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 |
|---|---|---|---|---|
2,648,526 | <p>I really cannot figure this question out. Can anyone help me please!?</p>
<blockquote>
<p>Prove that the length of the median $m_a$ of obtuse triangle $△ABC$ with the obtuse $∠CAB$ is smaller than $\dfrac{1}{2}|BC|$.</p>
</blockquote>
<p>Thank you very much!</p>
| Michael Rozenberg | 190,319 | <p>Let $\Phi$ be the circle with diameter $BC$.</p>
<p>Since $\measuredangle BAC>90^{\circ},$ we obtain that $A$ is placed inside the circle.</p>
<p>Now, let $M$ be the midpoint of $BC$ with the ray $MA\cap \Phi=\{A_1\}.$</p>
<p>Thus, $$\frac{1}{2}BC=A_1M>AM=m_a$$ and we are done!</p>
|
3,051,281 | <p>The question is:</p>
<p>Let <span class="math-container">$n \ge 2$</span> be an integer and consider a uniformly random permutation
(<span class="math-container">$a_1$</span>, <span class="math-container">$a_2$</span>, . . . , <span class="math-container">$a_n$</span>) of the set (1, 2, . . . , n). </p>
<p>For each <span class="math-container">$k$</span> with <span class="math-container">$1 \le k \le n$</span>, define the event</p>
<p><span class="math-container">$A_k$</span> = "<span class="math-container">$a_k$</span> is the largest element among the first <span class="math-container">$k$</span> elements
in the permutation".</p>
<p>Let <span class="math-container">$k$</span> and <span class="math-container">$l$</span> be two integers with <span class="math-container">$1 \le k \lt l \le n$</span>. Prove that the events <span class="math-container">$A_k$</span> and <span class="math-container">$A_l$</span> are independent.</p>
<p>I know that we should define Pr(<span class="math-container">$A_k$</span>), Pr(<span class="math-container">$A_k$</span>), Pr(<span class="math-container">$A_k \cap A_l$</span>) and than check relation between those but I keep getting into a dead end where those events are dependent.</p>
| Siong Thye Goh | 306,553 | <p>Guide:</p>
<ul>
<li><p>Choose particular value of <span class="math-container">$x$</span>, for example choose <span class="math-container">$x=0$</span>. You should be able to solve one of the unknown directly.</p></li>
<li><p>Can you choose another value of <span class="math-container">$x$</span> to make <span class="math-container">$\cos x = 0$</span>? </p></li>
</ul>
|
3,051,281 | <p>The question is:</p>
<p>Let <span class="math-container">$n \ge 2$</span> be an integer and consider a uniformly random permutation
(<span class="math-container">$a_1$</span>, <span class="math-container">$a_2$</span>, . . . , <span class="math-container">$a_n$</span>) of the set (1, 2, . . . , n). </p>
<p>For each <span class="math-container">$k$</span> with <span class="math-container">$1 \le k \le n$</span>, define the event</p>
<p><span class="math-container">$A_k$</span> = "<span class="math-container">$a_k$</span> is the largest element among the first <span class="math-container">$k$</span> elements
in the permutation".</p>
<p>Let <span class="math-container">$k$</span> and <span class="math-container">$l$</span> be two integers with <span class="math-container">$1 \le k \lt l \le n$</span>. Prove that the events <span class="math-container">$A_k$</span> and <span class="math-container">$A_l$</span> are independent.</p>
<p>I know that we should define Pr(<span class="math-container">$A_k$</span>), Pr(<span class="math-container">$A_k$</span>), Pr(<span class="math-container">$A_k \cap A_l$</span>) and than check relation between those but I keep getting into a dead end where those events are dependent.</p>
| Shubham Johri | 551,962 | <p>If any one of <span class="math-container">$\lambda_1,\lambda_2$</span> is non-zero, the <span class="math-container">$LHS$</span> will be periodic with period <span class="math-container">$2\pi$</span> while the <span class="math-container">$RHS$</span> is not. Therefore, <span class="math-container">$\lambda_{1,2}=0$</span>.</p>
|
1,164,471 | <p>I'm having trouble with the following problem:</p>
<p>A man found that $3$ out of $10$ inspected bottles were defective. What is the probability that the $2$ first defective bottles were found in the first $7$ inspected bottles? The probability that a bottle is defective is $0.1988$.</p>
<p>Let $A$ be the event that the $2$ first defective bottles were found in the first $7$ inspected bottles and $B$ the event that $3$ out of $10$ inspected bottles were defective. I need to find $P(A|B)=\frac{P(A∩B)}{P(B)}$. Using binomial distribution,$P(B)=C^{10}_3\times0.1988^3\times0.8012^7=0.20$</p>
<p>I'm stuck finding $P(A∩B)$. How can I proceed?</p>
<p>Thanks.</p>
| Krish | 177,430 | <p>Define a map $\phi:U(n) \to Gal(Q(\xi_n)/Q), a \mapsto \sigma_a$ where $\sigma_a:Gal(Q(\xi_n)/Q) \to Gal(Q(\xi_n)/Q)$ is defined by $\xi_n \to \xi_n^a.$ (Here $U(n) = (\mathbb Z/n\mathbb Z)^{\times}$ and $\xi_n$ is a primitive root of unity.) Then $\phi$ will be an isomorphism.</p>
<p><em>Note:</em> See <a href="http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/cyclotomic.pdf" rel="nofollow">this</a> for more clarification.</p>
|
453,502 | <p>Let $I$ be a generalized rectangle in $\Bbb R^n$ </p>
<p>Suppose that the function $f\colon I\to \Bbb R$ is continuous. Assume that $f(x)\ge 0$, $\forall x \in I$</p>
<p>Prove that $\int_{I}f=0 \iff$ the function $f\colon I\to \Bbb R$ is identically $0$. </p>
<hr>
<p>My idea is that</p>
<p>For $(\impliedby)$</p>
<p>Since $f\colon I\to \Bbb R$ is identically zero, $$f(I)=0$$</p>
<p>Then $$\int_{I}f=\int 0=0$$</p>
<p>For $(\implies)$,</p>
<p>Since $f$ is continuous, the function is integrable. </p>
<p>i.e $\int _{I} f $ exists. </p>
<hr>
<p>I need the show that $\int f=0$ but how? </p>
<p>Hopefully, other solution is true. Please check this. And how to continue this? Thank you:) </p>
| Pedro | 23,350 | <p>One direction is obvious. If $f$ is identically $0$, the integral is zero. Now suppose $f$ is continuous but <strong>not</strong> identically zero. Then there exists a point in $I$ such that $f(\xi)>0$. By continuity there exists a neighborhood of $\xi$ where $f(x)>0$ whenever $x\in(\xi-\delta,\xi+\delta)$. Can you take it from here?</p>
<p>Since $f\geqslant 0$, it follows that $$\int_I f \geq \int_{(\xi-\delta,\xi+\delta)} f>0$$</p>
<p>Thus, we have proven the contrapositive.</p>
|
2,166,847 | <p>Let $$v_n=\dfrac 1 {n+1}\sum_{k=0}^n \dfrac 1 {k+1}$$
We wanna study the sum $$S=\sum_{n=0}^{+\infty}(-1)^n v_n$$</p>
<p>The problem says we should first find $\omega(x)$ s.t.
$$v_n=\int_0^1 x^n\omega(x)dx$$
Then we'll have $S=\int_0^1\dfrac {\omega(x)} {1+x}dx$, but I can't find such $\omega(x)$. What's the idea of constructing such integral?</p>
| Jack D'Aurizio | 44,121 | <p>After we get
$$ S = -\int_{0}^{1}\frac{\log(1-x)}{1+x}\,dx = -\int_{0}^{1}\frac{\log x}{2-x}\,dx =-\left.\frac{d}{d\alpha}\int_{0}^{1}\frac{x^{\alpha}\,dx}{2-x}\,\right|_{\alpha=0^+}$$
we also have
$$ S = \text{Li}_2\left(\frac{1}{2}\right) = \color{red}{\frac{\pi^2}{12}-\frac{\log(2)^2}{2}}$$
by the <a href="https://math.stackexchange.com/questions/1056052/proof-of-dilogarithm-reflection-formula-zeta2-logx-log1-x-operatorname">dilogarithm reflection formula</a>.</p>
|
1,934,268 | <p>Let's say I have an arbitrary matrix $A$. </p>
<p>An eigenvector of $A$ would be a vector $\vec{x}$, such that </p>
<ol>
<li>$A\vec{x} = \lambda \vec{x}$</li>
<li>$\vec{x} \in N(A-\lambda I)$ i.e. $(A - \lambda I)\vec{x} = \vec{0}$</li>
</ol>
<p>where $\lambda$ is an eigenvalue of $\vec{x}$</p>
<hr>
<p>Now are all $\vec{x} \in N(A-\lambda I)$ eigenvectors of $A$, i.e. do all $\vec{x}$ in the nullspace of $A - \lambda I$ also automatically satisfy condition $(1)$, that $A\vec{x} = \lambda \vec{x}$?</p>
<p>Written more concretely is the following statement true:</p>
<p>$$\vec{x} \in N(A-\lambda I) \implies A\vec{x} = \lambda \vec{x} \ \ \ \ \ \forall \vec{x} \in N(A-\lambda I)$$</p>
<p>If that was the case then the eigenvectors, would $span$ the entire nullspace. </p>
<p>Furthemore what relation does $N(A-\lambda I)$ have to $N(A)$ and the eigenvectors? If $\vec{x} \in N(A-\lambda I)$, does that imply that $\vec{x} \in N(A)$? Can we deduce anything about the eigenvectors of $A$ from $N(A)$ or can we only deduce them from $N(A - \lambda I)$?</p>
| egreg | 62,967 | <p>The answer to your title question is "just some", because there is a single exception: the null vector.</p>
<p>By definition, an eigenvalue of $A$ relative to $\lambda$ is a <em>nonzero</em> vector $x$ such that $Ax=\lambda x$ (forgive me if I don't use arrows for vectors). Now
$$
Ax=\lambda x
\iff
Ax-(\lambda I)x=0
\iff
(A-\lambda I)x=0
\iff
x\in N(A-\lambda I)
$$
so the only exception is the vector that is not an eigenvector by definition, that is the zero vector.</p>
<p>Of course, if your definition allows $0$ to be an eigenvector, then the answer to the title question is "all". However, most books on the subjects don't consider $0$ an eigenvector, so check carefully yours.</p>
<hr>
<p>About relationships between the eigenvalues/eigenvectors and $N(A)$, I'm afraid there's none: consider simply the case of an invertible matrix, which can have any set of nonzero eigenvalues compatible with its order, but has $N(A)=0$.</p>
|
3,435,511 | <p>I'm looking to write down formally that a multiset of elements contains at least two elements that differs in value.
e.g.,
S1 = {1,1,1,1,1,1} and S2={1,1,1,1,0,1}
S1 has all identical elements, S2 has at least two elements that differs in value.</p>
| Robert Israel | 8,508 | <p>Write <span class="math-container">$$\ln(1-e^{-k}) = - \sum_{j=1}^\infty e^{-jk}/j$$</span>
and take <span class="math-container">$k = x + t$</span> so your integral becomes
<span class="math-container">$$ \sum_{j=1}^\infty e^{-jx} \int_0^\infty \frac{\sqrt{t(t+2x)}}{j(x+t)} e^{-jt} \; dt$$</span>
Now <span class="math-container">$$ \int_0^\infty \frac{\sqrt{t(t+2x)}}{j(x+t)} e^{-jt}\; dt= \frac{\sqrt{2\pi}}{2 j^{5/2} \sqrt{x}} - \frac{9 \sqrt{2\pi}}{16 j^{7/2} x^{3/2}} + \frac{345\sqrt{2\pi}}{256 j^{9/2} x^{5/2}} + \ldots$$</span>
so it seems to me the asymptotic form should be
<span class="math-container">$$ e^{-x} \left( \frac{\sqrt{2\pi}}{2 \sqrt{x}} - \frac{9 \sqrt{2\pi}}{16 x^{3/2}} + \frac{345\sqrt{2\pi}}{256 x^{5/2}} + \ldots\right)$$</span></p>
|
4,030,877 | <p><strong>Question:</strong> Ten different candies are given to three children <code>A</code>, <code>B</code> and <code>C</code>. each child has at least one. How many different ways are there?</p>
<p>I use two different methods to solve this problem:</p>
<pre><code> f[1, n_] := n;
f[n_, 1] := 1;
f[n_, m_] := f[n, m] = m (f[n - 1, m - 1] + f[n - 1, m])
f[10, 3]
</code></pre>
<pre><code> Select[Tuples[Range[3], 10], Length[Union[#]] == 3 &] // Length
</code></pre>
<p>But the result of the first method is not equal to <code>StirlingS2[10, 3] 3!</code>. I want to know how to solve this problem correctly?</p>
| Ben | 887,799 | <p><span class="math-container">$$\sum_{i=0}^n x^i = \frac{1-x^{n+1}}{1-x}$$</span></p>
<p>This means that the value of the right is exactly equal to that on the right and most importantly the right converges if and only if the left converges.</p>
<p>For any x, <span class="math-container">$\frac{1-x^{n+1}}{1-x} = \frac{1}{1-x} - \frac{x^{n+1}}{1-x}$</span> so <span class="math-container">$\frac{1-x^{n+1}}{1-x}$</span> converges if and only if both <span class="math-container">$\frac{1}{1-x}$</span> and <span class="math-container">$\frac{x^{n+1}}{1-x}$</span> converge. <span class="math-container">$\frac{1}{1-x}$</span> obviously converges for all x as it is independent of n so <span class="math-container">$\frac{1-x^{n+1}}{1-x}$</span> converges if and only if <span class="math-container">$\frac{x^{n+1}}{1-x}$</span> converges</p>
<p><span class="math-container">$$ \lim_{n \rightarrow \infty} \frac{x^{n+1}}{1-x} = \frac{1}{1-x}\lim_{n \rightarrow \infty} x^{n+1}$$</span></p>
<p>So <span class="math-container">$\frac{x^{n+1}}{1-x}$</span> converges if and only if <span class="math-container">$x^{n+1}$</span> converges which converges for <span class="math-container">$|x| \lt 1$</span>. So to answer the last question, it isn't a sufficient condition but in this specific case the condition for the terms to approach 0 happens to be the same as the condition for the closed form of the sum to converge</p>
|
3,091,464 | <p>If I had an array of 7 numbers and I wanted all numbers to be equally spaced within it but needed to start at 5 and end at 35 how would I do this?</p>
<p>For instance if I were given:</p>
<ul>
<li>Numbers in array: <strong>7</strong></li>
<li>Largest number in array: <strong>35</strong></li>
<li>Smallest number in array: <strong>5</strong></li>
</ul>
<p>How do I find what number to increment the array by if I wanted to create an array of equally spaced numbers?</p>
<p>An example answer would be: Given the above three numbers I would say increment each array position by 5 and all 7 numbers will be equally spaced all the way to 35. The end result would look like this: [5,10,15,20,25,30,35]</p>
<p><strong>Question 1</strong>: What would the formula be to repeat this any time I have array length, smallest, and largest number?</p>
<p><strong>Question 2</strong>: In the above example if I were given the number 30 what is the formula to find its position in the given array? The formula should produce a result like this: If the array started at index one it would be position 6.</p>
<p>Any help would be great, thanks!</p>
| RRL | 148,510 | <p>Correct.</p>
<p>As the extension is bounded, where <span class="math-container">$|\hat{f}(x)| \leqslant M$</span> on <span class="math-container">$[0,1]$</span>, we have as <span class="math-container">$\epsilon \to 0$</span>,</p>
<p><span class="math-container">$$\left|\int_0^1\hat{f} - \int_\epsilon^{1-\epsilon} f \right| = \left|\int_0^1\hat{f} - \int_\epsilon^{1-\epsilon} \hat{f} \right|\leqslant \left|\int_0^\epsilon \hat{f}\right| + \left|\int_{1-\epsilon}^1 \hat{f} \right| \leqslant 2M\epsilon \to 0$$</span></p>
|
186,555 | <p>I'm a high school student who is trying to figure out a complete course of self-study for each year of high school. How can I self-learn grades of math without devoting too much time? This is a complex issue for me, as other students at my competitive high school have tutors and the like. Please recommend textbooks that have detailed explanations and progressive practice problems, for self-study for each area such as:</p>
<p>Algebra</p>
<p>Geometry</p>
<p>Trigonometry and Analytic Algebra</p>
<p>Pre-calculus</p>
<p>BC Calculus</p>
<p>Other people have skipped grades of math due to help from tutors and parents. Can I cover all of geometry and trigonometry in 8 months without going insane and be able to skip a grade?</p>
| Robert Mastragostino | 28,869 | <p>You probably won't find one book that covers everything, but in all honesty if you're a good math student you could probably skip through trigonometry with ease. I'd recommend <a href="http://books.google.ca/books?id=Gs2pHYioDbQC&printsec=frontcover&dq=algebra+demystified&source=bl&ots=cxxr2HGerP&sig=bNwwuV-dnbesSAuk-J66ha9w7c0&hl=en&sa=X&ei=Y084UJvgF-yDyAHw-YCgBg&ved=0CDMQ6AEwAA#v=onepage&q=algebra%20demystified&f=false" rel="nofollow">Algebra Demystified</a> and <a href="http://books.google.ca/books?id=klIfOHAAmJkC&printsec=frontcover&dq=trigonometry+demystified&source=bl&ots=mM1mkn_hXI&sig=S6u-ylSj34Et5htIcQ9xHsBwUcI&hl=en&sa=X&ei=jk84UJ-pB4ftygHOhID4Dg&sqi=2&ved=0CC4Q6AEwAA#v=onepage&q=trigonometry%20demystified&f=false" rel="nofollow">Trigonometry Demystified</a>. They have a good amount of practice questions, and go through the motivations of the topics involved (especially in trigonometry). You could also find more than enough around the internet, say through <a href="http://www.khanacademy.org" rel="nofollow">Khan Academy</a> which has plenty of videos and more importantly a decent practice section.</p>
<p>While I'm not sure what the curriculum is where you are, in my high school we did:</p>
<p>Grade 9: Cartesian coordinates and linear equations</p>
<p>Grade 10: Quadratic Equations (factoring, expanding, completing the square and the quadratic formula) and basic trigonometry (working with definitions of $\sin$, $\cos$ and $\tan$).</p>
<p>Grade 11: Functions, factoring higher-degree polynomials, proving trigonometric identities, shifting/stretching functions, graphs of trig functions.</p>
<p>Grade 12: More functions, inverses, exponentials and logarithms, inverse trig functions, secant,cosecant and cotangent functions, how to divide polynomials the long way and with synthetic division, the remainder theorem and rational root theorem.</p>
<p>I don't think that's <em>quite</em> everything, but it's more than enough. Stroll through google books (you can see quite a bit of the books on there, including whole sections and practice problems) and the internet at large if you want more. Also see this wonderful <a href="http://hbpms.blogspot.ca/2008/05/stage-1-elementary-stuff.html" rel="nofollow">list of legitimately free textbooks and/or course notes</a>.</p>
|
186,555 | <p>I'm a high school student who is trying to figure out a complete course of self-study for each year of high school. How can I self-learn grades of math without devoting too much time? This is a complex issue for me, as other students at my competitive high school have tutors and the like. Please recommend textbooks that have detailed explanations and progressive practice problems, for self-study for each area such as:</p>
<p>Algebra</p>
<p>Geometry</p>
<p>Trigonometry and Analytic Algebra</p>
<p>Pre-calculus</p>
<p>BC Calculus</p>
<p>Other people have skipped grades of math due to help from tutors and parents. Can I cover all of geometry and trigonometry in 8 months without going insane and be able to skip a grade?</p>
| Martin Peters | 185,067 | <p>Have a go at the book <em>Mathematical omnibus. Thirty lectures on classic mathematics</em> by Dmitry Fuchs and Sergei Tabachnikov. This book will engage your mind and point you to very interesting mathematics.</p>
<p>Another good book to try out is this one: <a href="http://imaginary.org/sites/default/files/5to15_en_gb.pdf" rel="nofollow">Vladimir Arnold: <em>Problems for children from 5 to 15</em></a>. </p>
|
61,697 | <p>The hamiltonian flow box theorem, as stated in Abraham and Marsden's Foundations of Mechanics, says that:</p>
<p>Given an hamiltonian system $(M,\omega,h)$ with $dh(x_0)\neq 0$ for some $x_0$ in $M$, there is a symplectic chart $(U,\phi)$ on $M$ centered at $x_0$ such that $\phi_{\ast}h(x)=h(x_0)+\omega_0(\phi_{\ast}X_h(x_0),x)$, where $\omega_0$ is the canonical symplectic form.</p>
<p>*<em>Question:</em>*I know some different proofs of this theorem, but I would know if, at your knowledge, in the literature there is a proof which uses the Moser's trick as in the proof of the Darboux' theorem.</p>
<hr>
<p>In Abraham and Marsden there is a proof using the contact structure associated to the symplectic one and its canonical transormations.
I know even that it has an extension in a theorem of Cartan which says:
Given a $2n$-dimensional symplectic manifold $(M,\omega)$, it is possible to extend to a system of symplectic coordinates on $(M,\omega)$ any set of local functions $f_1,\ldots,f_k,g_1,\ldots,g_l$ on $M$ such that $f_1,\ldots,f_k$ are independent and in involution, $g_1,\ldots,g_l$ are independent and in involution, and $\{f_i,g_j\}=\delta_{ij}$ for any $i,j$.</p>
| agt | 12,617 | <p>Warning: I have posted this as an answer and not as comment, not to gain in reputation, but just in order to have enough space to write to Michael what I had understood of his answer that has been very much useful to me.</p>
<p>Let $(M_0,\omega_1)$ and $(M_1,\omega_1)$ be symplectic manifolds of the same dimension.
If $h_i$ is a smooth function on $M_i$ with $dh_i(x_i)\neq 0$ for some $x_i\in M_i$, and $i=0,1$, then there exists a local diffeomorphism $\phi$ from $M_0$ to $M_1$ such that $\phi(x_0)=x_1, \phi_{\ast}\omega_0=\omega_1$, and $d\phi_{\ast}h_0=dh_1$.</p>
<p>For the result from linear algebra reported in Michael's answer, there a local diffeomorphism $\psi$ from $M_0$ to $M_1$ such that $\psi(x_0)=x_1, \psi_{\ast}\omega_0(x_1)=\omega_1(x_1)$, and $d\psi_{\ast}h_0(x_1)=dh_1(x_1)$.</p>
<p>So we have now a manifold $M$ with symplectic forms $\Omega_0$ and $\Omega_1$ coinciding at a point $x_0$, and smooth regular functions $H_0$ and $H_1$ with $dH_0(x_0)=dH_1(x_1)$.
With no loss of generality we can assume also $H_0(x_0)=H_1(x_0)$.</p>
<p>Let us introduce the following time dependent forms on $M$:<br>
$H_t=H_0+t\tilde{H}=H_0+t(H_1-H_0)$ and $\Omega_t=\Omega_0+t\tilde{\Omega}=\Omega_0+t(\Omega_1-\Omega_0)$.</p>
<p>In order to construct the required local diffeomorphism using the Moser's trick, we need a time dependent local vector field $X_t$ around $x_0$ satisfying:
$di_{X_t}\Omega_t+\tilde{\Omega}=0$, $i_{X_t}dH_t+\tilde{H}=0$, $X_t(x_0)=0$, for $t\in\[0,1\]$.
Really the third condition follows from the second one because $\tilde{H}(x_0)=0$ and $H_t$ is regular.</p>
<p>Let $\alpha$ be a local primitive of $\tilde{\Omega}$ vanishing at $x_0$.
The first condition becomes $i_{X_t}\Omega_t=-\alpha+df_t$, and determines a unique $X_t$ for each smooth function $f=\{f_t\}_t$ on a neighborhood of $M\times\[0,1\]$.</p>
<p>Finally the second condition becomes the following one only on $f=\{f_t\}_t$:
$\mathcal{L}(Y_t).(f_t)=g_t\equiv\tilde{H}-i_{Y_t}\alpha$.
Where $Y_t$ is the Hamiltonian vector field corresponding to $H_t$ w.r.t. $\Omega_t$.</p>
<p>A solution for this equation $\mathcal{L}(Y_t+0\frac{\partial}{\partial t}).f=g$ could be constructed using the method of characterics, considering that $Y\equiv Y_t+0\frac{\partial}{\partial t}$ is non singular because such is $dH_t$.</p>
|
2,970,773 | <p>Let's assume I am given a positive integer <span class="math-container">$n$</span>, as well as an upper limit <span class="math-container">$L$</span>.</p>
<p>How could one find all, or at least one, possible solutions for <span class="math-container">$a$</span> and <span class="math-container">$b$</span> such that <span class="math-container">${a \mod b = n}$</span> where as <span class="math-container">$0 <=a, b <= L$</span>?</p>
| user87690 | 87,690 | <p>The space is assumed to be regular, so if <span class="math-container">$\{B_n\}_{n ∈ ℕ}$</span> is a basis, then for every <span class="math-container">$x ∈ B_m$</span>, there exists <span class="math-container">$n$</span> such that <span class="math-container">$x ∈ B_n ⊆ \overline{B_n} ⊆ B_m$</span>. Nobody claims that <span class="math-container">$\overline{B_n} ⊆ B_m$</span> for every <span class="math-container">$n, m$</span>, that's nonsense. The Urysohn lemma is used only for those pairs <span class="math-container">$n, m$</span> such that <span class="math-container">$\overline{B_n} ⊆ B_m$</span>, and regularity asserts that there is enough such pairs. I don't understand the example with unit half-open intervals – they are not open, and they just do not form a basis of the topology of <span class="math-container">$ℝ$</span>.</p>
|
3,412,217 | <p>The category of abelian groups <span class="math-container">$\mathbb{Ab}$</span> has a monoidal (closed) structure <span class="math-container">$(\otimes, \mathbb{Z})$</span>. Moreover, it is monadic over the category of sets via the free abelian group monad <span class="math-container">$$\mathbb{Z}[\_]: \text{Set} \to \text{Set}.$$</span></p>
<blockquote>
<p><strong>Q:</strong> I wonder (and I believe) if it is possible to recover the monoidal structure from the monad and the underlying monoidal structure on the category of sets.</p>
</blockquote>
<p>The question might seem vague, an accepted answer would look like "<em>of course it is possible, <span class="math-container">$A \otimes B$</span> is some algebra structure on the coequalizer of some diagram of the form <span class="math-container">$\mathbb{Z}[\mathsf{U}A \times \mathsf{U}B] \rightrightarrows \mathbb{Z}[\mathsf{U}A \times \mathsf{U}B]$</span></em>", or something similar where the cartesian product of Set and the monad appear.</p>
| Z Ahmed | 671,540 | <p>If GCD od <span class="math-container">$a$</span> and <span class="math-container">$b$</span> is 1.
Let the pairs of integers <span class="math-container">$(x_0,y_0)$</span>, be the first solution of <span class="math-container">$ax+by=n~(*)$</span>. Then
Let <span class="math-container">$x=x_0+bm, y=y_0-am$</span> are the possible solutions. Since both <span class="math-container">$x,y \ge 0$</span>
you can easily find the possible values of <span class="math-container">$m$</span> and hence the number of solutions.
The number of solutions of this Eq (*). are <span class="math-container">$N=[\frac{n}{ab}]$</span> where <span class="math-container">$[.]$</span> means integer part. Also, <span class="math-container">$N$</span> may be one more than this if the remainder(<span class="math-container">$r$</span>) when <span class="math-container">$n$</span> is divided by <span class="math-container">$ab$</span> satisfies the relation <span class="math-container">$ax+by=r$</span> one pair of non-negative integers <span class="math-container">$x$</span> and <span class="math-container">$y$</span>.</p>
|
2,265,819 | <p>I need to parameterize the intersection of $$4x^2 + y^2 + z^2 = 9\tag{1}$$ and $$z=x^2+y^2\tag{2}$$. </p>
<p>First, I'll solve (2) for $y^2$ and substitute the result into (1):</p>
<p>$$3x^2+z+z^2 = 9 \tag{3}$$</p>
<p>Next, I'll make the substitution $u=\sqrt{3}x$, such that we can complete the square in (3) by adding $1/4$ to each side and arrive at</p>
<p>$$\frac{u^2}{r^2} + \frac{(z+\frac{1}{2})^2}{r^2} = 1$$</p>
<p>where $r^2 = 9 + \frac{1}{4} = \frac{37}{4}$</p>
<p>Now I'll write a parameterization: </p>
<p>$$u = r\cos \phi \implies x(\phi) = \frac{1}{\sqrt{3}}r\cos\phi$$
$$z(\phi) = r\sin \phi -\frac{1}{2}$$
$$y(\phi) = \pm \sqrt{z-x^2} = \pm\left(\sqrt{r\sin \phi - \frac{1}{2} - \left(\frac{1}{\sqrt{3}}r\cos\phi\right)^2}\right)$$</p>
<p>such that we have two branches:</p>
<p>$$\mathbf{r}(\phi)_1 = \big<x(\phi), y(\phi), z(\phi)\big>$$
$$\mathbf{r}(\phi)_2 = \big<x(\phi), -y(\phi), z(\phi)\big>$$</p>
<p>Is this correct?</p>
| Yiorgos S. Smyrlis | 57,021 | <p><strong>Step 1.</strong> Clearly $\mathrm{Im}\,f(x+iy)\ne 0$, whenever $y\ne 0$. In particular,
$\mathrm{Im}\,f(z)$ maintains sign in the upper half plane and the lower half plane, and it has different signs in each half plane. Without loss of generality, $\mathrm{Im}\,f(x+iy)> 0$, if $y>0$, and $\mathrm{Im}\,f(x+iy)< 0$, if $y<0$.</p>
<p><strong>Step 2.</strong> Assume that $f'(x)=0$, for some $x\in\mathbb R$. Without loss of generality, assume that $x=0$. Otherwise, replace $f$ by $f-f(0)$. So $f'(0)=0$, and $f\not\equiv 0$, implies the existence of a $k\in\mathbb N$, such that $f(z)=z^kg(z)$, where $g$ entire and $g(0)\ne 0$. Thus, there exists an $a\in\mathbb R$, such that, if $f=u+iv$, then
$$
0\ne a=f^{(k)}(0)=\partial_x^kf(0)=\partial_x^ku(0)=\partial_x^ku(0)=\partial_y^kv(0).
$$
Clearly $k$ is odd, otherwise $v$ would have a local minimum, which is impossible, since $v<0$, in the lower half plane. In fact, $a>0$, since $v$ is increasing in $y$ at $y=0$. Hence $f(x)>0$ and $f$ strictly increasing for $x>0$, and $x$ sufficiently small and $f(x)<0$ and $f$ also strictly increasing, for $x<0$ and $|x|$ sufficiently small. But this means, that the equation
$$
f(x+iy)=\eta
$$
would have $k$ solution for $|\eta|$ is a small disk centred at $0$. But all the $k$ roots have to be negative, which means that $k=1$, since $f$ is strictly increasing. Hence, $f'(0)\ne 0$.</p>
<p>In you need to show that $f$ is linear, then, observe that</p>
<p>a. $f$ has to be strictly increasing, in $x\in\mathbb R$, and by Picard's Little Theorem, $f$ is a polynomial, since, the neighbourhoods of infinity do not contain absolutely small real numbers. But the only root in real, and there is only one real real root and single, and hence $f$ is a linear function of the form
$$
f(z)=az+b
$$
where $a,b\in\mathbb R$ and $a\ne 0$.</p>
|
4,418,518 | <p><span class="math-container">$X$</span> is Banach Space and <span class="math-container">$A\subseteq X$</span>.
<span class="math-container">$A$</span> is dense <span class="math-container">$G_{\delta}$</span> set. We have to show that <span class="math-container">$A-A=X$</span>.</p>
<p>Since <span class="math-container">$A$</span> is dense we can write <span class="math-container">$\bar{A}= X$</span>. So, it is enough to show that <span class="math-container">$A-A= \bar{A}$</span>. Here <span class="math-container">$A-A = \{y-z | y,z\in A\}$</span>.</p>
<p>Consider <span class="math-container">$\tau$</span> is the induced topology by <span class="math-container">$X$</span>. Since <span class="math-container">$X$</span> is normable, then <span class="math-container">$\tau$</span> has a bounded convex nbd of 0.</p>
<p>Let <span class="math-container">$V$</span> be a bounded, balanced, convex nbd of 0. So, <span class="math-container">$A= \cap_{r\in \mathbb{Q} } rV$</span>.</p>
<p>From this I can claim that <span class="math-container">$A-A\subseteq \bar{A}$</span>. How to show other side?</p>
<p>Any alternative methods also appreciated.</p>
| Gerd | 960,044 | <p>As <span class="math-container">$A$</span> is a dense <span class="math-container">$G_\delta$</span> set the complement <span class="math-container">$X\setminus A$</span> is of first category. Let <span class="math-container">$x_0 \in X$</span> and set <span class="math-container">$h:X \to X$</span>, <span class="math-container">$h(x)=x+x_0$</span>. Then <span class="math-container">$h$</span> is a homeomorphism, and with <span class="math-container">$X\setminus A$</span> also <span class="math-container">$X\setminus h(A)=h(X\setminus A)$</span> is of first category in <span class="math-container">$X$</span>. Thus <span class="math-container">$(X\setminus A) \cup (X\setminus h(A)) \not=X$</span> which means <span class="math-container">$A \cap h(A) \not= \emptyset$</span>. Choose <span class="math-container">$a \in A \cap h(A)$</span>. Then <span class="math-container">$a=h(b)$</span> for some <span class="math-container">$b \in A$</span> which means <span class="math-container">$x_0=a-b$</span>. Thus each <span class="math-container">$x_0 \in X$</span> is a difference of elements of <span class="math-container">$A$</span>.</p>
|
4,448,525 | <p>To do a proof by induction, it is necessary to prove</p>
<p>(1) the proposition <span class="math-container">$P(n)$</span> is true for <span class="math-container">$n=1$</span></p>
<p>(2) if the proposition <span class="math-container">$P(n)$</span> is true for <span class="math-container">$n=k$</span>, then <span class="math-container">$P(n)$</span> is true for <span class="math-container">$n=k+1$</span></p>
<p>My questions are</p>
<p>(1) What is the significance of showing the proposition <span class="math-container">$P(n)$</span> is true for <span class="math-container">$n=1$</span>? Why is it necessary to do this step?</p>
<p>(2) But based on which properties or axioms of number can we be sure that <span class="math-container">$P(n)$</span> is true for <span class="math-container">$n=k+1$</span> if <span class="math-container">$P(n)$</span> is true for <span class="math-container">$n=k$</span>?</p>
<p>(3) Can we use proof by induction when the proposition only involves natural numbers, i.e. no non-integer? For example, we cannot use proof by induction to show that <span class="math-container">$cos^2(x)+sin^2(x)=1$</span></p>
<p>(4) Also under which condition proof by contradiction cannot be used? Any example?</p>
<p>Thanks.</p>
| Robert Shore | 640,080 | <p>Fundamentally, a proof by induction is using the well-ordering property of the natural numbers, i.e., the fact that every <em>non-empty</em> set of natural numbers has a least element. And it can always be thought of as a proof by contradiction. You’re assuming the set of counterexamples of <span class="math-container">$P(x)$</span> is non-empty and showing that the set can have no least element, which contradicts the well-ordering property. That contradiction shows the set of counterexamples must be empty.</p>
<p>Let’s see why that’s the case. Assume you’ve completed your proof by induction. Let’s also assume the set of counterexamples were non-empty. Then it has to have a least element. What is that least element?</p>
<p>It can’t be <span class="math-container">$1$</span>. Why not? Because you separately proved <span class="math-container">$P(1)$</span>, so you directly proved that <span class="math-container">$1$</span> isn’t a counterexample. That’s the reason you have to separately prove the statement for <span class="math-container">$x=1$</span>.</p>
<p>But the least element also can’t be <span class="math-container">$m$</span> for any <span class="math-container">$m \gt 1$</span>. Because if <span class="math-container">$m \gt 1$</span>, then for some natural number <span class="math-container">$n, m=n+1$</span>. And since we’re assuming <span class="math-container">$m$</span> is the <em>least</em> counterexample to <span class="math-container">$P(x)$</span>, it follows from <span class="math-container">$n \lt m$</span> that <span class="math-container">$n$</span> is <em>not</em> a counterexample; in other words, that <span class="math-container">$P(n)$</span> is true. And in our inductive step we showed that if <span class="math-container">$P(n)$</span> is true, we also know that <span class="math-container">$P(n+1)=P(m)$</span> is true.</p>
<p>But that means that <span class="math-container">$m$</span> isn’t a counterexample after all, contradicting our assumption that it was the least counterexample. And that, in turn, means that <em>no</em> element can fill the role of least possible counterexample. But every <em>non-empty</em> set of positive integers has a least element.</p>
<p>Therefore, the fact (which we just proved) that the set of counterexamples has no least element can only mean the set of counterexamples is empty; in other words, <span class="math-container">$P(x)$</span> holds for all positive integers <span class="math-container">$x$</span>.</p>
|
1,958,152 | <blockquote>
<p>I want to know why $\frac{\log4}{\log b}$ can't be simplified to $\frac4b$. </p>
</blockquote>
<p>I am a high school student. Please do not quote some theories that are too advanced for me. Thank you!</p>
| E.H.E | 187,799 | <p>$$\frac{\log 4}{\log b}=\frac{\log(4^{1/4})^4}{\log(b^{1/b})^b}=\frac{4}{b}\frac{\log\sqrt[4]{4}}{\log\sqrt[b]{b}}$$
so
$$\frac{\log 4}{\log b}\neq \frac{4}{b}\quad{\text{(except when b=4 and b=2})}$$</p>
|
3,543,060 | <p>The ODE is defined on <span class="math-container">$[0,b]$</span> with Neumann boundary conditons.
<span class="math-container">$$y''(x)=\frac{c_1y(x)}{c_2+y(x)}$$</span>
<span class="math-container">$$y'(0)=0; y'(b)=0$$</span></p>
<p>How to solve the above ODE? Any help is appreciated!</p>
| user577215664 | 475,762 | <p><span class="math-container">$$y''(x)=\frac{c_1y(x)}{c_2+y(x)}$$</span>
Multiply both sides by <span class="math-container">$2y'$</span>:
<span class="math-container">$$2y'y''(x)=\frac{2c_1y(x)y'}{c_2+y(x)}$$</span>
<span class="math-container">$$(y'^2)'=\frac{2c_1y(x)y'}{c_2+y(x)}$$</span>
Integrate:
<span class="math-container">$$y'^2+K= 2c_1\int \frac{y(x)}{c_2+y(x)}dy$$</span>
<span class="math-container">$$y'^2+K= 2c_1y-2c_1c_2\int \frac{dy}{c_2+y(x)}$$</span>
<span class="math-container">$$y'^2+K= 2c_1y-2c_1c_2 \ln |{c_2+y(x)}|$$</span>
Unfortunately, it's not easy to integrate.
<span class="math-container">$$y'^2= 2c_1 \left (y-c_2 \ln |{c_2+y(x)}| \right )+C$$</span></p>
|
29,601 | <p>I know the order of the group is the number of elements in the set. For example the group of $U_{10}$ (units of congruence class of 20) has order 4. </p>
<p>Major Edit, kinda changed the question.
Lets say my element $a$ has a finite order $n$. Then what is the order of $a, a^2, a^3...a^{11}$?</p>
| Fredrik Meyer | 4,284 | <p>The order of an element in a group is the order of the subgroup it generates. Equivalently, it is the least integer n such that $a^n$ is the identity. If the order of the group is n, then the order of any element in the group actually divides n (this is Lagranges theorem).</p>
|
223,087 | <p>Given a list of numbers in decimal form, what is the most efficient way to determine if there are any consecutive 1s in the binary forms of those numbers? My solution so far:</p>
<pre><code>dim = 3;
declist = Range[0, 2^dim - 1];
consecutiveOnes[binary_] := AnyTrue[Total /@ Split[binary], # > 1 &];
consecutiveOnes[#] & /@ IntegerDigits[declist, 2]
</code></pre>
<p>which gives <code>{False, False, False, True, False, False, True, True}</code>, in accordance with the binary representations <code>{{0}, {1}, {1, 0}, {1, 1}, {1, 0, 0}, {1, 0, 1}, {1, 1, 0}, {1, 1, 1}}</code>.</p>
<p>For <code>dim=15</code> this takes ~600ms on my machine, which seems a little high, and I just want to see if there is a cleaner way to do it. I've tried using BlockMap with Times but it was much slower.</p>
<p>Two "extras":</p>
<ol>
<li><p>I guess as a comment, it is also acceptable if your method simply returns all decimal numbers up to some max number for which the binary representations have no consecutive 1s. In other words, I'm just going to run <code>Pick</code> on the <code>declist</code> with the negated results of this function, so if your solution just cuts out the middle man, that is great/acceptable.</p></li>
<li><p>I also care about the possibility of "wrapping around", i.e. if the first and last binary digits are both 1s. Obviously I could just append the first digit to the end of each list, but perhaps this is not the most efficient way to proceed.</p></li>
</ol>
<p><strong>Addendum</strong>: Some great solutions! I took the liberty of implementing and speed testing them, with some minor modifications - hopefully I have not distorted your codes too badly:</p>
<pre><code>dim = 15;
declist = Range[0, 2^dim - 1];
m1[range_] :=
FromDigits[#, 2] & /@
DeleteCases[IntegerDigits[range, 2], {___, 1, 1, ___}];
m2helper[num_] := NoneTrue[Total /@ Split[num], # > 1 &];
m2[range_] := Pick[declist, m2helper[#] & /@ IntegerDigits[range, 2]];
m3helper[num_] :=
NestWhile[Quotient[#, 2] &, num, # > 0 && BitAnd[#, 3] != 3 &] > 0
m3[range_] := Pick[declist, Not[m3helper[#]] & /@ range];
m41 = (4^(Ceiling[dim/2]) - 1)/3;
m42 = 2 m41;
m4helper = Function[{n},
Evaluate[
Nor[BitAnd[BitAnd[n, m42], BitShiftLeft[BitAnd[n, m41], 1]] > 0,
BitAnd[BitAnd[n, m42], BitShiftRight[BitAnd[n, m41], 1]] >
0]], {Listable}];
m4[range_] := Pick[declist, m4helper[range]];
Clear[m5];
m5[0] = {0};
m5[1] = {0, 1};
m5[n_?(IntegerQ[#] && # > 1 &)] :=
m5[n] = Join[m5[n - 1], 2^(n - 1) + m5[n - 2]]
m6[range_] :=
Pick[range, Thread[BitAnd[range, BitShiftRight[range, 1]] == 0]];
aa = m1[declist] // RepeatedTiming;
bb = m2[declist] // RepeatedTiming;
cc = m3[declist] // RepeatedTiming;
dd = m4[declist] // RepeatedTiming;
ee = m5[dim] // AbsoluteTiming;
ff = m6[declist] // RepeatedTiming;
Column[{aa[[1]], bb[[1]], cc[[1]], dd[[1]],ee[[1]],ff[[1]]}]
aa[[2]] == bb[[2]] == cc[[2]] == dd[[2]] == ee[[2]]==ff[[2]]
</code></pre>
<p>yields</p>
<pre><code>0.0464
0.619
0.322
0.0974
0.00024
0.0086
True
</code></pre>
<p>So the direct construction method seems clearly the fastest - still this does "skip" the actual pruning step, which is not required for me, but maybe is in other use cases. If the actual pruning list is desired, it seems like the direct <code>BitAnd</code>+<code>BitShiftRight</code> method is fastest, followed by the <code>SelectCases</code>/<code>DeleteCases</code>. But if other people have other methods, certainly share them!</p>
| vi pa | 58,013 | <pre><code>dim = 3;
declist = Range[0, 2^dim - 1];
consecutiveOnes[decimal_]:=If[StringCases["11"][IntegerString[decimal,2]]=={},False,True]
consecutiveOnes/@declist
</code></pre>
<p><a href="https://i.stack.imgur.com/Qp43v.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Qp43v.png" alt="enter image description here"></a></p>
<p><strong>Edit</strong>: A little improvement</p>
<pre><code>consecutiveOnes[decimal_]:=StringMatchQ[IntegerString[decimal,2],___~~"11"~~___]
</code></pre>
|
1,032,874 | <p>Prove that $(1+2+3+\cdots + n)^2 = 1^3+2^3+3^3+\cdots + n^3$ for every $n \in \mathbb{N}$.</p>
<p>Proof. We will use mathematical induction.
If $n = 1$, then we have $(1)^2= 1^3 = 1$. We must show that $S_n$ implies $S_{n+1}$. Assume that for $n \in \mathbb{N}$, the statement $(1+2+3+\cdots + n)^2 = 1^3+2^3+3^3+\cdots + n^3$ holds. Then</p>
<p>$\begin{align} &\left[\displaystyle\sum_{n \in \mathbb{N}}^n n\right]^2+(n+1)^3 \\ =& \displaystyle\sum_{n \in \mathbb{N}}^n n^3+(n+1)^3 \\ =& \displaystyle\sum_{n \in \mathbb{N}}^n n^3 + n^3 + 3n^2 +3n + 1 \end{align}$</p>
<p>How do I finish carrying out the induction? I can't seem to manipulate the right hand side properly get the form $\left[\displaystyle\sum_{n \in \mathbb{N}}^n n+1\right]^2$. </p>
| TonyK | 1,508 | <p>Both sides are polynomials of degree $4$. Both sides take the values $0,1,9,36,100$ for $n=0,1,2,3,4$. So they are the same polynomial!</p>
|
1,032,874 | <p>Prove that $(1+2+3+\cdots + n)^2 = 1^3+2^3+3^3+\cdots + n^3$ for every $n \in \mathbb{N}$.</p>
<p>Proof. We will use mathematical induction.
If $n = 1$, then we have $(1)^2= 1^3 = 1$. We must show that $S_n$ implies $S_{n+1}$. Assume that for $n \in \mathbb{N}$, the statement $(1+2+3+\cdots + n)^2 = 1^3+2^3+3^3+\cdots + n^3$ holds. Then</p>
<p>$\begin{align} &\left[\displaystyle\sum_{n \in \mathbb{N}}^n n\right]^2+(n+1)^3 \\ =& \displaystyle\sum_{n \in \mathbb{N}}^n n^3+(n+1)^3 \\ =& \displaystyle\sum_{n \in \mathbb{N}}^n n^3 + n^3 + 3n^2 +3n + 1 \end{align}$</p>
<p>How do I finish carrying out the induction? I can't seem to manipulate the right hand side properly get the form $\left[\displaystyle\sum_{n \in \mathbb{N}}^n n+1\right]^2$. </p>
| Martín-Blas Pérez Pinilla | 98,199 | <p>Interesting graphical proof with applet in <a href="https://www.geogebratube.org/student/m37497" rel="nofollow">The Sum of the Cubes of the First N Natural Numbers</a>.</p>
|
2,668,538 | <p>Firstly I'm aware of the proofs/reasons regarding .999...=1. I'm not asking for anyone to reference or reiterate them but rather to look at my proof in isolation and help me understand my own mistakes and fallacies.</p>
<p>Another disclaimer I suppose; it's difficult to call this 'proof' my 'own' as it's extremely obvious and simple. Nonetheless I can't find the inconsistency in it.</p>
<p>A 'proof' .999... =/= 1</p>
<p>Suppose .999... = 1 then there is some number x = 1 / .999... and x should obviously be 1. If we take the values for x in the equation x = 1 / .999.. starting with x = 1 / .9 and as x approaches 1 / .999... we get a value for x where x =/= 1. x equals 1.01 (with a repeat bar over the zero and then an additional repeat bar over the zero and one together - I don't know how to write a double repeat bar like that) I'm also going to call the value .01 repeating bar over zero with an additional repeat bar over both the zero and one ε for simplicity's sake.</p>
<p>Reason: 1 / .9 = 1.1 (repeating) 1 / .99 = 1.01 (repeating) 1 / .999 = 1. 001 (repeating) etc.</p>
<p>When you reach an infinite series of .9's the zeros from the value of 1 / .999... becomes infinite before reaching the first 1 then when you reach that first 1 another series of infinite zeroes occurs before another 1 and then that pattern repeats infinitely.</p>
<p>I then find the value for x for x = 1 / .999... is actually x = 1 + ε (the value I defined as the .01 double repeating bars I referred to earlier)</p>
<p>Subtracting 1 from both sides from the equation 1 + ε = 1 / .999... gives the literal value for what I'm referring to as the infinitesimal; ε = .01 (repeating bar over zero then additional repeat bar over both zero and one)</p>
<p>So then the equation ends up not being .999... = 1 but rather .999... + ε = 1.</p>
<p>If I do some basic checks I find ε = 1 - .999... to be true and I have to show I can derive .999... + ε = 1 from 1 + ε = 1 / .999... and due to ε + ε = ε (#) I can show that 1*.999.. + ε*.999... = 1 = .999... + ε</p>
<p>(#) I'm suggesting that in the infinite series ε 'adding' any more of the exact same ε to that infinite series doesn't change the 'value' of that infinite series, you still end up with the same infinite number of the exact same steps. Thus ε * .999... = ε as well as ε * <em>n</em> = ε</p>
<p>So there you have it, 'my' 'proof' .999... + ε = 1 and that ε = .01 repeating bar over zero with an additional repeating bar over both the zero and one.</p>
<p>I fail to find where it's incorrect, any help would be appreciated! And again I'm aware of the reasons for .999...=1 so please don't just simply reiterate them, I'd like to know where I went wrong in the logic of what I wrote, thank you!</p>
| Andrew Li | 344,419 | <p>The very assertion that:</p>
<p>$$\dfrac{1}{0.\bar 9} = 1 + 0.\bar 01$$</p>
<p>Proves that $1 = 0.\bar 9$. This is because:</p>
<p>$$0.\bar 01 = 0$$</p>
<p>Conceptually speaking, there is no way to stick a $1$ at the end of an infinite number of $0$s because there is no end to an infinite number of $0$s. There will just be more $0$s—by definition an infinite number of them, before you ever reach a one. You can't write an infinite amount of $0$s then write a one—then you wouldn't have an infinite amount of $0$s! Thus there <em>is no "final" $1$</em> so $0.\bar 01$ is just an infinite number of $0$s which is $0$.</p>
<p>The same exactly logic applies to the problem of $1-0.\bar 9$. You end up with your result of $0.\bar 01$ again, which by the same reasoning is $0$. Thus you can again conclude that since the difference between $1$ and $0.\bar 9$ is $0$ that $1 = 0.\bar 9$.</p>
|
4,630,145 | <p>We need to determine all <span class="math-container">$\mathbb{Q}$</span>-homomorphisms from <span class="math-container">$\mathbb{Q}(\sqrt[5]2)$</span> to <span class="math-container">$\mathbb{Q}(\sqrt[5]2, \omega)$</span>, where <span class="math-container">$\omega=e^{2\pi i/5}$</span>.</p>
<p>I can figure it out that there are 5 injective homomorphisms because there are <span class="math-container">$5$</span> roots of <span class="math-container">$x^5-2$</span> in <span class="math-container">$\mathbb{Q}(\sqrt[5]2, \omega)$</span>.</p>
<p>But how do I proceed after this to find all of them?</p>
| simo210 | 1,142,316 | <p>First, notice that <span class="math-container">$\mathbb{Q}(\sqrt[5]{2},\omega)$</span> is the splitting field over <span class="math-container">$\mathbb{Q}$</span> of the polynomial <span class="math-container">$f(X)=X^5-2$</span>.</p>
<p>In general, it's a well known fact that the set of <span class="math-container">$\mathbb{Q}$</span>-homomorphisms from a simple, algebraic extension <span class="math-container">$\mathbb{Q}(\alpha)$</span> to <span class="math-container">$\mathbb{C}$</span> is in bijection with the set of complex roots of <span class="math-container">$f_{\alpha,\mathbb{Q}}$</span> minimal polynomial of <span class="math-container">$\alpha$</span> over <span class="math-container">$\mathbb{Q}$</span>, and the bijection is given by sending <span class="math-container">$\sigma\space{\in}\space{Hom_{\mathbb{Q}}(\mathbb{Q}(\alpha),\mathbb{C})}$</span> to <span class="math-container">$\sigma(\alpha)$</span>.</p>
<p>Making this explicitly : you have a bijection from <span class="math-container">$Hom_{\mathbb{Q}}(\space\mathbb{Q}(\sqrt[5]{2}),\space \mathbb{Q}(\sqrt[5]{2},\omega)\space)$</span> to <span class="math-container">$\{\sqrt[5]{2},\omega\sqrt[5]{2},\omega^2\sqrt[5]{2},\omega^3\sqrt[5]{2},\omega^4\sqrt[5]{2}\}$</span> roots of <span class="math-container">$f$</span> over <span class="math-container">$\mathbb{C}$</span>. Thus, you get five homomorphisms, each of them determined by the image of <span class="math-container">$\sqrt[5]{2}$</span>, generator of <span class="math-container">$\mathbb{Q}(\sqrt[5]{2})$</span>.</p>
<p>Still, you'd get the same result by requesting only homomorphism of field from <span class="math-container">$\mathbb{Q}(\sqrt[5]{2})$</span> to <span class="math-container">$\mathbb{Q}(\sqrt[5]{2},\omega)$</span>. In fact, for every homorphism <span class="math-container">$\sigma:\mathbb{Q}(\sqrt[5]{2})\longrightarrow \mathbb{Q}(\sqrt[5]{2},\omega)$</span>, you have <span class="math-container">$\sigma(1)=1$</span> implies <span class="math-container">$\left.\sigma\right|_\mathbb{Q}=id_\mathbb{Q}$</span>.</p>
|
1,948,163 | <p>I don't know how to differentiate this function. $$u=\left(\frac{1}{t}-\frac{1}{\sqrt{t}}\right)^{2}$$</p>
<p>Should I use the quotient rule or just the power chain rule?</p>
| yngabl | 111,900 | <p>You want $\frac{du}{dt}$ so set $$p=\frac{1}{t}-\frac{1}{\sqrt{t}}$$ then $u=p^2$ and $\frac{du}{dt}=2p\frac{dp}{dt}$. Calculate $\frac{dp}{dt}$, put back with $p=\sqrt{u}$ and you will find the answer.</p>
|
725,746 | <p>I am not sure I understand the $N - \epsilon$ method for proving the equality of a limit.</p>
<p>I have a past mid-semester exam question that has:
$$\lim \limits_{x \to 1} (x^2 - 4x) = -3$$</p>
<p>Now it seems I want to take the $-3$ over $\rightarrow$ $|x^2 - 4x + 3| \lt \epsilon$ $$\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }|x-3||x-1| \lt \epsilon$$</p>
<p>I now want to set $n \geq N$ where $N \in \mathbb{N}$ is some function $N(\epsilon)$</p>
<p>I am unsure how to continue this problem. There is only one type of question that I know how to approach(given that it is all that the lecturer has went over), which is questions with limits of fractions, where N = $\frac{1}{\epsilon}$ so when I sub N in for n, $\epsilon$ goes to the top.</p>
<p>How would one approach these problems in general?</p>
| DeepSea | 101,504 | <p>You want to estimate $|x - 3|$. One way to get an upper bound for this is to use the triangle inequality: $|x - 3| =|(x - 1) + 2| \longleftarrow |x - 1| + |2| = |x - 1| + 2$. So if you let $|x - 1| \lt 1$, then $|x - 3| \lt 1 + 2 = 3$. So $|x - 3| \cdot |x - 1| \lt 3|x - 1|$. And you need $3|x - 1| \lt e$. All you need is once more that : $|x - 1| \lt \frac{e}{3}$. So this gives you a way to choose $d$. And you see that $d =$ min{$1, \frac{e}{3}$} fits the bill. You're done. </p>
|
725,746 | <p>I am not sure I understand the $N - \epsilon$ method for proving the equality of a limit.</p>
<p>I have a past mid-semester exam question that has:
$$\lim \limits_{x \to 1} (x^2 - 4x) = -3$$</p>
<p>Now it seems I want to take the $-3$ over $\rightarrow$ $|x^2 - 4x + 3| \lt \epsilon$ $$\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }|x-3||x-1| \lt \epsilon$$</p>
<p>I now want to set $n \geq N$ where $N \in \mathbb{N}$ is some function $N(\epsilon)$</p>
<p>I am unsure how to continue this problem. There is only one type of question that I know how to approach(given that it is all that the lecturer has went over), which is questions with limits of fractions, where N = $\frac{1}{\epsilon}$ so when I sub N in for n, $\epsilon$ goes to the top.</p>
<p>How would one approach these problems in general?</p>
| davidlowryduda | 9,754 | <p>Continue reasoning. The values of $x$ you are restricting to are somewhere near $1$. Let's say now and forever that we will demand the distance from $1$, which I'll call $\delta$, is at most $1$. (I'm not quite sure what you mean by $N$ - I suspect you would call my $\delta = \frac{1}{N}$). I do this so that I can say that $|x-3| < 3$ for all $x$ in my $\delta$ ball around $1$.</p>
<p>So then we're reduced to $3|x-1| < \epsilon$ being our goal, or rather $|x - 1| < \frac{\epsilon}{3}$. As we control exactly how far $x$ can be from $1$, we see that as long as $\delta < \frac{\epsilon}{3}$ (or is $1$ if $\epsilon$ is very large for some reason), we get the desired proof of the limit.</p>
|
3,525,017 | <p>a) Show that the generating function by length for binary strings where every block of 0s has length at least 2, each block of ones has length at least 3 is:</p>
<p><span class="math-container">$$\frac{(1-x+x^3)(1-x+x^2)}{1-2x+x^2-x^5}$$</span></p>
<p>b) Give a recurrence relation and enough initial conditions to determine coefficients of power series.</p>
<p>So for a), I came up with the block decomposition <span class="math-container">$((0^*00)^*(1^*111)^*)^*$</span> and found the generating function, using the fact that <span class="math-container">$0\leadsto x$</span>, <span class="math-container">$1\leadsto x$</span>, and <span class="math-container">$a^*\leadsto\frac{1}{1-a}$</span> where a is some binary string:</p>
<p><span class="math-container">$$\frac{(1-x)^2}{(1-x-x^2)(1-x-x^3)}$$</span></p>
<p>which, clearly, does not equal the expected result. Could someone clear up for me where I went wrong?</p>
<p>Also, for b), how would you find a recurrence relation, since the degree of the numerator and denominator are the same, so there would be no general <span class="math-container">$a_n$</span> term.</p>
<p>Thanks in advance for any help!</p>
| RobPratt | 683,666 | <p>Let <span class="math-container">$a_n$</span> be the number of strings that start with <span class="math-container">$0$</span>, and let <span class="math-container">$b_n$</span> be the number of strings that start with <span class="math-container">$1$</span>. Then <span class="math-container">$a_0=b_0=1$</span>, <span class="math-container">$a_1=0$</span>, and, by conditioning on the length <span class="math-container">$k$</span> of the current run, we see that
<span class="math-container">\begin{align}
a_n &= \sum_{k=2}^n b_{n-k} &&\text{for $n \ge 2$}\\
b_n &= \sum_{k=1}^n a_{n-k} &&\text{for $n \ge 1$}.
\end{align}</span>
Let <span class="math-container">$A(z)=\sum_{n=0}^\infty a_n z^n$</span> and <span class="math-container">$B(z)=\sum_{n=0}^\infty b_n z^n$</span>. Then the recurrence relations imply
<span class="math-container">\begin{align}
A(z)-a_0 -a_1z &=\frac{z^2}{1-z} B(z) \\
B(z)-b_0 &=\frac{z}{1-z} A(z)
\end{align}</span>
Solving for <span class="math-container">$A(z)$</span> and <span class="math-container">$B(z)$</span> yields
<span class="math-container">\begin{align}
A(z) &= \frac{1-2z+2z^2-z^3}{1-2z+z^2-z^3}\\
B(z) &= \frac{1-z}{1-2z+z^2-z^3}
\end{align}</span>
So the desired generating function (subtracting the <span class="math-container">$1z^0=1$</span> term for the empty string that is otherwise counted twice) is
<span class="math-container">$$A(z)+B(z)-1=\frac{1-z+z^2}{1-2z+z^2-z^3}.$$</span></p>
|
95,341 | <p>I know that the complement of the zero set of a polynomial $P: \mathbb{C}^n \rightarrow \mathbb{C}$ is connected in $\mathbb{C}^n$ (by the way, can anybody suggest a reference?).</p>
<p>Is it possible to extend the proof also to polynomials
$P: SL(N,\mathbb{C})^n \rightarrow \mathbb{C}$ ?</p>
<p>Thanks!</p>
| Marc Palm | 10,400 | <p>Let $X_{j}$ be elements of $SL(n, \mathbb{C})$ with entries ${X_{j}^{k,l}}$.</p>
<p>Express the polynomial $P: SL(n,\mathbb{C})^N \rightarrow \mathbb{C}$ as a polynomial $Q: \mathbb{C}^{n \cdot N} \rightarrow \mathbb{C}$ in variables ${X_{j}^{k,l}}$</p>
<p>Similar express the polynomial
$$ Q''(X_1, \cdots, X_n) = \prod\limits_{i=1}^N \left( \det(X_i) -1 \right)$$
in variables ${X_{j}^{k,l}}$. </p>
<p>You can then apply the well-known result for $\mathbb{C}^{N \cdot n^2}$ to $Q' \cdot Q''$, and get what you want.</p>
<p>Edit due the comment: Note that we have both $$\{ Q' Q'' = 0 \} \subset \mathbb{C}^{N \cdot n^2}$$ is a connected subset, and that it is really a subset of $SL_n(\mathbb{C})^N$
$$\{ Q' Q'' = 0 \} \subset \{ Q'' =0 \} = SL_n(\mathbb{C})^N.$$</p>
<p>I am not sure, what topology you are working in, but it holds in any topology, the original results for $\mathbb{C}^N$ holds in;)</p>
|
1,468,070 | <p>Let <span class="math-container">$\{a_n\}_1^\infty$</span> and <span class="math-container">$\{b_n\}_1^\infty$</span> be two sequences in <span class="math-container">$\mathbb{R}$</span> such that <span class="math-container">$\forall n \in \mathbb{N}$</span>, it is true that <span class="math-container">$a_n \leq b_n, a_n \leq a_{n+1}, \text{ and } b_{n+1} \leq b_n$</span>.</p>
<p>We want to show <span class="math-container">$\forall m,n \in \mathbb{N}$</span> it is true that <span class="math-container">$a_m \leq a_n$</span> and that there is a number <span class="math-container">$r \in \mathbb{R}$</span> such that <span class="math-container">$a_m \leq r \leq b_n$</span>.</p>
<p>I've proceeding as follows:</p>
<p>We have <span class="math-container">$a_{n} \leq b_{n} \implies a_{n+1} \leq b_{n+1}$</span> and thus <span class="math-container">$a_n \leq a_{n+1} \leq b_{n+1} \leq b_n$</span>.</p>
<p>Does this not imply that <span class="math-container">$a_m \leq a_n$</span>? Even without stating the obvious fact that the sets are upper and lower bounds of each other? It seems then that <span class="math-container">$r$</span> would follow..</p>
<p><strong>EDIT</strong>: Taking some of the ideas from below I have written a simple proof. Feedback is welcome and appreciated.</p>
<p>Since <span class="math-container">$a$</span> is monotonically increasing and <span class="math-container">$b$</span> is monotonically decreasing we have <span class="math-container">$\forall m,n \in \mathbb{N}, a_n \leq a_{\max(m,n)} \leq b_{\max(m,n)} \leq b_n \implies a_m \leq b_n$</span>. Take <span class="math-container">$r = a_{\max(m,n)} \text{ or } r = b_{\max(m,n)} \implies a_m \leq r \leq b_n$</span>.</p>
| Kevin Sheng | 150,297 | <p>You are asked to show $a_m \le r \le b_n$. Notice that the indices must be different, the result you give has the same index (i.e. $ a_n
\le r \le b_n$). Given $m,n \in \mathbb{N}$, we have without loss of generality $m\le n$. Then $a_m \le a_n \le b_n$ by your hypothesis, take $r=a_n$ and you are done.</p>
|
27,759 | <p>Suppose $N$ is an RSA modulus (ie, it's the product of two distinct primes), 256 bits long. What is the best method to factor it?</p>
<p>Trial division is out of the question, Pollard's Rho is probably out as well (without significant parallelization). I doubt there are any online tools or math libraries that can handle this number (I think Wolfram Alpha uses Pollard's Rho algorithm).</p>
<p>Moduli up to 768 bits have been factored, and RSA Corp's (now defunct) challenge list doesn't even address numbers as small as 256 bits, so it must be pretty easy... but how?</p>
| Jonas Meyer | 1,424 | <p>This would be easy for a diagonal matrix, because $\begin{bmatrix}a & 0\\ 0 & b\end{bmatrix}^3=\begin{bmatrix}a^3 & 0\\ 0 & b^3\end{bmatrix}$, which means that you could just take the cube root of each diagonal entry to solve the problem. While $A$ is not diagonal, it is symmetric and therefore diagonalizable. If you're comfortable with diagonalizing, find $S$ such that $SAS^{-1}=\begin{bmatrix}a & 0 \\ 0 & b \end{bmatrix}$. You know how to find a matrix whose cube gives the right hand side. Then notice how conjugation behaves with cubing: $(S^{-1}C'S)^3=S^{-1}C'^3S$. Therefore, you can take $C=S^{-1}\begin{bmatrix}\sqrt[3]{a} & 0 \\ 0 & \sqrt[3]{b} \end{bmatrix}S$. </p>
<p>(I was going to write more involving eigenvectors, but then other answers were posted covering this.)</p>
<p>An alternative approach using polynomial interpolation will work for all diagonalizable matrices having eigenvalues $-1$ and $5$, and does not require finding eigenvectors. For more on this and generalizations, see Chapter 1 of Higham's <em><a href="http://books.google.com/books?id=S6gpNn1JmbgC&dq" rel="nofollow">Functions of matrices</a></em>.</p>
<p>In this case, the <a href="http://en.wikipedia.org/wiki/Lagrange_polynomial" rel="nofollow">Lagrange interpolating polynomial</a> of the cube root function on the spectrum $\{5,-1\}$ of $A$ is $$p(t)=\sqrt[3]{5}\cdot\frac{t+1}{5+1}+\sqrt[3]{-1}\cdot\frac{t-5}{-1-5}=\frac{\sqrt[3]{5}+1}{6}\cdot t +\frac{\sqrt[3]{5}-5}{6},$$ so that a matrix cube root for $A$ can be obtained as $$p(A)=\frac{\sqrt[3]{5}+1}{6}\cdot A +\frac{\sqrt[3]{5}-5}{6}\cdot I=\frac{1}{2}\begin{bmatrix}\sqrt[3]5-1&\sqrt[3]5+1\\\sqrt[3]5+1&\sqrt[3]5-1\end{bmatrix}.$$</p>
|
276,818 | <p>I am attempting to numerically optimize a function of only 2 parameters <code>k, \[Theta]</code>.</p>
<p>The function is well-defined and the constraints are simple. However, the optimisation keeps returning <code>Indeterminate</code> or <code>not a real number</code> comment over the parameter space, i.e. <code>function value not a real number at {k,\[Theta]} = {0.918621,0.716689}</code>.</p>
<p>If I instead initialize {k,[Theta]} to these values and run the function, it does provide a value. I am unsure why the optimization fails. The <em>MWE</em> is provided below. Some things I have noticed is that the function value <code>val</code> can return some really small imaginary components that should be zero. If I force the value to be real via either <code>Chop</code> or <code>Re</code>, I keep getting an <code>Indeterminate</code> value.</p>
<p>How can I get around this?</p>
<p><strong>MWE:</strong></p>
<pre><code>s[0 | 0.] = 0;
s[x_] = x Log2[x];
SetAttributes[s, Listable]
A = {{0.0632, 0, 0, 0}, {0, 0.3065, 0, 0}, {0, 0, 0.0632, 0}, {0, 0, 0, 0.3065}};
B = {{0.3065, 0, 0, 0}, {0, 0.0632, 0, 0}, {0, 0, 0.3065, 0}, {0, 0, 0, 0.0632}};
T = {{2 k Cos[\[Theta]]^2 + k + 1, 0, 0, 6 k Cos[\[Theta]]^2 - k - 1}, {0, 2 k Sin[\[Theta]]^2 + 3 - 3 k, 0, 0}, {0, 0, 2 k Sin[\[Theta]]^2 + 3 - 3 k, 0}, {6 k Cos[\[Theta]]^2 - k - 1, 0, 0, 2 k Cos[\[Theta]]^2 + k + 1}};
C1 = A.T.A;
C2 = A.T.B;
C3 = B.T.A;
C4 = B.T.B;
C5 = KroneckerProduct[{{1, 0}, {0, 0}}, C1] + KroneckerProduct[{{0, 1}, {0, 0}}, C2] + KroneckerProduct[{{0, 0}, {1, 0}}, C3] + KroneckerProduct[{{0, 0}, {0, 1}}, C4];
val = Total[s[Eigenvalues[C5]]] - Total[s[Eigenvalues[C1]]] - Total[s[Eigenvalues[C4]]];
</code></pre>
<p>I then try to optimize <code>val</code> using:</p>
<pre><code>NMinimize[{val, 0<=k<=1, 0<\[Theta]<=2\[Pi]}, {k, \[Theta]}]
</code></pre>
<p>which does not work. Even <code>val /. {k->0.5, \[Theta]->0.1}</code> fails. What is the workaround?</p>
| Nasser | 70 | <p>You can't obtain the recurrence formula with V 13.1</p>
<p>Looking at "Q&A with Calculus Developers: Live with the R&D team" Nov 30, 2022 at
<a href="https://www.twitch.tv/videos/1666674721" rel="nofollow noreferrer">https://www.twitch.tv/videos/1666674721</a> for V 13.2, I did not see this feature being there (see around 7 minutes frame). Not even Maple can do this for series solutions. i.e. give explicit recurrence formula used for series solution.</p>
<p>I have a function myself that does this (but still in development) and it only handle linear 1st and second order odes'. For your ode, this is the recurrence formula it found</p>
<p><span class="math-container">$n=0$</span> gives
<span class="math-container">$$
a_{2} = -\frac{a_{1}}{2}
$$</span>
<span class="math-container">$n=1$</span> gives
<span class="math-container">$$
a_{3} = -\frac{a_{0}}{6}-\frac{2 a_{1}}{3}
$$</span>
<span class="math-container">$n=2$</span> gives
<span class="math-container">$$
a_{4} = \frac{a_{0}}{24}+\frac{a_{1}}{4}
$$</span>
For <span class="math-container">$3\le n$</span>, the recurrence equation is
<span class="math-container">$$
a_{n +2}= -\frac{5 n a_{n}}{\left(n +2\right) \left(n +1\right)}-\frac{\left(5 n -10\right) a_{n -2}}{\left(n +2\right) \left(n +1\right)}-\frac{\left(3 n -2\right) a_{n -1}}{\left(n +2\right) \left(n +1\right)}-\frac{a_{n +1}}{n +2}
$$</span></p>
<p>Now you are able to generate very fast all terms you want. Here is the Mathematica code</p>
<pre><code>a[2] = -a[1]/2
a[3] = -a[0]/6 - 2 a[1]/3
a[4] = a[0]/24 + a[1]/4
Table[a[n + 2] = -5*n*a[n]/((n + 2)*(n + 1)) - (5*n - 10)*
a[n - 2]/((n + 2)*(n + 1)) - (3*n - 2)*
a[n - 1]/((n + 2)*(n + 1)) - a[n + 1]/(n + 2), {n, 3, 10}];
sol = Sum[a[n]*x^n, {n, 0, 10}];
Collect[sol, {a[0], a[1]}]
</code></pre>
<p><img src="https://i.stack.imgur.com/vdTFj.png" alt="Mathematica graphics" /></p>
<p>Compare to</p>
<pre><code>AsymptoticDSolveValue[{y''[x] + (5 x^3 + 3 x^2 + 5 x + 1) y'[x] +
x y[x] == 0}, y[x], {x, 0, 10}]
</code></pre>
<p><img src="https://i.stack.imgur.com/GSoEe.png" alt="Mathematica graphics" /></p>
<p>Note that I generates all the <span class="math-container">$a_n$</span> using normal <code>Table</code> command. You could try to use <code>RecurrenceTable</code> instead to see if performance is better.</p>
<p>You can now solve for <span class="math-container">$a_0,a_1$</span> easily given the initial conditions.</p>
|
31,539 | <p>I want to learn a bit about Linear Programming. </p>
<p>After some research, I decided to solve the <a href="http://en.wikipedia.org/wiki/Cutting_stock_problem" rel="nofollow">Cutting Stock</a> problem as an example to learn. After doing some more research, I feel like I finally understand Linear Programming enough to use a 3rd party solver to solve the problem. Yet, I've only found a suitable solver for non-integer solutions. Using this type of solver is it possible to solve the tighter integer constraints problem? </p>
<p>If so, can someone please post some resources that I can use to read and perhaps implement an integer solver using an existing solver. </p>
<p>My understanding of this topic is EXTREMELY low, and I'm simply looking to learn more about it, so if the problem as I have stated it doesn't make sense, please let me know.</p>
<p>Thanks. </p>
| Timothy Chow | 3,106 | <p>I'd recommend you start by consulting the <a href="http://www.neos-guide.org/content/lp-faq" rel="nofollow">Linear Programming FAQ</a>.</p>
<p>As for a solver, I'd recommend <a href="http://www.gnu.org/software/glpk/" rel="nofollow">GLPK</a>, which is free and can handle integer linear programming.</p>
|
513,239 | <p>I've recently heard a riddle, which looks quite simple, but I can't solve it.</p>
<blockquote>
<p>A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "<em>Yes</em>", "<em>No</em>", or "<em>I don't know</em>," and after the girl answers it, he knows what the number is. What is the question?</p>
</blockquote>
<p>Note that the girl is professional in maths and knows EVERYTHING about these three numbers.</p>
<hr>
<p><strong>EDIT:</strong> The person who told me this just said the correct answer is:</p>
<blockquote class="spoiler">
<p> "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"</p>
</blockquote>
| alex | 98,428 | <p>Is it correct that your number equals 1 or it equals 2 and {any question she don't know answer for}?</p>
<p>Yep, it's a boring answer, but it always works for such kind of problems.</p>
|
513,239 | <p>I've recently heard a riddle, which looks quite simple, but I can't solve it.</p>
<blockquote>
<p>A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "<em>Yes</em>", "<em>No</em>", or "<em>I don't know</em>," and after the girl answers it, he knows what the number is. What is the question?</p>
</blockquote>
<p>Note that the girl is professional in maths and knows EVERYTHING about these three numbers.</p>
<hr>
<p><strong>EDIT:</strong> The person who told me this just said the correct answer is:</p>
<blockquote class="spoiler">
<p> "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"</p>
</blockquote>
| PearsonArtPhoto | 5,101 | <p>Most of the answers seem to rely on relatively complex mathematics, that I couldn't easily do in my head. I'm hoping this is the simpliest answer out there, or at least, one of the simplest.</p>
<p>If I take your number, and subtract 2 from it, then take the reciprocal, is it positive? In other words: $\frac{1}{x-2}$</p>
<ul>
<li>Yes- Number must have been 3</li>
<li>No- Number must have been 1.</li>
<li>I don't know- Is 1/0 positive or negative? It could be either, and thus I Don't Know is the appropriate answer.</li>
</ul>
|
513,239 | <p>I've recently heard a riddle, which looks quite simple, but I can't solve it.</p>
<blockquote>
<p>A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "<em>Yes</em>", "<em>No</em>", or "<em>I don't know</em>," and after the girl answers it, he knows what the number is. What is the question?</p>
</blockquote>
<p>Note that the girl is professional in maths and knows EVERYTHING about these three numbers.</p>
<hr>
<p><strong>EDIT:</strong> The person who told me this just said the correct answer is:</p>
<blockquote class="spoiler">
<p> "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"</p>
</blockquote>
| Kostya | 28,158 | <p>How about this one:</p>
<blockquote>
<p>Let's say your number is $n$.<br>
For every even number $x$ ($x>2$) is that true that $x+n$ is representable as a sum of $n$ primes?</p>
</blockquote>
|
513,239 | <p>I've recently heard a riddle, which looks quite simple, but I can't solve it.</p>
<blockquote>
<p>A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "<em>Yes</em>", "<em>No</em>", or "<em>I don't know</em>," and after the girl answers it, he knows what the number is. What is the question?</p>
</blockquote>
<p>Note that the girl is professional in maths and knows EVERYTHING about these three numbers.</p>
<hr>
<p><strong>EDIT:</strong> The person who told me this just said the correct answer is:</p>
<blockquote class="spoiler">
<p> "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"</p>
</blockquote>
| Xiaochuan | 5,476 | <p>The boy gives the three numbers different names: "Yes", "No" and "I don't know". So the question is, "what is the name of your number?"</p>
|
513,239 | <p>I've recently heard a riddle, which looks quite simple, but I can't solve it.</p>
<blockquote>
<p>A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "<em>Yes</em>", "<em>No</em>", or "<em>I don't know</em>," and after the girl answers it, he knows what the number is. What is the question?</p>
</blockquote>
<p>Note that the girl is professional in maths and knows EVERYTHING about these three numbers.</p>
<hr>
<p><strong>EDIT:</strong> The person who told me this just said the correct answer is:</p>
<blockquote class="spoiler">
<p> "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"</p>
</blockquote>
| chubakueno | 37,844 | <p>I am thinking of a positive integer. Is your number, raised to my number and then increased in $1$, a prime number?</p>
<p>$$1^n+1=2\rightarrow \text{Yes}$$
$$2^n+1=\text{possible fermat number}\rightarrow \text{I don't know}$$
$$3^n+1=2k\rightarrow \text{No}$$</p>
|
513,239 | <p>I've recently heard a riddle, which looks quite simple, but I can't solve it.</p>
<blockquote>
<p>A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "<em>Yes</em>", "<em>No</em>", or "<em>I don't know</em>," and after the girl answers it, he knows what the number is. What is the question?</p>
</blockquote>
<p>Note that the girl is professional in maths and knows EVERYTHING about these three numbers.</p>
<hr>
<p><strong>EDIT:</strong> The person who told me this just said the correct answer is:</p>
<blockquote class="spoiler">
<p> "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"</p>
</blockquote>
| Terry Li | 60,889 | <p>Assume the girl's number is X.</p>
<p>The boy asks:</p>
<blockquote>
<p>Is half day before OP posted this question past October X?</p>
</blockquote>
|
513,239 | <p>I've recently heard a riddle, which looks quite simple, but I can't solve it.</p>
<blockquote>
<p>A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "<em>Yes</em>", "<em>No</em>", or "<em>I don't know</em>," and after the girl answers it, he knows what the number is. What is the question?</p>
</blockquote>
<p>Note that the girl is professional in maths and knows EVERYTHING about these three numbers.</p>
<hr>
<p><strong>EDIT:</strong> The person who told me this just said the correct answer is:</p>
<blockquote class="spoiler">
<p> "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"</p>
</blockquote>
| Rigel | 189,549 | <p>Divide the other two numbers left to get a proper fraction. Ask her to subtract 1 from numerator and 2 from denominator. Ask her if the 'number' obtained is greater than 0.</p>
<p>If yes. She should have had 1 and thought of number 1.
If she says she doesn't know. The number she thought of must be 3
If no. It should be 2</p>
|
513,239 | <p>I've recently heard a riddle, which looks quite simple, but I can't solve it.</p>
<blockquote>
<p>A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "<em>Yes</em>", "<em>No</em>", or "<em>I don't know</em>," and after the girl answers it, he knows what the number is. What is the question?</p>
</blockquote>
<p>Note that the girl is professional in maths and knows EVERYTHING about these three numbers.</p>
<hr>
<p><strong>EDIT:</strong> The person who told me this just said the correct answer is:</p>
<blockquote class="spoiler">
<p> "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"</p>
</blockquote>
| Jack D'Aurizio | 44,121 | <p>Let $k$ be your your number. Is there a $3k$-colouring of the elements of $\mathbb{R}^2$ such that points having a unit distance have different colours?</p>
<p><em>Yes</em>$\,\to k=3.\quad$<em>No</em>$\,\to k=1.\quad$<em>I don't know</em>$\,\to k=2.$</p>
<p>Have a look at <a href="https://www.cut-the-knot.org/proofs/ChromaticNumber.shtml" rel="nofollow noreferrer">the chromatic number of the plane</a>.</p>
|
910,414 | <p>My question is about visualizing projective space, in particular the real projective plane $\mathbb{P}^2(\mathbb{R})$. I know there are different ways to define this space, but in each we can say that "two parallel lines intersect." If you type projective space into Wikipedia, or look it up in a textbook, a lot of the time you will see an image of two train tracks. The tracks are parallel, but as they go off into the distance (towards infinity) they appear to be approaching one another. </p>
<p>I believe I understand why we can say that two lines which are parallel in $\mathbb{R}^3$ intersect in $\mathbb{P}^2(\mathbb{R})$ (at least with the equivalence class construnction) but I do not see how this is related to that visual image of the two tracks appearing to approach a common point. </p>
<p>Maybe it is naive, but my question is: why do parallel lines, when looked at as in the image of the train tracks, appear to approach a common point? What is the mathematical reason for this?</p>
| Bananeen | 165,356 | <p>The mathematical reason for this is like this:
suppose you look at an infinitely large plane from a point 1 meter above that plane and you have two infinitely long parallel lines drawn on it. Further, suppose the distance between lines is also 1 meter. Then, when you look at these lines right under your point of view the light has to travel 1 meter from each of the lines to your eyes. Then if you start looking at the lines far off, the light has to travel for example 1 km from each line to your eyes. Therefore, in the first case you have an almost equiangular triangle with sides of 1 meter, in the second case you have an isosceles triangle with sides of 1 km and base of 1 meter. In this case points at the base seem to be almost the same, so we perceive the lines to "start intersecting" when we follow them to "infinity".
If that is confusing I would try to draw some pictures. Hope that helps</p>
|
174,339 | <p>This one is somewhat hard to explain, but I'll try my best to.<br>
I'm trying to generate a list of numbers (containing only pi/10, 0, -pi/10). Numbers are randomly selected from these 3, where the probability of getting 0 is always 70%. But the probability of getting pi/10 and -pi/10 depend on the previous outcomes. </p>
<p>Here's I'm going to explain it how it depends:<br/></p>
<p>Initially, the probability of getting pi/10 is 25%, and of getting -pi/10 is 5%.
But once I get -pi/10 as an outcome, probabilities will be switched. (i.e., the probability of getting pi/10 will be 5%, and of getting -pi/10 will be 25%.)<br/> Similarly, when pi/10 comes next comes as an outcome, probabilities will again get switched and so on...</p>
<p>So, here's what I've tried, but didn't get the desired result.</p>
<pre><code>rr := RandomReal[{0, 1}]
x1 = \[Piecewise]{{0, # < 0.7}, {-\[Pi]/10, 0.7 <= # < 0.75}, {\ [Pi]/10,True}} &@rr
x2 = \[Piecewise]{{0, # < 0.7}, {\[Pi]/10, 0.7 <= # < 0.75}, {-\[Pi]/10, True}} &@rr
</code></pre>
<p>And then I run a for-loop:<br/>
(Here, where I have used which function, my 3rd condition is when I get 0, and in that case, probabilites don't get switched, so it remains the previous "a". If you have a better way to that also, then that'll be great.)</p>
<pre><code>list{}
For[i = 1 ; a = 1, i <= 100, i++;
If[a == 0, x = x1, x = x2];
a = Which[x == \[Pi]/10, 0, x == -\[Pi]/10, 1, True, a];
AppendTo[list, x]]
list
</code></pre>
<p>Your help will be really appreciated!!</p>
<p>P.S.: There might be some syntax errors, if you could point that out too, I'll be thankful to you (as I'm new to Mathematica).</p>
| mef | 449 | <p>This answer relies on the built-in function $\texttt{DiscreteMarkovProcess}$. In order to use it, the states of the world must be specified. A state is composed of two items: the current value and the most recent nonzero value. (Again I use {-1, 0, 1} for the values instead of {-Pi/10, 0, Pi/10}.) </p>
<p>There are four possible states</p>
<pre><code>State: {current, most recent nonzero}
1 {-1, -1}
2 { 0, -1}
3 { 0, 1}
4 { 1, 1}
</code></pre>
<p>Note that third state cannot be reached directly from either of the first two states. Similarly, the second state cannot be reached from either of the last two states. </p>
<p>Let us construct the transition matrix and setup the Markov process object:</p>
<pre><code>row = {.25, .7, 0., .05};
mat = {row, row, Reverse[row], Reverse[row]};
(* start in the first state *)
process = DiscreteMarkovProcess[{1, 0, 0, 0}, mat];
</code></pre>
<p>In order to convert the state numbers 1 through 4 into the current values, it is convenient to define a helper function:</p>
<pre><code>fun = Function @@ {x, Piecewise[{{-1, x == 1}, {1, x == 4}}]}
</code></pre>
<p>We are now set to simulate from this process:</p>
<pre><code>ran = fun /@ RandomFunction[process, {1, 10^6}]["ValueList"][[1]];
</code></pre>
<p>On my MacBook this takes about 0.33 seconds. (This is faster than my other answer, which see for additional post processing.)</p>
|
763,381 | <p>There are 10 men and 7 women working as supervisors in a company. The company has recently decided to form a committee to represent all the employees. The committee has to consist of 3 members, all of whom must be supervisors. The members will be President, General Secretary and Coordinator respectively. Answer the following questions based on this information.</p>
<p>(a) How many ways can the committee be formed from the supervisors available? <br/>
(b) How many ways can the committee be formed if the General Secretary must be a female?<br/>
(c) How many ways can the committee be formed if it must have at least one man and at least one woman? <br/></p>
<p>(a) $\binom{17}{3}$ <br/>
(b) $\binom{16}{2}$ <br/>
(c) $\binom{10}{1}\binom{7}{2} + \binom{10}{2} \binom{7}{1}$. </p>
<p>Do you think my answers are correct?</p>
| JLBarba | 144,734 | <p>I guess the answer is that $f(x)$ has no roots.
<br>Firstly we can see that degree of $f(x)$ must be even.
Why?. Using the left hand side of the definition $dgr(f(x))=2m$.
<br>Then if we had one root $(r_0)$, we could build the increasing sequence $r_{n+1}=r_n^2+1$ of solutions. Which it's a contradiction with the fact of a finite degree of the polynomial.</p>
|
308,014 | <p>First of all, on this Matlab exercise sheet that I am currently working through what does the term 'the lower $2 \times 2$ block' mean in the question below?</p>
<p>$A =
\left[\begin{array}\
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9 \\
\end{array}\right]$,</p>
<p>Define a $2 \times 2$ matrix that is the lower $2 \times 2$ block in $A$.</p>
| Bravo | 24,451 | <p>It is clear from the wording; it is the submatrix $\left[\begin{array}\
5 & 6 \\
8 & 9 \\
\end{array}\right]$. In Matlab, you need to say $A(2:3,2:3)$ to get the submatrix.</p>
|
421,951 | <p>I did search for whether this question was already answered but couldn't find any.</p>
<p>Does a function have to be "continuous" at a point to be "defined" at the point?</p>
<p>For example take the simple function $f(x) = {1 \over x}$; obviously it is not continuous at $x = 0$. However it does have the $-$ and $+$ limits because $\lim_{x \to 0-} f(x) = -\infty $ and $\lim_{x \to 0+} f(x) = +\infty$.</p>
<p>Would you then say that $f(x) = {1 \over x}$ is "defined" at $x = 0$ or not? Please justify your answer.</p>
<p>I ask because at <a href="http://en.wikipedia.org/wiki/Continuous_function#Examples" rel="noreferrer">http://en.wikipedia.org/wiki/Continuous_function#Examples</a> there is the text: </p>
<blockquote>
<p>the function $f(x) = \frac {2x-1} {x+2}$ is defined for all real numbers $x \neq -2$ and is continuous at every such point. The question of continuity at $x = -2$ does not arise, since $x = -2$ is not in the domain of $f$. </p>
</blockquote>
<p>and the caption of the associated graph reads:</p>
<blockquote>
<p>The function is not defined for $x = -2$.</p>
</blockquote>
<p>Do I interpret this to mean that a function <em>can not</em> be defined at a point of discontinuity or merely that the function is <em>intentionally</em> not defined at the point of discontinuity only to achieve the status of being continuous (for whatever purpose) over the entire domain it is defined on?</p>
| paul garrett | 12,291 | <p>First, the <em>formal</em> question "must a function be defined at a point to be continuous there?" has answer "yes", from the definition of "continuity", since the latter definition refers to the value of the function at the point.</p>
<p>But this formal sense of the question does not go far enough, in my opinion. That is, I'd argue that at a (single) point where a function is <em>not</em> continuous, its point-wise value is "ambiguous"... I'll argue in favor of this informal qualification of the formal answer:</p>
<p>Even in calculus (and certainly in complex analysis) we do speak of "removable singularities/discontinuities", where either a function continuous at a point was whimsically redefined just at that point to make it discontinuous, or, more importantly, where the function had <em>no</em> formal definition at some particular point, but <em>could</em> be defined "by continuity" there. That is, given $f$ defined in a neighborhood of a point $x_o$, if there is a unique value $y_o$ such that the extension of $f$ defined by $f(x_o)=y_o$ is continuous at $x_o$, then this extension is "extension by continuity". Of course, there may fail to be such an extension.</p>
<p>But, in many practical circumstances, there <em>is</em> such an extension. There is at most one such, if there is any. </p>
<p>Thus, depending how we think about it, the value of such $f$ at $x_o$ is entirely determined (by continuity) by near-by values... even if $f$ was not originally "defined" there.</p>
<p>From another side: to redefine a function at a single point does not change its integrals against other functions. Thus, somehow, values at single points are partly irrelevant to the interaction with other objects.</p>
<p>Another: for pointwise evaluation purposes, we might like the map $f\rightarrow f(x_o)$ to be <em>continuous</em>, under some metric on functions themselves. The simplest metric to put on functions on an interval $[a,b]$ is $d(f,g)=\sup_{x\in[a,b]}|f(x)-g(x)|$, and this makes pointwise evaluation continuous.</p>
<p>With other (otherwise reasonable) metrics and such on "functions", pointwise evaluation is often <em>not</em> a continuous map from functions to values.</p>
<p>In that, vein, when we take limits of sequence of functions, we might hope that $\lim_n f_n(x_o)=(\lim_n f_n)(x_o)$. One must stipulate in what sense the limit of functions is taken. If we use the sup-norm metric, this property does hold, and, in fact, the "limit function" is itself continuous.</p>
<p>So, in effect, _in_practice_, a purported pointwise value at a point where a function is not continuous either can be corrected so that the function <em>is</em> continuous, or else has no useful specific value.</p>
|
421,951 | <p>I did search for whether this question was already answered but couldn't find any.</p>
<p>Does a function have to be "continuous" at a point to be "defined" at the point?</p>
<p>For example take the simple function $f(x) = {1 \over x}$; obviously it is not continuous at $x = 0$. However it does have the $-$ and $+$ limits because $\lim_{x \to 0-} f(x) = -\infty $ and $\lim_{x \to 0+} f(x) = +\infty$.</p>
<p>Would you then say that $f(x) = {1 \over x}$ is "defined" at $x = 0$ or not? Please justify your answer.</p>
<p>I ask because at <a href="http://en.wikipedia.org/wiki/Continuous_function#Examples" rel="noreferrer">http://en.wikipedia.org/wiki/Continuous_function#Examples</a> there is the text: </p>
<blockquote>
<p>the function $f(x) = \frac {2x-1} {x+2}$ is defined for all real numbers $x \neq -2$ and is continuous at every such point. The question of continuity at $x = -2$ does not arise, since $x = -2$ is not in the domain of $f$. </p>
</blockquote>
<p>and the caption of the associated graph reads:</p>
<blockquote>
<p>The function is not defined for $x = -2$.</p>
</blockquote>
<p>Do I interpret this to mean that a function <em>can not</em> be defined at a point of discontinuity or merely that the function is <em>intentionally</em> not defined at the point of discontinuity only to achieve the status of being continuous (for whatever purpose) over the entire domain it is defined on?</p>
| Alexander Shamov | 37,460 | <p>Surely, this is not a precise mathematical question - or, at least, as a precise mathematical question it is rather trivial: there are lots of discontinuous functions that are perfectly well-defined everywhere, if you want them to.</p>
<p>However, there is a deeper perspective coming from algebraic geometry and functional analysis, in which one works not with a single function, but rather with a ring $A$ and thinks of it as a ring of functions. Then one just defines the "points where these functions are defined" to correspond to maximal (or prime, if you really want them) ideals of that ring. This is what people call <a href="http://en.wikipedia.org/wiki/Spectrum_of_a_ring" rel="nofollow">spectrum of a ring</a> and <a href="http://en.wikipedia.org/wiki/Spectrum_of_a_commutative_Banach_algebra" rel="nofollow">Gelfand spectrum of a Banach algebra</a> (let's stick to Banach algebras for the sake of simplicity, instead of polynormed algebras or even worse). The Gelfand spectrum of a Banach algebra carries a natural compact Hausdorff topology, generated by the functions in the algebra, and this topology is just designed to make these functions continuous. However, this construction depends only on $A$ being a Banach algebra, so if originally it was also an algebra of functions defined on some space, you may find that the spectrum is much richer than that space. In this case the spectrum provides the natural compactification, and in some cases it may be so large and complicated that it actually looks a little frightening at first sight (in particular, the spectra of $\ell^\infty$ or $L^\infty$ certainly do), but it is a really, really useful thing.</p>
<p>An alternative, but closely related, point of view is to fix a ring $A$ and also a field (or ring) $\mathbb{K}$, for your purpose usually $\mathbb{R}$ or $\mathbb{C}$, and define a $\mathbb{K}$-point of $A$ to be a ring homomorphism $A \to \mathbb{K}$ (see <a href="http://en.wikipedia.org/wiki/Functor_of_points" rel="nofollow">functor of points</a>). If $\mathbb{K}$ carries a topology, then so does this space of $\mathbb{K}$-points, and again, this topology is designed to make the functions continuous. In ths case of complex Banach algebras the space of $\mathbb{C}$-points is the same as the Gelfand spectrum.</p>
|
867,938 | <p>I was reading the book Linear Algebra Done Right by Axler. In the chapter on inner product space (Ch.6), he defines the norm of x on $R^n$ space as:</p>
<p>$||x|| = \sqrt{x_1^2 + ... + x_n^2}$</p>
<p>and says:</p>
<p>"The norm is not linear on $R^n$. To inject linearity into the discussion, we introduce the dot product."</p>
<p>I don't see why the norm is not linear.</p>
<p>If I check a multiplicity for $R^2$, using a scalar of 3 for example</p>
<p>$3||x|| =? ||3x||$ </p>
<p>$3 \sqrt{x_1^2 + x_2^2} =? \sqrt{(3 x_1)^2 + (3 x_2)^2} $</p>
<p>$3 \sqrt{x_1^2 + x_2^2} =? \sqrt{9(x_1^2 + x_2)^2} $</p>
<p>$3 \sqrt{x_1^2 + x_2^2} = 3 \sqrt{x_1^2 + x_2^2} $</p>
<p>Why is the norm not linear? </p>
<p>Axler then says:</p>
<p>"Also, if $y \in R^n$ is fixed, then clearly the map from $R^n$ to $R$ that sends $x \in R^n$ to $x \cdot y$ is linear."</p>
<p>Why is the dot product linear if the norm isn't?</p>
<p>Regards,
Madeleine.</p>
| Community | -1 | <p>Does</p>
<p>$$\|x+y\|=\|x\|+\|y\|\quad?$$
Find a counterexample and recall the triangle inequality.</p>
|
1,287,692 | <p>To show whether or not the 3 planes
$$x+y-2z=5\tag 1$$
$$x-y+3z=6 \tag2$$
$$x+5y-12z=12 \tag 3$$ all have a common line of intersection.</p>
<p>Can I do $(3)-(2)$ to get the line $6y-15z=6$ and $(1)-(2)$ to get the line $2y-5z=-1$ which is $6y-15z=-3$ , and say that as these aren't the same line, they don't have a common line of intersection?</p>
<p>Another thing that is confusing me is that if instead of eliminating $x$, I chose to eliminate $z$, I would get different lines in terms of $x$ and $y$. But how can I get the equations of two different lines by eliminating from the same pair of plane equations? There's only one line of intersection between any pair of planes, so surely I should only be able to get one unique line if I eliminate a variable from a pair of planes?
Any help would be appreciated</p>
| Dinesh Lama | 349,528 | <p>first by solving 2 planes find y and z, where u have to consider z as t, hence u'l get parametric equation of y and z w.r.t t. now put this value in any plane which will give u parametric equation of x in terms of t only. Now substitute this values in any sphere, than u'll get quadratic equation in terms of t, if it is line than it will have two values, by which using which u can find the exact two points of intersection. Try it, it works.
tutorial is <a href="https://www.youtube.com/watch?v=sQ0Z9yDvvsc" rel="nofollow">here</a> and <a href="http://www.ambrsoft.com/TrigoCalc/Sphere/SpherLineIntersection_.htm" rel="nofollow">here</a>.</p>
|
2,768,308 | <p>If there is an easier way to see this I would be thrilled to know that way. I think the trick is to make sure the maximum of $f_n$ occurs to the right of a fixed $x_0$ so that we can avoid the function converging to $\frac{1}{e}$.</p>
<p><strong>Attempt.</strong></p>
<p>For x=1 and x=0, $f_n(x)=0$. Fix an $x_0\in(0,1)$. Since the max of each $f_n$ occurs at $x=\frac{n}{n+1}$, choose $N_1$ such that $n>N$ imples $\frac{n}{n+1} >x_0$. Then for $n>N, \exists \delta>0$ such that $\frac{n}{n+1}-\delta>x_0$.</p>
<p>So $|f_n(x_0)|=|nx_0^n(1-x_0)|\leq |n(\frac{n}{n+1}-\delta)^n(1-(\frac{n}{n+1}-\delta))| = |n(\frac{n}{n+1}-\delta)^n(\frac{1}{n+1}+\delta)|$</p>
<p>Now,
$|n(\frac{n}{n+1}-\delta)^n(\frac{1}{n+1}+\delta)| \leq |n(\frac{n}{n+1}-\delta)^n(\frac{1}{n+1})| = |\frac{n}{n+1}(\frac{n}{n+1}-\delta)^n|$</p>
<p>and since $\frac{n}{n+1} \leq 1$, $|f_n(x_0)| \leq (\frac{n}{n+1}-\delta)^n = (1-(\frac{1}{n+1}+\delta))^n \leq (1-\delta)^n$</p>
<p>So given an $\epsilon>0$, choose $n>\frac{log(\epsilon)}{log(1-\delta)}$</p>
| user539887 | 539,887 | <p>Fix $x \in (0,1)$. It suffices to prove that
$$
\lim\limits_{n \to \infty} (n x^n) = 0.
$$
We have
$$
\tag{$*$}
\frac{(n+1) x^{n+1}}{n x^n} = \Bigl( 1 + \frac{1}{n} \Bigr) x,
$$
which is $< 1$ for $n > x/(1-x)$. Consequently, the sequence $(n x^n)$, being eventually decreasing and bounded below (by $0$), converges to some $a \in [0, 1)$. But $\lim\limits_{n \to \infty} ((n + 1) x^{n+1}) = \lim\limits_{n \to \infty} (n x^{n}) = a$, so it follows from $(*)$ that $a = a x$, which is possible only if $a = 0$. </p>
|
4,584,609 | <p>Which is bigger</p>
<p><span class="math-container">$$ \int_0^{\frac{\pi}{2}}\frac{\sin x}{1+x^2}dx$$</span> or <span class="math-container">$$ \int_0^{\frac{\pi}{2}}\frac{\cos x}{1+x^2}dx~?$$</span></p>
<p>I let <span class="math-container">$x=\frac{\pi}{2}-t$</span> in the second integral, and I obtain this
<span class="math-container">$$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1+(\frac{\pi}{2}-x)^2}dx$$</span>
But it is still to decide which is the bigger.</p>
| dezdichado | 152,744 | <p>Actually, this is rather crude:
<span class="math-container">$$\int_0^{\frac{\pi}{2}}\dfrac{\sin x}{1+x^2}dx<\int_0^{\frac{\pi}{2}}\dfrac{x}{1+x^2}dx = \dfrac{1}{2}\ln\dfrac{4+\pi^2}{4}\approx 0.62 < 0.72\approx \dfrac{3}{2}\arctan\frac{\pi}{2} - \frac{\pi}{4}=\int_0^{\frac{\pi}{2}}\dfrac{1-x^2/2}{1+x^2}dx <\int_0^{\frac{\pi}{2}}\dfrac{\cos x}{1+x^2}dx $$</span></p>
|
41,707 | <p>Is there a slick way to define a partial computable function $f$ so that $f(e) \in W_{e}$ whenever $W_{e} \neq \emptyset$? (Here $W_{e}$ denotes the $e^{\text{th}}$ c.e. set.) My only solution is to start by defining $g(e) = \mu s [W_{e,s} \neq \emptyset]$, where $W_{e,s}$ denotes the $s^{\text{th}}$ finite approximation to $W_{e}$, and then set
$$
f(e) = \begin{cases}
\mu y [y \in W_{e, g(e)}] &\text{if } W_{e} \neq \emptyset \\
\uparrow &\text{otherwise},
\end{cases}
$$
but this is ugly (and hence not slick).</p>
| lhf | 589 | <p><a href="http://en.wikipedia.org/wiki/Linear_dynamical_system" rel="nofollow noreferrer">Linear discrete dynamical systems</a>, aka <a href="http://en.wikipedia.org/wiki/Recurrence_relation" rel="nofollow noreferrer">recurrence relations</a>, are best studied in a matrix formulation <span class="math-container">$x_{n+1} = A x_n$</span>. The solution of course is <span class="math-container">$x_n = A^n x_0$</span>, but the point is to exploit the properties of <span class="math-container">$A$</span> to allow the computation of <span class="math-container">$A^n$</span> without performing all multiplications. As an example, take the <a href="http://en.wikipedia.org/wiki/Fibonacci_number" rel="nofollow noreferrer">Fibonacci numbers</a>. The <a href="http://en.wikipedia.org/wiki/Fibonacci_number#Closed-form_expression" rel="nofollow noreferrer">formula</a> for them comes directly from this matrix formulation (plus diagonalization).</p>
<p>Don't forget the origins of matrix multiplication: linear change of coordinates. See, for instance, section 3.4 of Meyers's book (page 93) at <a href="http://web.archive.org/web/20110714050059/matrixanalysis.com/Chapter3.pdf" rel="nofollow noreferrer">http://web.archive.org/web/20110714050059/matrixanalysis.com/Chapter3.pdf</a>.</p>
<p>See also <a href="http://en.wikipedia.org/wiki/Matrix_multiplication#Application_Example" rel="nofollow noreferrer">http://en.wikipedia.org/wiki/Matrix_multiplication#Application_Example</a>.</p>
|
41,707 | <p>Is there a slick way to define a partial computable function $f$ so that $f(e) \in W_{e}$ whenever $W_{e} \neq \emptyset$? (Here $W_{e}$ denotes the $e^{\text{th}}$ c.e. set.) My only solution is to start by defining $g(e) = \mu s [W_{e,s} \neq \emptyset]$, where $W_{e,s}$ denotes the $s^{\text{th}}$ finite approximation to $W_{e}$, and then set
$$
f(e) = \begin{cases}
\mu y [y \in W_{e, g(e)}] &\text{if } W_{e} \neq \emptyset \\
\uparrow &\text{otherwise},
\end{cases}
$$
but this is ugly (and hence not slick).</p>
| Niel de Beaudrap | 439 | <p>Matrix multiplcation plays an important role in quantum mechanics, and all throughout physics. Examples include the <a href="http://en.wikipedia.org/wiki/Moment_of_inertia#Moment_of_inertia_tensor" rel="noreferrer">moment of inertia tensor</a>, continuous-time descriptions of the evolution of physical systems using <a href="http://en.wikipedia.org/wiki/Hamiltonian_%28quantum_mechanics%29" rel="noreferrer">Hamiltonians</a> (especially in systems with a finite number of basis states), and the most general formulation of <a href="http://en.wikipedia.org/wiki/Lorentz_transformation#Matrix_form" rel="noreferrer">the Lorentz transformation from special relativity</a>.</p>
<p>General relativity also makes use of <strong><em>tensors</em></strong>, which are a generalization of the sorts of objects which row-vectors, column-vectors, and matrices all are. Very roughly speaking, row- and column-vectors are 'one dimensional' tensors, having only one index for its coefficients, and matrices are 'two dimensional' tensors, having two indices for its coefficients, of two different 'kinds' representing rows and columns — input and output, if you prefer. Tensors allow three or more indices, and to allow more than one index to have the same 'kind'.</p>
|
41,707 | <p>Is there a slick way to define a partial computable function $f$ so that $f(e) \in W_{e}$ whenever $W_{e} \neq \emptyset$? (Here $W_{e}$ denotes the $e^{\text{th}}$ c.e. set.) My only solution is to start by defining $g(e) = \mu s [W_{e,s} \neq \emptyset]$, where $W_{e,s}$ denotes the $s^{\text{th}}$ finite approximation to $W_{e}$, and then set
$$
f(e) = \begin{cases}
\mu y [y \in W_{e, g(e)}] &\text{if } W_{e} \neq \emptyset \\
\uparrow &\text{otherwise},
\end{cases}
$$
but this is ugly (and hence not slick).</p>
| Freddie | 11,471 | <p>Hey Alex, a central theme of Machine Learning is about finding structures (preferably linear ones) in the data space; the intrinsic dimentionalities of your observations if you may (see Eigenfaces).</p>
<p>I understand this may not be about matrix multiplication per se; instead, this is about what, many times, happens right before it. It begins with the <a href="http://en.wikipedia.org/wiki/Spectral_theorem" rel="noreferrer">spectral theorem</a>: A = SΛS' (inverse when A is non-symmetric); it is Literally the basis of so many things (see what I did there?).</p>
|
991,372 | <p>Consider $f_{k}(x)=(\frac{1}{ x}\frac{d}{d x})^k(\frac{x}{\sinh x})$, $x>0$, $k=0,1,2,\cdots.$ Then, Is $f_{k}(x)$ always a bounded function? </p>
<p>The only thing one need to care is the behavior when $x$ is near $0$, and prove it's bounded. I try the case $k=1,2,$ and I think it's true for general k as well. Does anyone know how to prove this or disprove it?</p>
| Blue | 409 | <p>Consider $A$'s relationship with various circles.</p>
<p>Since $B^\prime$ and $C^\prime$ are points of tangency of lines through $A$ with the incircle, we have
$$|AB^\prime| = |AC^\prime| \qquad (\star)$$
Further, by the <a href="http://en.wikipedia.org/wiki/Power_of_a_point#Theorems" rel="nofollow">"Power of a Point" Theorem</a>,
$$|AA_B||AC^\prime| = \operatorname{pow}(A\;,\;\bigcirc GB^\prime C^\prime) = |AA_C||AB^\prime|$$
Also,
$$|AB_A||AC^\prime| = \operatorname{pow}(A\;,\;\bigcirc A^\prime G C^\prime) = |AG||AA^\prime| = \operatorname{pow}(A\;,\;\bigcirc A^\prime B^\prime G) = |AC_A||AB^\prime|$$
Together with $(\star)$, these imply
$$|AA_B| = |AA_C| \qquad\text{and}\qquad |AB_A| = |AC_A|$$</p>
<p>This makes $\square A_B A_C B_A C_A$ an isosceles trapezoid, whose bases, $A_BA_C$ and $B_AC_A$, have the bisector of $\angle A$ as their common perpendicular bisector. The incenter, $I$, of $\triangle ABC$ lies on that bisector, so we can write
$$|IA_B| = |IA_C| \qquad\text{and}\qquad |IB_A| = |IC_A|$$
Identical arguments considering points $B$ and $C$ lead to corresponding relations
$$|IB_C| = |IB_A| \qquad |IC_B| = |IA_B| \qquad\text{and}\qquad|IC_A|=|IC_B|\qquad|IA_C|=|IB_C|$$
Necessarily, $I$ is equidistant from all six points $A_B$, $A_C$, $B_C$, $B_A$, $C_A$, $C_B$. $\square$</p>
|
3,796,937 | <p>Prove that <span class="math-container">$2^n+1$</span> is not a cube for any <span class="math-container">$n\in\mathbb{N}$</span>.</p>
<p>I managed to prove this statement but I would like to know if there any other approaches different from mine.</p>
<p>If existed <span class="math-container">$k\in\mathbb{N}$</span> such that <span class="math-container">$2^n+1=k^3$</span> then <span class="math-container">$k=2l+1$</span> for some <span class="math-container">$l\in\mathbb{N}$</span>. Then <span class="math-container">$(2l+1)^3=2^n+1 \iff 4l^3+6l^2+3l=2^{n-1}$</span>. As I am looking for an integer solution, from the Rational Root Theorem <span class="math-container">$l$</span> would need to be of the form <span class="math-container">$2^j$</span> for <span class="math-container">$j=1,...,n-1$</span>. But then</p>
<p><span class="math-container">$$4(2^j)^3+6(2^j)^2+3\times2^j=2^{n-1} \iff 2^{2j+2}+3(2^{j+1}+1)=2^{n-1-j}$$</span></p>
<p>the LHS is odd which implies that <span class="math-container">$j=n-1$</span>. Absurd.</p>
<p>Thank you in advance.</p>
| J. W. Tanner | 615,567 | <p>Here is a different approach.</p>
<p>Modulo <span class="math-container">$7$</span>, there aren't so many cubes, so that can be a good setting to investigate such problems:</p>
<p><span class="math-container">$2^n+1\equiv 2, 3, $</span> or <span class="math-container">$5\pmod7$</span>, but <span class="math-container">$m^3\equiv0, 1, $</span> or <span class="math-container">$6\pmod 7$</span>.</p>
|
1,204,721 | <p>How can I prove that A is non-invertible nxn matrix if sum of all elements in rows of A equal to zero?</p>
<p>$$a_{11} + a_{12} + \cdots + a_{1n} = 0$$</p>
<p>$$a_{21} + a_{22} + \cdots + a_{2n} = 0$$</p>
<p>$$\vdots$$</p>
<p>$$a_{n1} + a_{n2} + \cdots + a_{nn} = 0$$</p>
| DeepSea | 101,504 | <p>The equation $A\cdot X = 0=A\cdot 0$ has a non-zero solution, namely $X = (1,1,..,1)$. This shows that the property that: " $AB = AC \Rightarrow B = C$" fails, thus $A$ is not invertible.</p>
|
1,204,721 | <p>How can I prove that A is non-invertible nxn matrix if sum of all elements in rows of A equal to zero?</p>
<p>$$a_{11} + a_{12} + \cdots + a_{1n} = 0$$</p>
<p>$$a_{21} + a_{22} + \cdots + a_{2n} = 0$$</p>
<p>$$\vdots$$</p>
<p>$$a_{n1} + a_{n2} + \cdots + a_{nn} = 0$$</p>
| Nicolas | 213,738 | <p>Let $A=\left(C_{1},\ldots,C_{n}\right)$
where $C_{j}$
is the $j$ -th column of $A$
. Then your condition is $C_{1}+\ldots+C_{n}=0$
that is $C_{n}=-\left(C_{1}+\ldots+C_{n-1}\right)$
i.e. the columns of $A$
are linearly dependant. Thus the determinant is zero.</p>
|
339,289 | <blockquote>
<p>Prove that if $f$ is defined for $x\ge 0$ by $f(x)=\sqrt x$, then $f$
is continuous at every point of its domain.</p>
</blockquote>
<p>Definition of a continuous function is:</p>
<p>Let $A\subseteq\mathbb{R}$ and let $f:A\to\mathbb{R}$. Denote $c\in A$.</p>
<p>Then $f(x)$ is continuous at $c$ iff for every $\varepsilon>0$, $\exists$ $\delta>0$ such that </p>
<p>$|x-c|<\delta\implies |f(x)-f(c)|<\varepsilon.$</p>
<p>My attempt:</p>
<p>We know that the function $f: x\to \mathbb{R}$, where $x\in [0,\infty)$ is defined to be $f(x)=\sqrt x$. So, for $0\le x<\infty$, then
$|f(x)-f(c)|=|\sqrt x - f(c)|$ and I can't continue since I don't necessarily know what $c$ is in this case.</p>
| Ian | 83,396 | <p>user62089's answer is not quite right. At $c \neq 0$, user62089's work proves that $\delta = \varepsilon \sqrt{c}$ is sufficient. At $c=0$ you have a slightly different situation: you want $\delta>0$ so that $\sqrt{x}<\varepsilon$ for $x<\delta$. $\delta=\varepsilon^2$ works for this purpose. </p>
|
1,410,164 | <p>I've found the following identity.</p>
<blockquote>
<p>$$\int_0^1 \frac{1}{1+\ln^2 x}\,dx = \int_1^\infty \frac{\sin(x-1)}{x}\,dx $$ </p>
</blockquote>
<p>I could verify it by using CAS, and calculate the integrals in term of <a href="http://mathworld.wolfram.com/ExponentialIntegral.html">exponential</a> and <a href="http://mathworld.wolfram.com/SineIntegral.html">trigonometric integrals</a>, then using identities between them. However, I think there is a more elegent way to prove it.</p>
<blockquote>
<p>How could we prove this identity?</p>
</blockquote>
<p>Also would be nice to see some references.</p>
| Noam Shalev - nospoon | 219,995 | <p>Define $I(a)=\int_0^{\infty} e^{-(x+1)a}\frac{\sin(x)}{x+1}dx$.
Then $\displaystyle I'(a)=-\int_0^{\infty} e^{-(x+1)a}\sin(x)dx=-\frac{e^{-a}}{a^2+1}$, and since $\lim_{a\to \infty} I(a)=0$, we have $\ \displaystyle I(0)=\int_0^{\infty} \frac{\sin(x)}{x+1}=\int_0^{\infty}\frac{e^{-a}}{a^2+1}da=\int_0^1\frac{dx}{1+\ln^2 x}.$</p>
|
794,182 | <p>I am trying to find an example of when $(A \bigcup B)^\circ \supset A^\circ \bigcup B^\circ$. Where $^\circ$ denotes the interior of a set. It has been previously shown in <a href="http://web.pdx.edu/~erdman/PTAC/problemtext_pdf.pdf" rel="nofollow">the text</a> that:</p>
<blockquote>
<p>The interior of the set $\mathbb{Q}$ of rational numbers is empty.</p>
</blockquote>
<p>My intuition is telling me that $(\mathbb{R} - \mathbb{Q} \bigcup \mathbb{Q})^\circ \supset (\mathbb{R} - \mathbb{Q})^\circ \bigcup \mathbb{Q}^\circ$. Since, $(\mathbb{R} - \mathbb{Q} \bigcup \mathbb{Q})^\circ = \mathbb{R} = (-\infty, \infty)$, but I don't exactly know what the size of $(\mathbb{R} - \mathbb{Q})^\circ \bigcup \mathbb{Q}^\circ$ is, all I can think of reducing it down to:</p>
<p>$$(\mathbb{R} - \mathbb{Q})^\circ \bigcup \emptyset$$
$$(\mathbb{R} - \mathbb{Q})^\circ$$</p>
<p>Is there a way I can show $\mathbb{R}^\circ \supset (\mathbb{R} - \mathbb{Q})^\circ$? Is it even true?</p>
| user139388 | 139,388 | <p>$1 \in R(f)$ if and only if
$$
7x - x^2
= 12 e,
$$
or
$$
x
= \frac{7 \pm \sqrt{49-48 e}}{2}.
$$
Since you are dealing with a function with a real domain, and these values of $x$ do not lie in $\mathbb{R}$, there is no pre-image of $1$.</p>
|
794,182 | <p>I am trying to find an example of when $(A \bigcup B)^\circ \supset A^\circ \bigcup B^\circ$. Where $^\circ$ denotes the interior of a set. It has been previously shown in <a href="http://web.pdx.edu/~erdman/PTAC/problemtext_pdf.pdf" rel="nofollow">the text</a> that:</p>
<blockquote>
<p>The interior of the set $\mathbb{Q}$ of rational numbers is empty.</p>
</blockquote>
<p>My intuition is telling me that $(\mathbb{R} - \mathbb{Q} \bigcup \mathbb{Q})^\circ \supset (\mathbb{R} - \mathbb{Q})^\circ \bigcup \mathbb{Q}^\circ$. Since, $(\mathbb{R} - \mathbb{Q} \bigcup \mathbb{Q})^\circ = \mathbb{R} = (-\infty, \infty)$, but I don't exactly know what the size of $(\mathbb{R} - \mathbb{Q})^\circ \bigcup \mathbb{Q}^\circ$ is, all I can think of reducing it down to:</p>
<p>$$(\mathbb{R} - \mathbb{Q})^\circ \bigcup \emptyset$$
$$(\mathbb{R} - \mathbb{Q})^\circ$$</p>
<p>Is there a way I can show $\mathbb{R}^\circ \supset (\mathbb{R} - \mathbb{Q})^\circ$? Is it even true?</p>
| Mikotar | 149,637 | <p>Your problem is that you are confounding the domain of a function with its range. A domain is the set of valid inputs of a function. The range of a function is its set of possible outputs. In your case, that set of inputs requires two things: (i) that $\frac{7x−x^2}{12}$ is positive, and (ii) that $ln(\frac{7x−x^2}{12})$ is positive. These constraints can be simplified so that the only costraint is that $\frac{7x−x^2}{12} \geq 1$. Solving the resulting inequality ($x^2-7x+12 \leq 0$) gives bounds on $x$: $x \in [3,4]$.</p>
<p>What remains is to show that this set of inputs never outputs a value of one. The easiest way to do this is to consider the following: the values of the function at $x=3$ and $x=4$ are $0$. Furthermore, the function is continuous as differentiable over the range of interest. The derivative shows a local maximum at 3.5, and the value at this point (estimated as $0.0029608...$) is less than one, so the function's range does not include 1.</p>
|
827,899 | <p>I came across this problem on a HackerRank challenge.</p>
<p>The function $f(n)$ is</p>
<ul>
<li>$1$ if $n = 0$</li>
<li>$2f(n - 1)$, if $n$ is odd</li>
<li>$f(n -1) + 1$, if $n$ is even</li>
</ul>
<p>I solved the problem using a recursive function and it worked just well. However, I am assuming that a program would work faster if instead of recursion we use an explicit function.</p>
<p>The site gave this solution:</p>
<p>$f(n) = Pow(2, (n+1)/2 + 1) - 1 - (2 \% n)$</p>
<p>Can someone help me arrive at this explicit formation?</p>
| Joel | 85,072 | <p>First I would start by writing out the sequence: $$f(0)=1, f(1) = 2, f(2) = 3, f(3) = 6, f(4) = 7, f(5) = 14, f(6) = 15, ...$$</p>
<p>Now notice that $f(0) = 2 - 1 = 2^{0+1} - 1$, $f(2) = 4-1 = 2^{1+1} - 1$, $f(4) = 8-1 = 2^{2+1} - 1$, and $f(6) = 16 - 1 = 2^{3+1}-1$. So we conjecture that $$f(2n) = 2^{n+1} - 1$$</p>
<p>Now it is also clear that $$f(2n-1) = 2f(2n-2) = 2^{n+1}-2$$</p>
<hr>
<p>From here we just need to verify the conjectures with induction. We see that it certainly holds for $f(k)$ up to $k=6$. Now suppose it holds for $f(k)$ where $k=2n$.</p>
<p>Consider $$f(k+1) = f(2n+1) = 2f(2n) = 2(2^{n+1} - 1) = 2^{n+2} - 2$$ so we have verified this for the odd case.</p>
<p>Now if the conjecture holds for $f(k)$ where $k=2n+1$ then we have $$f(k+1) = f(2n+2) = f(2n+1) + 1 = 2^{n+2}-2+1 = 2^{n+2} - 1$$</p>
<p>Thus we see by induction that $$f(2n) = 2^{n+1}-1$$ and $$f(2n+1) = 2^{n+2} - 2$$ as we conjectured.</p>
<hr>
<p>Now if you want this in one equation, we write $$f(k) = 2^{[(k+1)/2]+1} - 1.5 + (0.5)(-1)^{k}$$</p>
|
827,899 | <p>I came across this problem on a HackerRank challenge.</p>
<p>The function $f(n)$ is</p>
<ul>
<li>$1$ if $n = 0$</li>
<li>$2f(n - 1)$, if $n$ is odd</li>
<li>$f(n -1) + 1$, if $n$ is even</li>
</ul>
<p>I solved the problem using a recursive function and it worked just well. However, I am assuming that a program would work faster if instead of recursion we use an explicit function.</p>
<p>The site gave this solution:</p>
<p>$f(n) = Pow(2, (n+1)/2 + 1) - 1 - (2 \% n)$</p>
<p>Can someone help me arrive at this explicit formation?</p>
| Jika | 143,855 | <p>I tried to solve it as follow:</p>
<p>For $n$ odd, let $n=2k+1$ for some $k$. Then:
$$
f_n=f_{2k+1}=2f_{2k}=2\left(f_{2k-1}+1\right)=2f_{2k-1}+2=2^2f_{2k-2}+2=\cdots
$$
I continue until $f_0$. And I did the same thing for $n$ even. So I get the following:</p>
<p>Let, $n\div2=p$ and $n\mod2=q=\begin{cases}0\, \text{if}\, n\, \text{is even}\\1\,\text{ else}\end{cases}$, then: $n=2\times p+q$
$$
f_n=
\begin{cases}
\sum\limits_{i=1}^{p+q}2^i,\text{if }n\, \text{is odd}\\
1+2^{p}+\sum\limits_{i=1}^{p-1}2^i,\text{if }n\, \text{is even}
\end{cases}.
$$</p>
|
154,818 | <p>Let $G$ be a finitely generated group. The <em>weight</em> $w(G)$ of $G$ is defined to be the minimum number of elements of $G$ whose normal closure in $G$ is the whole of $G$ (this is sometimes also called the <em>normal rank</em>). Obviously, $d(G^{\operatorname{ab}})\leq w(G) \leq d(G)$, where $d(G)$ is the rank of $G$.</p>
<p>A minimal generating set for $G$ does not necessarily contain a minimal normal generating set. The question is: does there always exist such a minimal generating set, i.e. one that realises the rank and contains a subset realising the weight? </p>
<p>If not, which conditions on $G$ would guarantee the existence of such a generating set?</p>
| Luc Guyot | 84,349 | <p>The answer is positive for finitely generated soluble groups. (And also for finitely generated simple groups, but it's easy!)</p>
<blockquote>
<p><strong>Claim.</strong> Let <span class="math-container">$G$</span> be a finitely generated soluble group. Then <span class="math-container">$G$</span> has a generating set with <span class="math-container">$d(G)$</span> elements which contains a subset with <span class="math-container">$w(G)$</span> elements whose normal closure is <span class="math-container">$G$</span>.</p>
<p><em>Proof.</em> Let <span class="math-container">$W(G)$</span> be the intersection of the maximal normal subgroups of <span class="math-container">$G$</span>.
It follows from the definition of <span class="math-container">$W(G)$</span> that a vector in <span class="math-container">$G^n$</span> with <span class="math-container">$n \ge w(G)$</span> normally generates <span class="math-container">$G$</span> if and only if its image under the natural map <span class="math-container">$G \rightarrow \overline{G} \Doteq G/W(G)$</span> normally generates <span class="math-container">$\overline{G}$</span>. Let <span class="math-container">$S$</span> be a generating vector of <span class="math-container">$G$</span> of length <span class="math-container">$d(G)$</span>. By [1, Folgerung 2.10 and Satz 6.4], the group <span class="math-container">$\overline{G}$</span> is Abelian and we have <span class="math-container">$w(G) = w(G_{ab}) = w(\overline{G})$</span>. Thus we can find a <a href="https://en.wikipedia.org/wiki/Nielsen_transformation" rel="nofollow noreferrer">Nielsen transformation</a> <span class="math-container">$\psi \in \text{Aut}(F_{d(G)})$</span> such that the last <span class="math-container">$d(G) - w(G)$</span> components of <span class="math-container">$S \cdot \psi$</span> lie in <span class="math-container">$W(G)$</span> [2, Theorem 1.1]. By the previous remark, this means that the first <span class="math-container">$w(G)$</span> components of <span class="math-container">$S \cdot \psi$</span> normally generate <span class="math-container">$G$</span>.</p>
</blockquote>
<hr />
<p>[1] R. Baer, "Der reduzierte Rang einer Gruppe", 1964.<br />
[2] D. Oancea, "A note on Nielsen equivalence in finitely generated abelian groups", 2011.</p>
|
4,186,303 | <p>Given the following definitions, taken from Wikipedia:</p>
<blockquote>
<p><strong>Support of a function</strong>. When <span class="math-container">$X$</span> is a topological space and <span class="math-container">$f : X \to \mathbb C$</span> is a continuous function, the <em>support</em> of <span class="math-container">$f$</span> is defined topologically as the closure of the subset of <span class="math-container">$X$</span> where <span class="math-container">$f$</span> is non-zero.
<span class="math-container">${\displaystyle \operatorname {supp} (f):={\overline {\{x\in X\,|\,f(x)\neq 0\}}}={\overline {f^{-1}\left(\left\{0\right\}^{c}\right)}}.}$</span></p>
</blockquote>
<blockquote>
<p><strong>Support of a distribution</strong>. Suppose that <span class="math-container">$u$</span> is a distribution and that <span class="math-container">$U$</span> is an open set in Euclidean space such that, for all test functions <span class="math-container">$f$</span> such that <span class="math-container">${\rm supp}\ f \subseteq U$</span>, <span class="math-container">${\displaystyle u(f)=0}$</span>. Then <span class="math-container">$u$</span> is said to vanish on <span class="math-container">$U$</span>.
Now, if <span class="math-container">$u$</span> vanishes on an arbitrary family <span class="math-container">$U_{\alpha}$</span> of open sets, then for any test function <span class="math-container">$f$</span> supported in <span class="math-container">$\bigcup \limits_\alpha U_{\alpha }$</span>, (...) <span class="math-container">$u(f)=0$</span> as well.
Hence we can define the support of <span class="math-container">$u$</span> as <strong>the complement of the largest open set on which <span class="math-container">$u$</span> vanishes.</strong></p>
</blockquote>
<hr />
<p><strong>Question</strong>. For a given function <span class="math-container">$g(x)$</span> and a distribution <span class="math-container">$u(f)$</span>, assuming <span class="math-container">$g, f \in {\cal D}(\mathbb R^n)$</span>, how can we define the support of the product <span class="math-container">$g(x)\ u(f)$</span>? <span class="math-container">${\rm supp}\ g \ \cap {\rm supp} \ u $</span>?</p>
| David C. Ullrich | 248,223 | <p>Say <span class="math-container">$U$</span> is the largest open set in which <span class="math-container">$u$</span> vanishes; let <span class="math-container">$O$</span> be the complement of the support of <span class="math-container">$g$</span>, so <span class="math-container">$O=(g^{-1}(0))^o$</span>.</p>
<p>The main step is</p>
<blockquote>
<blockquote>
<p><strong>Proposition.</strong> <span class="math-container">$gu$</span> vanishes in <span class="math-container">$U\cup O$</span>.</p>
</blockquote>
</blockquote>
<p>Proof: Say <span class="math-container">$\phi\in C^\infty_c(U\cup O)$</span>. Say <span class="math-container">$\psi_1,\psi_2$</span> is a <em>partition of unity</em>: <span class="math-container">$0\le\psi_j\le 1$</span>, <span class="math-container">$supp(\psi_1)\subset U$</span>, <span class="math-container">$supp(\psi_2)\subset O$</span>, and <span class="math-container">$\psi_1+\psi_2=1$</span>. Let <span class="math-container">$\phi_j=g\psi_j$</span>.
Then <span class="math-container">$\phi_1$</span> is supported in <span class="math-container">$U$</span>, so <span class="math-container">$\phi_1g=0$</span>, hence by definition <span class="math-container">$gu(\phi)=u(\phi_1g)=0$</span>. Similarly <span class="math-container">$gu(\phi_2)=0$</span>, since <span class="math-container">$g\phi_2$</span> is supported in <span class="math-container">$O$</span> and <span class="math-container">$u$</span> vanishes in <span class="math-container">$O$</span>. So <span class="math-container">$gu(\phi)=0+0=0$</span>.</p>
<p>So now you just have to figure out why that's the <em>largest</em> open set in which <span class="math-container">$gu$</span> vanishes, and then see what that is qed. having just determined that we didn't learn the <em>definition</em> of <span class="math-container">$gu$</span> before asking the question I'll be stopping here.</p>
|
2,465,803 | <p>Given a random permutation <span class="math-container">$\sigma \in S_n$</span> from <span class="math-container">$[n] \to [n]$</span> in a uniform probability space, what is the probability that <span class="math-container">$\sigma $</span> has exactly <span class="math-container">$k$</span> fixed points for a given <span class="math-container">$k$</span> between <span class="math-container">$1$</span> and <span class="math-container">$n$</span>?</p>
<p>In other words: what is the probability that <span class="math-container">$\exists x_1 ,...,x_k \in [n] : \sigma (x_i) = x_i $</span> for <span class="math-container">$\ i\in \{1,...,k\}$</span> and for every <span class="math-container">$y \notin \{x_1 , ... , x_k\}$</span> we get <span class="math-container">$\sigma(y) \neq y$</span>.</p>
<p>I saw that <span class="math-container">$\lim_{n \to \infty } prob(A_0) = e^{-1}$</span> using Inclusion–exclusion principle and i belive that for a given k : <span class="math-container">$\lim_{n \to \infty} prob(A_k) = \frac{e^{-1}}{k!}$</span> but I am not sure how to show it.</p>
<p>*<span class="math-container">$A_k$</span> stands for the event "k".</p>
| drhab | 75,923 | <p>There are $\binom{n}{k}$ possibilities for the $k$ fixed points.</p>
<p>If they are established then there are $!(n-k)$ <a href="https://en.wikipedia.org/wiki/Derangement" rel="noreferrer">derangements</a> for the other points. </p>
<p>That gives $\binom{n}{k}\left[!(n-k)\right]$ permutations with exactly $k$ fixed points on a total of $n!$ permutations.</p>
<p>Also we have the formula: $$!(n-k)=(n-k)!\sum_{i=0}^{n-k}\frac{(-1)^i}{i!}$$ and we end up with probability: $$\frac1{k!}\sum_{i=0}^{n-k}\frac{(-1)^i}{i!}$$</p>
|
2,465,803 | <p>Given a random permutation <span class="math-container">$\sigma \in S_n$</span> from <span class="math-container">$[n] \to [n]$</span> in a uniform probability space, what is the probability that <span class="math-container">$\sigma $</span> has exactly <span class="math-container">$k$</span> fixed points for a given <span class="math-container">$k$</span> between <span class="math-container">$1$</span> and <span class="math-container">$n$</span>?</p>
<p>In other words: what is the probability that <span class="math-container">$\exists x_1 ,...,x_k \in [n] : \sigma (x_i) = x_i $</span> for <span class="math-container">$\ i\in \{1,...,k\}$</span> and for every <span class="math-container">$y \notin \{x_1 , ... , x_k\}$</span> we get <span class="math-container">$\sigma(y) \neq y$</span>.</p>
<p>I saw that <span class="math-container">$\lim_{n \to \infty } prob(A_0) = e^{-1}$</span> using Inclusion–exclusion principle and i belive that for a given k : <span class="math-container">$\lim_{n \to \infty} prob(A_k) = \frac{e^{-1}}{k!}$</span> but I am not sure how to show it.</p>
<p>*<span class="math-container">$A_k$</span> stands for the event "k".</p>
| Marko Riedel | 44,883 | <p>By way of enrichment here is an alternate formulation using
combinatorial classes. The class of permutations with fixed points
marked is</p>
<p><span class="math-container">$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SET}(\mathcal{U} \times
\textsc{CYC}_{=1}(\mathcal{Z}) +
\textsc{CYC}_{=2}(\mathcal{Z}) +
\textsc{CYC}_{=3}(\mathcal{Z}) +
\cdots).$$</span></p>
<p>This gives the generating function
<span class="math-container">$$G(z, u) =
\exp\left(uz + \frac{z^2}{2} +
\frac{z^3}{3} +
\frac{z^4}{4} +
\frac{z^5}{5} + \cdots\right)$$</span></p>
<p>which is</p>
<p><span class="math-container">$$G(z, u) =
\exp\left(uz-z+\log\frac{1}{1-z}\right)
= \frac{\exp(uz-z)}{1-z}.$$</span></p>
<p>Now for <span class="math-container">$k$</span> fixed points we get</p>
<p><span class="math-container">$$[u^k] \frac{\exp(uz-z)}{1-z}
= [u^k] \frac{\exp(uz)\exp(-z)}{1-z}
= \frac{z^k}{k!} \frac{\exp(-z)}{1-z}.$$</span></p>
<p>This is the EGF of permutations having <span class="math-container">$k$</span> fixed points. We extract
the count by computing (the factor <span class="math-container">$n!$</span> is canceled because we require
the average)</p>
<p><span class="math-container">$$[z^n] \frac{z^k}{k!} \frac{\exp(-z)}{1-z}
= \frac{1}{k!} [z^{n-k}] \frac{\exp(-z)}{1-z}.$$</span></p>
<p>We find</p>
<p><span class="math-container">$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{k!} \sum_{q=0}^{n-k} \frac{(-1)^q}{q!}.}$$</span></p>
<p>We can identify this as choosing the <span class="math-container">$k$</span> fixed points and combining
them with a derangement of the rest:</p>
<p><span class="math-container">$$\frac{1}{n!} {n\choose k} (n-k)!
\sum_{q=0}^{n-k} \frac{(-1)^q}{q!}$$</span></p>
<p>which is the combinatorial class</p>
<p><span class="math-container">$$\textsc{SET}_{=k}(\mathcal{Z}) \times
\textsc{SET}(\textsc{CYC}_{\ge 2}(\mathcal{Z})).$$</span></p>
|
2,307,753 | <p>I know that $-\int \tan(t)dt$ = $\ln |\cos t|$ (letting $C=0$). So I would think that $e^{-\int \tan(t)dt}$ would be equal to $e^{\ln |\cos t|} = |\cos t|$. However, my math textbook and Wolfram Alpha both say that $e^{-\int \tan(t)dt}=e^{\ln (\cos t)} = \cos t$. Why can the absolute value be ignored when taking the indefinite integral in this case?</p>
<p>Context: Finding an integrating factor for $x' = x\tan(t) + \sin(t)$. But Wolfram Alpha also gave me this answer without any differential equations context.</p>
| G.H.lee | 445,037 | <p>(integrating factor : $ \mu (t)$)</p>
<p>Then </p>
<p>$\ln \left | \mu (t) \right | = - \int \tan(t)dt = \ln \left | \cos t \right |$</p>
<p>so $|\mu (t)| =|\cos t|$</p>
<p>It means $\mu (t) =\pm \cos t$</p>
<p>But $ \mu (t)$ should be differentiable. (When $\cos t \neq 0$) </p>
<p>so </p>
<p>Case 1: $\mu (t) = \cos t $ $(t \in R)$</p>
<p>or</p>
<p>Case2: $\mu (t) = - \cos t$ $(t \in R)$</p>
<p>A solution is $x = \frac{\int \mu (t) \sin (t) dt}{\mu (t)} + \frac{C}{\mu (t)} = \mathbf{\frac{-\cos 2t +4C}{4\cos t}}$ ($C$ is constant) in both cases.</p>
<p>So, It is not important if $\mu (t) = \cos t $ or $\mu (t) = - \cos t$ </p>
|
2,307,753 | <p>I know that $-\int \tan(t)dt$ = $\ln |\cos t|$ (letting $C=0$). So I would think that $e^{-\int \tan(t)dt}$ would be equal to $e^{\ln |\cos t|} = |\cos t|$. However, my math textbook and Wolfram Alpha both say that $e^{-\int \tan(t)dt}=e^{\ln (\cos t)} = \cos t$. Why can the absolute value be ignored when taking the indefinite integral in this case?</p>
<p>Context: Finding an integrating factor for $x' = x\tan(t) + \sin(t)$. But Wolfram Alpha also gave me this answer without any differential equations context.</p>
| John Wayland Bales | 246,513 | <p>Your question is "Why can the absolute value be ignored?"</p>
<p>We have</p>
<p>\begin{eqnarray} e^{-\int\tan t\,dt}&=&e^{\ln\vert\cos(t)\vert+c_1}\\
&=&e^{c_1}\vert\cos(t)\vert
\end{eqnarray}</p>
<p>But $e^{c_1}>0$ for all values of $c_1$ and $\vert\cos(t)\vert\ge0$ when in fact solutions such as $-\cos(t)$ also exist. But the formulation $y=e^{c_1}\vert\cos(t)\vert$ excludes those solutions.</p>
<p>We can solve the dilemma by getting rid of the absolute value sign and replacing the positive constant $e^{c_1}$ by a constant $C$ which can take on values less than or equal to $0$. So we write</p>
<p>$$ y=C\cdot\cos(t)$$</p>
<p>Keep in mind that Wolfram also assumes that complex solutions are allowed.</p>
|
8,764 | <p>Q is a rational field. Q[x] is polynomial ring over Q 。(x) is maximal ideal of Q[x].
Take Q[x]/(x) as a module over Q[x]. Then what is Q[x]-module Q[x]/(x) localize at 0??</p>
<p>I think the result is
Q[x]/(x) \ otimes_{Q[x]}Q(x) but on the other hand, from another way, I know it should be Q[1/x]/Q.</p>
<p>But how can I prove they are isomorphism?</p>
| Rado | 1,645 | <p>Isn't $Q[x]/(x)$ just $Q$. In general $Q[x]/(x-\alpha)$ is just evaluation of all polynomials at alpha.</p>
<p>Then localization does nothing as we already have a full field of fractions.</p>
|
1,031,033 | <p>I have an inequality to prove and I can't get a hold of it... I hope someone can help with it or point me in the right direction. I tried it based on previous one, but without success... The prev. ones also seemed easy after I grasped them, so I'm sure that it's just a bit hard for me to get the first thought on the right path...</p>
<p>$$x\in\mathbb{R}, x\geq 0$$
$$1+x^3\geq x+x^2$$</p>
| Harald Hanche-Olsen | 23,290 | <p><strong>Hint:</strong> Both sides are multiples of $1+x$.</p>
|
1,031,033 | <p>I have an inequality to prove and I can't get a hold of it... I hope someone can help with it or point me in the right direction. I tried it based on previous one, but without success... The prev. ones also seemed easy after I grasped them, so I'm sure that it's just a bit hard for me to get the first thought on the right path...</p>
<p>$$x\in\mathbb{R}, x\geq 0$$
$$1+x^3\geq x+x^2$$</p>
| amWhy | 9,003 | <p>$$\begin{align} 1+x^3\geq x+x^2 &\iff x^3-x \geq x^2 - 1\\ \\
& \iff x(x^2-1) \geq x^2 - 1 \quad\forall x\geq 0\end{align}$$</p>
|
1,031,033 | <p>I have an inequality to prove and I can't get a hold of it... I hope someone can help with it or point me in the right direction. I tried it based on previous one, but without success... The prev. ones also seemed easy after I grasped them, so I'm sure that it's just a bit hard for me to get the first thought on the right path...</p>
<p>$$x\in\mathbb{R}, x\geq 0$$
$$1+x^3\geq x+x^2$$</p>
| Dr. Sonnhard Graubner | 175,066 | <p>we have to prove that<br>
$1+x^3\geq x(1+x)$ for all $x$ with $x>0$,
then we have
$1+x^3-(1+x)x\geq 0$<br>
$(1+x)(1+x^2-x)-x(1+x)\geq 0$
this is equivalent to<br>
$(1+x)(1+x^2-2x)\geq 0$
or
$(x-1)^2(x+1)\geq 0$
which is true.
Sonnhard.</p>
|
246,827 | <p>Let $A$ and $B$ be two positive definite $n \times n$ matrices. It is, of course, not true that $AB+BA$ is necessarily positive definite. </p>
<p>Consider, though, the results of the following numerical experiment. I generated $A$ by letting its eigenvalues be random in $[0,1]$, and selecting its eigenvectors by generating a random matrix of standard Gaussians and applying Gram-Schmidt to it. The matrix $B$ is generated in the same way.</p>
<p>I did this 1000 times and checked what proportion of times the matrix $AB+BA$ has at least one negative eigenvalue [1]. Here are the results for different dimensions $n$:</p>
<ul>
<li>$n=2, ~~~~94.8 \%$</li>
<li>$n=3, ~~~~89.4 \%$. </li>
<li>$n=4, ~~~~78 \%$. </li>
<li>$n=5, ~~~~72.7 \%$.</li>
<li>$n=10, ~~~40.3 \%$.</li>
<li>$n=15, ~~~20.1 \%$. </li>
<li>$n=20, ~~~11.4 \%$. </li>
<li>$n=50, ~~~0.3\%$. </li>
<li>$n=100, ~~0 \%$. </li>
</ul>
<p>This suggests that, as a function of $n$, examples with $AB+BA$ not psd tend to get rarer and rarer. Is it possible to give a proof of this? </p>
<p>It may be more natural to consider a different random model of randomly generated psd matrices; I only generated them in the way I described above because it seemed easiest. </p>
<p>[1] Actually, I checked if there is an eigenvalue less then $-1 \cdot 10^{-5}$ in MATLAB to account for rounding error. </p>
| Robert Israel | 13,650 | <p><span class="math-container">$\text{tr}(AB+BA) = 2 \operatorname{tr}(A^{1/2} B A^{1/2}) > 0$</span>, so that may produce some bias toward positive eigenvalues. In particular if you generate your "random" matrices in such a way that the eigenvalues of <span class="math-container">$AB+BA$</span> will tend to be concentrated very close together, this may produce the results you observed.</p>
<p>But I tried a different experiment: <span class="math-container">$A = X^T X$</span> and <span class="math-container">$B = Y^T Y$</span> where <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are random <span class="math-container">$n \times n$</span> matrices with integer entries in <span class="math-container">$[-100,100]$</span>.<br />
For the case <span class="math-container">$n=10$</span>, I found that it was very rare (0 occurrences in 3000 trials) for <span class="math-container">$AB + BA$</span> to be positive definite.</p>
|
1,415,214 | <p>I'm supposed to calculate the derivative of $\frac{d}{dx}\int_{x^{2}}^{x^{8}}\sqrt{8t}dt$ the answer I got is $8x^7\cdot \sqrt{8x^8}$ but when I put this into the grading computer it is marked wrong. I've tried solving in several different ways and always got the same answer so I used my graphing calculator to test.</p>
<p>I solved for $\frac{d}{dx}\int_{x^{2}}^{x^{8}}\sqrt{8t}dt$ with my graphing calculator's derivative and integral functions, with $x=3$. The result was $40083336.292$. The I tried solving with my formula, again plugging in $3$ for $x$. The result was $4008379.039$.</p>
<p>Very similar, but not the same. Is my formula wrong or is it a bug in the grading program?</p>
<p>Thanks for any help. If I solved this wrongly please explain where I went wrong. Thanks.</p>
| Aaron Maroja | 143,413 | <p><strong>Hint:</strong> Remember that you apply the Fundamental Theorem of Calculus as in </p>
<p>$$\int_{x^2}^{x^8} \sqrt{8t} dt = \int_{c}^{x^8} \sqrt{8t} dt - \int_{c}^{x^2} \sqrt{8t} dt $$</p>
<p>then </p>
<p>$$\frac{d}{dx} \left(\int_{x^2}^{x^8} \sqrt{8t} dt\right) = 8x^7\sqrt{8 x^8} - 2x\sqrt{8 x^2} $$</p>
|
352,243 | <p>In the page 10 of the paper "Filling Riemannian manifolds" by Gromov <a href="https://projecteuclid.org/euclid.jdg/1214509283" rel="noreferrer">(ProjetEuclid link)</a>, the author proves the following inequality (1.2) relating the systole and the filling radius of manifolds.
<span class="math-container">$$\operatorname{Sys}_1(M)\leq 6\operatorname{FillRad}(M)$$</span></p>
<p>During the proof, he uses an argument something like this:</p>
<p><i>"<span class="math-container">$M$</span> is a closed <span class="math-container">$n$</span>-dimensional manifold and <span class="math-container">$[M]$</span> is the fundamental class of it. In some ambient space <span class="math-container">$X$</span>, the fundamental class <span class="math-container">$[M]$</span> is null-homologous. Therefore, one can find <span class="math-container">$n+1$</span>-dimensional singular chain <span class="math-container">$c$</span> inside <span class="math-container">$X$</span> such that <span class="math-container">$\partial c=[M]$</span>. Moreover, using a <span class="math-container">$\textbf{piecewise linear approximation}$</span> of <span class="math-container">$c$</span>, one can construct a polyhedron <span class="math-container">$P$</span> such that <span class="math-container">$M\subset P \subset X$</span> and the fundamental class <span class="math-container">$[M]$</span> is null-homologous inside of this <span class="math-container">$P$</span>."</i></p>
<p>So, somehow he is constructing a triangulation of realization of the singular chain <span class="math-container">$c$</span> (i.e., union of continuous images of <span class="math-container">$\Delta_{n+1}$</span>) using "piecewise linear approximation". But I don't understand this step. How do we guarantee this realization of singular chain is triangulable? It is not necessarily manifold. What is precise meaning of his "piecewise linear approximation" of singular chain?</p>
<p><em>Edit:</em> I changed the title of this question from "When a topological space (not necessarily manifold) has a triangulation?" to "A question on the Gromov's proof of <span class="math-container">$\operatorname{Sys}_1(M)\leq 6\operatorname{FillRad}(M)$</span>". Even though the original question is interesting in its own right, but I want to focus on one specific topic: Understanding Gromov's argument.</p>
| Max Alekseyev | 7,076 | <p>Yes, such construction is always possible.</p>
<p>Consider two sets of pairs of values:
<span class="math-container">$$\big\{ (2+t,m-t)\quad :\quad t=0\,..\,\lfloor\frac{m-1}{4}\rfloor-1\big\},$$</span>
where differences of elements modulo <span class="math-container">$m$</span> are: <span class="math-container">$2,4,\dots,2\cdot\lfloor\frac{m-1}{4}\rfloor$</span>, and
<span class="math-container">$$\big\{ (\lfloor\frac{m+1}{2}\rfloor+t+1,\lfloor\frac{m+1}{2}\rfloor-t)\quad :\quad t=0\,..\,\lfloor\frac{m+1}{4}\rfloor-1\big\},$$</span>
where differences of elements modulo <span class="math-container">$m$</span> are: <span class="math-container">$1, 3, \dots, 2\cdot\lfloor\frac{m+1}{4}\rfloor-1.$</span> Together they give all differences from <span class="math-container">$1$</span> to <span class="math-container">$\lfloor\frac{m-1}{2}\rfloor$</span>.</p>
<p>Notice that all elements forming pairs in these sets are <span class="math-container">$\ne0,1$</span> and are distinct modulo <span class="math-container">$m$</span>.</p>
<p>Therefore, it's enough to assign the values from these sets to the corresponding pairs <span class="math-container">$(a_j,a_{n+2-j})$</span>, giving <span class="math-container">$2\cdot\lfloor\frac{m-1}{2}\rfloor$</span> nonzero <span class="math-container">$a_j$</span>'s. If <span class="math-container">$m$</span> is even we need to assign one more yet unassigned nonzero value (which is <span class="math-container">$\lfloor\frac{3m}{4}\rfloor+1$</span>) to any of yet unassigned <span class="math-container">$a_j$</span> with <span class="math-container">$j>\frac{n}2$</span>. The other unassigned <span class="math-container">$a_j$</span> are set to zero.</p>
<hr />
<p><strong>ADDED.</strong> Here is a <a href="https://sagecell.sagemath.org/?z=eJxtkc-KwyAQxu99ioFS0E6k1ZYeAnmSEIJUswirIdb2suy7r_8gHtbD4Hz85ptxVHqB5-pewb-fYZbEdZb2B4hHf7SbFQwwEoGhsyxQWFYPAYwDL92XJsQyTi-XO51yxRZhYjFJIgurUtVhw4C82_71wNYjh1xznc7kWzuSTSimaxmJ0gKNfS-mSGagSrxqhdwdZXV0JYvXKPCcpHnMPk_qE1uUHRTYIE-4Gs2UCgGOIOcfg-K3gRwKRqJG2Q7z5k1mAXsSw3BtnVnpVkrI7WzTJgDrZDl4Hd7egYzJof0o_ug4p38tnm9E&lang=sage&interacts=eJyLjgUAARUAuQ==" rel="nofollow noreferrer">SageMath code</a> implementing the above construction. There is a sample call <code>construct_a(16,11)</code>, which constructs vector <span class="math-container">$(a_1,\dots,a_n)$</span> for parameters <span class="math-container">$n=16$</span> and <span class="math-container">$m=11$</span>.</p>
|
103,164 | <p>I want to solve an equation which contains an infinite continued fraction $F(n)$. Then I must (obviously) truncate this continued fraction at $n=2000$.</p>
<p>The problem here is that <em>Mathematica</em> does not display the $2000$ terms of this fraction on the screen. The screen closes directly.</p>
<p>Please, how do I display this fraction up to $n=2000$?</p>
<p>The code is as follows:</p>
<pre><code>F[n_] := 1/(1 + I A x - (n + 1)^2/(4 (n + 1)^2 - 1)
x^2 A (1 - I 2 B x) With[{nplus1 = n + 1}, Hold[F[nplus1]]])
Fold[(#1 /. Hold[F[#2]] :> F[#2]) &, F[1], Range[1, 2000]]
</code></pre>
<p>$A$ and $B$ are real constants.</p>
| J. M.'s persistent exhaustion | 50 | <p>As Mr. Wizard supposed in a comment, one can indeed use <code>ContinuedFractionK[]</code> here:</p>
<pre><code>With[{A = 3., B = 2., x = 0.1},
1/(1 + I A x + ContinuedFractionK[-n^2/(4 n^2 - 1) x^2 A (1 - I 2 B x),
1 + I A x, {n, 2, 2000}])]
0.9197103744410972 - 0.28251974414934944 I
</code></pre>
<hr>
<p>However, if what you want is to approximate the value of the infinite CF and not just a truncation, one can certainly do better. Michael displays one approach, using forward evaluation, but a possible danger is that the convergent's numerators and denominators might grow large even if the actual CF's value itself is perfectly modest. (As an easier example, consider the CF for $\phi$, where the numerators and denominators of the convergent are the Fibonacci numbers, which have an exponential growth rate.)</p>
<p>One might thus consider a modification of the forward method where the numerators and denominators are suitably scaled. I am aware of two, but there certainly might be others. Here, for instance, is <a href="http://books.google.com/books?id=WGHWBoX89doC&pg=PA181" rel="nofollow">Steed's algorithm</a>:</p>
<pre><code>With[{A = 3., B = 2., x = 0.1},
b = 1 + I A x; f = h = d = 1/b; k = 2;
While[a = -x^2 A (1 - I 2 B x) k^2/(4 k^2 - 1);
d = 1/(b + a d);
h *= d b - 1; f += h; k++;
Abs[h] > $MachineEpsilon Abs[f]];
f]
0.9197103744410973 - 0.28251974414934944 I
</code></pre>
<p>and here is the <a href="http://dx.doi.org/10.1364/AO.15.000668" rel="nofollow">Lentz-</a><a href="http://dx.doi.org/10.1016/0021-9991%2886%2990046-X" rel="nofollow">Thompson-Barnett</a> algorithm:</p>
<pre><code>With[{A = 3., B = 2., x = 0.1},
b = 1 + I A x; f = c = b; d = 0; k = 2;
While[a = -x^2 A (1 - I 2 B x) k^2/(4 k^2 - 1);
d = 1/(b + a d); c = b + a/c;
h = c d; f *= h; k++;
Abs[h - 1] > $MachineEpsilon Abs[f]];
1/f]
0.9197103744410975 - 0.2825197441493497 I
</code></pre>
<p>Both are useful, in that Steed is often faster than Lentz-Thompson-Barnett, but the latter is more easily modified for "problematic" cases. Note that in both methods, the number of numerators and denominators generated is much less than the 2000 that was desired by the OP.</p>
|
3,623,724 | <p>The game works as follows: assume that there are two players, 1 and 2. 1 decides to lie or tell the truth. If 1 tells the truth, 2 needs to decide to take her medications or not. Later, in case 2 did not take her medications or 1 lied to her, 2 needs to decide to go to the hospital or not.
If 1 lied to her and she went to the hospital, then the utility of the 1 player is 0 and the second player's utility is 1. If 1 lied to her but she did not go to the hospital, then the first player's utility is 1 and the second player's utility is 0. If 1 told her the truth and she did not take her medications, then her going to hospital generates 1 utility to the first players and 0 utility to the second; not going to the hospital gives both of them 0. If 2 takes her medications after 1 told the truth, then they both get 2.
Okay, I used backward induction and this is what I got: if player 1 lies, 2 goes to the hospital. If player 1 tells the truth, player 2 is better off taking the meds. My question is: when stating my SPNE, should I mention that if 2 player decides not to take meds, then she is indifferent between going and not going to the hospital? Because it's clear that the decision to take the meds clearly dominates the decision not to take meds, should I even mention it when stating SPNE??
THANKS FOR YOUR HELP!!</p>
| Community | -1 | <p>Yes, you should say what each player should do in every subgame of the game. Hagen is essentially analyzing the strategic form of the game, rather than the extensive form.</p>
<p>In the extensive form, working backwards,</p>
<ul>
<li>If 1 told the truth and 2 did not take the meds, 2 is indifferent between going to the hospital and not, so going to the hospital with any probability maximizes her payoff, for payoffs of <span class="math-container">$(p1,0)$</span> where <span class="math-container">$p=pr[hospital]$</span>.
<ul>
<li>If 1 lied, 2 strictly prefers the hospital over not, for payoffs of (1,1). If 1 told the truth, taking meds gives 2 a payoff of 2 while not taking the meds gives a payoff of 0, so 2 takes the meds.</li>
<li>So if 2 tells the truth, the expected payoff is 1, while if 1 lies, the expected payoff is 1. So 1 strictly prefers to tell the truth.</li>
</ul></li>
</ul>
<p>The SPNE is 1 tells the truth; 2 goes to the hospital if 1 lied and takes the meds if 1 told the truth; if 2 did not take the meds after being told the truth, any randomization over hospital and not is part of an SPNE.</p>
<p>You should listen to your professor. The "stories" that frame games are just stories, you are supposed to learn and apply the equilibrium concepts correctly, not intuitively. The point is to understand what kinds of answers different equilibrium concepts give, and then judge the quality of the answers across many games. If your intuition says there's a problem, then you come up with new equilibrium concepts or refinements to explain why, and that's what game theory "is". But you have to apply the concepts correctly to get the right answer. You might think it is silly in this game, but there are many games where off-path behavior determines the equilibrium, and it really matters.</p>
|
11,651 | <p>Mathematics can come across as a sterile, dead subject - a catalogue of techniques long-ago decided, and forever relearned by each successive generation of students.</p>
<p>This is <em>approximately</em> true for elementary and secondary mathematics, and for the standard progression of undergraduate courses (eg, Calculus 1,2,3, discrete math / combinatorics, ODE + Vector Calc, Analysis and Algebra).</p>
<p>Of course, the subject is alive and kicking, with many thousands of active researchers learning, creating, refining, and publishing every day. But the vast majority of fresh research requires considerable expertise to understand, and are therefore inaccessible to younger students of the subject.</p>
<p><strong>What, then, are some recent results that are interesting and accessible to students at (say) a secondary school level, which might exemplify that the subject remains active?</strong></p>
<p>A couple of examples that come to mind (which could be fleshed out as answers) are the recent progress against the Twin Prime Conjecture, and the surprising observation that primes ending in $X$ 'favor' being followed by a prime ending in $Y$, for various $(X,Y)$ pairs.</p>
<p>Where it's appropriate, please include links to any media treatment of the result.</p>
<p>Let's roughly define 'recent' as being within the lifetime of some collection of students.</p>
| Joseph Malkevitch | 1,865 | <p>There are elementary results in mathematics being produced regularly either in the sense that the ideas don't need a lot of background to understand them or sometimes that even the "details" of the proofs are comprehensible. An example, is that of the de Bruijn graph, which has ties to the theory of Eulerian circuits in directed graphs and has found many applications, in particular, to assisting with genome sequencing.</p>
<p><a href="https://en.wikipedia.org/wiki/De_Bruijn_graph" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/De_Bruijn_graph</a></p>
|
803,488 | <p>Imagine Rock Paper Scissors, but where winning with a different hand gives a different reward.</p>
<ul>
<li><p>If you win with Rock, you get \$9. Your opponent loses the \$9.</p>
</li>
<li><p>If you win with Paper, you get \$3. Your opponent loses the \$3.</p>
</li>
<li><p>If you win with Scissors, you get \$5. Your opponent loses the \$5.</p>
</li>
<li><p>If you tie, you get $0</p>
<p>My first intuition would be that you should play Rock with a probability of <code>9/(9+3+5)</code>, Paper with <code>3/(9+3+5)</code> and Scissors with <code>5/(9+3+5)</code> however this seems wrong, as it doesn't take into consideration the risk you expose yourself to (if you play <code>Paper</code>, you have an upside of \$3 but a downside of \$5).</p>
</li>
</ul>
<p>So I put the question to you, in such a game -- what is the ideal strategy.</p>
<p>Edit: By "ideal" strategy, I mean playing against an adversarial player who knows your strategy.</p>
| Peter Franek | 62,009 | <p>If "optimal" means Nash equilibrium (i.e. a state that is stable wrt. small perturbations of strategies), than it can be computed. If you assume that $x_1$ is the probability of first player to play Rock, $x_2$ his probability to play Scissors and $1-x_1-x_2$ his probability to play Paper, and similarly for $y_i$, then the Payoff of the first player is
$$f(x_1, x_2, y_1, y_2) = x_1 (9y_2 - 3 (1-y_2-y_3)) + x_2 (-9 y_1 + 5(1-y_2-y_3)) + (1-x_1-x_2)(3y_1-5y_2)$$
or something like that. The condition on Nash is that all partial derivatives vanish; you can probably easily compute the probabilities and check, whether you guessed the right solutions (the solution should be unique in this case with $x_i$ and $y_i$ nonzero).</p>
<p>However, in different circumstances, optimal may meen different things; if they are good friends and know that it's a zero sum game, they can also play both Rock all the time.</p>
|
3,162,547 | <p>I am trying to convert instances of nested 'for' loops into a summation expression. The current code fragment I have is:</p>
<pre><code>for i = 1 to n:
for j = 1 to n:
if (i*j >= n):
for k = 1 to n:
sum++
endif
</code></pre>
<p>Basically, the 'if' conditional is confusing me. I know that the loops prior will be called n^2 times, but the third loop is only called when <span class="math-container">$i*j >= n$</span>. How would I write the third summation to account for this, and then evaluate the overall loop's time complexity?</p>
| Community | -1 | <p>At the basic level, for loops are also rewritable as conditionals. Anyways, on to the meat of the problem. The conditional can be removed if you replace 1 in the previous loop with <code>ceil(n/i)</code>. So it can be rewritten as:</p>
<p><code>for i = 1 to n:
for j = ceil(n/i) to n:
for k = 1 to n:
sum++</code></p>
<p>Then the inner 2 loops compress to:</p>
<p><code>sum+=n*(n-ceil(n/i)+1)</code></p>
<p>Which then gets called n times, but the sum can be wriiten as:</p>
<p><span class="math-container">$$n\sum_{i=1}^{n}(n-\lceil\frac{n}{i}\rceil+1)$$</span></p>
<p>As to the time complexity of the code as wriiten..., it'll take, okay I don't quite know that part. </p>
|
2,371,668 | <p>For a finite group $G$ and a subgroup $H,$ Lagrange's theorem says that $|G|=|G:H||H|,$ where $|G:H|$ is the number of cosets of $H$ in $G.$<br>
My question is for any subgroup $H$ can we find another subgroup with order $|G:H|$?</p>
| hamam_Abdallah | 369,188 | <p><strong>hint</strong></p>
<p>in your inequality, $\ge $ must be $\le $.</p>
<p>For every real $x\ge 0 $ we have
$$ -x^2+2x-1\le 0$$
thus
$$-x+2\sqrt {x}-1\le 0$$</p>
<p>$$\implies 2\sqrt {x}-x\le 1$$
$$\implies \sqrt {x}-\frac {x}{2}\le \frac {1}{2} $$
$$\implies \sqrt {a+b}-\frac {a+b}{2}\le \frac {1}{2} .$$</p>
<p>nearly done.</p>
|
3,488,245 | <p>Define <span class="math-container">$f:\mathbb{R} \to \mathbb{R}$</span> given by <span class="math-container">$f(x)=x^2$</span>. If <span class="math-container">$f^{-1}(x)$</span> is interpreted as a function, it is undefined, since <span class="math-container">$f$</span> is not injective. If <span class="math-container">$f^{-1}$</span> is interpreted as a multifunction, it returns <span class="math-container">$\{-\sqrt{x},\sqrt{x}\}$</span>. When is each of these interpretations normally made?</p>
| Community | -1 | <p>This is why we include domains and codomains in function definitions. Given an invertible mapping <span class="math-container">$f:A\to B$</span>, the inverse mapping is <span class="math-container">$f^{-1}:B\to A$</span>. Given any mapping <span class="math-container">$f:A\to B$</span>, the preimage mapping is <span class="math-container">$f^{-1}:\mathcal P(B)\to\mathcal P(A)$</span>, where <span class="math-container">$\mathcal P(X)$</span> refers to the power set of <span class="math-container">$X$</span>.</p>
<p>Therefore, <span class="math-container">$f^{-1}(2)$</span> would refer to the inverse function, and <span class="math-container">$f^{-1}(\{2\})$</span> would refer to the pre-image function. </p>
<p>In practice, few people work with domains where there is any risk of not understanding which function is meant. I will note that my foundations professor who had noble but futile dreams of fixing mathematical notation, thought that <span class="math-container">$f_>$</span> and <span class="math-container">$f^<$</span> were distinct but still clear notations indicating the image and pre-image functions.</p>
|
23,378 | <p>Franel uses the convergence of</p>
<p>$ \frac{\zeta(s+1)}{\zeta(s)} = \sum \frac{c(n)}{n^s}$</p>
<p>as an equivalent to the Riemann hypothesis.</p>
<p>Does anybody have a citation for this result and/or hints for computing $c(n)$?</p>
<p>Thanks for any insight.</p>
<p>Cheers, Scott</p>
| François G. Dorais | 2,000 | <p>Since
$$\zeta(s+1) = \sum_{n=1}^\infty \frac{1/n}{n^s}$$
and
$$\frac{1}{\zeta(s)} = \sum_{n=1}^\infty \frac{\mu(n)}{n^s}$$
where $\mu$ is the <a href="http://en.wikipedia.org/wiki/M%C3%B6bius_function">Möbius function</a>, we have
$$c(n) = \sum_{d \mid n} \frac{d}{n}\mu(d) = \frac{1}{n}\prod_{p \mid n} (1-p)$$
using <a href="http://en.wikipedia.org/wiki/Dirichlet_convolution">Dirichlet convolution</a>.</p>
|
3,132,039 | <p>I am trying to create a set of 10-tuples, where each tuple consists of elements taken from a set {defective, nondefective). Each sample must have precisely 1 defective part and 9 nondefective parts. My attempt</p>
<p>Let the sample be a 10-tuple <span class="math-container">$$S:=(X,\Omega,F)$$</span> <span class="math-container">$$X=\{i \ \epsilon \ Z:1\leq i\leq10\}$$</span> </p>
<p><span class="math-container">$$\Omega=\{\text{defective, nondefective}\} $$</span></p>
<p><span class="math-container">$$F:X\to\Omega$$</span></p>
<p><span class="math-container">$$Q=\{x \ \epsilon \ S:(\exists i \ \epsilon \ X) F(i)=\text{defective} \ \land \ (\forall n \ \epsilon \ X)n\ne i \to \ F(n)=\text{nondefective}\}$$</span></p>
<p>Assuming my notation is correct, the issue I see is that my quantifier for i does not preclude the existence of multiple defective parts being part of the tuple. What quantifier can I use that specifies existence and uniqueness? </p>
| Ben Dyer | 164,207 | <p>Left exact on all open sets implies left exact on stalks. This follows from exactness of direct limits for categories of modules, cf. <a href="https://math.stackexchange.com/questions/121122/why-do-direct-limits-preserve-exactness">Why do direct limits preserve exactness?</a>.</p>
|
8,237 | <p>In this example, I want a series of four buttons to change the value of a variable that is used dynamically to drive a plot. I am trying to figure out why using Table around the buttons causes a problem.</p>
<p>This works:</p>
<pre><code>{Button["1", freq = 1], Button["2", freq = 2], Button["3", freq = 3],
Button["4", freq = 4]}
Dynamic[Plot[Sin[freq * x], {x, 0, 2 \[Pi]}]]
</code></pre>
<p>Why doesn't this work:</p>
<pre><code>Table[Button[ToString[i], freq2 = i], {i, 4}]
Dynamic[Plot[Sin[freq2 * x], {x, 0, 2 \[Pi]}]]
</code></pre>
| Mr.Wizard | 121 | <p>As already stated this is an evaluation problem because of the Hold attribute that <code>Button</code> has, and the mechanism of <code>Table</code> which is akin to <code>Block</code>. You need a way to get the value into the expression. In this case I would use <code>Function</code> and <code>Array</code>:</p>
<pre><code>Array[Button[#, freq2 = #] &, 4]
Dynamic @ Plot[Sin[freq2*x], {x, 0, 2 π}]
</code></pre>
<p>(<code>ToString</code> is unnecessary.)</p>
<p>See the answers to this question for other options:</p>
<p><a href="https://mathematica.stackexchange.com/q/7756/121">Function in Table</a></p>
|
3,477,152 | <p>I am trying to prove that</p>
<p><span class="math-container">$$\sum_{n=1}^\infty \frac{n}{\sqrt{n+1}}$$</span></p>
<p>diverges without checking the limit, bounds or doing any other lengthy steps, as it should be seen as divergent "immediately", but I have no clue about how I would quickly prove this.</p>
<p>So far I thought about using the P-series convergence test where it only converges for <span class="math-container">$p>1$</span> but it does not seem to make any sense for this one. I also thought about comparing it to other series but nothing comes to my mind.</p>
| Lonidard | 206,444 | <p>The sequence <span class="math-container">$a_n = \frac{n}{\sqrt{n+1}}$</span> is asymptotic to <span class="math-container">$b_n ={\sqrt{n+1}}$</span>:
<span class="math-container">$$
\lim_{n\to \infty} \frac{a_n}{b_n} = \lim_{n\to \infty} \frac{n}{n+1} = 1
$$</span>
What can you say about the following series?
<span class="math-container">$$\sum_{n=1}^{\infty}\sqrt{n+1}$$</span></p>
|
2,434,373 | <p><strong>Question</strong></p>
<blockquote>
<p>Using Calculus, find points <span class="math-container">$A$</span> and <span class="math-container">$B$</span> on the parabola <span class="math-container">$y=1-x^2$</span> such that an equilateral triangle is formed by the <span class="math-container">$x$</span>-axis and the tangents <span class="math-container">$A$</span> and <span class="math-container">$B$</span>.</p>
</blockquote>
<p>If I'm not mistaken, I think it would look roughly something like this.</p>
<p><a href="https://i.stack.imgur.com/pHAEY.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pHAEY.jpg" alt="enter image description here" /></a></p>
| BBSysDyn | 6,786 | <p>Dr. Silver graciously responded and forwarded the paper <em>Monte-Carlo Simulation Balancing</em>, found at</p>
<p><a href="http://www.machinelearning.org/archive/icml2009/papers/500.pdf" rel="nofollow noreferrer">http://www.machinelearning.org/archive/icml2009/papers/500.pdf</a></p>
<p>There is a derivation there, looks similar to what I did.</p>
|
2,434,373 | <p><strong>Question</strong></p>
<blockquote>
<p>Using Calculus, find points <span class="math-container">$A$</span> and <span class="math-container">$B$</span> on the parabola <span class="math-container">$y=1-x^2$</span> such that an equilateral triangle is formed by the <span class="math-container">$x$</span>-axis and the tangents <span class="math-container">$A$</span> and <span class="math-container">$B$</span>.</p>
</blockquote>
<p>If I'm not mistaken, I think it would look roughly something like this.</p>
<p><a href="https://i.stack.imgur.com/pHAEY.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pHAEY.jpg" alt="enter image description here" /></a></p>
| A.D | 248,459 | <p>There is a small error in the second to last step:</p>
<p>$\nabla_\theta [ h(a,s,\theta) - log \sum_b h(b,s,\theta) \frac{\exp(h(b,s,\theta)}{\sum_b \exp(h(b,s,\theta))}]$</p>
<p>$\phi(a,s) - \nabla_\theta log \sum_b h(b,s,\theta) \frac{\exp(h(b,s,\theta)}{\sum_b \exp(h(b,s,\theta))}$</p>
<p>$\nabla_\theta \log(\sum_b \exp(h(b,s,\theta))) = \frac{\nabla_\theta \sum_b \exp(h(b,s,\theta))}{\sum_b \exp(h(b,s,\theta))} = \frac{\sum_b \exp(h(b,s,\theta))\hspace{0.2cm} \nabla_\theta [h(b,s,\theta)]}{\sum_b \exp(h(b,s,\theta))}$
(chain rule twice!)</p>
<p>Which gives:</p>
<p>$\phi(a,s) - \sum_b \frac{\exp(h(b,s,\theta)}{\sum_b \exp(h(b,s,\theta))}\hspace{0.2cm}\phi(b,s) $</p>
<p>$\phi(a,s) - \mathbb{E}_{b\sim \pi} \phi(b,s)$</p>
|
1,661,941 | <p>Evaluate </p>
<p>$\binom{n}0-2\binom{n}1 + 3\binom{n}2 +···+(−1)^n(n+1)\binom{n}n$ </p>
<p>which is the same as:</p>
<p>$\sum_{k=0}^n=(-1)^k(k+1)\binom{n}k$</p>
<p><strong>My attempt:</strong></p>
<p>Using the Binomial Theorem, we get:</p>
<p>$(1+x)^n=\binom{n}0+\binom{n}1x+\binom{n}2x^2+...+\binom{n}nx^n$</p>
<p>$(n+1)(1+x)^n = \binom{n}0+2\binom{n}1x+3\binom{n}2x^2+...+(n+1)\binom{n}nx^n$</p>
<p>Letting $x=-1$ we get:</p>
<p>$(n+1)(1-1)^n = \binom{n}0-2\binom{n}1+3\binom{n}2+...+(n+1)\binom{n}n(-1)^n$ </p>
<p>Therefore,</p>
<p>$\binom{n}0-2\binom{n}1 + 3\binom{n}2 +···+(−1)^n(n+1)\binom{n}n$ = $(n+1)(0)^n=0$</p>
<p>or </p>
<p>$\sum_{k=0}^n=(-1)^k(k+1)\binom{n}k$ = $0$</p>
| Andreas Caranti | 58,401 | <p>Let $n \ge 1$.</p>
<p>Start with $$(1+x)^n=\binom{n}{0}+\binom{n}{1} x+\binom{n}{2} x^2+ \dots +\binom{n}nx^n.$$</p>
<p>Multiply by $x$ to get
$$
x (1+x)^n=\binom{n}{0} x+\binom{n}{1} x^2+\binom{n}{2} x^3+ \dots +\binom{n}{n} x^{n+1}.
$$
Differentiate to get
$$
(1 + x)^n + n x (1 + x)^{n-1}
=
\binom{n}0+2\binom{n}1x+3\binom{n}2x^2+ \dots +(n+1)\binom{n}nx^n.
$$
Now set $x = -1$. Note that the case $n = 1$ is special.</p>
|
1,661,941 | <p>Evaluate </p>
<p>$\binom{n}0-2\binom{n}1 + 3\binom{n}2 +···+(−1)^n(n+1)\binom{n}n$ </p>
<p>which is the same as:</p>
<p>$\sum_{k=0}^n=(-1)^k(k+1)\binom{n}k$</p>
<p><strong>My attempt:</strong></p>
<p>Using the Binomial Theorem, we get:</p>
<p>$(1+x)^n=\binom{n}0+\binom{n}1x+\binom{n}2x^2+...+\binom{n}nx^n$</p>
<p>$(n+1)(1+x)^n = \binom{n}0+2\binom{n}1x+3\binom{n}2x^2+...+(n+1)\binom{n}nx^n$</p>
<p>Letting $x=-1$ we get:</p>
<p>$(n+1)(1-1)^n = \binom{n}0-2\binom{n}1+3\binom{n}2+...+(n+1)\binom{n}n(-1)^n$ </p>
<p>Therefore,</p>
<p>$\binom{n}0-2\binom{n}1 + 3\binom{n}2 +···+(−1)^n(n+1)\binom{n}n$ = $(n+1)(0)^n=0$</p>
<p>or </p>
<p>$\sum_{k=0}^n=(-1)^k(k+1)\binom{n}k$ = $0$</p>
| Brian M. Scott | 12,042 | <p>A combinatorial evaluation is also possible. Let </p>
<p>$$f(n)=\sum_{k=0}^n(-1)^k\binom{n}k(k+1)\;,$$</p>
<p>the sum in question. Then</p>
<p>$$\begin{align*}
f(n)&=\sum_{k=0}^n(-1)^k\binom{n}{n-k}(k+1)\\
&=\sum_{k=0}^n(-1)^{n-k}\binom{n}k(n-k+1)\\
&=(-1)^n\sum_{k=0}^n(-1)^k\binom{n}k(n+1-k)\;,
\end{align*}$$</p>
<p>and I’ll evaluate that last summation combinatorially. </p>
<p>As usual let $[n]=\{1,\ldots,n\}$, and let $A=[n]\cup\{0\}$; I’ll count the one-element subsets of $A$ that are equal to $\{k\}$ for each $k\in[n]$. For each $k\in[n]$ let</p>
<p>$$F_k=\big\{\{\ell\}:\ell\in A\setminus\{k\}\big\}\;;$$</p>
<p>if $\varnothing\ne J\subseteq[n]$, then</p>
<p>$$\left|\bigcap_{k\in J}F_k\right|=n+1-|J|\;,$$</p>
<p>so by the <a href="https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle#Statement" rel="nofollow">inclusion-exclusion principle</a> we have</p>
<p>$$\left|\bigcup_{k\in[n]}F_k\right|=\sum_{\varnothing\ne J\subseteq[n]}(-1)^{|J|+1}(n+1-J)=\sum_{k=1}^n(-1)^{k+1}\binom{n}k(n+1-k)\;.$$</p>
<p>This is the number of one-element subsets of $A$ that are different from at least one of the sets $\{k\}$ with $k\in[n]$. There are altogether $n+1$ one-element subsets of $A$, so there are</p>
<p>$$n+1-\sum_{k=1}^n(-1)^{k+1}\binom{n}k(n+1-k)=\sum_{k=0}^n(-1)^k\binom{n}k(n+1-k)$$</p>
<p>one-element subsets of $A$ that are equal to $\{k\}$ for <strong>every</strong> $k\in[n]$. </p>
<p>If $n\ge 2$, this is obviously $0$, since a one-element subset of $A$ cannot be simultaneously $\{1\}$ and $\{2\}$. If $n=1$, it’s $1$: the only subset that meets the requirement is $\{1\}$. And if $n=0$, it’s vacuously $1$. Thus,</p>
<p>$$f(n)=(-1)^n\sum_{k=0}^n(-1)^k\binom{n}k(n+1-k)=\begin{cases}
(-1)^0(1)=1,&\text{if }n=0\\
(-1)^1(1)=-1,&\text{if }n=1\\
(-1)^n(0)=0,&\text{if }n>1\;.
\end{cases}$$</p>
|
1,829,086 | <p>So far, I've tried out to reformulate: $$\int{\frac{1}{\cos(x)}}dx$$
to: $$\int{\frac{\sin(x)}{\cos(x)\sin(x)}}dx$$</p>
<p>which is basically: $$\int{\frac{\tan(x)}{\sin(x)}}dx$$
But I'm not sure if this is the right way to go, or if I try something else.</p>
<p>Any tips or methods would be very helpful.</p>
| haqnatural | 247,767 | <p><span class="math-container">$$\int { \frac { 1 }{\cos(x) } } dx=\int { \frac { \cos { x }\, dx }{ \cos ^{ 2 }{ x } } } =\int { \frac { d\left( \sin { x } \right) }{ 1-\sin^{ 2 }{ x } } } =\frac { 1 }{ 2 } \int { \left( \frac { 1 }{ 1-\sin { x } } +\frac { 1 }{ 1+\sin { x } } \right) d\left( \sin { x } \right) } =\frac { 1 }{ 2 } \ln { C\left| \frac { 1+\sin { x } }{ 1-\sin { x } } \right| } $$</span></p>
|
1,829,086 | <p>So far, I've tried out to reformulate: $$\int{\frac{1}{\cos(x)}}dx$$
to: $$\int{\frac{\sin(x)}{\cos(x)\sin(x)}}dx$$</p>
<p>which is basically: $$\int{\frac{\tan(x)}{\sin(x)}}dx$$
But I'm not sure if this is the right way to go, or if I try something else.</p>
<p>Any tips or methods would be very helpful.</p>
| user84413 | 84,413 | <p>$\displaystyle\int\frac{1}{\cos x}dx=\int\frac{1}{\sin(x+\frac{\pi}{2})}dx=\int\frac{1}{\sin t}dt\;\;$ with $t=x+\frac{\pi}{2}.\;\;$ Now let $u=\tan\frac{t}{2}$ to get</p>
<p>$\displaystyle\int\frac{1}{\frac{2u}{1+u^2}}\cdot\frac{2}{1+u^2}du=\int\frac{1}{u}du=\ln|u|+C=\ln\big|\tan\left(\frac{x}{2}+\frac{\pi}{4}\right)\big|+C$,</p>
<p>which can be rewritten as $\displaystyle\ln\left\vert\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}\right\vert+C=\ln\left\vert\frac{\cos\frac{x}{2}+\sin\frac{x}{2}}{\cos\frac{x}{2}-\sin\frac{x}{2}}\right\vert+C=\ln\left\vert\sec x+\tan x\right\vert+C$</p>
|
2,451,281 | <p>$$y=x+\frac{1}{x-4}$$ I tried to find range of this function as below
$$y(x-4)=x(x-4)+1\\yx-4y=x^2-4x+1\\x^2-x(4+y)+(4y+1)=0\\\Delta \geq 0 \\(4+y)^2-4(4y+1)\geq 0 \\(y-4)^2-4\geq 0\\|y-4|\geq 2\\y\geq 6 \cup y\leq2$$ this usuall way. but I am interested in finall answer... is the other idea to find function range ?</p>
| Peter Szilas | 408,605 | <p>Let $z:= x-4,$ and $z \ne 0$.</p>
<p>$y = z + 1/z +4$. </p>
<p>1) $z \gt 0:$</p>
<p>AM -GM: </p>
<p>$z+1/z \ge 2.$</p>
<p>Hence: $y \in [6,\infty)$.</p>
<p>2) $ z\lt 0$.</p>
<p>Above reasoning for $-z$ (positive),i.e.</p>
<p>$-z + 1/(-z) \ge 2$,</p>
<p>leads to:</p>
<p>$y \in (-\infty, 2]$.</p>
<p>Altogether :</p>
<p>Range $(y) = (-\infty, 2]\cup[4,\infty)$.</p>
|
728,916 | <p>I have been developing an RPG and I finally want to program a better development curve. I have decided to use the following equations</p>
<p>Linear, Quadratic and Cubic</p>
<p>The Linear one is easy. I have no problems with that. But I'm stuck with the other 2.</p>
<p><strong>Quadratic</strong></p>
<p>The formula for this equation is $y = ax^2 + bx + c$
What I want to do is find out what $a$ and $b$ are. Given that I know what $y$, $x$ and $c$ are. How would I figure out what this is?</p>
<p><strong>Cubic</strong></p>
<p>The formula for this equation is $y = ax^3 + bx^2 + cx + d$
What I want to do is find out what $a, b$ and $c$ are. Given that I know what $y, x$ and $d$ are. How would I figure out what this is?</p>
<p>Can anyone help me with this? This is the first time learning about curves and I am falling flat.</p>
<p><strong>Bonus Question</strong>
In the Cubic equation, there is a part that looks like the middle of an S. How would I extend that middle in the equation?</p>
<p><strong>Bonus Question 2</strong>
I'd love to know about more graph based curves built using equations. Do you know of any?</p>
<p>Thanks for any help!</p>
| dtldarek | 26,306 | <p>There are two aspects of your questions:</p>
<ul>
<li>How to do it in an abstract, ideal setting.</li>
<li>How to do it in practice.</li>
</ul>
<p>First, if you have $n$ unknowns, you need $n$ data points (modulo special cases), like pairs $(x,y)$. Then you get a system of <em>linear</em> equations, which you can easily solve.
I would start with <a href="http://en.wikipedia.org/wiki/System_of_linear_equations" rel="nofollow">Wikipedia</a>, but there are libraries which would solve them for you (however, please <em>do</em> start with some theory, it would be much easier for you later, as for the software, start with what <a href="http://en.wikipedia.org/wiki/Basic_Linear_Algebra_Subprograms" rel="nofollow">BLAS</a> is).</p>
<p>To give a concrete example, suppose that you would like your model to have the following formula $$y = ax^2+bx+c.$$</p>
<p>There are three unknowns, namely $a,b,c$, so you need three independent data points. Let's make it $(0,1), (2,4), (8,16)$ where the pair means $(x,y)$.
We put this into some engine, e.g. <a href="http://www.wolframalpha.com/input/?i=1%3Da%2a0%5E2%2Bb%2a0%2Bc%2C+4%3Da%2a2%5E2%2Bb%2a2%2Bc%2C+16%3Da%2a8%5E2%2Bb%2a8%2Bc" rel="nofollow">Wolfram Alpha</a> and obtain $a = \frac{1}{16}$, $b = \frac{11}{8}$ and $c = 1$.</p>
<p>Now, how to do it practice is a whole other question. In particular you will often have multiple data points which contain some noise, and you would like to model the general behavior. Some relevant topics are <a href="http://en.wikipedia.org/wiki/Curve_fitting" rel="nofollow">curve fitting</a> and <a href="http://en.wikipedia.org/wiki/Regression_analysis" rel="nofollow">regression analysis</a>. Be aware that this area is <em>huge</em> and there are whole courses that consider only a small part of it. If you want to dwell more on this, I would recommend machine learning first to gain a wider perspective (e.g. there are some great online resources on this). If you are <em>serious</em>, but want to stick to only a few methods, I would recommend processing your data with <a href="http://en.wikipedia.org/wiki/Kalman_filter" rel="nofollow">Kalman filters</a> (probably an overkill).</p>
<p>Finally, if this is not a system-critical part, some trail-and-error might give you better results than using some mathematical approach you have no experience with.</p>
<p>I hope this helps $\ddot\smile$</p>
|
233,397 | <p>could anyone please clarify me the meaning of the term 'hypothesis'? </p>
<p>with relation to terms 'reasoning' and 'assumption' ?</p>
<p>Many thanks</p>
| glebovg | 36,367 | <p>A hypothesis, in mathematics is just another word for conjecture, and conjectures are based on heuristic arguments, calculations, similar results etc. They are almost never simply guesses. In spite of induction and computer science tags, an induction hypothesis is different (a little). You try a few cases (typically refereed to as base cases) and than assume that a statement holds for some arbitrary integer, say $k$ -- this is the induction hypothesis, which you then use to prove that $k + 1$ case is also true. However, if you strongly believe in something (or you are given a homework problem) you usually consider one or two base cases.</p>
|
898,002 | <p>Let's say I have a continuous piecewise function of a single variable, so that $y = f(x)$ if $x < c$ and $y = g(x)$ if $x>=c$. Is it right to say that the derivative of the function at $x=c$ exists iff $f'(c-)=g'(c+)$, where $f'$ and $g'$ are obtained using derivative rules?</p>
<p>This would seem reasonable to me, and I fail to find an example where this does not hold. However, my calculus professors have always taught me that the only way to evaluate a derivative of such a point is using the limit definition of the derivative.</p>
| humble | 169,868 | <p>First you have to find RHD at $x=c$ this can be done by taking $\lim_{x\to a+}f'(x)$ provided</p>
<ol>
<li>function is right continuous at $x=a$</li>
<li>there exists some right neighbourhood where $f(x)$ is differentiable</li>
<li>$\lim_{x\to a+}$ exists it is equal to + or - infinity</li>
</ol>
<p>Similarly find LHD and if both are same then the derivative you get from here is your and.</p>
<p>Note, if any one of the above conditions fail you can't use this.</p>
|
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