unsure of: is T bounded? How does closure of T come into play? How does the assumption of convergence to 0 in norm take part in this argument?</p>
| Dalamar | 94,110 | <p>Perhaps you could work by contradiction and assume the two norms are not equivalent, that one is unbounded with respect to the other. Then you can find a sequence $(x_n)$ such that $||x_n||_1 \leq c$ (with $c$ a positive constant) and $||x_n||_2 \geq n$ $\forall n$. Now the sequence $y_n = x_n/ \sqrt n$ clearly converges to $0$ in the first norm and diverges in the other, so that if the norms are not equivalent you can always violate your hypothesis of implied convergence. </p>
|
1,751,825 | <p>Suppose there are two norms on a Banach Space: (X, $||•||_1$) and (X, $||•||_2$) where, for any sequence {$x_n$} in X, $||x_n||_1$->0 implies $||x_n||_2$->0. Show that these norms are equivalent. </p>
<p>I'm fairly certain that making an identity between (X, $||•||_1$) and (X, $||•||_2$) is a reasonable start to this problem. From there, the closed graph theorem and bounded inverse theorem are likely used (since they are introduced in the same section as this exercise) although I'm not really sure as to how each are used. </p>
<p>Let the identity mapping mentioned above be T, then I know that with X being complete, we have that T is closed (unless my knowledge of the closed graph theorem is mistaken). Also, the bounded inverse theorem would imply that if T is bounded linear and bijective then it's inverse is also bounded. The bounded was would likely be helpful with showing equivalency.</p>
<p>The details I'm unsure of: is T bounded? How does closure of T come into play? How does the assumption of convergence to 0 in norm take part in this argument?</p>
| Rick Sanchez | 332,412 | <p>First show, using the sequence condition that $\Vert x\Vert_2 \leq c\Vert x\Vert_1$ for some $c>0$. Then, we have that the identity map $$I:X_1 \rightarrow X_2$$ is a bounded linear bijection between Banach spaces. Open mapping theorem (bounded inverse) says its inverse is also bounded...</p>
|
43,650 | <p>Consider the following problem:</p>
<blockquote>
<p>Let $f:{\mathbb R}^3 \to{\mathbb R}$ be $$f(x,y,z)=x+4z$$ where $$x^2+y^2+z^2\leq 2.$$ Find the minimum of $f$. </p>
</blockquote>
<p>This is similar to the question <a href="https://math.stackexchange.com/q/41385/9464">here</a>. However, since this is not an analytic function with complex variable, one may not be able to use the "Maximum modulus principle". </p>
<p>What I think is that one may rewrite the inequality constraint $x^2+y^2+z^2\leq 2$ as
$$x^2+y^2+z^2=2-\delta\qquad \delta\in[0,2]$$
then one can use the "Lagrange Multiplier Method" with the parameter $\delta$. Or one can do it on the xOz plane with the geometric meaning of $C$ in $C=x+4z$.</p>
<p>Is there any other way better than the idea above to solve this problem?</p>
| p.s. | 17,433 | <p>The other answers cover how to solve this with Lagrange multipliers, but I'd like to point out how the problem generalizes, using vector notation. If $\mathbf{x}=(1,0,4)$ and $r=\sqrt{2}$, then problem can be expressed:
$$
\min_{\|\mathbf{y}\|_2 \le r} \ \langle \mathbf{x}, \mathbf{y}\rangle = -r\|\mathbf{x}\|_{2}
$$
Where the argmin is $-r\mathbf{x}/\|\mathbf{x}\|_2$. This is related to a problem that anyone familiar with analysis should know by heart:
$$
\max_{\|\mathbf{y}\|_2 \le 1} \ \langle \mathbf{x}, \mathbf{y}\rangle = \|\mathbf{x}\|_{2}
$$
Where the argmax is $\mathbf{x}/\|\mathbf{x}\|_2$. The latter problem is particularly important because it's the definition of a dual norm, and it shows that the 2-norm is self-dual.</p>
|
1,775,166 | <p>If $$T:V\rightarrow V$$ is a bijective linear transformation then can $T^2$ also be a bijective linear transformation? Can $T^n$ also be bijective linear transformation?</p>
| Patrick Abraham | 337,503 | <p>Let $T$ be a linear bijective transformation with $T:V \rightarrow V$.</p>
<p>Thus $\exists\ \hat{T} : T \circ \hat{T} = id$</p>
<p>Base: $T^n$ is bijective may n be, without loss of generality, equal to 2. $n=2$</p>
<p>$ id = T \circ id \circ \hat{T} = T \circ T \circ \hat{T} \circ \hat{T} = T^2 \circ \hat{T^2}$</p>
<p>Let n be arbitrary but fixed, with $id=T^n\circ\hat{T^n}$</p>
<p>$n\rightarrow n+1$</p>
<p>$T^{n+1}\circ\hat{T^{n+1}}=T\circ T^n\circ \hat{T^{n}}\circ\hat{T} \stackrel{\text{base}}{=} T \circ id \circ \hat{T} = T \circ \hat{T} = id$</p>
<p>Thus $T^n$ is bijective. $\square$</p>
|
3,114,873 | <p>Does the function <span class="math-container">$\frac{1-2xy}{x^2 +y^2}$</span> have a max or min value for <span class="math-container">$(x,y)=/=0$</span>?</p>
<p>What I've tried so far is to take the the partial derivatives:</p>
<p><span class="math-container">$$\frac{\partial f}{\partial x} = \frac{2(-x+x^2*y - y^3)}{(x^2 + y^2)^2}$$</span>
<span class="math-container">$$\frac{\partial f}{\partial y} = \frac{2(x^3 -x*y^2 +y)}{(x^2 + y^2)^2}$$</span></p>
<p>However I can't see what satisfy will <span class="math-container">$\nabla f(a,b)=0$</span></p>
<p>It looks like the function has a singular point in <span class="math-container">$(0,0)$</span> since it doesn't exist there, but I am told to ignore that point.</p>
<p>And seeing there is no boundary to f, the max/min can't be there either.</p>
<p>So, how can I then show that this function has a max/min other than in <span class="math-container">$(0,0)$</span>? </p>
| Mark | 470,733 | <p>Look at the permutation <span class="math-container">$\sigma=(123...n)$</span>. I say <span class="math-container">$ S_{n-1},\sigma S_{n-1},\sigma^2 S_{n-1},..., \sigma^{n-1} S_{n-1}$</span> are <span class="math-container">$n$</span> distinct cosets. Can you prove it? I'll give a hint. Suppose <span class="math-container">$\sigma^i S_{n-1}=\sigma^j S_{n-1}$</span> when <span class="math-container">$0\leq i<j<n$</span>. What can you say about <span class="math-container">$\sigma^{j-i}$</span>? </p>
|
1,040,846 | <p>I want to prove, that $a_n$ is a null sequence if $$\lim_{n \to \infty}|\frac{a_{n+1}}{a_n}|= c < 1$$</p>
<p>That means that $\forall \epsilon > 0: \exists N \in \mathbb{N}: n \ge N: |\frac{a_{n+1}}{a_n} - c| < \epsilon$</p>
<p>How can I get rid of the $a_{n+1}$ and the $c$, to show $\forall \epsilon > 0: \exists N \in \mathbb{N}: n \ge N: |a_n| < \epsilon$</p>
<p>?</p>
| lhf | 589 | <p>$$
\lim_{n \to \infty}|\frac{a_{n+1}}{a_n}|= c < 1
$$
implies that $|a_{n+1}| \le b |a_{n}|$ for $n\ge N$ and $b=(c+1)/2 < 1$.
Therefore,
$|a_{n}| \le b^{n-N} |a_{N}|$ for $n\ge N$. Since $|b|<1$, we have $b^n\to 0$.</p>
|
4,403,249 | <p><span class="math-container">$$f(x,y)=(x^2+y^2)^{\left|x\right|} $$</span> find limit at <span class="math-container">$(0,0)$</span></p>
<p>moving into parametric form <span class="math-container">$f(r\cos\phi,r\sin\phi)=(r^2)^{r\left|\cos\phi\right|}= e^{2r\left|\cos\phi\right|\ln r}$</span></p>
<p><span class="math-container">$\bigl|r\left|\cos\phi\right|\ln r\bigr|\leq r\times \ln(\frac{1}{r})$</span> which goes to <span class="math-container">$0$</span></p>
<p>so <span class="math-container">$$f(x,y) \leq e^{2\sqrt{x^2+y^2}\ln(\frac{1}{\sqrt{x^2+y^2}})}$$</span> which goes to <span class="math-container">$1$</span> as <span class="math-container">$(x,y) \to (0,0)$</span></p>
<p>I am stuck how conclude that limit is actually <span class="math-container">$1$</span>? I get that <span class="math-container">$f(x,y)$</span> is less than some function which goes to <span class="math-container">$1$</span>.</p>
| A. P. | 1,027,216 | <p>We get,
<span class="math-container">\begin{align*}
\lim_{(x,y)\to (0,0)} (x^{2}+y^{2})^{|x|}&=\lim_{r\searrow 0}(r^{2})^{|r\cos \theta|}, \tag{1}\\
&=\lim_{r\searrow 0} r^{|2r\cos \theta|}, \\
&=\lim_{r\searrow 0} e^{\log r^{|2r\cos \theta|}}, \tag{2}\\
&=e^{\displaystyle \lim_{r\searrow 0} \log r^{|2r\cos \theta|}} , \\
&=e^{\displaystyle\lim_{r\searrow 0}|2r\cos \theta|\log r}, \tag{3}\\\
&=e^{0}, \\
&=1.
\end{align*}</span>
where <span class="math-container">$(1)$</span> is change of variables to polar coordinates <span class="math-container">$x=r\cos \theta, y=r\sin \theta$</span> with <span class="math-container">$r>0$</span> and <span class="math-container">$\theta\in [0,2\pi[$</span> and <span class="math-container">$(2)$</span> is just <span class="math-container">$x^{y}=e^{\log x^{y}}$</span>. Finally, <span class="math-container">$(3)$</span> is given because <span class="math-container">$\displaystyle\lim_{r\searrow 0}|2r\cos \theta|\log r=0$</span> because it is bounded. Hence,
<span class="math-container">$$\boxed{\lim_{(x,y)\to (0,0)}(x^{2}+y^{2})^{|x|}=1}$$</span></p>
<hr />
<p>More details:</p>
<p>Notice that,
<span class="math-container">$$-1\leqslant \cos \theta \leqslant 1,$$</span> <span class="math-container">$$-|2r|\log r\leqslant |2r|\cos \theta \leqslant |2r|\log r,$$</span> <span class="math-container">$$0\leqslant \lim_{r\searrow 0}|2r|\cos\theta \leqslant 0.$$</span>
Hence, <span class="math-container">$$\lim_{r\searrow 0} |2r\cos \theta|\log r=0.$$</span></p>
|
291,676 | <p>I have a question about a problem I encountered:</p>
<p>$\exists$ a,b $\epsilon$ $\mathbb{R}$+ such that $\sqrt{a+b}=\sqrt{a}+\sqrt{b}$</p>
<p>Any tips for going about solving this?</p>
<p>I tried:</p>
<p>$\sqrt{a+b}=\sqrt{a}+\sqrt{b}$</p>
<p>$a+b=a+b$</p>
<p>I have a feeling this isn't a legal operation...</p>
| Git Gud | 55,235 | <p>$\textbf{Hint:}$ Suppose such $a,b\in \mathbb{R}^+$ do exist, then square both sides of $\sqrt{a+b}=\sqrt{a}+\sqrt{b}$.</p>
|
3,063,130 | <blockquote>
<p>Suppose function <span class="math-container">$f(z)$</span> is holomorphic on <span class="math-container">$\mathbb{D}(0,2)$</span> and <span class="math-container">$N>0$</span>
is an integer such that: <span class="math-container">$$ |f^{(N)}(0)| = N! \sup\{|f(z)|: |z|=1\} $$</span>
show that <span class="math-container">$f(z) = cz^N$</span>, <span class="math-container">$c \in \mathbb{C}$</span>.</p>
</blockquote>
<p>I have shown that since <span class="math-container">$f(z)$</span> is holomorphic in <span class="math-container">$\mathbb{D}(0,2)$</span>, then it has a power series expansion around zero. </p>
<p><span class="math-container">$$ f(z) = \sum_{n=0}^{\infty} a_n z^n $$</span></p>
<p>Calculating the <span class="math-container">$N$</span>-th derivative I got:</p>
<p><span class="math-container">$$ |f^{(N)}(0)| = N! a_N$$</span></p>
<p>from which I conclude that <span class="math-container">$f(z)$</span> is bounded by <span class="math-container">$a_N$</span> on the unit circle. Therefore by maximum principle for holomorphic function we may also conclude that it is bounded in the unit disc. But I still can't figure out how to show that <span class="math-container">$f(z) = cz^N$</span>. </p>
| Kenny Wong | 301,805 | <p>By Cauchy's formula,
<span class="math-container">$$ f^{(N)} (0) = \frac{N!}{2\pi i} \oint_{|z| = 1} \frac{f(z) \ dz}{z^{N+1}} = \frac{N!}{2\pi } \int_{0}^{2\pi} e^{-iN\theta}f(e^{i\theta}) \ d\theta.$$</span></p>
<p>Thus
<span class="math-container">$$ |f^{(N)}(0)| \leq N!\sup_{\theta \in [0, 2\pi)} |f(e^{i\theta})|,$$</span></p>
<p>with equality if and only if</p>
<p><span class="math-container">$$ e^{-iN\theta} f(e^{i\theta}) = c$$</span></p>
<p>for some constant <span class="math-container">$c \in \mathbb C$</span>.</p>
<p>As the equality <em>does</em> hold by assumption, we have <span class="math-container">$f(z) = cz^N$</span> when <span class="math-container">$|z| = 1$</span>.</p>
|
2,076,656 | <p>As a homework, I was asked to solve this <strong>equation</strong>, $$(3x-4\lfloor x\rfloor=0),x\in \Bbb R$$ For $x\in \Bbb Z:\lfloor x\rfloor=x \implies x=0$ But for $x\not\in\Bbb Z : \lfloor x\rfloor=\frac 34x$ So now we know that $\frac 34x\in\Bbb Z$ and $x\in\Bbb R-\Bbb Z$, so maybe ? define a function such that : $$f:\begin{Bmatrix}(\Bbb R-\Bbb Z)\to \Bbb Z \\ x\mapsto \frac34x\end{Bmatrix}$$ Even if trying this did walk me into $x=\frac43$ I'm still left with no <em>rigorous</em> proof (An explanation is would be nice if possible). Any help would be appreciated. Thank you for your time.</p>
| Asinomás | 33,907 | <p>rewrite this as $4\lfloor x \rfloor =3x$, so that $x=\frac{n}{3}$ for some integer $n$.</p>
<p>We now have $4\lfloor\frac{n}{3}\rfloor=n$</p>
|
2,076,656 | <p>As a homework, I was asked to solve this <strong>equation</strong>, $$(3x-4\lfloor x\rfloor=0),x\in \Bbb R$$ For $x\in \Bbb Z:\lfloor x\rfloor=x \implies x=0$ But for $x\not\in\Bbb Z : \lfloor x\rfloor=\frac 34x$ So now we know that $\frac 34x\in\Bbb Z$ and $x\in\Bbb R-\Bbb Z$, so maybe ? define a function such that : $$f:\begin{Bmatrix}(\Bbb R-\Bbb Z)\to \Bbb Z \\ x\mapsto \frac34x\end{Bmatrix}$$ Even if trying this did walk me into $x=\frac43$ I'm still left with no <em>rigorous</em> proof (An explanation is would be nice if possible). Any help would be appreciated. Thank you for your time.</p>
| Steven Alexis Gregory | 75,410 | <h2><span class="math-container">$$3x=4\lfloor x\rfloor$$</span></h2>
<p>Suppose that, for some integer, <span class="math-container">$n$</span>, we know that <span class="math-container">$n \le x \lt n+1$</span>. Then
<span class="math-container">$\lfloor x\rfloor = n$</span> and
<span class="math-container">$$x = \dfrac 43n$$</span> </p>
<p>But we still need to make sure that </p>
<p><span class="math-container">\begin{array}{cccC}
n &\le &x &\lt &n+1 \\
n &\le &\dfrac 43n &\lt &n+1 \\
3n &\le &4n &\lt &3n+3 \\
0 &\le &n &\lt &3 \\
\end{array}</span></p>
<p>So <span class="math-container">$n \in \{0,1,2\}$</span>; that is, <span class="math-container">$x \in \left\{ 0,\dfrac 43,\dfrac 83 \right\}$</span></p>
|
4,520,435 | <p>(<strong>Note:</strong> I've posted <a href="https://math.stackexchange.com/a/4539959/231608">my own answer</a>, slightly redefining trirationals to be composed of reals instead of integers and addressing the problems pointed out here. Please take note of this while reading my original question.)</p>
<p><strong>Definition:</strong> By <em>trirational number</em>, I mean a number that can represent a ratio of <strong>three</strong> integers (e.g. <span class="math-container">$2:3:5$</span>) in the same way rational numbers represent a ratio of two integers. I will express trirational numbers in this form: <span class="math-container">$a\unicode{x25B6}b\unicode{x25B6}c$</span>.</p>
<p>Let <span class="math-container">$t$</span> be a trirational number that represents the triple integer ratio <span class="math-container">$a:b:c$</span>. We'll define a trirational number to be</p>
<p><span class="math-container">$$t=a\unicode{x25B6}b\unicode{x25B6}c\overset{\text{def}}{=}a*b^{\omega}*c^{\omega^2}$$</span></p>
<p>where <span class="math-container">$\omega$</span> and <span class="math-container">$\omega^2$</span> are the two primitive <strong>third roots of unity</strong>. This is analogous to how a rational number representing the ratio <span class="math-container">$a:b$</span> can be expressed as <span class="math-container">$a*b^{-1}$</span> where <span class="math-container">$-1$</span> is of course the primitive second root of unity. Since <span class="math-container">$\omega$</span> and <span class="math-container">$\omega^2$</span> have non-zero imaginary parts, <span class="math-container">$t$</span> is not confined to the real number line.</p>
<p>In order for <span class="math-container">$t$</span> to be a proper representation of <span class="math-container">$a:b:c$</span>, it must represent the fact that <span class="math-container">$a:b:c=xa:xb:xc$</span> for any integer <span class="math-container">$x$</span>. This is satisfied given that <span class="math-container">$1+\omega+\omega^2=0$</span>, and therefore</p>
<p><span class="math-container">$$(xa)\unicode{x25B6}(xb)\unicode{x25B6}(xc)=(xa)*(xb)^{\omega}*(xc)^{\omega^2}=x^{1+\omega+\omega^2}t=t$$</span></p>
<p>The ternary operation used to generate trirational numbers is analogous to division in certain ways:</p>
<ul>
<li><span class="math-container">$x\unicode{x25B6}x\unicode{x25B6}x=1$</span></li>
<li><span class="math-container">$x\unicode{x25B6}1\unicode{x25B6}1=x$</span></li>
<li><span class="math-container">$0\unicode{x25B6}x\unicode{x25B6}y=0$</span></li>
<li><span class="math-container">$x\unicode{x25B6}0\unicode{x25B6}y$</span> and <span class="math-container">$x\unicode{x25B6}y\unicode{x25B6}0$</span> are both undefined.</li>
<li>Just as <span class="math-container">$(x/y)^{-1}=y/x$</span>, we have
<span class="math-container">$(x\unicode{x25B6}y\unicode{x25B6}z)^{\omega}=z\unicode{x25B6}x\unicode{x25B6}y$</span>, <span class="math-container">$\\(x\unicode{x25B6}y\unicode{x25B6}z)^{\omega^2}=y\unicode{x25B6}z\unicode{x25B6}x$</span>, and <span class="math-container">$\\(x\unicode{x25B6}y\unicode{x25B6}z)^{-1}=~(x\unicode{x25B6}y\unicode{x25B6}z)^{\omega+\omega^2}=zy\unicode{x25B6}xz\unicode{x25B6}yx$</span></li>
</ul>
<p>Other properties:</p>
<ul>
<li><span class="math-container">$x\unicode{x25B6}y\unicode{x25B6}y=x/y$</span>, meaning all rational numbers are trirational numbers. Take note that both <span class="math-container">$x\unicode{x25B6}y\unicode{x25B6}1\neq x/y$</span> and <span class="math-container">$x\unicode{x25B6}1\unicode{x25B6}y\neq x/y$</span> unless <span class="math-container">$y=1$</span>.</li>
<li><span class="math-container">$y\unicode{x25B6}x\unicode{x25B6}y=(x/y)^{\omega}$</span>, meaning two rational numbers <span class="math-container">$\frac{a}{c}$</span> and <span class="math-container">$\frac{b}{c}$</span> can be combined into a trirational number <span class="math-container">$a\unicode{x25B6}b\unicode{x25B6}c$</span> via <span class="math-container">$\frac{a}{c}*(\frac{b}{c})^{\omega}=(a\unicode{x25B6}c\unicode{x25B6}c)*(c\unicode{x25B6}b\unicode{x25B6}c)=a\unicode{x25B6}b\unicode{x25B6}c$</span></li>
<li><span class="math-container">$y\unicode{x25B6}y\unicode{x25B6}x=(x/y)^{\omega^2}$</span>, meaning two rational numbers <span class="math-container">$\frac{a}{b}$</span> and <span class="math-container">$\frac{c}{b}$</span> can be combined into a trirational number <span class="math-container">$a\unicode{x25B6}b\unicode{x25B6}c$</span> via <span class="math-container">$\frac{a}{b}*(\frac{c}{b})^{\omega^2}=(a\unicode{x25B6}b\unicode{x25B6}b)*(b\unicode{x25B6}b\unicode{x25B6}c)=a\unicode{x25B6}b\unicode{x25B6}c$</span></li>
</ul>
<hr />
<p><strong>Problems with generalization:</strong></p>
<p>At first it seemed likely to me that a generalized <span class="math-container">$n$</span>-rational number representing a ratio of <span class="math-container">$n$</span> integers <span class="math-container">$a_1:a_2:a_3:...:a_n$</span> would be of the form
<span class="math-container">$$a_1*a_2^{\omega_1}*a_3^{\omega_2}*...*a_n^{\omega_{n-1}}$$</span></p>
<p>where <span class="math-container">${\omega_1}$</span> up to <span class="math-container">${\omega_{n-1}}$</span>, along with <span class="math-container">$1$</span>, are the <span class="math-container">$n$</span>th roots of unity. But problems arise even for <span class="math-container">$n=4$</span>.</p>
<p>The fourth roots of unity in the complex plane are <span class="math-container">$1$</span>, <span class="math-container">$i$</span>, <span class="math-container">$-1$</span>, and <span class="math-container">$-i$</span>, which give us this candidate form for 4-rational numbers:</p>
<p><span class="math-container">$$a*b^i*c^{-1}*d^{-i}=\frac{a}{c}*(\frac{b}{d})^i$$</span></p>
<p>This might seem promising because <span class="math-container">$1+i-1-i=0$</span>, but it still fails as a representation of ratios of four integers. For example, <span class="math-container">$2:5:4:10$</span> in the above form would simplify to <span class="math-container">$\frac{1}{2}*(\frac{1}{2})^i$</span> even though <span class="math-container">$1:1:2:2$</span> is not an equivalent ratio. Even worse, something like <span class="math-container">$2:7:2:7$</span> in this form ends up as <span class="math-container">$1$</span>!</p>
<p>It looks like we'll need to look in higher dimensions for a solution.</p>
<hr />
<p><strong>Updates:</strong></p>
<p>As per @Stinking Bishop's advice I'll just quickly define trirational multiplication:</p>
<p><span class="math-container">$$(a_1\unicode{x25B6}b_1\unicode{x25B6}c_1)*(a_2\unicode{x25B6}b_2\unicode{x25B6}c_2)\overset{\text{def}}{=}a_1a_2\unicode{x25B6}b_1b_2\unicode{x25B6}c_1c_2$$</span></p>
<p>Unfortunately, I still don't have a general definition of trirational addition. I actually can't find anything online about adding ratios of three numbers. I even tried looking into projective spaces as @Qiaochu Yuan's suggested, but it seems addition is also a problem there.</p>
<p>So in most cases, I need to convert the trirational addends to complex numbers before I could add them. These are the only (trivial) exceptions:</p>
<p><span class="math-container">$$(a_1\unicode{x25B6}b\unicode{x25B6}c)+(a_2\unicode{x25B6}b\unicode{x25B6}c)\overset{\text{def}}{=}(a_1+a_2)\unicode{x25B6}b\unicode{x25B6}c$$</span>
<span class="math-container">$$(a_1\unicode{x25B6}b_1\unicode{x25B6}b_1)+(a_2\unicode{x25B6}b_2\unicode{x25B6}b_2)\overset{\text{def}}{=}(a_1b_2+a_2b_1)\unicode{x25B6}b_1b_2\unicode{x25B6}b_1b_2$$</span></p>
<p>Aside from those, I usually don't even know how to convert the complex sum in the general case back to trirational form!</p>
<p>As for generalizing trirational numbers (i.e. <span class="math-container">$n$</span>-rational numbers), I now see that it's impossible to do for <span class="math-container">$n>3$</span> in the complex plane using roots of unity due to simplification problems:</p>
<p>For any <span class="math-container">$n$</span>th root of unity, if <span class="math-container">$n$</span> is <strong>even</strong> then you get <span class="math-container">$n$</span>th root pairs that are negatives of each other, so terms could cancel out: <span class="math-container">$a^{\omega_0}*a^{-\omega_0}=1$</span>. This is fine for <span class="math-container">$n=2$</span> (the rational numbers), but for <span class="math-container">$n>2$</span> it could break the number's representation of ratios of <span class="math-container">$n$</span> numbers, as we saw above with the fourth roots of unity.</p>
<p>If <span class="math-container">$n$</span> is <strong>odd</strong> then you get <span class="math-container">$n$</span>th root pairs that are conjugates of each other so they devolve into real numbers: <span class="math-container">$a^{\omega_1}*a^{\overline {\omega_1}}=a^{2*\operatorname {Re} (\omega_1)}$</span>. This is okay for <span class="math-container">$n=3$</span> (the trirationals) but for <span class="math-container">$n>3$</span> it could once again break the representation. A representation of say a ratio of five numbers like <span class="math-container">$a:b:c:c:b$</span> that evaluates to a real number is not a very good representation, and is not what a 5-rational number should be.</p>
<p>Later I will try to find a solution for 4-rational and 5-rational numbers outside of the complex plane.</p>
<hr />
<p><strong>Questions:</strong></p>
<ul>
<li><p>While division can be interpreted as a scaling down of a number (when the divisor is a scalar <span class="math-container">$d>1$</span>) or a rotation in the opposite direction (when the divisor has a non-zero imaginary part), is there also a geometric interpretation that could help visualize the ternary <span class="math-container">$\unicode{x25B6}\unicode{x25B6}$</span> operation?</p>
</li>
<li><p>Aside from the trivial case of rational numbers (where <span class="math-container">$a/b=a\unicode{x25B6}b\unicode{x25B6}b$</span>), are there trirational numbers that, when you're given their polar or rectangular form, you can derive or even estimate the trirational form <span class="math-container">$a\unicode{x25B6}b\unicode{x25B6}c$</span> via a simple algorithm?</p>
</li>
<li><p>Does anyone have other ideas for trirational number addition? Perhaps there's another clue I'm missing? In real life we can combine two collections of the same three kinds of things in different ratios, so my intuition tells me a general trirational addition formula should exist.</p>
</li>
</ul>
| Francis Ocoma | 231,608 | <p>If we redefine <span class="math-container">$a\unicode{x25B6}b\unicode{x25B6}c$</span> such that <span class="math-container">$a\in\Bbb{R}$</span> and <span class="math-container">$b,c\in\Bbb{R}_{>0}$</span> instead of being integers, we get the following nifty conversions between the complex polar form and the trirational form:</p>
<p><span class="math-container">$$re^{i\theta}=re^{\frac{\theta}{\sqrt{3}}}\unicode{x25B6}e^{\frac{2\theta}{\sqrt{3}}}\unicode{x25B6}1=re^{\frac{-\theta}{\sqrt{3}}}\unicode{x25B6}1\unicode{x25B6}e^{\frac{-2\theta}{\sqrt{3}}}=r\unicode{x25B6}e^{\frac{\theta}{\sqrt{3}}}\unicode{x25B6}e^{\frac{-\theta}{\sqrt{3}}}$$</span></p>
<p><span class="math-container">$$a\unicode{x25B6}b\unicode{x25B6}c=\frac{a}{\sqrt{bc}}*e^{i\operatorname{ln}\left(\frac{b}{c}\right)\frac{\sqrt{3}}{2}}$$</span></p>
<p>Through this (and Euler's formula) we can convert trirationals to complex rectangular forms, add them, then convert them back to trirational form. A general addition formula for the trirational form can then be derived from this, though it's quite complicated so I will leave it as an exercise to the reader.</p>
<p>Since the denominators <span class="math-container">$b$</span> and <span class="math-container">$c$</span> are always positive, the additive inverse is <span class="math-container">$(-a)\unicode{x25B6}b\unicode{x25B6}c$</span> while the multiplicative inverses are <span class="math-container">${\operatorname{sgn}(a)}bc\unicode{x25B6}{|a|}c\unicode{x25B6}{|a|}b$</span> as well as <span class="math-container">$a^{-1}\unicode{x25B6}b^{-1}\unicode{x25B6}c^{-1}$</span> for any non-zero trirational number.</p>
<p>I think it's clear that this definition of trirational numbers is a field isomorphic to <span class="math-container">$\Bbb{C}$</span>.</p>
<hr />
<p>As for generalizations, an <span class="math-container">$n$</span>-rational number <span class="math-container">$r\unicode{x25B6}d_{1}\unicode{x25B6}d_{2}\unicode{x25B6}\dots\unicode{x25B6}d_{n-1}$</span>, where the numerator <span class="math-container">$r\in\Bbb{R}$</span> and the denominators <span class="math-container">$d_{1},\dots,d_{n-1}\in\Bbb{R}_{>0}$</span> is equal to the following:</p>
<p><span class="math-container">$$r*d_{1}^{\omega_{n}}*d_{2}^{\omega_{n}^{2}}*\dots*d_{n-1}^{\omega_{n}^{n-1}}$$</span></p>
<p>where <span class="math-container">$\{1,\omega_{n},\dots,\omega_{n}^{n-1}\}$</span> is a cyclic group with <strong>affinely independent</strong> elements from an algebra with <span class="math-container">$n-1$</span> dimensions and where</p>
<p><span class="math-container">$$\sum_{x=0}^{n-1}\omega_{n}^{x}=0$$</span></p>
<p>The affine independence of this cyclic group means that for <span class="math-container">$n>2$</span>, none of the elements in the group would be negatives of each other. This solves the problem we saw with roots of unity in the complex plane for any even <span class="math-container">$n>3$</span>, where opposite elements caused havoc in the ratio representation.</p>
<p>The convex hull of this cyclic group is a regular <span class="math-container">$(n-1)$</span>-simplex centered at the origin. I wrote about algebras that contain such cyclic groups in <a href="https://math.stackexchange.com/questions/4528468/highly-polytopic-algebras">this question</a>, which I eventually <a href="https://math.stackexchange.com/a/4539050/231608">answered myself</a>, detailing things like the orthonormal basis and matrix representations of numbers in such algebras. This is vital when studying <span class="math-container">$n$</span>-rational numbers, especially their addition and subtraction.</p>
<p>Regarding division, there are two kinds of multiplicative inverse for a non-zero <span class="math-container">$n$</span>-rational number. Here's one:</p>
<p><span class="math-container">$$(r\unicode{x25B6}d_1\unicode{x25B6}d_2\unicode{x25B6}\dots\unicode{x25B6}d_{n-1})^{-1}=r^{-1}\unicode{x25B6}d_1^{-1}\unicode{x25B6}d_2^{-1}\unicode{x25B6}\dots\unicode{x25B6}d_{n-1}^{-1}$$</span></p>
<p>i.e. we simply take each term's multiplicative inverse. Or we could do it like this:</p>
<p><span class="math-container">$$(r\unicode{x25B6}d_1\unicode{x25B6}d_2\unicode{x25B6}\dots\unicode{x25B6}d_{n-1})^{-1}={\operatorname{sgn}(r)}\left(\prod_{a=1}^{n-1}d_{a}\right)\unicode{x25B6}{|r|}\frac{\prod_{a=1}^{n-1}d_{a}}{d_1}\unicode{x25B6}{|r|}\frac{\prod_{a=1}^{n-1}d_{a}}{d_{2}}\unicode{x25B6}\dots\unicode{x25B6}{|r|}\frac{\prod_{a=1}^{n-1}d_{a}}{d_{n-1}}$$</span></p>
<p>i.e., each term is replaced by the product of all the other terms, except the numerator retains its original sign, and it is the absolute value of <span class="math-container">$r$</span> that's used in the denominators. This ensures that the resulting denominators are positive and that every non-zero <span class="math-container">$n$</span>-rational number has a multiplicative inverse.</p>
<p>Multiplication is component-wise, so multiplying an <span class="math-container">$n$</span>-rational number to its first inverse will immediately turn all terms to <span class="math-container">$1$</span>, while multiplying it to its second inverse will render all terms equal to each other and make the whole thing reducible to <span class="math-container">$1$</span>.</p>
|
3,224,765 | <p>The following question was asked on a high school test, where the students were given a few minutes per question, at most:</p>
<blockquote>
<p>Given that,
<span class="math-container">$$P(x)=x^{104}+x^{93}+x^{82}+x^{71}+1$$</span>
and,
<span class="math-container">$$Q(x)=x^4+x^3+x^2+x+1$$</span>
what is the remainder of <span class="math-container">$P(x)$</span> divided by <span class="math-container">$Q(x)$</span>?</p>
</blockquote>
<hr>
<p>The given answer was:</p>
<blockquote>
<p>Let <span class="math-container">$Q(x)=0$</span>. Multiplying both sides by <span class="math-container">$x-1$</span>:
<span class="math-container">$$(x-1)(x^4+x^3+x^2+x+1)=0 \implies x^5 - 1=0 \implies x^5 = 1$$</span>
Substituting <span class="math-container">$x^5=1$</span> in <span class="math-container">$P(x)$</span> gives <span class="math-container">$x^4+x^3+x^2+x+1$</span>. Thus,
<span class="math-container">$$P(x)\equiv\mathbf0\pmod{Q(x)}$$</span></p>
</blockquote>
<hr>
<p>Obviously, a student is required to come up with a “trick” rather than doing brute force polynomial division. How is the student supposed to think of the suggested method? Is it obvious? How else could one approach the problem?</p>
| Spitemaster | 604,925 | <p>If it's not obvious, an examination of the question quickly reveals the trick. Say</p>
<p><span class="math-container">$$P(x)=x^n$$</span></p>
<p>Then begin long division by <span class="math-container">$Q(x)$</span>:</p>
<p><span class="math-container">$$x^n-x^n-x^{n-1}-x^{n-2}-x^{n-3}-x^{n-4}$$</span>
<span class="math-container">$$x^{n-5}$$</span>
<span class="math-container">$$\dots$$</span>
<span class="math-container">$$x^{n-5k}$$</span></p>
<p>While it may not be obvious just by looking at the question, anyone who attempts the naive solution has (at least) a reasonable chance of running across a way of solving it.</p>
|
990,418 | <p>What is the maximum value of $\sin A+\sin B+\sin C$ in a triangle $ABC$. My book says its $3\sqrt3/2$ but I have no idea how to prove it. </p>
<p>I can see that if $A=B=C=\frac\pi3$ then I get $\sin A+\sin B+\sin C=\frac{3\sqrt3}2$. And also <a href="http://www.wolframalpha.com/input/?i=max+sin(a)%2Bsin(b)%2Bsin(c)+for+a%2Bb%2Bc%3Dpi" rel="nofollow noreferrer">according to WolframAlpha</a> maximum is attained for $a=b=c$. But this does not give me any idea for the proof.</p>
<p>Can anyone help? </p>
| Christian Blatter | 1,303 | <p>The function $f(\alpha,\beta,\gamma):=\sin\alpha+\sin\beta+\sin\gamma$ assumes a maximum on the simplex $$S:=\bigl\{(\alpha,\beta,\gamma)\>\bigm|\>\alpha\geq0, \ \beta\geq0,\ \gamma\geq0,\ \alpha+\beta+\gamma=\pi\bigr\}\ .$$
On the other hand, if $\alpha>\beta\geq0$ one has
$$\sin\alpha+\sin\beta=2\sin{\alpha+\beta\over2}\cos{\alpha-\beta\over2}<2\sin{\alpha+\beta\over2}\ ,$$
and $\bigl({\alpha+\beta\over2},{\alpha+\beta\over2},\gamma\bigr)\in S$.</p>
<p>This allows to conclude that ${\rm argmax}_S f=\bigl({\pi\over3},{\pi\over3},{\pi\over3}\bigr)$, so that$$f(\alpha,\beta,\gamma)\leq {3\sqrt{3}\over2}\qquad\bigl((\alpha,\beta,\gamma)\in S\bigr)\ .$$</p>
|
2,692,957 | <p>I am interested in the isometries of the hyperbolic plane in the Beltrami–Klein disk model. (<a href="https://en.wikipedia.org/wiki/Beltrami%E2%80%93Klein_model" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Beltrami%E2%80%93Klein_model</a>) The Wikipedia article does not say anything about the structure of the isometries in this model.</p>
<p>Since the isometries in the upper half-plane model are well-known, I did a change of variables to go from upper half-plane to the Beltrami–Klein disk. After some calculations, I determined that every isometry in the Beltrami-Klein disk model must be a projective transformation $\mathbb R \mathbb P^2 \to \mathbb R \mathbb P^2$ which maps the "unit disk" $\{ [x:y:1] \in \mathbb R \mathbb P^2 ~|~ x^2 + y^2 < 1 \}$ to itself. (The unit disk is, after all, the disk in the Beltrami-Klein disk model of hyperbolic space.)</p>
<p>Some questions I have:</p>
<ol>
<li>Is every projective transformation of this form an isometry of the hyperbolic plane? (Is there an easy way to see this?)</li>
<li>The set of such projective transformations is a subgroup of $SL(3,\mathbb R)$. Is there a nice characterization of this group?</li>
</ol>
<p>Update: I just realized that every projective transformation which maps the unit disk to itself must be an element of $SO(2,1)$. Here is the reason: such a projective transformation must map the unit circle to itself, so the corresponding linear transformation $\mathbb R^3 \to \mathbb R^3$ must preserve the quadratic form $x^2+y^2-z^2$. Therefore, if the answer to Question 1 is "yes," then the answer to Question 2 should be $SO(2,1)$.</p>
| Lee Mosher | 26,501 | <p>As you have discovered, the answers to your questions are "yes". You can read about this in <a href="http://library.msri.org/books/gt3m/" rel="nofollow noreferrer">Thurston's notes</a> which explains in detail the relations between various models of the hyperbolic plane, including a step-by-step way to get between any two of those models such as the upper half plane model and the Beltrami-Klein disc model.</p>
<p>There is also a more direct but more abstract way to go between those two models. You need a theorem of differential geometry which says that any two complete, simply connected, Riemannian manifolds of dimension 2 and of curvature -1 are isometric. </p>
<p>One of the two models is the upper half plane $\mathbb{H}$ with its metric $\frac{dx^2+dy^2}{y^2}$. The group $PSL(2,\mathbb{R})$ acts on $\mathbb{H}$ by orientation preserving isometries, and this action is transitive on positively oriented orthonormal frames, hence it is the entire isometry group.</p>
<p>The other one is the hyperboloid model. The underlying space of the hyperboloid model is one sheet of the two-sheeted hyperbola $x^2+y^2-z^2=-1$. The metric on that space is the restriction of the Lorentz matrix $dx^2 + dy^2 - dz^2$: although that is not a Riemannian metric on $x,y,z$ space, when you restrict it to the tangent planes of the one-sheeted hyperboloid it does become a Riemannian metric on that hyperboloid, and it is complete and of sectional curvature $-1$. One can show pretty easy that the action of $SO(2,1)$ restricts to an isometry action on the one-sheeted hyperboloid. That action is transitive on oriented orthonormal frames, hence it is the entire isometry group.</p>
<p>The differential geometry theorem I quoted above then produces an isometry between the upper half plane model and the one-sheeted hyperboloid model (as said, Thurston's notes gives a more explicit construction of an isometry).</p>
<p>Finally, to get from the hyperboloid model to the Beltrami-Klein disc model, one places that disc in the $z=1$ plane, and one uses rays through the origin to project from the one-sheeted hyperboloid to the disc.</p>
|
3,082,337 | <p>Forgive my ignorance.<br>
Is the condition <span class="math-container">$x\in\mathbb{R}$</span> necessary to the set statement <span class="math-container">$\{x \in\mathbb{R} \vert x> 0\}$</span>?<br>
In other words, if <span class="math-container">$x$</span> is greater than zero, then is it not, by definition, a real number?<br>
Thank you very much!</p>
| Community | -1 | <p>One more point. It might not even make sense to say <span class="math-container">$x>0$</span> unless <span class="math-container">$0$</span> and <span class="math-container">$>$</span> are defined. </p>
<p>For example the set <span class="math-container">$\{ x \in \Bbb C : x >0\}$</span> doesn't make sense as <span class="math-container">$>$</span> is not defined (consistent with addition and multiplication) on <span class="math-container">$\Bbb C$</span>. </p>
<p>Or even worse, consider the set of all states in the USA. Does asking if a state is "greater than zero" make sense? </p>
|
4,090,609 | <p>Not any even rank real vector bundle over a smooth manifold has a complex structure, because there are even dimensional manifolds that have no almost complex structures. I am curious about the following special case: Suppose <span class="math-container">$M$</span> is a real compact, connected smooth manifold and <span class="math-container">$E\to M$</span> is an orientable real <span class="math-container">$2$</span>-plane bundle. Also suppose, for some connected open subset <span class="math-container">$U$</span>, <span class="math-container">$E|_U$</span> has a complex structure (so that it can be seen as a complex line bundle). In this case can we extend the complex structure on <span class="math-container">$E|_U$</span> to <span class="math-container">$E$</span>?</p>
| Georges Elencwajg | 3,217 | <p>Here are some results concerning the possibility of giving the structure of a complex vector bundle to a differentiable real vector bundle <span class="math-container">$E$</span> of rank <span class="math-container">$2r$</span> on the differentiable manifold <span class="math-container">$M$</span> .<br />
The principal tool here consists of the Stiefel-Whitney classes <span class="math-container">$w_i(E)\in H^i(M,\mathbb Z/2)$</span>.
<strong>First necessary condition: odd Stiefel-Whitney classes</strong><br />
If <span class="math-container">$E$</span> has a complex structure, then all odd Stiefel-Whitney classes vanish: <span class="math-container">$$w_{2k+1}(E)=0$$</span> In particular <span class="math-container">$w_1(E)=0$</span>, which is equivalent to <span class="math-container">$E$</span> being orientable.<br />
So we retrieve the obvious necessary condition that <span class="math-container">$E$</span> be orientable.<br />
<strong>Second necessary condition: even Stiefel-Whitney classes</strong><br />
Let <span class="math-container">$\mathcal E$</span> be a complex vector bundle with underlying real bundle <span class="math-container">$\mathcal E_\mathbb R=E$</span>.<br />
Then <span class="math-container">$\mathcal E$</span> has Chern classes <span class="math-container">$c_j(\mathcal E)\in H^{2j}(M,\mathbb Z)$</span> and we have <span class="math-container">$w_{2j}(E)=\operatorname {nat}(c_j(\mathcal E))$</span>, where <span class="math-container">$\operatorname {nat}$</span> is the natural change of coefficients morphism <span class="math-container">$\operatorname {nat}:H^{2j}(M,\mathbb Z)\to H^{2j}(M,\mathbb Z/2)$</span>.<br />
This gives the necessary condition that the even Stiefel-Whitney classes of <span class="math-container">$E$</span> should be liftable from <span class="math-container">$\mathbb Z/2$</span> to <span class="math-container">$\mathbb Z$</span> coefficients.<br />
An example of a non liftable Stiefel-Whitney class <span class="math-container">$w_2(E)$</span> of a <span class="math-container">$4$</span>-dimensional vector bundle <span class="math-container">$E$</span> has been given by Bertram Arnold <a href="https://mathoverflow.net/a/389637/450">here</a>, and that bundle can thus not be given a complex structure.<br />
<strong>Warning</strong><br />
These necessary conditions are not sufficient for a real vector bundle to admit of a complex structure.<br />
Many partial results are known but the general problem is (as far as I know) open.<br />
<strong>Bibliography</strong><br />
The basics of characteristic classes are contained in Hatcher's freely available online document <a href="https://pi.math.cornell.edu/%7Ehatcher/VBKT/VB.pdf" rel="nofollow noreferrer">Vector Bundles and K-Theory</a>.</p>
|
960,026 | <p>I'm having some trouble finding out how to calculate the probability of the following problem:</p>
<p><img src="https://i.stack.imgur.com/1VdCX.png" alt="enter image description here"></p>
<p>I'm confused as to how I should be calculating the probability of B. I'm not sure how to factor in the possibility of A winning in one state but B winning in another. </p>
| Mark Fischler | 150,362 | <p>The answer your instructor or book is looking for is that there is a 50% chance that B wins Kentucky, and if he does not, then there is a 28% chance that he wins Tennessee. The total probability that B wins at least one is then 50% plus (1 - 50%) times 28%, for a total of 64%.</p>
<p>There are some subtleties for the purist; for example, the conditions given are not unabiguously consistent. But it is clear you are not meant to overthink the matter. </p>
|
2,685,822 | <p>How can we prove that $L = \lim_{n \to \infty}\frac{\log\left(\frac{n^n}{n!}\right)}{n} = 1$</p>
<p>This is part of a much bigger question however I have reduced my answer to this, I have to determine the limit of $\log(n^{n}/n!)/n$ when $n$ goes to infinity.</p>
<p>Apparently the answer is 1 by wolfram alpha but I have no clue how to get it. Any idea how I could proceed (without sirlings approximation as well).</p>
| Atmos | 516,446 | <p>Applying Stirling's formula
$$
n! \underset{(+\infty)}{\sim} \sqrt{2n \pi}\left(\frac{n}{e}\right)^n
$$
Hence
$$
\frac{n^n}{n!}\underset{(+\infty)}{\sim} \frac{n^n}{\sqrt{2n \pi}\left(\frac{n}{e}\right)^n}=\frac{e^n}{\sqrt{2n \pi}}
$$
Then
$$
\ln\left(\frac{n^n}{n!}\right)\underset{(+\infty)}{=}n\ln\left(e\right)-\frac{1}{2}\ln\left(2n\pi\right)+o\left(1\right)
$$
So with $ln(e)=1$
$$
\frac{\ln\left(\frac{n^n}{n!}\right)}{n}=1-\frac{\ln(2n\pi)}{2n}+o\left(1\right)$$</p>
<p>here's your result</p>
|
718,590 | <p>Is it true that in general complete metric space $(M,d)$, a closed ball of radius $r$ centered at $p\in M$ is always compact? That is, the ball is the set of all points $\left\{x:d(x,p)\leq r\right\}$.</p>
| wckronholm | 10,449 | <p>No. If $(M,d)$ is an infinite set endowed with the discrete metric $d$, then $M$ is the closed ball around any one of its points of radius $1$. However, $M$ is not compact.</p>
<p>Given any metric space $(M,d)$, you can define a new metric space $(M,d')$ by $d'(x,y) = \min\{d(x,y),1\}$. These metric spaces have the same topology. But $M$ is now bounded, so if $M$ is not compact with the metric $d$ then it is still not compact with the metric $d'$.</p>
|
718,590 | <p>Is it true that in general complete metric space $(M,d)$, a closed ball of radius $r$ centered at $p\in M$ is always compact? That is, the ball is the set of all points $\left\{x:d(x,p)\leq r\right\}$.</p>
| Dan Rust | 29,059 | <p>Any metric space $(M,d)$ can be given a new metric which induces the same topology, but in which the entire space has finite diameter by setting $$d'(x,y)=\frac{d(x,y)}{1+d(x,y)}.$$ It follows that if $M$ is not compact, then all balls with radius larger than $1$ in this new metric $d'$ must also be non-compact.</p>
|
470,506 | <p>When is this true?
$$\lim_{r\to 0}\int_{-K}^K f(rx)dx=\int_{-K}^K \lim_{r\to 0} f(rx)dx$$
Is it true without the hypothesis of continuity of
$f$? </p>
<p>Thank you.</p>
| detnvvp | 85,818 | <p>The problem is here:</p>
<p>"As there are infinite sequences converging to points of the form $\left(\frac{1}{2^k},0\right)$, and $J$ is finite, $J$ can contain points from only a finite number of them."</p>
<p>Here $J$ is finite, but the balls you mention will contain finally all the terms of all the sequences converging to that limit point.</p>
|
1,046,961 | <p>Find all continuous functions $f:\mathbb{R} \to \mathbb{R}$ such that for all $x \in \mathbb{R}$, $f(x) + f(2x) = 0$ <br/>
I'm thinking; <br/>
Let $f(x)=-f(2x)$ <br/>
Use a substitution $x=y/2$ for $y \in \mathbb{R}$. <br/>
That way $f(y)=-f(y/2)=-f(y/4)=-f(y/8)=....$ <br/>
Im just not sure if this is a good approach. Opinions please..</p>
| Henry | 6,460 | <p>Hints:</p>
<ul>
<li>What is $f(0)$?</li>
<li>What does continuity when $x=0$ mean?</li>
<li>What bounds does this put on $f(x)$ for general $x$?</li>
</ul>
|
4,240,794 | <p>I was trying to obtain the square root of a matrix through the eigenvalues and eigenvectors, and there is something that doesn't add up in some of the demonstrations that I observed after getting stuck.</p>
<p>So, being <span class="math-container">$Q$</span> the eigenvector column matrix of <span class="math-container">$A$</span> and <span class="math-container">$\Lambda$</span> a diagonal matrix with the eigenvalues of <span class="math-container">$A$</span>, we can put <span class="math-container">$A$</span> as:</p>
<p><span class="math-container">$$ A = Q \Lambda Q^{-1} $$</span></p>
<p>From this is easily deducible that:</p>
<p><span class="math-container">$$ A^n = (Q \Lambda Q^{-1})(Q \Lambda Q^{-1})...(Q \Lambda Q^{-1}) = Q \Lambda^n Q^{-1} $$</span></p>
<p>But the demonstration of this only seems valid for all the natural numbers and I cannot see how it can extend directly for the integer or rational numbers.</p>
| Robert Lee | 695,196 | <p>You have a slight mistake in your calculations. You should have
<span class="math-container">\begin{align*}
&-2 xe^{-2x}- e^{-2x} = -2 ye^{-2y}- e^{-2y} \\
\color{blue}{\implies} &-(2x+1) e^{-2x}\cdot \frac{1}{e} = -(2y+1) e^{-2y}\cdot \frac{1}{e} \\
\color{blue}{\implies} &-(2x+1) e^{-(2x+1)} = -(2y+1) e^{-(2y+1)}
\end{align*}</span>
So although the question does reduce to checking when <span class="math-container">$xe^{x}$</span> is invertible, the argument, in this case, is <span class="math-container">$-(2x+1)$</span>, and not just <span class="math-container">$-2x$</span>.</p>
<hr />
<p>The function <span class="math-container">$xe^{x}$</span> is not one-two-one for <em>all</em> real values of <span class="math-container">$x$</span>. However, it <em>is</em> one-two-one if you restrict the domain.</p>
<p>If you take <span class="math-container">$-(2x+1),-(2y+1) \ge -1 \color{blue}{\implies} x,y >0$</span> then the principal branch of the <a href="https://en.wikipedia.org/wiki/Lambert_W_function" rel="nofollow noreferrer">Lambert-W function</a> <span class="math-container">$W_0$</span> does give you an inverse for <span class="math-container">$xe^{x}$</span>, and thus, on the interval <span class="math-container">$\left[0,\infty\right)$</span> your function is one-two-one.</p>
<p>Similarly, if your domain is <span class="math-container">$(-\infty, 0]$</span> the function <span class="math-container">$W_{-1}$</span> gives you an inverse. But if you're taking the <em>entire</em> real number line as your domain, a simple plot of the function combined with the horizontal-line test tells you the function is not one-to-one on this domain.</p>
<p><a href="https://i.stack.imgur.com/Y2AWW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Y2AWW.png" alt="enter image description here" /></a></p>
|
2,411,568 | <p>I would like to find all $(a,b,c)\in\mathbb{N}^3$ such that $a\times 1111+b\times 111+c\times 11=9002$.</p>
<p>It is obvious that $a+b+c \equiv 2 \pmod{10}$. I made a little computer program, that basically looped over all possiblities and I found no answer, so I just have to prove it...<br>
I am thinking of an inductive solution, something like:<br>
If no solution is found for $a=a_{1}$, then there is no solution for $a=a_{1}+1$</p>
| Joffan | 206,402 | <p>Working $\bmod 11$, you have:<br>
$9002\equiv 4$<br>
$11\equiv 0$<br>
$111\equiv 1$<br>
$1111\equiv 0$ </p>
<p>So we need $b\equiv 4\bmod 11$ (so assuming $a,b,c$ are non-negative, we have $b\in \{4,15,\ldots,81\}$). Any of these $b$ values produces one or more solutions; you can assume $a=0$, solve directly for $c$ and then find solutions with larger $a$ (if the initial $c$ is big enough).</p>
|
2,411,568 | <p>I would like to find all $(a,b,c)\in\mathbb{N}^3$ such that $a\times 1111+b\times 111+c\times 11=9002$.</p>
<p>It is obvious that $a+b+c \equiv 2 \pmod{10}$. I made a little computer program, that basically looped over all possiblities and I found no answer, so I just have to prove it...<br>
I am thinking of an inductive solution, something like:<br>
If no solution is found for $a=a_{1}$, then there is no solution for $a=a_{1}+1$</p>
| Barry Cipra | 86,747 | <p>Rewrite the equation as</p>
<p>$$1000a+100(a+b)+11(a+b+c)=9002$$</p>
<p>and now look for solutions of $1000A+100B+11C=9002$ with $0\le A\le B\le C$ (or $0\lt A\lt B\lt C$ if you want <em>positive</em> integer solutions). From $11C\equiv2$ mod $100$, we see $-C\equiv99C\equiv18$ mod $100$, so $C\equiv82$ mod $100$. Thus, since $882\cdot11=9702\gt9002$, we have</p>
<p>$$C\in\{82,182,282,382,482,582,682,782\}$$</p>
<p>From these we get</p>
<p>$$10A+B=
\begin{cases}
81\\70\\59\\48\\37\\26\\15\\4
\end{cases}$$</p>
<p>and appropriate values for $0\le A\le B$ can be easily worked out, and then values of $(a,b,c)$ obtained. For example $(A,B,C)=(7,11,82)$ corresponds to $(a,b,c)=(7,4,71)$.</p>
|
1,532,401 | <blockquote>
<p>If $X\sim \exp(\lambda)$, then $E[X\mid X>20]=20+E[X]$</p>
</blockquote>
<p>Can anyone give an intuitive proof of this property? In the context that $X$ is exponential and thereby memoryless. I am having trouble in visualizing continuous memoryless distributions. Any intuitive example for the same is also appreciated. </p>
| Ross Millikan | 1,827 | <p>The intuitive answer is that memoryless means that if you are waiting for an event that has not yet happened, it doesn't matter how long you have been waiting. If you get to $x=20$ and it hasn't happened, the expected time until it happens is the same from now as it was at the start. Since you have already waited $20$ so far, the expected total time is $20+E(X)$. The fact that we are now at $20$ and it hasn't happened is the definition of the $|X\gt 20]$ in your expected value, so $E[X|X \gt 20]=20+E[X]$ </p>
<p>To do it mathematically, the pdf of the <a href="https://en.wikipedia.org/wiki/Exponential_distribution" rel="nofollow">exponential distribution</a> is $\lambda e^{-\lambda x}$. The chance it doesn't happen by $x=20$ is $1-\int_0^{20}\lambda e^{-\lambda x}dx=e^{-20 \lambda}$ Plug this into your formula and observe that $E[X|X \gt 20]=20+E[X]$</p>
|
1,532,401 | <blockquote>
<p>If $X\sim \exp(\lambda)$, then $E[X\mid X>20]=20+E[X]$</p>
</blockquote>
<p>Can anyone give an intuitive proof of this property? In the context that $X$ is exponential and thereby memoryless. I am having trouble in visualizing continuous memoryless distributions. Any intuitive example for the same is also appreciated. </p>
| Michael Hardy | 11,667 | <p>Memorylessness says the conditional distribution of $X-20$ given $X>20$ is the same as the distribution of $X$.</p>
<p>Therefore the conditional expected value of $X-20$ given $X>20$ is the same as the expected value of $X$:</p>
<p>$$
\operatorname{E} (X - 20 \mid X>20) = \operatorname{E}(X).
$$</p>
<p>But
$$
\operatorname{E} (X - 20 \mid X>20) = \operatorname{E} (X \mid X>20) - \operatorname{E} (20 \mid X>20) = \operatorname{E} (X \mid X>20) - 20.
$$
Hence
$$
\operatorname{E} (X) = \operatorname{E} (X-20 \mid X>20) = \operatorname{E} (X \mid X>20) - 20.
$$
So
$$
\operatorname{E} (X) + 20 = \operatorname{E} (X \mid X>20).
$$</p>
|
248,658 | <p>Based on the next relation:</p>
<p>$$\det\begin{bmatrix}A & B \\ C & D\end{bmatrix} = \det(A)\det(D - CA^{-1}B),$$</p>
<p>I have that for computing the eigenvalues of the block matrix:</p>
<p>$$\det\begin{bmatrix}A-\lambda I & B \\ C & D-\lambda I\end{bmatrix} = \det(A-\lambda I)\det((D-\lambda I) - C(A-\lambda I)^{-1}B) = 0$$</p>
<p>So $\det(A - \lambda I) = 0$ says that the eigenvalues of $A$ are eigenvalues of the block matrix? But from some numerical simulations I have found that this is not true, what am I missing here? Maybe is because the first relation requires $A$ nonsingular and $A-\lambda I$ is not? </p>
<p>Then, this leads me to another question, why this expression holds
$$\det\begin{bmatrix}A-\lambda I & 0 \\ C & D-\lambda I\end{bmatrix} = \det(A-\lambda I)\det((D-\lambda I) = 0$$
for stating that the eigenvalues of the block matrix are the eigenvalues of $A$ and $D$ if $A-\lambda I$ is singular?</p>
<p>Many thanks in advance.</p>
| Hans Engler | 9,787 | <p>If $\det(A - \lambda I) = 0$, then you cannot form $(A - \lambda I)^{-1}$, which appears also in the formula. Hence the formula does not apply, unless $B = 0$ or $C = 0$, the case of a block tridiagonal matrix.</p>
|
2,907,378 | <p>In my Math book I'm solving a case where this is the situation:</p>
<p>"The demand curve for good X is linear. At a price (p) of 300 the demand is 600 units. At a price of 680 the demand is 220 units. Also the supply curve for good X is linear. If the price is 400 then the supply equals 200 units, whereas for a price of 800 the supply will be 1000 units."</p>
<p>I'm asked to formulate the system of equations.</p>
<p>For the demand function I did:</p>
<p><code>SLOPEqd = delta y / delta x = -1</code> (<code>delta y 680-300, delta x 220-600</code>)</p>
<p>Which I could use and verify by the coordinates and I got the formula <code>Demand(x) = -x + 900</code>)</p>
<p>If I do the same for the supply function I end up wrong:</p>
<p><code>SLOPEqs = delta y / delta x = 0.5</code> (<code>delta y 800-400, delta x 1000-200</code>)</p>
<p>If I try to verify this for the first supply point (800, 1000) I would get:</p>
<p><code>Supply(x) = 0,5x + 600</code> but this is incorrect for the second point (400, 200). In the answers I found the slope for the supply function should be <code>2</code> instead of <code>0,5</code>. Why is this? I think I'm overlooking something super obvious, but please enlighten me.</p>
<p>(sorry, I haven't figured out how to write the equations as fancy as I see them in other questions)</p>
| David K | 139,123 | <p>For supply you have computed the following:</p>
<blockquote>
<p><code>SLOPEqs = delta y / delta x = 0.5</code> (<code>delta y 800-400, delta x 1000-200</code>)</p>
<p>If I try to verify this for the first supply point (800, 1000) I would get:</p>
<p><code>Supply(x) = 0,5x + 600</code></p>
</blockquote>
<p>Presumably, this is because you think the point <span class="math-container">$(x,y) = (800,1000)$</span> should lie on the graph of this function.</p>
<p>Now look carefully. In one place you have <span class="math-container">$\Delta y = 800 - 400,$</span>
in another you have <span class="math-container">$x = 800.$</span> But the <span class="math-container">$800$</span> comes from the exact same given quantity (a price on the supply curve) in both places. So is it a value of <span class="math-container">$x$</span> or a value of <span class="math-container">$y$</span>?</p>
<p>If you decide prices should be <span class="math-container">$x$</span> in both places, then you'll have
<span class="math-container">$\Delta x = 800 - 400 = 400,$</span>
<span class="math-container">$\Delta y = 1000 - 200 = 800,$</span> and slope <span class="math-container">$\Delta y/ \Delta x = 2.$</span>
Then when you try <span class="math-container">$(x,y) = (800,1000)$</span> you'll get
<span class="math-container">$$ S(x) = y = 2x - 600. $$</span></p>
<p>On the other hand, if you say price should be <span class="math-container">$y$</span> in both places, you'll still have the slope <span class="math-container">$0.5,$</span> but now you're trying to fit a function to the point
<span class="math-container">$(x,y) = (1000,800),$</span> from which you get
<span class="math-container">$$ y = 0.5x + 300 = 0.5 S(y) + 300. $$</span></p>
<p>As noted in another answer, the conventions of economics are confusing in the regard that they tend to plot the supply and demand <em>curves</em> with quantity on the horizontal axis and price on the vertical axis,
yet the supply and demand <em>functions</em> give quantity as a function of price.
My recommendation to minimize the confusion about this is:</p>
<p>Never use <span class="math-container">$x$</span> or <span class="math-container">$y$</span> to denote a price or quantity (of supply or demand).
Always use <span class="math-container">$P$</span> for price and <span class="math-container">$Q$</span> for quantity (possibly with subscripts) when you are naming variables.</p>
|
65,220 | <p>Si $f$ es una función continua y si $f(m/2^n)=0$ para todo entero $m$ y todo natural $n$, ¿cómo demuestro que $f(x)=0$ para todo numero real $x$?</p>
<hr/>
<p><em>[translation by mixedmath]</em></p>
<p>If $f$ is a continuous function and if $f\left(\dfrac{m}{2^n}\right) = 0$ for all integers $m$ and all natural $n$, how do I show that $f(x) = 0$ for all real $x$?</p>
| Brian M. Scott | 12,042 | <p>Let $D = \left\{\frac{m}{2^n}:m\in\mathbb{Z} \land n\in\mathbb{N}\right\}$. (Members of $D$ are called <em>dyadic rationals</em>.) Since $f$ is continuous, you know that $\lim\limits_{n\to\infty}f(x_n)=f(x)$ whenever $\lim\limits_{n\to\infty}x_n=x$. Let $x$ be any real number. If we can find a sequence $\langle x_n:n\in\mathbb{N} \rangle$ of dyadic rationals such that $\lim\limits_{n\to\infty}x_n=x$, we’ll be done, because then $f(x) = \lim\limits_{n\to\infty}f(x_n)=0$.</p>
<p>One way to do this is to use the binary expansion of $x$. Suppose that $\displaystyle x = \sum\limits_{k=m}^{\infty} b_k2^{-k}$ for some integer $m\le 1$, where each $b_k$ is either $0$ or $1$. For each $n\in \mathbb{N}$ let $\displaystyle x_n = \sum\limits_{k=m}^n b_k2^{-k}$. Then $$x_n = \frac{\sum_{k=m}^n b_k 2^{n-k}}{2^n} \in D,$$ and it's clear that $\lim\limits_{n\to\infty}x_n=x$.</p>
<p>(I’m sorry that I can’t translate this into Spanish.)</p>
|
65,220 | <p>Si $f$ es una función continua y si $f(m/2^n)=0$ para todo entero $m$ y todo natural $n$, ¿cómo demuestro que $f(x)=0$ para todo numero real $x$?</p>
<hr/>
<p><em>[translation by mixedmath]</em></p>
<p>If $f$ is a continuous function and if $f\left(\dfrac{m}{2^n}\right) = 0$ for all integers $m$ and all natural $n$, how do I show that $f(x) = 0$ for all real $x$?</p>
| André Nicolas | 6,312 | <p><strong>Added</strong>: The question changed while the answer below was being typed. A short note is added at the end to answer the altered question.</p>
<p>The wording of the original question does not accurately convey the intention. I believe that the OP <em>wants</em> a proof that every real number $x$ is the limit of a sequence
$(a_k)$, where each $a_k$ is of the form $m/2^n$, with $m$ an integer and $n$ a non-negative integer. (Numbers of the form $m/2^n$ are called <em>dyadic rationals</em>.)</p>
<p>It is enough to show that for any positive integer $k$, we can find a number $a_k$ of the required form such that $|x-a_k|<1/k$.</p>
<p>We prove the result for $x>0$. A small modification takes care of the case $x<0$.</p>
<p>Let $n$ be the smallest integer such that $2^n>k$. Consider the numbers $0/2^n$, $1/2^n$, $2/2^n$, $3/2^n$, and so on. </p>
<p>Let $m$ be the smallest non-negative integer such that $(m+1)/2^n>x$. Then $m/2^n \le x$. Let $a_k=m/2^n$. Since
$$\frac{m+1}{2^n} -\frac{m}{2^n}=\frac{1}{2^n} <\frac{1}{k},$$
it follows that
$$0\le x-\frac{m}{2^n}=x-a_k <\frac{1}{k}.$$ </p>
<p><strong>Added Note</strong>: The question was changed and now asks one to show that if $f$ is a continuous function which is $0$ on the dyadic rationals, then $f$ is $0$ everywhere.</p>
<p>Let $x$ be a real number. We show that $f(x)=0$. Let $(a_k)$ be a sequence of dyadic rationals with limit $x$. Such a sequence exists by the calculation above. Then
$$f(x)=\lim_{k\to \infty} f(a_k) =0.$$</p>
|
41,450 | <p>I want to perform a Pearson's $\chi^2$ test to analyse contingency tables; but because I have small numbers, it is recommended to perform instead what is called a Fisher's Exact Test.</p>
<p>This requires generating all integer matrices with the same column and row totals as the one given, and compute and sum all p-values from the corresponding distribution which are lower than the one from the data.</p>
<p>See <a href="https://en.wikipedia.org/wiki/Fisher%27s_exact_test" rel="nofollow noreferrer">Wikipedia</a> and <a href="http://mathworld.wolfram.com/FishersExactTest.html" rel="nofollow noreferrer">MathWorld</a> for relevant context.</p>
<p>Apparently <em>R</em> offers that, but couldn't find it in <em>Mathematica</em>, and after extensive research couldn't find an implementation around, so I did my own.</p>
<p>The examples in the links are with 2x2 matrices, but I did a n x m implementation and, at least for the <em>MathWorld</em> example, numbers match.</p>
<p>I have one question: The code I wrote uses <code>Reduce</code>; although it seemed to me generating all matrices was more a combinatorial problem. I pondered using <a href="http://reference.wolfram.com/mathematica/ref/FrobeniusSolve.html" rel="nofollow noreferrer"><code>FrobeniusSolve</code></a>, but still seemed far from what's needed. Am I missing something or is <code>Reduce</code> the way to go?</p>
<p>The essential part of the code, which I made available in github <a href="https://github.com/carlosayam/fisher-exact/blob/master/package/fisher_exact.m" rel="nofollow noreferrer">here</a>, is that for a matrix like</p>
<p>$$
\left(
\begin{array}{ccc}
1 & 0 & 2 \\
0 & 1 & 2 \\
\end{array}
\right)$$</p>
<p>with row sums 3, 3 and column sums 1, 1, 4, it creates a system of linear equations like:</p>
<p>$$
\begin{array}{c}
x_{1,1}+x_{1,2}+x_{1,3}=3 \\
x_{2,1}+x_{2,2}+x_{2,3}=3 \\
\end{array}
$$
$$
\begin{array}{c}
x_{1,1}+x_{2,1}=1 \\
x_{1,2}+x_{2,2}=1 \\
x_{1,3}+x_{2,3}=4 \\
\end{array}
$$</p>
<p>subject to the constrains $ x_{1,1}\geq 0$, $x_{1,2}\geq 0$, $x_{1,3}\geq 0$, $x_{2,1}\geq 0$, $x_{2,2}\geq 0$, $ x_{2,3}\geq 0 $ and feeds this into <code>Reduce</code> to solve this system over the <code>Integers</code>. <code>Reduce</code> returns all the solutions, which is what we need to compute Fisher's exact p-value.</p>
<p><em>Note:</em> I just found <a href="https://mathematica.stackexchange.com/questions/26174/recommended-settings-for-git-when-using-with-mathematica-projects">this advice</a> on how to use github better for Mathematica projects. For the time being, I leave it as-is. Hope easy to use and test.</p>
<p>You can test the above mentioned code like</p>
<pre><code>FisherExact[{{1, 0, 2}, {0, 0, 2}, {2, 1, 0}, {0, 2, 1}}]
</code></pre>
<p>It has some <em>debugging</em> via <code>Print</code> which shows all the <em>generated</em> matrices and their p-value. The last part (use of <code>Select</code>) to process all found matrices didn't seem very <em>Mathematica</em> to me, but it was late and I was tired - feedback is welcome.</p>
<p>I would give my tick to the answer with more votes after a couple of days if anyone bothers to write me two lines :)</p>
<p>Thanks in advance!</p>
| Romke Bontekoe | 1,178 | <p>Maybe you are willing to consider a Bayesian approach to this perennial problem. Beware though: Bayesians have no random variables, no p-values, no null hypotheses, etc. They have probabilities, or ratios thereof.</p>
<p>The (out of print) book "Rational Descriptions, Decisions and Designs" by Miron Tribus (1969!) has an excellent chapter on contingency tables. From this book I have copied the solutions below. His solutions are exact and work for small counts as well as non-square tables.
He considers two mutially exclusive hypotheses: "the rows and columns are independent" vs "the rows and columns are dependent", under a variety of different types of knowledge. </p>
<p>Here I give only two cases: </p>
<p>-- Knowledge type 1A, with no specific prior knowledge on the (in-)dependence and no controls,</p>
<p>-- Knowledge type 1B, also with no specific prior knowledge but with a control on the counts of the rows (see examples below). </p>
<p>Tribus computes the "evidence in favor of the hypothesis of independence of rows and columns" for these types. (The references in the code are to chapters and pages in his book.)</p>
<p>The evidence for type 1A is:</p>
<pre><code>(* Evidence rows-cols independent: 1A VI-38 p. 194 *)
evidence1A[table_] :=
Module[{r, s, nidot, ndotj, ntot, ev, prob},
(* Table dimensions r=nr of rows, s=nr of cols *)
{r, s} = Dimensions[table];
(* Margin and Total counts *)
nidot = Total[table, {2}] ;(* sum in r-direction *)
ndotj = Total[table, {1}] ;(* sum in s-direction *)
ntot = Total[table, 2]; (* overall total *)
(* evidence of VI-38 p.194 *)
ev = Log[ ((ntot + r*s - 1)! * ntot!)/
((ntot + r - 1)!*(ntot + s - 1)!)] -
Log[ (r*s - 1)!/((r - 1)!*(s - 1)!) ] +
(Total[ Log[ nidot!]] - Log[ntot!]) +
(Total[ Log[ ndotj!]] - Log[ntot!]) -
(Total[Log[table!], 2] - Log[ntot!]);
(* probability from evidence: III-13 p.84 *)
prob = (1 + Exp[-ev])^-1 ;
{ev // N, prob // N} (* output *)
] (* End of Module *)
</code></pre>
<p>Tribus tests this using an interesting example of eye-color vs hair-color correlation of soldiers in conscription military service (!). Note that this is a 3x4 table.</p>
<pre><code>(* Soldier table VI-1 p.183: eye color vs. hair color *)
soldier = {
(* blonde,brown,black,red *)
(* blue *) {1768, 807, 189, 47},
(* green *) {946, 1387, 786, 53},
(* brown *) {115, 438, 288, 16}};
(* Tribus p.197 gives 560 Napiers *)
(* prob that the table is row-col independent *)
evidence1A[soldier]
(* output: {-560.661, 3.22157*10^-244} *)
</code></pre>
<p>The probability of independence of rows and columns is 3.22*10^-244, and thus virtually zero. As expected.</p>
<hr>
<p>The case 1B applies to tests with a pre-set count for the columns. In Tribus' tobacco test flavor example: 250 packages with mixed cigarettes + pipe tobacco vs. 150 packages with only cigarettes. </p>
<pre><code>(* Tobacco problem p.198 : solution is independent of s *)
tobacco = {
(* cigaret+pipe tobacco: mixed, not mixed *)
(* no change *) {72, 119},
(* change aroma *) {178, 31}
(* fixed counts : {250,150} *)
};
</code></pre>
<p>The evidence for this problem is:</p>
<pre><code>(* Evidence rows-cols independent: 1B VI-54 p. 200 *)
(* solution is independent of s *)
evidence1B[table_] :=
Module[ {r, s, nidot, ndotj, ntot, ev, prob},
(* Table dimensions r=nr of rows, s=nr of cols *)
{r, s} = Dimensions[table];
(* Margin and Total counts *)
nidot = Total[table, {2}] ;(* sum in r-direction *)
ndotj = Total[table, {1}] ;(* sum in s-direction *)
ntot = Total[table, 2]; (* overall total *)
(* evidence Eq.VI-54 p.200 *)
ev = Log[(r - 1)!/(ntot + r - 1)!] +
Total[Log[(ndotj + r - 1)!/(r - 1)!]] +
(Total[Log[nidot!]] - Log[ntot!]) -
(Total[Log[table!], 2] - Log[ntot!]) ;
(* probability from evidence: III-13 p.84 *)
prob = (1 + Exp[-ev])^-1 ;
{ev // N, prob // N} (* output *)
] (* End of Module *)
</code></pre>
<p>Tribus' solution:</p>
<pre><code>(* Tribus p.200 : 1.45 10^-21 *)
evidence1B[tobacco]
(* output: {-47.9818, 1.45138*10^-21} *)
</code></pre>
<p>Also here the probability for rows and columns to be independent is pretty small 1.45*10^-21.</p>
<hr>
<p>Your example of a 3x3 table:</p>
<pre><code>caya = {{1, 0, 2}, {0, 0, 2}, {2, 1, 0}, {0, 2, 1}};
evidence1A[caya]
(* output: {-2.62276, 0.0676881} *)
evidence1B[caya]
(* output: {-1.7158, 0.152413} *)
</code></pre>
<p>The probabilities for independence of rows and columns are small-ish. But they are not very small. Depending on the details of your problem, such probability values can signal: <em>inconclusive</em>.</p>
|
89,987 | <p>In the sequence: </p>
<p>$$\lim _{n\rightarrow \infty }{\frac {n+1}{2\,n+3}}\neq 3$$</p>
<p>I know how to prove that the limit is actually $1/2$, but is there another way to prove that 1 is NOT the limit? </p>
<p>I tried to prove by negation showing that if 1 is the limit I can't find an $N$ that for every epsilon etc etc but got a bit lost. I know this is trivial but would appreciate the help.
Thanks</p>
| Will Jagy | 10,400 | <p>Yes, the value of the fraction is always below 1/2. So it stays far away from 3, indeed never gets within 5/2 of it. Note that it is not necessary to know that there is any limit to do answer your question. Anyway, take $\varepsilon = 2$ and no $n$ ever gets you within $\varepsilon$ of 3.</p>
|
4,610,313 | <p>I am working on AoPS Vol. 2 exercises in Chapter 1 and attempting to solve the below problem:</p>
<blockquote>
<p>Given that <span class="math-container">$\log_{4n} 40\sqrt{3} = \log_{3n} 45$</span>, find <span class="math-container">$n^3$</span> (MA<span class="math-container">$\Theta$</span> 1991).</p>
</blockquote>
<p>My approach is to isolate <span class="math-container">$n$</span> and then cube it. Observe:
<span class="math-container">\begin{align*}
\frac{\log 40\sqrt{3}}{\log 4n} = \frac{\log 45}{\log 3n} \\
\log 40\sqrt{3}\log 3n = \log 45\log 4n\\
\log 40\sqrt{3} \cdot (\log 3 + \log n) = \log 45 \cdot (\log 4 + \log n)\\
\log n \cdot (\log 40\sqrt{3} - \log 45) = \log 45\log 4 - \log 40\sqrt{3}\log 3
\end{align*}</span></p>
<p>Dividing through and putting the coefficients as powers, we have:
<span class="math-container">\begin{align*}
\log n &= \frac{\log 45^{\log 4} - \log \left[(40\sqrt{3})^{\log 3}\right]}{\log\left(\frac{40\sqrt{3}}{45}\right)}
=\frac{\log \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right) }{\log\left(\frac{40\sqrt{3}}{45}\right)} \\
&=\log \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right)^{\log\left(\frac{40\sqrt{3}}{45}\right)^{-1}}
\end{align*}</span></p>
<p>which shows that</p>
<p><span class="math-container">\begin{align*}
n^3 = \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right)^{3\cdot\log\left(\frac{40\sqrt{3}}{45}\right)^{-1}}
\end{align*}</span></p>
<p>Somehow it feels like this answer may be simplified further. Are the steps shown so far correct and can the answer be expressed in a better way?</p>
| lone student | 460,967 | <p>I would go the following way.</p>
<p>You have :</p>
<p><span class="math-container">$$\begin{align}&\log_{4n} 40\sqrt{3} = \log_{3n} 45\\
\implies &\log_{4n} \left(3\cdot 40^2\right) = \log_{3n} 45^2=k
\end{align}$$</span></p>
<p>This leads to :</p>
<p><span class="math-container">$$\begin{align}&\begin{cases}(4n)^k=3\cdot 40^2\\
(3n)^k=45^2 \end{cases}\\
\implies &\left(\frac 43\right)^k=\frac {3\cdot 40^2}{45^2}=\left(\frac {4}{3}\right)^3\\
\implies &k=3\\
\implies &n^3=\frac {45^2}{3^3}=75\thinspace.\end{align}$$</span></p>
|
265,801 | <p>Consider a 12-sided fair die. What is the distribution of the number T of rolls required to roll a 1, a 2, a 3, and a 4?</p>
<p>Taking inspiration from the Coupon Collector's Problem, I believe that the expected number of rolls to achieve the goal would be</p>
<p>$$E[T] = \sum\limits_{i=0}^3 \frac{12}{4-i} = 25$$</p>
<p>Similarly, the variance would be</p>
<p>$$Var[T] = \sum\limits_{i=0}^3 \frac{1-\frac{4-i}{12}}{(\frac{4-i}{12})^2} = 180$$</p>
<p>But applying Chebyshev here does not yield very useful bounds. My question is therefore, how would you compute, for example, $P(T=16)$ or $P(T<30)$?</p>
<p>Ideally this could be generalized to a set of k required numbers, not just 4 as in the example.</p>
| Henry | 6,460 | <p>Personally I would call $P(n,b,k,j)$ the probability that after rolling $n$ dice with $b$ sides, $j$ of the target $k$ sides had been found, with the formula</p>
<p>$$P(n+1,b,k,j)= \frac{b-k+j}{b} P(n,b,k,j) + \frac{k+1-j}{b} P(n,b,k,j-1)$$</p>
<p>starting from $P(0,b,k,j)=0$ for $j \not = 0$ and $P(0,b,k,0)=1$.</p>
<p>Then $$\Pr(T \le t) = P(t,b,k,k)$$ and $$\Pr(T = t) = P(t,b,k,k)-P(t-1,b,k,k).$$</p>
<p>So in your example it is not difficult to calculate $\Pr(T = 16) \approx 0.0380722$ and $\Pr(T \le 30) \approx 0.7305294$. Your expected value of $T$ and variance appear to be correct.</p>
|
265,801 | <p>Consider a 12-sided fair die. What is the distribution of the number T of rolls required to roll a 1, a 2, a 3, and a 4?</p>
<p>Taking inspiration from the Coupon Collector's Problem, I believe that the expected number of rolls to achieve the goal would be</p>
<p>$$E[T] = \sum\limits_{i=0}^3 \frac{12}{4-i} = 25$$</p>
<p>Similarly, the variance would be</p>
<p>$$Var[T] = \sum\limits_{i=0}^3 \frac{1-\frac{4-i}{12}}{(\frac{4-i}{12})^2} = 180$$</p>
<p>But applying Chebyshev here does not yield very useful bounds. My question is therefore, how would you compute, for example, $P(T=16)$ or $P(T<30)$?</p>
<p>Ideally this could be generalized to a set of k required numbers, not just 4 as in the example.</p>
| joriki | 6,622 | <p>You can get this distribution by conditioning on the number $N$ of rolls up to $4$ required:</p>
<p>\begin{align}
\textsf{Pr}(T\le t)
&=\left(\frac23\right)^t\sum_{n=0}^t\binom tn2^{-n}\textsf{Pr}(N\le n)
\\
&=\left(\frac23\right)^t\sum_{n=0}^t\binom tn2^{-n}\frac{4!}{4^n}\left\{n\atop4\right\}
\\
&=\left(\frac23\right)^t\sum_{n=0}^t\binom tn8^{-n}\sum_{j=0}^4(-1)^j\binom4jj^n
\\
&=\left(\frac23\right)^t\sum_{j=0}^4(-1)^j\binom4j\sum_{n=0}^t\binom tn\left(\frac j8\right)^n
\\
&=\left(\frac23\right)^t\sum_{j=0}^4(-1)^j\binom4j\left(1+\frac j8\right)^t
\\
&=\sum_{j=0}^4(-1)^j\binom4j\left(\frac23+\frac j{12}\right)^t
\\
&=1-4\left(\frac{11}{12}\right)^t+6\left(\frac56\right)^t-4\left(\frac34\right)^t+\left(\frac23\right)^t\;,
\end{align}</p>
<p>where $\left\{n\atop4\right\}$ is a Stirling number of the second kind and the distribution $P(N\le n)$ is given at <a href="https://math.stackexchange.com/questions/379525">Probability distribution in the coupon collector's problem</a>. </p>
<p>The expectation comes out right as</p>
<p>\begin{align}
E[T]&=\sum_{t=0}^\infty\textsf{Pr}(T\gt t)\\
&=\sum_{t=0}^\infty\left(4\left(\frac{11}{12}\right)^t-6\left(\frac56\right)^t+4\left(\frac34\right)^t-\left(\frac23\right)^t\right)\\
&=12\left(4\cdot\frac11-6\cdot\frac12+4\cdot\frac13-1\cdot\frac14\right)\\
&=25\;.
\end{align}</p>
<p>Here are plots of the <a href="http://www.wolframalpha.com/input/?dataset=&i=plot%201-4%5Cleft%28%5Cfrac%7B11%7D%7B12%7D%5Cright%29%5Et%2B6%5Cleft%28%5Cfrac56%5Cright%29%5Et-4%5Cleft%28%5Cfrac34%5Cright%29%5Et%2B%5Cleft%28%5Cfrac23%5Cright%29%5Et%20for%20t%3D0..60" rel="nofollow noreferrer">cumulative distribution function</a> and the <a href="http://www.wolframalpha.com/input/?dataset=&i=plot%20%284%2F11%29%5Cleft%28%5Cfrac%7B11%7D%7B12%7D%5Cright%29%5Et-%286%2F5%29%5Cleft%28%5Cfrac56%5Cright%29%5Et%2B%284%2F3%29%5Cleft%28%5Cfrac34%5Cright%29%5Et-%281%2F2%29%5Cleft%28%5Cfrac23%5Cright%29%5Et%20for%20t%3D1..60" rel="nofollow noreferrer">probability mass function</a>.</p>
<p>In your examples,</p>
<p>$$
\textsf{Pr}(T=16)=\textsf{Pr}(T\gt15)-\textsf{Pr}(T\gt16)=\frac{293289532268461}{7703510787293184}\approx3.8\%
$$</p>
<p>and</p>
<p>$$
\textsf{Pr}(T\lt30)=\textsf{Pr}(T\le29)=\frac{292029927548835623394780280045}{412111655902378135987760922624}\approx70.9\%\;.
$$</p>
<p>For general $k$ instead of $4$ numbers required and general $m$ instead of $12$ numbers on the die, the result is</p>
<p>$$
\textsf{Pr}(T\le t)=\sum_{j=0}^k(-1)^j\binom kj\left(1-\frac{k-j}m\right)^t\;.
$$</p>
|
1,288,461 | <p>Here's the proof I was given:</p>
<p>Proposition.- If $A$ is countable then $\mathbb{R} \setminus A $ is dense.</p>
<p>Proof:
Suppose otherwise, then there exists real numbers $a$ and $b$, with $a < b$, such that there is no $a < x < b$ with $x \in \mathbb{R} \setminus A $. To put it in another way, if $a < x < b$ then $x \in \mathbb{R} \setminus A$, as $ \mathbb{R} \setminus A$ is the complement of $A$, we see that if $a < x < b$ then $x \in A$. Thus,$(x,y) \subseteq A$. But $A$ is countable and $(x,y)$ is not, this leads to a contradiction. QED.</p>
<p>Is it correct to assume that $(x,y)$ is uncountable because it is not dense? That all countable sets are dense? (That non-dense sets are uncountable?)</p>
<p>What would be the rule that governs why $(x,y)$ is not countable?
Is there anything wrong with this proof?</p>
<p>Thank you</p>
| hamid kamali | 208,328 | <p>$2a_n-a_{n-1}=4$. From here: $2a_nx^n-a_{n-1}x^n=4x^n$. By summation:
$$
\begin{align}
\sum_{n=1}^\infty (2a_nx^n-a_{n-1}x^n)&=\sum_{n=1}^\infty 4x^n\\
&=\frac{4x}{1-x}\,\,\,\, for\,\, |x|\lt 1
\end{align}
$$ For the left side of above equation, we can write:
$$
\begin{align}
\sum_{n=1}^\infty (2a_nx^n-a_{n-1}x^n)&=2(-a_0+\sum_{n=0}^\infty a_nx^n)-x\sum_{n=0}^\infty a_nx^n\\
&=-\frac43 +(2-x)\sum_{n=0}^\infty a_nx^n
\end{align}
$$
So: $$(2-x)\sum_{n=0}^\infty a_nx^n=\frac{4x}{1-x}+\frac43$$ Then, the radius of convergence of the power series $\sum_{n=0}^\infty a_nx^n$ is $1$.</p>
|
268,308 | <p>In composing a proof that is reliant on proven theorems, does one simply assume the reader's familiarity with said theorems, or does one explicitly include their logic in the new logic? </p>
| Calvin Lin | 54,563 | <p>With all proofs (in whatever format), you must first identify your intended audience. That will allow you to decide what you can / cannot assume, and how to include in the right amount of detail.</p>
<p>Are they students in your field that are interested in understanding the buildup of the theory? Are they 'professionals' in related fields who believe in the truth of your statements and only want to understand implications of your results? Are they high schoolers who want to know why quintics can't be 'solved' and only know the basics of group theory?</p>
<p>Having additional proofs is preferable. You can helps others along by labeling them as Theorem 1, Proof (or even placed in the Appendix), so if readers are familiar, they can skip it.</p>
|
2,072,866 | <p>My notes say that if $F$ is conservative (i.e. $F=\nabla f$) then $\text{curl }(F)=0$. But I feel this is not quite right.</p>
<p>There is a theorem that says that if $f(x,y,z)$ has continuous second order partial derivatives then $\text{curl }(\nabla f)=0$. (I'm fine with this theorem, it makes sense.) </p>
<p>Obviously the 'proof' for my question is
$$\text{curl }(F)=\text{curl }(\nabla f)=0$$</p>
<p>But I don't see why it is necessarily true that $f$ has continuous second order partial derivatives. Is there something I am missing?</p>
| nullUser | 17,459 | <p>The proof is essentially $\partial_{ij}f = \partial_{ji}f$, and the only hypotheses needed are "whenever this holds". Recall that the curl of $F$ is zero iff $\partial_i F_j = \partial_j F_i$ for all $i,j$. When $F$ is conservative, we have that $\partial_i F_j = \partial_i \partial_j f$ and $\partial_j F_i = \partial_j \partial_i f$.</p>
<p>Thus, if $F$ has first order partial derivatives and $F$ is conservative with potential $f$ which satisfies $\partial_{ij}f = \partial_{ji}f$ for all $i,j$, then $F$ is curl free.</p>
<p>Since the version of Schwartz theorem usually quoted in books requires $f$ to be $C^2$, most authors will subsequently choose to assume $F$ to be $C^1$, which guarantees that $f$ is $C^2$.</p>
|
3,011,216 | <p>If <span class="math-container">$F(x)$</span> is a continuous function in <span class="math-container">$[0,1]$</span> and <span class="math-container">$F(x) = 1$</span> for all rational numbers then <span class="math-container">$F(1/\sqrt2) = 1$</span>. (True/ False)</p>
<p>I think the statement is true because since <span class="math-container">$F$</span> is continuous and <span class="math-container">$1/\sqrt2$</span> is in neighbourhood of some rational number so <span class="math-container">$F(1/\sqrt2)$</span> should be one so that limit can exist. Is this reason valid to prove the statement true.</p>
| mlerma54 | 406,626 | <p>The statement is true, although I would express it: there is a sequence <span class="math-container">$\{r_n\}_{n\in\mathbb{N}}$</span> of rational numbers in the interval <span class="math-container">$[0,1]$</span> such that <span class="math-container">$r_n \to 1/\sqrt{2}$</span>, hence by continuity
<span class="math-container">$$
F(1/\sqrt{2}) =
F(\lim_{n\to\infty} r_n) =
\lim_{n\to\infty} F(r_n) =
\lim_{n\to\infty} 1 = 1 \,.
$$</span></p>
|
2,798,338 | <p>Let $G$ be a group of order 20, Prove that $G$ has a normal subgroup of order 5.</p>
<p>Obviously by Sylow theorem there is a subgroup of order 5, and since all Sylow p-subgroups are conjugate the only problem is to show that there is only one sungroup of order 5.</p>
<p>any help would be appreciated</p>
| thesmallprint | 438,651 | <p>$|G|=20=5\cdot 2^2$. Now, let $n_p(G)$ be the number of Sylow $p$-subgroups of $G$.</p>
<p>Then, Sylow III says</p>
<ul>
<li>$n_5(G)\equiv 1\bmod 5$</li>
<li>$n_5(G)$ divides $4$</li>
</ul>
<p>So, $n_5(G)=1$. This means there is only one Sylow $5$-subgroup of $G$; which in turn, is a normal subgroup of $G$.</p>
|
1,915,273 | <p>$$\int_0^r\sqrt{x-x^2}.dx$$</p>
<p>I only have basic calculus and would like to know how would one go about integrating an expression of this form.
I have tried substituting say $u=x-x^2$ but I'm still left with an $x$.</p>
<p>The method is not in my book and I can't find a similar example anywhere.</p>
<p>A hint in the right direction is sufficient.</p>
| Jan Eerland | 226,665 | <p>$$\mathcal{I}(\text{r})=\int_0^\text{r}\sqrt{x-x^2}\space\text{d}x=\int_0^\text{r}\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}\space\text{d}x=$$</p>
<hr>
<p>Substitute $u=x-\frac{1}{2}$ and $\text{d}u=\text{d}x$.</p>
<p>This gives a new lower bound $u=0-\frac{1}{2}=-\frac{1}{2}$ and upper bound $u=\text{r}-\frac{1}{2}$:</p>
<hr>
<p>$$\int_{-\frac{1}{2}}^{\text{r}-\frac{1}{2}}\sqrt{\frac{1}{4}-u^2}\space\text{d}u=$$</p>
<hr>
<p>Substitute $u=\frac{\sin(s)}{2}$ and $\text{d}u=\frac{\cos(s)}{2}\space\text{d}s$.</p>
<p>Then $\sqrt{\frac{1}{4}-u^2}=\sqrt{\frac{1}{4}-\frac{\sin^2(s)}{4}}=\frac{\cos(s)}{2}$ and $s=\arcsin(2u)$.</p>
<p>This gives a new lower bound $s=\arcsin\left(-1\right)=-\frac{\pi}{2}$ and upper bound $s=\arcsin\left(2\text{r}-1\right)$:</p>
<hr>
<p>$$\frac{1}{4}\int_{-\frac{\pi}{2}}^{\arcsin\left(2\text{r}-1\right)}\cos^2(s)\space\text{d}s=$$</p>
<hr>
<p>Use $\cos^2(s)=\frac{1+\cos(2s)}{2}$:</p>
<hr>
<p>$$\frac{1}{8}\int_{-\frac{\pi}{2}}^{\arcsin\left(2\text{r}-1\right)}\cos(2s)\space\text{d}s+\frac{1}{8}\int_{-\frac{\pi}{2}}^{\arcsin\left(2\text{r}-1\right)}1\space\text{d}s=$$</p>
<hr>
<p>For $\int_{-\frac{\pi}{2}}^{\arcsin\left(2\text{r}-1\right)}\cos(2s)\space\text{d}s$ substitute $p=2s$ and $\text{d}p=2\space\text{d}s$.</p>
<p>This gives a new lower bound $s=2\cdot-\frac{\pi}{2}=-\pi$ and upper bound $s=2\arcsin\left(2\text{r}-1\right)$:</p>
<hr>
<p>$$\frac{1}{16}\int_{-\pi}^{2\arcsin\left(2\text{r}-1\right)}\cos(p)\space\text{d}p+\frac{1}{8}\int_{-\frac{\pi}{2}}^{\arcsin\left(2\text{r}-1\right)}1\space\text{d}s=$$</p>
<hr>
<p>Use:</p>
<ul>
<li>$$\int1\space\text{d}s=s+\text{C}$$</li>
<li>$$\int\cos(p)\space\text{d}p=\sin(p)+\text{C}$$
<hr></li>
</ul>
<p>$$\frac{1}{16}\left[\sin(p)\right]_{-\pi}^{2\arcsin\left(2\text{r}-1\right)}+\frac{1}{8}\left[s\right]_{-\frac{\pi}{2}}^{\arcsin\left(2\text{r}-1\right)}=$$
$$\frac{\sin(2\arcsin\left(2\text{r}-1\right))-\sin(-\pi)}{16}+\frac{\arcsin\left(2\text{r}-1\right)+\frac{\pi}{2}}{8}=$$
$$\frac{\sin(2\arcsin\left(2\text{r}-1\right))}{16}+\frac{\arcsin\left(2\text{r}-1\right)+\frac{\pi}{2}}{8}$$</p>
|
1,915,273 | <p>$$\int_0^r\sqrt{x-x^2}.dx$$</p>
<p>I only have basic calculus and would like to know how would one go about integrating an expression of this form.
I have tried substituting say $u=x-x^2$ but I'm still left with an $x$.</p>
<p>The method is not in my book and I can't find a similar example anywhere.</p>
<p>A hint in the right direction is sufficient.</p>
| user84413 | 84,413 | <p>Let $\sqrt{x}=\sin t$, so $x=\sin^2 t,\; dx=2\sin t\cos t \,dt$.</p>
<p>Then $\displaystyle\int\sqrt{x-x^2}dx=\int(\sin t\cos t)(2\sin t\cos t)dt=\int2\sin^2 t\cos^2t dt=\frac{1}{2}\int\sin^2 2t dt$</p>
<p>$\;\;\;=\displaystyle\frac{1}{4}\int(1-\cos 4t)dt=\frac{1}{4}\left[t-\frac{1}{4}\sin 4t\right]+C=\frac{1}{4}\left[t-\sin t\cos t(\cos^2 t-\sin^2 t)\right]+C$</p>
<p>$\displaystyle=\frac{1}{4}\left[\sin^{-1}\sqrt{x}-\sqrt{x}\sqrt{1-x}((1-x)-x)\right]+C=\frac{1}{4}\left[\sin^{-1}\sqrt{x}-\sqrt{x-x^2}(1-2x)\right]+C$,</p>
<p>so $\displaystyle\int_0^r\sqrt{x-x^2}dx=\color{blue}{{\frac{1}{4}\left(\sin^{-1}\sqrt{r}-\sqrt{r-r^2}(1-2r)\right)}}$</p>
|
3,636,563 | <p>How to find the values of <span class="math-container">$x,y$</span> and <span class="math-container">$z$</span> if
<span class="math-container">$3x²-3(1126)x=96y²+24(124)y=8z²-4(734)z $</span>?</p>
<p>I dont have any idea!! I think we can have many values of <span class="math-container">$x,y$</span> and <span class="math-container">$z$</span> at a time or it is a no solution??</p>
| Aditya Sriram | 636,845 | <p>Let us assume that the antiderivative of <span class="math-container">$f$</span> is <span class="math-container">$F$</span>.
By the Fundamental Theorem of Calculus,
<span class="math-container">$$\therefore \int_{\frac{1}{k}}^{\frac{1}{k-1}}f(x)dx =F(\frac{1}{k-1})-F(\frac{1}{k})$$</span>
<span class="math-container">$$\therefore \sum_{k=2}^{n}\int_{\frac{1}{k}}^{\frac{1}{k-1}}f(x)dx=\sum_{k=2}^{n}F(\frac{1}{k-1})-F(\frac{1}{k})=F(1)-F(\frac{1}{n-1})$$</span>
<span class="math-container">$$\therefore \lim_{n \to \infty}F(1)-F(\frac{1}{n-1})=F(1)-F(0) = \int_{0}^{1}f(x)dx$$</span> Which proves your result. Hope this helps.</p>
|
3,603,066 | <p>My question: How does one distinguish between two embeddings of the same graph on the plane? </p>
<p>For instance, are two such embeddings considered the same if the degree sequence of their faces are the same? Or does even the slightest change, such as drawing an edge more crooked in one embedding than the other, constitute a different embedding? (Or perhaps none of these?)</p>
| Allawonder | 145,126 | <p>Let the rate per annum be <span class="math-container">$r.$</span> Then we have that after the first year, the height of the tree is <span class="math-container">$3(1+r).$</span> After the second year that's <span class="math-container">$3(1+r)^2,$</span> and after the twelfth year the tree is <span class="math-container">$$3(1+r)^{12}=15.$$</span> You are to solve this for a positive value of <span class="math-container">$r.$</span> Note that this is equivalent to <span class="math-container">$(1+r)^{12}=5,$</span> so that <span class="math-container">$1+r=5^{1/12},$</span> or that <span class="math-container">$$r=5^{1/12}-1,$$</span> which is positive provided we take the positive root out of the twelve. You may approximate this with a calculator.</p>
|
6,155 | <p>I have a category $C$, which is equipped with a symmetric monoidal structure (tensor product $\otimes$, unit object $1$). My category also has finite coproducts (I'll write them using $\oplus$, and write $0$ for the initial object), and $\otimes$ distributes over $\oplus$.</p>
<p>By an <em>exponential monad</em>, I mean a monad $(T,\eta,\mu)$ on $C$, where the functor $T:C\to C$ is equipped with some structure maps of the form
$$\nu \colon 1 \to T(0)$$
and
$$\alpha\colon T(X)\otimes T(Y) \to T(X\oplus Y).$$
The structure maps are isomorphisms, and are suitably "coherent" with respect to the two monoidal structures $\otimes$ and $\oplus$.</p>
<p>The simplest example is: $C$ is the category of $k$-vector spaces, and $T=\mathrm{Sym}$ is the commutative $k$-algebra monad (i.e., $\mathrm{Sym}(X)$ is the symmetric algebra $\bigoplus \mathrm{Sym}^q(X)$).</p>
<p>Now, I'm sure I can work out all the formalism that I need for this, if I have to. My question is: is there a convenient place in the literature I can refer to for this? Alternately, is there suitable categorical language which makes this concept easy to talk about?</p>
<p>I'd also like to have a good formalism for talking about a "grading" on $T$. This means a decomposition of the functor $T=\bigoplus T^q$, where $T^q\colon C\to C$ are functors, which have "nice" properties (for instance, $T^m(X\oplus Y)$ is a sum of $T^p(X)\otimes T^{m-p}(Y)$). The motivating example again comes from the symmetric algebra: $\mathrm{Sym}=\bigoplus \mathrm{Sym}^q$.</p>
| Michael Ortiz | 1,938 | <p>There is a nice discussion about multiplicative sequences, &c., in Lawson and Michelsohn's book "Spin Geometry". It discusses things like the Todd genus, the A-hat genus, and so on, but also the Chern character and the ring homomorphism from K-theory to ordinary cohomology. It is a readable exposition and perhaps "connects the dots" in a way that would be helpful to you.</p>
|
181,327 | <p>If $q$ and $w$ are the roots of the equation $$2x^2-px+7=0$$ </p>
<p>Then $q/w$ is a root of ? </p>
<p>P.s:- It is an another question of <a href="https://math.stackexchange.com/questions/181305/how-do-i-transform-the-equation-based-on-this-condition">How do I transform the equation based on this condition?</a></p>
| Emily | 31,475 | <p>The methods to be used will be highly dependent on the character of $f$. If it is non-convex, there are many such algorithms; see <a href="https://mathoverflow.net/questions/32533/is-all-non-convex-optimization-heuristic">this MO post</a> for instance. </p>
<p>You might also wish to look into evolutionary algorithms, such as genetic algorithms and simulated annealing. These algorithms are often much slower, but have the feature that they can sometimes "bump" you out of local extrema. They are also fairly easy to implement.</p>
<p>You can also hybridize approaches: combine an evolutionary algorithm with a standard convex optimization approach on a locally convex subdomain.</p>
<p>Finally, with an equality constraint, you essentially reduce the dimensionality of your problem by 1. That is, one variable is completely determined by the others.</p>
<p>$$x_k = B - \sum_{i=1,\ i\neq k}^n x_i.$$</p>
<p>And then, depending on the character of your function, you might be able to use any sort of algorithm.</p>
<p>But to answer your questions, 1.) there are many common and smart ways, but they depend on your function $f$. I would start with the most simple, conventional approach, and then see whether it is effective. 2.) I don't think you're missing anything. Numerical optimization is a <strong>big topic</strong> and there are many ways to go about it.</p>
|
2,021,734 | <p>Let $p$ be a prime number and $\mu_{p^{\infty}}$ denote $$\{ z\in \mathbb{C} : \exists k \ge 1 : z^{p^k}=1 \}$$ find all of its subgroups.</p>
<p>I was able to prove that its finite subgroups are of the form $\mu_{p^n}$, where $n$ is a positive integer (using Lagrange's Theorem and the fact that in, $\mu_{\infty}=\{ z\in \mathbb{C} : \exists k \ge 1 : z^k=1 \} $, a subgroup of order $n$ is $\mu_n=\{ z\in \mathbb{C} : z^n=1\}$). However I am struggling to find its infinite subgroups, if there are any. Any hints on that?</p>
| Couchy | 87,768 | <p>To expand on Andreas Caranti's answer, consider a sequence $a_1,a_2,\ldots\in\mathbb N$ so that we have the subgroup </p>
<p>$$H=\bigcup_{k=1}^\infty\mu_{p^{a_k}},$$
(where any subgroup $H\leq \mu_{p^\infty}$ can be written in such a way) then since for any $k\leq a_j$ we have $\mu_{p^{k}}\leq\mu_{p^{a_j}}$, then for any $n$ we have
$$\mu_{p^n}=\bigcup_{k=1}^n\mu_{p^k}$$
from which it follows that
$$H=\bigcup_{k=1}^\infty\mu_{p^{a_k}}=\bigcup_{k=1}^\infty\bigcup_{i=1}^{a_k}\mu_{p^i}=\bigcup_{k=1}^\infty\mu_{p^k}=\mu_{p^\infty}.$$</p>
|
179,322 | <p>I'm interested in classes C of $R^1$-valued random variables which possess the following properties:</p>
<p>1) the sum of two independent random variables from class C belongs to class C;</p>
<p>2) for any $\lambda \in R^1$, $\xi \in C$ we have $\lambda\xi \in C$; </p>
<p>3) any random variable from class $C$ has tails which are heavier than the Gaussian tails;</p>
<p>4) for any $\xi \in C$ we have $\mathrm{E}\xi=0, \mathrm{E}\xi^2<\infty$. </p>
<p>If we omit the restriction $\mathrm{E}\xi=0$ and restriction No. 2 then there is an obvious example:
a class of gamma-distributed random variables with one parameter of the distribution fixed. Can anyone suggest other examples? </p>
<p>Thanks in advance. </p>
<p>Best wishes, Ievgen. </p>
| Donu Arapura | 4,144 | <p>Let me supplement Vivek's nice answer with a few additional comments. People coming from differential geometry typically normalize the $n$th Chern class -- in the Chern-Weil formula -- by a factor of $1/(2\pi i)^n$ to get it to be integral. However, if you want compatibility with algebraic definitions, then you don't want to do this, so then $c_n$ would take values in $H^{2n}(X,(2\pi i)^n\mathbb{Z})$. For example, in additional to the etale first Chern class explained above, you can use algebraic de Rham cohomology. Here $c_1$ is given by the induced map on $H^2$ associated to
$$ d\log:\mathcal{O}_X^*[-1]\to \Omega_X^\bullet$$</p>
<p><strong>Continued--</strong> To expand slightly, one can see by a diagram chase that, under the Grothendieck isomorphism $H^2(X,\mathbb{C})\cong H^2(X,\Omega_X^\bullet)$, the image of this $c_1$ lands in $H^2(X,2\pi i\mathbb{Z})$. So by the usual tricks involving the splitting principle, one gets the a similar statement for higher Chern classes. In case, you prefer to avoid Chern classes, it follows (with some work), that the image of the cycle map $CH^n(X)_\mathbb{Q}\to H^{2n}(X, \mathbb{C})$ lands in $H^{2n}(X, \mathbb{Q}(n))$. Again this is consistent with what happens on the etale side.</p>
|
1,119,421 | <p>We know that p → q is not equivalent to q → p. But suppose we make a proof system that has all the rules of logical identities plus the rule (“commutativity of implies”) p → q ≃ q → p. (We are using the new symbol ≃ because these are not really equivalent.) </p>
<p>This new proof system is not sound. The point of this question is to show that you can prove any contradiction in this system.</p>
<p>Prove true ≃ false in this proof system. </p>
<p>Hint: You might want to start your chain of ≃ in the middle with false → true ≃ true → false. Note that the logical identities do not include a rule ¬true ≡ false, so if you want to use this, you should derive it from the other logical identities.</p>
<hr>
<p>How do I go about starting this proof? Can someone explain the hint to me? </p>
<p>Thanks!</p>
<p>EDIT: Here's a list of identities we're allowed to use.</p>
<p><strong>Commutativity</strong></p>
<p>p∧q ≡ q∧p</p>
<p>p∨q ≡ q∨p </p>
<p>p↔q ≡ q↔p</p>
<p><strong>Associativity</strong> </p>
<p>p∧(q∧r) ≡ (p∧q)∧r </p>
<p>p∨(q∨r) ≡ (p∨q)∨r</p>
<p><strong>Distributivity</strong> </p>
<p>p∨(q∧r) ≡ (p∨q)∧(p∨r)
p∧(q∨r) ≡ (p∧q)∨(p∧r)</p>
<p><strong>De Morgan</strong> </p>
<p>¬(p∧q) ≡ ¬p∨¬q </p>
<p>¬(p∨q) ≡ ¬p∧¬q</p>
<p><strong>Negation</strong>
¬(¬p) ≡ p</p>
<p><strong>Excluded Middle</strong></p>
<p>p∨¬p ≡ true</p>
<p><strong>Contradiction</strong></p>
<p>p∧¬p ≡ false</p>
<p><strong>Implication</strong></p>
<p>p→q ≡ ¬p∨q </p>
<p><strong>Contrapositive</strong></p>
<p>p→q ≡ ¬q→¬p </p>
<p><strong>Equivalence</strong></p>
<p>p↔q ≡ (p→q)∧(q→p)</p>
<p><strong>Idempotence</strong></p>
<p>p∨p ≡ p </p>
<p>p∧p ≡p</p>
<p><strong>Simplification I</strong></p>
<p>p ∧ true ≡ p</p>
<p>p ∨ true ≡ true</p>
<p>p ∧ false ≡ false</p>
<p>p ∨ false ≡ p</p>
<p><strong>Simplification II</strong> </p>
<p>p∨(p∧q) ≡ p</p>
<p>p∧(p∨q) ≡ p</p>
| ajotatxe | 132,456 | <p>Since $\Bbb Z_6$ is cyclic and generated by $1$, the value $f(1)$ determines the homomorphism. On the other hand, $f(1)$ must have order $6$ or a divisor of $6$, so $f(1)\in\{0,3,6,9,12,15\}$. This yields six homomorphisms.</p>
|
80,432 | <p>I know Mathematica's if format is </p>
<pre><code>If[test, then result, else alternative]
</code></pre>
<p>For example, this</p>
<pre><code>y:=If[RandomReal[]<0.2, 1, 3.14]
</code></pre>
<p>would take a random real number between $0$ and $1$, and evaluate it. If it's less than $0.2$, it'll map <code>y</code> to $1$, otherwise, it'll map <code>y</code> to $3.13$.</p>
<p>I would like to extend this to multiple intervals, short of writing out multiply nested <code>If</code> statements, how can this be done? Can this be done automatically?</p>
<p>For example, if <code>x := RandomReal[]</code>, I want to map to <code>1</code>, if <code>x < 0.1</code>, 3 if <code>0.1 <= x < 0.2</code>, 19.1 if <code>0.2 <= x < 0.34</code>, or 7.7 if <code>x >= 0.34</code>. </p>
| Mr.Wizard | 121 | <p>There are several functions and methods available which different strengths and limitations that can guide your choice. Among them:</p>
<h2><a href="http://reference.wolfram.com/language/ref/Which.html" rel="noreferrer"><code>Which</code></a></h2>
<p>Here combined with <a href="http://reference.wolfram.com/language/ref/Function.html" rel="noreferrer"><code>Function</code></a> and <a href="http://reference.wolfram.com/language/ref/Slot.html" rel="noreferrer"><code>Slot</code></a> to pass a single <code>RandomReal[]</code> value among the tests:</p>
<pre><code>y := Which[
# < 0.1, 1,
0.1 <= # < 0.2, 3,
0.2 <= # < 0.34, 19.1,
True, 7.7
] & @ RandomReal[]
</code></pre>
<p>Which, like <code>If</code>, is a flow control construct. <code>True</code> is used to create a default case. Related to it is:</p>
<h2><a href="http://reference.wolfram.com/language/ref/Switch.html" rel="noreferrer"><code>Switch</code></a></h2>
<p>The syntax of <code>Switch</code> allows us to do without the <code>Function</code> since the first argument is only evaluated once, but we must introduce <a href="http://reference.wolfram.com/language/ref/Pattern.html" rel="noreferrer"><code>Pattern</code></a> and <a href="http://reference.wolfram.com/language/ref/Condition.html" rel="noreferrer"><code>Condition</code></a>:</p>
<pre><code>y := Switch[RandomReal[],
x_ /; x < 0.1, 1,
x_ /; 0.1 <= x < 0.2, 3,
x_ /; 0.2 <= x < 0.34, 19.1,
_, 7.7
]
</code></pre>
<p>Here the pattern <code>_</code> (see <a href="http://reference.wolfram.com/language/ref/Blank.html" rel="noreferrer"><code>Blank</code></a>) is used to create the default case. <code>Switch</code> is also a flow control function, unlike:</p>
<h2><a href="http://reference.wolfram.com/language/ref/Piecewise.html" rel="noreferrer"><code>Piecewise</code></a></h2>
<p><code>Piecewise</code> is intended as a mathematical function and it therefore often behaves better within a mathematical framework; if the expression is going to be manipulated mathematically it is probably your best starting point. It has an optional nice looking "2D" input form:</p>
<p><img src="https://i.stack.imgur.com/DGaRT.png" alt="enter image description here"></p>
<pre><code>y := \[Piecewise] {
{1, # < 0.1},
{3, 0.1 <= # < 0.2},
{19.1, 0.2 <= # < 0.34},
{7.7, True}
} & @ RandomReal[]
</code></pre>
<h2><a href="http://reference.wolfram.com/language/ref/Interval.html" rel="noreferrer"><code>Interval</code></a> and <a href="http://reference.wolfram.com/language/ref/IntervalMemberQ.html" rel="noreferrer"><code>IntervalMemberQ</code></a></h2>
<p>The three methods above are general; however for the given example there are more specialized approaches. One is to use <code>Interval</code>, though it is important to understand that it represents an interval closed on both ends. Here is one formulation also using <a href="http://reference.wolfram.com/language/ref/Pick.html" rel="noreferrer"><code>Pick</code></a>; the interval range <code>{-∞, ∞}</code> is used for the default case:</p>
<pre><code>With[
{intv =
Interval /@ Append[Partition[{0, 0.1, 0.2, 0.34}, 2, 1], {-∞, ∞}]},
y := First @ Pick[{1, 3, 19.1, 7.7}, IntervalMemberQ[intv, RandomReal[]]]
]
</code></pre>
<h2><a href="http://reference.wolfram.com/language/ref/Interpolation.html" rel="noreferrer"><code>Interpolation</code></a></h2>
<p>Faster when applicable is an <a href="http://reference.wolfram.com/language/ref/InterpolatingFunction.html" rel="noreferrer"><code>InterpolatingFunction</code></a> as generated by <a href="http://reference.wolfram.com/language/ref/Interpolation.html" rel="noreferrer"><code>Interpolation</code></a>:</p>
<pre><code>intfn = Interpolation[{{0.1, 0.2, 0.34, 1*^99}, {1, 3, 19.1, 7.7}}\[Transpose],
InterpolationOrder -> 0];
y := intfn @ RandomReal[]
</code></pre>
<p>In the example above I had to use an arbitrary "large number" <code>1*^99</code> for the default case as <code>Infinity</code> is not accepted.</p>
<p>A <a href="http://reference.wolfram.com/language/ref/Plot.html" rel="noreferrer"><code>Plot</code></a> of <code>intfn</code>:</p>
<pre><code>Plot[intfn[x], {x, 0, 1}]
</code></pre>
<p><img src="https://i.stack.imgur.com/hifXy.png" alt="enter image description here"></p>
|
3,490,198 | <p>For all <span class="math-container">$a,b \in \mathbb{Z}$</span> and for all <span class="math-container">$m,n \in \mathbb{N}\setminus \left\lbrace0\right\rbrace$</span>,</p>
<p>is <span class="math-container">$a^{48m+1}+b^{48n+1} \equiv 0 \pmod{39} \iff a+b \equiv 0 \pmod{39}$</span>?</p>
<p>I think the answer is yes, but I can't prove it. Is there somebody who can help me? Thank you.</p>
| lab bhattacharjee | 33,337 | <p>Hint</p>
<p>Using <a href="http://mathworld.wolfram.com/FermatsLittleTheorem.html" rel="nofollow noreferrer">http://mathworld.wolfram.com/FermatsLittleTheorem.html</a></p>
<p>For <span class="math-container">$3\mid x(x^2-1),$</span></p>
<p>and <span class="math-container">$13|x(x^{12}-1)$</span></p>
<p><span class="math-container">$[3,13]$</span> will divide <span class="math-container">$x(x^{[2,12]}-1)$</span> and hence <span class="math-container">$x(x^{12m}-1)$</span> for any integer <span class="math-container">$m$</span></p>
|
1,285,870 | <p>Is the collection of all cardinalities a set or a proper class?
Does anybody ever think about the problem?</p>
| Trevor Wilson | 39,378 | <p>The collection of all cardinal numbers is a proper class. If it were a set $X$, then $(\sup X)^+$ would be a cardinal number greater than every element of $X$, which is a contradiction.</p>
<p>(The supremum of a set of cardinals is given by its union, and the cardinal successor $\kappa^+$ of a cardinal $\kappa$ is defined as the <a href="http://en.wikipedia.org/wiki/Hartogs_number" rel="nofollow">Hartogs number</a> of $\kappa$.)</p>
|
1,368,180 | <p>Here is a tricky compass and straightedge construction problem. </p>
<blockquote>
<p>Given triangle $\triangle ABC$ and point $D$ on segment $\overline{AB}$, construct point $P$ on line $\overleftrightarrow{CD}$ such that $\angle APB = \angle BAC$. </p>
</blockquote>
<p>This configuration appears frequently in Olympiad geometry problems and the diagram is impossible to draw precisely unless you know how to do the construction.</p>
| math_lover | 189,736 | <p>Construct isosceles triangle QAB with QA equals QB and angle QAB equals 90- BAC/2. The circumcircle of QAB is the locus of points X ( above the line AB) such that angle AXB equals BAC.. The desired point P is the intersection of this circle with CD.</p>
|
760,739 | <p>Find the limit of the sequence given by
$$\frac{10+12n+20n^4}{7n^4 + 5n^3 - 20}$$</p>
<p>I think the answer is $\frac{20}{7}$ after dividing, but is that right? </p>
| user2345215 | 131,872 | <p>Yes it is, because
$$\frac{20n^4+12n+10}{7n^4 + 5n^3 - 20}=\frac{20+\frac{12}{n^3}+\frac{10}{n^4}}{7+\frac5n-\frac{20}{n^4}}\to\frac{20+0+0}{7+0-0}=\frac{20}7$$
by the arithmetic of limits.</p>
<p>You should see this method works in general for a fraction of two polynomials, just divide $n^d$, where $d$ is the degree of the denominator.</p>
|
1,681,872 | <p>My task is this:</p>
<p>Use cylinder coordinates to calculate:$$\iiint\limits_{A}z\sqrt{x^2 + y^2}dA, \enspace A = \left\{(x,y,z):x^2 + (y - 1)^2 \leq 1,\: 0 \leq z \leq 2\right\}.$$</p>
<p>My works so far is this;</p>
<p>Switching to cylindrical coordinates we get:$$A = \left\{(r,\theta,z):0\leq r\leq 1,\: 0\leq \theta \leq 2\pi,\: 0\leq z \leq 2\right\}.$$
Now my book tells me that if you want the center in another point $(a,b,c)$, you should use the substitution: $$x = a + r\cos(\theta),\: y = b + r\sin(\theta),\: z = c + z.$$</p>
<p>With this in mind we change to cylindrical and add the bounderies (don't forget the jacobian).$$\int\limits_{0}^{2\pi}\int\limits_{0}^{1}\int\limits_{0}^{2}zr\sqrt{r^2\cos^2(\theta) + (1 + r\sin(\theta))^2}dz\:dr\:d\theta \:=\:2\int\limits_{0}^{2\pi}\int\limits_{0}^{1} r\sqrt{r^2\cos^2(\theta) + (1 + r\sin(\theta))^2}dr \:d\theta.$$</p>
<p>Now this is the part where i get stuck, if i did my calculations right teh expression under the root becomes $r^2 + 1 + 2r\sin(\theta).$ I'm not sure where to go from here so any tips and tricks would be appreciated. I would very much like to see how this is done with this substitution, but alternative solution that leads me to the right answer would also be of great value. Finally, don't show me the calculations down to the answer as i would like to do that myself.</p>
<p>Thanks in advance!</p>
| cpiegore | 268,070 | <p>It's actually easier to evaluate the integral as it stands. I entered</p>
<p>$\int_{0}^{2}\int_{0}^{2}\int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}z\sqrt{x^2+y^2} dxdydz$</p>
<p>in my calculator. It returned $64/9.$ You may want to look at <a href="https://math.stackexchange.com/questions/1677123/integrate-iint-limits-d-frac2a-sqrt4a2-x2-y2dxdy-d-x-a2y2-le/1677292#1677292">this post</a> where another person was having a very similar problem with a double integral. </p>
<p>As a general rule, a change of variables in a multiple integral is most effective when the same (or very similar) expressions occur in <strong>both</strong> the integrand and the region of integration. </p>
|
1,681,872 | <p>My task is this:</p>
<p>Use cylinder coordinates to calculate:$$\iiint\limits_{A}z\sqrt{x^2 + y^2}dA, \enspace A = \left\{(x,y,z):x^2 + (y - 1)^2 \leq 1,\: 0 \leq z \leq 2\right\}.$$</p>
<p>My works so far is this;</p>
<p>Switching to cylindrical coordinates we get:$$A = \left\{(r,\theta,z):0\leq r\leq 1,\: 0\leq \theta \leq 2\pi,\: 0\leq z \leq 2\right\}.$$
Now my book tells me that if you want the center in another point $(a,b,c)$, you should use the substitution: $$x = a + r\cos(\theta),\: y = b + r\sin(\theta),\: z = c + z.$$</p>
<p>With this in mind we change to cylindrical and add the bounderies (don't forget the jacobian).$$\int\limits_{0}^{2\pi}\int\limits_{0}^{1}\int\limits_{0}^{2}zr\sqrt{r^2\cos^2(\theta) + (1 + r\sin(\theta))^2}dz\:dr\:d\theta \:=\:2\int\limits_{0}^{2\pi}\int\limits_{0}^{1} r\sqrt{r^2\cos^2(\theta) + (1 + r\sin(\theta))^2}dr \:d\theta.$$</p>
<p>Now this is the part where i get stuck, if i did my calculations right teh expression under the root becomes $r^2 + 1 + 2r\sin(\theta).$ I'm not sure where to go from here so any tips and tricks would be appreciated. I would very much like to see how this is done with this substitution, but alternative solution that leads me to the right answer would also be of great value. Finally, don't show me the calculations down to the answer as i would like to do that myself.</p>
<p>Thanks in advance!</p>
| Kuifje | 273,220 | <p>Your can use a simpler (?) transformation with cylindrical coordinates. Given that <span class="math-container">$x^2+(y-1)^2=1$</span> is a cylinder shifted 1 unit along the <span class="math-container">$y$</span> axis, you should not consider points in the <span class="math-container">$y<0$</span> plane (<span class="math-container">$\pi \le \theta\le 2\pi$</span>), and the polar equation of the cylinder is not <span class="math-container">$r=1$</span> but <span class="math-container">$r=2\sin\theta$</span>. Therefore, it should be
<span class="math-container">$$
A=\{(r,\theta,z)\;|\; 0 \le \theta \le \pi, 0 \le r \le 2\sin\theta, 0\le z \le 2\}
$$</span></p>
<p>Once you have done that, the integral is easy and equals:
<span class="math-container">$$
\int_0^{\pi} \int_0^{2\sin\theta} \int_0^{2} z\, r\, r \; dz drd\theta = \frac{64}{9}
$$</span></p>
|
698,460 | <p>I've literally tried everything including proofs by cases. If I were to try a an odd integer, then I would get $8k^3 + 4k^2 + 8k^2 + 4k + 1$, which obviously couldn't convert to a $\mod 7$, and I'm honestly not sure what I can do! Does anyone here have any tips?</p>
<p>Note, this is a soft question.</p>
| NasuSama | 67,036 | <p>The following might help you out when determining the integrals without complete computation.</p>
<p><strong>Hint</strong>: Think $e^{-inx}$ of Euler's identity. We get $\cos(nx) - i\sin(nx)$. Now, look at the integral carefully. If we multiply each term by term and then rewrite the whole integral as two integrals, what do you know about each expression in the integral?</p>
<p><strong>Hint</strong>: The integral of even function over interval with symmetric bounds is zero, whereas the integral of odd function over interval with symmetric bounds is not zero.</p>
<p><strong>Hint</strong>: The product of two even functions is an even function. The product of two odd functions is an even function. Also, the product of an even function and an odd function is an odd function.</p>
<p><strong>Hint</strong>: $\cos$ is an even function, whereas $\sin$ is an odd function.</p>
<p>See the <a href="https://math.stackexchange.com/questions/438691/how-does-knowing-a-function-as-even-or-odd-help-in-integration">question asked here in MSE site</a> and reference about <a href="http://en.wikipedia.org/wiki/Even_and_odd_functions#Basic_properties" rel="nofollow noreferrer">even and odd functions</a>.</p>
|
718,664 | <p>I'm trying to find a small primitive root modulo $p^k$, where $p$ is prime. My strategy is to test small numbers $g=2,3,\ldots$ until I find a primitive root modulo $p$. That is, until $ord_pg=\phi(p)=p-1$. There are results that suggest such a search won't take long.</p>
<p>To go from modulo $p$ to modulo $p^k$, there is the following <a href="http://en.wikipedia.org/wiki/Primitive_root_modulo_n#Finding_primitive_roots" rel="noreferrer">well-known</a> theorem:</p>
<blockquote>
<p>Thm: If $g$ is a primitive root modulo $p$, then $$\begin{cases} g+p & g^{p-1}\equiv 1\pmod{p^2}\\g & g^{p-1}\not\equiv 1\pmod{p^2}\end{cases}$$
is a primitive root modulo $p^k$, for all $k\in\mathbb{N}.$</p>
</blockquote>
<p>However, I was unable to find an example that falls into the first case (except $p=2$, trivially). Hence, I have two related questions:</p>
<ol>
<li>Are there any odd primes $p$, and primitive roots $g$, such that $g^{p-1}\equiv 1\pmod{p^2}$?</li>
<li>Are there any odd primes $p$, and minimal primitive roots $g$, such that $g^{p-1}\equiv 1\pmod{p^2}$?</li>
</ol>
| Álvaro Lozano-Robledo | 14,699 | <p>For your first question, consider $p=71$ and $g=11$. Here, $11$ is a primitive root modulo $71$, but $11^{70}\equiv 1 \bmod 71^2$. Unfortunately, this is not an example of your second question, because $7\bmod 71$ is also a primitive root.</p>
<p>I found this example looking at lists of <a href="https://oeis.org/A174422" rel="noreferrer">Wieferich primes</a> with respect to a certain base. There is probably an example for your second question farther down the list; I see no reason why there couldn't be one.</p>
<p>I just checked the first $103$ odd primes $p$ and none of them satisfy $(2)$, with $g$ the smallest positive primitive root modulo $p$.</p>
|
4,589,828 | <p>I'm stuck in proving the function <span class="math-container">$$f(x) = \vert x + e^x\vert $$</span> has a minimum.</p>
<p>This is what I did:</p>
<p><span class="math-container">$$f'(x) = (1+e^x)\text{sgn}(1+e^x)$$</span></p>
<p>But this function is never zero, because the exponential is always positive. So when I study the sign of the derivative, it is always increasing.</p>
<p>Yet the plot of the funciton shows a "sort of" a minimum.</p>
<p><strong>Adds</strong></p>
<p>Since there is an absolute value I calculated the difference quotient for <span class="math-container">$f(x) > 0$</span> and <span class="math-container">$f(x) < 0$</span> obtaining the function is continuous when <span class="math-container">$x+e^x > 0$</span> and when <span class="math-container">$x+e^x < 0$</span></p>
<p>Yet I also understood @Martin R. comment but how to find that point where <span class="math-container">$f(x)$</span> is discontinuous?</p>
| Angelo | 771,461 | <p>You could also proceed in the following way.</p>
<p>By letting <span class="math-container">$\;t=x-\sqrt{1-x^2}\;,\;$</span> we get that</p>
<p><span class="math-container">$x=\dfrac12\left(t\pm\sqrt{2-t^2}\right)\;,\quad\mathrm dx=\dfrac12\left(\!\!1\mp\dfrac t{\sqrt{2-t^2}}\!\!\right)\mathrm dt\;\;.$</span></p>
<p><span class="math-container">$\displaystyle\int\frac1{x-\sqrt{1-x^2}}\,\mathrm dx=\frac12\!\int\frac1t\left(\!\!1\mp\frac t{\sqrt{2-t^2}}\!\!\right)\mathrm dt=$</span></p>
<p><span class="math-container">$\displaystyle\quad=\frac12\!\int\frac1t\,\mathrm dt\;\mp\;\frac12\!\int\frac{\mathrm dt}{\sqrt{2-t^2}}\;\;.$</span></p>
<p>Moreover ,</p>
<p><span class="math-container">$\mp\dfrac{\mathrm dt}{\sqrt{2-t^2}}=-\dfrac{\mathrm dt}{2x-t}=-\dfrac{t’(x)\,\mathrm dx}{x+\sqrt{1-x^2}}=$</span></p>
<p><span class="math-container">$\quad=-\dfrac{1+\frac x{\sqrt{1-x^2}}}{\sqrt{1-x^2}\left(\!1+\frac x{\sqrt{1-x^2}}\!\right)}\,\mathrm dx=-\dfrac{\mathrm dx}{\sqrt{1-x^2}}\;\;.$</span></p>
<p>Hence ,</p>
<p><span class="math-container">$\displaystyle\int\frac1{x-\sqrt{1-x^2}}\,\mathrm dx=\frac12\!\int\frac1t\,\mathrm dt\;\mp\;\frac12\!\int\frac{\mathrm dt}{\sqrt{2-t^2}}=$</span></p>
<p><span class="math-container">$\displaystyle\quad=\frac12\ln|t|-\frac12\!\int\!\dfrac{\mathrm dx}{\sqrt{1-x^2}}=$</span></p>
<p><span class="math-container">$\quad=\dfrac12\ln\bigg|x-\sqrt{1-x^2}\bigg|-\dfrac12\arcsin x+C\;\;.$</span></p>
|
2,760,605 | <p>$||x-1|-2|=|x-3|$.
Find the value of $x$.<br>
In my attempt I got the critical values of the expression as $1$ and $3$.
But I’m not sure is we can just not consider the $2$ in the LHS.</p>
<p>My steps are Case 1: when $x>3$, $x-1-2=x-3$, $0=0$</p>
<p>Case 2: $1<x<3$, $X-1-2=-x+3 \implies 2x=6 \implies x= 3$</p>
<p>Case 3: $x<1$, $-x+1-2=-x+3$. Not possible.</p>
| adfriedman | 153,126 | <p>First, write
$$\ln\left(\frac{n}{(n!)^{1/n}}\right)
=\ln(n) - \tfrac{1}{n}\ln(n!) = \ln(n) - \tfrac1n \sum_{k=1}^n \ln(k)
= \ln(n) - \tfrac1n S_n$$</p>
<p>Consider upper and lower approximations of the integral of $\ln(x)$:</p>
<p>$$S_n = \sum_{k=1}^n \ln(k) \leq \int_1^{n+1}\ln(x)\;dx \leq \sum_{k=2}^{n+1}\ln(k) = S_n + \ln(n+1)$$
Hence
$$S_n \leq \int_1^{n+1}\ln(x)\;dx = \left[ x\ln(x)-x \right]_1^{n+1} = (n+1)\ln(n+1) - n$$ and $$\int_1^{n+1}\ln(x)\;dx - \ln(n+1) = n\ln(n+1) - n \leq S_n$$</p>
<p>It follows:
$$\ln(n)-\tfrac1n \big((n+1)\ln(n+1) - n\big) \leq \ln(n) - \tfrac1n S_n \leq \ln(n) - \tfrac1n\big(n\ln(n+1) - n\big) $$
which simplifies to
$$\ln\Bigg(\underbrace{\frac{n}{(n+1)^{1+\tfrac1n}}}_{\to \; 1}\Bigg) + 1
\leq \ln(n)- \tfrac1n S_n \leq \ln\Bigg(\underbrace{\frac{n}{n+1}}_{\to\; 1}\Bigg) + 1$$</p>
<p>By the squeeze theorem $\ln\left(\frac{n}{(n!)^{1/n}}\right) = \ln(n) - \tfrac1n S_n \to 1$ as the log terms on both sides approach $\ln(1) = 0$, hence
$$\frac{n}{(n!)^{1/n}} = \exp\left(\ln\left(\frac{n}{(n!)^{1/n}}\right)\right) \to \exp(1) = e$$</p>
|
2,835,038 | <p>Assuming that sum of probabilities for all possible events that can occur should sum to 1, how does one denote this for a conditional probability? Is it $P(A|E_1) + P(A|E_2) + ... = 1$, where $E_i$ is a specific event to be conditioned on? Or is the answer something else entirely? </p>
| Shirish Kulhari | 458,802 | <p>If $E_1, E_2, \ldots$ is a collection of disjoint events whose union equals the entire sample space (exhaustive), then $P(A\cap E_1)+P(A\cap E_2)+\ldots = P(A|E_1)P(E_1)+P(A|E_2)P(E_2)+\ldots = P(A)$.</p>
|
2,871,947 | <p>I can prove that commutator is minimal subgroup such that factor group of it is abelian. I had encountered one statement as</p>
<blockquote>
<p>If <span class="math-container">$H$</span> is a subgroup containing commutator subgroup then <span class="math-container">$H$</span> is
normal.</p>
</blockquote>
<p>I.e. we have to show that <span class="math-container">$\forall g\in G$</span> such that <span class="math-container">$gHg^{-1}=H$</span> with fact that <span class="math-container">$G'\subset H$</span></p>
<p>It is for elements in <span class="math-container">$G'$</span> to show condition for normality. </p>
<p>But how to do for elements not in <span class="math-container">$G'$</span> but in <span class="math-container">$H$</span>, that is in <span class="math-container">$H\setminus G'$</span>?</p>
| Angina Seng | 436,618 | <p>$G'$ is certainly normal in $G$, and $G/G'$ is Abelian. Every
subgroup of an Abelian group is normal. But $H/G'$ is a subgroup
of $G/G'$ so $H/G'$ is normal in $G/G'$. Therefore, by the third
isomorphism theorem for groups, $H$ is normal in $G$.</p>
|
780,588 | <p>I need help with this proof:</p>
<p>Let $V, W$ be K-vectorspaces.
Let $T_1, T_2$ be subspaces of V with $T_1 \subseteq T_2$.
Let $f \in hom_K(V,W)$.</p>
<p>Show the following: If $ f(T_1) = f(T_2)$ then exists a subspace $U$ of $ker(f)$ so that $ U \bigoplus T_1 = T_2$</p>
<p>I'm sorry, but I really have no clue how to begin at least one direction of this proof, so I'm grateful for any kind of help!</p>
| Walid Taamallah | 145,898 | <p>Not a full answer: We have that $f(T_2)=f(T_1)$ hence in particular we have the first inclusion $f(T_2)\subset f(T_1)$. For each $t_2\in T_2$ we have that $f(t_2)\in f(T_2)\implies f(t_2)\in f(T_1)$, then there exists $t_1\in T_1$ such that
$$f(t_2)=f(t_1)\implies f(t_2-t_1)=0\implies t_2-t_1\in \ker f$$
Hence there exists $u\in \ker(f)$ such that
$$t_2-t_1=u\implies t_2=u+t_1$$
Hence
$$T_2=\ker f+T_1$$
Now using the other inclusion $f(T_1)\subset f(T_2)$ we show in a similar way that
$$T_1=\ker f+T_2$$.</p>
|
2,558,646 | <p>Okay, so this is bugging me now.</p>
<p>I know this:</p>
<p>while tg angle = Y / X</p>
<p>for y > 0, angle is an element of < 0, PI ></p>
<p>for y < 0, angle is an element of < PI, 2PI ></p>
<p>for y = 0, angle is 0 or PI or 2PI.</p>
<p>Okay, but how to determine what out of those 3 values angle actually is ?</p>
<p>So far this is what I know:</p>
<p>If X is negative, angle should be PI, and if X is positive, angle should be either 0 or 2PI... And thats bugging me now, how do I know if its 2PI or 0 ?</p>
<p>EDIT:</p>
<p>I am asking this because if I assume that angle is 0, while calculating root of complex number I would use:</p>
<p>r * cis ((0 + 2*k*PI) / n)</p>
<p>and if I assume that angle is 2PI, it would be</p>
<p>r * cis ((2PI + 2*k*PI) / n)</p>
<p>What should I assume my angle is when y = 0 and x > 0 ?:/</p>
| kimchi lover | 457,779 | <p>One way to start is to see if $f(x)/L_f(x)\sim 1$, where $L_f(x)$ is the lead term of $f(x)$.</p>
|
705,662 | <p>So my function is $f(z) = \sqrt(r)e^{i\theta/2}$.</p>
<p>I want to find its contour integral over the upper half-disk of $z = e^{i\theta}$, where $0\leq\theta\leq\pi$.</p>
<p>I know that:</p>
<p>$\int f(z) dz$ = $\int_a^b f(z(t)) * z'(t) dt$</p>
<p>But in the above problem, the function I want to find the contour integral of isn't in terms of z, it's $f(z) = \sqrt(r)e^{i\theta/2}$</p>
<p>I know this is a branch of $z^{1/2}$ but not sure where to go from here...</p>
<p>Feel like I'm pretty close but missing something</p>
<p>Thanks for the help,
Mariogs</p>
| Yiorgos S. Smyrlis | 57,021 | <p>The function $f(z)=z^{1/2}$, with $f(1)=1$, is analytic in $\mathbb C\smallsetminus \{ti: t\in (-\infty,0]\}$, and
$$
f(z)=\Big(\frac{2}{3}z^{3/2}\Big)'.
$$
So, for $\gamma: [0,\pi]\to\mathbb C$, with $\gamma(t)=\mathrm{e}^{it}$, and $g(z)=\frac{2}{3}z^{3/2}$,
$$
\int_\gamma f(z)\,dz=g\big(\gamma(\pi)\big)-g\big(\gamma(0)\big)=\frac{2}{3}\big(\mathrm{e}^{3\pi i}-1\big)=-\frac{4}{3}.
$$</p>
|
706,250 | <p>For all $n\ge3\in \mathbb N$, if $n \equiv 3 \pmod{4}$ then ${3^n} \equiv 2 \pmod{5}$.</p>
<p>I tried to set $n = 3+4k$ but it doesn't help.</p>
<p>Any hints first please?</p>
| Bill Dubuque | 242 | <p>Prime $\,P\nmid A\,\Rightarrow\,A^{\large J+K(P-1)} = A^{\large J}(\color{#c00}{\large A^{P-1}})^{\large K}\equiv A^{\large J} \color{#c00}1^{\large K}\equiv A^{\large J}\!\pmod P,\ $ by $\rm\,\color{#c00}{little\ Fermat}$.</p>
|
1,116,215 | <p>In C. Adam's Topology, it is written that $f(A) - f(B) \subset f(A - B)$ for any function, but $f(A) - f(B) = f(A - B)$ iff f is bijective. I can come the half way, i.e., $f(A) - f(B) \subset f(A - B)$; but for the $f(A - B) \subset f(A) - f(B)$, I don't know both of how to prove correctness of it for bijective and falseness of it for non-one-to-one function;</p>
<p>Suppose $y\in f(A) - f(B)$, then $y$ belongs to $f(A)$ but it does not belong to $f(B)$; thus there exist(s) one or more $x_i$ such that $y=f(x_i)$ and non of $x_i$ must be in $B$ since it results in $f(x_i)=y\in f(B)$, a contradiction; so for arbitrary $y$, if $y\in f(A) - f(B) \implies y\in f(A - B)$, thus $f(A) - f(B) \subset f(A - B)$.</p>
<p>I appreciate it if please help me to prove(one-to-one)/disprove(non-one-to-one) in case of $f(A - B) \subset f(A) - f(B)$. </p>
| Mauro ALLEGRANZA | 108,274 | <p>For the <em>bijective</em> case, consider :</p>
<blockquote>
<p><span class="math-container">$x \in f(A - B)$</span>;</p>
</blockquote>
<p>this means that there is <span class="math-container">$y_1 \in A - B$</span> such that : <span class="math-container">$x=f(y_1)$</span>, and <span class="math-container">$y_1 \in A - B$</span> implies : <span class="math-container">$y_1 \in A$</span> and <span class="math-container">$y_1 \notin B$</span>.</p>
<p>From : <span class="math-container">$x=f(y_1)$</span> for <span class="math-container">$y_1 \in A$</span> we have that : <span class="math-container">$x \in f(A)$</span>.</p>
<p>Assume now that : <span class="math-container">$x \in f(B)$</span>; this implies that there exists <span class="math-container">$y_2 \in B$</span> such that : <span class="math-container">$x=f(y_2)$</span>.</p>
<p>If <span class="math-container">$y_2 = y_1$</span>, we have that : <span class="math-container">$y_1 = y_2 \in B$</span>, contradicting the above fact that : <span class="math-container">$y_1 \notin B$</span>.</p>
<p>Thus we have <span class="math-container">$y_2 \ne y_1$</span> and :</p>
<blockquote>
<p><span class="math-container">$f(y_1)= x = f(y_2)$</span>.</p>
</blockquote>
<p>But according to the definition of <a href="http://en.wikipedia.org/wiki/Bijection" rel="nofollow noreferrer">bijection</a> :</p>
<blockquote>
<p>for a pairing between <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> to be a <em>bijection</em>, four properties must hold:</p>
<p>[...] <em>iv)</em> no element of <span class="math-container">$Y$</span> may be paired with more than one element of <span class="math-container">$X$</span>.</p>
</blockquote>
<p>Thus we have a contradiction, because <span class="math-container">$x$</span> is paired with two distict elements <span class="math-container">$y_1$</span> and <span class="math-container">$y_2$</span>.</p>
<p>Conclusion : we have to discard the assumption that : <span class="math-container">$x \in f(B)$</span>, and thus : <span class="math-container">$x \notin f(B)$</span>.</p>
<p>From : <span class="math-container">$x \in f(A)$</span> and <span class="math-container">$x \notin f(B)$</span> we conclude with : <span class="math-container">$x \in f(A) - f(B)$</span>, and thus we have :</p>
<blockquote>
<blockquote>
<p><span class="math-container">$f(A - B) \subset f(A) - f(B)$</span>.</p>
</blockquote>
</blockquote>
|
3,866,673 | <p>I’m working on my linear algebra assignment and now struggling to solve this problem, say, on the complex field, find the Jordan canonical form of
<span class="math-container">$$A=\begin{bmatrix}\quad &\quad &\quad & a_1\\ \quad & \quad & a_2 & \quad \\ \quad & \dots & \quad & \quad \\ a_n &\quad & \quad & \quad \\ \end{bmatrix}.$$</span>
Usually I’ll work with characteristic polynomial in order to obtain the eigenvalues when being asked to find JCF. However in this problem, at least I believe, finding characteristic polynomial can be a really tough task. See <span class="math-container">$\lambda I- A$</span> does not have a friendly look and there is no way to get the determinant and find all the roots.</p>
<p>I do believe the problem is rather difficult and I really do not have any direction in my mind. I would be very appreciated if anyone could offer some help.</p>
| 5xum | 112,884 | <p>The same way</p>
<p><span class="math-container">$$1+2+3+\cdots + 100 = 1+2+3+\cdots +99+100$$</span></p>
|
812,379 | <p>Is the above true? (I think it is!) if so, please can somebody explain why? I don't see it!</p>
| lhf | 589 | <p>By definition, $P$ is a prime ideal when $xy\in P \implies x\in P$ or $y\in P$. By induction, $x_1 x_2 \cdots x_n \in P \implies x_i \in P$, for some $i$.</p>
<p>In particular, since $0 \in P$, if $a^n=0$, then $a^n \in P$, and so $a\in P$.</p>
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