qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
1,625,292 | <p>Define a class $K$ of ordinals inductively as follows:</p>
<ul>
<li><p>$0=\emptyset\in K$.</p></li>
<li><p>For all $\alpha\in K$, the succesor of $\alpha$ is also an element of $K$.</p></li>
<li><p>For every function $f\colon \mathbb N\to K$, the ordinal that immediately follows after all ordinals $f(0),f(1),\dots f(n),\dots$ is also an element of $K$.</p></li>
</ul>
<p>Call an arbitrary ordinal $\beta$ fett iff $\beta\not\in K$.</p>
<p><b>Question</b>. Do fett ordinals exist? In other words:</p>
<blockquote>
<p>Is there an ordinal $\beta$ such that $\beta$ is <b>not</b> an element of the class $K$?</p>
</blockquote>
| Noah Schweber | 28,111 | <p>Yes, such ordinals exist - for example, $\omega_1$, the first uncountable ordinal. </p>
<p>The crucial issue here is <em>cofinality</em>: the cofinality of an ordinal $\alpha$ is the least $\beta$ such that there is a function $f:\beta\rightarrow\alpha$ whose range is unbounded in $\alpha$. It's easy to show that $K$ consists exactly of those ordinals of <em>countable cofinality</em> - that is, whose cofinality is $\omega$ - which are also not larger than any ordinal of uncountable cofinality.</p>
<p>Now, $\omega_1$ is <em>not</em> of countable cofinality, since the union of countably many countable sets is countable. So $\omega_1$ - and any larger ordinal - is not in $K$.</p>
|
1,625,292 | <p>Define a class $K$ of ordinals inductively as follows:</p>
<ul>
<li><p>$0=\emptyset\in K$.</p></li>
<li><p>For all $\alpha\in K$, the succesor of $\alpha$ is also an element of $K$.</p></li>
<li><p>For every function $f\colon \mathbb N\to K$, the ordinal that immediately follows after all ordinals $f(0),f(1),\dots f(n),\dots$ is also an element of $K$.</p></li>
</ul>
<p>Call an arbitrary ordinal $\beta$ fett iff $\beta\not\in K$.</p>
<p><b>Question</b>. Do fett ordinals exist? In other words:</p>
<blockquote>
<p>Is there an ordinal $\beta$ such that $\beta$ is <b>not</b> an element of the class $K$?</p>
</blockquote>
| Asaf Karagila | 622 | <p>First of all, note that the class of <em>all</em> the ordinals satisfies this definition. You want to say that $K$ is the smallest class of ordinals satisfying that the three requirements hold.</p>
<p>And I claim that at least under the axiom of choice $\omega_1$ is such a class of ordinals, and therefore $\omega_1$ is a Fett ordinal. It is in fact, the least Fett ordinal, therefore making it the Boba Fett of ordinals.</p>
<p>Let's see why.</p>
<ol>
<li>$\varnothing\in\omega_1$, easy peasy.</li>
<li>If $\alpha\in\omega_1$, then $\alpha+1$ is also countable, so $\alpha+1\in\omega_1$ as well.</li>
<li>Finally, if $f\colon\Bbb N\to\omega_1$ is any function, then $\sup\operatorname{rng}(f)$ is the countable union of countable ordinals, therefore a countable ordinal. So we have all three requirements.</li>
</ol>
<p>However, the axiom of choice is needed here. Not only that it is consistent that $\omega_1$ is not <em>not</em> a Fett ordinal without the axiom of choice, namely it <em>can</em> be the countable union of countable ordinals; it is possible that <em>every</em> ordinal is a successor or a countable union of smaller ordinals. In that case, no ordinal is a Fett ordinal.</p>
<p>So you need the axiom of choice to conclude the existence of Boba Fett.</p>
|
4,032,767 | <blockquote>
<p>Prove that there exists no bijective function <span class="math-container">$f: \Bbb{N} \to \Bbb{N}$</span> such that <span class="math-container">$$f(mn)=f(m)+f(n)+3f(m)f(n)$$</span> for <span class="math-container">$m,n \geqslant1.$</span></p>
</blockquote>
<p>This was a problem from a Putnam practice book and I couldn't seem to figure out how to show this. Initially, I started to wonder that if the function were bijective then it has to be injective as well as surjective. So trying to determine if its injective let <span class="math-container">$a,b \in \mathbb{N}$</span> then if it were injective <span class="math-container">$$f(mn)=f(ab) \implies f(m)+f(n)+3f(m)f(n) =f(a)+f(b)+3f(a)f(b)$$</span> but this doesn't seem to help since I don't have explicitly <span class="math-container">$f$</span>. How should I look at this problem?</p>
| peters onyilo Agnes | 889,933 | <p>1.Show that the function is well defined
2.Prove that there exist a kernel.
3. Prove that identity maps to element of the kernel
If you can't prove 1 and 2 the function or mapping is not bijective.
In other words, you can prove that the function is both injective and surjective( i.e if the inverse of the function maps back to every element in the domain) as required.</p>
|
1,821,411 | <p>$f:[a,b]\rightarrow R$ that is integrable on [a,b]</p>
<p>So we need to prove:</p>
<p>$$\int_{-b}^{-a}f(-x)dx=\int_{a}^{b}f(x)dx$$</p>
<p>1.) So we'll use a property of definite integrals: (homogeny I think it's called?)</p>
<p>$$\int_{-b}^{-a}f(-x)dx=-1\int_{-b}^{-a}f(x)dx$$</p>
<p>2.) Great, now using the fundamental theorem of calculus:</p>
<p>$$-1\int_{-b}^{-a}f(x)dx=(-1)^2\int_{-a}^{-b}f(x)dx=\int_{-a}^{-b}f(x)dx$$</p>
<p>This is where I'm stuck. For some reason I think it might be smarter to skip step 2, to leave it asL</p>
<p>$$-1\int_{-b}^{-a}f(x)dx$$ </p>
<p>because graphically, we've "flipped" the graph about the x-axis, but we're still calculating the same area. Proving that using properties seems to have stumped me.</p>
<p>I prefer hints over solutions, thanks.</p>
| Vineet Mangal | 346,869 | <p>There is a small error in your first step, you didn't change the sign of the limits. Always keep in mind that after making any substitution, don't forget to change the limits. Since you have made a substitution $x=-t$, so change the limits accordingly. So corrected step 1 is $$-\int_{b}^af(x)dx$$
Now you can reverse the limits to get the correct answer i.e.
$$\int_{a}^bf(x)dx$$</p>
|
3,888,146 | <p>When we give a proof that the tangent is the sine to cosine ratio of an oriented angle,</p>
<p><span class="math-container">$$\bbox[5px,border:2px solid #C0A000]{\tan \alpha=\frac{\sin\alpha}{\cos \alpha}}$$</span>
with <span class="math-container">$\cos \alpha \neq 0$</span>, we take the tangent <span class="math-container">$t$</span> in <span class="math-container">$A(1,0)\equiv S$</span> to the circle of center in <span class="math-container">$O(0,0)$</span> ad radius <span class="math-container">$r=1$</span>. See the image</p>
<p><a href="https://i.stack.imgur.com/LPQPV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LPQPV.png" alt="https://www.youmath.it/images/stories/funzioni-elementari/definizione-tangente.png" /></a></p>
<blockquote>
<p>The name tangent has been given because we consider the tangent to the circle of radius <span class="math-container">$1$</span> at point <span class="math-container">$A\equiv S$</span> or for another reason?</p>
</blockquote>
| user247327 | 247,327 | <p>Saying that the diameter lies on a given line segment means that the center is on that line. If A and B are points on the circle then the line segment between them is a chord and its perpendicular bisector also passes through the center of the circle.</p>
<p>So construct the perpendicular bisector of AB. The point where it crosses the given line is the center of the circle. Set the leg of compasses at that point and draw the circle through A and B.</p>
|
3,755,288 | <p>I'm trying to solve this:</p>
<blockquote>
<p>Which of the following is the closest to the value of this integral?</p>
<p><span class="math-container">$$\int_{0}^{1}\sqrt {1 + \frac{1}{3x}} \ dx$$</span></p>
<p>(A) 1</p>
<p>(B) 1.2</p>
<p>(C) 1.6</p>
<p>(D) 2</p>
<p>(E) The integral doesn't converge.</p>
</blockquote>
<p>I've found a lower bound by manually calculating <span class="math-container">$\int_{0}^{1} \sqrt{1+\frac{1}{3}} \ dx \approx 1.1547$</span>. This eliminates option (A). I also see no reason why the integral shouldn't converge. However, to pick an option out of (B), (C) and (D) I need to find an upper bound too. Ideas? Please note that I'm not supposed to use a calculator to solve this.</p>
<p>From <strong>GRE problem sets by UChicago</strong></p>
| Jack D'Aurizio | 44,121 | <p>Since <span class="math-container">$\sqrt{t^2+1/3}$</span> is a convex function on <span class="math-container">$[0,1]$</span>, you may simply use the Hermite-Hadamard inequality to derive that</p>
<p><span class="math-container">$$ \sqrt{2+\frac{1}{3}}\leq 2\int_{0}^{1}\sqrt{t^2+1/3}\,dt \leq \sqrt{3} $$</span>
so <span class="math-container">$(C)$</span> is the correct option.</p>
|
4,095,831 | <p>If <span class="math-container">$\frac{dy}{dx}=y\sec^2x$</span> and <span class="math-container">$y=5$</span> when <span class="math-container">$x=0$</span>, then <span class="math-container">$y=$</span>?
A) <span class="math-container">$e^{\tan x} + 4$</span></p>
<p>B) <span class="math-container">$e^{\tan x} + 4$</span></p>
<p>C) <span class="math-container">$5e^{\tan x}$</span></p>
<p>D) <span class="math-container">$\tan x +5$</span></p>
<p>E) <span class="math-container">$\tan x + 5e^x$</span></p>
<p>This is how I solved it:</p>
<p><span class="math-container">$\frac{dy}{y}=\sec^2x dx$</span></p>
<p><span class="math-container">$\int \frac{dy}{y}= \int \sec^2x dx$</span></p>
<p><span class="math-container">$\ln|y|= \tan x +C$</span></p>
<p><span class="math-container">$e^{\ln|y|}=e^{\tan x}+C$</span></p>
<p><span class="math-container">$|y|=e^{\tan x} + C$</span></p>
<p><span class="math-container">$y=\pm (e^{\tan x} + C)$</span></p>
<p>Since <span class="math-container">$y=5$</span> we choose <span class="math-container">$y=e^{\tan x} + C$</span></p>
<p>But the answer is C)</p>
<p>I know that I can find <span class="math-container">$dy/dx$</span> for each answer choice and C) is the only one that fits.. however I don't know what I did wrong in my solution.</p>
| jjagmath | 571,433 | <p>Here is where you made a mistake</p>
<blockquote>
<p><span class="math-container">$\ln|y|= \tan x +C$</span></p>
<p><span class="math-container">$e^{\ln|y|}=e^{\tan x}+C$</span></p>
</blockquote>
<p>If <span class="math-container">$\ln|y|= \tan x +C$</span>, after exponentiation you get <span class="math-container">$e^{\ln|y|}=e^{\tan x+C}$</span></p>
|
3,752,162 | <p>I already knew that normal subgroups where important because they allow for quotient space to have a group structure.
But I was told that normal subgroups are also important in particular because they are the only subgroups that can occur as kernels of goup homomorphisms. Why is this property a big deal in algebra?</p>
| halrankard | 688,699 | <p>I suppose this is more of opinion-based question. If I interpret your question as "Why is it a big deal that the subgroup determined by the kernel of a homomorphism must be normal?", then the answer is "It's not really a big deal. That's trivial." Instead, if I interpret your question as "Why is it a big a deal that normal subgroups arise as kernels of homomorphisms?" then that's another story.</p>
<p>There are many situations where you can prove something interesting about (normal) subgroups by being clever with kernels of homomorphisms. Here are two simple examples.</p>
<p><strong>Fact 1:</strong> If <span class="math-container">$G$</span> is a group and <span class="math-container">$H$</span> is a subgroup of index <span class="math-container">$n$</span>. Then <span class="math-container">$H$</span> contains a normal subgroup <span class="math-container">$K$</span> of <span class="math-container">$G$</span> with index at most <span class="math-container">$n!$</span>.</p>
<p><strong>Proof:</strong> Let <span class="math-container">$X$</span> be the set of left cosets of <span class="math-container">$H$</span>. Then any element of <span class="math-container">$g$</span> determines a permutation of <span class="math-container">$X$</span> in which each coset is multiplied on the left by <span class="math-container">$g$</span>. Define the function <span class="math-container">$f:G\to S_{X}$</span> sending <span class="math-container">$g$</span> to its associated permutation. (Here <span class="math-container">$S_{X}$</span> is the group of permutations of <span class="math-container">$X$</span>.) It is easily checked that <span class="math-container">$f$</span> is a homomorphism. Let <span class="math-container">$K$</span> be the kernel. Then <span class="math-container">$K$</span> is a normal subgroup of index at most <span class="math-container">$|S_{X}|=n!$</span>, and it is easy to check that <span class="math-container">$K$</span> is contained in <span class="math-container">$H$</span>.</p>
<p>In the previous proof, the definition of <span class="math-container">$f$</span> implies that <span class="math-container">$K$</span> is the intersection of all conjugates of <span class="math-container">$H$</span>. So <span class="math-container">$K$</span> is the "largest" normal subgroup of <span class="math-container">$G$</span> that is contained in <span class="math-container">$H$</span>.</p>
<p><strong>Comment (added later).</strong> Continuing from the last sentence, set <span class="math-container">$K=\bigcap_{g\in G}gHg^{-1}$</span>. So <span class="math-container">$K$</span> is the largest normal subgroup of <span class="math-container">$G$</span> that is contained in <span class="math-container">$H$</span>. And the proof above shows that it has index at most <span class="math-container">$n!$</span>. But one my try the following more direct approach. Choose left coset representatives <span class="math-container">$g_{1},\ldots,g_{n}$</span> for <span class="math-container">$H$</span>. It is not hard to see that <span class="math-container">$K=\bigcap_{t=1}^{n}g_{t} Hg_{t}^{-1}$</span>. Also, any conjugate of <span class="math-container">$H$</span> still has index <span class="math-container">$n$</span>. So <span class="math-container">$K$</span> is an intersection of <span class="math-container">$n$</span> subgroups of <span class="math-container">$G$</span> each of index <span class="math-container">$n$</span>. Now the general formula <span class="math-container">$[G:H_1\cap H_2]=[G:H_1]\cdot [G:H_2]$</span> tells us that <span class="math-container">$K$</span> has index at most <span class="math-container">$n^n$</span>. So this proves the Fact without using homomorphisms but with a worse bound on the index.</p>
<p><strong>Fact 2:</strong> Suppose <span class="math-container">$G$</span> has order <span class="math-container">$2n$</span> where <span class="math-container">$n$</span> is odd. Then <span class="math-container">$G$</span> has a normal subgroup of size <span class="math-container">$n$</span>.</p>
<p><strong>Proof:</strong> Any element of <span class="math-container">$g$</span> determines a permutation of <span class="math-container">$G$</span> via multiplication on the left. So we get a map <span class="math-container">$\varphi:G\to S_{G}$</span> which is a homomorphism. Let <span class="math-container">$\psi:S_G\to C_2$</span> be the map that sends a permutation in <span class="math-container">$S_G$</span> to <span class="math-container">$0$</span> if and only if it is even. (Here <span class="math-container">$C_2$</span> is the cyclic group of size <span class="math-container">$2$</span>.) Compose these to get a homomorphism <span class="math-container">$f: G\to C_2$</span>. By Cauchy's Theorem, <span class="math-container">$G$</span> has an element <span class="math-container">$x$</span> of order <span class="math-container">$2$</span>. So <span class="math-container">$\varphi(x)$</span> is a permutation of order two with no fixed points. So <span class="math-container">$\varphi(x)$</span> is a product of <span class="math-container">$n$</span> disjoint transpositions. Since <span class="math-container">$n$</span> is odd, we have <span class="math-container">$\psi(\varphi(x))=1$</span>. So <span class="math-container">$f$</span> is surjective. If <span class="math-container">$K$</span> is the kernel of <span class="math-container">$f$</span> then <span class="math-container">$G/K$</span> is isomorphic to <span class="math-container">$C_2$</span>. So <span class="math-container">$K$</span> has index <span class="math-container">$2$</span>, i.e. size <span class="math-container">$n$</span>.</p>
<p>The last fact tells us there are no simple groups of size <span class="math-container">$2n$</span> if <span class="math-container">$n$</span> is odd and bigger than <span class="math-container">$1$</span>.</p>
<p>By the way, the last proof used Cauchy's Theorem for <span class="math-container">$p=2$</span>. I know it's got nothing to do with your main question, but here's a cute proof. Suppose <span class="math-container">$G$</span> is a group with even order. We aim to find an element of order <span class="math-container">$2$</span>. Let <span class="math-container">$X$</span> be the set of elements of order greater than <span class="math-container">$2$</span>. Then no element in <span class="math-container">$X$</span> is equal to it's inverse, so we can partition <span class="math-container">$X$</span> into sets of size <span class="math-container">$2$</span> by putting each element with its inverse. So <span class="math-container">$|X|$</span> is even which means <span class="math-container">$|G\setminus X|$</span> is even. Since <span class="math-container">$G\setminus X$</span> contains the identity, it must contain at least one more element, which has order 2.</p>
|
1,773,776 | <blockquote>
<blockquote>
<p>Question: Prove that the directrix is tangent to the circles that are drawn on a focal chord of a parabola as
diameter.</p>
<p>Here is a picture; <a href="https://i.stack.imgur.com/F8dga.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/F8dga.png" alt="enter image description here"></a></p>
</blockquote>
</blockquote>
<hr>
<p>What I have attempted;</p>
<p>Let the parabola be $y^2=4ax$ </p>
<p>Hence the focus will be at $(a,0)$</p>
<p>Let the focal chord be $y = m(x-a) $</p>
<p>Subbing in $y^2=4ax$</p>
<p>$$y^2=4ax$$</p>
<p>$$ \Leftrightarrow (m(x-a))^2 = 4ax $$</p>
<p>$$ \Leftrightarrow m^2 (x^2-2ax+a^2) = 4ax $$</p>
<p>$$\Leftrightarrow m^2x^2 - 2am^2x + m^2a^2 - 4ax = 0 $$</p>
<p>$$ \Leftrightarrow m^2x^2 -(2am^2+4a)x + m^2a^2 = 0 $$</p>
<p>If $x_1$ and $x_2$ are roots then</p>
<p>$$ x_1 + x_2 = \frac{2am^2+4a}{m^2}$$</p>
<p>$$ \therefore x_1 + x_2 = 2a + \frac{4a}{m^2} $$</p>
<p>and $$ x_1 \cdot x_2 = \frac{m^2a^2}{m^2} $$</p>
<p>$$ \therefore x_1 \cdot x_2 = a^2 $$</p>
<p>Corresponding </p>
<p>$$y_1 + y_2 = m(x_1 - a + x_2 - a)$$</p>
<p>$$y_1 + y_2 = m(x_1 + x_2 - 2a)$$</p>
<p>$$y_1 + y_2 = m(2a + \frac{4a}{m^2} - 2a)$$</p>
<p>$$ \therefore y_1 + y_2 = \frac{4a}{m} $$</p>
<p>$$ y_1 \cdot y_2 = m^2(x_1-a)(x_2-a) $$</p>
<p>$$y_1 \cdot y_2 = m^2(x_1x_2 - a(x_1+x_2) + a^2) $$</p>
<p>$$ y_1 \cdot y_2 = m^2( a^2 - a^2(2 + \frac{4}{m^2}) + a^2) $$</p>
<p>$$ y_1 \cdot y_2 = m^2 (\frac{-4a^2}{m^2}) $$</p>
<p>$$ y_1 \cdot y_2 = -4a^2 $$</p>
<p>Now consider </p>
<p>$$ (x_1 - x_2)^2 = (x_1+x_2)^2 - 4x_1x_2 $$</p>
<p>$$ (x_1 - x_2)^2 = a^2(2 + \frac{4}{m^2})^2 - 4a^2 $$</p>
<p>$$ (x_1 - x_2)^2 = a^2(4+\frac{16}{m^2} + \frac{16}{m^4}) - 4a^2 $$</p>
<p>$$ (x_1 - x_2)^2= \frac{16a^2}{m^2} + \frac{16a^2}{m^4} $$</p>
<p>and</p>
<p>$$ (y_1 - y_2)^2 = (y_1+y_2)^2 - 4y_1y_2 $$</p>
<p>$$(y_1 - y_2)^2 = (\frac{4a}{m})^2 -4 \cdot -4a^2 $$</p>
<p>$$(y_1 - y_2)^2 = \frac{16a^2}{m^2} + 16a^2 $$</p>
<p>Therefore </p>
<p>$$ (x_1 - x_2)^2 + (y_1 - y_2)^2 = \frac{16a^2}{m^2} + \frac{16a^2}{m^4} + \frac{16a^2}{m^2} + 16a^2 $$</p>
<p>$$ (x_1 - x_2)^2 + (y_1 - y_2)^2 = 16a^2(\frac{1}{m^4} + \frac{2}{m^2} + 1) $$</p>
<p>$$ (x_1 - x_2)^2 + (y_1 - y_2)^2 = 16a^2(\frac{1}{m^2} + 1)^2 $$</p>
<p>Hence diameter of the circle is given as</p>
<p>$$ D = \sqrt{16a^2(\frac{1}{m^2} + 1)^2} $$</p>
<p>$$ \therefore D = 4a(\frac{1}{m^2} + 1) $$</p>
<p>Distance from centre of directrix is the $x$ coordinate $+a$ </p>
<p>$$= a + \frac{2a}{m^2} + a $$</p>
<p>$$= 2a + \frac{2a}{m^2} $$</p>
<p>$$= 2a(1+\frac{1}{m^2}) $$</p>
<p>So distance is $2a(1+\frac{1}{m^2}) $</p>
<p>Also notice that the radius of the circle is given as $R = 2a(\frac{1}{m^2} + 1) $</p>
<p>Which equals the distance from centre to the directrix hence the directrix must be tangent to the circle. </p>
<hr>
<p>Could someone please check my proof and tell me if I am correct or not (correct my working and tell me where i went wrong) or also provide me with an alternative way of approaching this question?</p>
| Nikunj | 287,774 | <p>An alternate way (probably shorter) would be to take two points as $(at_1^2,2at_1), (at_2^2,2at_2)$ as the ends of the focal chord. </p>
<p>As this chord passes through focus, we obtain $t_1t_2=-1$
Now, the equation of circle can be written in diametric form as:</p>
<p>$$(x-at_1^2)\left(x-\frac{a}{t_1^2}\right)+(y-2at_1)\left(y+\frac{2a}{t_1}\right)=R^2$$
Where $R$ can be written using $t_1$ and $t_2$.</p>
<p>Now you can differentiate this at $x=-a$ and check for the slope of the tangent.</p>
|
1,773,776 | <blockquote>
<blockquote>
<p>Question: Prove that the directrix is tangent to the circles that are drawn on a focal chord of a parabola as
diameter.</p>
<p>Here is a picture; <a href="https://i.stack.imgur.com/F8dga.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/F8dga.png" alt="enter image description here"></a></p>
</blockquote>
</blockquote>
<hr>
<p>What I have attempted;</p>
<p>Let the parabola be $y^2=4ax$ </p>
<p>Hence the focus will be at $(a,0)$</p>
<p>Let the focal chord be $y = m(x-a) $</p>
<p>Subbing in $y^2=4ax$</p>
<p>$$y^2=4ax$$</p>
<p>$$ \Leftrightarrow (m(x-a))^2 = 4ax $$</p>
<p>$$ \Leftrightarrow m^2 (x^2-2ax+a^2) = 4ax $$</p>
<p>$$\Leftrightarrow m^2x^2 - 2am^2x + m^2a^2 - 4ax = 0 $$</p>
<p>$$ \Leftrightarrow m^2x^2 -(2am^2+4a)x + m^2a^2 = 0 $$</p>
<p>If $x_1$ and $x_2$ are roots then</p>
<p>$$ x_1 + x_2 = \frac{2am^2+4a}{m^2}$$</p>
<p>$$ \therefore x_1 + x_2 = 2a + \frac{4a}{m^2} $$</p>
<p>and $$ x_1 \cdot x_2 = \frac{m^2a^2}{m^2} $$</p>
<p>$$ \therefore x_1 \cdot x_2 = a^2 $$</p>
<p>Corresponding </p>
<p>$$y_1 + y_2 = m(x_1 - a + x_2 - a)$$</p>
<p>$$y_1 + y_2 = m(x_1 + x_2 - 2a)$$</p>
<p>$$y_1 + y_2 = m(2a + \frac{4a}{m^2} - 2a)$$</p>
<p>$$ \therefore y_1 + y_2 = \frac{4a}{m} $$</p>
<p>$$ y_1 \cdot y_2 = m^2(x_1-a)(x_2-a) $$</p>
<p>$$y_1 \cdot y_2 = m^2(x_1x_2 - a(x_1+x_2) + a^2) $$</p>
<p>$$ y_1 \cdot y_2 = m^2( a^2 - a^2(2 + \frac{4}{m^2}) + a^2) $$</p>
<p>$$ y_1 \cdot y_2 = m^2 (\frac{-4a^2}{m^2}) $$</p>
<p>$$ y_1 \cdot y_2 = -4a^2 $$</p>
<p>Now consider </p>
<p>$$ (x_1 - x_2)^2 = (x_1+x_2)^2 - 4x_1x_2 $$</p>
<p>$$ (x_1 - x_2)^2 = a^2(2 + \frac{4}{m^2})^2 - 4a^2 $$</p>
<p>$$ (x_1 - x_2)^2 = a^2(4+\frac{16}{m^2} + \frac{16}{m^4}) - 4a^2 $$</p>
<p>$$ (x_1 - x_2)^2= \frac{16a^2}{m^2} + \frac{16a^2}{m^4} $$</p>
<p>and</p>
<p>$$ (y_1 - y_2)^2 = (y_1+y_2)^2 - 4y_1y_2 $$</p>
<p>$$(y_1 - y_2)^2 = (\frac{4a}{m})^2 -4 \cdot -4a^2 $$</p>
<p>$$(y_1 - y_2)^2 = \frac{16a^2}{m^2} + 16a^2 $$</p>
<p>Therefore </p>
<p>$$ (x_1 - x_2)^2 + (y_1 - y_2)^2 = \frac{16a^2}{m^2} + \frac{16a^2}{m^4} + \frac{16a^2}{m^2} + 16a^2 $$</p>
<p>$$ (x_1 - x_2)^2 + (y_1 - y_2)^2 = 16a^2(\frac{1}{m^4} + \frac{2}{m^2} + 1) $$</p>
<p>$$ (x_1 - x_2)^2 + (y_1 - y_2)^2 = 16a^2(\frac{1}{m^2} + 1)^2 $$</p>
<p>Hence diameter of the circle is given as</p>
<p>$$ D = \sqrt{16a^2(\frac{1}{m^2} + 1)^2} $$</p>
<p>$$ \therefore D = 4a(\frac{1}{m^2} + 1) $$</p>
<p>Distance from centre of directrix is the $x$ coordinate $+a$ </p>
<p>$$= a + \frac{2a}{m^2} + a $$</p>
<p>$$= 2a + \frac{2a}{m^2} $$</p>
<p>$$= 2a(1+\frac{1}{m^2}) $$</p>
<p>So distance is $2a(1+\frac{1}{m^2}) $</p>
<p>Also notice that the radius of the circle is given as $R = 2a(\frac{1}{m^2} + 1) $</p>
<p>Which equals the distance from centre to the directrix hence the directrix must be tangent to the circle. </p>
<hr>
<p>Could someone please check my proof and tell me if I am correct or not (correct my working and tell me where i went wrong) or also provide me with an alternative way of approaching this question?</p>
| Jean Marie | 305,862 | <p>Here is a simple alternative way, fully geometrical.</p>
<p>Have a look at the following picture, with $M_1,M_2$ on parabola with focus $F$ and directrix $D$, $H_1, H_2, H$ the orthogonal projections on D of $P_1, P_2, F$ resp. </p>
<p>Let $r_k:=M_kH_k=M_kF \ (k=1,2)$. Let $C$ be the midpoint of $M_1M_2$, i.e., the center of the circle with diameter $M_1M_2$. The radius of this circle is $r=\dfrac{r_1+r_2}{2}$.</p>
<p>In trapezoid $M_1,H_1,H_2,M_2$, consider line segment $[FH]$ joining midpoints $C$ and $H$ of line segments $M_1M_2$ and $H_1H_2$ resp. The length of $[FH]$ is the mean $\dfrac{r_1+r_2}{2}$ of the lengthes $r_1$ and $r_2$ of $[M_1H_1]$ and $[M_2H_2]$ resp., i.e., the radius of the circle, as desired.</p>
<p><a href="https://i.stack.imgur.com/SmxOT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SmxOT.jpg" alt="enter image description here"></a></p>
|
20,942 | <p>If I had a recursive function (<code>f(n) = f(n-1) + 2*f(n-2)</code> for example), how would I derive a formula to solve this? For example, with the Fibonacci sequence, Binet's Formula can be used to find the nth term.</p>
| Klaus | 524 | <p><a href="http://en.wikipedia.org/wiki/Formal_power_series" rel="nofollow">Formal power series</a> can be also be used to solve recurrence relations.</p>
<p>Let $a_n=f(n)$ and</p>
<p>$$\begin{eqnarray}
S=\sum a_nx^n&=&a_0+a_1x+\sum_2 a_{n-1}x^n+2\sum_2 a_{n-2}x^n\\
&=&a_0+a_1x+\sum_1 a_n x^{n+1}+2\sum_0 a_n x^{n+2}\\
&=&a_0+a_1x+x(S-a_0)+2x^2S
\end{eqnarray}
$$</p>
<p>For the example, suppose that $a_0=0, \ a_1=1$. Then</p>
<p>$$\begin{eqnarray}
S&=&\frac{-x}{2x^2+x-1}\\
&=&\frac{1}{3}\left(\frac{1}{x+1}-\frac{1}{2x-1}\right)\\
&=&\frac{1}{3}\sum_0 (-(-1)^n+2^n)x^n
\end{eqnarray}
$$</p>
<p>Thus $a_n=\frac{1}{3}\left(-(-1)^n+2^{n}\right)$.</p>
|
4,188,656 | <p><strong>Problem</strong>: How many strings are there of length <span class="math-container">$ n $</span> over <span class="math-container">$ \{ 1,2,3,4,5,6 \} $</span> s.t. the sum of all characters in the string divide by <span class="math-container">$ 3 $</span>.</p>
<p><strong>Attempt</strong>:
Initially I thought about solving this using generating functions ( <span class="math-container">$ (x^0 + x^3 +...)^n = (\frac{1}{1-x^3})^n $</span> ), but that is little bit problematic since we don't have an equation equal to some number and maybe this way is too difficult.
Then I thought about using recurrence relation and I done like this:<br />
Let <span class="math-container">$ a_n $</span> denote the number of strings of length <span class="math-container">$ n $</span> s.t. the sum of all characters in the string divide by <span class="math-container">$ 3 $</span>. Let's look at the first character, if it divides by 3 then there are two choices - <span class="math-container">$ 3,6 $</span> and the rest of the <span class="math-container">$ n-1 $</span> string is legal, similarly if the first character does not divide by 3 then there are four choices - <span class="math-container">$ 1,2,4,5 $</span> and the rest of the <span class="math-container">$ n-1 $</span> string is legal, so we have the recurrence relation <span class="math-container">$ a_n = a_{n-1} + a_{n-1} = 2*a_{n-1} $</span>.<br />
Obviously this recurrence relation is wrong <strong>but why?</strong> I keep repeating mistakes like this when creating recurrence relations, and I'd like to know where my mistake is here. ( besides having wrong values for different <span class="math-container">$ n $</span>, <strong>where was the fallacy in logic that led me to the creation of this recurrence relation?</strong> maybe is it that if the first character is from <span class="math-container">$\{ 3,6 \} $</span> then the rest of the string is not necessarily legal? ).</p>
| peek-a-boo | 568,204 | <p>No, let <span class="math-container">$\Omega$</span> be any subset of <span class="math-container">$\Bbb{R}$</span> with finite, positive Lebesgue measure, and let <span class="math-container">$A=\Omega$</span>. For each <span class="math-container">$\alpha\in A$</span>, let <span class="math-container">$f_{\alpha}(x)=1$</span> if <span class="math-container">$x=\alpha$</span> and <span class="math-container">$0$</span> otherwise. Then,
<span class="math-container">\begin{align}
\sup_{\alpha\in A}\int_{\Omega}f_{\alpha}\,d\lambda &= 0<\lambda(\Omega)=\int_{\Omega} 1\,d\lambda =\int_{\Omega}\sup_{\alpha\in A}f_{\alpha}\,d\lambda.
\end{align}</span></p>
|
1,719,840 | <p>If there is a group $G$ with order $a$, having a subgroup $H_1$ with order $b$, and $H_2$ with order $c$, and $bc=a$, $H_1 \cap H_2 = e $. Is $H_1 H_2 =G$? </p>
| Community | -1 | <p>The answer to your question is yes: if $G$ is a finite group, and $H$ and $K$ are subgroups such that $|G| = |H||K|$ and $H \cap K = 1$, then $G = HK$.</p>
<p>The easiest way to see this is to use the identity
$$|HK| = \frac{|H||K|}{|H \cap K|}$$
Note that this identity holds even if $HK$ is merely a subset (not a subgroup) of $G$.</p>
<p>In your case, $|H \cap K| = 1$ and $|H||K| = |G|$, so the identity becomes
$$|HK| = |G|$$
Therefore, $HK$ is a subset of $G$ with the same cardinality as $G$, so $HK = G$.</p>
<p><hr>
For completeness, here is a simple proof of the identity.</p>
<p>Recall that $H \times K$ is the direct product of $H$ and $K$. It is the set of all ordered pairs of the form $(h,k)$ where $h \in H$ and $k \in K$, and it has cardinality $|H \times K| = |H||K|$. ($H \times K$ inherits a group operation from $H$ and $K$, but we won't need to use it here.)</p>
<p>Also recall that $HK$ is the set $\{hk : h \in H, k \in K\}$ which is not necessarily a subgroup of $G$, but it is certainly a subset of $G$.</p>
<p>Define the map $f : H \times K \to HK$ by $f(h,k) = hk$. Clearly this map is surjective, so its image is $HK$, which has cardinality $|HK|$.</p>
<p>Now, how many elements of $H \times K$ are mapped to a given $hk \in HK$? </p>
<p>Note that if $d \in H \cap K$, then $(hd, d^{-1}k) \in H \times K$, and $f(hd, d^{-1}k) = hk$. This shows that there are at least $|H \cap K|$ elements of $H \times K$ which are mapped to $hk$. </p>
<p>It's easy to verify that these are all of the elements which are mapped to $hk$. Suppose that $f(h', k') = h'k' = hk$. Then $h^{-1}h' = k(k')^{-1} \in H \cap K$, call this element $d$. Thus $h' = hd$ and $k' = d^{-1}k$.</p>
<p>This shows that for each $hk \in HK$, there are exactly $|H \cap K|$ elements of $H \times K$ which are mapped to $hk$. Consequently,
$$|H \times K| = |H \cap K||HK|$$
Since $|H \times K| = |H||K|$, this is equivalent to
$$|H||K| = |H \cap K| |HK|$$
Dividing by $|H \cap K|$ gives us the desired identity.</p>
|
1,342,747 | <p>I am studying H. L. Royden's Real Analysis which includes some introduction to Measure Theory; and I encountered $(a,\infty]$ instead of $(a,\infty)$ for the first time! </p>
<p>What is the difference(s) between $(a,\infty)$ and $(a,\infty]$?</p>
| Plutoro | 108,709 | <p>A lot of times these two mean the same thing, but it is important to consider the superset of which this is an interval. Sometimes, (especially in measure theory, which is why I mention it) it is useful to work in the extended reals, which includes a point at $\infty$, so $(a,\infty)$ means every number greater than $a$ accept infinity and $(a,\infty]$ would, as the notation suggests, include that point.</p>
|
1,669,096 | <p>How do I show that <span class="math-container">$\ell^{ \infty}$</span> is a normed linear space,
where <span class="math-container">$\ell^{ \infty}$</span> is define as <span class="math-container">$$\|\{a_n\}_{n=1}^{\infty}\|_{\ell^\infty}=\sup_{1 \leq k \leq \infty} |a_k|?$$</span>
There are three properties that I need to check in order for this to be a normed linear space. Nonnegativity, positive homogeneity, and the triangle inequality.</p>
<p>I am having trouble working with the supremum. Any idea will be greatly appreciated thanks</p>
| carmichael561 | 314,708 | <p>For the triangle inequality, suppose that <span class="math-container">$\{a_n\},\{b_n\}\in \ell^{\infty}$</span>. For each index <span class="math-container">$n$</span> we have
<span class="math-container">$$|a_n+b_n|\leq |a_n|+|b_n|\leq \sup_{m}|a_m|+\sup_m|b_m|=\|a\|_{\infty}+\|b\|_{\infty}.$$</span>
Then taking the supremum over all <span class="math-container">$n$</span> shows that
<span class="math-container">$$ \|a+b\|_{\infty}\leq \|a\|_{\infty}+\|b\|_{\infty}. $$</span></p>
|
3,084,479 | <p><span class="math-container">$h\in \mathbb{R}$</span>, because we have defined the Trigonometric Functions only on <span class="math-container">$\mathbb{R}$</span> so far.</p>
<p>I have a look at <span class="math-container">$e^{ih}=\sum_{k=0}^{\infty}\frac{(ih)^k}{k!}=1+ih-\frac{h^2}{2}+....$</span> </p>
<p><strong>How can one describe the nth term of the sum?</strong></p>
<p>Then I look at <span class="math-container">$\frac{e^{ih}-1}{h}=\frac{(1-1)}{h}+i-\frac{h}{2}+...=i-\frac{h}{2}+....$</span> </p>
<p><strong>Again how can I describe that the nth term of the sum?</strong> </p>
<p>Because <span class="math-container">$\frac{e^{ih}-1}{h}=\sum_{k=1}^{\infty}\frac{\frac{(ih)^k}{h}}{k!}<\sum_{k=0}^{\infty}\frac{\frac{(ih)^k}{h}}{k!}=\sum_{k=1}^{\infty}\frac{(ih^{-1})^k}{k!}=e^{ih^{-1}}$</span></p>
<p>and <span class="math-container">$ih^{-1}$</span> is a complex number and the exponential-series converges absolutely for all Elements in <span class="math-container">$\mathbb{C}$</span>, I have found a convergent majorant. And I can apply the properties of Limits on <span class="math-container">$\frac{e^{ih}-1}{h}\forall, h\in \mathbb{R}$</span>.</p>
<p><strong>How can I now prove formally (i.e by chosing an explicit <span class="math-container">$\delta$</span>) that</strong> </p>
<p><span class="math-container">$$\forall_{\epsilon>0}\exists_{\delta>0}\forall_{h\in\mathbb{R}}|h-0|=|h|<\delta\Longrightarrow |(\frac{e^{ih}-1}{h}=i-\frac{h}{2}+...)-i|<\epsilon$$</span></p>
<p><strong>I am also seeking for advice how to argue in such cases more intuitively (i.e by not always ginving an explicit <span class="math-container">$\delta$</span> ).</strong></p>
| Community | -1 | <p>Show that <span class="math-container">$\lim_{h\rightarrow 0}\frac{e^{ih}-1}{h}=i$</span>:</p>
<p>As we know: <span class="math-container">$ \cos \theta + i \sin \theta = e^{i \theta}$</span></p>
<p>So, <span class="math-container">$$\lim \limits_{h\rightarrow 0} [\frac { \cos h + i \sin h - 1 } {h} ] = \lim \limits_{h \rightarrow 0} [ \frac {1 - \frac{h^2}{2!} + \phi_1 (h^4) + i( h - \frac{h^3}{3!}... \phi_2(h^5) )-1} {h} ] = \lim \limits_{h \rightarrow0} \frac { i(1 - higher \space powers\space of\space h) }{1} = i$$</span></p>
<p>Alternatively,</p>
<p><span class="math-container">$$e^{ih} = 1 + ih + \frac{-h^2}{2!}...$$</span> will give the same result, just take terms of <span class="math-container">$i$</span> to one side and none <span class="math-container">$i$</span> ones to other, then take <span class="math-container">$i$</span> common and cancel 1 from numerator, then take <span class="math-container">$h$</span> common and cancel it from denominator, you'll get the desired result!</p>
|
1,335,698 | <p>For this problem do I use the distance formula that I would use between two regular points? </p>
<p>$d=\sqrt{(x_2−x_1)^2+(y_2−y_1)^2}$</p>
<p>The distance between points $u$ and $v$ on the $x$-axis is given by $|u-v|$. Solve $|x-5|+|x-6|=1$ (think geometrically).</p>
| Mythomorphic | 152,277 | <p><img src="https://i.stack.imgur.com/uYnuy.png" alt="enter image description here"></p>
<p>For $x\le5$,</p>
<p>$$|x-6|-|x-5|=1$$
So by adding the original equation,
$$2|x-6|=2$$
$$x=5/7\text{ (rej.)}$$</p>
<p>Similarly </p>
<p>For $x\ge6$, we get $x=6/4\text {(rej.)}$</p>
<p>For $5<x<6$, Let $x=5+k$, where $0<k<1$.</p>
<p>We have</p>
<p>$$|5+k-5|+|5+k-6|=1$$
$$k+(1-k)=1$$
which is true for all $k$.</p>
<p>So the solution is $5\le x\le6$.</p>
|
306,788 | <p>I understand determinants but I cannot understand the following question, can someone explain it to me ?</p>
<p>Suppose that a $4 x 4$ matrix with rows $v_1,v_2,v_3$ and $v_4$ has determinant det A = -1. Find the following determinants: </p>
<p>$$det \begin{bmatrix}v_1\\6v_2\\v_3\\v_4 \end{bmatrix}=$$
$$det \begin{bmatrix}v_2\\v_1\\v_4\\v_3 \end{bmatrix}=$$
$$det \begin{bmatrix}v_1\\v_2+7v_1\\v_3\\v_4 \end{bmatrix}=$$</p>
<p>I know how to find the determinate of a matrix but do not understand how this works.</p>
| Sean Ballentine | 62,751 | <p>Split the determinant across multiplication and notice that:</p>
<p>$\begin{bmatrix} v_1 \\ 6v_2 \\ v_3 \\ v_4 \end{bmatrix} = \begin{bmatrix} 1 \ 0\ 0\ 0 \\ 0\ 6\ 0\ 0 \\ 0\ 0\ 1\ 0\\ 0\ 0\ 0\ 1 \end{bmatrix}$ $\begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{bmatrix}$</p>
<p>$\begin{bmatrix} v_2 \\ v_1 \\ v_4 \\ v_3 \end{bmatrix} = \begin{bmatrix} 0 \ 1\ 0\ 0 \\ 1\ 0\ 0\ 0 \\ 0\ 0\ 0\ 1\\ 0\ 0\ 1\ 0 \end{bmatrix}$ $\begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{bmatrix}$</p>
<p>$\begin{bmatrix} v_1 \\ v_2 +7V_1 \\ v_3 \\ v_4 \end{bmatrix} = \begin{bmatrix} 1 \ 0\ 0\ 0 \\ 7\ 1\ 0\ 0 \\ 0\ 0\ 1\ 0\\ 0\ 0\ 0\ 1 \end{bmatrix}$ $\begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{bmatrix}$</p>
|
2,418,916 | <blockquote>
<p>Find how many terms there are in this geometric sequence:</p>
<p><span class="math-container">$-1, 2, -4, 8, ..., -16777216$</span></p>
</blockquote>
<p>My attempt:</p>
<p><span class="math-container">$a_k=a.r^{k-1}$</span></p>
<p>And in this sequence:</p>
<p><span class="math-container">$a=-1$</span>, <span class="math-container">$r=-2$</span></p>
<p>So</p>
<p><span class="math-container">$a_k=(-1){(-2)}^{k-1}$</span></p>
<p><span class="math-container">$-16777216=(-1){(-2)}^{k-1}$</span></p>
<p><span class="math-container">$16777216={(-2)}^{k-1}$</span></p>
<p><span class="math-container">$log(16777216)=log({(-2)}^{k-1})$</span></p>
<p><span class="math-container">$log(16777216)=(k-1)log{(-2)}$</span></p>
<p><span class="math-container">$k-1={{log(16777216)} \over {log{(-2)}}}$</span></p>
<p>But <span class="math-container">$-2$</span> is negative, and logarithm not defined for negative numbers?,
So what can I do ?</p>
<p>Thanks</p>
| Vasili | 469,083 | <p>$-16777216=(-1){(-2)}^{k-1}={(-1)}^k2^{k-1}=-2^{k-1}$ (exponent can't be negative so minus has to come from -1)</p>
|
2,418,916 | <blockquote>
<p>Find how many terms there are in this geometric sequence:</p>
<p><span class="math-container">$-1, 2, -4, 8, ..., -16777216$</span></p>
</blockquote>
<p>My attempt:</p>
<p><span class="math-container">$a_k=a.r^{k-1}$</span></p>
<p>And in this sequence:</p>
<p><span class="math-container">$a=-1$</span>, <span class="math-container">$r=-2$</span></p>
<p>So</p>
<p><span class="math-container">$a_k=(-1){(-2)}^{k-1}$</span></p>
<p><span class="math-container">$-16777216=(-1){(-2)}^{k-1}$</span></p>
<p><span class="math-container">$16777216={(-2)}^{k-1}$</span></p>
<p><span class="math-container">$log(16777216)=log({(-2)}^{k-1})$</span></p>
<p><span class="math-container">$log(16777216)=(k-1)log{(-2)}$</span></p>
<p><span class="math-container">$k-1={{log(16777216)} \over {log{(-2)}}}$</span></p>
<p>But <span class="math-container">$-2$</span> is negative, and logarithm not defined for negative numbers?,
So what can I do ?</p>
<p>Thanks</p>
| Oscar Lanzi | 248,217 | <p>To use logarithms, take the absolute values of the terms. Then you have</p>
<p>$|-1|×|-2|^{k-1}=|-16777216|$</p>
<p>$1×2^{k-1}=16777216$</p>
<p>where all numbers are positive and the logarithms can be manipulated without trouble. When you find $k$ you must check against the original equation with the negative signs included to verify that the proposed solution is valid.</p>
|
482,801 | <p>What does it mean that the characteristic function <span class="math-container">$f(x)=1_{[b \le x \lt \infty]}$</span> is right continuous with left limits? Here <span class="math-container">$x ,b \in \mathbb{R}$</span>.</p>
| Anthony Carapetis | 28,513 | <p>It means that at every point $x_0\in \mathbb{R}$, both one-sided limits $$ \lim_{x \nearrow x_0} f(x) \textrm{ and } \lim_{x \searrow x_0} f(x) $$ exist, and furthermore that $f(x_0) = \lim_{x \searrow x_0} f(x)$. In this example the function is fully continuous (both one-sided limits are equal to the function) everywhere except at $b$, where it fails to be left-continuous, but still has a limit.</p>
|
61,106 | <p>Let <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> be Poisson random variables with means <span class="math-container">$\lambda$</span> and <span class="math-container">$1$</span>, respectively. The difference of <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> is a <a href="http://en.wikipedia.org/wiki/Skellam_distribution" rel="nofollow noreferrer">Skellam random variable</a>, with probability density function
<span class="math-container">$$\mathbb P(X - Y = k) = \mathrm e^{-\lambda - 1} \lambda^{k/2} I_k(2\sqrt{\lambda}) =: S(\lambda, k),$$</span>
where <span class="math-container">$I_k$</span> denotes the <a href="http://en.wikipedia.org/wiki/Bessel_function#Modified_Bessel_functions_:_I.CE.B1.2C_K.CE.B1" rel="nofollow noreferrer">modified Bessel function of the first kind</a>. Let <span class="math-container">$F(\lambda)$</span> denote the probability that <span class="math-container">$X$</span> is larger than <span class="math-container">$Y$</span>: <span class="math-container">$$F(\lambda) := \mathbb P(X > Y) = \sum_{k=1}^{\infty} S(\lambda, k) = \mathrm e^{-\lambda - 1} \sum_{k=1}^\infty \lambda^{k/2} I_k(2\sqrt{\lambda}).$$</span> According to Mathematica, the graph of the function <span class="math-container">$F$</span> looks like</p>
<p><img src="https://i.stack.imgur.com/m0mPi.png"><br/><sub>(source: <a href="https://cims.nyu.edu/~lagatta/F.png" rel="nofollow noreferrer">nyu.edu</a>)</sub><br/></p>
<p>My question:<li>Is there a closed-form expression for the function <span class="math-container">$F$</span>?</li><li>If not, what are <span class="math-container">$\lim_{\lambda \to 0} F'(\lambda)$</span> and <span class="math-container">$F'(1)$</span>? What is the asymptotic behavior as <span class="math-container">$\lambda \to \infty$</span>?</p>
| Michael Lugo | 143 | <p>Numerically, $\lim_{\lambda \to 0} F^\prime(\lambda) = 1/e$. Heuristically this should be true because to have $X > Y$ when $\lambda$ is very small, the most likely case will be $X = 1, Y = 0$ by far; that occurs with probability $\lambda e^{-\lambda} e^{-1}$.</p>
<p>Maple gives an explicit formula for $F^\prime(1)$ involving sums; evaluating gives $F^\prime(1) = 0.3085083225$, and the Inverse Symbolic Calculator says this is somehow to $I_0(2) e^{-2}$. I'm not sure how to prove this at all but maybe knowing the answer helps?</p>
|
118,074 | <p>Just based on some reading, I know that every Möbius transformation is a bijection from the Riemann sphere to itself. </p>
<p>I'm curious about the converse. For any holomorphic bijection on the sphere, why is it necessarily a Möbius transformation? Is there a proof or reference of why this converse is true? Thanks.</p>
<p>(I would appreciate an explanation at the level of someone whose just self-studying complex analysis for the first time.)</p>
| Mariano Suárez-Álvarez | 274 | <p>Suppose $f$ is an holomorphic bijection of the sphere to itself. There is a Moebius transformation $g$ which maps $f(\infty)$ to $\infty$. Let $h=g\circ f$, which is again an holomorphic bijective map of the sphere to itself, and which maps $\infty$ to $\infty$. It follows that $h(\mathbb C)\subseteq\mathbb C$, because of injectivity, and the restriction $h|_{\mathbb C}:\mathbb C\to\mathbb C$ is a injective entire function.</p>
<p>Now, the big theorem of Picard, or several others, imply that $h|_\mathbb C$ is necessarily linear. It follows that $h$ itself is linear, and then $f=g^{-1}\circ h$ is a Moebius transformation.</p>
|
3,896,817 | <blockquote>
<p>Solve <span class="math-container">$x^2 \equiv 12 \pmod {13}$</span></p>
</blockquote>
<p>By guessing I can say that the solutions are <span class="math-container">$5$</span> and <span class="math-container">$8$</span>, but is there another way to find the solution besides guessing?</p>
| cosmo5 | 818,799 | <p><strong>Hint :</strong></p>
<p>As <span class="math-container">$x^2 \equiv a^2 \pmod {n}$</span> is easiest to handle, and <span class="math-container">$12\equiv25 \pmod {13}$</span>, we have</p>
<p><span class="math-container">$$ x^2 \equiv (\pm 5)^2 \pmod {13}$$</span></p>
|
3,896,817 | <blockquote>
<p>Solve <span class="math-container">$x^2 \equiv 12 \pmod {13}$</span></p>
</blockquote>
<p>By guessing I can say that the solutions are <span class="math-container">$5$</span> and <span class="math-container">$8$</span>, but is there another way to find the solution besides guessing?</p>
| Daniel Schepler | 337,888 | <p>In the special case of trying to find a square root of <span class="math-container">$-1$</span> modulo a prime <span class="math-container">$p \equiv 1 \pmod{4}$</span>, we can use the following algorithm: first, select some random integer <span class="math-container">$a$</span> with <span class="math-container">$1 < a < p - 1$</span>. Now, if <span class="math-container">$p-1 = 2^k \cdot q$</span> with <span class="math-container">$q$</span> odd, then form <span class="math-container">$a^q \operatorname{mod} p$</span>. If you get something other than <span class="math-container">$1$</span> or <span class="math-container">$-1$</span> at this point, then repeatedly square until you get <span class="math-container">$-1$</span>. Then, the number just before <span class="math-container">$-1$</span> will be a square root of <span class="math-container">$-1$</span>. (The reason this works: by Fermat's little theorem, <span class="math-container">$a^{p-1} \equiv 1 \pmod{p}$</span>, so after taking the square <span class="math-container">$k$</span> times, you get to <span class="math-container">$1$</span>. Since <span class="math-container">$p$</span> is prime, the number just before you reach <span class="math-container">$1$</span> for the first time is forced to be <span class="math-container">$-1$</span>.)</p>
<p>On the other hand, if <span class="math-container">$a^q \equiv \pm 1 \pmod{p}$</span> already, then go back and choose another random value of <span class="math-container">$a$</span> to try. Since <span class="math-container">$q < \frac{p - 1}{2}$</span>, it is impossible for every value of <span class="math-container">$a$</span> to satisfy <span class="math-container">$a^q \equiv \pm 1 \pmod{p} \Leftrightarrow a^{2q} \equiv 1 \pmod{p}$</span>.</p>
<p>So, in the case of <span class="math-container">$p = 13$</span>, where <span class="math-container">$p - 1 = 2^2 \cdot 3$</span> and so <span class="math-container">$q = 3$</span>, let us first start with a trial of <span class="math-container">$a = 2$</span>. Then, <span class="math-container">$a^q = 2^3 = 8$</span> and <span class="math-container">$8 \not\equiv \pm 1 \pmod{13}$</span>, so we're good to go. From here, <span class="math-container">$8^2 = 64 \equiv -1 \pmod{13}$</span>, and so we find 8 as a square root of <span class="math-container">$-1 \pmod{13}$</span>.</p>
|
2,011,003 | <p>I stumbled upon this logic question in a math class recently. </p>
<p>My teacher told us that a statement that is not tested/is empty is true. For example, that if I stated that: "if the team A wins the game, I am gonna buy you a coke", and then team B goes on and wins the game, the statement would be true, independent of me buying a coke. Could anybody elaborate how this can be the case, and why?</p>
<p>It came up as an explanation to why the the empty-set is both an open and a closed set. </p>
| Bananach | 70,687 | <p>Without going into a formal treatment, what your teacher means is:</p>
<p>The statement $$\forall x\in X: A (x) \Rightarrow B (x) $$ is true if $A (x) $ is false for all $ x\in X$ , no matter what $ B $ is.</p>
<p>That your teacher is right follows from the DEFINITION of the right arrow $\Rightarrow $, that is, it follows from the fact that the truth table of this symbol always assigns "True" when the first argument is false.</p>
<p>There is no epistemological claims behind what your teacher said. </p>
|
264,745 | <p>When I was learning statistics I noticed that a lot of things in the textbook I was using were phrased in vague terms of "this is a function of that" e.g. a statistic is a function of a sample from a distribution. I realized that while I know the definition of a function as a relation and I have an intuitive notion of what "function of" means, it's unclear to me how you transform this into a rigorous definition of "function of". So what is the actual definition of "function of"?</p>
| Ittay Weiss | 30,953 | <p>The modern approach is, as you say, to view a function as a relation. Thus $f\subseteq A\times B$ is a function if it satisfies that if $(a,b)\in f$ and $(a,b')\in f$ then $b=b'$. It is then common to write $f(a)=b$ instead of $(a,b)\in f$.</p>
<p>This is a way to formalize the notion of $f$ defining its output as a function of its input. If you like then, this is the actual definition of 'function of'. </p>
<p>It is helpful to keep in mind the long history of the development of the notion of function. During the early days of the calculus a function $f:\mathbb R \to \mathbb R$ was vaguely defined to mean something like: f is a process that transforms the input $x$ to some output $f(x)$ and moreover $f$ does so in a very smooth way (almost always differentiable). </p>
<p>This historical approach to function, while not rigorous, is more in-line with $y$ being a function of $x$. The modern approach of a function as a relation, while very rigorous, is more static. This may be viewed as a shortcoming of this rigorous definition. However, the formalization of function is simple enough and easily allows abuse of concepts to actually think of a function as some process while it is formally not. </p>
<p>This situation is somewhat similar to the definition of a random variable. A random variable is nothing but a function with a particular domain and codomain. Thus, according to the relational definition, it is a very static thing. Nonetheless, we think of a random variable as a highly variable thing, even as if it's value is not yet known or is uncertain. However, this formalization of random variable within the rigorous confines of measure theory is highly useful, allowing one to correctly argue about uncertain events. This goes to show just how powerful the modern axiomatization is - there is enough flexibility in the interpretation of the notion of function to accomodate many situations. </p>
|
3,090,448 | <p>I have the following question to complete.</p>
<p>Let <span class="math-container">$X$</span> be an inner product space. Let <span class="math-container">$(e_{j})_{j\geq1}$</span> be an orthonormal sequence in <span class="math-container">$X$</span>. Show that,
<span class="math-container">\begin{align}
\sum_{j=1}^{\infty}|(x|e_{j})(y|e_{j})|\leq\|x\|\|y\|,
\end{align}</span>
for all <span class="math-container">$x,y\in X$</span>.</p>
<p>I have tried to use the Cauchy-Schwartz Inequality.</p>
<p><span class="math-container">\begin{align}
\sum_{j=1}^{\infty}|(x|e_{j})(y|e_{j})|\leq\|x\|\|y\|\sum_{j=1}^{\infty}\|e_{j}\|^{2}.
\end{align}</span>
However, that remaining sum, as far as I know, does not converge.</p>
<p>I tried to use Parseval's Identity, but that didn't work either. Can someone offer me a hint?</p>
| Kavi Rama Murthy | 142,385 | <p>Use Cauchy -Schwraz inequlairty for sequences rather than inner product. <span class="math-container">$|\sum \langle x,e_i \rangle \langle y,e_i \rangle| \leq (\sum |\langle x,e_i \rangle|^{2})^{1/2}(\sum |\langle y,e_i \rangle|^{2})^{1/2} \leq \|x\|\|y\|$</span>.</p>
|
1,664,081 | <p>Solve the equation $2^x - 3^{x-1}=-(x+2)^2$</p>
<p>How I got this question? I created this question so I know the answer. The answer is 5. But I have no idea how to solve it. Take note that I cannot do logarithm, guess and check and modulus. Does anybody know how to solve this? I have no idea how to start.</p>
<p>$2^x - 3^{x-1}$ is infactorisable. Even if I did $2^x - (2+2^0)^{x-1}$, it is STILL infactorisable. I was hoping to solve by hand.</p>
<p>So, I do I solve the equation?</p>
| SS_C4 | 242,290 | <p>Well, from $2^x - 3^{x-1} = -(x+2)^2$, $2^x = 3^{x-1} - (x+2)^2$. </p>
<p>LHS is always even, and $3^n$ is always odd. Therefore $(x+2)^2$ has to be odd, $\Rightarrow$ x is odd. </p>
<p>Also, from the first equation, $2^x < 3^{x-1}$. This is true for $x > 2$.
Since x is odd, the new condition for $x$ is $x \ge 3$ and x is odd. So,
$$x \in {\{3,5,7,9,\dots\}}$$</p>
<p>Checking, we see that $5$ is an answer.</p>
<p>(But I'm not sure how to prove that this is the only answer. Maybe I could bring the graph part here).</p>
|
1,712,289 | <p>If from twice the greater of two numbers 17 is subtracted, the result is half the other number. If from half the greater number 1 is subtracted, the result is two-thirds of the smaller number.</p>
<p>$$2x - 17 = \frac{ 1 }{2}y$$</p>
<p>$$\frac{ x }{2} - 1 = \frac{ 2 }{3}y$$</p>
<p>$$-17 - 4 = \frac{ 1 }{2}y - \frac{ 8 }{3}y$$</p>
<p>I'm so, off. I need a little help.</p>
| Nikunj | 287,774 | <p><strong>Half</strong> of the greater number, this means $\frac{x}{2}$ in the second equation, rest seems all right!</p>
<p>Now, if you multiply the lower equation by $4$ and subtract it from the first, you get:
$$-13=\frac{y}{2}-\frac{8y}{3}$$
$$\implies y=6$$ you can find that $$x=10$$</p>
|
1,712,289 | <p>If from twice the greater of two numbers 17 is subtracted, the result is half the other number. If from half the greater number 1 is subtracted, the result is two-thirds of the smaller number.</p>
<p>$$2x - 17 = \frac{ 1 }{2}y$$</p>
<p>$$\frac{ x }{2} - 1 = \frac{ 2 }{3}y$$</p>
<p>$$-17 - 4 = \frac{ 1 }{2}y - \frac{ 8 }{3}y$$</p>
<p>I'm so, off. I need a little help.</p>
| Tom | 255,814 | <h3>This can be solved using substitution</h3>
<p>Here's a step by step solution:</p>
<ol>
<li><p>Solve for x in the first equation, <br><br><span class="math-container">$x=\frac{17}{2}+\frac{y}{4}$</span></p>
</li>
<li><p>Replace all occurrences of x in the second equation,<br><br><span class="math-container">$\frac{\left(\frac{17}{2}+\frac{y}{4}\right)}{2}-1=\frac{2y}{3}$</span></p>
</li>
<li><p>Solve for y,<br><br><span class="math-container">$y=6$</span></p>
</li>
<li><p>Take the solution for x and replace y with 6,<br><br><span class="math-container">$x=\frac{17}{2}+\frac{6}{4}$</span></p>
</li>
<li><p>Solve,<br><br><span class="math-container">$x=10$</span></p>
</li>
</ol>
<hr />
<p><strong>Answer</strong></p>
<p><span class="math-container">$y=6$</span><br>
<span class="math-container">$x=10$</span></p>
|
3,016,169 | <p>I want to prove or disprove that <span class="math-container">$C^1([a,b], \mathbb{R}^n)$</span> equipped with the norm <span class="math-container">$||x||=\underset{t\in[a,b]}{\sup}|x(t)|_{\mathbb{R}^n}+\underset{t\in[a,b]}{\sup}|\dot{x}(t)|_{\mathbb{R}^n}$</span> is a reflexive Banach space. </p>
<p>I figured out that it is indeed a Banach space, but I'm stuck on the reflexive part. I know that a Banach space <span class="math-container">$X$</span> is reflexive if the mapping <span class="math-container">$F: X \rightarrow X''$</span> is surjective, where <span class="math-container">$X''$</span> is the second dual space of <span class="math-container">$X$</span>. </p>
<p>However, I have the feeling that this space is not reflexive, but I'm not sure how to prove that.</p>
<p>Is it indeed true that this space is not reflexive? </p>
| Robert Israel | 8,508 | <p>We can embed <span class="math-container">$Y = C([a,b],\mathbb R)$</span> into <span class="math-container">$X = C^1([a,b], \mathbb R^n)$</span> by
<span class="math-container">$T(f)(x) = \int_a^x f(t)\; dt\; {\bf u}$</span> where <span class="math-container">$\bf u$</span> is some nonzero vector in <span class="math-container">$\mathbb R^n$</span>. Then <span class="math-container">$T^{**}$</span> embeds <span class="math-container">$Y^{**}$</span> into <span class="math-container">$X^{**}$</span>. But <span class="math-container">$Y^{**}$</span> is not separable, therefore <span class="math-container">$X^{**}$</span> is not separable, but <span class="math-container">$X$</span> is separable, so <span class="math-container">$X$</span> is not reflexive.</p>
|
1,292,889 | <p>I've read the paper <a href="http://web.mit.edu/leozhou/www/gauss.pdf" rel="noreferrer">Least square fitting of a Gaussian function to a histogram</a> by Leo Zhou on how to perform a Least Square Fitting of a gaussian function to a histogram.</p>
<p>The Gaussian function used to fit the data is:
$$f(y)=A\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)$$</p>
<p>However, the method described in the paper doesn't work if the dataset has a vertical offset (i.e. all the points are shifted on the Y-axis by some amount $K$).</p>
<p>I was wondering how to perform LSF (or any other kind of fitting) to estimate the parameters $A$, $\mu$, $\sigma$ and $K$ of the function</p>
<p>$$f(y)=A\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)+K$$</p>
<p>(without prior knowledge of the parameter $K$, of course).</p>
| JJacquelin | 108,514 | <p>The usual methods of non-linear regression involve iterative process starting from guessed values of the parameters. </p>
<p>There is a straight forward method (not iterative, no need for guessed values) which general principle is explain in the paper : <a href="https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales" rel="nofollow noreferrer">https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales</a> (in French, presently no translation available). </p>
<p>The application to the "y-shifted gaussian" case is shown below :</p>
<p>$$y=h+c\: e^{-\frac{(x-a)^2}{b}}$$
Rank the data in increasing order of $x_k$</p>
<p>$(x_1\:,\:y_1)$ , $(x_2\:,\:y_2)$ , … , $(x_k\:,\:y_k)$ , … , $(x_n\:,\:y_n)$ </p>
<p>$S_1=0$</p>
<p>$S_k=S_{k-1}+\frac{1}{2}(y_k+y_{k-1})(x_k-x_{k-1})$</p>
<p>$T_1=0$</p>
<p>$T_k=T_{k-1}+\frac{1}{2}(x_k y_k+x_{k-1} y_{k-1})(x_k-x_{k-1})$</p>
<p>Compute matrix $(M)=$
$$(M)=
\begin{pmatrix}
\displaystyle\sum _{k=1}^n S_k^2& \displaystyle\sum _{k=1}^n S_kT_k & \displaystyle\sum _{k=1}^n S_k(x_k^2-x_1^2)& \displaystyle \sum _{k=1}^n S_k(x_k-x_1) \\
\displaystyle\sum _{k=1}^n S_kT_k& \displaystyle\sum _{k=1}^n T_k^2 & \displaystyle\sum _{k=1}^n T_k(x_k^2-x_1^2)& \displaystyle\sum _{k=1}^n T_k (x_k-x_1)\\
\displaystyle\sum _{k=1}^n S_k(x_k^2-x_1^2) &\displaystyle\sum _{k=1}^n T_k(x_k^2-x_1^2) &\displaystyle\sum _{k=1}^n (x_k^2-x_1^2)^2& \displaystyle\sum _{k=1}^n (x_k^2-x_1^2)(x_k-x_1)\\
\displaystyle\sum _{k=1}^n S_k(x_k-x_1) &\displaystyle\sum _{k=1}^n T_k(x_k-x_1) & \displaystyle\sum _{k=1}^n (x_k^2-x_1^2)(x_k-x_1) &\displaystyle\sum _{k=1}^n (x_k-x_1)^2&\\
\end{pmatrix}$$
and vector $(V)=$
$$(V)=\begin{pmatrix}
\displaystyle\sum _{k=1}^n S_k(y_k-y_1) \\
\displaystyle\sum _{k=1}^n T_k(y_k-y_1) \\
\displaystyle\sum _{k=1}^n (x_k^2-x_1^2)(y_k-y_1) \\
\displaystyle\sum _{k=1}^n(x_k-x_1)(y_k-y_1)
\displaystyle\end{pmatrix}
$$
Then, compute $A,B$ from :
$$ \begin{pmatrix}
A\\
B\\
C\\
D\\
\end{pmatrix}=(M)^{-1}(V)$$</p>
<p>The approximates of $a,b$ are :</p>
<p>$$a=-\frac{A}{B}$$
$$b=-\frac{2}{B}$$</p>
<p>Then, compute $t_k$ :
$$t_k=e^{-\frac{(x_k-a)^2}{b}}$$
and compute the approximates of $c$ and $h$ :
$$\begin{pmatrix}
c\\
h
\end{pmatrix}=
\begin{pmatrix}
\displaystyle\sum _{k=1}^n t_k^2&\displaystyle \sum _{k=1}^n t_k\\
\displaystyle\sum _{k=1}^n t_k& n
\end{pmatrix}^{-1}
\begin{pmatrix}
\displaystyle\sum _{k=1}^n t_k y_k \\
\displaystyle\sum _{k=1}^n y_k
\end{pmatrix}
$$
The approximate of $y_k$ is $Y_k=h+c\:e^{-\frac{(x_k-a)^2}{b}}$</p>
<p>A numerical example is detailed below:</p>
<p><img src="https://i.stack.imgur.com/92Sep.jpg" alt="enter image description here"></p>
|
1,553,391 | <p>Let $E$ be a measurable set of finite measure and $1\leq p_1 < p_2 \leq \infty$ . Then $L^{p_2} (E) \subseteq L^{p_1} (E)$ Furthermore $||f||_{p_1} \leq c \cdot ||f||_{p_2}$ for all $f$ in $L^{p_2}(E)$ where $c =[m(E)]^{\frac{p_2-p_1}{p_1p_2}} $ if $p_2<\infty$ and $c=[m(E)]^{\frac{1}{p1}}$ if $p_2 =\infty$ </p>
<p>I need help with the second part ($p_2= \infty$)</p>
<p>Here is the proof of the first part: $p_2 < \infty$</p>
<p>Define $p = \frac{p_2}{p_1}> 1 $ and $q$ be the conjugate of $p$.
Let $f \in L^{p_2} (E) $ then we can prove that $f^{p_1} \in L^p(E)$ and let $g=\chi_E \in L^q(E)$ ( since $m(E)<\infty$)
By Holder inequality we have:
$\int_E |f|^{p_1} = \int_E |f|^{p_1} g =||f^{p_1}||_p ||g||_q \leq \Big[\int_E (|f|^{p_1})^{\frac{p_2}{p_1}} \Big]^{\frac{p_1}{p_2}} \cdot \Big[\int_E |g|^q \Big]^{\frac{1}{q}}= ||f||_{p_2}^{p_1} \Big[ m(E)\Big]^{\frac{1}{q}}$</p>
<p>$\Rightarrow \text{ by power to }\frac{1}{p_1}$ we get $||f||_{p_1} \leq \Big[m(E)\Big]^{\frac{p_2-p_1}{p_1p_2}} ||f||_{p_2}$</p>
<p>Can anyone help me with the second part: $p_2= \infty$?</p>
| Hamit | 277,958 | <p>$L^{p}(E)$s are not comparable unless you have finite measure space for whole space. </p>
|
1,379,513 | <p>A hot dog stand has 12 different toppings available. How many different kinds of hot dogs can be made, assuming the order of the toppings does not make a difference. I believe the correct answer is 882050, with the maximum varieties per number of toppings selected being 665280 when there six toppings. I am also not sure about how to create a formula that would arrive at this result.</p>
| quid | 85,306 | <p>For each topping you can decide to use it or not to use it. </p>
<p>So for each topping you have $2$ ways. Thus in total you have $2^{12}$ ways. </p>
|
1,379,513 | <p>A hot dog stand has 12 different toppings available. How many different kinds of hot dogs can be made, assuming the order of the toppings does not make a difference. I believe the correct answer is 882050, with the maximum varieties per number of toppings selected being 665280 when there six toppings. I am also not sure about how to create a formula that would arrive at this result.</p>
| Luca C. | 258,065 | <p>Also, you can view it like this:
$$
\sum_{k=0}^{12} C_{12,k} = \sum_{k=0}^{12} \binom{12}{k} = 4096
$$</p>
<p>i.e. for each number k of toppings, you get k-combinations between those elements, from the starting 12. We are using combinations without repetition because order of selection does not matter.</p>
<p>Numerically:
$$\sum_{k=0}^{12} \binom{12}{k} = 1 + 12 + 66 +220 +495 +792 +924 +792 +495 +
220 +66 +12 +1 = 4096
$$</p>
|
4,107,232 | <h2>Problem</h2>
<p>Robert is playing a game with numbers. If he has the number <span class="math-container">$x$</span>, then in the next move, he can do one of the following:</p>
<ul>
<li>Replace <span class="math-container">$x$</span> by <span class="math-container">$\lceil{\frac{x^2}{2}}\rceil$</span></li>
<li>Replace <span class="math-container">$x$</span> by <span class="math-container">$\lfloor{\frac{x}{3}}\rfloor$</span></li>
<li>Replace <span class="math-container">$x$</span> by <span class="math-container">$9x+2$</span></li>
</ul>
<p>He starts with the number <span class="math-container">$0$</span>. How many integers less than or equal to <span class="math-container">$7000$</span> can he achieve using the above functions?</p>
<p><em>[It is permitted to use a number greater than <span class="math-container">$7000$</span> in the way of achieving the desired numbers.]</em></p>
<h2>My Approach</h2>
<p>Call the functions <span class="math-container">$f_1,f_2,f_3$</span> respectively. <span class="math-container">$2$</span> is easily achievable from <span class="math-container">$0$</span> (using <span class="math-container">$f_3$</span>). I've found that all the integers from <span class="math-container">$0$</span> to <span class="math-container">$10$</span> are achievable (Though we achieve them in a long way). The numbers get messy when we get ahead further. I can't prove that any number is unachievable. I've noticed that base-<span class="math-container">$3$</span> numbers can help for <span class="math-container">$f_2$</span> and <span class="math-container">$f_3$</span>.</p>
<hr />
<p>How to get ahead further?</p>
<p><strong>Update:</strong> Mr. Mike showed that all integers are <em>achievable</em> by this process through codes. Mr. Calvin also gave a partial proof for that. So, a complete proof is needed currently.</p>
| Calvin Lin | 54,563 | <p>This is not a valid solution.<br />
Ravi pointed out that there is an error.</p>
<hr />
<p><strong>Claim:</strong> For any integer <span class="math-container">$n$</span>, there exists integers <span class="math-container">$K , L \geq 0$</span> such that <span class="math-container">$$ n\times 3^K \leq 2 \times 10 ^{2^L} \leq (n+1) \times 3^K.$$</span></p>
<p><strong>Proof:</strong> Working mod <span class="math-container">$\log 3$</span>, we want to show that there exists a <span class="math-container">$L > 0$</span> such that</p>
<p><span class="math-container">$$ \frac{\log n - \log 2}{\log 10} \leq 2^L \leq \frac{\log (n+1) - \log 2}{\log 10} \quad \pmod{ \log 3}$$</span></p>
<p>(I am unable to complete this proof. It requires us to show that <span class="math-container">$\frac{1}{ \log 3} $</span> in base 2 has all finite binary strings.)</p>
<p><strong>Corollary:</strong> <span class="math-container">$ b^K a^{L-1} c^2 (0) = n$</span>, where</p>
<ul>
<li><span class="math-container">$a(x) = \lceil \frac{ x^2 }{ 2 } \rceil $</span></li>
<li><span class="math-container">$b(x) = \lfloor \frac{x}{3} \rfloor $</span></li>
<li><span class="math-container">$c(x) = 9x+2$</span>.</li>
</ul>
<hr />
<p>Notes</p>
<ul>
<li>As conjectured and established via computer by Steven and Mike respectively, after using <span class="math-container">$ c(0) = 2, c(2) = 20$</span>, it seems like we don't need the <span class="math-container">$c(x)$</span> function anymore.</li>
<li>In addition, since <span class="math-container">$ab(x) \approx b^2a(x) \approx \frac{x^2}{18}$</span> (but the floor and ceiling functions could get in the way of equality), if there was a sequence to get to <span class="math-container">$n$</span> using just <span class="math-container">$a(x), b(x)$</span>, then it might be reasonable that we could collate <span class="math-container">$a(x), b(x)$</span> separately.</li>
<li>The above 2 comments could motivate the given solution. However, that's not how I came up with it.</li>
<li>Working in base 3 is suggested by functions <span class="math-container">$b(x), c(x)$</span>, and <span class="math-container">$bbc(x) = x$</span>.</li>
<li>(for me at least) Viewing <span class="math-container">$b(x)$</span> as truncating in base 3 and <span class="math-container">$c(x)$</span> as appending 02 in base 3, made it much easier to think about these function.</li>
<li>Based on initial iterations (esp because I avoided <span class="math-container">$a(x)$</span> as that made numbers huge), my guesses for achievable numbers were like A) <span class="math-container">$6k, 6k+2$</span>, B) <span class="math-container">$2k$</span>, C) Trenary numbers involving only 0 and 2 (maybe with additional conditions).</li>
<li>It is clear that if we only used <span class="math-container">$b(x), c(x)$</span>, then the base 3 representations are limited to digits of 0 and 2 (and in fact, 2's must be separated by 0's). The followup question is "Can we introduce a digit of 1 in base 3 using <span class="math-container">$a(x)$</span>"?</li>
<li>We could do that with <span class="math-container">$ a(20) = 200 = 21102_3$</span>, and so I thought that the set of achievable numbers were Numbers in base 3 whose starting digit was 2.</li>
<li>Looking at <span class="math-container">$a^2 (20) = 20000 = 1000102202_3$</span>, I realized that would give us <span class="math-container">$1$</span> (and <span class="math-container">$10_3, 100_3, \ldots)$</span>.</li>
<li>With that realization, we simply want the inequality in the claim.</li>
<li>Of course, there could be other ways of reaching <span class="math-container">$n$</span>. One possible approach could be to show that we can reach all even numbers, and then by applying <span class="math-container">$b(x)$</span> we can reach all numbers.</li>
</ul>
|
71,608 | <p>Consider the following question:</p>
<p>Is there a family $\mathcal{F}$ of subsets of $\aleph_\omega$ that satisfies the following properties?</p>
<p>(1) $|\mathcal{F}|=\aleph_\omega$</p>
<p>(2) For all $A\in \mathcal{F}$, $|A|<\aleph_\omega$</p>
<p>(3) For all $B\subset \aleph_\omega$, if $|B|<\aleph_\omega$, then there exists some $B'\in \mathcal{F}$ such that $B\subset B'$.</p>
<p>I am not sure if there is anything special about $\aleph_\omega$, but this was the example that came up. </p>
<p>Any help?</p>
| Andreas Blass | 6,794 | <p>There is no such family $\mathcal F$. Suppose, toward a contradiction, that you had such an $\mathcal F$ and list it in a sequence of order-type $\aleph_\omega$. For each $n\in\omega$, let $\mathcal F_n$ consist of the first $\aleph_n$ members of the sequence that have cardinality at most $\aleph_n$. Notice that $\mathcal F$ is the union of these subfamilies $\mathcal F_n$. The union of all the sets in $\mathcal F_n$ has cardinality at most $\aleph_n$, so we can choose some $a_n$ that is in $\aleph_\omega$ but not in this union. Then $\{a_n:n\in\omega\}$ is a countable subset of $\aleph_\omega$ not covered by any element of $\mathcal F$.</p>
|
3,191,402 | <p>I have tried to answer by taking change the variable <span class="math-container">$\theta$</span> to <span class="math-container">$\theta/2$</span>, so the integration is now over unit circle, then I have taken <span class="math-container">$z=e^{i\theta}$</span>. Now I tried to use residue formula for integration, but I failed.</p>
| Doug M | 317,176 | <p><span class="math-container">$\int_0^{\pi} (a+\cos x)^n \ dx = \frac 12 \int_0^{2\pi} (a+\cos x)^n \ dx\\
\cos x = \frac 12 (e^{ix} + e^{-ix})\\
\frac 1{2^{n+1}} \int_0^{2\pi} (2a+e^{ix} + e^{-ix})^n \ dx
z = e^{ix}\\
dx = \frac {1}{iz}\ dz$</span></p>
<p><span class="math-container">$\frac 1{2^{n+1}i} \oint_{|z| = 1} \frac {1}{z}(2a+z + z^{-1})^n \ dz$</span></p>
<p>Now the trick. When we expand <span class="math-container">$(2a+z + z^{-1})^n$</span> we only care about the constant term. The rest of the terms will evaluate to <span class="math-container">$0.$</span></p>
<p><span class="math-container">$(2a+z + z^{-1})^n = \sum_\limits{k=0}^n {n\choose k} (z+z^{-1})^k(2a)^{n-k}$</span></p>
<p>There is only a constant term for <span class="math-container">$(z+z^{-1})^k$</span> if <span class="math-container">$k$</span> is even, and it will equal <span class="math-container">${k\choose \frac{k}{2}}$</span></p>
<p><span class="math-container">$\frac{\pi}{2^n}\sum_\limits{k=0}^{\lfloor \frac n2\rfloor} {n\choose 2k}{2k\choose k}(2a)^{n-2k}$</span></p>
|
3,191,402 | <p>I have tried to answer by taking change the variable <span class="math-container">$\theta$</span> to <span class="math-container">$\theta/2$</span>, so the integration is now over unit circle, then I have taken <span class="math-container">$z=e^{i\theta}$</span>. Now I tried to use residue formula for integration, but I failed.</p>
| Robert Israel | 8,508 | <p>If your integral is <span class="math-container">$J_n$</span>, the exponential generating function of the sequence <span class="math-container">$J_n$</span> is</p>
<p><span class="math-container">$$ \eqalign{g(x) &= \sum_{n=0}^\infty \int_0^\pi \frac{(a+\cos(\theta))^n x^n}{n!}\; d\theta \cr
&= \int_0^\pi \exp((a + \cos(\theta) x)\; d\theta\cr
&= \pi e^{ax} I_0(x)}$$</span>
where <span class="math-container">$I_0$</span> is a modified Bessel function. Since
<span class="math-container">$$ e^{ax} = \sum_{j=0}^\infty \frac{a^j x^j}{j!}$$</span>
and
<span class="math-container">$$ I_0(x) = \sum_{k=0}^\infty \frac{x^{2k}}{4^k k!^2}$$</span>
we get
<span class="math-container">$$ J_n = \pi n!\sum_{k=0}^{\lfloor n/2 \rfloor} \frac{a^{n-2k}}{4^k (n-2k)! k!^2}$$</span></p>
|
1,987,026 | <p>So I know to get a probability like $P(2\leq X\leq 4)$, you simply do $P(X\leq4) - P(X\leq1)$, but when there is a question like $P(2<X<4)$ what am I supposed to do? </p>
<p>Not just limited to in between two values, I also don't know what to do if it's just $P(X<2)$, so far all our examples have been greater/less than or equal to, for $P(X\leq 2)$ you just do the cumulative distribution up to $2$ but what do I do if that $2$ is not included in the range?</p>
| BGM | 297,308 | <p>The complement of $\{X \geq 2\}$ is $\{X < 2\}$, e.g. we have $\Pr\{X \geq 2\} = 1 - \Pr\{X < 2\}$</p>
<p>The first example you shown is imprecise. The general version should be
$$ \Pr\{2 \leq X \leq 4\} = \Pr\{X \leq 4\} - \Pr\{X < 2\}$$
Only, when you are given that $\Pr\{1 < X < 2\} = 0$, e.g. $X$ is discrete with integral support, then you have your claim.</p>
<p>In your question $\Pr\{X > 2\} = 1 - \Pr\{X \leq 2\} = 1 - F_X(2)$ so it can be expressed in terms of the CDF $F$ conveniently. </p>
<p>If you want to express, say $\Pr\{X < 2\} = \Pr\{X \leq 2\} - \Pr\{X = 2\}$ in terms of the CDF, then it depends on whether it has a mass on $2$. If it has no mass on $2$, then it is simply $F_X(2)$; otherwise you need to subtract from that. Or equivalently
$$\Pr\{X < 2\} = \lim_{x\to 2^-}F_X(x)$$</p>
|
1,150,805 | <p>An unfair 3-sided die is rolled twice. The probability of rolling a 3 is $0.5$, the probability of rolling a 1 is $0.25$, and the probability of rolling a 2 is $0.25$. Let $X$ be the outcome of the first roll and $Y$ the outcome of the second.</p>
<ul>
<li><p>Find the Joint Distribution of $X$ and $Y$ in a Table.</p>
<p>The outcome of $X = \{1,2,3\}$.</p>
<p>The outcome of $Y = \{1,2,3\}$.</p>
<p>Would I just make a table of all the roll possibilities?</p></li>
<li><p>Find the Probability $\mathrm{P}(X+Y \geq 5)$.</p>
<p>The only roll that will make this is a 3 or a 2.
Should I just take the same of every possible roll to find this probability?</p></li>
</ul>
| Mike Pierce | 167,197 | <p>Let $X_n = \{1, 2, \dotsc, n\}$ be the finite set of $n$ elements (unique up to a bijective morphism). Denote the power set of a set $X_i$ as $\mathcal{P}(X_i)$. Let's write out the first few power sets of these finite sets:
$$\begin{align}
\mathcal{P}(X_0) = \mathcal{P}(\{\}) &= \{\{\}\} \\
\mathcal{P}(X_1) = \mathcal{P}(\{1\}) &= \{\{\},\{1\}\} \\
\mathcal{P}(X_2) = \mathcal{P}(\{1,2\}) &= \{\{\},\{1\},\{2\},\{1,2\}\} \\
\mathcal{P}(X_3) = \mathcal{P}(\{1,2,3\}) &= \{\{\},\{1\},\{2\},\{1,2\},
\{3\},\{1,3\},\{2,3\},\{1,2,3\},\} \\
\end{align}$$</p>
<p>We can see that each $\mathcal{P}(X_i)$ consists of elements $s$ and $s\cup \{i\}$ for each $s \in \mathcal{P}(X_{i-1})$. In other words:
$$
\mathcal{P}(X_i) \;=
\bigcup_{s \in \mathcal{P}(X_{i-1})} \Big\{s, s\cup\{i\}\Big\} \;\;=\;\;
\mathcal{P}(X_{i-1}) \cup \big\{s\cup\{i\} \mid s \in \mathcal{P}(X_{i-1})\big\}
$$</p>
|
127,493 | <p>How many number less than $k$ contain the digit $3$?
For instance:</p>
<p>How many number contain the digit $3$ in the following list?</p>
<pre><code>Table[n, {n, 33}]
</code></pre>
<p>$\lbrace 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, \
20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33\rbrace$</p>
<p>I tried: </p>
<pre><code>numbers[k_] := Count[Table[n, {n, k}], 3]
</code></pre>
<p>but it doesn't work.</p>
<p>Then I want to find the limit</p>
<pre><code>Limit[numbers[k]/k, k -> Infinity]
</code></pre>
<p>( <a href="https://www.youtube.com/watch?v=UfEiJJGv4CE" rel="nofollow">See Numberphile video here.</a>)</p>
| Haohu Shen | 43,438 | <p>1.So if you meet such situations again which I mean you can't get the result you want in MMA,try to solve the same problem in other alternatives such as Maxima,Wolfram Alpha to see if there is a same result first.</p>
<p>2.If there is an difference among these platforms,try to debug your calculation process to make sure if it's MMA's bug indeed by some tools.Rubi is effective in this example.</p>
<p>3.Maybe you should seek some ways to change the descriptions for your question in MMA or something.Just think widely.</p>
<p>4.If there is no result you want eventually,you should check the problem by maths definitions which I refer to the necessary and sufficient conditions of integrablity and the existence of the result expression by elementary functions.</p>
|
4,315,283 | <p>Let <span class="math-container">$X$</span> be a continuous random variable, having pdf <span class="math-container">$f(x)$</span> and let be <span class="math-container">$Y=f(X)$</span>.
I have already proved these two following results:</p>
<ol>
<li>If <span class="math-container">$f(x)$</span> is strictly increasing, <span class="math-container">$Y$</span> has the same distribution as <span class="math-container">$X$</span> if and only if</li>
</ol>
<p><span class="math-container">$$
f(x)=
\begin{cases}
x, \qquad x \in \,[0, \sqrt{2}] \\
0, \qquad otherwise
\end{cases}
$$</span></p>
<ol start="2">
<li>If <span class="math-container">$f(x)$</span> is strictly decreasing, <span class="math-container">$Y$</span> has the same distribution as <span class="math-container">$X$</span> if and only if</li>
</ol>
<p><span class="math-container">$$
f(x)=
\begin{cases}
\frac{a^{2}}{x}, \qquad x \in \,[ae^{-\frac{1}{2a^2}}, ae^{\frac{1}{2a^2}}] \\
0, \qquad otherwise
\end{cases}
$$</span></p>
<p>Now, I should investigate about the random variable that has pdf partially increasing and partially decreasing:</p>
<p><span class="math-container">$$
f(x)=
\begin{cases}
x, \qquad x \in [0,1] \\
\frac{1}{x}, \qquad x \in \,[1, e^{\frac{1}{2}}]
\end{cases}
$$</span></p>
<p>What can I say about the distribution of the random variable having aforementioned pdf? Do the last pdf have some simmetry (<span class="math-container">$\frac{1}{x}$</span> is the reciprocal of <span class="math-container">$x$</span>)?</p>
| Tobsn | 414,776 | <p>Assume that <span class="math-container">$f$</span> is continuously differentiable and bijective on its support. Then by the classical transformation rule <span class="math-container">$Y$</span> has density <span class="math-container">$p$</span> with
<span class="math-container">\begin{equation}
p(x)=x|(f^{-1})'(x)|
\end{equation}</span>
if <span class="math-container">$x\in\operatorname{im}(\operatorname{supp}(f))$</span> and <span class="math-container">$f'(f^{-1}(x))\ne 0$</span>. Elsewhere it's zero. This can be generalised in various ways. For non-bijective functions, you can split up the function in bijective parts. Also the differantiability assumption can be relaxed, it becomes somewhat technical then however.</p>
|
2,725,317 | <blockquote>
<p>For which $p,q\in \mathbb R$ is the following system stable?$$\frac{\mathrm dx}{\mathrm dt} = \begin{bmatrix} p & -q \\ q & p \end{bmatrix}x(t)$$ </p>
</blockquote>
<p>If I'm correct about this, isn't it just when the eigenvalues are $< 1?$ Or is there something more fancy to it? Any and all help appreciated.</p>
| Robert Lewis | 67,071 | <p>Assuming that $p, q \in \Bbb R$ are constants, the system</p>
<p>$\dfrac{d \vec x(t)}{dt} = \begin{bmatrix} p & -q \\ q & p \end{bmatrix} \vec x(t) \tag 1$</p>
<p>has a well-known explicit solution. Suppose we set</p>
<p>$A = \begin{bmatrix} p & -q \\ q & p \end{bmatrix}, \tag 2$</p>
<p>and let</p>
<p>$\vec x_0 = \vec x(t_0) \tag 3$</p>
<p>be an initialization of $\vec x(t)$; then we have</p>
<p>$\vec x(t) = e^{A(t - t_0)} \vec x_0 = \exp(A(t - t_0)) \vec x_0; \tag 4$</p>
<p>equation (4) is easily validated by direct differentiation; indeed,</p>
<p>$\dot{\vec x}(t) = \dfrac{d \exp(A(t - t_0))}{dt} \vec x_0 = A \exp(A(t - t_0)) \vec x_0 = A \vec x(t); \tag 5$</p>
<p>we also see from (4) that</p>
<p>$\vec x(t_0) = \exp(A(t_0 - t_0)) \vec x_0 = \exp(A(0)) \vec x_0 = \exp(0) \vec x_0 = I \vec x_0 = \vec x_0; \tag 6$</p>
<p>thus (4) is consistent with our choice of initial condition. We compute the matrix $\exp(A(t - t_0))$ as follows:</p>
<p>set </p>
<p>$J = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}; \tag 7$</p>
<p>then we have</p>
<p>$A = \begin{bmatrix} p & -q \\ q & p \end{bmatrix} = pI + qJ; \tag 8$</p>
<p>it follows that</p>
<p>$\exp(A(t - t_0)) = \exp((pI + qJ)(t - t_0)) = \exp(p(t - t_0)I + q(t - t_0)J); \tag 9$</p>
<p>now since $I$ and $J$ commute, </p>
<p>$IJ = JI, \tag{10}$</p>
<p>so do $p(t - t_0)I$ and $q(t - t_0)J$</p>
<p>$(p(t - t_0)I)(q(t - t_0)J) = (q(t - t_0)J)(p(t - t_0)I); \tag{11}$</p>
<p>this allows us to conclude that, just as if $I$ and $J$ were ordinary real or complex numbers,</p>
<p>$\exp((p(t - t_0)I + q(t - t_0)J) = \exp(p(t - t_0)I) \exp(q(t - t_0)J); \tag{12}$</p>
<p>indeed, a demonstration of (12) given (11) is virtually identical the the proof that</p>
<p>$\exp(a + b) = \exp(a) \exp (b), \; a, b \in \Bbb C; \tag{13}$</p>
<p>for more information, see <a href="https://en.wikipedia.org/wiki/Matrix_exponential" rel="nofollow noreferrer">this wikipedia page</a>.</p>
<p>So far we have</p>
<p>$\exp(A(t - t_0)) = \exp(p(t - t_0)I + q(t - t_0)J) = \exp(p(t - t_0)I) \exp(q(t - t_0)J); \tag{14}$</p>
<p>the next step is the evaluation of $\exp(p(t - t_0)I)$ and $\exp(q(t - t_0)J)$; the first of these two is easy, using the power series:</p>
<p>$\exp(p(t - t_0)I) = \displaystyle \sum_0^\infty \dfrac{(p(t - t_0)I)^n}{n!} = \sum_0^\infty \dfrac{(p(t - t_0))^n}{n!}I^n$
$= \left (\displaystyle \sum_0^\infty \dfrac{(p(t - t_0))}{n!} \right ) I = \exp(p(t - t_0))I; \tag {15}$</p>
<p>in order to find $\exp(q(t - t_0)J)$ we note that</p>
<p>$J^2 = -I; \; J^3 = -J; \; J^4 = -J^2 = I; \; J^5 = J^4J = J; \; J^6 = J^4J^2 = -I, \tag{16}$</p>
<p>and so forth, just as $i = \sqrt{-1}$ satisfies</p>
<p>$i^2 = -1; \; i^3 = -i; \; i^4 = 1; \; i^5 = i; \; i^6 = -1, \tag{17}$</p>
<p>etc.; it is easy to see that $i$ and $J$ follow essentially the same pattern, so it follows that the power series for $\exp(q(t - t_0)J$ will, just as</p>
<p>$e^{iq(t - t_0))} = \cos q(t - t_0) + i \sin q(t - t_0), \tag{18}$</p>
<p>yield</p>
<p>$\exp(q(t - t_0)J) = \cos q(t - t_0) I + \sin q(t - t_0) J$
$= \begin{bmatrix} \cos q(t - t_0) & -\sin q(t - t_0) \\ \sin q(t - t_0) & \cos q(t - t_0) \end{bmatrix} = R(t); \tag{19}$</p>
<p>we see the matrix above, which we have chosen to name $R(t)$, is in fact an ordinary $2$-dimensional rotation matrix, and as such is orthogonal; it is easy to see that</p>
<p>$R(t)R^T(t) = R^T(t)R(t) = I, \; \forall t \in \Bbb R. \tag{20}$</p>
<p>If we now assemble (4), (14), (15) and (19) together we may write</p>
<p>$\vec x(t) = \exp(p(t - t_0)) R(t) \vec x_0 \tag{21}$ </p>
<p>as the solution to (1) with $x(t_0) = x_0$; this equation allows us to address stability questions concerning (1) directly in terms of $p$: first of all, suppose that</p>
<p>$p = 0; \tag{22}$</p>
<p>then (21) becomes</p>
<p>$\vec x(t) = R(t) \vec x_0, \tag{23}$</p>
<p>which has several relvant consequences: first of all, it implies</p>
<p>$\Vert x(t) \Vert = \Vert x_0 \Vert, \tag{24}$</p>
<p>since $R(t)$ is orthogonal; thus the solutions may be seen to be circles surrounding the origin; as such the system trajectories maintain a fixed distance from $(0, 0)$ and hence the system is <em>stable</em>; furthermore, (1) manifests another form of stability in the event that $p = 0$: suppose that we have two solutions $\vec x(t)$ and $\vec y(t)$ of (1), with</p>
<p>$\vec x(t_0) = \vec x_0, \; \vec y(t_0) = \vec y_0; \tag{25}$</p>
<p>then</p>
<p>$\dfrac{d}{dt}(\vec x(t) - \vec y(t)) = \dot{\vec x}(t) - \dot{\vec y}(t) = A \vec x(t) - A \vec y(t) = A(\vec x(t) - \vec y(t)); \tag{26}$</p>
<p>also,</p>
<p>$\vec x(t_0) - \vec y(t_0) = \vec x_0 - \vec y_0; \tag{27}$</p>
<p>based upon (25)-(27) we see that $\vec x(t) - \vec y(t)$ is a solution initialized at $\vec x_0 - \vec y_0$; thus, in accord with (23), (24) we may write</p>
<p>$(\vec x(t) - \vec y(t)) = R(t) (x_0 - y_0), \tag{28}$</p>
<p>$\Vert (\vec x(t) - \vec y(t) \Vert = \Vert x_0 - y_0 \Vert, \tag{29}$</p>
<p>which shows that the system (1) with $p = 0$ enjoys an even more comprehensive stability property: not only is $(0, 0)$ a stable equilibrium, but any two system points remain the same distance apart for all time; indeed, it is almost as if the flow of (1) is in fact a <em>rigid motion</em> of the plane; in any event, it is clear from (29) that any two solutions remain as close as they started out forever, an enhancement upon the simple stability of the origin. </p>
<p>We now turn to the case $p < 0$; we then see from (21) that</p>
<p>$\Vert \vec x(t) \Vert = \exp(p(t - t_0)) \Vert \vec x_0 \Vert \to 0 \; \text{as} \; t \to \infty \tag{30}$ </p>
<p>for every solution $\vec x(t)$; thus the origin is <em>globally asymptotically stable</em>; also, it is easily seen that (25)-(27) in this case imply, <em>via</em> (21), that</p>
<p>$\Vert \vec x(t) - \vec y(t) \Vert = \exp(p(t - t_0)) \Vert \vec x_0 - \vec y_0 \Vert \to 0 \; \text{as} \; t \to \infty, \tag{31}$</p>
<p>i.e., all trajectories converge to one another with the passage of time. In this case, the origin is a <em>spiral stable point</em> if $q \ne 0$, and a <em>stable node</em> when $q$ vanishes.</p>
<p>When $p > 0$ the origin is of course unstable and all trajectories diverge from one another; the reader may easily work out the details along lines analogous to the above.</p>
<p>Finally, we have presented the above discussion in terms of $p$ even though the problem statement spoke in terms of eigenvalues of the matrix $A$; but the characteristic polynomial of the matrix $A$ is</p>
<p>$\det(A - \lambda I) = \lambda^2 - 2p \lambda + (p^2 + q^2); \tag{32}$</p>
<p>that the zeroes of (32) are $p \pm iq$ is an easy calculation, so in dealing with $p$ we are actually addressing the real part of the eigenvalues of $A$, as per request.</p>
|
1,156,907 | <p>I don't know anything about measure theory, I'm studying real analysis and this showed up in the book I'm reading as a way to characterize integrable functions. The author defined that a subset $X \subset \mathbb{R}$ has measure zero if for each $\epsilon > 0$ we can find infinitely countable open intervals $I_n$ such that $X \subset \bigcup_{n=1}^{\infty}I_n$ and $\sum_{n=1}^{\infty} |I_n| < \epsilon $ where $|I|$ is the size of $I$, as in, if $I = (a,b)$, then $|I| = b - a$.</p>
<p>Now, the author gives the following proof that the countable union of measure-zero sets has measure zero:</p>
<p>"Let $Y =\bigcup_{i=1}^{\infty} X_i $, where each $X_i$ has measure zero. Now, given $\epsilon > 0 $ we can, for each $n$, write $X_n \subset \bigcup_{i=1}^{\infty} I_{n_i}$ where each $I_{n_i}$ is an open subset and $\sum_{i=1}^{\infty}|I_{n_i}| < \epsilon / 2^n$. Therefore, $Y \subset \bigcup_{n,j=1}^{\infty} I_{n_J}$ where $\sum_n \sum_j |I_{n_J}| < \sum_{i=1}^{\infty} \epsilon /2^n = \epsilon$. Therefore, $m(Y) = 0$"</p>
<p>I'm really confused about the ending. It is very intuitive, but it's not rigorous enough for me, I wanna see formally why this holds:</p>
<p>$\sum_n \sum_j |I_{n_J}| < \sum_{i=1}^{\infty} \epsilon /2^n$</p>
<p>Like maybe looking at the definition of a series, the limit of the sequence of partial sums. Any help?</p>
| Community | -1 | <p>Given two sequences $x_n$ and $y_n$, we have the relation
$$
x_n\leq y_n\Rightarrow \sum_n x_n\leq \sum_n y_n.
$$
Therefore, if we set $x_n=\sum^{\infty}_{j=1}|I_{n_j}|$ and $y_n=\frac{\epsilon }{2^n}$, applying the above gives
$$
\sum_n\sum^{\infty}_{j=1}|I_{n_j}|<\sum_n\frac{\epsilon}{2^n}
$$</p>
|
1,156,907 | <p>I don't know anything about measure theory, I'm studying real analysis and this showed up in the book I'm reading as a way to characterize integrable functions. The author defined that a subset $X \subset \mathbb{R}$ has measure zero if for each $\epsilon > 0$ we can find infinitely countable open intervals $I_n$ such that $X \subset \bigcup_{n=1}^{\infty}I_n$ and $\sum_{n=1}^{\infty} |I_n| < \epsilon $ where $|I|$ is the size of $I$, as in, if $I = (a,b)$, then $|I| = b - a$.</p>
<p>Now, the author gives the following proof that the countable union of measure-zero sets has measure zero:</p>
<p>"Let $Y =\bigcup_{i=1}^{\infty} X_i $, where each $X_i$ has measure zero. Now, given $\epsilon > 0 $ we can, for each $n$, write $X_n \subset \bigcup_{i=1}^{\infty} I_{n_i}$ where each $I_{n_i}$ is an open subset and $\sum_{i=1}^{\infty}|I_{n_i}| < \epsilon / 2^n$. Therefore, $Y \subset \bigcup_{n,j=1}^{\infty} I_{n_J}$ where $\sum_n \sum_j |I_{n_J}| < \sum_{i=1}^{\infty} \epsilon /2^n = \epsilon$. Therefore, $m(Y) = 0$"</p>
<p>I'm really confused about the ending. It is very intuitive, but it's not rigorous enough for me, I wanna see formally why this holds:</p>
<p>$\sum_n \sum_j |I_{n_J}| < \sum_{i=1}^{\infty} \epsilon /2^n$</p>
<p>Like maybe looking at the definition of a series, the limit of the sequence of partial sums. Any help?</p>
| Anthony Peter | 58,540 | <p>We're exploiting Zeno's Paradox. Since each set has measure $0$, we can cover it by intervals whose total length is less than any positive real number. Since the union is countable, we can enumerate our sets of measure $0$ as $\{I_1, I_2, I_3, \ldots, \}$. Let $\mu(S) = (b-a)$ for $S=(a,b)$.</p>
<p>Let $\epsilon > 0$. Let $I_j$ be covered by an open covering so that $$\mu(A_j) < \frac{\epsilon}{2^j}$$ $$I_j \subset A_j$$
Then, since $\mu$ is countably sub-additive, $$\mu\left(\bigcup_{j=1}^{\infty} I_j \right) \leq \sum_{j=1}^{\infty} \mu(I_j) \leq \sum_{j=1}^{\infty} \mu(A_j) = \sum_{j=1}^{\infty} \frac{\epsilon}{2^j} = \epsilon.$$</p>
<p>Thus, our set has measure 0. </p>
<p>This is nice because we're able to do a countable number of ''things'' with only an epsilon of room. </p>
|
4,350,015 | <p>I'm trying to learn calculus through self study and I happened upon the following exercise:</p>
<p><span class="math-container">$$ \int_{0}^{2} \sqrt{4 - x^2}\cdot\operatorname{sgn}(x-1) \,dx $$</span></p>
<p>Seeing the sgn I thought: well this is easy and concluded that since <span class="math-container">$ \int_{0}^{2} \operatorname{sgn}(x-1) \,dx = 0$</span>, the answer should be zero. But of course then I remembered that <span class="math-container">$ \int f(x)g(x) \,dx \neq \int f(x) \,dx \int g(x) \,dx$</span> and got lost.</p>
<p>I know how to solve <span class="math-container">$ \int_{0}^{2} \sqrt{4 - x^2}dx$</span>, since that geometrically corresponds to a quarter circle of radius <span class="math-container">$2$</span> it's just <span class="math-container">$\pi$</span>, but how do I approach the product of these things?</p>
| Taladris | 70,123 | <p>To expand on @José Carlos Santos' answer, you can compute <span class="math-container">$\int \sqrt{4-x^2}\; dx$</span> by trigonometric substitution: let <span class="math-container">$x=2\sin(t)$</span> with <span class="math-container">$-\frac{\pi}{2}\le t\le \frac{\pi}{2}$</span> . Then <span class="math-container">$\sqrt{4-x^2}=2\cos(t)$</span> and <span class="math-container">$dx=2\cos(t)\; dt$</span> so</p>
<p><span class="math-container">$$\int \sqrt{4-x^2}\; dx = \int 4\cos^2(t)\; dt = 2 \int 1+\cos(2t)\; dt
=
2\left(t + \frac{\sin(2t)}{2} \right)+ C$$</span></p>
<p>Substituting back, we get</p>
<p><span class="math-container">$$\int \sqrt{4-x^2}\; dx = 2 \arcsin\left(\frac x 2\right) + \frac{x\sqrt{4-x^2}}{2} + C$$</span></p>
<p>Therefore,</p>
<p><span class="math-container">$$\int_{0}^{2} \sqrt{4 - x^2}\cdot\operatorname{sgn}(x-1) \,dx
=
-\int_0^1\sqrt{4-x^2}\,\mathrm dx+\int_1^2\sqrt{4-x^2}\,\mathrm dx
=
-\left(2\arcsin(1/2)+\frac{\sqrt{3}}{2}
\right)
+
\left(2\arcsin(1)-2\arcsin(1/2)-\frac{\sqrt{3}}{2}
\right)
=\frac{\pi}{3}-\sqrt{3}$$</span></p>
|
1,793,854 | <p>I am messed up on solving this question. What should I do first in order to get the answer ?</p>
<p><a href="https://i.stack.imgur.com/hE4rG.png" rel="nofollow noreferrer">This is the trigonometric function</a></p>
<p>$$ \lim \limits_{x \rightarrow 0} \frac{(a+x)\sec(a+x) - a \sec(a)}{x} $$</p>
| Claude Leibovici | 82,404 | <p>One solution, if allowed, uses Taylor series.</p>
<p>Built a round $x=0$, $$\sec(a+x)=\sec (a)+x \tan (a) \sec (a)+O\left(x^2\right)$$ $$(x+a)\sec(a+x)=a \sec (a)+x (\sec (a)+a \tan (a) \sec (a))+O\left(x^2\right)$$</p>
|
4,198,805 | <p>A <a href="https://en.wikipedia.org/wiki/Sober_space" rel="nofollow noreferrer">sober space</a> is a topological space such that every irreducible closed subset is the closure of exactly one point. Looking for examples I convinced myself that the following is true.</p>
<blockquote>
<p>Every finite <span class="math-container">$T_0$</span> topological space is sober.</p>
</blockquote>
<p>As I could not find this mentioned anywhere, can someone provide a proof to have it as a reference?</p>
| diracdeltafunk | 19,006 | <p>This is true. It suffices to show that every finite irreducible space has a generic point, since <span class="math-container">$T_0$</span> implies that generic points are unique. So, let <span class="math-container">$X$</span> be a finite irreducible space. Then <span class="math-container">$X$</span> is the union of the closures of its points, but this is a finite union of closed sets, so irreducibility says that one of these closures is <span class="math-container">$X$</span>!</p>
|
751,053 | <p>T : Rn → Rm is a linear transformation where n,m>= 2</p>
<p>Let V be a subspace of Rn
and let W ={T(v ) | v ∈ V}
. Prove completely that W
is a subspace of Rm. </p>
<p>For this question how do I show that the subspace is non empty, holds under scaler addition and multiplication! I have never proved subspaces with transformations before!</p>
<p>I think I know this subspace is non empty when v is the 0 vector so is T(v) right?</p>
<p>b)Assume now that nullity(T) = 0 and that A =
{
a1; a2; . . . ; ap
}
is a basis for V as
in part (a). Prove completely that the set C =
{
T(a1); T(a2); . . . . ; T(ap)
}
is a basis
for W as in part (a) </p>
<p>Overhere Im lost!
since the nullity is 0 we know the matrix representation of the transformation is invertible. but then what</p>
| Community | -1 | <ul>
<li>We have $T(0_{\Bbb R^n})=0_{\Bbb R^m}\in W$ since $T$ is linear so $W\ne\emptyset$.</li>
<li>Let $y_1,y_2\in W$ and $\alpha\in\Bbb F$ then there's $x_1,x_2\in V$ s.t.
$$T(x_1)=y_1\qquad T(x_2)=y_2$$
but since $V$ is a linear subspace then $\alpha x_1+x_2\in V$ and by linearity of $T$ we have
$$T(\alpha x_1+ x_2)=\alpha T(x_1)+T(x_2)=\alpha y_1+y_2\in W$$
hence $W$ is invariant by linear combination. Conclude.</li>
</ul>
|
878,373 | <p>Often I have heard about the link between Algebra (in particular Representations of Groups and Algebras) and some "indefinite" field of Physics.</p>
<p>I have a good preparation in Algebra and Representation Theory (in particular about Representations of Lie Algebras), and I'm fascinated with Physics. My idea is try to understand this link and eventually study it with more depth.</p>
<p>Hence I'm looking for an introductory book that emphasizes the applications of Algebra in Physics from a comprehensible and mathematical point of view.</p>
<p>Does anyone have an idea for a book with these requisites?</p>
<p>Thank you! </p>
| user44670 | 44,670 | <p>I'm sorry, I was very stupid. We can just apply Ito's formula and get:
\begin{align*}
& \int_a^b f(t)dW_t=f(b)W_b-f(a)W_a-\int_a^b W_t f'(t)dt.
\end{align*}
This yields that indeed we can find the desired pathwise upper bound of $\int_a^b f(t)dW_t$ in terms of $||f||$, $||f'||$, $\sup_{t \in [a,b]}|W(t)|$.</p>
|
2,010,069 | <p>I am looking on the solution to this problem presented in the book <em>"Fifty Challenging Problems in Probability with Solutions"</em> by Mosteller (p.18-19).</p>
<blockquote>
<p>On average, how many times must a die be thrown until one gets a 6?</p>
</blockquote>
<p>There are many ways to solve this problem as this is simple example of a geometric distribution, but I don't quite get the trick the author did it with the $qm$.
<br> <strong>I am looking for explanation/interpretation of this trick (transition (2) (3) )</strong>?
<br>
<br>
The first expression is clear, it is just the expansion of the expected value definition</p>
<p><br>
(p be the probability of a 6 on a given trial)
<br>
$$ m = p + 2 pq + 3pq^2 + 4pq^3 + ... \quad \quad \quad (1)$$
<br>
Further the "trick" with qm has been used
<br>
$$qm =\ \ \ \ \ \ \ \ pq + 2pq^2 + 3pq^3 + ... \quad \quad \quad (2)$$
so that
$$m - qm = p + pq + pq^2 + ... \quad \quad \quad (3)$$
$$m(1-q) = 1 \quad \quad \quad (4)$$
$$m=\frac{1}{p} \quad \quad \quad (5)$$</p>
| Bobbie D | 317,218 | <p>Using the property of the transpose $\langle A^Tw,v\rangle = \langle w, Av\rangle$, I get:</p>
<p>$$\pmatrix{a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23}}\pmatrix{x \\ y \\ z} = \pmatrix{\langle A^T\pmatrix{1 \\ 0}, \pmatrix{x \\ y \\ z}\rangle \\ \langle A^T\pmatrix{0 \\ 1}, \pmatrix{x \\ y \\ z}\rangle} = \pmatrix{\langle \pmatrix{1 \\ 0}, A\pmatrix{x \\ y \\ z}\rangle \\ \langle \pmatrix{0 \\ 1}, A\pmatrix{x \\ y \\ z}\rangle} = \pmatrix{\operatorname{proj}_{e_1}(Ax) \\ \operatorname{proj}_{e_2}(Ax)}$$</p>
<p>That last bit should be absolutely clear -- The first component of $Ax$ is the projection of $Ax$ onto $e_1 = \pmatrix{1 \\ 0}$ and likewise for the second component.</p>
|
2,010,069 | <p>I am looking on the solution to this problem presented in the book <em>"Fifty Challenging Problems in Probability with Solutions"</em> by Mosteller (p.18-19).</p>
<blockquote>
<p>On average, how many times must a die be thrown until one gets a 6?</p>
</blockquote>
<p>There are many ways to solve this problem as this is simple example of a geometric distribution, but I don't quite get the trick the author did it with the $qm$.
<br> <strong>I am looking for explanation/interpretation of this trick (transition (2) (3) )</strong>?
<br>
<br>
The first expression is clear, it is just the expansion of the expected value definition</p>
<p><br>
(p be the probability of a 6 on a given trial)
<br>
$$ m = p + 2 pq + 3pq^2 + 4pq^3 + ... \quad \quad \quad (1)$$
<br>
Further the "trick" with qm has been used
<br>
$$qm =\ \ \ \ \ \ \ \ pq + 2pq^2 + 3pq^3 + ... \quad \quad \quad (2)$$
so that
$$m - qm = p + pq + pq^2 + ... \quad \quad \quad (3)$$
$$m(1-q) = 1 \quad \quad \quad (4)$$
$$m=\frac{1}{p} \quad \quad \quad (5)$$</p>
| Jair Taylor | 28,545 | <p>It is easiest to see this in one dimension first.</p>
<p>Our goal is to show that any linear transformation $T: \mathbb{R}^n \rightarrow \mathbb{R}$ can be represented in the form $Tu = \beta^Tu$ for some $n$-dimensional vector $\beta$. Say that $u \in \mathbb{R}^n$; let $e_1, \ldots, e_n$ be the standard basis vectors for $\mathbb{R}^n$ (where $e_i$ has a $1$ in the $i$th position, and $0$'s elsewhere.) Then we can write $u = \sum_i u_i e_i$ where $u_i$ is the $i$-th coordinate of $u$. Since $T$ is linear, we have $$Tu =T \sum_i u_i e_i = \sum_i u_i Te_i = \sum_i u_i \beta_i,$$ where $\beta_i = Tb_i$. This means that with respect to the standard basis, $Tu = \beta \cdot u$ where $\beta$ is the vector $(\beta_1, \ldots, \beta_n)$. Thus every linear map from $\mathbb{R}^n$ to $\mathbb{R}$ can be represented by taking the dot product with a fixed vector.</p>
<p>Now for the multi-dimensional case: If $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ then $T$ is equivalent to the $m$-tuple of functions $(T_1, T_2, \ldots, T_m)$ where $T_jx$ is the $j$-th coordinate of $Tx$. Then for each $j$ there is a vector $\beta^j$ with $T_jx = \beta^j \cdot x$ and the result follows. (note that the $j$ in $\beta^j$ is just a superscript here meaning the $j$-th one, and doesn't have anything to with the $j$-th power.)</p>
|
119,481 | <p>I need to prove the following trigonometric identity:
$$ \frac{\sin^2(\frac{5\pi}{6} - \alpha )}{\cos^2(\alpha - 4\pi)} - \cot^2(\alpha - 11\pi)\sin^2(-\alpha - \frac{13\pi}{2}) =\sin^2(\alpha)$$</p>
<p>I cannot express $\sin(\frac{5\pi}{6}-\alpha)$ as a function of $\alpha$. Could it be a textbook error?</p>
| Pedro | 23,350 | <p>Some important translations: </p>
<p>$$\tag 1\sin(x\pm 2 \pi) = \sin x $$
$$\tag {1'}\cos(x\pm 2 \pi) = \cos x $$
$$\tag 2\cot(x\pm \pi)= \cot x$$
$$\tag {2'}\tan(x\pm \pi)= \tan x$$</p>
<p>$$\tag 3 \sin \left(\frac \pi 2 -x \right)=\cos x$$
$$\tag 4 \cos \left(\frac \pi 2 -x \right)=\sin x$$
$$\tag 5 \sin(\pi-x)=\sin x$$ and $$\tag 6\cos(\pi-x)=-\cos x$$</p>
<p>and $$\tag 7 \sin(-x)=-\sin x$$
$$\tag 8 \cos (-x) = \cos x$$ </p>
<p>Although taking $\alpha =0$ reveals that the equality doesn't hold, assume that there is no typo, then, we could move on as follows.</p>
<p>You have that</p>
<p>$$ \frac{\sin^2 \left(\frac{5\pi}{6} - \alpha \right)}{\cos^2(\alpha - 4\pi)} - \cot^2(\alpha - 11\pi)\sin^2 \left(-\alpha - \frac{13\pi}{2}\right) =\sin^2(\alpha)$$</p>
<p>Using the above, we can write</p>
<p>$$\eqalign{
& {\sin ^2}\left( {\frac{{5\pi }}{6} - \alpha } \right) = {\left[ { - \sin \left( {\alpha - \frac{{5\pi }}{6}} \right)} \right]^2} = {\sin ^2}\left( {\alpha - \frac{{5\pi }}{6}} \right) \cr
& {\cot ^2}(\alpha - 11\pi ) = {\cot ^2}\left( {\alpha - 10\pi } \right) = \cdots = {\cot ^2}\alpha \cr
& {\sin ^2}\left( { - \alpha - \frac{{13\pi }}{2}} \right) = {\left[ { - \sin \left( {\alpha + \frac{{13\pi }}{2}} \right)} \right]^2} = \sin {\left( {\alpha + \frac{{13\pi }}{2}} \right)^2} \cr
& {\cos ^2}(\alpha - 4\pi ) = {\cos ^2}(\alpha - 2\pi ) = {\cos ^2}\alpha \cr} $$</p>
<p>so that we have</p>
<p>$$\frac{{{{\sin }^2}\left( {\alpha - \frac{{5\pi }}{6}} \right)}}{{{{\cos }^2}\alpha }} - {\cot ^2}\alpha {\sin ^2}\left( {\alpha + \frac{{13\pi }}{2}} \right) = {\sin ^2}\alpha $$</p>
<p>Now</p>
<p>$${\sin ^2}\left( {\alpha - \frac{{5\pi }}{6}} \right) = {\sin ^2}\left( {\alpha - \frac{{3\pi }}{6} - \frac{{2\pi }}{6}} \right) = {\sin ^2}\left( {\alpha - \frac{\pi }{3} - \frac{\pi }{2}} \right) = {\left( { - 1} \right)^2}{\sin ^2}\left( {\frac{\pi }{2} - \left( {\alpha - \frac{\pi }{3}} \right)} \right) = {\cos ^2}\left( {\alpha - \frac{\pi }{3}} \right)$$</p>
<p>and</p>
<p>$$\eqalign{
& {\sin ^2}\left( {\alpha + \frac{{13\pi }}{2}} \right) = {\sin ^2}\left( {\alpha + \frac{{12\pi }}{2} + \frac{\pi }{2}} \right) = {\sin ^2}\left( {\alpha + 6\pi + \frac{\pi }{2}} \right) = {\sin ^2}\left( {\alpha + \frac{\pi }{2}} \right) \cr
& = {\sin ^2}\left( {\frac{\pi }{2} - \left( { - \alpha } \right)} \right) = {\cos ^2}\left( { - \alpha } \right) = {\cos ^2}\alpha \cr} $$</p>
<p>so that you have</p>
<p>$$\frac{{{{\cos }^2}\left( {\alpha - \frac{\pi }{3}} \right)}}{{{{\cos }^2}\alpha }} - {\cot ^2}\alpha {\cos ^2}\alpha = {\sin ^2}\alpha $$</p>
<p>Now, solving for $${{{\cos }^2}\left( {\alpha - \frac{\pi }{3}} \right)}$$</p>
<p>gives</p>
<p>$${\cos ^2}\left( {\alpha - \frac{\pi }{3}} \right) = {\cos ^2}\alpha {\sin ^2}\alpha + {\cos ^6}\alpha \frac{1}{{{{\sin }^2}\alpha }}$$</p>
<p>There is some typo in your excercise, since letting $\alpha =0$ gives $1/4$ on the LHS and is not defined for the RHS. When you discover what the typo is, move on with the listed translations. </p>
|
1,878,734 | <p>Is it true that if an isomorphism $f$ maps a cyclic group $G$ to group $H$ that $H$ must also be cyclic? It seems intuitive but until I can actually prove it I'm always a bit dubious to believe it. </p>
| florence | 343,842 | <p>Suppose $G$ is cyclic and $\phi: G\to H$ is surjective. Let $G$ be generated by $a\in G$, i.e. $G = \langle a\rangle$. Let $b\in H$. Then there exists some $x\in G$ so that $b = \phi(x)$, as $\phi$ is surjective. Further, since $G$ is cyclic, we have $x = a^n$ for some $n$. Then $b = \phi(x) = \phi(a^n) =\phi(a)^n$. This can be done for any element of $H$, so we see that $H = \langle \phi(a) \rangle$.</p>
|
2,793,983 | <p>For example I find myself wanting to write $x$ is an element of the integers from $1$ to $50$,</p>
<p>Is this the quickest way? </p>
<p>$x\in \left[ 1,50\right] \cap \mathbb{N} $</p>
<p>Also is this standard on here? $\mathbb{N} = \{0, 1, 2,\dotsc \}$,
$\mathbb{ℤ}_+ = \{1, 2, \dotsc \}$.</p>
| user21820 | 21,820 | <p>As others have said, you should always define non-standard notation, but here is one that you can consider (and is actually valid syntax in some programming languages):</p>
<blockquote>
<p>$[a\,..b]$ represents the integers from $a$ to $b$ inclusive.</p>
</blockquote>
<p>This is also compatible with the convention for square/round-brackets to denote closed/open interval endpoints:</p>
<blockquote>
<p>$[a\,..b)$ represents the half-open interval from $a$ to less than $b$.</p>
</blockquote>
<p>Though mixed-bracket interval notation might best be avoided.</p>
|
1,985,552 | <p>Where $p_n \rightarrow p$. I'm trying to prove that for $E=\{ p_n : n \in \mathbb{N}$ and $lim_{n\rightarrow \infty} p_n =p \}$, then $Cl(E)=E \cup \{p \}$ and $Cl(E)$ is compact. </p>
<p>Also, I'm currently using the definition of limit points as p is a limit point if $\forall r>0, (E \cap N_r(p)) \backslash \{p\} \neq \emptyset$. </p>
<p>Here's a rough outline for what I have:</p>
<p><strong>Proving $Cl(E)=E \cup \{p \}$:</strong></p>
<p>By a theorem, we konw that $\{ p_n \}$ converges to $p \in E$ iff every neighborhood of p contains $p_n$ for all but finitely many n. So I'm thinking this theorem shows that $\forall r>0, (E \cap N_r(p)) \backslash \{p\} \neq \emptyset$.However, I'm not sure I'm supposed to prove that p is the only limit point. </p>
<p><strong>Proving compact</strong></p>
<p>I'm also still unsure how to prove $Cl(E)$ is compact just by the definition. Since $d(p_n, p) < \epsilon$, does this imply that all open covers $\mathcal{G}$ is somehow bounded by this and we therefore have finite subcovers? </p>
| Jack D'Aurizio | 44,121 | <p>The answer is given by the coefficient of $x^{30}$ in $A(x)\cdot B(x)\cdot C(x)\cdot D(x)$ where
$$A(x)=(x^1+x^2+x^2+x^4+x^5),\quad B(x)=(x^4+x^5+x^6+x^7+x^8+x^9)\\C(x)=(x^6+x^7+x^8+x^9+x^{10}+x^{11}),\quad D(x)=(x^{10}+x^{11}+x^{12}+x^{13}+x^{14}+x^{15})$$
That is also the coefficient of $x^9$ in $E(x)\cdot F(x)\cdot G(x)\cdot H(x)$ where
$$ E(x)=(1+x+x^2+x^3+x^4),\quad F(x)=G(x)=H(x)=(1+x+x^2+x^3+x^4+x^5)$$ hence it is enough to compute
$$ [x^9]\frac{(1-x^5)(1-x^6)^3}{(1-x)^4} $$
and since $$\frac{1}{(1-x)^4}=\sum_{n\geq 0}\binom{n+3}{3}x^n $$
the answer is given by
$$\binom{9+3}{3}-\binom{4+3}{3}-3\binom{3+3}{3} = 220-35-60 = \color{red}{\large 125}.$$</p>
|
1,985,552 | <p>Where $p_n \rightarrow p$. I'm trying to prove that for $E=\{ p_n : n \in \mathbb{N}$ and $lim_{n\rightarrow \infty} p_n =p \}$, then $Cl(E)=E \cup \{p \}$ and $Cl(E)$ is compact. </p>
<p>Also, I'm currently using the definition of limit points as p is a limit point if $\forall r>0, (E \cap N_r(p)) \backslash \{p\} \neq \emptyset$. </p>
<p>Here's a rough outline for what I have:</p>
<p><strong>Proving $Cl(E)=E \cup \{p \}$:</strong></p>
<p>By a theorem, we konw that $\{ p_n \}$ converges to $p \in E$ iff every neighborhood of p contains $p_n$ for all but finitely many n. So I'm thinking this theorem shows that $\forall r>0, (E \cap N_r(p)) \backslash \{p\} \neq \emptyset$.However, I'm not sure I'm supposed to prove that p is the only limit point. </p>
<p><strong>Proving compact</strong></p>
<p>I'm also still unsure how to prove $Cl(E)$ is compact just by the definition. Since $d(p_n, p) < \epsilon$, does this imply that all open covers $\mathcal{G}$ is somehow bounded by this and we therefore have finite subcovers? </p>
| Felix Marin | 85,343 | <p>$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&x_{1} + x_{2} + x_{3} + x_{4} = 30\implies y_{1} + y_{2} + y_{3} + y_{4} = 9
\\[2mm] &\
\mbox{where}\quad
0 \leq y_{1} \leq 4\,,\quad 0 \leq y_{2} \leq 5\,,\quad
0 \leq y_{3} \leq 5\,,\quad 0\leq y_{4}\leq 5
\end{align}
<hr>
\begin{align}
&\sum_{y_{1} = 0}^{4}\sum_{y_{2} = 0}^{5}\sum_{y_{3} = 0}^{5}
\sum_{y_{4} = 0}^{5}\bracks{y_{1} + y_{2} + y_{3} + y_{4} = 9} =
\sum_{y_{1} = 0}^{4}\sum_{y_{2} = 0}^{5}\sum_{y_{3} = 0}^{5}
\sum_{y_{4} = 0}^{5}\oint_{\verts{z} = 1^{-}}
{1 \over z^{10 - y_{1} - y_{2} - y_{3} - y_{4}}}\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
\oint_{\verts{z} = 1^{-}}
{1 \over z^{10}}\sum_{y_{1} = 0}^{4}z^{y_{1}}
\pars{\sum_{y = 0}^{5}z^{y}}^{3}\,{\dd z \over 2\pi\ic} =
\oint_{\verts{z} = 1^{-}}
{1 \over z^{10}}{z^{5} - 1 \over z - 1}\pars{z^{6} - 1 \over z - 1}^{3}
\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
\oint_{\verts{z} = 1^{-}}
{\pars{1 - z^{5}}\pars{1 - z^{6}}^{3} \over z^{10}\pars{1 - z}^{4}}
\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
\sum_{m = 0}^{3}\sum_{n = 0}^{\infty}{3 \choose m}{-4 \choose n}
\pars{-1}^{m + n}
\oint_{\verts{z} = 1^{-}}\bracks{z^{6m + n - 10} - z^{6m + n - 5}}
\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
\sum_{m = 0}^{3}\sum_{n = 0}^{\infty}{3 \choose m}{n + 3 \choose 3}\pars{-1}^{n}
\pars{-1}^{m + n}
\braces{\vphantom{\large A}\bracks{n = 9 - 6m} - \bracks{n = 4 - 6m}}
\\[5mm] = &\
\sum_{m = 0}^{1}{3 \choose m}{12 - 6m \choose 3}\pars{-1}^{m} -
\sum_{m = 0}^{0}{3 \choose m}{7 - 6m \choose 3}\pars{-1}^{m} =
{12 \choose 3} - {3 \choose 1}{6 \choose 3} - {7 \choose 3}
\\[5mm] = &\
220 - 3 \times 20 - 35 =
\bbox[#ffe,10px,border:1px groove navy]{\ds{125}}
\end{align}</p>
|
4,483,507 | <p>How can I change <span class="math-container">$\dfrac{-(3-\sqrt{3})}{(3+\sqrt{3})}$</span> to <span class="math-container">$\dfrac{1-\sqrt{3}}{1+\sqrt{3}}$</span>?</p>
<p>Background:</p>
<p>I tried solving <span class="math-container">$\tan(345°)$</span> with the trigonometric angle <em><strong>sum/difference</strong></em> identity.</p>
<p>I used <span class="math-container">$\tan(45°-30°)$</span> to find <span class="math-container">$\tan(15°)$</span>.</p>
<p>I then used <span class="math-container">$\tan(360°-15°)$</span> to get <span class="math-container">$\tan(345°)$</span>.</p>
<p>I used the identity:
<span class="math-container">$$\tan(\theta \pm \phi)=
\frac{\tanθ \pm tan \phi}{1 \mp \tanθ×\tan\phi}$$</span></p>
<p>My answer was <span class="math-container">$\frac{-(3-√3)}{(3+√3)}$</span> but the answer given was <span class="math-container">$\frac{(1-√3)}{(1+√3)}$</span>.</p>
<p>I used a calculator to find out these are one and the same.</p>
<blockquote>
<p><a href="https://i.stack.imgur.com/mfP7E.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mfP7E.jpg" alt="enter image description here" /></a></p>
</blockquote>
<p>However, without using a calculator, How would I be able to transform my answer to the given answer.</p>
| Suzu Hirose | 190,784 | <p>The logic of the proof is correct, but it reads like word salad. For example,</p>
<blockquote>
<p>Let <span class="math-container">$d$</span> be the standard Euclidean metric.</p>
</blockquote>
<p>You've put that in twice, and after you've already used <span class="math-container">$d$</span>. You're considering a metric space so one assumes that you already have a metric on <span class="math-container">$\mathbb{R}^2$</span>. You even use the properties of <span class="math-container">$d$</span> before the above.</p>
<blockquote>
<p><strong>Thus,</strong> for <span class="math-container">$(x,y)\in U$</span> there exists <span class="math-container">$r>0$</span> such that <span class="math-container">$B_d((x,y),r) \subset U$</span>. So, <span class="math-container">$U$</span> is open in <span class="math-container">$\mathbb{R}^2$</span>.</p>
</blockquote>
<p>Why "thus"? Why do you repeat "for <span class="math-container">$(x,y)\in U$</span>" like that? You've already said that <span class="math-container">$(x,y)$</span> is in <span class="math-container">$U$</span> at the beginning. Anyway here is my suggested rewrite of part of the proof:</p>
<blockquote>
<p>Define <span class="math-container">$C=\left\{(x,y)|x^2+y^2=1\right\}$</span>, <span class="math-container">$O=(0,0)$</span>.</p>
</blockquote>
<blockquote>
<p><strong><span class="math-container">$V$</span> is open:</strong> Let <span class="math-container">$X\equiv(x,y)\in V$</span>, then <span class="math-container">$d(X, O) >1$</span> by definition. Let <span class="math-container">$s=\frac{1}{2}(d(X,O)-1)$</span>. Since <span class="math-container">$d(X,O)>1$</span>, <span class="math-container">$s>0$</span>. Let <span class="math-container">$B\equiv B_d(X,s)$</span>. If <span class="math-container">$q\in B$</span> then <span class="math-container">$d(q,O)+s>d(q,O)+d(q,X)\geq d(X,O)$</span> and so <span class="math-container">$$d(q,O)\geq d(X,O)-s=\frac12(d(X,O)+1)>1,$$</span> hence <span class="math-container">$q\not\in C$</span>, thus <span class="math-container">$B \cap C=\emptyset$</span>, hence <span class="math-container">$B \subset V$</span>. So, <span class="math-container">$V$</span> is open in <span class="math-container">$\mathbb{R}^2$</span>.</p>
</blockquote>
|
1,627,619 | <p>Could anyone please check my solution to the following problem?</p>
<blockquote>
<p><strong>Problem:</strong> Let $f(x, y) = (x^2 + y^2)e^{-(x^2 + y^2)}$. Find global extrema of $f$ on $M = {\mathbf R}^2$.</p>
</blockquote>
<p><strong>Proposed solution:</strong> Taking partial derivatives of $f$, we conclude that critical points are $[0,0]$ and points of the unit circle $C = \{[x,y] \in {\mathbf R}^2:\ x^2 + y^2 = 1\big\}$.</p>
<p>We can reason immediately that the global minimum is attained at $[0,0]$ as the function is nonnegative. We observe that the value of $f$ on $C$ is $e^{-1}$.</p>
<p>To prove that $f$ attains global maximum on $C$, we let $r := x^2 + y^2$. We observe that for any two points $[x_1,y_1]$ and $[x_2,y_2]$, $r_1 = r_2$ (ie. the function is constant on circles). Now let $r \to \infty$. Then
$re^{-r} \to 0$. From the definition of limit, it follows that for any $\varepsilon > 0$, we find $\delta > 0$ such that</p>
<p>$$\forall r \in P(\infty, \delta) = ({1 \over \delta}, \infty): re^{-r} < \varepsilon.$$</p>
<p>Let $\varepsilon = (2e)^{-1}$. Then there's $\delta$ from the definition above and we know that for $r \in ({1 \over \delta}, \infty)$, the value of f is less than the value of f on $C$. Restricting ourselves to the compact set</p>
<p>$$C' = \big\{[x, y] \in {\mathbf R}^2:\ x^2 + y^2 \le {1 \over \delta}\big\},$$</p>
<p>we can now argue that $f$ on $C'$ is indeed maximized on $C$, as it is a continuous function on a compact set, it's value around the boundary is at most $(2e)^{-1}$ and all critical points have been considered.</p>
<p>Therefore, the maximum value of $f$ is attained on $C$ with respect to $M$ as well.</p>
| Travis Willse | 155,629 | <p>This looks correct to me, but one can treat this a little more efficiently: If $f$ achieves a global extremum at $(x, y)$, then the map $g: [0, \infty) \to \Bbb R$ defined by $$g(r) := r^2 e^{-r^2}$$ achieves a global extremum at $\sqrt{x^2 + y^2}$ and vice versa. Since $g$ is differentiable, to determine the latter it's enough to find the value of $g$ at the solutions of $g'(x) = 0$ and the value $g(0)$ at the endpoint $x = 0$. As was observed in the question, we know that $g$ achieves a minimum at $x = 0$ because evaluating gives that $g(0) = 0$ and $g(r) \geq 0$ for all $r$ (as $g$ is a product of nonnegative functions), leaving us just to solve $g'(r) = 0$, evaluate $M_a := g(r_a)$ for each solution $r_a$, and determine the maximum of the values $M_a$.</p>
|
60,152 | <p>What is the motivation behind topology?</p>
<p>For instance, in real analysis, we are interested in rigorously studying about limits so that we can use them appropriately. Similarly, in number theory, we are interested in patterns and structure possessed by algebraic integers and algebraic prime numbers.</p>
<p>Some googling and wiki-ing gave me that topology studies about deformation of objects in some space i.e. how an object in some space behaves under a continuous map. However, when I started reading the subject it starts of by defining what a topology is i.e. a set of subsets of a set with certain properties. I fail to see the connection immediately. I would appreciate if someone could give a short bird's eye view of topology.</p>
<p>Thanks,
Adhvaitha</p>
| Qiaochu Yuan | 232 | <p><a href="https://math.stackexchange.com/questions/31859/what-concept-does-an-open-set-axiomatise">This math.SE question</a> may be relevant, but not pedagogically optimal. </p>
<p>Pedagogically I think the simplest answer is to axiomatize topological spaces via the <a href="http://en.wikipedia.org/wiki/Kuratowski_closure_axioms" rel="nofollow noreferrer">Kuratowski closure axioms</a>. Instead of specifying what properties open sets or closed sets satisfy, the Kuratowski closure axioms specify a closure operator $S \mapsto \text{cl}(S)$ on subsets $S$ of a set $X$ and axioms it ought to satisfy. You should think of closure as axiomatizing <strong>abstract limits</strong> (that is, $\text{cl}(S)$ roughly corresponds to the set of all possible limits of sequences of elements of $S$, but it actually doesn't; we need to replace sequences by <a href="http://en.wikipedia.org/wiki/Filter_(mathematics)" rel="nofollow noreferrer">filters</a> or <a href="http://en.wikipedia.org/wiki/Net_(mathematics)" rel="nofollow noreferrer">nets</a> in full generality). Then the closed sets are precisely the ones for which $S = \text{cl}(S)$ and the open sets are the complements of the closed sets as usual.</p>
<p>It's worth mentioning that the notion of a topological space is <em>absurdly general.</em> For many applications you'll only need to think about the topology of much more restricted types of spaces (e.g. manifolds, CW-complexes...). Nevertheless, because it is so general, it can be fruitfully applied to many areas of mathematics, and because the set of axioms used is relatively sparse, proofs in full generality are generally relatively short. </p>
|
457,557 | <p>Use a triple integral to find the volume of the solid: The solid enclosed by the cylinder $$x^2+y^2=9$$ and the planes $$y+z=5$$ and $$z=1$$<br>
This is how I started solving the problem, but the way I was solving it lead me to 0, which is incorrect. $$\int_{-3}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\int_{1}^{5-y}dzdxdy=\int_{-3}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\left(4-y\right)dxdy=\int_{-3}^3\left[4x-xy\right]_{-\sqrt{9-y^2}}^\sqrt{9-y^2}dy= {8\int_{-3}^3{\sqrt{9-y^2}}dy}-2\int_{-3}^3y{\sqrt{9-y^2}}dy$$<br>
If this is wrong, then that would explain why I'm stuck. If this is correct so far, that's good news, but the bad news is that I'm still stuck. If someone could help me out, that would be wonderful, thanks!</p>
| apnorton | 23,353 | <p>Ok. So you have the triple integral:
$$\begin{align}
\int_{-3}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\int_1^{5-y} \;dz\;dx\;dy
&= \int_{-3}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}4-y\;dx\;dy \\
&=\int_{-3}^34x-xy\Bigg|_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\;dy \\
&=\int_{-3}^38\sqrt{9-y^2}-2y\sqrt{9-y^2}\;dy \\
&= 8\int_{-3}^33\sqrt{1-\left(\frac{y}{3}\right)^2}\;dy-2\int_{-3}^3y\sqrt{9-y^2}\;dy
\end{align}$$</p>
<hr>
<p>Now, I'm going to break this up. For the left-hand integral, we must use trig-substitution. Let $\cos(t) = \frac{y}{3}$. This implies that $dy = -3\sin(t)\;dt$. The limits of integration change as well, to $t=\arccos\left(\frac{-3}{3}\right) = \pi$ to $t = \arccos\left(\frac{3}{3}\right) = 0$.</p>
<p>So, the integral becomes:
$$\begin{align}
24\int_\pi^0\sqrt{1-\cos^2(t)}(-3\sin t)\;dt &= -72\int_\pi^0\sin^2(t)\;dt\\
&=72\int_0^\pi\frac{1}{2}-\frac{\cos(2t)}{2}\;dt\\
&=36\int_0^\pi1-\cos(2t)\;dt\\
&=36\left(t - \frac{\sin(2t)}{2}\right)\Bigg|_0^\pi\\
&=\boxed{36\pi}
\end{align}$$</p>
<hr>
<p>Now, for the left-hand integral, we apply $u$-substitution. If we set $u = 9-y^2$, then $du = -2y\;dy$. The limits are transformed to $u = 9-(-3)^2 = 0$ to $u = 9-(3)^2 = 0$</p>
<p>So, the integral becomes:
$$\begin{align}
-2\int_{-3}^3y\sqrt{9-y^2}\;dy &= \int_0^0\sqrt{u}\;du\\
&= \boxed{0}
\end{align}$$</p>
<p>Well, that wasn't exciting. <code>:)</code></p>
<hr>
<p>So, putting it all together, we end up with:</p>
<p>$$V = 36\pi + 0 = \boxed{36\pi}$$</p>
|
1,917,790 | <p>Can anyone help me to solve this? </p>
<blockquote>
<p>Determine the value or values of $k$ such that $x + y + k = 0$ is tangent to the circle $x^2+y^2+6x+2y+6=0$.</p>
</blockquote>
<p>I don't know how to calculate the tangent.</p>
| Michael Hoppe | 93,935 | <p>The tangent's gradient is $-1$. From $$(x+3)^2+(y+1)^2=4$$ we know that the normal's gradient is $(y+1)/(x+3)$, so we must have $(y+1)/(x+3)=1$, that is $y+1=x+3$, hence $2(x+3)^2=4$. We conclude $x=-3\pm\sqrt2$ and $y=x+2=-1\pm\sqrt2$.</p>
<p>So the points where the tangents touch the circle are $(-3\pm\sqrt2,-1\pm\sqrt2)$. From here is is well known (isn't it?) how to derive the tangent's equations:
$$(x+3)(\pm\sqrt2)+(y+1)(\pm\sqrt2)=4.$$
Now expand and compare coefficients. No need for discriminants here.</p>
<p>Edit: In fact we do not have to know the coordinates of the two points, but solely $x+3$ and $ y+1$, which makes the calculation even simpler.</p>
|
1,740,458 | <p>I saw somewhere on Math Stack that there was a way of finding integrals in the form $$\int \frac{dx}{a+b \cos x}$$ <strong>without</strong> using Weierstrass substitution, which is the usual technique. </p>
<p>When $a,b=1$ we can just multiply the numerator and denominator by $1-\cos x$ and that solves the problem nicely. </p>
<p>But I remember that the technique I saw was a nice way of evaluating these even when $a,b\neq 1$.</p>
| user5713492 | 316,404 | <p>Kepler found the substitution when he was trying to solve the equation
$$\ell=mr^2\frac{d\nu}{dt}=\text{constant}$$
where $\ell$ is the orbital angular momentum, $m$ is the mass of the orbiting body, the true anomaly $\nu$ is the angle in the orbit past periapsis, $t$ is the time, and $r$ is the distance to the attractor. This is Kepler's second law, the law of areas equivalent to conservation of angular momentum. Then Kepler's first law, the law of trajectory, is
$$r=\frac{a(1-e^2)}{1+e\cos\nu}$$
where $a$ and $e$ are the semimajor axis and eccentricity of the ellipse. So to get $\nu(t)$, you need to solve the integral
$$\int\frac{d\nu}{(1+e\cos\nu)^2}$$
So as to relate the area swept out by a line segment joining the orbiting body to the attractor Kepler drew a little picture
<a href="https://i.stack.imgur.com/TlBAu.png" rel="noreferrer"><img src="https://i.stack.imgur.com/TlBAu.png" alt="Figure 1"></a></p>
<p>The attractor is at the focus of the ellipse at $O$ which is the origin of coordinates, the point of periapsis is at $P$, the center of the ellipse is at $C$, the orbiting body is at $Q$, having traversed the blue area since periapsis and now at a true anomaly of $\nu$. To perform the integral given above, Kepler blew up the picture by a factor of $1/\sqrt{1-e^2}$ in the $y$-direction to turn the ellipse into a circle. </p>
<p><a href="https://i.stack.imgur.com/MVCCe.png" rel="noreferrer"><img src="https://i.stack.imgur.com/MVCCe.png" alt="Figure 2"></a>
The orbiting body has moved up to $Q^{\prime}$ at height
$$y=\frac{a\sqrt{1-e^2}\sin\nu}{1+e\cos\nu}$$But still $$x=\frac{a(1-e^2)\cos\nu}{1+e\cos\nu}$$
Now he could get the area of the blue region because sector $CPQ^{\prime}$ of the circle centered at $C$, at $-ae$ on the $x$-axis and radius $a$ has area $$\frac12a^2E$$ where $E$ is the eccentric anomaly and triangle $COQ^{\prime}$ has area $$\frac12ae\cdot\frac{a\sqrt{1-e^2}\sin\nu}{1+e\cos\nu}=\frac12a^2e\sin E$$ so the area of blue sector $OPQ^{\prime}$ is $$\frac12a^2(E-e\sin E)$$
and then we can go back and find the area of sector $OPQ$ of the original ellipse as $$\frac12a^2\sqrt{1-e^2}(E-e\sin E)$$
So if doing an integral with a factor of $\frac1{1+e\cos\nu}$ via the eccentric anomaly was good enough for Kepler, surely it's good enough for us. </p>
<p>In the original integer,
$$\int\frac{dx}{a+b\cos x}=\frac1a\int\frac{dx}{1+\frac ba\cos x}=\frac1a\int\frac{d\nu}{1+\left|\frac ba\right|\cos\nu}$$
where $\nu=x$ is $ab>0$ or $x+\pi$ if $ab<0$. Now we see that $e=\left|\frac ba\right|$, and we can use the eccentric anomaly,
$$\sin E=\frac{\sqrt{1-e^2}\sin\nu}{1+e\cos\nu}$$
$$\cos E=\frac{\cos\nu+e}{1+e\cos\nu}$$
$$d E=\frac{\sqrt{1-e^2}}{1+e\cos\nu}d\nu$$
and the integral reads
$$\begin{align}\int\frac{dx}{a+b\cos x}&=\frac1a\int\frac{d\nu}{1+e\cos\nu}=\frac12\frac1{\sqrt{1-e^2}}\int dE\\
&=\frac1a\frac1{\sqrt{1-e^2}}E+C=\frac{\text{sgn}\,a}{\sqrt{a^2-b^2}}\sin^{-1}\left(\frac{\sqrt{a^2-b^2}\sin\nu}{|a|+|b|\cos\nu}\right)+C\\&=\frac{1}{\sqrt{a^2-b^2}}\sin^{-1}\left(\frac{\sqrt{a^2-b^2}\sin x}{a+b\cos x}\right)+C\end{align}$$
Of course it's a different story if $\left|\frac ba\right|\ge1$, where we get an unbound orbit, but that's a story for another bedtime.</p>
|
484,367 | <p>I've been trying to find a tight upper bound for the series</p>
<p>$$S (x) = e^{-x} \sum_{k=0}^{\infty} \frac{x^k}{k!} \sqrt{k+1}$$</p>
<p>So far, I've managed to get a reasonable bound for small values of $x$ by using the inequality $\sqrt{k+1} \leq \sqrt{\frac{k^{2}}{4} + k + 1} = \frac{k}{2} + 1 ~\forall~k \geq 0$, but it becomes very loose when $x$ is large. I've also tried taking a Taylor series approximation to $\sqrt{k+1}$, but this leads to a complicated infinite sum of weighted Bell polynomials which, as far as I'm aware, doesn't have a closed form. Any suggestions would be greatly appreciated!</p>
| Raymond Manzoni | 21,783 | <p>An asymptotic expansion for your series $\;\displaystyle S (x) := e^{-x} \sum_{k=0}^{\infty} \sqrt{k+1}\frac{x^k}{k!} \;$
seems to be, as $\,x\to +\infty$ :
$$S(x)\sim\sqrt{x}\left(1+\frac3{8\;x}+\frac 1{128\;x^2}+\frac 9{1024\;x^3}+O\left(\frac 1{x^4}\right)\right)$$
I have no proof for that sorry... (the ideas used are similar to those from this <a href="https://math.stackexchange.com/questions/115410/whats-the-sum-of-sum-limits-k-1-infty-fractkkk">thread</a>).</p>
|
3,046,979 | <p>Recently I've been trying to tackle the integral <span class="math-container">$\int_0^{\frac{\pi}{2}} \ln(\sin(\theta))d\theta$</span> using the Beta function
<span class="math-container">$$\frac{B(\frac{x}{2},\frac{1}{2})}{2}=\int_0^{\frac{\pi}{2}} \sin^{x-1}(\theta)d\theta=\frac{\sqrt{\pi}}{2}\left(\Gamma\left(\frac{x+1}{2}\right)\right)^{-1}$$</span>
Differentiating both sides
<span class="math-container">$$\int_0^{\frac{\pi}{2}} \ln(\sin(\theta))\sin^{x-1}(\theta)d\theta=-\frac{\sqrt{\pi}}{4}\frac{\psi(\frac{x+1}{2})}{\Gamma(\frac{x+1}{2})}$$</span>
However, at <span class="math-container">$x=1$</span> <span class="math-container">$$\int_0^{\frac{\pi}{2}} \ln(\sin(\theta))d\theta\ne\frac{\gamma\sqrt{\pi}}{4}$$</span></p>
<p>Where did I go wrong?</p>
| Jack D'Aurizio | 44,121 | <p>Cleaner approach: for any <span class="math-container">$\alpha\geq 0$</span>,</p>
<p><span class="math-container">$$ \int_{0}^{\pi/2}\left(\sin\theta\right)^{\alpha}\,d\theta = \int_{0}^{1}\frac{u^\alpha}{\sqrt{1-u^2}}\,du = \frac{1}{2}\int_{0}^{1}v^{\frac{\alpha-1}{2}}(1-v)^{-1/2}\,dv = \frac{\Gamma\left(\frac{\alpha+1}{2}\right)}{\Gamma\left(\frac{\alpha+2}{2}\right)}\cdot\frac{\sqrt{\pi}}{2}$$</span>
Now we differentiate both sides with respect to <span class="math-container">$\alpha$</span>. In order to differentiate the RHS, we multiply it by its logarithmic derivative. We get:</p>
<p><span class="math-container">$$ \int_{0}^{\pi/2}\left(\sin\theta\right)^{\alpha}\log\sin\theta\,d\theta = \frac{\sqrt{\pi}}{4}\cdot \frac{\Gamma\left(\frac{\alpha+1}{2}\right)}{\Gamma\left(\frac{\alpha+2}{2}\right)}\cdot\left[\psi\left(\tfrac{\alpha+1}{2}\right)-\psi\left(\tfrac{\alpha+2}{2}\right)\right]. $$</span>
Now we evaluate at <span class="math-container">$\alpha=0$</span>, recalling that <span class="math-container">$\Gamma(1)=1,\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$</span> and
<span class="math-container">$$ \sum_{n\geq 0}\frac{1}{(n+1)\left(n+\tfrac{1}{2}\right)}=\frac{\psi(1)-\psi(1/2)}{1-1/2}= 4\log 2.$$</span>
The final outcome is:
<span class="math-container">$$ \int_{0}^{\pi/2}\log\sin\theta\,d\theta = -\frac{\pi}{2}\log 2.$$</span></p>
|
3,179,505 | <p>Help me please , I am not able to solve this problem.I have tried in many ways to figure out such as Ration test , Integral test , Comparison test , Limit Comparison Test , Root Test but i can't find the way out . This is my first question and i'm not good at English. If there is something wrong or you are not comfortable with my language usage I'm so sorry.</p>
| 5xum | 112,884 | <p>Taking the discrete metric on <span class="math-container">$X_1\times X_2$</span> will probably be enough to find a counterexample...</p>
|
4,257,962 | <p>By definition - A real number is algebraic if it is a root of a non-zero polynomial equation with rational coefficients. What does non-zero polynomial equation mean?</p>
<p>Well, an equation f(x) = x -5, becomes zero when x = 5, so this is a zero polynomial equation. Is the definition saying that the equation should not equal zero in any case?</p>
<p>Can someone clarify this?</p>
| Tito Eliatron | 84,972 | <p>Let <span class="math-container">$P_0(x):=0$</span> be the ZERO-POLYNOMIAL.</p>
<p>If we not avoid this pathological case, then every real number would be algebraic, since every real number is a "solution" of the equation <span class="math-container">$P_0(x)=0$</span> (even <span class="math-container">$\pi$</span> or <span class="math-container">$e$</span>).</p>
|
3,970,959 | <blockquote>
<p><span class="math-container">$S = \frac{1}{1001} + \frac{1}{1002}+ \frac{1}{1003}+ \dots+\frac{1}{3001}$</span>.</p>
</blockquote>
<blockquote>
<p>Prove that <span class="math-container">$\dfrac{29}{27}<S<\dfrac{7}{6}$</span>.<br></p>
</blockquote>
<p>My Attempt:<br>
<span class="math-container">$S<\dfrac{500}{1000} + \dfrac{500}{1500}+ \dfrac{500}{2000}+ \dfrac{500}{2500}+\dfrac{1}{3000} =\dfrac{3851}{3000}$</span></p>
<p>(Taking 250 terms together involves many fractions and is difficult to calculate by hand.)</p>
<p>Using AM-HM inequality gave me <span class="math-container">$S > 1$</span>, but the bounds are weak.</p>
<p><a href="https://math.stackexchange.com/questions/688432">Prove that $1<\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\dots+\frac{1}{3001}<\frac43$</a> <br><br>
<a href="https://math.stackexchange.com/questions/605111">Inequality with sum of inverses of consecutive numbers</a> <br><br>
The answers to these questions are nice, but the bounds are weak.</p>
<p>Any help without calculus and without calculations involving calculators would be appreciated.</p>
<p>(I encountered this question when I was preparing for a contest which neither allows calculators nor calculus(Only high-school mathematics.))</p>
| TonyK | 1,508 | <h2>Upper bound <span class="math-container">$\dfrac{7}{6}$</span></h2>
<p>Your question describes a method of breaking the range up into sub-ranges, and then estimating the sum over a sub-range by noting that each term is <span class="math-container">$\le$</span> the first term. But we can do much better if we estimate this sum using a linear approximation instead. Then we only need three sub-ranges to establish the upper bound of <span class="math-container">$\frac76$</span>.</p>
<p>Consider a 'graph' of the function <span class="math-container">$$f(k)=\frac{1}{k}$$</span> for integers <span class="math-container">$k\ge 1$</span>. If we look at any interval <span class="math-container">$m\le k\le n$</span> with <span class="math-container">$1\le m\le n-2$</span>, we can compare it with the straight-line function joining points <span class="math-container">$(m,\frac{1}{m})$</span> and <span class="math-container">$(n,\frac{1}{n})$</span>. We can define this function explicitly if you want, as:
<span class="math-container">$$g(k)=\frac{1}{n-m}\left(\frac{k-m}{n}+\frac{n-k}{m}\right)$$</span>
Then <span class="math-container">$f(m)=g(m),f(n)=g(n)$</span>; and for <span class="math-container">$m<k<n,f(k)<g(k)$</span> (because <span class="math-container">$f$</span> is a convex function).</p>
<p>So <span class="math-container">$$\sum_{k=m}^nf(k)<\sum_{k=m}^ng(k)$$</span>
But the <span class="math-container">$g(k)$</span> are in arithmetic progression, so the RHS is half the sum of the first and the last terms multiplied by the number of terms, i.e.
<span class="math-container">$$\frac{n-m+1}{2}\left(\frac{1}{m}+\frac{1}{n}\right)$$</span></p>
<p>We divide the range into three equal sub-ranges of size <span class="math-container">$667$</span>. This gives us:
<span class="math-container">$$\begin{align}
\sum_{k=1001}^{3001}\frac{1}{k} & =\sum_{k=1001}^{1667}\frac{1}{k}+\sum_{k=1668}^{2334}\frac{1}{k}+\sum_{k=2335}^{3001}\frac{1}{k}\\
& <\frac{667}{2}\left(\frac{1}{1001}+\frac{1}{1667}\right)+\frac{667}{2}\left(\frac{1}{1668}+\frac{1}{2334}\right)+\frac{667}{2}\left(\frac{1}{2335}+\frac{1}{3001}\right)\\
& <\frac{667}{2}\left(\frac{1}{1000.5}+\frac{1}{1667.5}+\frac{1}{1667.5}+\frac{1}{2334.5}+\frac{1}{2334.5}+\frac{1}{3001.5}\right)\\
& =\frac13+\frac15+\frac15+\frac17+\frac17+\frac19\\
& =\frac{356}{315}\\
& <\frac{7}{6}\end{align}$$</span></p>
<p>where we have used the fact that <span class="math-container">$\frac{1}{m}+\frac{1}{n}<\frac{1}{m-\frac12}+\frac{1}{n+\frac12}$</span> if <span class="math-container">$m<n$</span>, again by convexity of <span class="math-container">$f(k)$</span>.</p>
<p>Note that if the last term were <span class="math-container">$\frac{1}{3000}$</span>, we could do this with just two sub-ranges, and we would get exactly <span class="math-container">$\frac{7}{6}$</span> as a bound. So perhaps I have missed an easier way.</p>
<h2>Lower bound <span class="math-container">$\dfrac{29}{27}$</span></h2>
<p>This time we use the convexity of the function to estimate the sum over a sub-range from below. The sum of <span class="math-container">$\frac{1}{k}$</span> over the range <span class="math-container">$m\le k\le n$</span> is greater than the number of elements multiplied by the middle element if the range contains an odd number of elements. So we get
<span class="math-container">$$\begin{align}
\sum_{k=1001}^{3001}\frac{1}{k} & =\sum_{k=1001}^{1667}\frac{1}{k}+\sum_{k=1668}^{2334}\frac{1}{k}+\sum_{k=2335}^{3001}\frac{1}{k}\\
& > 667\left(\frac{1}{1334}+\frac{1}{2001}+\frac{1}{2668}\right)\\
& = \frac12+\frac13+\frac14\\
& = \frac{13}{12}\\
& >\frac{29}{27}
\end{align}$$</span></p>
|
3,970,959 | <blockquote>
<p><span class="math-container">$S = \frac{1}{1001} + \frac{1}{1002}+ \frac{1}{1003}+ \dots+\frac{1}{3001}$</span>.</p>
</blockquote>
<blockquote>
<p>Prove that <span class="math-container">$\dfrac{29}{27}<S<\dfrac{7}{6}$</span>.<br></p>
</blockquote>
<p>My Attempt:<br>
<span class="math-container">$S<\dfrac{500}{1000} + \dfrac{500}{1500}+ \dfrac{500}{2000}+ \dfrac{500}{2500}+\dfrac{1}{3000} =\dfrac{3851}{3000}$</span></p>
<p>(Taking 250 terms together involves many fractions and is difficult to calculate by hand.)</p>
<p>Using AM-HM inequality gave me <span class="math-container">$S > 1$</span>, but the bounds are weak.</p>
<p><a href="https://math.stackexchange.com/questions/688432">Prove that $1<\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\dots+\frac{1}{3001}<\frac43$</a> <br><br>
<a href="https://math.stackexchange.com/questions/605111">Inequality with sum of inverses of consecutive numbers</a> <br><br>
The answers to these questions are nice, but the bounds are weak.</p>
<p>Any help without calculus and without calculations involving calculators would be appreciated.</p>
<p>(I encountered this question when I was preparing for a contest which neither allows calculators nor calculus(Only high-school mathematics.))</p>
| Anindya Prithvi | 811,225 | <p><span class="math-container">$$\sum_{n=1}^{2001}\frac{1}{1000\left(1+\frac{n}{1000}\right)} =\int_{0}^{2}\frac{1}{\left(1+x\right)}dx \approx \ln3$$</span></p>
<p><span class="math-container">$$\frac{29}{27} < \ln3 <\frac{7}{6}$$</span></p>
<p>Used:</p>
<ol>
<li>Summation to integration when step size is small</li>
</ol>
|
3,080,230 | <p>On <a href="https://en.wikipedia.org/wiki/Net_(mathematics)#Properties" rel="nofollow noreferrer">Wikipedia</a> it states that a space <span class="math-container">$X$</span> is compact if and only if every net has a convergent subnet. It then states that a net in the product topology has a limit if and only if each projection has a limit. I understand why both of these facts are true. However it then states this leads to a slick proof of Tychonoff's theorem, and I don't quite see how.</p>
<p>In particular, it seems to me that the first fact implies that every compact space is sequentially compact. Since every sequence is also a net, it has a convergent subnet, which gives a convergent subsequence. This is obviously not true, since <span class="math-container">$\{0,1\}^\mathbb{R}$</span> is not sequentially compact, but it is compact by Tychonoff's theorem.</p>
| SmileyCraft | 439,467 | <p>The problem in the reasoning in the OP is that a net can admit no subnet that is a sequence. This is because a subnet needs to be final. For example if the index set of the subnet is <span class="math-container">$\omega_1$</span>, the first uncountable ordinal, then there exists no final function <span class="math-container">$h:\mathbb{N}\to\omega_1$</span>, since <span class="math-container">$\cup h(n)$</span> is an upper bound on <span class="math-container">$h(n)$</span>.</p>
<p>To prove Tychonoff's theorem with nets, the main idea is to use Zorn's lemma on partial clusterpoints. We define a partial cluster point of a sequence <span class="math-container">$\{f_n\}$</span> in <span class="math-container">$\Pi X_{\alpha\in A}$</span> as a tuple <span class="math-container">$(f,I)$</span> where <span class="math-container">$f\in\Pi X_{\alpha\in A}$</span> and <span class="math-container">$I\subseteq A$</span> such that <span class="math-container">$f$</span> is a cluster point of <span class="math-container">$\{f_n|_I\}$</span>. We can then define the partial ordening <span class="math-container">$(f,I)\leq(g,J)$</span> iff <span class="math-container">$I\subseteq J$</span> and <span class="math-container">$g|_I=f$</span>.</p>
<p>Then there obviously exists some partial clusterpoint; consider <span class="math-container">$(f,\emptyset)$</span>. There is a canonical notion of union of clusterpoints which gives us an upper bound for any chain. Hence, by Zorn's lemma there exists a maximal cluster point <span class="math-container">$(f,I)$</span>. To show that <span class="math-container">$I=A$</span>, consider a hypothetical element <span class="math-container">$a\in A\setminus I$</span>. Then the <span class="math-container">$a$</span>-coordinate of the subnet <span class="math-container">$\{g_n|_I\}$</span> of <span class="math-container">$\{f_n|_I\}$</span> converging to <span class="math-container">$f$</span> must have a cluster point <span class="math-container">$p$</span>. Then setting <span class="math-container">$f^+(a)=p$</span> while <span class="math-container">$f^+|_I=f$</span> we find a greater partial cluster point <span class="math-container">$(f^+,I\cup\{a\})$</span>.</p>
|
56,847 | <p>What are the angles formed at the center of a tetrahedron if you draw lines to the vertices?</p>
<p>I'm trying to make these:</p>
<p><img src="https://i.stack.imgur.com/FRUi8.jpg" alt="caltrop"> </p>
<p>I need to know what angles to bend the metal.</p>
| Mark Bennet | 2,906 | <p>(Assuming the tetrahedron is supposed to be regular) Take the tetrahedron with vertices $(1,1,1), (1,-1,-1), (-1,1,-1), (-1,-1,1)$, which has centre at the origin, and use the dot product formula:</p>
<p>$a\cdot b = |a| |b|\cos\theta$ </p>
<p>which gives $\cos\theta=-\frac13$</p>
|
254,030 | <p>I am following a course in basic algebra, and we have covered rings & groups in class, but I am having trouble visualising them. Are there applications of group &/or ring theory that can be more easily visualized than the abstract object? For instance, are there objects, or properties of objects, that behave as elements of a group in physics, chemistry, or other fields?</p>
| still_learning | 42,808 | <p>Dihedral groups arise frequently in art and nature. Many of the
decorative designs used on floor coverings, pottery, and buildings have
one of the dihedral groups as a group of symmetry. Corporation logos
are rich sources of dihedral symmetry. Chrysler’s logo has D5 as a
symmetry group, and that of Mercedes-Benz has D3. The ubiquitous
five-pointed star has symmetry group D5. The phylum Echinodermata
contains many sea animals (such as starfish, sea cucumbers, feather
stars, and sand dollars) that exhibit patterns with D5 symmetry.
Chemists classify molecules according to their symmetry. Moreover,
symmetry considerations are applied in orbital calculations, in determining
energy levels of atoms and molecules, and in the study of molecular
vibrations.</p>
<p>Source : Contemporary Abstract Algebra, Gallian, Chapter 2</p>
|
258,704 | <p>How can I solve a system of linear congruences as such?</p>
<p><span class="math-container">$$\begin{align*}
3x+2y+28z &= 9 \pmod {29} \\
5x+27y+z &= 9 \pmod {29} \\
2x+y+z &= 6 \pmod {29}
\end{align*}$$</span></p>
<p>I tried it this way as a system of equations, but no luck:</p>
<pre><code>eqn1 = FullSimplify[{3*x + 2*y + 28*z == 9 + (29*i) &&
5*x + 27*y + z == 9 + (29*j) && 2*x + y + z == 6 + (29*k)}]
Table[FindInstance[eqn1, {x, y, z, i, j, k}, Integers, 1] ]
</code></pre>
<p>Additionally, how can I solve these linear congruences:</p>
<p><span class="math-container">$$ 3x = 5 \pmod 6 $$</span></p>
<p>Tried this: No luck! <code>Reduce[3*x - 5 == 6, x, Modulus -> 6]</code></p>
<p>and</p>
<p><span class="math-container">$$ x^2 + x = 2 \pmod 8 $$</span></p>
<p>and</p>
<p>Find Multiplicative inverse of [5] in z42 ? which would mean <span class="math-container">$$ 5x + 42y =1 $$</span></p>
<p>and lastly:</p>
<p>Solve these systems in z11:</p>
<p><span class="math-container">$$ [2][x]+[7][y] = [4] $$</span>
<span class="math-container">$$ [3][x]+[2][y] = [9] $$</span></p>
<p>I'm pretty sure Mathematica can input and solve these.</p>
| cvgmt | 72,111 | <pre><code>Solve[{3*x + 2*y + 28*z == 9 + (29*i), 5*x + 27*y + z == 9 + (29*j),
2*x + y + z == 6 + (29*k)}, {x, y, z}, {i, j, k}, Integers,
GeneratedParameters -> c]
</code></pre>
<blockquote>
<p><code>{{x -> ConditionalExpression[ 24 + 29 c[1], (c[1] | c[2] | c[3]) ∈ Integers], y -> ConditionalExpression[ 23 + 29 c[2], (c[1] | c[2] | c[3]) ∈ Integers], z -> ConditionalExpression[ 22 + 29 c[3], (c[1] | c[2] | c[3]) ∈ Integers]}} </code></p>
</blockquote>
<pre><code>FindInstance[{3*x + 2*y + 28*z == 9 + (29*i),
5*x + 27*y + z == 9 + (29*j), 2*x + y + z == 6 + (29*k)}, {x, y, z,
i, j, k}, Integers, 1]
</code></pre>
<blockquote>
<p><code>{{x -> 24, y -> 23, z -> 22, i -> 25, j -> 26, k -> 3}}</code></p>
</blockquote>
<pre><code>Solve[x + x^2 == 2 + (8*i), {x}, {i}, Integers,
GeneratedParameters -> c]
</code></pre>
<blockquote>
<p><code>{{x -> ConditionalExpression[1 - 8 c[1], c[1] ∈ Integers]}, {x -> ConditionalExpression[6 - 8 c[1], c[1] ∈ Integers]}}</code></p>
</blockquote>
|
3,534,377 | <p>I've followed this tutorial (<a href="http://web.eecs.utk.edu/~jplank/plank/papers/CS-96-332.pdf" rel="nofollow noreferrer">http://web.eecs.utk.edu/~jplank/plank/papers/CS-96-332.pdf</a>) which introduces Reed-Solomon coding and therefore covers finite fields. My problem is with one of the examples of how the logarithm and anti logarithm tables are looked up to perform a division. I have included an image of the examples and table below:</p>
<p><a href="https://i.stack.imgur.com/02QcZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/02QcZ.png" alt="Excerpt of log tables and examples"></a></p>
<p>The tutorial is written for programmers and so <span class="math-container">$gflog[i]$</span> is really <span class="math-container">$log_2(i)$</span> and <span class="math-container">$gfilog[i]$</span> is <span class="math-container">$log_2^{-1}(i)$</span>.</p>
<p>I am able to follow all but the last example. I used an online finite field calculator which indicates the result should be <span class="math-container">$10$</span>. The result in the example is <span class="math-container">$14$</span>, at first I assume this to be a simple mistake of looking up the result in the incorrect row (i.e, looking up the <span class="math-container">$log_2(i)$</span> row instead of the <span class="math-container">$log_2^{-1}(i)$</span> row) as the
table correctly shows the result should be <span class="math-container">$10$</span>. </p>
<p>But upon closer inspection, their arithmetic for the indexing into the table seems to be off for the final example, <span class="math-container">$gfilog[4-10] = gfilog[9]$</span>. Shouldn't the result be <span class="math-container">$gfilog[10]$</span>?</p>
<p>Now my what I <em>suspect</em> is happening is the arithmetic on the indexes is regular integer arithmetic. And in this case, <span class="math-container">$4-10$</span> results in <span class="math-container">$-6$</span>. Since this is an invalid index, and the log tables are over <span class="math-container">$GF(2^4)$</span>, the correct position of <span class="math-container">$-6$</span> is determined by <span class="math-container">$-6 mod 16$</span> which gives <span class="math-container">$10$</span>. However! the <span class="math-container">$10$</span> is just the <em>rank</em> of the correct index and <em>not</em> the index itself. Therefore, the 10th position on the table is really at the 9th index. Hence <span class="math-container">$gfilog[9] = 10$</span>.</p>
<p>Is my understanding of this correct? I've tried this out with a few other values and it seems to work out. This is all very new to me, so kindly elaborate a bit in your responses as my background is primarily in programming :)</p>
| Bram28 | 256,001 | <p>As already pointed out in the comments, <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are sets, not statements.</p>
<p>Still, there are obvious connections between sets and logical statements. For example, the set <span class="math-container">$A \cap B$</span> is the set of all objects that are elements of <span class="math-container">$A$</span> <em>and</em> of <span class="math-container">$B$</span>. Or, to make the connection between <span class="math-container">$\cap$</span> and <span class="math-container">$\land$</span> even more clear:</p>
<p><span class="math-container">$A \cap B = \{x \mid x \in A \land x \in B \}$</span></p>
<p>Using this format, we can also say:</p>
<p><span class="math-container">$(A \cap B') \cup (B \cap A') = \{ x \mid (x \in A \land \neg (x \in B)) \lor (\neg (x \in A) \land x \in B) \}$</span></p>
<p>If we can use the logical biconditional, you can simplify this to:</p>
<p><span class="math-container">$(A \cap B') \cup (B \cap A') = \{ x | (x \in A \leftrightarrow \neg (x \in B) \}$</span></p>
|
3,534,377 | <p>I've followed this tutorial (<a href="http://web.eecs.utk.edu/~jplank/plank/papers/CS-96-332.pdf" rel="nofollow noreferrer">http://web.eecs.utk.edu/~jplank/plank/papers/CS-96-332.pdf</a>) which introduces Reed-Solomon coding and therefore covers finite fields. My problem is with one of the examples of how the logarithm and anti logarithm tables are looked up to perform a division. I have included an image of the examples and table below:</p>
<p><a href="https://i.stack.imgur.com/02QcZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/02QcZ.png" alt="Excerpt of log tables and examples"></a></p>
<p>The tutorial is written for programmers and so <span class="math-container">$gflog[i]$</span> is really <span class="math-container">$log_2(i)$</span> and <span class="math-container">$gfilog[i]$</span> is <span class="math-container">$log_2^{-1}(i)$</span>.</p>
<p>I am able to follow all but the last example. I used an online finite field calculator which indicates the result should be <span class="math-container">$10$</span>. The result in the example is <span class="math-container">$14$</span>, at first I assume this to be a simple mistake of looking up the result in the incorrect row (i.e, looking up the <span class="math-container">$log_2(i)$</span> row instead of the <span class="math-container">$log_2^{-1}(i)$</span> row) as the
table correctly shows the result should be <span class="math-container">$10$</span>. </p>
<p>But upon closer inspection, their arithmetic for the indexing into the table seems to be off for the final example, <span class="math-container">$gfilog[4-10] = gfilog[9]$</span>. Shouldn't the result be <span class="math-container">$gfilog[10]$</span>?</p>
<p>Now my what I <em>suspect</em> is happening is the arithmetic on the indexes is regular integer arithmetic. And in this case, <span class="math-container">$4-10$</span> results in <span class="math-container">$-6$</span>. Since this is an invalid index, and the log tables are over <span class="math-container">$GF(2^4)$</span>, the correct position of <span class="math-container">$-6$</span> is determined by <span class="math-container">$-6 mod 16$</span> which gives <span class="math-container">$10$</span>. However! the <span class="math-container">$10$</span> is just the <em>rank</em> of the correct index and <em>not</em> the index itself. Therefore, the 10th position on the table is really at the 9th index. Hence <span class="math-container">$gfilog[9] = 10$</span>.</p>
<p>Is my understanding of this correct? I've tried this out with a few other values and it seems to work out. This is all very new to me, so kindly elaborate a bit in your responses as my background is primarily in programming :)</p>
| R. Burton | 614,269 | <p>For any set <span class="math-container">$S$</span>, there is a predicate <span class="math-container">$P_S$</span> such that <span class="math-container">$x\in S\iff P_S(x)$</span> (in other words, <span class="math-container">$S=\{x:P_S(x)\}$</span>). For instance, if we are working in the domain of integers we can assign the set of even numbers to the predicate <span class="math-container">$E(n):=2\mid n$</span>. If <span class="math-container">$S_E$</span> is the set of even numbers, then <span class="math-container">$n\in S_E$</span> has exactly the same meaning as <span class="math-container">$E(n)$</span>. In some theories, this is actually the <em>definition</em> of a predicate/set.<span class="math-container">$^1$</span></p>
<p>Using this correspondence, it is possible to convert set-theoretic statements into logical ones and vice-versa. In this sense sets <em>can</em> be interpreted as [equivalence classes of] statements, provided that certain rules are observed.</p>
<p>There are many different ways to do this depending on the particular set-theory and logic you choose, the choice of logical connectives, and the presence or absence of quantifiers.</p>
<p>For propositional logic,<span class="math-container">$^2$</span> I would use the following:</p>
<p>Logical negation <span class="math-container">$\neg$</span> corresponds to the complement of a set <span class="math-container">$'$</span>.</p>
<p>Logical disjunction <span class="math-container">$\lor$</span> corresponds to the union of sets <span class="math-container">$\cup$</span>.</p>
<p>Logical conjunction <span class="math-container">$\land$</span> corresponds to the intersection of sets <span class="math-container">$\cap$</span>.</p>
<p>Predicate application <span class="math-container">$P_X(x)$</span> corresponds to set membership <span class="math-container">$x \in X$</span>.</p>
<p>For the example provided, Let <span class="math-container">$P_A$</span> and <span class="math-container">$P_B$</span> be predicates such that <span class="math-container">$a\in A\iff P_A(a)$</span> and <span class="math-container">$b\in B\iff P_B(b)$</span>. Then...</p>
<p><span class="math-container">$$x\in(A\cap B')\cup (B\cap A')\iff((P_A\land\neg P_B)\lor(P_B\land\neg P_A))(x)$$</span></p>
<p>You can simplify the logical statement using <em>exclusive disjunction</em> to <span class="math-container">$P_A\oplus P_B$</span>. The corresponding set is the <em>symmetric difference</em> <span class="math-container">$A\triangle B=(A\cup B)\setminus(A\cap B)$</span>.</p>
<p>In general, if two statements are logically equivalent (for example <span class="math-container">$P\land Q\equiv\neg (\neg P\lor\neg Q)$</span>, then the corresponding sets will be equivalent as well (<span class="math-container">$S_P\cap S_Q=(S_P'\cup S_Q')'$</span>). The connection between sets and logical statement in general is an important topic in the study of <em>Boolean algebras</em>.</p>
<p>Hugely helpful Hasse diagram: <a href="https://commons.wikimedia.org/wiki/File:Logical_connectives_Hasse_diagram.svg" rel="nofollow noreferrer">https://commons.wikimedia.org/wiki/File:Logical_connectives_Hasse_diagram.svg</a></p>
<hr>
<p>Notes for nit-pickers:</p>
<p><span class="math-container">$^1$</span> Actually, it is more common to refer to predicates [up to logical equivalence] as <em>classes</em> rather than sets - particularly in axiomatic set theory where the identification of predicates with sets is the same as <em>unrestricted comprehension</em>, which leads to paradoxes like <em>Russel's Paradox</em>.</p>
<p><span class="math-container">$^2$</span> Technically, this is predicate logic. However, it is much closer to propositional logic than true first-order logic in its application. In an alternative formulation, you could assign a proposition <span class="math-container">$\varphi_S$</span> to each set <span class="math-container">$S$</span> using a formula <span class="math-container">$\varphi_S:=\forall x.P_S(x)$</span> for the appropriate predicate <span class="math-container">$P_S$</span>. In this case, I would say that the set <span class="math-container">$S$</span> is the domain/universe in which <span class="math-container">$\varphi_S$</span> is <em>true</em>.</p>
|
959,219 | <p>let $a,b,c>0$, and such
$$a^2+b^2+c^2<2ab+2bc+2ca$$</p>
<p>show that
$$a^4+b^4+c^4+6(a^2b^2+b^2c^2+a^2c^2)+4abc(a+b+c)<4(ab+bc+ac)(a^2+b^2+c^2)$$</p>
<p>I know this indentity:
$$a^2+b^2+c^2-2(ab+bc+ac)
=-(\sqrt{a}+\sqrt{b}+\sqrt{c})(-\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{a}-\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b}-\sqrt{c})$$</p>
| James Harrison | 135,585 | <p>$$
\because a,b,c > 0\\
(a+b+c)^4 = a^4 + b^4 + c^4 + 4(a^3 b + a^3 c + b^3 a + b^3 c + c^3 a + c^3 b) + 6(a^2 b^2 + a^2 c^2 + b^2 c^2) + 12abc(a+b+c) > a^4 + b^4 + c^4 + 6(a^2 b^2 + b^2 c^2 + a^2 c^2 ) + 4abc(a+b+c)$$</p>
<p>From your starting condition we get:</p>
<p>$$
a^2 + b^2 + c^2 - 2(ab+ac+bc) <0 \\
a^2 + b^2 + c^2 < 2(ab+ac+bc) \\
(a+b+c)^2 < 4(ab+ac+bc) \\
\therefore (a+b+c)^4 < 16(ab+ac+bc)^2 \\
\because 2(ab+ac+bc) < a^2 + b^2 + c^2 \\
\therefore (a+b+c)^4 < 16(ab+ac+bc)^2 < 4(a^2+b^2+c^2)^2
$$</p>
<p>Now just show that $16(ab+ac+bc)^2 < 4(ab+ac+bc)(a^2+b^2+c^2) < 4(a^2 + b^2 + c^2)^2$</p>
|
3,236,067 | <p>I am having some trouble understanding where some linear boundary conditions are derived from </p>
<p>The following is an extract from my lecture notes on boundary value problems for second-order Linear ODE's</p>
<blockquote>
<p>In this section we are going to consider the different situation when some conditions are specified at the endpoints, or boundaries, of an interval of the independent variable, that is, at <span class="math-container">$x = x_1$</span> and <span class="math-container">$x=x_2$</span> with <span class="math-container">$x_1 < x_2$</span>. This problem is known as a <span class="math-container">$ \textbf{Boundary Value Problem} $</span> and the conditions are called boundary conditions. We are then interested in finding the solution <span class="math-container">$y(x)$</span> to the ODE (which we consider to be Linear) inside the interval <span class="math-container">$x_1 \le x \le x_2$</span>. we will consider only linear boundary conditions, where the left-hand sides of the conditions are linear combinations of the function and its derivatives at the same point and the right hand sides are given by constants, for example </p>
<p><span class="math-container">$$ y(x_1) = b_1, \ y(x_2)=b_2 \ \ or \ \ y'(x_1)=b_1, \ y'(x_2) = b_2$$</span> </p>
<p>or more generally </p>
<p><span class="math-container">$$ \tag{1} \alpha y'(x_1) + \beta y(x_1)=b_1, \gamma y'(x_2)+\delta y(x_2) =b_2,$$</span></p>
<p>where <span class="math-container">$\alpha , \beta , \gamma , \delta$</span> are given real constants such that <span class="math-container">$|\alpha |+ | \beta | > 0, | \gamma | + | \delta|>0. $</span></p>
</blockquote>
<p>My question is this, where do the conditions <span class="math-container">$(1)$</span> derive from? </p>
| Community | -1 | <p>There is no derivation. Those are the three types of boundary conditions generally seen.</p>
<p><span class="math-container">$$ y(x_{1}) = b_{1}, y(x_{2}) = b_{2} $$</span></p>
<p>is known as a Dirichlet Boundary Condition</p>
<p><span class="math-container">$$ y^{'}(x_{1}) = b_{1}, y^{'}(x_{2}) = b_{2} $$</span></p>
<p>is known as a Neumann Boundary Condition. </p>
<p><span class="math-container">$$ \alpha y'(x_1) + \beta y(x_1)=b_1, \gamma y'(x_2)+\delta y(x_2) =b_2,$$</span></p>
<p>The last type is called Robin Boundary Conditions. </p>
<p>There is an <strong>interpretation</strong>. </p>
<p>Dirichlet Conditions mean you are holding the boundaries at a specific temperature.</p>
<p>Neumann Conditions means that the boundaries are being given energy at a specific rate. </p>
<p>Robin is a linear combination of the above.</p>
|
3,017,928 | <p>Doing some self study from the text <em>Basic Mathematics</em> by Serge Lang
I ran into an exercise question which I can't seem to wrap my head around.
The question is:</p>
<p>Express the following expressions in the form <span class="math-container">$2^m3^na^rb^s$</span> ,where <span class="math-container">$m,n,r,s$</span> are positive integers.</p>
<p><span class="math-container">$8a^2b^3(27a^4)(2^5ab)$</span></p>
<p>After some research I found that the final answer is expressed as</p>
<p><span class="math-container">$2^83^3a^7b^4$</span></p>
<p>I've attempted to use distribution as a means of solving it but end up stuck and confused.
I'm entirely lost as to how that answer is derived. </p>
| Xavier Stanton | 620,797 | <p>I also have another way to do it. If you distribute the <span class="math-container">$8 a^2 b^3$</span> with the <span class="math-container">$27 a^4$</span>, you will get <span class="math-container">$216 a^6 b^3$</span>. Then, distribute that value with <span class="math-container">$2^5 a b$</span> and you will get <span class="math-container">$6,912 a^7 b^4$</span>. The number <span class="math-container">$6,912$</span> can be written exponentially as <span class="math-container">$(2^8)(3^3)$</span>. Bring that value down with the <span class="math-container">$a^7 b^4$</span>, and you get your answer.</p>
|
615,375 | <p>I would appreciate if somebody could help me with the following problem</p>
<p>Q: Let $f:[0,1]\longrightarrow \Bbb R$ be a continuously function such that
$$m\leq f(x)\leq M, m+M=1 (m:\text{minimum of} f(x), M:\text{maximum of} f(x) )$$
Prove that for every $x \in[0,1]$, there exists $c\in[0,1]$ such that $f(c)=1-f(x)$ </p>
| Mathronaut | 53,265 | <p>Note that, $m= 1-M \le 1-f(x) \le 1-m =M$, Now apply Intermediate Value theorem to get $c\in [0,1]$ st $f(c)=1-f(x)$</p>
|
61,047 | <p>I can add the value of a slider to the right of it using the Appearance-->Labelled option, but what if I want to add text after the automatic label. How can I do that?</p>
<p>Normally I want to do this to show the units of the value. For example, if the slider label is "4.7", I might want it to read "4.7 meters".</p>
| Nasser | 70 | <p>There are many ways to do this. The most basic is to use <code>Control</code>, added few versions earlier just for this purpose. Here is an example. <code>Control</code> can be inserted inside <code>Row</code> or <code>Column</code> or <code>Grid</code> for example</p>
<pre><code>Manipulate[x,
Row[{Control[{{x, 1, "x="}, 0, 1, .1}],
Spacer[5],
Dynamic[x],
Spacer[2],
"meters"}]
]
</code></pre>
<p><img src="https://i.stack.imgur.com/cAsXE.png" alt="Mathematica graphics"></p>
|
1,363,213 | <p>I am given a chessboard of size $8*8$. In this chessboard there are two holes at positions $(X1,Y1)$ and $(X2,Y2)$. Now I need to find the maximum number of rooks that can be placed on this chessboard such that no rook threatens another. </p>
<p>Also no two rooks can threaten each other if there is hole between them.</p>
<p>How can I tackle this problem? Please help</p>
<p><strong>NOTE : A hole can occupy only a single cell on the chess board.</strong></p>
| hmakholm left over Monica | 14,366 | <p>For most positions of the holes there seems to be room for 10 rooks.</p>
<p>If the holes are in the same row (not on the edge, not touching each other, and with at least two rows above and below it), then place 7 rooks as</p>
<pre><code> R
R
R o R o R
R
R
</code></pre>
<p>There are then 3 rows and 3 columns left without anything in them yet; put the last 3 rooks there.</p>
<p>If the holes are in different rows and columns (but not on the edge of the board), place 6 rooks as</p>
<pre><code> R
R o R
R o R
R
</code></pre>
<p>There will be 4 rows and 4 columns yet unused; place the 4 last rooks there.</p>
<p>These solutions are clearly optimal for their hole placements, because each maximal horizontal or vertical run of available cells on the board contains a rook.</p>
<p>If the holes are too close to the edges (or to each other) for this to work, there's room for fewer rooks. Writing down exact rules for those cases seems to be tedious but doable.</p>
<hr>
<p>If you just want an algorithm to <strong>find the maximum number of rooks</strong> rather than actually place them, one way that seems to work is to count the number of different <em>horizontal</em> hole-less runs of cells on the board, and the number of different <em>vertical</em> hole-less runs of cells on the board. The minimum of these two numbers is obviuously an upper bound on the rooks, and it can always be achieved (this I can only show with a tedious case analysis).</p>
|
1,363,213 | <p>I am given a chessboard of size $8*8$. In this chessboard there are two holes at positions $(X1,Y1)$ and $(X2,Y2)$. Now I need to find the maximum number of rooks that can be placed on this chessboard such that no rook threatens another. </p>
<p>Also no two rooks can threaten each other if there is hole between them.</p>
<p>How can I tackle this problem? Please help</p>
<p><strong>NOTE : A hole can occupy only a single cell on the chess board.</strong></p>
| David K | 139,123 | <p>We can get an algorithm to determine the maximum number of rooks
by considering attacks along ranks (parallel to one axis of the board)
separately from attacks along files (parallel to the other axis).</p>
<p>So first consider only attacks along ranks.
A rank with a single hole in it can still hold only one rook if the hole
is at either end of the rank, but can hold two rooks if the hole
is not at an end of the rank.
So if the holes are on two different ranks, determine the number of rooks
($1$ or $2$) that can be placed on the rank of the first hole,
add the number of rooks that can be placed on the rank of the second hole,
and then add $6$ (one rook on each of the remaining ranks).
This is the maximum number of rooks that can be placed without
an attack along a rank.</p>
<p>If the holes are in the same rank, the number of rooks that can be
placed on that rank depends on the positions of the holes as follows:</p>
<ul>
<li>Two holes adjacent to each other at one end of the rank: $1$ rook.</li>
<li>Two holes adjacent to each other but not at either end of the rank: $2$ rooks.</li>
<li>A hole at each end of the rank: $1$ rook.</li>
<li>A hole at one end of the rank, the other hole neither at the other end nor adjacent to the first hole: $2$ rooks.</li>
<li>Two holes, neither of which is adjacent to either end of the rank nor to the other hole: $3$ rooks.</li>
</ul>
<p>Add $7$ to this number (one rook in each remaining rank) to obtain
the maximum number of rooks that can be placed without an attack
along a rank.</p>
<p>This is an algorithm for finding $R(S)$, the maximum number of rooks
that can be placed on a two-hole board $S$ without allowing
attacks along ranks.</p>
<p>Now perform the same steps again, but substituting "file" every time
the word "rank" appears, to find $F(S)$, the maximum number of rooks
that can be placed on a two-hole board $S$ without allowing
attacks along files.</p>
<p>The maximum number of rooks that can be placed on a board with two
holes in configuration $S$ is $\min\{R(S), F(S)\}$.</p>
<hr>
<p>With a little thought, the rank and file parts of this algorithm can be
consolidated into a single set of criteria giving the maximum
number of rooks that can be placed:</p>
<ul>
<li>Both holes on the same edge of the board, or holes on opposite edges of the board: $8$ rooks.</li>
<li>A hole on an edge of the board and a second hole adjacent to the first: $8$ rooks.</li>
<li>At least one hole on an edge of the board, but neither of the two cases above is true: $9$ rooks.</li>
<li>Two holes adjacent to each other, but neither is on any edge of the board: $9$ rooks.</li>
<li>None of the above cases is true: $10$ rooks.</li>
</ul>
<hr>
<p>The rank-and-file algorithm can be abstracted to give the maximum number of rooks placed on an $n\times n$ chessboard with $m$ holes in configuration $S$.</p>
<p>Examine each rank. If there are no holes then the rank can hold one rook
without attack along a rank. If there are holes, they "cut" the rank
into one or more connected sets of squares ("pieces").
(An example of "cutting into one piece" occurs when all the holes are
in consecutive positions at one end of the rank.)
The number of pieces is the number of rooks that can be placed on the rank without attack along a rank.</p>
<p>Add up the number of rooks for each rank to obtain $R(S)$.</p>
<p>Using a similar algorithm for each file, add up the number of rooks for each file to obtain $F(S)$.</p>
<p>The final answer is $\min\{R(S), F(S)\}$.</p>
|
773,324 | <p>I know Dijkstra's algorithm to find the shortest way between 2 nodes, but is there a way to find the shortest path between 3 nodes among $n$ nodes? Here are the details:</p>
<p>I have $n$ nodes, some of which are connected directly and some of which are connected indirectly, and I need to find the shortest path between 3 of them.</p>
<p>For example, given $n = 6$ nodes labelled A through F, and the following graph:</p>
<pre><code>A-->B-->C
A-->D-->E
D-->F
</code></pre>
<p>How can I find the shortest path between the three nodes (A,E,F)?</p>
<p>I am looking for a solution similar to Dijkstra's shortest path algorithm, but for 3 nodes instead of 2.
<br/>
Please Note : <br/>
1- The Starting Node is A <br/>
2- The Sequential is not important just the path needs to cover all these Nodes <br/>
3- Their is no return back to A <br/>
Please find the diagram Image
<img src="https://i.stack.imgur.com/M1wxF.png" alt="enter image description here">
Regards & Thanks<br />
Nahed</p>
| ml0105 | 135,298 | <p>I'd say Phicar's solution of Floyd-Warshall's all-pairs, shortest paths algorithm is your best choice. Of course, this problem is NP-Hard. It's the optimal Hamiltonian Path problem, which is equivalent to the Traveling Salesman Problem.</p>
<p>The Floyd-Warshall algorithm can be executed in polynomial time. However, while sequence of the vertices may not matter to you, it does matter in minimizing the total cost. So your reduction is really from SAT. You really have to try combinations until you get the minimum sized path. </p>
|
2,304,448 | <blockquote>
<p>Does any group homomorphism $\Bbb Z \to \Bbb Z/n$ have kernel isomorphic to $\Bbb Z$?</p>
</blockquote>
<p>Here $n$ is any natural number.</p>
| MichaelGaudreau | 570,438 | <p>I like Tomek's answer a lot. But here is another answer that is essentially the same but is phrased slightly differently. </p>
<p>First observe that the proof of the usual open mapping theorem still works if instead of <span class="math-container">$Y$</span> Banach and <span class="math-container">$T$</span> surjective we assume that <span class="math-container">$Y$</span> is a normed space and <span class="math-container">$T(X)$</span> is not meager in <span class="math-container">$Y$</span>. Then <span class="math-container">$T$</span> is an open mapping from <span class="math-container">$X$</span> to <span class="math-container">$Y$</span>. In particular, <span class="math-container">$T(X)$</span> is open in <span class="math-container">$Y$</span>. Now use the following assertion (whose proof is an easy exercise): If <span class="math-container">$V$</span> is a subspace of a normed space <span class="math-container">$W$</span>, and <span class="math-container">$V$</span> contains an open subset of <span class="math-container">$W$</span>, then <span class="math-container">$V=W$</span>. </p>
|
1,712,256 | <p>I'm given some equations.</p>
<p>The first one, $x^3+2x^2-8x+1$ wants me to find the tangent line at $x=2$.</p>
<p>The second, (x^1.5) - (x^1/2) wants me to find the tangent line at $x=4$.</p>
<p>How would I go about solving this algebraically? I have to be able to prove the answers are $y=12x-23$ and $y=2.75x-5$ respectively, but I don't know how to start.</p>
| Brian Tung | 224,454 | <p>If you have a function $f(x)$, then at a given point $x_0$, the tangent line has slope $f'(x_0)$ and goes through the point $(x_0, f(x_0))$. The general equation of a line through a point $(x_0, y_0)$ and slope $m$ is</p>
<p>$$
y-y_0 = m(x-x_0)
$$</p>
<p>So in the first problem, $x_0 = 2$ and $f(x) = x^3+2x^2-8x+1$. Figure out what $f(x_0)$ and $f'(x_0)$ are, and proceed.</p>
|
512,118 | <p>Suppose $X$ is a metric space, $z$ is in $X$ and $(x_n)$ is a sequence in $X$. </p>
<p>Then what does it mean to say that, $z$ is in the "<em>closure of every tail of $(x_n)$</em>."</p>
<p>What does "<em>closure</em>" of every tail, mean ?</p>
| Prahlad Vaidyanathan | 89,789 | <p>It means that, for any $n \in \mathbb{N}$, consider the set
$$
S_n = \{x_n, x_{n+1}, x_{n+2}, \ldots\}
$$
Then, $z\in \overline{S_n}$ for all $n$ (here, $S_n$ is a tail of the sequence)</p>
|
2,003 | <p>I use some custom shortcut keys in <code>KeyEventTranslations.tr</code>. One is for the <code>Delete All Output</code> function: </p>
<pre><code>Item[KeyEvent["w", Modifiers -> {Control}],
FrontEnd`FrontEndExecute[FrontEnd`FrontEndToken["DeleteGeneratedCells"]]]
</code></pre>
<p>or simply:</p>
<pre><code>Item[KeyEvent["w", Modifiers -> {Control}], "DeleteGeneratedCells"]
</code></pre>
<p>This works as expected, putting up the dialog: "Do you really want to delete all the output cells in the notebook?". Is there any way to set up <code>KeyEventTranslations.tr</code> that when I hit <kbd>Ctrl</kbd>+<kbd>w</kbd> the dialog is automatically acknowledged and I don't have to hit <kbd>Enter</kbd>? The same goes for the <code>Quit kernel</code> function, that also puts up a dialog.</p>
| Albert Retey | 169 | <p>This just adds another hack for the <code>Quit</code> without confirm. It's not especially nice and I also haven't tested it in <code>KeyEventTranslations.tr</code> but it works from a Button with <code>Evaluator -> None</code> in versions 6,7 and 8 on Windows:</p>
<pre><code>FrontEnd`FrontEndExecute[{
FrontEnd`NotebookPut[
Notebook[{
Cell["NotebookClose[EvaluationNotebook[]];Quit[];", "Input"]
},
"ClosingSaveDialog" -> False,
WindowSize -> {10, 10},
WindowMargins -> {{-100, Automatic}, {-100, Automatic}}
]
],
FrontEnd`FrontEndToken[FrontEnd`NotebookPutReturnObject[],
"EvaluateNotebook"],
FrontEnd`SetSelectedNotebook[FrontEnd`InputNotebook[]]
}]
</code></pre>
<p>it should be noted that <code>Visible -> False</code> for the Notebook does not work, thus the settings for <code>WindowSize</code> and <code>WindowMargins</code>. The <code>SetSelectedNotebook</code> seems to reset the focus to the button notebook when used from a button, but since none of these functions are documented I don't know how it will behave when used from a keyboard shortcut.</p>
|
449,413 | <p>I'm trying to construct a norm on the space $\mathcal{D}(\Omega) := \{ f \in C^\infty(\Omega) | f $ is compactly supported on $ \Omega \}$ where $\Omega$ is an open subset of $\mathbb{R}$. I want this norm to include, somehow, the $L^\infty$-norms of <strong>all</strong> the derivatives of the smooth function to which it is applied. Specifically, I want to be able to encapsulate the statement</p>
<p>"<em>$f_n$$^{(m)} \rightarrow f^{(m)} ($as $n \rightarrow \infty)$ uniformly for all non-negative values of $m$.</em>"</p>
<p>as the statement</p>
<p>"$\|f_n - f\| \rightarrow 0$ $(n \rightarrow \infty)$.",</p>
<p>where $\|\cdot\|$ denotes my desired norm.</p>
<p>So far I've considered trying to write it as something along the lines of</p>
<p>\begin{align}\|f\| := \sum\limits_{m \in \mathbb{N}} \frac{1}{m!} \|f^{(m)}\|_\infty, \end{align}
where $\|\cdot\|_\infty$ denotes the usual $L^\infty$-norm. I am very unsure of the validity of this sort of "definition", as I can't see how to prove that this (or a related/similar) series converges. Any help on this would be greatly appreciated! Thanks in advance.</p>
| youler | 25,895 | <p>I'm not an expert in functional analysis, but I think this answers your question.</p>
<p>The topology of $\mathcal{D}(\Omega)$ is not induced by a norm. There is a discussion of this in chapter 1 of Rudin's Functional Analysis book. Briefly, a topological vector space whose topology is induced by a norm is locally bounded (proof: the open unit ball is bounded). So we want to show that $\mathcal{D}(\Omega)$ is not locally bounded. There is a result stating that a locally bounded topological vector space satisfying the Heine-Borel property (every closed and bounded set is compact) is finite-dimensional. So it only remains to show the Heine-Borel property for either $\mathcal{D}(\Omega)$ or $C^\infty(\Omega)$.</p>
<p>If you like, I can provide more details from Rudin's book.</p>
|
439,745 | <blockquote>
<p>Prove:$|x-1|+|x-2|+|x-3|+\cdots+|x-n|\geq n-1$</p>
</blockquote>
<p>example1: $|x-1|+|x-2|\geq 1$</p>
<p>my solution：(substitution)</p>
<p>$x-1=t,x-2=t-1,|t|+|t-1|\geq 1,|t-1|\geq 1-|t|,$</p>
<p>square,</p>
<p>$t^2-2t+1\geq 1-2|t|+t^2,\text{Since} -t\leq -|t|,$</p>
<p>so proved.</p>
<p><em>question1</em> : Is my proof right? Alternatives?</p>
<p>one reference answer: </p>
<p>$1-|x-1|\leq |1-(x-1)|=|1-x+1|=|x-2|$</p>
<p><em>question2</em> : prove:</p>
<p>$|x-1|+|x-2|+|x-3|\geq 2$</p>
<p>So I guess:( I think there is a name about this, what's that? wiki item?)</p>
<p>$|x-1|+|x-2|+|x-3|+\cdots+|x-n|\geq n-1$</p>
<p>How to prove this? This is <em>question3.</em> I doubt whether the two methods I used above may suit for this general case.</p>
<p>Of course, welcome any interesting answers and good comments.</p>
| Christian Blatter | 1,303 | <p><strong>(Edited)</strong></p>
<p>Given a real-valued data set ${\bf y}=(y_k)_{1\leq k\leq n}$ the function
$$f(x):=\sum_{k=1}^n |x-y_k|$$
assumes its minimum at the <em>median</em> $\mu$ of ${\bf y}$. When $y_1\leq y_2\leq \ldots\leq y_n$ and $n=2m+1$ then $\mu=y_{m+1}$ (the "middle value"), and if $n=2m$ then $f$ is actually constant on the interval $[y_m,y_{m+1}]$, and one defines $\mu:=(y_m+y_{m+1})/2$.</p>
<p><em>Proof.</em> When there are $r$ more $y_k$ to the right of $x$ than to the left of $x$ then increasing $x$ by $\Delta x$ will decrease $f$ by $r\cdot\Delta x$, and similarly, when there are $r$ more $y_k$ to the left of $x$ than to the right of $x$ then increasing $x$ by $\Delta x$ will increase $f$ by $r\cdot \Delta x$.$\qquad\square$</p>
<p>It follows that your sum $s(x)$ is minimal at $x={n+1\over2}$. It is then easily computed to $s_{\min}=n^2/4$ when $n$ is even and $s_{\min}=(n^2-1)/4$ when $n$ is odd.</p>
<p>When $n=1$ we have $s_{\min}=0=n-1$, and in the case $n=2$ we have $s_{\min}=1=n-1$. For $n\geq3$ we can argue as follows:
$$s_{\min}\geq{n^2-1\over 4}={(n-2)^2-1\over4}+n-1\geq n-1\ .$$</p>
|
1,250,132 | <p>Below is part of a solution to a critical points question. I'm just not sure how the equation on the left becomes the equation on the right. Could someone please show me the steps in-between? Thanks.</p>
<blockquote>
<p>$$\frac{-1}{x^2}+2x=0 \implies 2x^3-1=0$$</p>
</blockquote>
| Nikita | 509,540 | <p>For $x\neq 0$, we have $0= - \frac{1}{x^2} + 2x =- \frac{1}{x^2} + \frac{x^2}{x^2} \times 2x = - \frac{1}{x^2} + \frac{2x^3}{x^2} = \frac{2x^3 -1}{x^2}$.<p>
So we have $\frac{2x^3 -1}{x^2} = 0$ which will give us $2x^3 -1 = 0.$<p>
And we are done.</p>
|
1,569,728 | <p>Find one $z\in \mathbb{C}$ in the inequality $|z-25i|\le 15$ that has the largest argument ($\arg (z)$)</p>
<p>The inequality is equivalent to $x^2+(y-25)^2\le 15^2$ that represents the set of points in the circle of radius $15$ and center coordinate $C(0,25)$.</p>
<p>In this set, how to find one complex number which has the largest argument?</p>
| robjohn | 13,854 | <p><strong>Geometric Approach</strong></p>
<p>The image below shows how to compute the point in the given region with the greatest argument.</p>
<p><img src="https://i.stack.imgur.com/5nKt0.png" alt="enter image description here"></p>
<hr>
<p><strong>Calculus Approach</strong></p>
<p>The argument is an increasing function of $\frac yx$ when $y\gt0$. Therefore, we need to maximize $\frac yx$ on the boundary, which can be parametrized as
$$
(x,y)=(15\cos(\theta),25+15\sin(\theta))\tag{1}
$$
That is, maximize
$$
\frac{25+15\sin(\theta)}{15\cos(\theta)}=\frac53\sec(\theta)+\tan(\theta)\tag{2}
$$
whose derivative is
$$
\frac53\tan(\theta)\sec(\theta)+\sec^2(\theta)\tag{3}
$$
multiply by $\cos^2(\theta)$ to get
$$
\frac53\sin(\theta)+1\tag{4}
$$
which vanishes when $\sin(\theta)=-\frac35$, and since $x\lt0$, $\cos(\theta)=-\frac45$. Plug these values back into $(1)$ to get
$$
(x,y)=(-12,16)\tag{5}
$$</p>
|
1,569,728 | <p>Find one $z\in \mathbb{C}$ in the inequality $|z-25i|\le 15$ that has the largest argument ($\arg (z)$)</p>
<p>The inequality is equivalent to $x^2+(y-25)^2\le 15^2$ that represents the set of points in the circle of radius $15$ and center coordinate $C(0,25)$.</p>
<p>In this set, how to find one complex number which has the largest argument?</p>
| the_candyman | 51,370 | <p>Having posed that $z = x+iy$, then:
$$|z-25i|\le 15 \Rightarrow x^2 +(y-25)^2 \le 15^2.$$</p>
<p>Recall that:</p>
<p>$$\arg(z) =
\begin{cases}
f(y) & x > 0 \\
f(y) + \pi & x<0 \wedge y \ge 0 \\
f(y) - \pi & x<0 \wedge y < 0\\
\frac{\pi}{2} & x = 0 \wedge y > 0 \\
-\frac{\pi}{2} & x = 0 \wedge y < 0 \\
\text{undefined} & x = 0 \wedge y = 0
\end{cases},$$</p>
<p>where
$$f(y) = \arctan\left(\frac{y}{\sqrt{15^2 - (y-25)^2}}\right).$$</p>
<p>Notice that $x = \pm \sqrt{15^2 - (y-25)^2}$.</p>
<p>We want to find the maximum of $\arg(z)$ with respect to $y$. The $\arctan$ is monotonically increasing, then we can work on $f(y)$:</p>
<p>$$\frac{\partial }{\partial y} f(y) = \frac{25(y-16)}{\left(\sqrt{15^2 - (y-25)^2}\right)^3}.$$</p>
<p>Imposing it equal to $0$, you get $y = 16$ and $x=12$.</p>
<p>For $z = 12 + i16$, the argument is:</p>
<p>$$\arg(z) = \arctan\left(\frac{16}{\sqrt{15^2 - (16-25)^2}}\right) \simeq 0.9273.$$</p>
<p>When $x$ is negative, that is $x = - \sqrt{15^2 - (y-25)^2}$, then we have another candidate $z = -12 + 16i$. In this case $\arg(z) = \arctan\left(\frac{16}{\sqrt{15^2 - (16-25)^2}}\right) + \pi \simeq 4.0689$.</p>
<p>Last case to consider is $x = 0$ and $y = 40$. Here, we get that $\arg(z) = \frac{\pi}{2} \simeq 1.5708.$ </p>
<p>Finally, we conclude that $z = -12+16i$ is the maximum of $\arg(z)$, and $z$ stays in the circle you described.</p>
|
4,090,259 | <p>I began watching Gilbert Strang's lectures on Linear Algebra and soon realized that I lacked an intuitive understanding of matrices, especially as to why certain operations (e.g. matrix multiplication) are defined the way they are. Someone suggested to me 3Blue1Brown's video series (<a href="https://youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab" rel="nofollow noreferrer">https://youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab</a>) and it has helped immensely. However, it seems to me that they present matrices in completely different ways: 3Blue1Brown explains that they represent linear transformations, while Strang depicts matrices as systems of linear equations. What's the connection between these two different ideas?</p>
<p>Furthermore, I understand why operations on matrices are defined the way they are when we think of them as linear maps, but this intuition breaks when matrices are thought of in different ways. Since matrices are used to represent all sorts of things (linear transformations, systems of equations, data, etc.), how come operations that are seemingly defined for use with linear maps the same across all these different contexts?</p>
| user408858 | 408,858 | <p>First of all, simply by using matrix multiplication, you can rewrite <span class="math-container">$$Ax$$</span></p>
<p>into some equations</p>
<p><span class="math-container">$$a_{11}x_1+\cdots+a_{1n}x_n$$</span></p>
<p><span class="math-container">$$\cdots$$</span></p>
<p><span class="math-container">$$a_{m1}x_1+\cdots+a_{mn}x_n$$</span></p>
<p>This means, that a matrix does nothing more than being a representation of the coefficients of these linear equations, which is more convenient to write down.</p>
<p>If you think of it in a more abstract way, you could even just write down</p>
<p><span class="math-container">$$a_{11},\ldots, a_{1n},a_{21},\ldots,a_{2n},\ldots,a_{m1},\ldots,a_{mn}$$</span></p>
<p>into one vector or text file, where <span class="math-container">$a_{ij}$</span> are just some variables for anything. The total amount of variables is <span class="math-container">$m\cdot n$</span>.</p>
<p>Now, if you have given a matrix <span class="math-container">$A$</span> and you vary <span class="math-container">$x$</span> you can think of <span class="math-container">$A$</span> as a linear transformation of <span class="math-container">$x$</span>. For example the matrix</p>
<p><span class="math-container">$$\begin{pmatrix}
1 & 0 \\
0 & 0
\end{pmatrix}$$</span></p>
<p>will map the plane <span class="math-container">$\mathbb{R}^2$</span> onto a plane which has a <span class="math-container">$45°$</span> angle with the <span class="math-container">$x_1$</span>-axis. So in this case it might be better to think of <span class="math-container">$A$</span> as a map, which transforms <span class="math-container">$x$</span> in a linear way.</p>
<p>However, if you want to think of systems of linear equation with a given <span class="math-container">$b$</span>, you can rewrite <span class="math-container">$$Ax=b$$</span></p>
<p>into some equations</p>
<p><span class="math-container">$$a_{11}x_1+\cdots+a_{1n}x_n=b_1$$</span></p>
<p><span class="math-container">$$\cdots$$</span></p>
<p><span class="math-container">$$a_{m1}x_1+\cdots+a_{mn}x_n=b_m$$</span></p>
<p>Now, you are more likely to have an interest in <em>not</em> transforming the <span class="math-container">$x$</span>, but finding an <span class="math-container">$x$</span> that is a solution to the equations. In this case you are more likely to think of a system of equations, which needs a solution.</p>
|
1,221,914 | <blockquote>
<p>Let <span class="math-container">$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 ,\ n\ge 1$</span> have <span class="math-container">$n$</span> roots <span class="math-container">$x_1,x_2,\ldots,x_n \le -1$</span> and <span class="math-container">$a_0^2+a_1a_n=a_n^2+a_0a_{n-1}$</span>.
Find all such <span class="math-container">$P(x)$</span>. (Poland 1990).</p>
</blockquote>
<p>I used Viete Theorem and get: <span class="math-container">$$(x_1x_2\cdots x_n)^2 \pm \sum_{j=1}^n\prod_{i \not=j} x_i=1\pm(x_1x_2 \cdots x_n)(x_1+x_2+\cdots+x_n)$$</span> but not succeeded.</p>
| Mark Fischler | 150,362 | <p>Unless I misunderstand the problem,
$P(x) = x^2+2x+1$ meets the condition: Both roots are $-1$ and
$$a_0^2+a_1a_2 = a_2^2 + a_0a_1=3$$
Was the problem to find <em>all</em> such $P(x)$?</p>
|
1,036,636 | <p>The following statement makes sense intuitively, but is there a way to prove it mathematically? (This is something we make use of in applied optimization in calculus.)</p>
<blockquote>
<p>If $f$ is continuous on an interval $I$ and $x_0$ is the <strong>only</strong> relative (local) extremum, then $x_0$ is actually an <strong>absolute (global)</strong> extremum on $I$.</p>
</blockquote>
| David Holden | 79,543 | <p>suppose I is closed. if $x_0$ is a local maximum, then if it is not an absolute maximum $\exists x_1. f(x_1) \gt f(x_0)$. if $x_0 \ne \sup\{x | x \in I\}$ we may assume w.l.o.g $x_1 \gt x_0$ </p>
<p>since $I'=[x_0,x_1]$ is compact $f$ attains a minimum value on $I'$, say at $x'$, contradicting the assumption that the extremum at $x_0$ is unique.</p>
<p>other cases can be dealt with by slight modification of the same argument</p>
|
14,735 | <p>This question is somewhat related to <a href="https://mathematica.stackexchange.com/questions/4576/changing-the-order-of-elements-in-a-chart-legend">this</a> one.</p>
<p>Let's say this is the <code>BarChart</code> i want to make:</p>
<pre><code>BarChart[Range[5], ChartStyle -> "Rainbow", BarOrigin -> Left,
ChartLegends -> Map[ToString, Range[5]]]
</code></pre>
<p><img src="https://i.stack.imgur.com/dQQw3.jpg" alt="BarChart reflected"></p>
<p>The <code>BarChart</code> runs from the bottom and the legend starts at the top. The related question (linked above) gives a solution to reverse the legend, but i'd rather have the axis (and thus the purple bar) on top. It should look like this (made with <code>ImageReflect[]</code>):</p>
<p><img src="https://i.stack.imgur.com/KU79U.jpg" alt="BarChart reflected"></p>
<p>I tried <code>AxesOrigin</code>, but that doesn't work. My question seems pretty trivial, but i haven´t found a solution for it yet.</p>
<p><strong>Question: is there a way to specify the origin of a <code>BarChart</code>?</strong></p>
| Artes | 184 | <p>The appropriate function for symbolic representation of complex functions and numbers is <code>ComplexExpand</code>, e.g. </p>
<pre><code>ComplexExpand @ Table[(-1)^(k/3), {k, 3}]
</code></pre>
<blockquote>
<pre><code>{1/2 + (I Sqrt[3])/2, -(1/2) + (I Sqrt[3])/2, -1}
</code></pre>
</blockquote>
<p>For this specific task <code>ExpToTrig</code> yields the expected result, but for more general cases I recommend using <code>ComplexExpand</code> instead of <code>ExpToTrig</code>, for <code>F</code> (defined in the question) it yields the same :</p>
<pre><code>ComplexExpand @ Eigenvalues @ F == ExpToTrig @ Eigenvalues @ F
</code></pre>
<blockquote>
<pre><code>True
</code></pre>
</blockquote>
<p>Consider for example this matrix :</p>
<pre><code>m = Array[GCD, {3, 3}];
</code></pre>
<p>it yields eigenvalues of <code>m</code> in terms of <code>Root</code> objects, to get the result in terms of radicals one could add this option <code>Cubics->True</code> to <code>Eigenvalues</code>, <code>Eigensystem</code> etc. (this <a href="https://mathematica.stackexchange.com/questions/11498/mathematica-wont-give-eigenvectors-but-wolfram-alpha-will-what-am-i-doing-wron/11500#11500">answer</a> would be helpful as well).</p>
<p>Let's compare how <code>ExpToTrig</code> and <code>ComplexExpand</code> deal with <code>Eigenvalues</code> in this case :</p>
<pre><code>ExpToTrig @ Eigenvalues[ m, Cubics -> True] // TraditionalForm
</code></pre>
<p><img src="https://i.stack.imgur.com/t3gs7.gif" alt="enter image description here"></p>
<p>Therefore we can't even be sure that the eigenvalues are real numbers until we don't evaluate e.g. : </p>
<pre><code># ∈ Reals & /@ Eigenvalues[ m]
</code></pre>
<blockquote>
<pre><code>{True, True, True}
</code></pre>
</blockquote>
<p>we can see that <code>ExpToTrig</code> <strong>is not really helpful here</strong>, unlike <code>ComplexExpand</code> yielding symbolic eigenvalues, manifestly real: </p>
<pre><code>ComplexExpand @ Eigenvalues[ m, Cubics -> True] // TraditionalForm
</code></pre>
<p><img src="https://i.stack.imgur.com/5tNGk.gif" alt="enter image description here"></p>
|
36,735 | <p>In Peter J. Cameron's book "Permutation Groups" I found the following quote</p>
<blockquote>
<p>It is a slogan of modern enumeration theory that the ability to count a set is closely related to the ability to pick a random element from that set (with all elements equally likely).</p>
</blockquote>
<p>Indeed, one can count and sample uniformly from labeled trees, spanning trees, spanning forests, dimer models, young tableaux, plane partitions etc. However one can't do either of these very efficiently with groups, for example. My question is if one can make this into a rigorous statement, perhaps through complexity theory. That is, if I have an algorithm to produce a uniform sample from a set of objects, can I somehow come up with an efficient way to count them or vice-versa? </p>
<p>Does this slogan have a standard name? Are there any references?</p>
| Aaron Meyerowitz | 8,008 | <p>I suspect not. You did not ask about approximate counting or approximately uniform sampling (and of course that isn't what Cameron means). There are many situations where the desired set to enumerate sits in a larger set where we can easily and uniformly find a random element (for example: permutations of 1..n avoiding certain patterns (plus some side conditions perhaps) or lists of 17 integers under n^10 every 8 relatively prime). The enumeration problem does not seem aided by the fact that we can repeatedly pick a random element of the larger set and stop when we land in the target set.</p>
|
194,664 | <p>How to generate a list of fixpoint free permutations of n elements in mathematica?</p>
| Ulrich Neumann | 53,677 | <p>The problem is symmetric in a , {x,a} and {x,-a} solve the equation.
Try </p>
<pre><code>pic= ContourPlot[(Sin[a ]/a ) - x == 0, {x, 0, 1}, {a, 0, Pi},FrameLabel -> {x, a}]
</code></pre>
<p><a href="https://i.stack.imgur.com/pyJex.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pyJex.png" alt="enter image description here"></a></p>
<p>to see the solution of your equation. I don't think that an analytical solution exists.</p>
<p>Now you can get the solutionpoints from <code>pic</code></p>
<pre><code>xa = pic[[1, 1]];
</code></pre>
<p>and interpolate</p>
<pre><code>ip = Interpolation[xa] (*ip[x]=a[x]*)
Show[{pic, Plot[ip[x], {x, 0, 1},PlotStyle -> {Thickness[.01], Opacity[.3], Red}]}]
</code></pre>
<p><a href="https://i.stack.imgur.com/o0wKA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/o0wKA.png" alt="enter image description here"></a></p>
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.