qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
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386,018 | <p>I've been working with Spivak's Differential Geometry exercises and I found myself confused with this one: "Let $C\subset \mathbb{R} \subset \mathbb{R}^2$ be the Cantor set. Show that $\mathbb{R}^2 - C$ is homeomorphic to the surface shown at the top of the next page."</p>
<p><img src="https://i.stack.imgur.com/rARxC.jpg" alt="enter image description here"></p>
<p>Well, for me it's obviously homeomorphic: if we think of $\mathbb{R}$ inside $\mathbb{R}^2$ as the $x$-axis for instance, the set $\mathbb{R}^2 - C$ will be the plane minus the pieces of $\mathbb{R}$ that makes up Cantor set, so that we'll have in the end the plane with the $x$-axis with a lot of holes like the surface shown above. For me it's kind of intuitive that $\mathbb{R}^2-C$ can be continuously deformed into the surface above, but how do we really come tho <em>prove</em> this fact?</p>
<p>Since Spivak says that the pre-requisites are just his Calculus on Manifolds and basic notions of metric spaces I feel that we wouldn't need any advance topology for this job, but I'm a little confused really. </p>
<p>Can someone give a little advice on how do we solve problems like this?</p>
<p>Thanks very much in advance!</p>
| Matt E | 221 | <p>Remember that $C$ is obtained via the "middle thirds process": One takes the decreasing chain of closed sets $[0,1] = C_0 \supset C_1 \supset C_2 \supset \cdots,$ where $C_{n+1}$ is obtained from $C_n$ by removing the remaining "middle third" open intervals, and sets $C = \bigcap_{n=0}^{\infty} C_n$.</p>
<p>Thus $\mathbb R^2 \setminus C$ is equal to the union $\bigcup_{n=1}^{\infty} \mathbb R^2 \setminus C_n$. </p>
<p>Now, $C_n$ is a union of closed intervals. Let $D_n = C_n\times [0,1] \subset \mathbb R^2$. Note that deleting $\mathbb R^2 \setminus C_n$ is homemomorphic to $\mathbb R^2 \setminus D_n$, in way that is compatible with the inclusions as we pass from $n$ to $n+1$. (Exercise! And see below for a little more on this.)</p>
<p>Thus $\mathbb R^2 \setminus C$ is homemomorphic to the union $\mathbb R^2 \setminus D_n$. If you think about this union, you begin with $\mathbb R^2$ minus a closed square, then add in the "middle third" of that square, than the "middle thirds" of the two smaller remaining rectangles, and so on.</p>
<p>This gives you Spivak's picture, in the form described in <a href="https://math.stackexchange.com/a/386107/221">user75064's answer</a>, and with closed rectangles being removed rather than round disks. A final homeomorphism (intuivitely fairly evident, but a bit harder to write down) converts the rectangles to disks, and gives Spivak's picture.</p>
<hr>
<p>If you want to think about how to make all of this rigorous, my suggestion is to begin with proving that $\mathbb R^2$ minus a closed interval is homeomorphic to $\mathbb R^2$ minus a closed rectangle. </p>
<p>It is not so hard to prove this by a well-chosen explicit homeomorphism, but not completely trivial either. </p>
<p>The other steps, which involve removing the complements of rectangles by the complements of disks are harder to write down explicitly, but are (in my view) actually less deep than the preceding step, and relying on intuition for this kind of "deformation of shape" homeomorphism is not so bad. (When your topology
becomes a bit more sophisticated, you will prove, or learn the techniques for proving, pretty general results of this kind, which give existence results for the relevant homeomorphisms without writing down explicit formulas.)</p>
|
1,488,501 | <p>Let $\varphi:\mathbb{R}\backslash\{3\}\to \mathbb{R}$ a periodic function so that forall $x\in \mathbb{R}$ $$\varphi(x+4)=\frac{\varphi(x)-5}{\varphi(x)-3}$$ Find the period the $\varphi$.</p>
| Winther | 147,873 | <p>With $f(x) = \frac{x-5}{x-3}$ the functional equation can be written</p>
<p>$$\varphi(x+4) = f\circ \varphi(x)$$</p>
<p>A direct calculation shows that
$$f\circ f\circ f\circ f = \text{Id} \implies \varphi(x+16) = \varphi(x)$$
which implies that the period satisfy $T = \frac{16}{m}$ for some $m\in\mathbb{N}$. We can rule out $4\mid m$ as a possibillity since </p>
<p>$$f\circ \varphi(x) = \varphi(x+4) = \varphi\left(x+ T \cdot \frac{m}{4}\right) = \varphi(x)\implies \varphi(x) = 2\pm i \not\in\mathbb{R}$$</p>
<p>and we can also rule out $m \equiv 2\pmod 4$ by the same type of argument.</p>
<p>The remaining values $m\equiv \pm 1\pmod{4}$ are all possible and we can prove this by explcitly constructing solutions with the desired period. The form of the equation looks similar to the addition formula for $\tan(x+y)$ so taking the ansatz $\varphi(x) = A\tan(kx) + B$ in the functional equation and solving for $A,B,k$ we find the following family of solutions</p>
<p>$$\varphi(x) = \tan\left(\frac{\pi x}{T}\right)+2$$</p>
<p>where $T = \frac{16}{4n\pm 1}$ with $n\in\mathbb{N}$.</p>
|
1,765,530 | <p>How many $5$-digit numbers (including leading $0$'s) are there with no digit appearing exactly $2$ times? The solution is supposed to be derived using Inclusion-Exclusion.</p>
<p>Here is my attempt at a solution:</p>
<p>Let $A_0$= sequences where there are two $0$'s that appear in the sequence.</p>
<p>...</p>
<p>$A_{9}$=sequences where there are two $9$'s that appear in the sequence.</p>
<p>I want the intersection of $A_0^{'}A_1^{'}...A_9^{'}$= $N-S_1+S_2$ because you can only have at most two digits who are used exactly two times each in a $5$ digit sequence.</p>
<p>$N=10^5$, $S_1=10\cdot \binom{5}{2}\cdot[9+9\cdot 8 \cdot 7]$, and $S_2=10 \cdot 9 \cdot \binom{5}{4} \cdot8$.</p>
<p>The $S_1$ term comes from selecting which of the ten digits to use twice, selecting which two places those two digits take, and then either having the same digit used three times for the other three places, or having different digits used for the other three digits.</p>
<p>The $S_2$ term comes from selecting which two digits are used twice, selecting where those four digits go, and then having eight choices for the remaining spot.</p>
<p>So my answer becomes $10^5 -10 \cdot \binom{5}{2} \cdot [9+9 \cdot 8 \cdot 7]+10 \cdot 9 \cdot \binom{5}{4} \cdot 8$.</p>
<p>Am I doing this correctly?</p>
| André Nicolas | 6,312 | <p>The <em>structure</em> of the analysis is the same as yours. We count the <em>bad</em> strings, where some digit appears exactly twice. </p>
<p>We first count the strings where say the digit $0$ appears exactly twice. <strong>Where</strong> the $0$'s are can be chosen in $\binom{5}{2}$ ways. For each of these ways, the remaining $3$ places can be filled in $9^3$ ways.</p>
<p>So our first estimate of the number of bad strings is $(10)\binom{5}{2}(9^3)$.</p>
<p>But we have double-counted, for example, the strings where $0$ and $1$ appear exactly twice. There are $\binom{5}{2}$ ways to choose where the $0$'s go, and for each there are $\binom{3}{2}$ ways to choose where the $1$'s go. The remaining spot can be filled in $8$ ways. Multiply by $\binom{10}{2}$ for the number of ways to choose the two digits that appear twice. </p>
<p>Putting things together, we find that the number of bad strings is
$$(10)\binom{5}{2}(9^3)-\binom{10}{2}\binom{5}{2}\binom{3}{2}(8).\tag{1}$$</p>
<p>Finally, subtract (1) from $10^5$.</p>
|
2,865,122 | <p><a href="http://math.sfsu.edu/beck/complex.html" rel="nofollow noreferrer">A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka</a> Exer 3.8</p>
<blockquote>
<p>Suppose <span class="math-container">$f$</span> is holomorphic in region <span class="math-container">$G$</span>, and <span class="math-container">$f(G) \subseteq \{ |z|=1 \}$</span>. Prove <span class="math-container">$f$</span> is constant.</p>
</blockquote>
<p>(I guess we may assume <span class="math-container">$f: G \to \mathbb C$</span> s.t. image(f)=<span class="math-container">$f(G)$</span>. I guess it doesn't matter if we somehow have <span class="math-container">$f: A \to \mathbb C$</span> for any <span class="math-container">$A$</span> s.t. <span class="math-container">$G \subseteq A \subseteq \mathbb C$</span> as long as <span class="math-container">$G$</span> is a region and <span class="math-container">$f$</span> is holo there.)</p>
<p>I will now attempt to elaborate the following proof at a <a href="https://math.oregonstate.edu/%7Edeleenhp/teaching/winter17/MTH483/hw2-sol.pdf" rel="nofollow noreferrer">Winter 2017 course in Oregon State University</a>.</p>
<p><strong>Question 1</strong>: For the following elaboration of the proof, what errors if any are there?</p>
<p><strong>Question 2</strong>: Are there more elegant ways to approach this? I have a feeling this can be answered with Ch2 only, i.e. Cauchy-Riemann or differentiation/holomorphic properties instead of having to use Möbius transformations.</p>
<blockquote>
<p>OSU Pf (slightly paraphrased): Let <span class="math-container">$g(z)=\frac{1+z}{1-z}$</span>, and define <span class="math-container">$h(z)=g(f(z)), z \in G \setminus \{z : f(z) = 1\}$</span>. Then <span class="math-container">$h$</span> is holomorphic on its domain, and <span class="math-container">$h$</span> is imaginary valued by Exer 3.7. By a variation of Exer 2.19, <span class="math-container">$h$</span> is constant. QED</p>
</blockquote>
<p>My (elaboration of OSU) Pf: <span class="math-container">$\because f(G) \subseteq C[0,1]$</span>, let's consider the Möbius transformation in the preceding Exer 3.7 <span class="math-container">$g: \mathbb C \setminus \{z = 1\} \to \mathbb C$</span> s.t. <span class="math-container">$g(z) := \frac{1+z}{1-z}$</span>:</p>
<p>If we plug in <span class="math-container">$C[0,1] \setminus \{1\}$</span> in <span class="math-container">$g$</span>, then we'll get the imaginary axis by Exer 3.7. Precisely: <span class="math-container">$$g(\{e^{it}\}_{t \in \mathbb R \setminus \{0\}}) = \{is\}_{s \in \mathbb R}. \tag{1}$$</span> Now, define <span class="math-container">$G' := G \setminus \{z \in G | f(z) = 1 \}$</span> and <span class="math-container">$h: G' \to \mathbb C$</span> s.t. <span class="math-container">$h := g \circ f$</span> s.t. <span class="math-container">$h(z) = \frac{1+f(z)}{1-f(z)}$</span>. If we plug in <span class="math-container">$G'$</span> in <span class="math-container">$h$</span>, then we'll get the imaginary axis. Precisely: <span class="math-container">$$h(G') := \frac{1+f(G')}{1-f(G')} \stackrel{(1)}{=} \{is\}_{s \in \mathbb R}. \tag{2}$$</span></p>
<p>Now Exer 2.19 says that a real valued holomorphic function over a region is constant: <span class="math-container">$f(z)=u(z) \implies u_x=0=u_y \implies f'=0$</span> to conclude by Thm 2.17 that <span class="math-container">$f$</span> is constant or simply by partial integration that <span class="math-container">$u$</span> is constant. Actually, an imaginary valued holomorphic function over a region is constant too: <span class="math-container">$f(z)=iv(z) \implies v_x=0=v_y \implies f'=0$</span> again by Cauchy-Riemann Thm 2.13 to conclude by Thm 2.17 that <span class="math-container">$f$</span> is constant or simply by partial integration that <span class="math-container">$v$</span> is constant.</p>
<p><span class="math-container">$(2)$</span> precisely says that <span class="math-container">$h$</span> is imaginary valued over <span class="math-container">$G'$</span>. <span class="math-container">$\therefore,$</span> if <span class="math-container">$G'$</span> is a region (A) and if <span class="math-container">$h$</span> is holomorphic on <span class="math-container">$G'$</span> (B), then <span class="math-container">$h$</span> is constant on <span class="math-container">$G'$</span> with value I'll denote <span class="math-container">$Hi, H \in \mathbb R$</span>:</p>
<p><span class="math-container">$\forall z \in G',$</span></p>
<p><span class="math-container">$$Hi = \frac{1+f(z)}{1-f(z)} \implies f(z) = \frac{Hi-1}{Hi+1}, \tag{3}$$</span></p>
<p>where <span class="math-container">$Hi+1 \ne 0 \forall H \in \mathbb R$</span>.</p>
<p><span class="math-container">$\therefore, f$</span> is constant on <span class="math-container">$G'$</span> (Q4) with value given in <span class="math-container">$(3)$</span>.</p>
<p>QED except possibly for (C)</p>
<hr />
<blockquote>
<p>(A) <span class="math-container">$G'$</span> is a region</p>
</blockquote>
<p>I guess if <span class="math-container">$G \setminus G'$</span> is finite, then G' is a region. I'm thinking <span class="math-container">$D[0,1]$</span> is a region and then <span class="math-container">$D[0,1] \setminus \{0\}$</span> is still a region.</p>
<blockquote>
<p>(B) To show <span class="math-container">$h$</span> is holomorphic in <span class="math-container">$G'$</span>:</p>
</blockquote>
<p>Well <span class="math-container">$h(z)$</span> is differentiable <span class="math-container">$\forall z \in G'$</span> and <span class="math-container">$f(z) \ne 1 \forall z \in G'$</span> and <span class="math-container">$f'(z)$</span> exists in <span class="math-container">$G' \subseteq G$</span> because <span class="math-container">$f$</span> is differentiable in <span class="math-container">$G$</span> because <span class="math-container">$f$</span> is holomorphic in <span class="math-container">$G$</span>.</p>
<p><span class="math-container">$$h'(z) = g'(f(z)) f'(z) = \frac{2}{(1-w)^2}|_{w=f(z)} f'(z) = \frac{2 f'(z)}{(1-f(z))^2} $$</span></p>
<p>Now, <span class="math-container">$f'(z)$</span> exists on an open disc <span class="math-container">$D[z,r_z] \ \forall z \in G$</span> where <span class="math-container">$r_z$</span> denotes the radius of the open disc s.t. <span class="math-container">$f(z)$</span> is holomorphic at <span class="math-container">$z$</span>. So, I guess <span class="math-container">$\frac{2 f'(z)}{(1-f(z))^2} = h'(z)$</span> exists on an open disc with the same radius <span class="math-container">$D[z,r_z] \ \forall z \in G'$</span>, and <span class="math-container">$\therefore, h$</span> is holomorphic in <span class="math-container">$G'$</span>.</p>
<blockquote>
<p>(C) Possible flaw:</p>
</blockquote>
<p>It seems that on <span class="math-container">$G'$</span>, <span class="math-container">$f$</span> has value <span class="math-container">$\frac{Hi-1}{Hi+1}$</span> while on <span class="math-container">$G \setminus G'$</span>, <span class="math-container">$f$</span> has value <span class="math-container">$1$</span>.</p>
<p><span class="math-container">$$\therefore, \forall z \in G, f(z) = \frac{Hi-1}{Hi+1} 1_{G'}(z) + 1_{G \setminus G'}(z)$$</span></p>
<p>It seems then that we've actually show only that <span class="math-container">$f$</span> is constant on <span class="math-container">$G$</span> except for the subset of G where <span class="math-container">$f=1$</span>.</p>
| BCLC | 140,308 | <p>After discovering the term anti-holomorphic, I have an answer similar to zhw.'s: It starts the same but then doesn't reinvent the wheel or something:</p>
<blockquote>
<p>If <span class="math-container">$f: G \to \mathbb C$</span> is holomorphic on <span class="math-container">$G$</span> and anti-holomorphic on <span class="math-container">$G$</span>, then <span class="math-container">$f$</span> is constant on <span class="math-container">$G$</span>. <span class="math-container">$\tag{*}$</span></p>
</blockquote>
<p>We have <span class="math-container">$|f|=1$</span> on <span class="math-container">$G$</span> iff <span class="math-container">$f \overline f = 1$</span> on <span class="math-container">$G$</span>. Then just as in zhw.'s answer, <span class="math-container">$\overline f$</span> is holomorphic in <span class="math-container">$G$</span> i.e. <span class="math-container">$f$</span> is anti-holomorphic on <span class="math-container">$G$</span>. Therefore, <span class="math-container">$f$</span> is constant on <span class="math-container">$G$</span>.</p>
<p>Now, the non-wheel reinvention here is that the way one proves <span class="math-container">$(*)$</span> in the 1st place is what zhw. said. So yeah this exercise seems pretty easy if you know the concept of anti-holomorphic. It's not so advanced to teach an elementary complex analysis student, but maybe it's not something you'd teach an elementary complex analysis student.</p>
|
223,176 | <p>I made this problem: </p>
<p>$f(x)=e^{f^{\prime \prime}}$ </p>
<p>I have just been taught the first derivative, and was thinking about what if the derivative depended upon it own derivative. I understand that $e^x$ is its "own" derivative, but the problem I made I was thinking that the first derviative is not logical, because to know the first derivative you then must know the 2nd or 3rd derviative, it seems self-referenecing. </p>
<p>Is the problem I made, a real problem or just some abstract idea? </p>
| Henrique Tyrrell | 17,153 | <p>$\frac{d}{dx} [f(x)] = \frac{d}{dx}\exp (f''(x))$ and by the chain rule its derivative is
$\frac{d}{dx}f''(x) \times \frac{d}{d(f''(x))} \exp(f''(x)) = f'''(x) \times \exp(f''(x))$</p>
|
873,439 | <p>This is a follow-up <a href="https://math.stackexchange.com/questions/872921/prove-that-if-the-square-of-a-number-m-is-a-multiple-of-3-then-the-number-m/872927">question</a>.</p>
<p>The problem is:</p>
<blockquote>
<p>Given two natural numbers, $m$ and $n$, and $n \vert m^2$.</p>
<p>Find necessary and sufficient conditions for $n \vert m$.</p>
</blockquote>
<p>Here are what I find:</p>
<ul>
<li><p>Necessity</p>
<blockquote>
<ol>
<li>$m \geq n$ (trivial)</li>
<li>?</li>
</ol>
</blockquote></li>
<li><p>Sufficiency</p>
<blockquote>
<ol>
<li>$n$ is prime - follows directly from <strong>Matthias</strong>'s answer</li>
<li>$\color{#c00}{n \text{ is} ~square\!-\!free,}$ $\color{#c00}{\text{i.e., has no} ~repeated~prime~factor}$ (stated in <strong>JHance</strong>'s comment to my answer) - as pointed out by <strong>gammatester</strong>, it is wrong.</li>
<li>?</li>
</ol>
</blockquote></li>
</ul>
<p>Help me to complete this list, folks.</p>
| hola | 154,508 | <p>I think I found the solution. Do correct me if I am wrong.</p>
<blockquote>
<p>$n|m$ if and only if there is no such prime $p$ such that $p^2|n$.</p>
</blockquote>
<p><code>If</code> part then it is easy, for example $4|14^2$ but $4\nmid 14$.</p>
<p>For the <code>only if</code> part, we can show a contradiction.</p>
|
2,994,296 | <p>I'm trying to figure out how to prove, that <span class="math-container">$$\lim_{n\to \infty} \frac{n^{4n}}{(4n)!} = 0$$</span>
The problem is, that <span class="math-container">$$\lim_{n\to \infty} \frac{n^{n}}{n!} = \infty$$</span>
and I have no idea how to prove the first limit equals <span class="math-container">$0$</span>. </p>
| Community | -1 | <p>The ratio of successive terms is</p>
<p><span class="math-container">$$\frac{t_{n+1}}{t_n}=\frac{\dfrac{(n+1)^{4n+4}}{(4n+4)!}}{\dfrac{n^{4n}}{(4n)!}}=\left(\left(\frac{n+1}n\right)^n\right)^4\frac{(n+1)^4}{(4n+1)(4n+2)(4n+3)(4n+4)}.$$</span></p>
<p>The first factor is known to describe an increasing sequence that converges to <span class="math-container">$e^4$</span>, while the second is asymptotic to <span class="math-container">$\dfrac1{256}$</span> and can be bounded above by <span class="math-container">$\dfrac1{105}$</span>.</p>
<p>Finally,</p>
<p><span class="math-container">$$\frac{t_{n+1}}{t_n}<\frac{e^4}{105}<1.$$</span></p>
|
2,880,830 | <p><a href="https://i.stack.imgur.com/a4Wh2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/a4Wh2.png" alt="enter image description here"></a></p>
<p>Not sure how to do this one. If <span class="math-container">$S$</span> is a field, then I was considering that <span class="math-container">$\exists r_1,\ldots, r_n\in R$</span> s.t. <span class="math-container">$1 = r_1s_1+\cdots+r_ns_n$</span> so for <span class="math-container">$r = rr_1s_1+\cdots+rr_ns_n$</span>. Maybe that is somehow useful for taking inverses of elements. </p>
<p>The assumption that <span class="math-container">$S$</span> is an integral domain is necessary because otherwise we could have <span class="math-container">$S = \mathbb{Z}_p[x]/f(x)$</span> where <span class="math-container">$f(x)$</span> is not irreducible. This is still a finitely generated <span class="math-container">$\mathbb{Z}_p$</span>-module, but its not a field. </p>
<p>Any hints or solutions would be much appreciated. I feel like this isn't that hard and I'm missing something simple</p>
<p>Source: <a href="https://dornsife.usc.edu/assets/sites/363/docs/F17_510ab.pdf" rel="nofollow noreferrer">https://dornsife.usc.edu/assets/sites/363/docs/F17_510ab.pdf</a></p>
| Bernard | 202,857 | <p><strong>Hints</strong>:</p>
<p>$\Rightarrow$: If $R$ is a field, let $s\in S$, and consider multiplication by $s$ in $S$. Check this is an injective $R$-linear map. What can you conclude, knowing $S$ is a finite dimensional $R$-vector space?</p>
<p>$\Leftarrow$: If $S$ is a field, consider $r\in R$; you know $r^{-1}\in S$, hence it is a root of a monic polynomial in $R[X]$. Deduce from this polynomial equation that $r^{-1}$ is a polynomial in $r$, hence it belongs to $R$.</p>
|
3,844,448 | <p>Find all values of <span class="math-container">$h$</span> such that rank(<span class="math-container">$A$</span>) = <span class="math-container">$2$</span>.</p>
<p><span class="math-container">$A$</span> = <span class="math-container">$\begin{bmatrix}
1 & h & -1\\
3 & -1 & 0\\
-4 & 1 & 3
\end{bmatrix} $</span></p>
<p>I used row transformations to get</p>
<p><span class="math-container">$A$</span> = <span class="math-container">$\begin{bmatrix}
1 & h & -1\\
0 & -1-3h & 3\\
0 & 1+4h & -1
\end{bmatrix} $</span></p>
<p>But how do I solve to get the rank? I know the general idea is that rank(<span class="math-container">$A$</span>) = <span class="math-container">$2$</span> when dim(col(<span class="math-container">$A$</span>)) = dim(row(<span class="math-container">$A$</span>)) = <span class="math-container">$2$</span></p>
| Andrei | 331,661 | <p>You know that the first row is independent, but the second and third rows must be dependent. In other words, you can write <span class="math-container">$$-1-3h=C(1+4h)\\3=C(-1)$$</span>
Therefore <span class="math-container">$C=-3$</span>. Plug it into the first equation, to find <span class="math-container">$h$</span></p>
|
79,084 | <p>Let $X$ be a topological space (say a manifold). A result of R. Thom states that the pushforwards of fundamental classes of closed, smooth manifolds generate the rational homology of $X$. This work of Thom predates the development of bordism. Is there now a more elementary proof of this result that does not rely on spectral sequence techniques?</p>
| Lost | 18,632 | <p>Thank you everyone for helping with this question. I would like to attempt to provide my own answer (which came to me <em>after</em> reading all the comments): Let <span class="math-container">$B_* (M)$</span> be rational, oriented bordism and <span class="math-container">$H_* (M)$</span> be the rational homology of <span class="math-container">$M$</span>. I claim there is a map <span class="math-container">$$F:B_* (M) \rightarrow H_* (M)\otimes B_* (pt)$$</span> that is an isomorphism. The map <span class="math-container">$F$</span> sends <span class="math-container">$(P \rightarrow M) $</span> to <span class="math-container">$([P ] \otimes 1+1\otimes P')$</span> where <span class="math-container">$[P]\in H_* (M) $</span> represents the fundamental class of <span class="math-container">$P$</span> and <span class="math-container">$P' \in B_* (pt)$</span> is the bordism element represented by the abstract manifold <span class="math-container">$P$</span>. This map is clearly an isomorphism when <span class="math-container">$M= pt$</span>. The standard inductive argument over the number of cells implies this is an isomorphism in general. Note that no appeal to homotopy groups, spectral sequences or Thom spectra is being implicitly used. We do use the computation of 0 dimensional bordism groups.</p>
<p>Is this correct?</p>
|
95,819 | <p>I have a set of parametric equations in spherical coordinates that supposedly form circle trajectories. See below:</p>
<pre><code>r=C1
theta=C2*Sin[beta]*Sin[phi[t]]
phi=(C2*Sin[beta]*(Cos[theta[t]]/Sin[theta[t]])*Cos[phi[t]])+(C2*Cos[beta])
</code></pre>
<p>C1 and C2 are constants and beta is some angle, say 15 degrees, or (15/180)*Pi radians. </p>
<p>These are circle trajectories on the surface of a sphere, hence the constant r-component. </p>
<p>My question is this: How do I solve these trajectories and plot them on the surface of a sphere. This is what I have done already:</p>
<p>Step 1:</p>
<p>Solve the coupled differential equation with NDSolve. See below:</p>
<pre><code>Answer = NDSolve[
{theta'[t] == C2*Sin[beta]*Sin[phi[t]],
phi'[t] == (C2*Sin[beta]*Cos[theta[t]]*
Cos[phi[t]]/Sin[theta[t]]) + (C2*Cos[beta]),
theta[tMin] == StartingTheta,
phi[tMin] == StartingPhi},
{theta, phi},
{t, tMin, tMax}];
</code></pre>
<p>where I have defined StartingTheta=(45/180)*Pi and StartingPhi=(180/180)*Pi. tMin is 0 and tMax is, say, a 1000.</p>
<p>This will form ONE circle trajectory on the surface of the sphere. Changing StartingTheta will give another circle trajectory and so forth.</p>
<p>Step 2: Create the Sphere. I did that - see below: </p>
<pre><code>sphere = ParametricPlot3D[
{Cos[v]*Cos[u],
Sin[v]*Cos[u],
Sin[u]},
{v, 0, 2*Pi},
{u, -Pi/2, Pi/2}];
</code></pre>
<p>Step 3:</p>
<p>Evaluate. This is where I am struggling. See below what I have done so far:</p>
<pre><code>Trajectory = ParametricPlot3D[
Evaluate[
{????????????????}
/. Answer],
{t, tMin, tMax}
]
</code></pre>
<p>At the place where I inserted all the question marks is the problem, I am not sure what form I should be evaluating. I tried a few obvious expressions but I keep getting straight lines.</p>
<p>The next step I suppose is quite easy:</p>
<pre><code>Show[sphere,Trajectory]
</code></pre>
<p>If anybody out there is able to help me with what I should be evaluating in order to plot these circle trajectories on the surface of the sphere, it would be greatly appreciated. </p>
<p>Thanx!</p>
| Jack LaVigne | 10,917 | <p>I am quite certain that I don't understand your function but may be able to help with the format for plotting with <code>MeshFunction</code>.</p>
<pre><code>C2 = 1.;
beta = 15. Degree;
StartingTheta = 45. Degree;
StartingPhi = 180. Degree;
tMin = 0;
tMax = 1000.;
</code></pre>
<p>Using these parameters</p>
<pre><code>Answer = NDSolve[
{theta'[t] == C2*Sin[beta]*Sin[phi[t]],
phi'[t] == (C2*Sin[beta]*Cos[theta[t]]*
Cos[phi[t]]/Sin[theta[t]]) + (C2*Cos[beta]),
theta[tMin] == StartingTheta,
phi[tMin] == StartingPhi},
{theta, phi},
{t, tMin, tMax}]
</code></pre>
<p>results in two InterpolatingFunctions for <code>theta</code> (Answer[[1,1,2]]) and <code>phi</code> (Answer[[1,2,2]]).</p>
<p>The <code>theta</code> function is cyclic</p>
<pre><code>Plot[Answer[[1, 1, 2]][v], {v, 0, 6 π}, PlotLabel -> "theta"]
</code></pre>
<p><img src="https://i.stack.imgur.com/dBMkI.png" alt="Mathematica graphics"></p>
<p>The <code>phi</code> function starts low and continues to grow although it does have some periodicity.</p>
<pre><code>Plot[Answer[[1, 2, 2]][u], {u, 0, 6 π}, PlotLabel -> "phi"]
</code></pre>
<p><img src="https://i.stack.imgur.com/SPNTL.png" alt="Mathematica graphics"></p>
<p>Below is the format using these two functions as <code>MeshFunction</code>. <code>theta</code> is being applied to the parameter <code>u</code> and <code>phi</code> is applied to <code>v</code>.</p>
<pre><code>ParametricPlot3D[{Cos[v]*Cos[u], Sin[v]*Cos[u], Sin[u]}, {u, 0,
2*Pi}, {v, -Pi/2, Pi/2},
Mesh -> {16, 8},
MeshFunctions -> {
Answer[[1, 1, 2]][#4] &,
Answer[[1, 2, 2]][#5] &
},
MeshStyle -> {Red, Blue}
]
</code></pre>
<p><img src="https://i.stack.imgur.com/yzX1a.png" alt="Mathematica graphics"></p>
|
95,819 | <p>I have a set of parametric equations in spherical coordinates that supposedly form circle trajectories. See below:</p>
<pre><code>r=C1
theta=C2*Sin[beta]*Sin[phi[t]]
phi=(C2*Sin[beta]*(Cos[theta[t]]/Sin[theta[t]])*Cos[phi[t]])+(C2*Cos[beta])
</code></pre>
<p>C1 and C2 are constants and beta is some angle, say 15 degrees, or (15/180)*Pi radians. </p>
<p>These are circle trajectories on the surface of a sphere, hence the constant r-component. </p>
<p>My question is this: How do I solve these trajectories and plot them on the surface of a sphere. This is what I have done already:</p>
<p>Step 1:</p>
<p>Solve the coupled differential equation with NDSolve. See below:</p>
<pre><code>Answer = NDSolve[
{theta'[t] == C2*Sin[beta]*Sin[phi[t]],
phi'[t] == (C2*Sin[beta]*Cos[theta[t]]*
Cos[phi[t]]/Sin[theta[t]]) + (C2*Cos[beta]),
theta[tMin] == StartingTheta,
phi[tMin] == StartingPhi},
{theta, phi},
{t, tMin, tMax}];
</code></pre>
<p>where I have defined StartingTheta=(45/180)*Pi and StartingPhi=(180/180)*Pi. tMin is 0 and tMax is, say, a 1000.</p>
<p>This will form ONE circle trajectory on the surface of the sphere. Changing StartingTheta will give another circle trajectory and so forth.</p>
<p>Step 2: Create the Sphere. I did that - see below: </p>
<pre><code>sphere = ParametricPlot3D[
{Cos[v]*Cos[u],
Sin[v]*Cos[u],
Sin[u]},
{v, 0, 2*Pi},
{u, -Pi/2, Pi/2}];
</code></pre>
<p>Step 3:</p>
<p>Evaluate. This is where I am struggling. See below what I have done so far:</p>
<pre><code>Trajectory = ParametricPlot3D[
Evaluate[
{????????????????}
/. Answer],
{t, tMin, tMax}
]
</code></pre>
<p>At the place where I inserted all the question marks is the problem, I am not sure what form I should be evaluating. I tried a few obvious expressions but I keep getting straight lines.</p>
<p>The next step I suppose is quite easy:</p>
<pre><code>Show[sphere,Trajectory]
</code></pre>
<p>If anybody out there is able to help me with what I should be evaluating in order to plot these circle trajectories on the surface of the sphere, it would be greatly appreciated. </p>
<p>Thanx!</p>
| Maxwell | 34,479 | <p>Thanx for the answers I received concerning this question. @gpap - your answer was/is superb! Below is another take on it:</p>
<p>I used all the coding in the original question above, but this time I inserted the standard Cartesian-Spherical transformations where I had all the question marks. </p>
<pre><code>Trajectory = ParametricPlot3D[
Evaluate[
{1*Sin[theta[t]]*Cos[phi[t]],
1*Sin[theta[t]]*Sin[phi[t]],
1*Cos[theta[t]]}
/. Answer],
{t, tMin, tMax}, PlotStyle ->Green];
</code></pre>
<p>This formes ONE trajectory around the surface of the sphere. Changing the <code>StartingTheta</code> will generate another unique trajectory. The <code>beta</code>
angle involved is the off-set angle form the pole of the sphere. If <code>beta</code> was equal to zero, all these trajectories would have been concentric around the pole of the sphere. For good measure I added a new axis to show all the trajectories are concentric around this new axis. </p>
<p><a href="https://i.stack.imgur.com/ktWgb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ktWgb.png" alt="Sphere with trajectories on surface"></a></p>
|
1,967,847 | <blockquote>
<p>A vector space $V$ is called <strong>finite-dimensional</strong> if there is a finite subset of $V$ that is a basis for $V$. If there is no such finite subset of $V$, then $V$ is called <strong>infinite-dimensional</strong>.</p>
<hr>
<p>We now establish some results about finite-dimensional vector spaces that will tell about the number of vectors in a basis, compare two different bases, and give properties of bases. First, we observe that if $\{\mathbf{v}_1, \mathbf v_2,\dotsc, \mathbf v_k\}$ is a basis for a vector space $V$, then $\{c\mathbf v_1, \mathbf v_2, \dotsc, \mathbf v_k\}$ is also a basis when $c\neq 0$ (Exercise 35). Thus a basis for a nonzero vector space is never unique.<br>
<a href="https://i.stack.imgur.com/ghEh9.png" rel="nofollow noreferrer">Image.</a></p>
</blockquote>
<p>I am confused if $\mathbb R^2$ is a finite dimensional vector space. $[1 \ 0],[0 \ 1]$ will be the standard basis of $\mathbb R^2$. However, there are also $[2 \ 0],[0 \ 1]$ and I can find infinite many to be the basis of $\mathbb R^2$. So, $\mathbb R^2$ is an infinite dimensional vector space? </p>
| Junkai Dong | 366,631 | <p>I suppose you are mixing the terms "finite subset" and "finite number of subset".</p>
<p>The first one corresponds to the fact that in the subset there are only finitely many ${\it elements}$; the second one corresponds to the fact that there are finitely many ${\it subsets}$.</p>
<p>Hope I helped. I also mixed these two concepts at the start of learning linear algebra, so I think you met the same problem.</p>
|
201,688 | <p>There is a right angle corner with width 1 in both directions. One wants to find the largest area shape which can pass through this corner.
I know that this is a famous problem, but what is it called?</p>
| Bob Spaghetti | 60,810 | <p><a href="http://en.wikipedia.org/wiki/Moving_sofa_problem">The Moving sofa problem</a>, I believe.</p>
|
201,688 | <p>There is a right angle corner with width 1 in both directions. One wants to find the largest area shape which can pass through this corner.
I know that this is a famous problem, but what is it called?</p>
| Joseph O'Rourke | 6,094 | <p>A supplement to Ian's answer: Here is the largest-area sofa known,
due to Gerver:
<hr />
<img src="https://i.stack.imgur.com/FBlms.png" alt="GerverSofa"></p>
<hr />
<blockquote>
<p>Gerver, Joseph L. (1992). "On Moving a Sofa Around a Corner". <em>Geometriae Dedicata</em> 42 (3): 267–283. (<a href="http://link.springer.com/article/10.1007/BF02414066" rel="nofollow noreferrer">Springer link</a>.)</p>
</blockquote>
<p><strong>Added</strong> (triggered by @GeraldEdgar's remark).
The computational complexity of algorithms grows
exponentially in the dimension, about $n^5$ for
polyhedral objects with $n$ vertices moving in $\mathbb{R}^3$.
Here is an algorithm moving an $n{=}4500$-triangle piano
through a challenging apartment requiring several tricky maneuvers:
<hr />
<img src="https://i.stack.imgur.com/le9JF.png" alt="Piano"></p>
<hr />
<blockquote>
<p>Kuffner, James J., and Steven M. LaValle. "RRT-connect: An efficient approach to single-query path planning." <em>Robotics and Automation</em>, 2000. Proceedings. ICRA'00. IEEE International Conference on. Vol. 2. IEEE, 2000.
(<a href="http://ieeexplore.ieee.org/xpl/freeabs_all.jsp?arnumber=844730&abstractAccess=no&userType=inst" rel="nofollow noreferrer">IEEE link</a>.)</p>
</blockquote>
<p>Not surprisingly, the problem is also called <em>The Piano Mover's Problem</em>.</p>
|
2,963,886 | <p>What do these statements mean in discrete mathematics?</p>
<p><strong>Example 1:</strong> Let <span class="math-container">$P:\mathbb{Z}\times \mathbb{Z}\to \{T,F\}$</span>, where <span class="math-container">$P(x,y)$</span> denotes "<span class="math-container">$x+y=5$</span>".</p>
<p><strong>Example 2:</strong> Let <span class="math-container">$B=\{T,F\}$</span>. Let <span class="math-container">$P(p,q,r,\ldots )$</span> be a proposition. Then, <span class="math-container">$$P:=(p\rightarrow q)\rightarrow r:B\times B\times B\to B$$</span></p>
| fleablood | 280,126 | <p>Example 1: is a very abstract and rather obscure but necessary way to say:</p>
<p>"<span class="math-container">$x + y =5$</span> is either true or false"</p>
<p>"Let <span class="math-container">$P...:$</span>" means let <span class="math-container">$P$</span> be a way of writing a statement.</p>
<p>"Let <span class="math-container">$P:\mathbb Z\times \mathbb Z \to...."$</span> means let <span class="math-container">$P$</span> be a way of writing a statement about two integers. That is to say the <em>input</em> of <span class="math-container">$P$</span> will be <span class="math-container">$(x,y)$</span> a pair of integers. So <span class="math-container">$P(x,y)$</span> will be a statement about <span class="math-container">$x$</span> and <span class="math-container">$y$</span>.</p>
<p>"Let <span class="math-container">$P:\mathbb Z \times \mathbb Z \to \{T,F\}$</span> means the <em>output</em> of the statement will be either TRUE or FALSE. So given <span class="math-container">$x$</span> and <span class="math-container">$y$</span> the the statement <span class="math-container">$P(x,y)$</span> will either... be TRUE or FALSE.</p>
<p>So sayings <span class="math-container">$P(x,y)$</span> means "<span class="math-container">$x + y = 5$</span>" Means that we will evaluate any <span class="math-container">$P(x,y)$</span> as "Does <span class="math-container">$x + y = 5$</span> and we will determine that either ... yes, <span class="math-container">$x + y = 5$</span> or ... no, <span class="math-container">$x + y$</span> <em>doesn't</em> <span class="math-container">$=5$</span>></p>
<p>For example <span class="math-container">$P(2,3) = T$</span> because <span class="math-container">$2+3 = 5$</span>. And <span class="math-container">$P(4,2) = F$</span> because <span class="math-container">$4 + 2 \ne 5$</span>.</p>
<p>....</p>
<p>The idea is that we are trying to use the algebraic concept of <em>function</em> in a logical context.</p>
<p>In this case "{{A}}+{{B}} = 5" is a statement that takes to integers as input and outputs a single T/F value. In terms of <em>functions</em> this means <span class="math-container">$P$</span> is a function that maps an ordered pair of integers to a single T/F. So the <em>domain</em> of <span class="math-container">$P$</span> is <span class="math-container">$\mathbb Z\times \mathbb Z$</span> and the codomain of <span class="math-container">$P$</span> is <span class="math-container">$\{T,F\}$</span>.</p>
<p>Example 2:</p>
<p>We have three statements <span class="math-container">$p, q,$</span> or <span class="math-container">$r$</span> and each of those are either <span class="math-container">$TRUE$</span> or <span class="math-container">$FALSE$</span> and we want to know if the statement <span class="math-container">$(p\to q)\to r$</span> is <span class="math-container">$TRUE$</span> or <span class="math-container">$FALSE$</span>.</p>
<p>So <span class="math-container">$P$</span> is the the statement <span class="math-container">$(p\to q) \to r$</span>.</p>
<p>The input of <span class="math-container">$P$</span> is <span class="math-container">$(p,q,r)$</span> (not <span class="math-container">$(p,q,r,....)$</span> by the way; you typed that wrong) which is an ordered triple of TRUE/FALSE values. And the output is a single output TRUE/FALSE value.</p>
<p>So <span class="math-container">$P$</span> is a <em>function</em>. If <span class="math-container">$B$</span> is the set <span class="math-container">$\{T, F\}$</span> then the domain of <span class="math-container">$P$</span> is <span class="math-container">$B\times B\times B$</span>. That is the set of all possible triplets. That is <span class="math-container">$B\times B\times B = \{(T,T,T), (T,T,F), (T,F,T), (T,F,F), (F,T,T), (F,T,F), (T,F,T),(F,F,F)\}$</span>. The codomain of <span class="math-container">$P$</span> is <span class="math-container">$B = \{T,f\}$</span>.</p>
<p>And the function is defined as <span class="math-container">$P(p,q, r)$</span> as <span class="math-container">$(p\to q)\to r$</span>.</p>
<p>For example <span class="math-container">$P("bats\ are\ birds", "pigs\ eat\ corn", "Dogs\ teach\ algebra") = P(F, T,F) = (F\to T) \to F = T \to F = F$</span>.</p>
<p>If we wanted foralize:</p>
<p><span class="math-container">$P$</span> is the function:</p>
<p><span class="math-container">$(T,T,T) \mapsto T$</span></p>
<p><span class="math-container">$(T,T,F) \mapsto F$</span></p>
<p><span class="math-container">$(T,F,T) \mapsto T$</span></p>
<p><span class="math-container">$(T,F,F) \mapsto T$</span></p>
<p><span class="math-container">$(F,T,T) \mapsto T$</span></p>
<p><span class="math-container">$(F,T,F) \mapsto F$</span></p>
<p><span class="math-container">$(F,F,T) \mapsto T$</span></p>
<p><span class="math-container">$(F,F,F) \mapsto F$</span></p>
<p>[Math <em>does</em> get fairly abstract. But it <em>always</em> makes sense.] </p>
|
2,419,485 | <blockquote>
<p>In a certain family four girls take turns at washing dishes. Out of a total of four breakages, three were caused by the youngest girl, and she was thereafter called clumsy. Was she justified in attributing the frequency of her breakages to chance?</p>
</blockquote>
<p>I'm not sure how to solve the following question. The answer is <strong>13/64</strong> for a girl if the breakage was random and <strong>13/256</strong> for the youngest girl that broke 3 dishes.</p>
| Graham Kemp | 135,106 | <ul>
<li>count ways to select a girl and three dishes, multiply by the probability that each selection might happen.</li>
</ul>
<p>The probability that three from four dishes were randomly broken by the same girl is $\tbinom 41\tbinom 43{(\tfrac 14)}^3{(\tfrac 34)}^1$, that is $\tfrac {12}{64}$ or $3/16$.</p>
<p>The probability that three from four dishes were randomly broken by the youngest girl is $\tbinom 43{(\tfrac 14)}^3{(\tfrac 34)}^1$, that is $\tfrac {12}{256}$ or $3/64$. </p>
<p>It rather looks like the answer sheet contains a transcription typo due to all those 1s and 3s.</p>
<p>So, anyway, the conditional probability that the youngest did it, when given three from four dishes were randomly broken by the same girl, is $1/4$.</p>
|
479,594 | <p>I was wandering which is the best way to generate various combinations of $x_i$ such that $$\sum\limits_{i=1}^7 x_i = 1.0$$</p>
<p>where $ x_i \in \{0.0, 0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1.0\}$</p>
<p>I can generate these using brute-force, i.e checking through all $ 11^7$ combinations and only taking those which satisfies our constraint, however I am interested to know if there is another approach for this. Any ideas?</p>
| robjohn | 13,854 | <p>Subtract the formula for $a_{n-1}$ from that for $a_n$
$$
a_n-a_{n-1}=\frac14-\frac12a_{n-1}
$$
Multiply by $2^n$ and bring $a_{n-1}$ from the left to the right
$$
2^na_n=2^{n-2}+2^{n-1}a_{n-1}
$$
Using the formula for the sum of a geometric series, we get
$$
2^na_n=2^{n-1}+C
$$
Plug in $n=1$ to find that $C=-\frac12$. Thus,
$$
a_n=\frac12-\frac1{2^{n+1}}
$$</p>
|
3,089,326 | <p>9 person randomly enter 3 different rooms. What is the probability that</p>
<p>a)the first room has 3 person?</p>
<p>b)every room has 3 person?</p>
<p>c)the first room has n person, second room has 3 persons, third room 2 persons?</p>
<p>What i want to know is that which probability techniques i need to use when dealing with this type of questions, i have only learnt few techniques like combination and permutation only. </p>
| William Elliot | 426,203 | <p><span class="math-container">$\lim_{x\to a} f(x) = \lim_{x\to 0} f(x + a)
= \lim_{x\to 0} (f(x) + f(a))
= \lim_{x\to 0} f(x) + \lim_{x\to 0} f(a)
= f(0) + f(a) = f(0 + a) = f(a)$</span></p>
|
1,012,344 | <p>My knowledge of $C^*$-algebras is very little.</p>
<p>We call an element positive if $a=b^*b$ for some $b$ and make a relation on all positive elements by saying
$$
b \geqslant a \iff b-a \text{ is positive}.
$$
I can't figure out why this gives us a poset.</p>
| Tomasz Kania | 17,929 | <p>Only anti-symmetry seems to be non-trivial here. Let us apply the spectral theorem. </p>
<p>Suppose that $b\leqslant a$ and $a\leqslant b$. Since $a-b$ is positive, $C^*(a-b)$ is commutative and of course $b-a\in C^*(a-b)$. But now you can think of $a-b$ and $b-a$ as continuous functions on $\sigma(a-b)$ which are both positive (non-negative, strictly speaking). Since $a-b = -(b-a)$, necessarily $a-b=0$.</p>
|
3,493,519 | <p>Can I get a verification if this is the right way to approach this problem?</p>
<blockquote>
<p>Give an example of a linear map <span class="math-container">$T$</span> such that <span class="math-container">$\dim(\operatorname{null}T) = 3$</span> and <span class="math-container">$\dim(\operatorname{range}T) = 2$</span>.</p>
</blockquote>
<p>By the fundamental theorem of linear maps,
<span class="math-container">$$\dim V = \dim \operatorname{range}T + \dim\operatorname{null}T,$$</span>
thus <span class="math-container">$\dim V=5$</span>. Let <span class="math-container">$e_1,e_2,e_3,e_4,e_5$</span> be a basis for <span class="math-container">$\mathbb{R}^5$</span>.
Let <span class="math-container">$f_1,f_2$</span> be a basis for <span class="math-container">$\mathbb{R}^2$</span>. Define a linear map <span class="math-container">$T \in \mathcal{L}(\mathbb{R}^5,\mathbb{R}^2)$</span> by <span class="math-container">$$T(a_1e_1+a_2e_2+a_3e_3+a_4e_4+a_5e_5)=a_1f_1+a_2f_2.$$</span></p>
<p>Thus <span class="math-container">$\dim(\operatorname{null}T) = 3$</span> and <span class="math-container">$\dim(\operatorname{range}T) = 2$</span>.</p>
| Community | -1 | <p>Let's do an easy variation on your example, where <span class="math-container">$V=\Bbb R^5$</span> and <span class="math-container">$W=\Bbb R^2$</span>. </p>
<p>Define <span class="math-container">$T$</span> by <span class="math-container">$T(x_1,x_2, x_3, x_4, x_5)=(x_1,x_3)$</span>..</p>
<p>Then clearly <span class="math-container">$T$</span> is surjective. Hence <span class="math-container">$\operatorname {rank}T=2$</span>. This forces, by the "Fundamental theorem", as you call it, that <span class="math-container">$\operatorname {null}T=3$</span>.</p>
<p>It is easy to come up with other such variations. Here are some others, in a slightly different spirit: <span class="math-container">$T(v_1,v_2, v_3, v_4, v_5)=(av_1+bv_2, cv_3)$</span>, for any <span class="math-container">$a,b, c\ne0$</span> Actually,
we could let one of <span class="math-container">$a$</span> and <span class="math-container">$b$</span> be zero.</p>
<p>Actually, the collection of all such maps is in <span class="math-container">$1-1$</span> correspondence with the set of rank <span class="math-container">$2, 2×5$</span> matrices.</p>
<p>I think your approach was fine, and by coincidence, I defined the same map, essentially, without having read what you did carefully yet. Though I used, implicitly, the standard basis, and you, as @Arthur pointed out, did it in a more general setting.</p>
|
3,493,519 | <p>Can I get a verification if this is the right way to approach this problem?</p>
<blockquote>
<p>Give an example of a linear map <span class="math-container">$T$</span> such that <span class="math-container">$\dim(\operatorname{null}T) = 3$</span> and <span class="math-container">$\dim(\operatorname{range}T) = 2$</span>.</p>
</blockquote>
<p>By the fundamental theorem of linear maps,
<span class="math-container">$$\dim V = \dim \operatorname{range}T + \dim\operatorname{null}T,$$</span>
thus <span class="math-container">$\dim V=5$</span>. Let <span class="math-container">$e_1,e_2,e_3,e_4,e_5$</span> be a basis for <span class="math-container">$\mathbb{R}^5$</span>.
Let <span class="math-container">$f_1,f_2$</span> be a basis for <span class="math-container">$\mathbb{R}^2$</span>. Define a linear map <span class="math-container">$T \in \mathcal{L}(\mathbb{R}^5,\mathbb{R}^2)$</span> by <span class="math-container">$$T(a_1e_1+a_2e_2+a_3e_3+a_4e_4+a_5e_5)=a_1f_1+a_2f_2.$$</span></p>
<p>Thus <span class="math-container">$\dim(\operatorname{null}T) = 3$</span> and <span class="math-container">$\dim(\operatorname{range}T) = 2$</span>.</p>
| tf3 | 805,267 | <p>I was also attempting to solve this question but your approach seemed abstract to me, and upon little reflection, I was able to come up with the following more concrete example which helped increase my understanding of the concept.</p>
<p>The key understanding I got, which prompted me to write this answer, is as follows : since any linear map <span class="math-container">$T \in \mathcal{L}(\mathbb{R}^5,\mathbb{R}^2)$</span>, which has dim range <span class="math-container">$T = 2$</span> will definitely have dim null <span class="math-container">$T = 3$</span> (according to Fundamental theorem of Linear Maps), all we need to do is to look for a map <span class="math-container">$T \in \mathcal{L}(\mathbb{R}^5,\mathbb{R}^2)$</span> for which dim range <span class="math-container">$T = 2$</span>.</p>
<p>Define <span class="math-container">$T \in \mathcal{L}(\mathbb{R}^5,\mathbb{R}^2)$</span> by :
<span class="math-container">$$T(x_1, x_2, x_3, x_4, x_5) = (x_1 + x_2 +x_3 +x_4+x_5, x_1 + x_2)$$</span></p>
<p>The range has a basis <span class="math-container">$(1, 0),(0,1)$</span> and hence dim range <span class="math-container">$T = 2$</span> and null <span class="math-container">$T$</span> has basis <span class="math-container">$(1, -1, 0, 0, 0), (0, 0, 1, -1, 0), (0, 0, 1, 0, -1)$</span> and hence dim null <span class="math-container">$T =3$</span>.</p>
|
2,278,431 | <p>"Apply Green's Theorem to evaluate the line integral of F around positively oriented boundary"</p>
<p>$$F(x,y)=x^2yi+xyj$$</p>
<p>C: The region bounded by y=$x^2$ and y=4x+5</p>
| Robert Israel | 8,508 | <p>If $a(n) = n^n/(n! e^n)$, then</p>
<p>$$ \eqalign{\ln \left( \frac{a(n+1)}{a(n)} \right) &= n \log(1+1/n) - 1\cr
&= -\frac{1}{2n} + \frac{1}{3n^2} - \frac{1}{4n^3} + \ldots}$$
I claim this is negative. Indeed,
for $n \ge 1$, $$-\frac{1}{2m n^{2m}} + \frac{1}{(2m+1)n^{2m+1}} < 0 $$
Thus $a(n+1)/a(n) < 1$, and the sequence is decreasing.</p>
|
2,284,208 | <p>Hi I need help with the completion of this proof, I believe I am nearly at the end but I do not know how to end it</p>
<p>Proof: If v1 and v2 are a basis of v, then a<strong>v1</strong>+b<strong>v2</strong>=<strong>v</strong> for all v element of v.
Then if <strong>v1</strong>+<strong>v2</strong>, <strong>v1</strong>-<strong>v2</strong> is a basis, c(<strong>v1</strong>+<strong>v2</strong>)+d(<strong>v1</strong>-<strong>v2</strong>) = <strong>v</strong></p>
<p>Then by simplification I get <strong>v1</strong>(c+d)+<strong>v2</strong>(c-d) = v</p>
<p>Then this implies c+d=a and c-d=b
I do not know where to go from here, does this mean I can just assume this is a valid basis? May seem like a really silly question so sorry!</p>
<p>Thanks in advance!</p>
| Mark H | 81,870 | <p>You've shown that the new vectors have the same span as the originals (almost--you should find explicit expressions for $c$ and $d$). Now, you need to show that $\vec{v}_1 + \vec{v}_2$ and $\vec{v}_1-\vec{v}_2$ are linearly independent.</p>
|
2,284,208 | <p>Hi I need help with the completion of this proof, I believe I am nearly at the end but I do not know how to end it</p>
<p>Proof: If v1 and v2 are a basis of v, then a<strong>v1</strong>+b<strong>v2</strong>=<strong>v</strong> for all v element of v.
Then if <strong>v1</strong>+<strong>v2</strong>, <strong>v1</strong>-<strong>v2</strong> is a basis, c(<strong>v1</strong>+<strong>v2</strong>)+d(<strong>v1</strong>-<strong>v2</strong>) = <strong>v</strong></p>
<p>Then by simplification I get <strong>v1</strong>(c+d)+<strong>v2</strong>(c-d) = v</p>
<p>Then this implies c+d=a and c-d=b
I do not know where to go from here, does this mean I can just assume this is a valid basis? May seem like a really silly question so sorry!</p>
<p>Thanks in advance!</p>
| Jean Marie | 305,862 | <p>Here is another - more direct - proof: </p>
<p>Basis $\{v_1,v_2\}$ is sent onto system $\{v_1+v_2,v_1-v_2\}$ by a linear application $L$ with matrix </p>
<p>$$\begin{pmatrix}1&\ \ 1\\1&-1\end{pmatrix}$$</p>
<p>which is invertible. Thus $L$ is bijective ; therefore $\{v_1+v_2,v_1-v_2\}$ is a basis also.</p>
|
1,390,423 | <p>An acquaintance of mine proposed a scenario. Imagine parents who ground their child. Initially, the grounding is for 5 days, but for every day the child misbehaves while they're grounded, the parents will tack on an extra 2 days. The child is very predictable and has a 30% chance of misbehaving on any given day, no matter how long they've already been grounded.</p>
<p>What is the expected value for this poor kid's imprisonment? When I look at the problem, I come up with $\infty$ because there's a non-zero chance he'll keep adding days on forever. My acquaintance said this wasn't the case but can't prove it.</p>
<p>If the EV isn't $\infty$, can someone explain where I'm going wrong? My probability knowledge is very old and rusty, unfortunately.</p>
| true blue anil | 22,388 | <p>Expected # of days misbehaved = $0.3\times5 = 1.5$</p>
<p>Expected # of additional days tacked on = $2\times1.5 = 3$</p>
<p>so it is a G.P. with <em>a</em> = 5, <em>r</em> = $\dfrac35$ =0.6</p>
<p>$S_\infty = \dfrac{a}{(1-r)} = \dfrac{5}{0.4} = 12.5$</p>
|
1,390,423 | <p>An acquaintance of mine proposed a scenario. Imagine parents who ground their child. Initially, the grounding is for 5 days, but for every day the child misbehaves while they're grounded, the parents will tack on an extra 2 days. The child is very predictable and has a 30% chance of misbehaving on any given day, no matter how long they've already been grounded.</p>
<p>What is the expected value for this poor kid's imprisonment? When I look at the problem, I come up with $\infty$ because there's a non-zero chance he'll keep adding days on forever. My acquaintance said this wasn't the case but can't prove it.</p>
<p>If the EV isn't $\infty$, can someone explain where I'm going wrong? My probability knowledge is very old and rusty, unfortunately.</p>
| Community | -1 | <p>This answer comes in two parts:</p>
<ul>
<li>Show that the expectation is finite</li>
<li>Calculate the expectation.</li>
</ul>
<h2>Showing the expectation is finite</h2>
<p>Let's ignore the punishment for a moment and just consider the kids actions for all eternity. Each day, starting with day 1, the child is either good or bad, with probabilities 0.7 and 0.3 respectively.</p>
<p>Now, let's consider the following variation on counting days of the punishment: on day 0, we start a tally at 5. Each day, we either add or subtract one from the tally depending on whether the child was good or bad. This tally is allowed to reach zero or even go negative.</p>
<p>The question you ask is the expected day of the first zero, but that's a harder question: let's first ask about the probability that the tally on any particular day is positive. Let $T_n$ be the tally on day $n$. We have $T_0 = 5$.</p>
<p>It turns out that this tally is fairly easy to study. To get the tally on day $n$, let $G_n$ be the number of good days up to day $n$ and $B_n$ be the number of bad days. Then $G_n+B_n = n$, and the tally is $T_n = 5 + B_n - G_n$.</p>
<p>Thus, $T_n > 0$ if and only if $G_n \leq \lfloor (n+5)/2 \rfloor$.</p>
<p>It is easy to obtain the probability that $G_n=g$: it is</p>
<p>$$ \binom{n}{g} (0.7)^g (0.3)^{n-g} $$</p>
<p>This comes from there being $\binom{n}{g}$ different ways to pick which of the $n$ days were good, and each individual possibility has odds $(0.7)^g (0.3)^{n-g}$ of occurring.</p>
<p>Thus, the number of good days is given by a <em>binomial distribution</em>.</p>
<p>The probability that $G_n \leq \lfloor (n+5)/2 \rfloor$ converges to zero <em>very</em> quickly. There is no great loss if you justify everything in the rest of this section with "because this probability vanishes very quickly", but I will give a rigorous argument too.</p>
<p>The tail bounds listed at <a href="https://en.wikipedia.org/w/index.php?title=Binomial_distribution&oldid=660817458#Tail_Bounds" rel="nofollow noreferrer">Wikipedia</a> in fact say it decreases at least as fast as exponentially:</p>
<p>$$ P(G_n \leq (n+5)/2 ) < e^2 (e^{-2/25})^n $$</p>
<p>Now, the total length of the punishment is no bigger than the total number of days where the tally is positive. Since the odds that the tally is positive on the $n$-th days is strictly less than a Geometric series, the expected total is finite:</p>
<p>$$ \sum_{n=0}^{\infty} P(T_n > 0) <\sum_{n=0}^{\infty} e^2 (e^-{2/25})^n < 100$$</p>
<p>Thus, the expected length of the punishment must be finite too.</p>
<h2>Finding the average</h2>
<p>The answer by <a href="https://math.stackexchange.com/a/1390431/14972">mjq</a> starts with a decent idea; if $f(n)$ expresses the expected remainder of the punishment given that it's currently $n$ days, then</p>
<p>$$ f(n) = 1 + \frac{7}{10} f(n-1) + \frac{3}{10} f(n+1) \qquad \qquad f(0) = 0$$</p>
<p>There are standard ways to solve recursions of this form, which would be suitable for a new question. For those who know the methods, the main trick here is to write $f(n) = g(n) + An$ to account for that $1$, and then you have an ordinary linear difference equation.</p>
<p>The general solution to the recursion is</p>
<p>$$ f(n) = \frac{5n}{2} + A + B \left(\frac{7}{3} \right)^n $$</p>
<p>The condition $f(0) = 0$ lets us simplify to</p>
<p>$$ f(n) = \frac{5n}{2} + B \left( \left(\frac{7}{3} \right)^n - 1 \right)$$</p>
<p>Heuristically, exponential growth should be too fast, so we need $B=0$. A rigorous argument of this fact can be given by adapting the argument at the end of the previous section to show that $f(n) \in O(e^{2n/5})$.</p>
<p>So finally, the answer for the expected length of punishment is $f(5) = 25/2 = 12.5$.</p>
<h2>Argument for linearity</h2>
<p>Once we know the expectation is finite, there is an argument that it should be linear: if the punishment is $n$ days, then there is only one way the punishment can end: it can decrease to $m$ days, and then an $m$-day punishment can expire (where $0 \leq m \leq n$).</p>
<p>But that's the same as waiting for an $n-m$ day punishment to expire followed by an $m$-day punishment. Thus, we have to have $f(n) = f(n-m) + f(m)$. Consequently, $f(n) = n \cdot f(1)$.</p>
|
3,408,846 | <p>This is an example in Serge Lang "Introduction to Linear Algebra", page 48. I try to multiply these two <span class="math-container">$2$</span>x<span class="math-container">$3$</span> and <span class="math-container">$3$</span>x<span class="math-container">$2$</span> matrices but fail to obtain the result as mentioned in the text.</p>
<p>I have:
<span class="math-container">$
\left( \begin{array}{cc}
2 & 1 & 5\\
1 & 3 & 2
\end{array} \right)
%
\left( \begin{array}{c}
3 & 4 \\
-1 & 2 \\
2 & 1
\end{array} \right)
=\left( \begin{array}{c}
15 & 30 \\
4 & 2
\end{array} \right)
$</span></p>
<p>Serge's result is, however:</p>
<p><span class="math-container">$
\left( \begin{array}{cc}
2 & 1 & 5\\
1 & 3 & 2
\end{array} \right)
%
\left( \begin{array}{c}
3 & 4 \\
-1 & 2 \\
2 & 1
\end{array} \right)
=\left( \begin{array}{c}
15 & 15 \\
4 & 2
\end{array} \right)
$</span></p>
<p>Where did I do wrong?</p>
| user | 505,767 | <p>We have</p>
<p><span class="math-container">$$\left( \begin{array}{cc}
\color{red}{2} & \color{red}{1} & \color{red}{5}\\
1 & 3 & 2
\end{array} \right)
%
\left( \begin{array}{c}
3 & \color{red}{4} \\
-1 & \color{red}{2} \\
2 & \color{red}{1}
\end{array} \right)
=\left( \begin{array}{c}
15 & \color{red}{15} \\
4 & 12
\end{array} \right)$$</span></p>
<p>that is <span class="math-container">$2\cdot 4+1\cdot 2+5\cdot 1=8+2+5=15$</span>.</p>
|
3,454,725 | <p>I am reading real analysis book and encountered this symbol <span class="math-container">$\wedge$</span> and <span class="math-container">$\vee.$</span></p>
<p>The author says following:</p>
<ol>
<li><span class="math-container">$f\wedge g=\frac{1}{2}(f+g-
|f-g|)$</span>,</li>
<li><span class="math-container">$f\vee g=\frac{1}{2}(f+g+|f-g|)$</span>.</li>
</ol>
<p>What are the meanings of <span class="math-container">$\wedge$</span> and <span class="math-container">$\vee$</span>? For example, I want to know what <span class="math-container">$f\wedge g$</span> means like <span class="math-container">$f^{+}$</span> means the positive part of the function.</p>
<p>Edit: he also asserts that <span class="math-container">$\chi_{A\cap B}=\chi_A\wedge \chi_B$</span> and <span class="math-container">$\chi_{A\cup B}=\chi_A\vee\chi_B.$</span></p>
| azif00 | 680,927 | <p>Define <span class="math-container">$I = \{1,\dots,|\phi_M|\}$</span> and put
<span class="math-container">$$A = \{\{x_i\}:\, i\in I\} \cup \{\{x_i,x_j\}:\, (i,j)\in I^2\} \cup \cdots \cup \{\{\{x_{i_1}\},\dots,\{x_{i_k}\}\} :\, (i_1,\dots,i_k)\in I^k, \, k=|\phi_M|-1\} \cup \phi_M$$</span></p>
|
324,594 | <p>You have three buckets, two big buckets holding <code>8 litres</code> of water each and one small empty bucket that can hold <code>3 litres</code> of water. How will you split the <code>16 litres</code> of water to <code>four people</code> evenly? Each person has a container but once water is distributed to someone it cannot be taken back. </p>
<p>In this puzzle, we need to allocate 4 litres to each person. So I considered the initial state as below <br></p>
<blockquote>
<p>8 8 0 [0, 0, 0, 0]</p>
</blockquote>
<p>The values in the bracket are the water given to those 4 people.
I tried the below steps </p>
<blockquote>
<p>5 8 3 [0, 0, 0, 0] <br>
5 8 0 [3, 0, 0, 0] <br>
5 5 3 [3, 0, 0, 0] <br>
5 5 0 [3, 3, 0, 0] <br>
2 8 0 [3, 3, 0, 0] <br>
0 8 2 [3, 3, 0, 0] <br>
0 7 3 [3, 3, 0, 0] <br>
0 4 3 [3, 3, 3, 0] <br>
0 1 3 [3, 3, 3, 3] <br>
0 0 3 [4, 3, 3, 3] <br></p>
</blockquote>
<p>But I couldnt take it further by giving those 1 litre for rest of the 3 persons since there is a constraint that water given to person cant be retrieved back. <br>
Any help/hint towards final solution is greatly appreciated.</p>
| joriki | 6,622 | <pre><code>8 8 0 [0 0 0 0]
8 5 3 [0 0 0 0]
8 5 0 [3 0 0 0]
8 2 3 [3 0 0 0]
8 0 3 [3 2 0 0]
8 3 0 [3 2 0 0]
5 3 3 [3 2 0 0]
5 6 0 [3 2 0 0]
2 6 3 [3 2 0 0]
2 8 1 [3 2 0 0]
2 8 0 [3 2 1 0]
0 8 2 [3 2 1 0]
0 7 3 [3 2 1 0]
3 7 0 [3 2 1 0]
3 4 3 [3 2 1 0]
6 4 0 [3 2 1 0]
6 1 3 [3 2 1 0]
6 0 3 [3 2 1 1]
8 0 1 [3 2 1 1]
8 0 0 [4 2 1 1]
5 0 3 [4 2 1 1]
5 3 0 [4 2 1 1]
2 3 3 [4 2 1 1]
0 3 3 [4 4 1 1]
0 0 3 [4 4 4 1]
0 0 0 [4 4 4 4]
</code></pre>
|
324,594 | <p>You have three buckets, two big buckets holding <code>8 litres</code> of water each and one small empty bucket that can hold <code>3 litres</code> of water. How will you split the <code>16 litres</code> of water to <code>four people</code> evenly? Each person has a container but once water is distributed to someone it cannot be taken back. </p>
<p>In this puzzle, we need to allocate 4 litres to each person. So I considered the initial state as below <br></p>
<blockquote>
<p>8 8 0 [0, 0, 0, 0]</p>
</blockquote>
<p>The values in the bracket are the water given to those 4 people.
I tried the below steps </p>
<blockquote>
<p>5 8 3 [0, 0, 0, 0] <br>
5 8 0 [3, 0, 0, 0] <br>
5 5 3 [3, 0, 0, 0] <br>
5 5 0 [3, 3, 0, 0] <br>
2 8 0 [3, 3, 0, 0] <br>
0 8 2 [3, 3, 0, 0] <br>
0 7 3 [3, 3, 0, 0] <br>
0 4 3 [3, 3, 3, 0] <br>
0 1 3 [3, 3, 3, 3] <br>
0 0 3 [4, 3, 3, 3] <br></p>
</blockquote>
<p>But I couldnt take it further by giving those 1 litre for rest of the 3 persons since there is a constraint that water given to person cant be retrieved back. <br>
Any help/hint towards final solution is greatly appreciated.</p>
| Théophile | 26,091 | <p>Here is a solution using 24 steps (one fewer than joriki's solution). I believe this is minimal, since I wrote a shortest path algorithm to find it:</p>
<pre><code>8 8 0 [0 0 0 0]
8 5 3 [0 0 0 0]
8 5 0 [3 0 0 0]
8 2 3 [3 0 0 0]
8 0 3 [3 2 0 0]
8 3 0 [3 2 0 0]
5 3 3 [3 2 0 0]
5 6 0 [3 2 0 0]
2 6 3 [3 2 0 0]
2 8 1 [3 2 0 0]
2 8 0 [4 2 0 0]
2 5 3 [4 2 0 0]
0 7 3 [4 2 0 0]
3 7 0 [4 2 0 0]
3 4 3 [4 2 0 0]
6 4 0 [4 2 0 0]
6 1 3 [4 2 0 0]
8 1 1 [4 2 0 0]
8 1 0 [4 2 1 0]
5 1 3 [4 2 1 0]
5 1 0 [4 2 4 0]
2 1 3 [4 2 4 0]
0 1 3 [4 4 4 0]
0 4 0 [4 4 4 0]
0 0 0 [4 4 4 4]
</code></pre>
|
4,480,905 | <p>When <span class="math-container">$T$</span> is any linear operator acting on a vector space <span class="math-container">$V$</span>, and <span class="math-container">$n$</span> is a natural number, <span class="math-container">$T^n$</span> means <span class="math-container">$T$</span> applied <span class="math-container">$n$</span> times (composition) and that is also a linear operator. That is clear.</p>
<p>When <span class="math-container">$T$</span> is a nonzero linear operator acting on a vector space <span class="math-container">$V$</span>, then <span class="math-container">$T^0$</span> is the identity operator <span class="math-container">$T^0 = I$</span>. But I think that should also be true (true by definition), when <span class="math-container">$T$</span> is the zero operator i.e. the operator which sends all vectors to the zero vector.</p>
<p>Why? Because <span class="math-container">$T^0$</span> means that we are not applying any operator. So it makes sense to say: OK, all vectors stay unchanged when "applying" <span class="math-container">$T^0$</span> even when <span class="math-container">$T$</span> is the zero operator. I say "applying" because we're not actually applying anything.</p>
<p>Is that indeed so?</p>
<p>I am asking because this kind of disagrees with what we have for real numbers where <span class="math-container">$0^0$</span> is usually left undefined.</p>
<p><strong>EDIT:</strong><br />
What's the context of this question? I was reading a proof for the uniqueness of the Jordan Normal Form. There this expression comes up <span class="math-container">$2d(\phi^p) - d(\phi^{p-1}) - d(\phi^{p+1})$</span>, where <span class="math-container">$p$</span> is a positive integer, and <span class="math-container">$d$</span> is the defect of the linear operator in the brackets. The proof is very nice but convoluted and eventually it boils down to proving the uniqueness for a special linear operator which has only <span class="math-container">$0$</span> as a characteristic root (as an eigenvalue). So I had some doubts what happens exactly with the expression <span class="math-container">$\phi^{p-1}$</span> when <span class="math-container">$p = 1$</span>, and if we need to put some restrictions on the linear operator <span class="math-container">$\phi$</span>.</p>
| Randall | 464,495 | <p>We define <span class="math-container">$T^0=I$</span> for <strong>any</strong> linear operator <span class="math-container">$T:V \to V$</span> so that the usual laws of exponents hold for composition. So, you should take this as a definition which is convenient and not anything extraordinarly deep. Here's why.</p>
<p>For a <em>positive</em> integer <span class="math-container">$n$</span> we can define <span class="math-container">$T^n$</span> to mean <span class="math-container">$T \circ T \circ \cdots \circ T$</span> (<span class="math-container">$n$</span> factors), and <span class="math-container">$T^n$</span> is again a linear operator on <span class="math-container">$V$</span>. You can then check, thanks to associativity of function composition, that we have <span class="math-container">$T^n \circ T^m = T^{n+m}$</span> for positive integers <span class="math-container">$n, m$</span>. Doesn't that look nice and familiar?</p>
<p>If we extend this definition to include the possibility that <span class="math-container">$n=0$</span> by <strong>defining</strong> <span class="math-container">$T^0=I$</span>, then the law <span class="math-container">$T^n \circ T^m = T^{n+m}$</span> now holds for integers <span class="math-container">$n,m \geq 0$</span>.
After all, <span class="math-container">$T^n \circ T^0 = T^n \circ I = T^n$</span>, so we have <span class="math-container">$T^n \circ T^0 = T^{n+0}$</span> and all is well.</p>
<p>Even better, if <span class="math-container">$T$</span> happens to be invertible we can define <span class="math-container">$T^{-n}$</span> for an integer <span class="math-container">$n \geq 0$</span> as either <span class="math-container">$(T^n)^{-1}$</span> or as <span class="math-container">$(T^{-1})^n$</span> as they are both the same, which is a great exercise. Then the law of exponents
<span class="math-container">$$
T^n \circ T^m = T^{n+m}
$$</span>
holds for <strong>all</strong> integers <span class="math-container">$n, m$</span>. Not only is this algebraically reminiscient of real number exponents in a nice way, this makes the applications of polynomial relations to linear operators incredibly powerful.</p>
<p>But, note that none of this has anything to do with whether or not <span class="math-container">$T$</span> is the zero transformation. In that case, <span class="math-container">$T^0=I$</span> is still the case by definition. (Note also that this is all about composition of functions and not any sort of real number multiplication/exponentiation, so issues related to <span class="math-container">$0^0$</span> in the reals are actually not relevant.)</p>
|
3,836,542 | <p>In last lines in the image from Lawrece. C. Evans, Partial Differentail Equations, he states that in Hilbert space every bounded sequence contains a weakly convergent subsequence.</p>
<p>What is wrong in my counter example ?</p>
<p>Let <span class="math-container">$\mathcal H$</span> be an infinite dimensional Hilbert space with <em>orthonormal</em> basis <span class="math-container">$\{u_k\}_{k=1}^\infty$</span>. Take them as a sequence. Now this sequence is bounded. I don't understand what subsequence of it may converge <em>weakly</em> and to what it may converge.</p>
<p><a href="https://i.stack.imgur.com/fhEX5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fhEX5.png" alt="enter image description here" /></a></p>
| Will Jagy | 10,400 | <p>I get <span class="math-container">$$ \left(x^2 + n^2 \right)^2 - \left(2nx+1 \right)^2 = (x^2 - n^2)^2 - 4nx-1 $$</span></p>
<p>Once written as a difference of squares, note that <span class="math-container">$A^2 - B^2 = 0$</span> means <span class="math-container">$(A+B)(A-B) = 0,$</span> so that <span class="math-container">$A+B=0$</span> or <span class="math-container">$A-B = 0,$</span> possibly both.</p>
<p>Here <span class="math-container">$$ n = 2018 $$</span></p>
|
3,836,542 | <p>In last lines in the image from Lawrece. C. Evans, Partial Differentail Equations, he states that in Hilbert space every bounded sequence contains a weakly convergent subsequence.</p>
<p>What is wrong in my counter example ?</p>
<p>Let <span class="math-container">$\mathcal H$</span> be an infinite dimensional Hilbert space with <em>orthonormal</em> basis <span class="math-container">$\{u_k\}_{k=1}^\infty$</span>. Take them as a sequence. Now this sequence is bounded. I don't understand what subsequence of it may converge <em>weakly</em> and to what it may converge.</p>
<p><a href="https://i.stack.imgur.com/fhEX5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fhEX5.png" alt="enter image description here" /></a></p>
| sirous | 346,566 | <p>Another approach:</p>
<p><span class="math-container">$$(x^2-2018^2)^2=8072 x+1$$</span></p>
<p><span class="math-container">$$(x-2018)^2(x+2018)^2=4\times 2018 x+1$$</span></p>
<p>A:</p>
<p><span class="math-container">$ \begin{cases}(x-2018)^2=4\times 2018 x+1 \\ (x+2018)^2=1\end {cases}$</span></p>
<p><span class="math-container">$x=-2019$</span> and <span class="math-container">$x=-2017$</span> which are the solution of <span class="math-container">$(x+2018)^2=1$</span> can be solutions of this system if they suffice the first equation. This gives a family of solutions.</p>
<p>B:</p>
<p><span class="math-container">$ \begin{cases}(x+2018)^2=4\times 2018 x+1 \\ (x-2018)^2=1\end {cases}$</span></p>
<p>Similarly this system can gives another family of solutions.</p>
|
916,963 | <p>I am reading an introductory book on mathematical proofs and I don't seem to understand the mechanics of proof by contradiction. Consider the following example. </p>
<p>$\textbf{Theorem:}$ If $P \rightarrow Q$ and $R \rightarrow \neg Q$, then $P \rightarrow \neg R$.</p>
<p>$\textbf{Proof:}$ (by contradiction)
Assume $P$, then it follows that $Q$.
Now, assume $R$, then it follows that $\neg Q$. Contradiction, we have $Q$ and $\neg Q$ at the same time. Hence, $\neg R$.
Therefore, if $P \rightarrow Q$ and $R \rightarrow \neg Q$, then $P \rightarrow \neg R$, as desired.</p>
<p>What I don't understand in this proof, is that why having arrived at contradiction, we decide that our assumption that $R$ is necessarily false? It also could have been that our first assumption, namely, $P$, was false. Or both of them could be false.</p>
<p>So my question is: in general, when proving by contradiction, how do we know which assumption exactly is false? And how do we know that exactly one assumption must be wrong in order to proceed with the proof?</p>
| Jean-Claude Arbaut | 43,608 | <p>To answer <em>It also could have been that our first assumption, namely, P, was false. Or both of them could be false</em>: yes, it could, but if you assume P, then R must be false, what you write $P\rightarrow\neg R$. You could have concluded $R\rightarrow \neg P$, of course (and it's the <a href="https://en.wikipedia.org/wiki/Contraposition">contraposition</a>), by assuming $R$ instead of $P$.</p>
<p>Notice that both of these implications are true, even if $P$ and $R$ are false. You do not <em>know</em> which is false, because none is false <em>a priori</em>. You assume one is true, and you conclude something about the other.</p>
<hr>
<p>When you do that in practice, it's usually not a problem. Example, let's prove $\sqrt{2}$ is irrational, by contradiction.</p>
<p>So, we assume $\sqrt{2}$ is rational, and we will be led to something that is certainly wrong, hence the assumption is wrong.</p>
<p>Since $\sqrt{2}$ is assumed to be rational, we have $\sqrt{2}=\frac pq$ for some integers $p$, $q$ that have no common factor (otherwise, factor them out).</p>
<p>Hence $p^2=2q^2$, so $2$ divides $p$, and $p=2p'$, and you rewrite your equality $4p'\,^2=2q^2$, or $q^2=2p'\,^2$. But then $2$ divides also $q$. It's not possible, since $p$ and $q$ have no common factor.</p>
<p>Hence something must be wrong. What? The only thing we have assumed, that $\sqrt{2}$ is a rational number.</p>
|
14,508 | <p>Suppose that $f$ is a weight $k$ newform for $\Gamma_1(N)$ with attached $p$-adic Galois representation $\rho_f$. Denote by $\rho_{f,p}$ the restriction of $\rho_f$ to a decomposition group at $p$. When is $\rho_{f,p}$ semistable (as a representation of
$\mathrm{Gal}(\overline{\mathbf{Q}}_p/\mathbf{Q}_p)$?</p>
<p>To make things really concrete, I'm happy to assume that $k=2$ and that the $q$-expansion of $f$ lies in $\mathbf{Z}[[q]]$. </p>
<p>Certainly if $N$ is prime to $p$ then $\rho_{f,p}$ is in fact crystalline, while
if $p$ divides $N$ exactly once then $\rho_{f,p}$ is semistable (just thinking about the Shimura construction in weight 2 here, and the corresponding reduction properties of $X_1(N)$
over $\mathbf{Q}$ at $p$). For $N$ divisible by higher powers of $p$, we know that these representations are de Rham, hence potentially semistable. Can we say more? For example,
are there conditions on "numerical data" attached to $f$ (e.g. slope, $p$-adic valuation of $N$, etc.) which guarantee semistability or crystallinity over a specific
extension? Can we bound the degree and ramification of
the minimal extension over which $\rho_{f,p}$ becomes semistable in terms of numerical
data attached to $f$? Can it happen that $N$ is highly divisible by $p$ and yet $\rho_{f,p}$ is semistable over $\mathbf{Q}_p$?</p>
<p>I feel like there is probably a local-Langlands way of thinking about/ rephrasing this question, which may be of use... </p>
<p>As a possible example of the sort of thing I have in mind: if $N$ is divisible by $p$ and $f$ is ordinary at $p$ then $\rho_{f,p}$ becomes semistable over an abelian extension of
$\mathbf{Q}p$
and even becomes crystalline over such an extension provided that the Hecke eigenvalues
of $f$ for the action of $\mu_{p-1}\subseteq (\mathbf{Z}/N\mathbf{Z})^{\times}$ via the diamond operators
are not all 1.</p>
| Kevin Buzzard | 1,384 | <p>The right way to do this sort of question is to apply Saito's local-global theorem, which says that the (semisimplification of the) Weil-Deligne representation built from $D_{pst}(\rho_{f,p})$ by forgetting the filtration is precisely the one attached to $\pi_p$, the representation of $GL_2(\mathbf{Q}_p)$ attached to the form via local Langlands. Your suggestions about the $p$-adic valuation of $N$ and so on are rather "coarse" invariants---$\pi_p$ tells you everything and is the invariant you really need to study.</p>
<p>So now you can just list everything that's going on. If $\pi_p$ is principal series, then $\rho$ will become crystalline after an abelian extension---the one killing the ramification of the characters involved in the principal series. If $\pi_p$ is a twist of Steinberg by a character, $\rho_{f,p}$ will become semistable non-crystalline after you've made an abelian extension making the character unramified. And if $\pi_p$ is supercuspidal, $\rho_{f,p}$ will become crystalline after a finite non-trivial extension that could be either abelian or non-abelian, and figuring out which is a question about $\pi_p$ (it will be a base change from a quadratic extension if $p>2$ and you have to bash out the possibilities).</p>
<p>Seems to me then that semistable $\rho$s will show up precisely when $\pi_p$ is either unramified principal series or Steinberg, so the answer to your question is (if I've got everything right) that $\rho_{f,p}$ will be semistable iff either $N$ (the level of the newform) is prime to $p$, or $p$ divides $N$ exactly once and the component at $p$ of the character of $f$ is trivial. Any other observations you need should also be readable from this sort of data in the same way.</p>
<p>One consequence of this I guess is that $\rho_{f,p}$ is semi-stable iff the $\ell$-adic representation attached to $f$ is semistable at $p$.</p>
|
3,478,098 | <p><a href="https://i.stack.imgur.com/dcxhi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dcxhi.jpg" alt="enter image description here" /></a></p>
<p>Guys, Why is this weird statement true? It seems counterintuitive to me I cannot understand or lack creativity understanding it can you help me explain it? Guys please if possible make it visual.</p>
| Kavi Rama Murthy | 142,385 | <p>If there is a rational number <span class="math-container">$x$</span> in this interval with <span class="math-container">$q \leq N$</span> then <span class="math-container">$x \in Q_N$</span> and the definition of <span class="math-container">$\delta$</span> shows that <span class="math-container">$\delta \leq |x-a|$</span> This contradicts the fatc that <span class="math-container">$x \in (a-\delta, a+\delta)$</span>. </p>
|
1,575,764 | <p>$Q$ is a linear operator from $V \to V$ with $V$ being a finite dimensional complex inner-product-space. </p>
<p>Given: $Q^*=5Q$, $Q^*$ being the adjoint.</p>
<p>Show that $0$ is the only eigenvalue of $Q$. </p>
<p>I've been staring at this problem for quite a while now. I think the answer lies in the eigenvalues of $Q$ and $Q^*$ being the same, but I can't think of a way to prove $0$ is an eigenvalue let alone the only one. </p>
| Justpassingby | 293,332 | <p>If you have one nonzero eigenvalue, then its multiples by all powers of $5$ must also be eigenvalues.</p>
|
131,283 | <p>I came across a question which required us to find $\displaystyle\sum_{n=3}^{\infty}\frac{1}{n^5-5n^3+4n}$. I simplified it to $\displaystyle\sum_{n=3}^{\infty}\frac{1}{(n-2)(n-1)n(n+1)(n+2)}$ which simplifies to $\displaystyle\sum_{n=3}^{\infty}\frac{(n-3)!}{(n+2)!}$. I thought it might have something to do with partial fractions, but since I am relatively inexperienced with them I was unable to think of anything useful to do. I tried to check WolframAlpha and it gave $$\sum_{n=3}^{m}\frac{(n-3)!}{(n+2)!}=\frac{m^4+2m^3-m^2-2m-24}{96(m-1)m(m+1)(m+2)}$$
From this it is clear that as $m\rightarrow \infty$ the sum converges to $\frac{1}{96}$, however I have no idea how to get there. Any help would be greatly appreciated!</p>
| robjohn | 13,854 | <p>The <a href="http://en.wikipedia.org/wiki/Heaviside_cover-up_method">Heaviside Method</a> gives the partial fraction decomposition
$$
\begin{align}
&\frac{1}{(n-2)(n-1)n(n+1)(n+2)}\\[6pt]
&=\frac{1}{24(n-2)}-\frac{1}{6(n-1)}+\frac{1}{4n}-\frac{1}{6(n+1)}+\frac{1}{24(n+2)}\tag{1}
\end{align}
$$
Notice that
$$
\frac{1}{24}-\frac{1}{6}+\frac{1}{4}-\frac{1}{6}+\frac{1}{24}=0\tag{2}
$$
Therefore,
$$
\begin{align}
&\sum_{n=3}^m\left(\frac{1}{24(n-2)}-\frac{1}{6(n-1)}+\frac{1}{4n}-\frac{1}{6(n+1)}+\frac{1}{24(n+2)}\right)\\
&=\frac{1}{24}\sum_{n=1}^{m-2}\frac1n-\frac{1}{6}\sum_{n=2}^{m-1}\frac1n+\frac{1}{4}\sum_{n=3}^{m}\frac1n-\frac{1}{6}\sum_{n=4}^{m+1}\frac1n+\frac{1}{24}\sum_{n=5}^{m+2}\frac1n\tag{$\ast$}\\
&=\frac{1}{24}\left(\frac11+\frac12+\frac13+\frac14\right)\\
&-\frac{1}{6}\left(\frac12+\frac13+\frac14+\frac{1}{m-1}\right)\\
&+\frac{1}{4}\left(\frac13+\frac14+\frac{1}{m-1}+\frac{1}{m}\right)\\
&-\frac{1}{6}\left(\frac14+\frac{1}{m-1}+\frac{1}{m}+\frac{1}{m+1}\right)\\
&+\frac{1}{24}\left(\frac{1}{m-1}+\frac{1}{m}+\frac{1}{m+1}+\frac{1}{m+2}\right)\\
&=\frac{m^4+2m^3-m^2-2m-24}{96(m-1)m(m+1)(m+2)}\tag{3}
\end{align}
$$
where $\displaystyle\sum_{n=5}^{m-2}\frac1n$ is cancelled out of each summation in $(\ast)$ due to $(2)$.</p>
<p>As $m\to\infty$ the $(3)$ tends to $\dfrac{1}{96}$.</p>
|
3,185,226 | <p>I have a simple question that confuses me for a while:</p>
<blockquote>
<p><span class="math-container">$$f(X) = \text{tr} \left( [ \log(X) ]^2 \right)$$</span></p>
<p>where <span class="math-container">$X$</span> is an <span class="math-container">$m \times m$</span> symmetric positive definite (SPD) matrix and <span class="math-container">$\log(X)$</span> is the matrix logarithm of matrix <span class="math-container">$X$</span>. What is <span class="math-container">$\frac{\partial f}{\partial X}$</span>?</p>
</blockquote>
<p>Using the chain rule, I have</p>
<p><span class="math-container">$df = \text{tr}(2ZdZ)$</span>,</p>
<p>where <span class="math-container">$Z=\log (X)$</span>. I think we should have <span class="math-container">$dZ = X^{-1}dX$</span> as a scalor function, so we will have</p>
<p><span class="math-container">$\frac{\partial f}{\partial X} = 2\log(X)X^{-1}$</span>,</p>
<p>but I haven't found any related reference.</p>
<p>Any comment or hint will be appreciated!</p>
| Community | -1 | <p>Let <span class="math-container">$\phi:X\in U\mapsto \log(X)$</span>, where <span class="math-container">$U$</span> is the set of <span class="math-container">$n\times n$</span> complex matrices that have no eigenvalues in <span class="math-container">$\mathbb{R}^-=(-\infty,0]$</span> (we use the principal <span class="math-container">$\log$</span>). Its derivative is</p>
<p>cf. Higham, functions of matrices</p>
<p><span class="math-container">$D\phi_X:H\in M_n\mapsto \int_0^1(t(X-I)+I)^{-1}H(t(X-I)+I)^{-1}dt$</span>.</p>
<p><span class="math-container">$\textbf{Proposition 1}$</span>. Let <span class="math-container">$f:X\in M_n\mapsto tr(\log(X))$</span>.</p>
<p>Then its derivative is <span class="math-container">$Df_X(H)=tr(X^{-1}H)$</span> and its gradient is <span class="math-container">$\nabla(f)(X)=X^{-T}$</span>.</p>
<p><span class="math-container">$\textbf{Proof}$</span>. <span class="math-container">$Df_X(H)=tr(D\phi_X(H))=\int_0^1tr((t(X-I)+I)^{-1}H(t(X-I)+I)^{-1})dt=$</span></p>
<p><span class="math-container">$\int_0^1tr((t(X-I)+I)^{-2}H)dt=tr(\int_0^1(t(X-I)+I)^{-2}dtH)=tr(X^{-1}H)$</span>.</p>
<p><span class="math-container">$\textbf{Proposition 2}$</span>. Let <span class="math-container">$g:X\in M_n\mapsto tr((\log(X))^2)$</span>.</p>
<p>Then its derivative is <span class="math-container">$Dg_X(H)=2tr(\log(X)X^{-1}H)$</span> and its gradient is </p>
<p><span class="math-container">$\nabla(g)(X)=2X^{-T}\log(X^T)$</span>.</p>
<p><span class="math-container">$\textbf{Proof}$</span>. <span class="math-container">$Dg_X(h)=2tr(\log(X)D\phi_X(H))=$</span></p>
<p><span class="math-container">$2\int_0^1tr(\log(X)(t(X-I)+I)^{-1}H(t(X-I)+I)^{-1})dt$</span>.</p>
<p>The key is that <span class="math-container">$\log(X)$</span> and <span class="math-container">$(t(X-I)+I)^{-1}$</span> commute (both are polynomials in <span class="math-container">$X$</span>); then (as above) </p>
<p><span class="math-container">$Dg_X(H)=2tr(\log(X)\int_0^1(t(X-I)+I)^{-2}dtH)=2tr(\log(X)X^{-1}H)$</span>.</p>
<p><span class="math-container">$\textbf{Remark}$</span>.The derivative of the function <span class="math-container">$tr(A\log(X))$</span> (where <span class="math-container">$A\in M_n$</span> is fixed), is much more complicated!!</p>
|
4,281,654 | <p>My professor gave me an exercise where I had to show that the special linear group <span class="math-container">$SL(2,\mathbb{R})$</span> is a lie subgroup of <span class="math-container">$GL(2,\mathbb{R})$</span>. I was able to do this part. However, I was then asked to do the following:</p>
<p>All real <span class="math-container">$2\times 2$</span> matrices <span class="math-container">$\begin{pmatrix} a & b \\ c & d\end{pmatrix}$</span> can be identified with <span class="math-container">$(a,b,c,d) \in \mathbb{R}^4$</span>. In this way, <span class="math-container">$SL(2,\mathbb{R})$</span> can be thought of as a subset of <span class="math-container">$\mathbb{R}^4$</span>. In this correspondence, find all matrices in <span class="math-container">$SL(2,\mathbb{R})$</span> that are closest to the origin.</p>
<p>I really don't have any idea how to approach this problem. The only things I have ever seen like this are Lagrange multipliers, but those don't seem to apply here. For reference, though this exercise is not in the text, our course is using Introduction to Smooth Manifolds by Lee.</p>
| orangeskid | 168,051 | <p>Let's first solve the problem using Lagrange multiplier. We consider the system
<span class="math-container">$$\operatorname{grad} ((a^2 + b^2 + c^2 + d^2) - 2\lambda (a d - b c - 1)) = 0$$</span>
that is
<span class="math-container">$$ (a, b, c, d) = \lambda(d,-c,-b,a)$$</span>
Since <span class="math-container">$a d - b c = 1$</span>, we conclude that <span class="math-container">$\lambda = \pm 1$</span>. Two cases:</p>
<p><span class="math-container">$a = d$</span>, <span class="math-container">$b = -c$</span>, and <span class="math-container">$a^2 + b^2 = 1$</span>, so
<span class="math-container">$$\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) = \left( \begin{matrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{matrix} \right)$$</span>
and the distance squared is <span class="math-container">$2$</span></p>
<p><span class="math-container">$a = -d$</span>, <span class="math-container">$b=c$</span> implies <span class="math-container">$a d - b c \le 0$</span>, not possible.</p>
<p>So the minimal distance is <span class="math-container">$2$</span>. Let's generalize this:</p>
<p>Consider <span class="math-container">$A \in SL(n, \mathbb{R})$</span>. We have <span class="math-container">$\|A\|^2 = \operatorname{trace} (A A^t)$</span></p>
<p>Since <span class="math-container">$\det (A A^t) = 1$</span>, and <span class="math-container">$A A^t$</span> is positive definite, we conclude that <span class="math-container">$\operatorname{trace} (A A^t)\ge n$</span> ( <a href="https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#The_inequality" rel="nofollow noreferrer">inequality of means</a>). We have equality if and only if all the eigenvalues of <span class="math-container">$A A^t$</span> are <span class="math-container">$1$</span>, which is equivalent to <span class="math-container">$A A^t = 1$</span>. Therefore, the minimizers are exactly the elements in <span class="math-container">$SO(n, \mathbb{R})$</span>.</p>
<p><span class="math-container">${\bf Added:}$</span> We could use Lagrange multipliers in the general case, and conclude that any critical point for <span class="math-container">$\|A\|^2$</span> on the manifold <span class="math-container">$\det A = 1$</span> is an element of <span class="math-container">$SO(n, \mathbb{R})$</span>.</p>
<p>Let's see how we find the critical points of <span class="math-container">$\|A\|^2$</span> on the closed submanifold <span class="math-container">$\det A = d$</span>, <span class="math-container">$d \ne 0$</span>. Set up the Lagrange system and obtain that the following matrices are proportional
<span class="math-container">$$A^t \simeq \operatorname{adj} A$$</span>
where <span class="math-container">$ \operatorname{adj} A$</span> is the <a href="https://en.wikipedia.org/wiki/Adjugate_matrix" rel="nofollow noreferrer">adjugate matrix</a> of <span class="math-container">$A$</span>.</p>
<p>Now, if <span class="math-container">$d > 0$</span>, then we obtain that <span class="math-container">$A = \sqrt[n]{d} \cdot O$</span>, where <span class="math-container">$O \in SO(n, \mathbb{R})$</span>, while if <span class="math-container">$d< 0$</span>, we get
<span class="math-container">$A = \sqrt[n]{-d} \cdot O_-$</span>, where <span class="math-container">$O_-$</span> is orthogonal of determinant <span class="math-container">$-1$</span>.</p>
|
290,910 | <p>Which sequences of adjacent edges of a polyhedron could be considered to be a geodesic? The edges of a face most surely will not, but the "equator" of the octahedron eventually will. But for what reasons? How do the defining property of a geodesic - having zero geodesic curvature - apply to a sequence of edges?</p>
<p>(One crude guess: any sequence of edges that pairwise don't share a face? What does this have to do with curvature?)</p>
| Gerry Myerson | 8,269 | <p>Some thoughts are given by Konrad Polthier and Markus Schmies, <a href="http://page.mi.fu-berlin.de/polthier/articles/straightest/straightest_preprint.pdf" rel="nofollow">Straightest geodesics on polyhedral surfaces</a>. Form the abstract: </p>
<p>Geodesic curves are the fundamental concept in geometry to generalize the idea of straight lines to curved surfaces and arbitrary manifolds. On polyhedral surfaces we introduce the notion of discrete geodesic curvature of curves and define straightest geodesics. This allows a unique solution of the initial value problem for geodesics, and therefore a unique movement in a given tangential direction, a property not available in the well-known concept of locally shortest geodesics.</p>
|
1,016,682 | <p>is my proof correct?</p>
<p>Definition:</p>
<p>Let $X\subset\mathbb R$ and let $x'\in\mathbb R$, we say that $x'$ is an adherent point of $X$ iff $\forall\epsilon>0\exists x\in X \text{ s.t. }d(x′,x)≤ε$. the closure of X is denoted as $\overline X$ and is defined to be the set of all the adherent points of $X$.</p>
<p>show: $\overline{\overline X} = \overline X$</p>
<p>suppose $\exists z \in \overline{\overline X}~~and~~z \notin \overline X$</p>
<p>then, $\exists x' \in \overline X ~~s.t.~~ |z-x'|\leq \epsilon$</p>
<p>$|z-x'|\leq \frac{\epsilon}{2}$</p>
<p>but we also know that $\exists x \in X s.t. |x'-x|\leq \frac{\epsilon}{2}$</p>
<p>hence, $z-x+x-x' \leq \epsilon/2$</p>
<p>$z-x\leq0$</p>
<p>hence z is an adherent point of X ($z\in \overline X$). but this is a contradiction with the above condition on $z$
hence, $\overline{\overline X} = \overline X$</p>
| 5xum | 112,884 | <p>From $z\notin \overline{X}$, you cannot conclude that there exists such an $x'$ that $|z-x'|\leq \epsilon.$</p>
<p>Since $z\in \overline X$ is equivalent to $$\forall \epsilon \exists x \in X: d(z, x)\leq \epsilon$$</p>
<p>the statement $z\notin \overline X$ is equivalent to $$\exists \epsilon:\forall x\in X: d(z,x)>\epsilon,$$
which is not what you wrote.</p>
|
1,016,682 | <p>is my proof correct?</p>
<p>Definition:</p>
<p>Let $X\subset\mathbb R$ and let $x'\in\mathbb R$, we say that $x'$ is an adherent point of $X$ iff $\forall\epsilon>0\exists x\in X \text{ s.t. }d(x′,x)≤ε$. the closure of X is denoted as $\overline X$ and is defined to be the set of all the adherent points of $X$.</p>
<p>show: $\overline{\overline X} = \overline X$</p>
<p>suppose $\exists z \in \overline{\overline X}~~and~~z \notin \overline X$</p>
<p>then, $\exists x' \in \overline X ~~s.t.~~ |z-x'|\leq \epsilon$</p>
<p>$|z-x'|\leq \frac{\epsilon}{2}$</p>
<p>but we also know that $\exists x \in X s.t. |x'-x|\leq \frac{\epsilon}{2}$</p>
<p>hence, $z-x+x-x' \leq \epsilon/2$</p>
<p>$z-x\leq0$</p>
<p>hence z is an adherent point of X ($z\in \overline X$). but this is a contradiction with the above condition on $z$
hence, $\overline{\overline X} = \overline X$</p>
| user141240 | 141,240 | <p>The closure of a set is by definition the intersection of all closed sets containing the set.
But any intersection of closed sets is closed, so the closure is closed, and closure of a closed set is itself.</p>
|
3,963,661 | <p>I am trying to find that <span class="math-container">$L ={\{w\text{ | } w ∈ {\{a, b\}} * \text{is not a palindrome}\}}$</span></p>
<p>This is related to <a href="https://math.stackexchange.com/questions/1034989/w-w-%E2%88%88-a-b-is-not-a-palindrome-prove-this-language-is-not-regular">this previous question</a>, though in this case I want to explicitly prove it via pumping lemma, since I am unsure my solution is correct. This is what I have:</p>
<hr />
<ul>
<li><span class="math-container">$\text{for any } N, \text{ let } w \in L, |w|\ge N$</span></li>
</ul>
<p><span class="math-container">$w=a^Nb^Na^{N+1}$</span></p>
<ul>
<li><span class="math-container">$\text{For any partition } w =xyz , |y|\ge 1, |xy| \le N$</span></li>
</ul>
<p><span class="math-container">$x=a^r, y=a^s, z=a^{N-r-s}b^Na^{N+1},\text{where }s\ge 1$</span></p>
<ul>
<li><span class="math-container">$\exists i \text{ where } xy^iz\notin L$</span></li>
</ul>
<p><span class="math-container">$i=2\text{ ---> }xy^2z=a^{N+s}b^Na^{N+1} \text{ ,where we know that } s\ge 1$</span></p>
<p><span class="math-container">$\text{if }s=1 \text{ then }xy^2z\notin L \text{ and thus L is not regular}$</span></p>
<hr />
<p>I am really unsure about the last part. Does the pumping lemma hold if the "pumped word" doesn't belong to L for some specific values of <span class="math-container">$s$</span>?</p>
| Brian M. Scott | 12,042 | <p>There are several problems here. First, you can’t use just any <span class="math-container">$N$</span>: you need to specify that <span class="math-container">$N$</span> is at least as big as the pumping length. Next, the pumping lemma says that there is <strong>at least one</strong> partition <span class="math-container">$w=xyz$</span> such that <span class="math-container">$|y|\ge 1$</span>, <span class="math-container">$|xy|\le N$</span>, and <span class="math-container">$xy^iz\in L$</span> for each <span class="math-container">$i\ge 0$</span>; it does not say that this is the case for <strong>every</strong> partition <span class="math-container">$w=xyz$</span> with <span class="math-container">$|y|\ge 1$</span> and <span class="math-container">$|xy|\le N$</span>. And that is why it is indeed illegitimate to pick a specific value of <span class="math-container">$s$</span>: the pumping lemma just says that there is at least one pair of <span class="math-container">$r$</span> and <span class="math-container">$s$</span> such that <span class="math-container">$r+s\le N$</span>, <span class="math-container">$s\ge 1$</span>, and</p>
<p><span class="math-container">$$a^{r+is}a^{N-r-s}b^Na^{N+1}=a^{N+(i-1)s}b^Na^{N+1}\tag{1}$$</span></p>
<p>is in <span class="math-container">$L$</span> for each <span class="math-container">$i\ge 0$</span>. There is no guarantee that <span class="math-container">$s=1$</span>. And since <span class="math-container">$s=1$</span> is the only value of <span class="math-container">$s$</span> for which we can find an <span class="math-container">$i\ge 0$</span> making <span class="math-container">$(1)$</span> a palindrome, your choice of <span class="math-container">$w$</span> simply can’t be made to work.</p>
<p>Because we have no control over <span class="math-container">$|y|$</span> beyond the fact that <span class="math-container">$1\le|y|\le N$</span>, I see no way to use the pumping lemma to show directly that <span class="math-container">$L$</span> is not regular. You can use it to show that the language of palindromes over <span class="math-container">$\{a,b\}^*$</span>, which is the complement of <span class="math-container">$L$</span>, is not regular and then use the fact that a language is regular if and only if it’s complement is regular to conclude that <span class="math-container">$L$</span> cannot be regular. Alternatively, you can use the <a href="https://en.wikipedia.org/wiki/Myhill%E2%80%93Nerode_theorem" rel="nofollow noreferrer">Myhill-Nerode theorem</a>, as in the answer to <a href="https://math.stackexchange.com/questions/3371431/proving-a-language-is-not-regular-using-myhill-nerode-theorem">this question</a>.</p>
|
3,963,661 | <p>I am trying to find that <span class="math-container">$L ={\{w\text{ | } w ∈ {\{a, b\}} * \text{is not a palindrome}\}}$</span></p>
<p>This is related to <a href="https://math.stackexchange.com/questions/1034989/w-w-%E2%88%88-a-b-is-not-a-palindrome-prove-this-language-is-not-regular">this previous question</a>, though in this case I want to explicitly prove it via pumping lemma, since I am unsure my solution is correct. This is what I have:</p>
<hr />
<ul>
<li><span class="math-container">$\text{for any } N, \text{ let } w \in L, |w|\ge N$</span></li>
</ul>
<p><span class="math-container">$w=a^Nb^Na^{N+1}$</span></p>
<ul>
<li><span class="math-container">$\text{For any partition } w =xyz , |y|\ge 1, |xy| \le N$</span></li>
</ul>
<p><span class="math-container">$x=a^r, y=a^s, z=a^{N-r-s}b^Na^{N+1},\text{where }s\ge 1$</span></p>
<ul>
<li><span class="math-container">$\exists i \text{ where } xy^iz\notin L$</span></li>
</ul>
<p><span class="math-container">$i=2\text{ ---> }xy^2z=a^{N+s}b^Na^{N+1} \text{ ,where we know that } s\ge 1$</span></p>
<p><span class="math-container">$\text{if }s=1 \text{ then }xy^2z\notin L \text{ and thus L is not regular}$</span></p>
<hr />
<p>I am really unsure about the last part. Does the pumping lemma hold if the "pumped word" doesn't belong to L for some specific values of <span class="math-container">$s$</span>?</p>
| Humberto Longo | 925,330 | <p>Consider the string <span class="math-container">$w = xyz = a^pba^{p+p!} \in L$</span>, where <span class="math-container">$p$</span> is the pumping length provided by the Pumping Lemma for regular languages. Since <span class="math-container">$|xy| \leqslant p$</span>, <span class="math-container">$z = a^{p-i-j}ba^{p+p!}$</span>, with <span class="math-container">$|x| = i$</span>, <span class="math-container">$|y| = j$</span>, <span class="math-container">$i \geqslant 0$</span> and <span class="math-container">$j>0$</span>. However, <span class="math-container">$w'= xy^{\left(\frac{p!}{j}+1\right)}z = a^i(a^j)^{\left(\frac{p!}{j}+1\right)}a^{p-i-j}ba^{p+p!} = a^{p+p!}ba^{p+p!} \notin L$</span>. Therefore, given the contradiction, <span class="math-container">$L$</span> is not a regular language!</p>
|
3,865,388 | <p>Let <span class="math-container">$A$</span> be the set of all <span class="math-container">$2\times2$</span> boolean matrices and <span class="math-container">$R$</span> be a relation defined on <span class="math-container">$A$</span> as <span class="math-container">$M \mathrel{R} N$</span> if and only if <span class="math-container">$m_{ij} \leqslant n_{ij}$</span>, where <span class="math-container">$1 \leqslant i, j \leqslant 2$</span>. Is <span class="math-container">$(A,R)$</span> a lattice? Justify.</p>
<p>I am unable to solve the above question. I already know the following: There are 16 elements in set <span class="math-container">$A$</span>. Now when i try to calculate relation <span class="math-container">$R$</span> according to the given conditions i am getting 81 pairs in relation <span class="math-container">$R$</span>. Drawing a hasse diagram and calculating <strong>least upper bound</strong> and <strong>greatest lower bound</strong> for each point in the hasse diagram will go too lengthy. Is there a better method to do this?? Please help.</p>
| amrsa | 303,170 | <p>If you have two matrices <span class="math-container">$A=\{a_{ij}:0\leq i,j\leq1\}$</span> and <span class="math-container">$B=\{b_{ij}:0\leq i,j\leq1\}$</span>, then
<span class="math-container">$$A \wedge B = C = \{c_{ij}:0\leq i,j\leq1\}$$</span>
and
<span class="math-container">$$A\vee B = D =\{d_{ij}:0\leq i,j\leq1\},$$</span>
where
<span class="math-container">$$c_{ij}=\min\{a_{ij},b_{ij}\}$$</span>
and
<span class="math-container">$$d_{ij}=\max\{a_{ij},b_{ij}\}.$$</span>
This also forms a Boolean algebra, with
<span class="math-container">$$A' =\{1- a_{ij}:0\leq i,j\leq1\}.$$</span></p>
|
1,199,304 | <p>Let $M\neq \{0\}$ be a semi-simple left $R$ module .Prove that it contains a simple sub-module.</p>
<p>An $R-$ module $M$ is said to be semi-simple if every submodule of $M$ is a direct summand of M
<strong>My solution</strong></p>
<p>Since $M\neq \{0\}$; $\exists m\in M$ such that $m\neq 0$.Then I can consider the left $R-$ module $Rm$ ;By hypothesis it is a direct summand of $M$.Thus $M=N+Rm$ where $N$ is a sub module of $M$ </p>
<p>How to proceed next?</p>
| 1123581321 | 482,390 | <blockquote>
<p>If <span class="math-container">$M\not=0$</span> is a semisimple <span class="math-container">$R-$</span>module then there exists a simple <span class="math-container">$R-$</span>module <span class="math-container">$N\leq M$</span></p>
</blockquote>
<p>Let <span class="math-container">$x\in M-\{0\}$</span> and <span class="math-container">$\mathcal{X}=\{M':\ M'\leq M \text{ submodule with } x\not\in M'\}$</span>. It is <span class="math-container">$\mathcal{X}\not=\emptyset$</span> since <span class="math-container">$0\in\mathcal{X}$</span>. If <span class="math-container">$\mathcal{Y}\subseteq\mathcal{X}$</span> is a chain then the <span class="math-container">$R-$</span>submodule <span class="math-container">$M_0=\cup\mathcal{Y}$</span> does not contain <span class="math-container">$x$</span> (o <span class="math-container">$M_0\in\mathcal{X}$</span>) and it is "bigger" than every <span class="math-container">$M'\in\mathcal{Y}$</span>. Hence (Zorn's Lemma) there is a maximal <span class="math-container">$M'\in\mathcal{X}$</span>. Since <span class="math-container">$M$</span> is semisimple <span class="math-container">$\exists\ N\leq M$</span> st <span class="math-container">$M=M'\oplus N$</span>.</p>
<p><span class="math-container">$N$</span> is a simple <span class="math-container">$R-$</span>module. It is <span class="math-container">$x\notin M'$</span> so <span class="math-container">$M'\subsetneq M\Rightarrow N\not=0$</span>. Suppose that there is a submodule <span class="math-container">$0\subsetneq L\subsetneq N$</span>. Since <span class="math-container">$N$</span> is semisimple we can find <span class="math-container">$L'\subseteq N$</span> st <span class="math-container">$N=L\oplus L'$</span>. Then <span class="math-container">$0\subsetneq L'\subsetneq N$</span>. So <span class="math-container">$M'\subsetneq M'\oplus L\Rightarrow$</span> <span class="math-container">$\fbox{$x\in M'\oplus L$}$</span>. It is also, <span class="math-container">$M'\subsetneq M'\oplus L'\Rightarrow \fbox{$x\in M'\oplus L'$}$</span> and so
<span class="math-container">$$x\in (M'\oplus L)\cup (M'\oplus L')\subseteq M=M'\oplus N=M'\oplus L\oplus L'$$</span>
But it is <span class="math-container">$(M'\oplus L)\cap(M'\oplus L')=M'\Rightarrow x\in M'$</span> contradiction since <span class="math-container">$M'\in\mathcal{X}$</span></p>
|
3,203,100 | <p>This equation just came to my mind, I tried solving it but can't find any solution to this problem. Can anyone please tell what is the process to approach this problem? </p>
| Dr. Sonnhard Graubner | 175,066 | <p>Considering the equation <span class="math-container">$$x=n^x$$</span> and taking the logarithm on both sides we get
<span class="math-container">$$\frac{\ln(x)}{x}=\ln(n)$$</span> so <span class="math-container">$$n=e^{\frac{\ln(x)}{x}}$$</span> now you can use calculus for <span class="math-container">$$e^{\frac{\ln(x)}{x}}$$</span></p>
|
2,840,192 | <p>I have this problem and I was thinking to use the mean value theorem, but in the hypothesis i don't have the conditions (explicitly at least). Any ideas of how to start?</p>
<blockquote>
<p>Suppose that $\lim_{x\to a} f(x) = 0$, and that exist $\lim_{x\to a} \frac{1}{f'(x)}$. Show that exist an open interval with center in $a$ such that in that interval but without the $a$ $f(x)≠0$.</p>
</blockquote>
| Paramanand Singh | 72,031 | <p>Since the limit of $1/f'$ exists the derivative $f'$ is non-zero in some deleted neighborhood of $a$ and therefore by intermediate value property it maintains a constant sign on each side of $a$ in this deleted neighborhood. Thus $f$ is strictly monotone on each side of $a$ in the deleted neighborhood. And since $f(a) =0$ it follows that $f$ is non-zero in the deleted neighborhood of $a$. </p>
|
2,840,192 | <p>I have this problem and I was thinking to use the mean value theorem, but in the hypothesis i don't have the conditions (explicitly at least). Any ideas of how to start?</p>
<blockquote>
<p>Suppose that $\lim_{x\to a} f(x) = 0$, and that exist $\lim_{x\to a} \frac{1}{f'(x)}$. Show that exist an open interval with center in $a$ such that in that interval but without the $a$ $f(x)≠0$.</p>
</blockquote>
| Torsten Schoeneberg | 96,384 | <p>Here's a proof idea which uses the MVT, or rather just its special case known as Rolle's Theorem.</p>
<p>Assume that for every $\epsilon >0$, there is $x\in (a, a+\epsilon)$ with $f(x)=0$.</p>
<p>Then for every $\epsilon$, there are actually infinitely many such $x$ (why?). Choose two of them, $x_{1,\epsilon}$ and $x_{2,\epsilon}$. By Rolle's Theorem (fill in details), $f'(x) = 0$ somewhere in the interval $(x_{1,\epsilon}, x_{2,\epsilon})$ which is contained in $(a, a+\epsilon)$. This holding for every $\epsilon$ contradicts the existence of $\lim_{x\to a}\frac{1}{f'(x)}$ (why?).</p>
<p>So there is some $\epsilon>0$ such that $f$ has no zero between $a$ and $a+\epsilon$. Completely analogously, find an interval to the left of $a$ without zeros of $f$.</p>
|
1,796,792 | <p>Is $\log_27$ a rational number?</p>
| Rob | 274,944 | <p>Suppose $\log_2 7 = {a\over b}$ for two positive integers $a$ and $b$. </p>
<p>$\log_2 7 = { \ln 7 \over \ln 2} = {a \over b}$</p>
<p>Cross multiply,</p>
<p>$b \ln 7 = a \ln 2 \implies \ln ( 7^b ) = \ln (2^a)$</p>
<p>Take the $e^{( \ \ )}$ of both sides, </p>
<p>$7^b = 2^a$</p>
<p>This is impossible for integers $a$ and $b$ because $7^b$ is always going to be an odd number, while $2^a$ will always be an even number. They can never be equal, thus, $\log_2 7$ is not a rational number.</p>
|
376,796 | <p>This is more of a pedagogical question rather than a strictly mathematical one, but I would like to find good ways to visually depict the notion of curvature. It would be preferable to have pictures which have a reasonably simple mathematical formalization and even better if there is a related diagram that explains torsion.</p>
<h2>One common picture</h2>
<p><a href="https://i.stack.imgur.com/bSiYsm.png" rel="noreferrer"><img src="https://i.stack.imgur.com/bSiYsm.png" alt="enter image description here" /></a></p>
<p>I've often used the above schematic to think about the Riemann curvature tensor
<span class="math-container">$$R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z + \nabla_{[X,Y]} Z.$$</span></p>
<p>This diagram intuitively shows that the curvature involves the difference of covariant derivatives. However, it doesn't really explain why there is another term in the formula (i.e., <span class="math-container">$\nabla_{[X,Y]} Z$</span>). Also, it takes some work to translate the picture into a precise and correct mathematical formula.</p>
<p>One way to formalize this (suggested by Robert Bryant) is to consider a parallelogram with sides <span class="math-container">$\epsilon X$</span> and <span class="math-container">$\epsilon Y$</span> in <span class="math-container">$T_p M$</span>. Then the diagram depicts the parallel transport of <span class="math-container">$Z$</span> along the exponential of the sides of the parallelogram.
To understand the picture, you parallel transport the vector labelled <span class="math-container">$R(X,Y)Z$</span> back to <span class="math-container">$p$</span>, divide by <span class="math-container">$\epsilon^2$</span> and let <span class="math-container">$\epsilon$</span> go to <span class="math-container">$0$</span>.
This interpretation is conceptually simple, but has the disadvantage that the top and right hand sides of the parallelogram are not geodesics, so we cannot use this interpretation to draw a similar diagram for torsion.</p>
<p>There are other ways to formalize this diagram, and it would be interesting to hear other simple and correct explanations for this picture (or any variation of it).</p>
<h2>Another common picture</h2>
<p><a href="https://i.stack.imgur.com/MhGf1m.png" rel="noreferrer"><img src="https://i.stack.imgur.com/MhGf1m.png" alt="By Fred the Oyster, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=35124171" /></a></p>
<p>Another commonly used picture to explain curvature is a spherical triangle with two vertices on the equator and a third at a pole. This intuitively shows that curvature gives rise to holonomy, but also relies on the global geometry of the sphere. In other words, it doesn't really depict curvature as "local holonomy."</p>
| Joseph O'Rourke | 6,094 | <p>With advances in discrete differential geometry, it is now nearly
routine to compute curvature on meshed surfaces. Here are two
of many possible color-coded examples.</p>
<hr />
<img src="https://i.stack.imgur.com/motmV.png" width="300" />
<img src="https://i.stack.imgur.com/xPIQj.png" width="150" />
<hr />
<blockquote>
<p>Rusinkiewicz, Szymon. "Estimating curvatures and their derivatives on triangle meshes." In <em>Proceedings. 2nd International Symposium on 3D Data Processing, Visualization and Transmission</em>, 2004. 3DPVT 2004., pp. 486-493. IEEE, 2004. Fig. 4 (detail).
<a href="https://doi.org/10.1109/TDPVT.2004.1335277" rel="noreferrer">DOI</a>.</p>
</blockquote>
<hr />
<img src="https://i.stack.imgur.com/70ID4.png" width="500" />
<hr />
<blockquote>
<p>Gatzke, Timothy, Cindy Grimm, Michael Garland, and Steve Zelinka. "Curvature maps for local shape comparison." In <em>International Conference on Shape Modeling and Applications</em> 2005 (SMI'05), pp. 244-253. IEEE, 2005.
<a href="https://doi.org/10.1109/SMI.2005.13" rel="noreferrer">DOI</a>.</p>
</blockquote>
<p>(<em>Added in response to comment</em>:)</p>
<img src="https://i.stack.imgur.com/sjLHT.png" width="300" />
<p>Found at <a href="https://www.pinterest.com/pin/243898136048931116/" rel="noreferrer">this link</a>
(originator unknown.)</p>
|
376,796 | <p>This is more of a pedagogical question rather than a strictly mathematical one, but I would like to find good ways to visually depict the notion of curvature. It would be preferable to have pictures which have a reasonably simple mathematical formalization and even better if there is a related diagram that explains torsion.</p>
<h2>One common picture</h2>
<p><a href="https://i.stack.imgur.com/bSiYsm.png" rel="noreferrer"><img src="https://i.stack.imgur.com/bSiYsm.png" alt="enter image description here" /></a></p>
<p>I've often used the above schematic to think about the Riemann curvature tensor
<span class="math-container">$$R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z + \nabla_{[X,Y]} Z.$$</span></p>
<p>This diagram intuitively shows that the curvature involves the difference of covariant derivatives. However, it doesn't really explain why there is another term in the formula (i.e., <span class="math-container">$\nabla_{[X,Y]} Z$</span>). Also, it takes some work to translate the picture into a precise and correct mathematical formula.</p>
<p>One way to formalize this (suggested by Robert Bryant) is to consider a parallelogram with sides <span class="math-container">$\epsilon X$</span> and <span class="math-container">$\epsilon Y$</span> in <span class="math-container">$T_p M$</span>. Then the diagram depicts the parallel transport of <span class="math-container">$Z$</span> along the exponential of the sides of the parallelogram.
To understand the picture, you parallel transport the vector labelled <span class="math-container">$R(X,Y)Z$</span> back to <span class="math-container">$p$</span>, divide by <span class="math-container">$\epsilon^2$</span> and let <span class="math-container">$\epsilon$</span> go to <span class="math-container">$0$</span>.
This interpretation is conceptually simple, but has the disadvantage that the top and right hand sides of the parallelogram are not geodesics, so we cannot use this interpretation to draw a similar diagram for torsion.</p>
<p>There are other ways to formalize this diagram, and it would be interesting to hear other simple and correct explanations for this picture (or any variation of it).</p>
<h2>Another common picture</h2>
<p><a href="https://i.stack.imgur.com/MhGf1m.png" rel="noreferrer"><img src="https://i.stack.imgur.com/MhGf1m.png" alt="By Fred the Oyster, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=35124171" /></a></p>
<p>Another commonly used picture to explain curvature is a spherical triangle with two vertices on the equator and a third at a pole. This intuitively shows that curvature gives rise to holonomy, but also relies on the global geometry of the sphere. In other words, it doesn't really depict curvature as "local holonomy."</p>
| Sebastian | 4,572 | <p>This is a very similar picture to that in the answer by Gabe, but concerning the sectional curvature of a Riemannian metric. Consider a point <span class="math-container">$p\in M$</span>, and a plane <span class="math-container">$V\subset T_pM.$</span> For small radius <span class="math-container">$r$</span> consider the image under the exponential map of the circle of radius <span class="math-container">$r$</span> in <span class="math-container">$V$</span> centered at <span class="math-container">$0$</span>. This is a closed curve <span class="math-container">$C(r)$</span> in the manifold and its length behaves like <span class="math-container">$$L(C(r))\sim 2\pi r (1-\frac{1}{6}Kr^2\dots)$$</span> for <span class="math-container">$r$</span> small. It turns out that <span class="math-container">$K$</span> is the sectional curvature of the plane <span class="math-container">$V\subset T_pM.$</span> Of course, pictures for 2-dimensional Riemannian manifolds in 3-dimensional space are particularly useful to see what curvature does. To prove this expansion of the length just use the Taylor expansion of the metric in normal coordinates. If I am not mistaken, this was used by Riemann to define curvature (of a Riemannian manifold).</p>
|
4,338,382 | <blockquote>
<p>Let <span class="math-container">$a,b>0$</span>. Prove that: <span class="math-container">$$\frac{1}{a^2}+b^2\ge\sqrt{2\left(\frac{1}{a^2}+a^2\right)}(b-a+1)$$</span></p>
</blockquote>
<p>Anyone can help me get a nice solution for this tough question?
My approach works for 2 cases:</p>
<p>Case 1: <span class="math-container">$b-a+1>0$</span> then squaring both side, we get equivalent inequality: <span class="math-container">$$\frac{1}{a^4}+b^4+2\frac{b^2}{a^2}\ge2\left(\frac{1}{a^2}+a^2\right)(b^2+a^2+1-2ab-2a+2b)$$</span> Or: <span class="math-container">$$\frac{1}{a^4}+b^4\ge\frac{2}{a^2}(a^2+1-2ab-2a+2b)+2a^2(b^2+a^2+1-2ab-2a+2b)$$</span>
The rest is so complicated. Is there nice idea etc: AM-GM, C-S to prove this inequality.</p>
<p>Case 2: <span class="math-container">$b-a+1<0$</span> which is obviously true.</p>
<p>I hope we can find a better approach for the inequality. Thank you very much!</p>
| obscurans | 619,038 | <p>Alternate way, since who the knight is is important:</p>
<p><strong>Suppose A is a knight</strong>: then B is a spy, and C is a knave.</p>
<p><strong>Suppose C is a knight</strong>: then A is a knave, so B is <em>not</em> a spy, which is a contradiction.</p>
<p><strong>Suppose B is a knight</strong>: A is lying, but both non-knights can lie. It works either way if A is a spy or a knave.</p>
<p>So without B's statement, A or B could be knights. Which means B is lying, so it's the first case.</p>
|
3,501,332 | <blockquote>
<p>A cryptoanalist, while trying to decipher a message, found that the
most frequent blocks were RH and NI, which must correspond to TH and
HE, which are the most common in the english language. Supposing the
text was codified using a 2x2 block cipher, what was the used matrix?</p>
</blockquote>
<p>I know that</p>
<p><span class="math-container">$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}T&H\\H&E\end{bmatrix} =\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}19&7\\7&4\end{bmatrix} = \begin{bmatrix}R&N\\H&I\end{bmatrix}=\begin{bmatrix}17&13\\7&8\end{bmatrix} \pmod{26}$</span></p>
<p>I got as far as making a system of equations but that didn't get me anywhere. How do I solve this?</p>
| lonza leggiera | 632,373 | <p>The equation
<span class="math-container">$$
\begin{bmatrix}
a&b\\c&d
\end{bmatrix}
\begin{bmatrix}
19&7\\7&4
\end{bmatrix}
\equiv
\begin{bmatrix}
17&13\\7&8
\end{bmatrix}
\pmod{26}
$$</span>
gives you four equations in the four unknowns <span class="math-container">$\ a,b,c,d\ $</span>. Two of these equations,
<span class="math-container">\begin{align}
19a+7b&\equiv17\pmod{26}\\
7a+4b&\equiv13\pmod{26}\ ,
\end{align}</span>
contain only <span class="math-container">$\ a\ $</span> and <span class="math-container">$\ b\ $</span>, while the other two contain only <span class="math-container">$\ c\ $</span> and <span class="math-container">$\ d\ $</span>.</p>
<p>You can solve these equations by <a href="https://en.wikipedia.org/wiki/Gaussian_elimination" rel="nofollow noreferrer">Gaussian elimination</a>, just like you do for linear equations in real numbers, except that you use arithmetic modulo <span class="math-container">$26$</span> instead of normal arithmetic. In arithmetic modulo <span class="math-container">$26$</span>, besides not being able to divide by <span class="math-container">$0$</span>, you can't divide by <span class="math-container">$26$</span>, <span class="math-container">$2$</span>, or <span class="math-container">$13$</span> either (unless, in the latter two cases, you divide the modulus as well as both sides of the equation). Division by any other number, <span class="math-container">$\ m\ $</span> modulo <span class="math-container">$26$</span> requires you to compute its reciprocal, <span class="math-container">$\ m^{-1}\ $</span>, modulo <span class="math-container">$26$</span> and multiply by that. Reciprocals can be computed efficiently with the <a href="https://en.wikipedia.org/wiki/Euclidean_algorithm#Euclidean_division" rel="nofollow noreferrer">Euclidean algorithm</a>, but if you're not familiar with that algorithm, it's not all that tedious to find them by trial and error.</p>
<p>For the above equations, if you add the second to the first, you get
<span class="math-container">\begin{align}
26a+11b \equiv 11b&\equiv 30 \pmod{26}\\
&\equiv 4 \pmod{26}\ .
\end{align}</span>
The reciprocal of <span class="math-container">$11$</span> modulo <span class="math-container">$26$</span> is <span class="math-container">$19$</span> (check: <span class="math-container">$11\cdot19\equiv$$ -11\cdot7\equiv$$-77\equiv1\pmod{26}\ $</span>). So you get
<span class="math-container">$$
19\cdot 11b\equiv b\equiv19\cdot4\equiv-7\cdot4\equiv-2\equiv24\pmod{26}\ .
$$</span>
Now substituting <span class="math-container">$\ b\equiv-2\ $</span> back into the second of the above equations, you get
<span class="math-container">$$
7a-8\equiv 13\pmod{26}\ \text{, or}\\
7a\equiv 21 \pmod{26}\ .
$$</span>
Here, you don't even need to compute the reciprocal of <span class="math-container">$7$</span> modulo <span class="math-container">$26$</span> to solve for <span class="math-container">$\ a\ $</span>. Because <span class="math-container">$\ 21=7\cdot3\ $</span>, you can immediately deduce <span class="math-container">$\ a=3\ $</span>.</p>
<p>Can you now solve the other two equations to find <span class="math-container">$\ c\ $</span> and <span class="math-container">$\ d\ $</span>?</p>
|
275,539 | <p>Kind of leading on from my other question, how would I solve for $i$? Or how would I check that it is possible to have such an $i$?</p>
<p>First I had to check for all $2^i$ and clearly this doesn't happen as all $2^i$ are even and so I will just get even $x's$ such that $2^i \equiv x \mod 28$. So the next one I go onto is $3$.</p>
<p>Now how do I go about doing this? </p>
| Matt E | 221 | <p>If you know the Chinese Remainder Theorem, then you see that you can check this separately mod $4$ and mod $7$. Already $3^2 \equiv 1 \bmod 4,$ and so you
are left to finding the smallest power of $3$ that is $1 \bmod 7$. By Euler it is a factor of $7- 1,$ i.e. of $6$, and one easily rules out $1,2,$ and $3$. Thus
$i = 6$ is your answer. </p>
<hr>
<p>In this case there is not much difference between this approach and applying
Euler's theorem directly to $28$, since all the numbers involved are small. But
in general, if we have $N = mn$ with $m$ and $n$ coprime, then this method
means you only have to compute congruences mod $m$ and $n$, which might be quite a bit easier than working directly mod $N$. </p>
<p>Also, this method shows that the value of $i$ has to be a factor of the lcm of $\varphi(m)$ and $\varphi(n)$,
whereas applying Euler directly mod $N$, we only get that the value of $i$ is a factor of $\varphi(N) = \varphi(m)\varphi(n)$.
(E.g. since $\varphi(4) = 2$ and $\varphi(7) = 6$, we saw immediately that $i$ would be a factor of $6$, whereas since $\varphi(28) = 12$, applying Euler directly mod $28$ just gives that $i$ is a factor of $12$.)</p>
|
1,368,988 | <p>I was thinking about different ways of finding $\pi$ and stumbled upon what I'm sure is a very old method: dividing a circle of radius $r$ up into $n$ isosceles triangles each with radial side length $r$ and central angle $$\theta=\frac{360^\circ}{n}$$ Use $s$ for the side opposite to $\theta$.</p>
<p><img src="https://i.stack.imgur.com/vueMQ.png" alt="An example for n=8."></p>
<p>Then we can approximate the circumference as $sn$. By the law of cosines: $$s=\sqrt{2r^2-2r^2 \cos{\theta}}=r\sqrt{2-2\cos{\left(\frac{360^\circ}{n}\right)}}$$</p>
<p>We know $\pi=\text{circumference}/\text{diameter} \approx \frac{sn}{2r}=\frac{n}{2}\sqrt{2-2\cos\left(\frac{360^\circ}{n}\right)}$. This becomes exact at the limit:
$$\pi=\lim_{n \to \infty}\frac{n}{2}\sqrt{2-2\cos\left(\frac{360^\circ}{n}\right)}$$</p>
<p>Now for my question: <strong>How would you solve the opposite problem?</strong> To make my meaning more clear, above I used the definition of $\pi$ to determine a limit that gives its value. But if I had just been given the limit, what technique(s) could I have used to determine that it evaluates to $\pi$?</p>
| Mark Viola | 218,419 | <p>Since there were already two solid answers that provided efficient approaches, I thought that it would be instructive to see a different way forward. </p>
<p>Here, we will use the expansion of the cosine as </p>
<p>$$\cos x=1-\frac12 x^2+O(x^4) \tag 1$$</p>
<p>Letting $x=\frac{2\pi}{n}$ in $(1)$ yields </p>
<p>$$2-2\cos \left(\frac{2\pi}{n}\right)=\frac{4\pi^2}{n^2} +O(n^{-4})$$</p>
<p>Finally, we have</p>
<p>$$\begin{align}
\frac{n}{2}\sqrt{2-2\cos \left(\frac{2\pi}{n}\right)}&=\frac n2 \sqrt{\frac{4\pi^2}{n^2} +O(n^{-4})}\\\\
&=\frac n2 \frac{2\pi}{n}\left(1+O(n^{-2})\right)\\\\
&=\pi+O(n^{-3})\to \pi
\end{align}$$</p>
<p>as expected!</p>
|
2,934,028 | <blockquote>
<p>A particle moves along the top of the
parabola <span class="math-container">$y^2 = 2x$</span> from left to right at a constant speed of 5 units
per second. Find the velocity of the particle as it moves through
the point <span class="math-container">$(2, 2)$</span>. </p>
</blockquote>
<p>So I isolate <span class="math-container">$y$</span>, giving me <span class="math-container">$y=\sqrt{2x}$</span>. I then find the derivative of <span class="math-container">$y$</span>, which is <span class="math-container">$1/\sqrt{2x}$</span>. And <span class="math-container">$\sqrt{2x}=y$</span> so the derivative of also equal to <span class="math-container">$1/y$</span>. At <span class="math-container">$(2,2)$</span> the derivative is 0.5. Not sure where to go from there though. Any help is appreciated. </p>
| PabloG. | 597,199 | <p>Consider the curve in a parametric form <span class="math-container">$\vec{z}=(x(t),y(t))$</span>, in that case we know that </p>
<p><span class="math-container">$$
\vec{v}=\frac{\mathrm{d}\vec{z}}{\mathrm{d}t}=(x'(t),y'(t))
$$</span></p>
<p>Additionally, we know that the speed is <span class="math-container">$\|\vec{v}\|=5$</span>. Therefore, we need to find the direction of <span class="math-container">$\vec{v}$</span>. Given the equation of the curve </p>
<p><span class="math-container">$$
y^2=2x
$$</span>
we can write the parametric from in terms of <span class="math-container">$y$</span> alone:
<span class="math-container">$$
\vec{z}=\left(\frac{y(t)^2}{2},y(t) \right)
$$</span>
Moreover, we can pick the parameter <span class="math-container">$t$</span> such that <span class="math-container">$y(t)=ct$</span>, where <span class="math-container">$c$</span> is a constant. Then:
<span class="math-container">$$
\vec{z}=\left(\frac{c^2t^2}{2},ct \right)
$$</span>
Then the velocity is
<span class="math-container">$$
\vec{v}=(c^2t,c)
$$</span>
At the point <span class="math-container">$\vec{z}=(2,2)$</span>, <span class="math-container">$t=2/c$</span>. Therefore, </p>
<p><span class="math-container">$$
\vec{v}=(2c,c) \Rightarrow \|\vec{v}\|=\sqrt{4c^2+c^2}=5
$$</span>
Then, <span class="math-container">$c=\sqrt{5}$</span>. Lets check our solution,
<span class="math-container">$$
\begin{align}
\vec{z}(t)=&\left(\frac{5t^2}{2},\sqrt{5}t\right)\\
\vec{v}(t)=&\left(5t,\sqrt{5}\right)
\end{align}
$$</span>
at <span class="math-container">$t=\frac{2}{\sqrt{5}}$</span>:
<span class="math-container">$$
\begin{align}
\vec{z}\left(\frac{2}{\sqrt{5}}\right)=&\left(\frac{5}{2}\cdot\frac{4}{5},\sqrt{5}\frac{2}{\sqrt{5}}\right)=(2,2)\\
\vec{v}\left(\frac{2}{\sqrt{5}}\right)=&\left(5\frac{2}{\sqrt{5}},\sqrt{5}\right)=(2\sqrt{5},\sqrt{5})
\end{align}
$$</span></p>
|
200,903 | <p>My teacher was explaining quadratics in my class and it was a little bit unclear to me. The problem was <br> <br>
Suppose $at^2 + 5t + 4 > 0$, show that $a > 25/16$ . <br> <br></p>
<p>My teacher said that there are no solutions for this function when it is greater than $0$ and used $b^2-4ac \lt 0$, and this is the part that confused me. I understand why he used $b^2-4ac \lt 0$ but I cannot understand why there are no solutions by just looking at the function. Could someone explain this to me?</p>
| Community | -1 | <p>The question really means that $at^2+5t+4>0$ for all real values of $t$. This is only possible if $a>0$ and the quadratic curve does not intersect the horizontal axis. In this case there are no real solutions to the equation $at^2+5t+4=0$ so that the discriminant is negative.</p>
|
103,675 | <p>I have defined a recursive sequence</p>
<pre><code>a[0] := 1
a[n_] := Sqrt[3] + 1/2 a[n - 1]
</code></pre>
<p>because I want to calculate the <code>Limit</code> for this sequence when n tends towards infinity.</p>
<p>Unfortunately I get a <code>recursion exceeded</code> error when doing:</p>
<pre><code>Limit[a[n], n -> Infinity]
</code></pre>
<p>How can I calculate the <code>Limit</code> for this sequence using Mathematica?</p>
| IPoiler | 30,913 | <p>You can attempt to convert your sequence in terms of a function which is not recursive, then take the <code>Limit</code> of that function</p>
<pre><code>a[0] = 1;
a[n_] := a[n] = Sqrt[3] + 1/2 a[n - 1]
seq30 = Table[a[ic], {ic, 0, 30}];
func = FindSequenceFunction[seq30, n]
Limit[func, n -> Infinity]
</code></pre>
<blockquote>
<pre><code>2^(1 - n) (1 - 2 Sqrt[3] + 2^n Sqrt[3])
2 Sqrt[3]
</code></pre>
</blockquote>
|
1,903,473 | <p>Given three permutations $p_1,p_2,p_3$ of $\{1,2,\ldots,n^3+1\}$, prove that two of them have a common subsequence of length $n+1$.</p>
<p>I have tried to solve this using the pigenhole principle but I didnt progress too much, any help would be appreciated</p>
<p>edit: when I say subsequence I mean that there are $1<=r<q<=3$
and $1<=i(1)<i(2)<...<i(n+1)<=n^3+1$ and $1<=j(1)<j(2)<...<j(n+1)<=n^3+1$ so that $p_r(i1)=p_q(j1)$, $p_r(i2)=p_q(j2)$ and so on, not necessarily consecutive</p>
| Kaligule | 182,303 | <p>I don't think that statement is true.</p>
<p>I am not shure what you think a permutation of a set is, so lets just assume we talk about sequences.
Take $n=2$, so $n^3+1 = 9$, so our sequence is $l=(1,2,3,4,5,6,7,8,9)$. Now I have these three permutations:</p>
<ul>
<li>$p_1(l) = (1,2,3,4,5,6,7,8,9) = l$</li>
<li>$p_2(l) = (9,8,7,6,5,4,3,2,1)$</li>
<li>$p_3(l) = (1,3,5,7,9,2,4,6,8)$</li>
</ul>
<p>Clearly, none of those have any common subsequence of length $n+1=3$ (or even of length $2$).</p>
|
4,495,950 | <blockquote>
<p>Why does <span class="math-container">$-\frac{1}{17-x}$</span> equal <span class="math-container">$\frac{1}{x-17}$</span>?</p>
</blockquote>
<p>Is there any simple computation to make this seem a little bit more intuitive? Right now, I cannot wrap my head around the fact that I can just switch signs of the term in the denominator.</p>
| G Tony Jacobs | 92,129 | <p>You can break this down into a couple of basic facts. First of all, we have <span class="math-container">$\frac{1}{-1}=-1$</span>. Thus, if you want to see the <em>opposite</em>, or negative, of a fraction, we can multiply either the numerator or denominator by <span class="math-container">$-1$</span>.</p>
<p>Secondly, and this might be the part you're really asking about, <span class="math-container">$a-b$</span> and <span class="math-container">$b-a$</span> are opposites. Observe: <span class="math-container">$$-1(a-b) = -a - (-b) = -a+b = b-a$$</span>
Even better, think of numbers: What's <span class="math-container">$5-3$</span>? What's <span class="math-container">$3-5$</span>? To do the second computation, don't you just do the first one, and then include a minus sign, since the order was "wrong" for doing subtraction as an intuitive "take-away"?</p>
<p>Putting these together: to get the negative of a fraction, you can negate <em>either</em> the numerator or the denominator, and we choose the denominator. To negate a difference, you can just flip the subtraction statement around.</p>
<p>Does this help?</p>
|
2,512,294 | <p>So I've been given the following problem:</p>
<p>How many positive integers are there that can not be written as a sum of 5's and 7's? For example, 4 is one of those integers, but 19 is not because 19 = 5 + 7 + 7. How to solve this? </p>
| Piquito | 219,998 | <p>It is easy enough to conclude that a such positive integer $n$ must be of the form $$n=5a+7b+c$$ where $a,b$ are non-negative integers and $c$ are positive integers with $c\lt 35$, coprime with $35$ and such that $c$ is not solution of $5a+7b=c$. </p>
<p>There are $\phi(35)=(5-1)(7-1)=24$ positive integers coprimes with $35$ and less than it:$$1,2,3,4,6,8,9,11,12,13,16,17,18,19,22,23,24,26,27,29,31,32,33,34$$ of which the following twelve $$12,17,19,22,24,26,27,29,31,32,33,34$$ are solution of $5a+7b=c$ (this has been easily found taking sums among $5,10,15,20,25,30$ and $7,14,21,28$).</p>
<p>Consequently the asked numbers are given by all the positive integers of the form
$$\color{red}{5a+7b=c\text{ where } c=1,2,3,4,6,8,9,11,13,16,18,23}$$</p>
|
1,307,085 | <p>How does one solve this equation?</p>
<blockquote>
<p>$$\cos {x}+\sin {x}-1=0$$</p>
</blockquote>
<p>I have no idea how to start it.</p>
<p>Can anyone give me some hints? Is there an identity for $\cos{x}+\sin{x}$?</p>
<p>Thanks in advance!</p>
| Community | -1 | <p>Rewrite as:</p>
<p>$$\cos(x)+\sin(x)=1$$
Then square:
$$\cos^2(x)+2\cos(x)\sin(x)+\sin^2(x)=1$$
Note an important identity:
$$1+2\cos(x)\sin(x)=1$$
Then simplify and note another identity:
$$\sin(2x)=0$$</p>
<p>Can you take it from here?</p>
|
1,307,085 | <p>How does one solve this equation?</p>
<blockquote>
<p>$$\cos {x}+\sin {x}-1=0$$</p>
</blockquote>
<p>I have no idea how to start it.</p>
<p>Can anyone give me some hints? Is there an identity for $\cos{x}+\sin{x}$?</p>
<p>Thanks in advance!</p>
| Bhaskara-III | 246,676 | <p>starting from, $$\cos x+\sin x-1=0$$ $$(\sin x+\cos x)^2=1^2$$ completing square, i get $$\sin^2 x+\cos^2 x+2\sin \cos x=1$$ $$1+\sin 2x=1$$$$\sin 2x=0$$
$$2x=k\pi$$
$$x=\frac{k\pi}{2}$$ where, $k=0, \pm 1, \pm2, \ldots$</p>
|
1,044,507 | <p>Sample:
$$∀x ∈ R+,∃y ∈ R+, x < y ⇒ x > y$$</p>
<p>Say I tried <code>y = 5</code>. Do I need to check if the consequent is true for just the x values less than 5?</p>
<p>Secondly, Since there is no value y that makes the antecedent true, is this statement true since there are no counter examples? The implication is never used.</p>
<p>I'm looking for a small explanation, rather than a solution (don't treat it like a homework question looking for an answer).</p>
| hmakholm left over Monica | 14,366 | <p>Note that $\forall x$ is <em>before</em> $\exists y$. This means that the $y$ you use is allowed to depend on $x$ -- you don't have to select a single $y$ that must work with all possible $x$.</p>
<p>What the formula says is that <em>once</em> someone chooses an $x$, <em>then</em> it is possible to find a particular $y$ that makes $x<y\Rightarrow x>y$ true.</p>
<p>Since $x<y$ and $x>y$ can't be true at the same time, $x<y\Rightarrow x>y$ is true exactly when $x<y$ is <em>false</em>. So in other words the entire formula is true if "for each $x$ there is a $y$ that is not greater than $x$". But this is clearly true -- for example we might take $y$ to be $x$ itself, or $\frac12 x$.</p>
|
713,521 | <p>There are so many notations for differentiation. Some of them are:
$$
f^\prime(x) \qquad \frac{d}{dx}(f(x))\qquad \frac{dy}{dx}\qquad \frac{df}{dx}\qquad
D f(x)\qquad y^\prime\qquad D_x f(x)
$$
Why are there so many ways to say "the derivative of $f(x)$"? Is there a specific use for each notation? What is the difference between $\dfrac{d}{dx}$ and $\dfrac{dy}{dx}$? I am only asking this because I am worried that I might use the wrong notation sometimes. For example, I don't know when I should use $\dfrac{dy}{dx}$ instead of $D_xf(x)$, or vice versa. I thank you in advance for your answers.</p>
| Guy | 127,574 | <p>$f'(x)$ is equivalent to $\frac{d}{dx}(f(x))$. The difference is that in the first you aren't making explicit that you are differentiating <em>with respect to</em> $x$, while in the second that distinction is made clear. Although when we write $f'(x)$ is usually implied that the differentiation is with respect to $x$. $\frac{df}{dx}$ is also the same thing, in more compact notation. </p>
<p>$\dfrac{dy}{dx}$ and $y'$ are the same, but this time differentiation of $y$. It is not related to $f'(x)$ in any way, unless of course you have a relation in $y$ and $f(x)$.</p>
<p>I have not come across the notations $Df(x)$ and $D_xf(x)$ so cannot comment on that.</p>
|
846,108 | <p>How do you solve this equation: $2x+8=6x-12$ by using the guess and check method?</p>
<p>I divide $2x+8$ and I get $4$ then I divide $6x-12$ and I get $-2$ but I don't know what to do next or is it wrong?</p>
| afedder | 29,604 | <p>The answer is $x = 5$. You can get this solution using normal analytical methods (algebraic manipulation), i.e.,
$$\begin{align} 2x + 8 = 6x - 12 &\iff -4x + 8 = -12 \\&\iff -4x = -20 \\&\iff x = 5 \,\,. \end{align}$$
In terms of a "guess and check method", here's my strategy: factor $2$ out of the LHS and $6$ out of the RHS to obtain the equation $2(x + 4) = 6(x - 2)$. Now, we can divide $2$ from both sides to obtain $x + 4 = 3(x - 2)$. The solution is fairly easy to see in this form by guessing and checking. You will come to $x = 5$, as before.</p>
|
1,354,490 | <p>Prove this function is injective $f(x)=x+\mod(x,7)$.</p>
<p><strong>Attempt:</strong></p>
<p>I tried separating in two cases: $x \equiv y \pmod 7$ and $x \not \equiv y \pmod 7 $:</p>
<p>First case:
$$f(x)=f(y) \iff x+\mod(x,7)=y+ \mod (y,7)\implies x= y
$$</p>
<p>But I couldn't prove the second case.</p>
| Batominovski | 72,152 | <p>Note: $-7<\text{mod}(x,7)-\text{mod}(y,7)<7$ and $x-y\equiv \text{mod}(x,7)-\text{mod}(y,7)\pmod{7}$.</p>
<p>If $f(x)=f(y)$, then $-(x-y)=\text{mod}(x,7)-\text{mod}(y,7)$. What happens then?</p>
<p><strong>P.S.</strong> </p>
<p>(1) It's "injective," and not "inyective."</p>
<p>(2) We can use any other odd positive integer $m$ in place of $7$ and the claim still holds.</p>
|
1,354,490 | <p>Prove this function is injective $f(x)=x+\mod(x,7)$.</p>
<p><strong>Attempt:</strong></p>
<p>I tried separating in two cases: $x \equiv y \pmod 7$ and $x \not \equiv y \pmod 7 $:</p>
<p>First case:
$$f(x)=f(y) \iff x+\mod(x,7)=y+ \mod (y,7)\implies x= y
$$</p>
<p>But I couldn't prove the second case.</p>
| Asinomás | 33,907 | <p>Take two different numbers $7a+b$ and $7c+d$ with $0\leq b,d<7$. notice $f(7a+b)=7a+2b$ and $f(7c+d)=7c+2d$. The difference is therefore $7(a-b)+2(d-c)$, if this number was zero then $2(d-c)$ would have to be a multiple of $7$, the only possibility is $0$, of course this means $b=d$, but if $b=d$ then $f(7a+b)=7a+2b$ and $f(7c+d)=7c+2b$, which are different.</p>
|
3,490,329 | <blockquote>
<p>Show that a 2-dimensional subspace of the space of <span class="math-container">$2\times2$</span> matrices contains a non-zero symmetric matrix. </p>
</blockquote>
<p>I don't know if it should be written like the addition of two symmetric and skew-symmetric matrix or there is another way to show it. </p>
| JMoravitz | 179,297 | <p>Note that the space of <span class="math-container">$2\times 2$</span> matrices is <span class="math-container">$4$</span>-dimensional.</p>
<p>Note further that the subspace of symmetric <span class="math-container">$2\times 2$</span> matrices is <span class="math-container">$3$</span>-dimensional</p>
<hr>
<p>Now... suppose otherwise that your <span class="math-container">$2$</span>-dimensional space does not contain any non-zero symmetric matrices. It follows then that by taking a basis for your subspace and extending it by taking a basis for the subspace of symmetric matrices, that you will have five linearly independent matrices. (<em>Make sure you understand why</em>)</p>
<p>This is, however, a contradiction since it is impossible to find a number of linearly independent vectors from a vector space than the dimension of that space.</p>
<hr>
<p>In general, if you have a vector space <span class="math-container">$V$</span> of dimension <span class="math-container">$n$</span> and two subspaces <span class="math-container">$A,B$</span> of dimensions <span class="math-container">$a,b$</span> respectively such that <span class="math-container">$a+b>n$</span> then you will necessarily have <span class="math-container">$A\cap B\supsetneq \{0\}$</span> and have some nontrivial intersection between the spaces.</p>
|
4,109,827 | <p><span class="math-container">$$f(x,y)=\begin{cases}\dfrac{y^3}{x^2+y^2} &(x,y) \neq \ \mathbb{(0,0)}\\ 0 & (x,y)=(0,0) \\ \end{cases}$$</span></p>
<p>Evaluate <span class="math-container">$f_x(0,0)$</span> and <span class="math-container">$f_y(0,0)$</span> and <span class="math-container">$D_\overrightarrow{u}f(0,0)$</span></p>
<p>I tried directly taking the derivative to no avail (obviously) so then I tried to use the definition of partial derivative which also left me without a correct solution. Also I have proved that it is continuous (Sertoz Theorem) but how would I prove that it is also differentiable at <span class="math-container">$(0,0)$</span>?</p>
| M. Strochyk | 40,362 | <p>Using polar coordinates
<span class="math-container">$$\begin{gather}x = \rho\cos{\varphi}, \\ y= \rho\sin{\varphi}.\end{gather}$$</span>
we have
<span class="math-container">$${\frac{f(x,\,y) - f(0,\,0)}{\sqrt{x^2+y^2}}} ={\frac{f(\rho\cos{\varphi},\,\rho\sin{\varphi}) - 0}{\rho}} = {\frac{\rho^3\sin^3{\varphi}}{\rho^2\rho}} = \sin^3 \varphi$$</span></p>
<p>Since this can take arbitrary values when <span class="math-container">$\varphi$</span> changes, it means that the limit do not exists and so <span class="math-container">$Df(0,\,0)$</span> do not exists, i.e <span class="math-container">$f$</span> not differentiable in <span class="math-container">$(0,0)$</span>.</p>
|
1,829,030 | <p>The limit isn't too bad using l'hospital's rule, but I was wondering if there was a way to do it without l'hospital's. </p>
<p>Looking around the section limits without lhopital's, it seems usually evaluating without requires some clever factoring, while here the $\arctan$ seems to muck things up. </p>
<p>Here is the evaluation using l'hospital's in case someone visits with this question:
$$L=\lim_{x\rightarrow 0}(1+\arctan(\frac{x}{2}))^{\frac{2}{x}}\Rightarrow \log(L)=\lim_{x\rightarrow 0}\frac{2}{x}\log(1+\arctan(\frac{x}{2}))\\
\Rightarrow \log(L)=2\lim_{x\rightarrow 0}\frac{\log(1+\arctan(\frac{x}{2}))}{x}$$
and by l'hospital's
$$\log(L)=\lim_{x\rightarrow 0}\frac{1}{1+\arctan(\frac{x}{2})}*\frac{1}{1+(\frac{x}{2})^2}=1\\
\Rightarrow L=e$$</p>
| Pierpaolo Vivo | 302,446 | <p>Rewrite as
$$
\lim_{x\rightarrow 0}\exp\left[\frac{2}{x}\ln\left(1+\arctan(\frac{x}{2})\right)\right]
$$
and then use $\arctan(x/2)\sim x/2$ for $x\to 0$, and $\ln (1+x/2)\sim x/2$ for $x\to 0$. Therefore the limit is $=\exp[(2/x)(x/2)]=\mathrm{e}$.</p>
|
1,829,030 | <p>The limit isn't too bad using l'hospital's rule, but I was wondering if there was a way to do it without l'hospital's. </p>
<p>Looking around the section limits without lhopital's, it seems usually evaluating without requires some clever factoring, while here the $\arctan$ seems to muck things up. </p>
<p>Here is the evaluation using l'hospital's in case someone visits with this question:
$$L=\lim_{x\rightarrow 0}(1+\arctan(\frac{x}{2}))^{\frac{2}{x}}\Rightarrow \log(L)=\lim_{x\rightarrow 0}\frac{2}{x}\log(1+\arctan(\frac{x}{2}))\\
\Rightarrow \log(L)=2\lim_{x\rightarrow 0}\frac{\log(1+\arctan(\frac{x}{2}))}{x}$$
and by l'hospital's
$$\log(L)=\lim_{x\rightarrow 0}\frac{1}{1+\arctan(\frac{x}{2})}*\frac{1}{1+(\frac{x}{2})^2}=1\\
\Rightarrow L=e$$</p>
| haqnatural | 247,767 | <p>$$\lim _{ x\rightarrow 0 } \left( 1+\arctan \left( \frac { x }{ 2 } \right) \right) ^{ \frac { 2 }{ x } }=\lim _{ x\rightarrow 0 }{ \left[ \left( 1+\arctan \left( \frac { x }{ 2 } \right) \right) ^{ \frac { 1 }{ \arctan \left( \frac { x }{ 2 } \right) } } \right] } ^{ \frac { 2 }{ x } \arctan \left( \frac { x }{ 2 } \right) }=\\ =e^{ \lim _{ x\rightarrow 0 } \frac { 2 }{ x } \arctan \left( \frac { x }{ 2 } \right) }=e$$
beacuse it is well known that :$$\lim _{ x\rightarrow 0 } \frac { 2 }{ x } \arctan =\lim _{ x\rightarrow 0 } \frac { \arctan \left( \frac { x }{ 2 } \right) }{ \frac { x }{ 2 } } =1$$</p>
|
1,829,030 | <p>The limit isn't too bad using l'hospital's rule, but I was wondering if there was a way to do it without l'hospital's. </p>
<p>Looking around the section limits without lhopital's, it seems usually evaluating without requires some clever factoring, while here the $\arctan$ seems to muck things up. </p>
<p>Here is the evaluation using l'hospital's in case someone visits with this question:
$$L=\lim_{x\rightarrow 0}(1+\arctan(\frac{x}{2}))^{\frac{2}{x}}\Rightarrow \log(L)=\lim_{x\rightarrow 0}\frac{2}{x}\log(1+\arctan(\frac{x}{2}))\\
\Rightarrow \log(L)=2\lim_{x\rightarrow 0}\frac{\log(1+\arctan(\frac{x}{2}))}{x}$$
and by l'hospital's
$$\log(L)=\lim_{x\rightarrow 0}\frac{1}{1+\arctan(\frac{x}{2})}*\frac{1}{1+(\frac{x}{2})^2}=1\\
\Rightarrow L=e$$</p>
| Claude Leibovici | 82,404 | <p>From Taylor series, you can get more than just the limit. Consider $$y=\left(1+\arctan\left(\frac{x}{2}\right)\right)^{\frac{2}{x}}$$ and, as usual for this kind of problem, take logarithms $$\log(y)=\frac{2}{x}\,\log\left(1+\arctan(\frac{x}{2}) \right)$$ Now, use the classical $$\arctan(y)=y-\frac{y^3}{3}+O\left(y^4\right)$$ Replace $y$ by $\frac{x}{2}$ which makes $$\arctan(\frac{x}{2})=\frac{x}{2}-\frac{x^3}{24}+O\left(x^4\right)$$ $$\log(y)=\frac{2}{x}\,\log\left(1+\frac{x}{2}-\frac{x^3}{24}+O\left(x^4\right)\right)$$ Using again the expansion of $\log(1+y)$, we end with $$\log(y)=\frac{2}{x}\times\left(\frac{x}{2}-\frac{x^2}{8}+O\left(x^4\right)\right)=1-\frac{x}{4}+O\left(x^3\right)$$ Now, using $y=e^{\log(y)}$ and Taylor again $$y=e\left(1-\frac{x}{4}+\frac{x^2}{32}\right)+O\left(x^3\right)$$</p>
<p>If, for fun, you plot the original function and the last approximation for $0\leq x\leq 1$, you should notice that they almost coincide. So, suppose that you need to solve for $x$ equation $$\left(1+\arctan\left(\frac{x}{2}\right)\right)^{\frac{2}{x}}=2$$ solving the quadratic will give an estimate $$4-4 \sqrt{\frac{4}{e}-1}\approx 1.25331$$ while the exact solution is $\approx 1.33627$.</p>
|
2,135,191 | <p>In the sequence $a_{1}, a_{2}, a_{3}, ..., a_{100}$, the $k$th term is defined by $$a_{k} = \frac{1}{k} - \frac{1}{k+1}$$ for all integers $k$ from $1$ through $100$. What is the sum of $100$ terms of this sequence? </p>
<p>The answer given is $\frac{100}{101}$, but I am not sure how.</p>
<p>So far I am have plugged in the values of $k's$ and have the following values $$\frac{1}{2}, \frac{1}{6}, \frac{1}{12}, \frac{1}{12}, \frac{1}{20}, \frac{1}{30},....$$</p>
<p>The numerator makes sense to me as it's just $1$ and $100 \times 1 = 100$, however, I am not sure about the denominator. </p>
| Community | -1 | <p>Such a sum is called <strong>telescop sum</strong>. We have
$$\sum_{k=1}^{100} \left( \frac{1}{k}- \frac{1}{k+1} \right) \stackrel{\star}{=}1 - \frac{1}{101}.$$
In $(\star)$ we split the sum into two and make an index shift.</p>
|
2,849,643 | <p>Consider the following recurrence problem:
\begin{align}
d_{i-1} &= 2\varphi_{i+1}+4\varphi_i + 8d_i-7d_{i+1} - F \left( \delta_{i,N} + \delta_{i,N+1} \right) \, , \\
\varphi_{i-1} &= -7\varphi_{i+1}-16\varphi_{i} + 24 \left( d_{i+1}-d_{i} \right) + F \left( \delta_{i,N} + \delta_{i,N+1} \right) \, ,
\end{align}
where $d_i, i \in \{2, \cdots , N+1\}$ represent displacements, $\varphi_i$ inclinations, and $F$ is a known force acting at the nodes $N$ and $N+1$.</p>
<p>We require by the system symmetry that $d_{N+1}=d_N = d_\mathrm{C}$ and $\varphi_N = -\varphi_{N+1}= \varphi_\mathrm{C}$, where $d_\mathrm{C}$ and $\varphi_\mathrm{C}$ are still to be determined from the boundary conditions:</p>
<p>$d_1 = 0$ (zero displacement) and $2\varphi_1+\varphi_2 = 3d_2$ (zero torque)</p>
<p>In order to proceed, i have tried to first determine $d_{N-1}$ and $\varphi_{N-1}$ from the system above, and then $d_{N-2}$ and $\varphi_{N-2}$, etc... in a recursive way and then try to find out the general term of the resulting sequences. </p>
<p>For the term $N-1$, we obtain
\begin{align}
d_{N-1} &= d_\mathrm{C}+2\varphi_\mathrm{C}-F \, \\
\varphi_{N-1} &= -9\varphi_\mathrm{C}+3F \, .
\end{align}</p>
<p>Analogously, we get for the term $N-2$
\begin{align}
d_{N-2} &= d_\mathrm{C}-18\varphi_\mathrm{C}+4F \, \\
\varphi_{N-2} &= 89\varphi_\mathrm{C}-24F \, .
\end{align}</p>
<p>I was wondering whether there is a particular way to figure out the general term of such sequence.</p>
<p>Any help or suggestions are most welcome.</p>
| Empy2 | 81,790 | <p>Except for i=N and i=N+1, you can write the recursion for a 4-vector
$$\vec{v}_i=(d_i,d_{i-1},\phi_i,\phi_{i-1})^t$$
and a 4x4 matrix $A$. Write $d_i$ and $\phi_i$ in terms of the eigenvalues and eigenvectors of $A$, then deal with the boundary conditions at $0,N$ and $N+1$</p>
|
2,849,643 | <p>Consider the following recurrence problem:
\begin{align}
d_{i-1} &= 2\varphi_{i+1}+4\varphi_i + 8d_i-7d_{i+1} - F \left( \delta_{i,N} + \delta_{i,N+1} \right) \, , \\
\varphi_{i-1} &= -7\varphi_{i+1}-16\varphi_{i} + 24 \left( d_{i+1}-d_{i} \right) + F \left( \delta_{i,N} + \delta_{i,N+1} \right) \, ,
\end{align}
where $d_i, i \in \{2, \cdots , N+1\}$ represent displacements, $\varphi_i$ inclinations, and $F$ is a known force acting at the nodes $N$ and $N+1$.</p>
<p>We require by the system symmetry that $d_{N+1}=d_N = d_\mathrm{C}$ and $\varphi_N = -\varphi_{N+1}= \varphi_\mathrm{C}$, where $d_\mathrm{C}$ and $\varphi_\mathrm{C}$ are still to be determined from the boundary conditions:</p>
<p>$d_1 = 0$ (zero displacement) and $2\varphi_1+\varphi_2 = 3d_2$ (zero torque)</p>
<p>In order to proceed, i have tried to first determine $d_{N-1}$ and $\varphi_{N-1}$ from the system above, and then $d_{N-2}$ and $\varphi_{N-2}$, etc... in a recursive way and then try to find out the general term of the resulting sequences. </p>
<p>For the term $N-1$, we obtain
\begin{align}
d_{N-1} &= d_\mathrm{C}+2\varphi_\mathrm{C}-F \, \\
\varphi_{N-1} &= -9\varphi_\mathrm{C}+3F \, .
\end{align}</p>
<p>Analogously, we get for the term $N-2$
\begin{align}
d_{N-2} &= d_\mathrm{C}-18\varphi_\mathrm{C}+4F \, \\
\varphi_{N-2} &= 89\varphi_\mathrm{C}-24F \, .
\end{align}</p>
<p>I was wondering whether there is a particular way to figure out the general term of such sequence.</p>
<p>Any help or suggestions are most welcome.</p>
| Staufenberg | 140,351 | <p>Thanks again to Yuri for the helpful insights. I can say that I have now found the solution to my problem.</p>
<p>First of all, let us define $D_i = d_i-d_{i-1}$, in the same way as Yuri did, and let us get a recurrence equation that involves $\varphi_i$ only. </p>
<p>Actually, original problem (which is equivalent to the one stated above) was
\begin{align}
d_{i+1}-2d_i+d_{i-1} - \frac{1}{4} \left( \varphi_{i+1}-\varphi_{i-1}\right) + F \left( \delta_{i, N} + \delta_{i, N+1} \right) &=0\, , \\
d_{i+1}-d_{i-1} - \frac{1}{3} \left( \varphi_{i+1}+4\varphi_i+\varphi_{i-1} \right) &= 0 \, .
\end{align}</p>
<p>Accordingly, it can easily be shown that for $0 <i <N$,
\begin{align}
D_{i+1}-D_i &= \frac{1}{4} \left( \varphi_{i+1}-\varphi_{i-1} \right) \, , \\
D_{i+1}+D_i &= \frac{1}{3} \left( \varphi_{i+1}+4\varphi_i+\varphi_{i-1} \right) \, .
\end{align}</p>
<p>This leads to
\begin{align}
D_i = \frac{7}{24} \varphi_i + \frac{1}{24} \varphi_{i-2} + \frac{2}{3} \varphi_{i-1} =
\frac{1}{24} \varphi_{i+1}+\frac{7}{24} \varphi_{i-1} + \frac{2}{3} \varphi_i \, , \tag{1} \label{1}
\end{align}
or
$$
\frac{1}{24} \left( \varphi_{i+1}-\varphi_{i-2} \right)
+\frac{3}{8} \left( \varphi_i-\varphi_{i-1} \right) =0\, .
$$</p>
<p>Using Yuri's powerful approach, and posing $\varphi_i = A/p^i$ yields
$$
p^3+9p^2-9p-1=0 \, ,
$$
whose solutions are $p=1$ and $p=-5\pm 2 \sqrt{6}$.</p>
<p>Finally,
$$
\varphi_i = C_1 + C_2 \left(-5-2\sqrt{6}\right)^i + C_3 \left(-5+2\sqrt{6}\right)^i \, .
$$
(again, by noting that $\left(-5-2\sqrt{6}\right)\left(-5+2\sqrt{6}\right)=1$.)</p>
<p>Therefore,
$$
d_i
=D_i + D_{i-1} + \cdots + D_2 =
\sum_{n=2}^{n=i} D_n \, ,
$$
where the expression of $D_n$ follows readily from Eq. \eqref{1}, and using the fact that $d_1=0.$</p>
<p>The final expression read
$$
D_i = C_1 +
C_2 \left( -1+\frac{\sqrt{6}}{2} \right) \left( -5-2\sqrt{6} \right)^i +
C_3 \left( -1-\frac{\sqrt{6}}{2} \right) \left( -5+2\sqrt{6} \right)^i \, ,
$$
and
$$
d_i = (i-1)C_1 +
C_2 \left( -1+\frac{5\sqrt{6}}{12} \right)
\left[ 49+20\sqrt{6}-\left( -5-2\sqrt{6} \right)^{i+1} \right] +
C_3 \left( 1+\frac{5\sqrt{6}}{12} \right)
\left[ -49+20\sqrt{6}+\left( -5+2\sqrt{6} \right)^{i+1} \right] \, .
$$</p>
<p>Clearly, $d_1=0$.
Finally, at this point, the 3 unknown coefficients can readily be determined from the 3 boundary conditions.</p>
<blockquote>
<p>I have checked that the solution is very well behaved and is in full agreement with the numerical solution. </p>
</blockquote>
|
92,382 | <p>I was working on a little problem and came up with a nice little equality which I am not sure if it is well-known (or) easy to prove (It might end up to be a very trivial one!). I am curious about other ways to prove the equality and hence I thought I would ask here to see if anybody knows any or can think of any. I shall hold off from posting my own answer for a couple of days to invite different possible solutions.</p>
<blockquote>
<p>Consider the sequence of functions:
$$
\begin{align}
g_{n+2}(x) & = g_{n}(x) - \left \lfloor \frac{g_n(x)}{g_{n+1}(x)} \right \rfloor g_{n+1}(x)
\end{align}
$$
where $x \in [0,1]$ and $g_0(x) = 1, g_1(x) = x$. Then the claim is:
$$x = \sum_{n=0}^{\infty} \left \lfloor \frac{g_n}{g_{n+1}} \right \rfloor g_{n+1}^2$$</p>
</blockquote>
| Srivatsan | 13,425 | <p>I'll let you decide if it's trivial :-): </p>
<p>$$
\begin{align*}
\sum_{n=0}^{\infty} \left \lfloor \frac{g_n}{g_{n+1}} \right \rfloor g_{n+1}^2
&= \sum_{n=0}^{\infty} \left( \left \lfloor \frac{g_n}{g_{n+1}} \right \rfloor g_{n+1} \right) \cdot g_{n+1}
\\&= \sum_{n=0}^{\infty} \left( g_{n} - g_{n+2} \right) \cdot g_{n+1}
\\&= \sum_{n=0}^{\infty} \left(g_{n} g_{n+1} - g_{n+1} g_{n+2} \right)
\\&= g_0 g_1 = x,
\end{align*}
$$
by a telescopic cancelation.</p>
<hr>
<p><em>Technical note: Convergence.</em> The above manipulations are valid only after we check the convergence of the infinite series. Note that
$$
g_{n+2} = g_{n+1} \cdot \left \{ \frac{g_{n}}{g_{n+1}} \right\} ,
$$
where $\{ \cdot \}$ denotes fractional part. Hence, inductively we see that the sequence $(g_n)$ is nonnegative and monotone decreasing; therefore it converges. We claim that $g_n$ in fact converges to $0$. Towards a contradiction, assume that $\lim\limits_{n \to \infty} \ g_n > 0$. Then for large enough $n$, we have $1 < \frac{g_n}{g_{n+1}} < \frac{3}{2}$, and hence $ \left\{ \frac{g_n}{g_{n+1}} \right\} < \frac{1}{2}$. Therefore $g_{n+2}\lt \frac{g_{n+1}}{2}$, which is a contradiction to the previous sentence. :) Thus $g_n$ converges to $0$.</p>
<p>Finally, for any $N$, we have
$$
\left| g_0 g_1 - \sum_{n=0}^{N} (g_n g_{n+1} - g_{n+1} g_{n+2}) \right| = g_{N+1} g_{N+2} \to 0.
$$
That is, the series $\sum \limits_{n=0}^{\infty} \left(g_{n} g_{n+1} - g_{n+1} g_{n+2} \right)$ converges to $g_0 g_1$ as desired. </p>
|
3,373,528 | <p>Supposing <span class="math-container">$b > 0$</span> and <span class="math-container">$a < b$</span>, how could I prove:</p>
<p><span class="math-container">$$ \frac{a}{b} < \frac{a+1}{b+1} $$</span></p>
| Bman72 | 119,527 | <p>It's equivalent to
<span class="math-container">$$ab + {\color{red}a} < ab +{\color{red}b}$$</span></p>
|
3,373,528 | <p>Supposing <span class="math-container">$b > 0$</span> and <span class="math-container">$a < b$</span>, how could I prove:</p>
<p><span class="math-container">$$ \frac{a}{b} < \frac{a+1}{b+1} $$</span></p>
| Mike | 544,150 | <p>Hint: </p>
<p><span class="math-container">$\dfrac{a}{b} = \dfrac{a(b+1)}{b(b+1)} = \dfrac{ab+a}{b(b+1)}$</span>. However, <span class="math-container">$\dfrac{a+1}{b+1} = \dfrac{(a+1)(b)}{(b)(b+1)} = \dfrac{ab+b}{b(b+1)}$</span>. </p>
<p>If <span class="math-container">$a<b$</span> then what about <span class="math-container">$ab+a$</span> vs <span class="math-container">$ab+b$</span>?</p>
|
708,604 | <blockquote>
<p>Let $f: \mathbb{R^4} \to \mathbb{R}$ be a linear transformation defined by $f(a,b,c,d)=a+b+c+d$. Find a basis for the $Im(f)$.</p>
</blockquote>
<p>So, $Im(f)=\{f(a,b,c,d) \in \mathbb{R}: (a,b,c,d) \in \mathbb{R^4} \}$.</p>
<p>Then $Im(f)=\{a+b+c+d \in \mathbb{R}: a,b,c,d \in \mathbb{R} \}=\mathbb{R}$.</p>
<p>I know that $\mathbb{R}$ is the set of the real numbers on the real line. That real line can be any axis of the plane, space or other higher dimension.</p>
<p>So, its correct to say that a basis for $\mathbb{R}$ can be any single vector of a canonical basis of any $n$ dimensional space?</p>
<p>For instance, can $\{(1,0,0)\}$ be a basis for $\mathbb{R}$? Or $\{(0,1,0)\}$? Thanks</p>
| Martin Argerami | 22,857 | <p>It is simpler than that. $\mathbb R$ is a one-dimensional vector space. So any nonzero "vector" will be a basis. For example, $\{1\}$ is a basis for the image of $f$. So are $\{-2\}$, $\{\pi/7\}$ and $\{102.35\}$.</p>
|
129,293 | <p>I'm writing a survey that involves Levy processes and wanted to mention the different forms of the Levy-Khintchine formula found in literature.</p>
<p>The most common version seems to give the Levy symbol as</p>
<p>$$\Psi(u) = i\langle b,u \rangle - \frac{1}{2} \langle u,\Sigma u\rangle + \int_{\mathbb{R}^d} {(} e^{i\langle u,y \rangle}-1 - i\langle u,y \rangle\mathbf{1}_{|y|\le1}{)}\, dK(y)$$ </p>
<p>while in other versions it seems to be given as</p>
<p>$$\Psi(u) = i\langle b,u \rangle - \frac{1}{2} \langle u,\Sigma u\rangle + \int_{\mathbb{R}^d} {(} e^{i\langle u,y \rangle}-1 - \frac{ i\langle u,y \rangle}{1+|y|^2}{)} \, dK(y)$$</p>
<p>while at <a href="http://almostsure.wordpress.com/2010/09/15/processes-with-independent-increments/" rel="nofollow">almostsure blog</a> it is given as</p>
<p>$$\Psi(u) = i\langle b,u \rangle - \frac{1}{2} \langle u,\Sigma u\rangle + \int_{\mathbb{R}^d}{(} e^{i\langle u,y \rangle}-1 - \frac{ i\langle u,y \rangle}{1+|y|}{)} \, dK(y).$$</p>
<p>Are all of these correct and equivalent? If the last one is, does anyone know a published source I could cite that mentions it?</p>
| Tarasenya | 27,003 | <p>I consider this book: V. V. Petrov, Sums of independent random variables might be useful.</p>
|
4,177,639 | <p>I have an object with known coordinates in in 3D but on the ground (<code>z=0</code>). The object has a direction vector. My goal is to move this object on the ground (so <code>z</code> stays <code>0</code>) using its direction vector and via randomly-generated velocity vectors with one condition: I want to ensure that the generate velocity vector can only move the object within a "valid arc", defined in degrees with respect to the direction vector of the object. More specifically, the way I determine valid range is by ensuring that the new position is within 45 degrees of the old object's direction vector (-45 degrees to the left and +45 degrees to the right). Can someone write a pseudocode on how I can achieve this?</p>
<p>Here's my attempt to do this but this doesn't seem to be the correct way to help me achieve what I want:</p>
<pre><code>object_dir = object_position # the direction could be the same as the object's coortinates
while True:
vel_vec=[uniform(-max_vel, max_vel), uniform(-max_vel, max_vel)] # generate a random velocity vector
new_pos = object_dir + vel_vec # compute a new position (and/or object direction vector) for the object
if (compute_angle(new_pos, object_dir) < 45 or compute_angle(new_pos, object_dir) > 315):
break
</code></pre>
| Community | -1 | <p>In general, there is no analytical solution.</p>
<p>If <span class="math-container">$$b=a^{m/n}$$</span> where <span class="math-container">$m,n$</span> are naturals, the equation can be written as</p>
<p><span class="math-container">$$z^n+z^m=1$$</span> where <span class="math-container">$z:=a^{x/m}$</span>. There are explicit formulas for the cases <span class="math-container">$m,n\le4$</span>, and specific numerical methods for polynomials of higher degree. But in the latter case and when the exponent is irrational, direct resolution with an equation solver (Newton) is simpler.</p>
<hr />
<p>Another way to look at the problem is by setting <span class="math-container">$t:=a^x$</span> so that</p>
<p><span class="math-container">$$t^{\log b/\log a}+t=1$$</span> or <span class="math-container">$$t^\alpha+t=1.$$</span></p>
<p>From this</p>
<p><span class="math-container">$$\alpha=\frac{\log(1-t)}{\log(t)}=:f(t)$$</span> gives you an explicit relation between <span class="math-container">$t$</span> and <span class="math-container">$\alpha$</span>.</p>
<p>What you are after is the inverse, <span class="math-container">$t=f^{-1}(\alpha)$</span> and you can some how precompute/tabulate this inverse, by interpolation or other means.</p>
<p>We have the pretty good approximation</p>
<p><span class="math-container">$$f^{-1}(\alpha)=\frac{(\alpha-1)(\sqrt{0.25\alpha})^\alpha+1}{\alpha(\sqrt{0.25\alpha})^{\alpha-1}+1}$$</span> for <span class="math-container">$\alpha<1$</span>.</p>
|
510,080 | <p>This is not too obvious to me - what is the size of alternating group?</p>
<p>Following the hint in the comment, should it be $A_n = S_n/2$?</p>
<p>So I don't feel right up to here.....</p>
| Michael Joyce | 17,673 | <p>The key is to use the existence of the sign homomorphism $\text{sgn} : S_n \rightarrow \{ \pm 1 \}$. By definition $A_n$ is the kernel of $\text{sgn}$. Since $\text{sgn}$ is surjective, it follows immediately that $[S_n : A_n ] = 2$, so $|A_n| = |S_n| / 2 = n! / 2$.</p>
<p>Edit: As noted by Jared, one must assume that $n \geq 2$.</p>
|
3,371,922 | <p>The definition of the limit states that limit of <span class="math-container">$f(x)$</span> when <span class="math-container">$x$</span> approaches <span class="math-container">$c$</span> is <span class="math-container">$L$</span> iff for every <span class="math-container">$\epsilon > 0$</span> there exists <span class="math-container">$\delta > 0$</span> such that <span class="math-container">$|f(x) - L | < \epsilon$</span> and <span class="math-container">$0 < |x - c| < δ )$</span>.</p>
<p>This states that <span class="math-container">$f(x)$</span> can reach <span class="math-container">$L ( L- ε < f(x) < L + \epsilon)$</span>, while <span class="math-container">$x$</span> cannot reach <span class="math-container">$c ( 0 < |x-c|)$</span>.</p>
<p>The informal definition says that limit means the value the function approaches as the input approaches some value. (They use the same word.) Why can one reach its corespondent value (<span class="math-container">$L$</span>) while the other can't (<span class="math-container">$x$</span> to equal <span class="math-container">$c$</span>)? Why f(x) can equal L , and x can't equal c. What is the intuitive answer to this question?</p>
| user | 505,767 | <p>Indeed by the definition of limit </p>
<p><span class="math-container">$$\lim_{x\rightarrow c} f(x) = L \iff (\forall \varepsilon >0\, \exists \delta > 0: \forall x\in D\quad \color{red}{0<\vert x-c\vert <\delta} \implies \vert f(x)-L\vert <\varepsilon $$</span></p>
<p>it suffices that <span class="math-container">$f(x)$</span> approches <span class="math-container">$L$</span> as <span class="math-container">$x$</span> approaches <span class="math-container">$c$</span>, as for example</p>
<p><span class="math-container">$$\lim_{x\to 0} \frac{\sin x}{x}=1$$</span></p>
<p>for which we can define any value for <span class="math-container">$f(0)$</span> or also assume <span class="math-container">$f(x)$</span> not defined at <span class="math-container">$x=0$</span>.</p>
|
2,578,444 | <blockquote>
<p><span class="math-container">$\tan x> -\sqrt 3$</span></p>
</blockquote>
<p>How do I solve this inequality?</p>
<p>From the <a href="https://www.desmos.com/calculator/qb8bg1vbsf" rel="nofollow noreferrer">graph</a> it is evident that <span class="math-container">$\tan x>-\sqrt 3$</span> for <span class="math-container">$\left(\dfrac{2\pi}3 , \dfrac{5\pi} 3\right)$</span> <span class="math-container">$\forall x\in (0, 2\pi)$</span>.</p>
<p>Generalising this solution we get <span class="math-container">$\left(2n\pi +\dfrac{2\pi}3 , 2n\pi+\dfrac{5\pi} 3\right) \forall n \in \mathbb{Z}$</span> as the answer.</p>
<p>But the answer given is: <span class="math-container">$\left(n\pi - \dfrac \pi 3, n\pi + \dfrac \pi 2\right)$</span></p>
<p>Where have I gone wrong?</p>
| ℋolo | 471,959 | <p>First a side note: when you are working with graphs like this you should always work with the closest points to $0$, it will make your life easier</p>
<hr>
<p>So $\tan x>-\sqrt 3$, to find the values of $x$ first let's find where those 2 are equal, you will get $-\frac{\pi}3$(this is the closest point to $0$, ofc there are infinity many points like this).</p>
<p>Now $tan x$ is increasing function hence we found the lower bound, now we search for the first point where $tan x<-\sqrt 3$ <strong>and</strong> $x>-\frac{\pi}3$, but if $\tan x$ is strictly increasing we need to check only after a point it doesn't exists, the first point that doesn't exist and $x>-\frac{\pi}3$ is $\frac{\pi}2$, where after that point we are at $-\infty$, so this point have the 2 conditions that we look for. Hence we have the solution $(-\frac{\pi}3,\frac{\pi}2)$. Now $\tan x$ is periodic for $\pi$ so we get: $$x\in(-\frac{\pi}3+n\pi,\frac{\pi}2+n\pi)$$</p>
|
87,437 | <p>Let <span class="math-container">$R$</span> be a rectangular region of the integer lattice <span class="math-container">$\mathbb{Z}^2$</span>,
each of whose unit squares is labeled with a number
in <span class="math-container">$\lbrace 1, 2, 3, 4, 5, 6 \rbrace$</span>.
Say that such a labeled <span class="math-container">$R$</span> is <em>die-rolling Hamiltonian</em>,
or simply <em>rollable</em>,
if there is a Hamiltonian cycle obtained by rolling a unit die
cube over its edges so that, for each square <span class="math-container">$s \in R$</span>,
the cube lands on <span class="math-container">$s$</span> precisely once, and when it does so,
the top face of the cube matches the number in <span class="math-container">$s$</span>.
For example, the <span class="math-container">$4 \times 4$</span> "board" shown below
is rollable.
<br />
<img src="https://i.stack.imgur.com/6aR0a.jpg" alt="Dice Rolling" />
<br /></p>
<blockquote>
<p><b>Q</b>. Is it true that, if <span class="math-container">$R$</span> is die-rolling Hamiltonian, then the
Hamiltonian cycle is unique, i.e., there are never two distinct
die-rolling Hamiltonian cycles on <span class="math-container">$R$</span>?</p>
</blockquote>
<p>This "unique-rollability"
question arose out of a problem I posed in 2005, and was largely
solved two years later, in a paper entitled,
"On rolling cube puzzles" (complete citation below;
the <span class="math-container">$4 \times 4$</span> example above is from Fig. 17 of that paper).
Although the original question involved computational complexity,
the possible uniqueness of Hamiltonian cycles is independent
of those computational issues, so I thought it might be useful
to expose it to a different community, who might bring
different tools to bear.
It is known to hold for <span class="math-container">$R$</span> with side lengths at most 8.
If not every cell of <span class="math-container">$R$</span> is labeled, and unlabeled cells are forbidden
to the die, then there are examples with more than one Hamiltonian cycle.</p>
<p><b>Edit1</b>. Rolling a regular tetrahedron on the equilateral triangular (hexagonal) lattice
is not as interesting. See the Trigg article cited below.</p>
<p><b>Edit2</b>.
Serendipitously, <em>gordon-royle</em> posted a perhaps(?) relevantly related
question:
"<a href="https://mathoverflow.net/questions/87496/">Uniquely Hamiltonian graphs with minimum degree 4</a>."</p>
<hr />
<ul>
<li> The computational version is
<a href="http://cs.smith.edu/~jorourke/TOPP/P68.html#Problem.68" rel="nofollow noreferrer">
Open Problem 68</a> at
<a href="http://cs.smith.edu/~jorourke/TOPP/" rel="nofollow noreferrer">The Open Problems Project</a>.
</li>
<li>
"On rolling cube puzzles."
Buchin, Buchin, Demaine, Demaine, El-Khechen, Fekete, Knauer, Schulz, Taslakian.
<em>Proceedings of the 19th Canadian Conference on Computational Geometry</em>, Pages 141–144, 2007.
<a href="http://people.csail.mit.edu/schulz/papers/RollingFull.pdf" rel="nofollow noreferrer">PDF download.</a>
</li>
<li>
Charles W. Trigg. "Tetrahedron rolled onto a plane." <em>J. Recreational Mathematics</em>, 3(2):82–87, 1970.
</li>
</ul>
| domotorp | 955 | <p>UPDATE: I played around and came up with a construction (chance of containing a mistake is high!), below it I leave my original answer for explanation.</p>
<p>$\begin{array}{ccccccccccccccccccccccccc}
3&-&2& &2&-&1&-&5&-&6& &5&-&1&-&2&-&6&-&5&-&1&-&2\cr
|& &|& &|& & & & & &|& &|& & & & & & & & & & & &|\cr
1& &1& &3& &4&-&5&-&3& &4& &1&-&4&-&6&-&3&-&1&-&4\cr
|& &|& &|& &|& & & & & &|& &|& & & & & & & & & & \cr
4& &5& &5& &1&-&5&-&6&-&2& &5&-&4&-&2&-&3&-&5&-&4\cr
|& &|& &|& & & & & & & & & & & & & & & & & & & &|\cr
6& &6&-&4& &6&-&3&-&1&-&4&-&6&-&3&-&1&-&4&-&6& &6\cr
|& & & & & &|& & & & & & & & & & & & & & & &|& &|\cr
3& &2&-&4&-&5& &4&-&2& &5&-&6&-&2& &2&-&4& &5& &3\cr
|& &|& & & & & &|& &|& &|& & & &|& &|& &|& &|& &|\cr
1& &1& &1&-&5&-&6& &6& &3&-&6& &3&-&1& &1& &1& &1\cr
|& &|& &|& & & & & &|& & & &|& & & & & &|& &|& &|\cr
4& &5& &3& &2&-&3& &5&-&3&-&2& &3&-&5& &3& &2& &4\cr
|& &|& &|& &|& &|& & & & & & & &|& &|& &|& &|& &|\cr
6& &6& &6& &6& &6&-&5&-&1&-&2&-&6& &6& &6& &6& &6\cr
|& &|& &|& &|& & & & & & & & & & & &|& &|& &|& &|\cr
3&-&2&-&4&-&5&-&3&-&2&-&4&-&5&-&3&-&2&-&4&-&5&-&3\cr
|& & & & & & & &|& & & & & & & &|& & & & & & & &|\cr
1&-&2&-&6&-&5&-&1&-&2&-&6&-&5&-&1&-&2&-&6&-&5&-&1\cr
|& &|& &|& &|& & & & & & & & & & & &|& &|& &|& &|\cr
4& &4& &4& &4& &4&-&5&-&3&-&2&-&4& &4& &4& &4& &4\cr
|& &|& &|& &|& &|& & & & & & & &|& &|& &|& &|& &|\cr
6& &5& &1& &2&-&1& &5&-&1&-&2& &1&-&5& &1& &2& &6\cr
|& &|& &|& & & & & &|& & & &|& & & & & &|& &|& &|\cr
3& &3& &3&-&5&-&4& &4& &1&-&4& &1&-&3& &3& &3& &3\cr
|& &|& & & & & &|& &|& &|& & & &|& &|& &|& &|& &|\cr
1& &2&-&6&-&5& &6&-&2& &5&-&4&-&2& &2&-&6& &5& &1\cr
|& & & & & &|& & & & & & & & & & & & & & & &|& &|\cr
4& &4&-&6& &4&-&1&-&3&-&6&-&4&-&1&-&3&-&6&-&4& &4\cr
|& &|& &|& & & & & & & & & & & & & & & & & & & &|\cr
6& &5& &4& &3&-&5&-&4&-&2& &5&-&6&-&2&-&1&-&5&-&6\cr
|& &|& &|& &|& & & & & &|& &|& & & & & & & & & & \cr
3& &3& &1& &6&-&5&-&1& &6& &3&-&6&-&4&-&1&-&3&-&6\cr
|& &|& &|& & & & & &|& &|& & & & & & & & & & & &|\cr
1&-&2& &2&-&3&-&5&-&4& &5&-&3&-&2&-&4&-&5&-&3&-&2
\end{array}$</p>
<p>HOW THIS THING WORKS:</p>
<p>I think there are configurations that are not uniquely Hamiltonian. The example I have in mind should be around 14 times 14, except that I do not have a definite example, but I hope I can convince you that only a technical difficulty is missing.</p>
<p>Our goal is to prove a somewhat stronger statement, to exhibit an R that has two different die-rolling H-cycles in which the cube is in the same position over every field no matter which H-cycle you take. This allows us to define a nice graph on R. First we will give one H-cycle, then add the edges not contained in this H-cycle along which the cube could move, only allowing moves that take the cube into the same position over the field as it would have in the H-cycle. Denote the obtained graph by G. It is easy to see that G is a subgraph of the grid-graph, moreover, G cannot have cycles whose length is less than 10 and G has (at least) two H-cycles.</p>
<p>One such graph is given below on the 13 times 10 grid (130 vertices). Legend: X marks the squares contained in both H-cycles, while 1 and resp. 2 the squares contained in only one of them.</p>
<p>$\begin{array}{ccccccccccccc} %
X&X&X&X&X&X&X&X&X&X&X&X&X\cr
X& & & & & & & & & & & &X\cr
X& &X&X&X&X&X&X&X&X&X& &X\cr
X& &X& & & & & & & &X& &X\cr
1&1&1&1&X&X&X&X&X&2&2&2&2\cr
X& &X& & & & & & & &X& &X\cr
X& &X&X&X&X&X&X&X&X&X& &X\cr
X& & & & & & & & & & & &X\cr
X&X&X&X&X&X&X&X&X&X&X&X&X\
\end{array}
$</p>
<p>Unfortunately this graph is not yet good enough, we cannot give a good numbering of R to make the edges valid. However, we can play around (i.e. make more wiggly using more area) with the top and bottom parts (shared by both H-cycles) and I am sure that way we can ensure the validity of all edges. This is the only missing part which seems to be only technical.</p>
|
4,278,505 | <p>I would like to clear up a confusion which might be trivial. In a proof the author proved <span class="math-container">$T = T'$</span> as following:</p>
<p>The author showed if <span class="math-container">$x \in T$</span> then <span class="math-container">$x \in T'$</span>, the next line is -</p>
<blockquote>
<p>... proving that <span class="math-container">$T\subseteq T'$</span>. As a result <span class="math-container">$T = T'$</span>.</p>
</blockquote>
<p>Reasoning:</p>
<p>If for any <span class="math-container">$x$</span>, it exists in both set <span class="math-container">$T$</span> and <span class="math-container">$T'$</span>, then we can directly write <span class="math-container">$T = T'$</span>, and if <span class="math-container">$T = T'$</span> by definition we can say <span class="math-container">$T \subset T'$</span>.</p>
<p>Full Context:
<a href="https://i.stack.imgur.com/eAuN2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eAuN2.jpg" alt="enter image description here" /></a></p>
<p>Confusion:</p>
<p>Then why do author write first <span class="math-container">$T\subseteq T'$</span> then <span class="math-container">$T = T'$</span> (notice the sequence)?</p>
<p><span class="math-container">$T\subseteq T'$</span> means either <span class="math-container">$T\subset T'$</span> or <span class="math-container">$T= T'$</span>, it looks like at this point author does not know what should the case exactly but then based what he writes <span class="math-container">$T =T'$</span> in the next line?</p>
<p>If <span class="math-container">$T=T'$</span> is written for the reasoning given above then <span class="math-container">$T\subseteq T'$</span> is unnecessary, why do author write <span class="math-container">$T\subseteq T'$</span>?</p>
<p>What is the significance?</p>
<p>Another example is that often I found that we can say directly that <span class="math-container">$H_{i+1}= \rm Ker \; g$</span> but author(s) used <span class="math-container">$H_{i+1}\subseteq \rm Ker \; g$</span> (<a href="https://math.stackexchange.com/q/4256432/850314">see Question 1 in this post</a>), obviously <span class="math-container">$H_{i+1}= \rm Ker \; g$</span> does not contradict <span class="math-container">$H_{i+1}\subset \rm Ker \; g$</span> but why do author go for the weaker statement?</p>
| Michael | 179,940 | <p>When we prove all the elements of <span class="math-container">$T$</span> is also elements of <span class="math-container">$T'$</span>, it has two meanings:</p>
<ol>
<li><p><span class="math-container">$T=T'$</span> (note it does not contradict <span class="math-container">$T\subset T'$</span>),</p>
</li>
<li><p><span class="math-container">$T \subset T'$</span>,</p>
</li>
</ol>
<p>these two meanings can be written in a single line as-
<span class="math-container">$$T \subseteq T'$$</span></p>
<p>You constructed <span class="math-container">$T'$</span> by taking elements <span class="math-container">$(I+ E_{ik}), (I + E_{kj})$</span> from <span class="math-container">$T$</span> and multiplying those elements (to construct each element of the <span class="math-container">$T'$</span>) in the following way:</p>
<p><span class="math-container">$$(I+ E_{ik})(I + E_{kj}) (I + E_{ik})^{-1}(I + E_{kj})^{-1}$$</span></p>
<p>Above type of multiplication to construct each element of the <span class="math-container">$T'$</span> makes <span class="math-container">$T'$</span> a commutator subgroup, by definition. Since, commutator subgroup <span class="math-container">$T'$</span> is constructed by multiplying elements of <span class="math-container">$T$</span> in a specific way</p>
<p>(i.e. <span class="math-container">$(I+ E_{ik})(I + E_{kj}) (I + E_{ik})^{-1}(I + E_{kj})^{-1}$</span>), so <span class="math-container">$T'$</span> is the commutator subgroup of <span class="math-container">$T$</span>, i.e. <span class="math-container">$T' \subset T$</span>.</p>
<p>We have showed <span class="math-container">$T' \subset T$</span> and all the elements of <span class="math-container">$T$</span> are also elements of <span class="math-container">$T'$</span>, so we take the first meaning or possibility of <span class="math-container">$T \subseteq T'$</span>, thus, <span class="math-container">$T'=T$</span>.</p>
|
1,185,108 | <p>empty set is an subset of any sets maybe any collection of sets.</p>
<p>I wonder what about the case of the empty set being a member,not subset, of any collection (family) of sets.</p>
| Gal Porat | 221,699 | <p>This not true in general. For example, the empty set is a collection of sets that does not contain the empty set (because it does not contain any members).</p>
|
661,771 | <p>I am stuck on the following problem that says : </p>
<blockquote>
<p>Which of the following is a solution to the differential equation $y'=|y|^{\frac12},y(0)=0\,$ where square root means the
positive square root ? </p>
<ol>
<li><p>$y(t)=\frac{t^2}{4}$ </p></li>
<li><p>$y(t)=-\frac{t^2}{4}$ </p></li>
<li><p>$y(t)=\frac{t|t|}{4}$ </p></li>
<li><p>$y(t)=-\frac{t|t|}{4}$ </p></li>
</ol>
</blockquote>
<p>MY ATTEMPT: Taking $y>0,$ I get from the differential equation $y'=|y|^{\frac12},y(0)=0 \implies 2\sqrt y=x $. Now, after
looking at the options ,I am not sure which of the aforementioned options is correct . Can someone help? </p>
| Community | -1 | <p>Your proof consists of some correct steps done in the wrong order, which makes it something other than a valid proof. It looks more like scratchwork done in preparation for a proof. I rewrite it below, with some of the more important additions in bold. I will also change <span class="math-container">$t$</span> to <span class="math-container">$y$</span> throughout; when you wrote "<span class="math-container">$y$</span>" you probably meant the same thing as "<span class="math-container">$t$</span>".</p>
<hr />
<p>Suppose <span class="math-container">$D(x_0, r)$</span> is a closed ball. We show that <span class="math-container">$X\setminus D(x_0,r) $</span> is open. In other words, <strong>for every point <span class="math-container">$y\in X\setminus D(x_0,r)$</span></strong> we need to find an open ball contained in <span class="math-container">$X \setminus D$</span> with center <span class="math-container">$y$</span>.</p>
<p>Since <span class="math-container">$y \in X\setminus D(x_0,r)$</span>, it follows that <span class="math-container">$d(y,x_0) > r$</span>, so <span class="math-container">$d(y,x_0) - r > 0 $</span>. <strong>Let <span class="math-container">$r_1 = d(y,x_0)-r$</span>.</strong></p>
<p>I claim that the open ball <span class="math-container">$B(y, r_1)$</span> is contained in <span class="math-container">$X\setminus D(x_0,r)$</span>. To prove this, consider any <span class="math-container">$z \in B(y,r_1)$</span>. Notice by the triangle inequality</p>
<p><span class="math-container">$$ d(x_0,y) \leq d(x_0,z) + d(z,y) \implies d(z,x_0) \geq d(x_0,y) - d(z,y) > d(x_0,y) - r_1 =r.$$</span>
This shows <span class="math-container">$z\in X\setminus D(x_0,r)$</span>, which completes the proof.</p>
|
787,926 | <p>I need some help to solve this integral:</p>
<p>$$\int_0^1 dy\int_0^{1-y} \cos \left(\frac{x-y}{x+y} \right) \mathrm dx$$</p>
<p>Thank you.</p>
| Julián Aguirre | 4,791 | <p>Let $u_n\ne1$ be the $n$-th 5-smooth number. At least one of
$$
\frac32\,u_n, \frac43\,u_n, \frac65\,u_n
$$
is a 5-smooth number greater than $u_n$. Let $U_n$ be the smallest of them. Then
$$
u_n<u_{n+1}\le U_n.
$$
Check the numbers $u_n+1,u_n+2,\dots$ until you find a 5-smooth number. The algorithm can be made faster with the following tricks:</p>
<ol>
<li>Get a smaller $U_n$ by defining it as the smallest 5-smooth number among
$$
\frac32\,u_n, \frac43\,u_n, \frac54\,u_n, \frac65\,u_n, \frac98\,u_n,\frac{16}{15}\,u_n,\frac{25}{24}\,u_n,\frac{27}{25}\,u_n,\frac{81}{80}\,u_n,\dots
$$
Each fraction is the quotient of two coprime consecutive 5-smooth numbers.</li>
<li>To check if a number is 5-smooth, do not factorize. Just check if its only prime factors are $2$, $3$ and $5$. This can be done efficiently by repeated division.</li>
</ol>
<p>I have coded this in Mathematica and got that
$$
u_{1000}=51200000=2^{14}5^5.
$$</p>
|
427,564 | <p>I'm supposed to give a 30 minutes math lecture tomorrow at my 3-grade daughter's class. Can you give me some ideas of mathemathical puzzles, riddles, facts etc. that would interest kids at this age?</p>
<p>I'll go first - Gauss' formula for the sum of an arithmetic sequence.</p>
| john | 79,781 | <p>The <a href="http://en.wikipedia.org/wiki/Seven_Bridges_of_K%C3%B6nigsberg">Seven Bridges of Königsberg</a> is a nice one. From memory, third-grade is around the time when kids like to try those problems of "can you draw a house without raising your pen and only drawing each line once", etc. which is essentially what this problem is. The solution, I believe, is simple enough for them to understand.</p>
|
427,564 | <p>I'm supposed to give a 30 minutes math lecture tomorrow at my 3-grade daughter's class. Can you give me some ideas of mathemathical puzzles, riddles, facts etc. that would interest kids at this age?</p>
<p>I'll go first - Gauss' formula for the sum of an arithmetic sequence.</p>
| Little Endian | 67,892 | <p>When I was in second grade, we did an experiment where we tossed a pin n times and counted the number of times the pin fell across some line, given that it fell between two other lines. Years later I learned to compute this - but the idea was there. My vote is for doing something related to probability and statistics.</p>
<p>Another idea is to do something related to computer science. Find some simple algorithm and show how the time (number of steps) to complete changes as the the data set grows. One variation of this is to have the students themselves sort cards (make it a contest), time them for different size unorganized stacks and record their results. Then ask them about their process.</p>
<p>Whatever you do, don't go above their heads by doing something to mathy. These examples assume they are smart, but don't assume they know anything.</p>
|
427,564 | <p>I'm supposed to give a 30 minutes math lecture tomorrow at my 3-grade daughter's class. Can you give me some ideas of mathemathical puzzles, riddles, facts etc. that would interest kids at this age?</p>
<p>I'll go first - Gauss' formula for the sum of an arithmetic sequence.</p>
| Jacob Raccuia | 67,366 | <p>I don't know if it's too advanced but...</p>
<p>I'm 21 years old and I just recently learned about 'using your hands to find the multiples of 9'.</p>
<p>For those who don't know what I'm talking about, hold your two hands out in front of you.</p>
<p>Say we want to find 9 x 8. Starting from your left hand pinky, count off 8 fingers and put your 8th finger down. You now have seven fingers on your left hand side and two on your right. Put them together. Voila! 72.</p>
<p>Try again with 9 x 4. You now have three on your left hand side and six on the right!</p>
<p>While writing this, I just realized that this is taking one minus X and combining it with 10 - X. </p>
<p>9 * X = (X - 1) and (10 - X). It's fascinating me! </p>
|
427,564 | <p>I'm supposed to give a 30 minutes math lecture tomorrow at my 3-grade daughter's class. Can you give me some ideas of mathemathical puzzles, riddles, facts etc. that would interest kids at this age?</p>
<p>I'll go first - Gauss' formula for the sum of an arithmetic sequence.</p>
| Brian Rushton | 51,970 | <p>Knot theory can be fun for third graders; the trefoil knit can be formed with one person and a stick, and some knits can be formed with 2 people, so you could show a picture of a knot, and have them try and make it.</p>
<p>Or better yet, a lot of people play a game where you have a lot of people randomly grab hands and try to make a circle by unwinding. You could play this several times, and have people suspect that you always get the unknot. Then you can have them form a trefoil, and when they can't undo it, discuss why they can't.</p>
|
1,722,995 | <blockquote>
<blockquote>
<p>Question: Given the circle $x^2+y^2=25$ is inscribed in triangle $\triangle ABC$, where vertex $B$ lies on the first quadrant. Slope of $AB$ is $\sqrt 3$ and has a positive y-coordinate, and $|AB|=|AC|$. Find the equations for $AC$ and $BC$</p>
</blockquote>
</blockquote>
<p>I found out the equation for the straight line passing through $AB$: Let the line be $y=\sqrt 3 x+c$. Then</p>
<p>$3x^2+2\sqrt 3 cx+c^2+x^2=25$</p>
<p>$\Delta =0$ (discriminant)</p>
<p>$(2\sqrt 3 c)^2 - 4(4)(c^2-25)=0$</p>
<p>$c=10$</p>
<hr>
<p>However, I don't see any simple way to find out the equations of line for $AC$ and $BC$. While it seems like there is enough information, I have tried using similar triangles, etc, but I can't find out the coordinates of the vertices. Can anyone give me some hints? Thank you!</p>
| amd | 265,466 | <p>I found it fairly straightforward to describe these lines with vector equations. </p>
<p>The line $AB$ is given as having a slope of $\sqrt3$, so it can immediately be parametrized as $\vec P_{AB}=\vec A+t\,(1,\sqrt3)$. </p>
<p>The line $AC$ is the reflection of this line in the line $AO$. A parametrization for this line can be found by reflecting the above parametrization of $AB$. Doing so yields $$\vec P_{AC}=\vec A+t\,\left[2{\vec A\cdot(1,\sqrt3)\over\vec A\cdot\vec A}\vec A-(1,\sqrt3)\right].$$ </p>
<p>Finally, since the triangle is isosceles, the line $BC$ is perpendicular to $AO$ and tangent to the circle at the intersection of $AO$ with the circle. This suggests the equation $$\vec P_{BC}\cdot\vec A = \left(-5\frac{\vec A}{\|\vec A\|}\right)\cdot\vec A = -5\|\vec A\|.$$ </p>
<p>There’s not enough information to determine a unique value for $A$. It must be farther from the $y$-axis than the point at which $AB$ intersects the circle, $\left(-\frac52\sqrt3,\frac52\right)$, because there’s only one tangent to the circle there and moving any closer takes $B$ out of the first quadrant. On the other hand, when $B$ is on the $y$-axis, the triangle is equilateral, putting $A$ at $(-5\sqrt3,-5)$, so $A$ must be closer to the $y$-axis than this. Combining these constraints gives $$A=\frac52[(-\sqrt3,1)-s\,(\sqrt3,3)]$$ for $0<s<1$. I’ll leave it to you to grind through the algebra and convert these equations into the required forms. </p>
<p>chenbai’s suggestion of using an angle as the parameter seems like a less messy way to go, though. We still derive the equations for $AC$ and $BC$ in the same way. With a slope of $\sqrt3$, $AB$ makes an angle of $\frac\pi3$ with the positive $x$-axis. $AC$ is the reflection of $AB$ with respect to $AO$, so we just need to find the angle of this line, which results in the equation $${y-y_A\over x-x_A}=\tan\left(2\alpha - \frac\pi3\right)$$ for $AC$. We need to express the coordinates of $A$ as a function of $\alpha$, but that’s just a matter of writing the equation for $AB$ in polar form. The bounds for $\alpha$ can be found by making the same considerations as in the vector-based solution above. </p>
<p>Similarly, $BC$ is orthogonal to $AO$, so its equation will be $${y+5\sin\alpha\over x+5\cos\alpha}=\tan\left(\alpha+\frac\pi2\right).$$ The plusses on the left side are because we want the opposite side of the circle from $A$.</p>
|
2,469,798 | <p>Let $S = \left\{x \in \mathbb{Q} \mid 1 \leqslant {x}^2 \leqslant 29 \right\}$</p>
<p>What can we say about the supremum and infimum of this set? Would it be non-existent?</p>
<p>Would it be correct to say the following?</p>
<p>Suppose $ \sup S < \sqrt{29} $ then $ \exists x \in S $ such that $ x > \sup S$ </p>
<p>Therfore $ \sqrt{29} \leqslant \sup S$</p>
<p>But on the other hand $\sup S > \sqrt{29}$ then $\exists x \in \mathbb{Q}$ such that $ x < \sup S$</p>
<p>Therefore $\sup S \leqslant \sqrt{29}$</p>
<p>Therefore $\sup S $ can only be equal to $\sqrt{29}$</p>
<p>However $\sqrt{29} \notin \mathbb{Q}$ therefore $\sqrt{29} \notin S$</p>
<p>Therefore there can not be a supremum to this set.</p>
<p>(I am assuming the supremum of a set must belong to the set. Is this correct?)</p>
| fleablood | 280,126 | <p>The supremum, or least upper bound, of a set need not be elements of the set (and for open sets they <em>never</em> are).</p>
<p>Example. Let $T = (0,1)$. Then $\sup T = 1 \not \in S$. That's it.</p>
<p>So in your set, $\sup S = \sqrt{29} \not \in S$. That's it.</p>
<p>....</p>
<p>Okay, but listen up, this is <em>important</em>. If your universal space is $\mathbb Q$ or some other space. Then $\sqrt{29}$ will not exist in your <em>space</em>. So we would say in the space of the rationals $S$ does not have a least upper bound. </p>
<p>We say $\mathbb Q$ does not have the "least upper bound property" because it is possible for there to exist sets that are bounded above but that do not have a supremum. However if your <em>space</em> is a <em>different</em> set of points and is such a space where <em>every</em> set that is bounded above <em>must</em> have a least upper bound. Then we say that space has the "least upper bound property".</p>
<p>Now the <em>entire</em> point of the real numbers and the entire reason for the existence of irrational number is that $\mathbb Q$ does <em>not</em> have the least upper bound property, but $\mathbb R \supset \mathbb Q$ is an extension of $\mathbb Q$ that <em>DOES</em> have the least upper bound property.</p>
<p><strong>And the irrational numbers are nothing more or less than the least upper bounds of bounded sets of rational numbers that do not have a rational least upper bound.</strong></p>
<p>So your set $S$ does not have a rational least upper bound. So it <em>must</em> have an <em>irrational</em> least upper bound. And it does. It is $\sqrt{29}$.</p>
<p>And every irrational number (indeed every real number) is the least upper bound of some set of rational numbers.</p>
<p>.....</p>
<p>Indeed we can actually say: Let $x \in \mathbb R$ and let $S =\{q\in \mathbb Q| q<x\}$; then $\sup S = x$. </p>
<p>Period. That is always true.</p>
|
3,041,907 | <p>I am unable to isolate the variable <span class="math-container">$x$</span> of this inequality <span class="math-container">$y \leq \sqrt{2x-x^2}$</span> ( where <span class="math-container">$0 \leq y \leq 1 $</span>)</p>
<p>Is it correct doing this: <span class="math-container">$y^2 \leq 2x-x^2$</span>?
I found that <span class="math-container">$y^2 \leq x \leq 2-y^2$</span> and <span class="math-container">$0 \leq x \leq 2$</span>. Is it correct?</p>
<p>From here I am not sure how to proceed.
Thanks in advance for any help.</p>
| Bernard | 202,857 | <p>First, you have to determine the domain of validity of this inequation:
<span class="math-container">$$x^2-2x\ge 0\iff x(2-x)\ge 0\iff x\in[0,2].$$</span>
Next</p>
<ul>
<li>if <span class="math-container">$y\le 0$</span>, the inequation is satisfied for all <span class="math-container">$x\in[0,2]$</span>,</li>
<li>if <span class="math-container">$y>0$</span>, the inequation is equivalent (on <span class="math-container">$[0,2]$</span>) to <span class="math-container">$y^2\le 2x-x^2$</span>, i.e. to
<span class="math-container">$$p_y(x)=x^2-2x+y^2\le 0.$$</span>
As a quadratic inequation in <span class="math-container">$x$</span>, it is satisfied if and only if it has real roots, <span class="math-container">$\xi_0\le\xi_1$</span> and <span class="math-container">$\;x\in[0,2]\cap[\xi_0,\xi_1]$</span>. It has real roots if and only if its reduced discriminant <span class="math-container">$\;\Delta'=1-y^2\ge 0$</span>, i.e. <span class="math-container">$0<y\le1$</span> in the present case. So we need to place the numbers <span class="math-container">$\;0,2,\xi_0=1-\sqrt{1-y^2},\xi_1=1+\sqrt{1-y^2},2\;$</span> w.r.t. each other.</li>
</ul>
<p>Now <span class="math-container">$p_y(0)=p_y(2)=y^2\ge 0$</span>, so neither <span class="math-container">$0$</span> nor <span class="math-container">$2$</span> separates the roots. Furthermore,by Vieta's relations, <span class="math-container">$\dfrac{\xi_0+\xi_1}2=1$</span>, so that <span class="math-container">$\;0\le\xi_0 <1<\xi_1 \le 2\;$</span>, and finally the solutions are
<span class="math-container">$$[\xi_0,\xi_1]\quad \text{if }\;0<y\le 1.$$</span></p>
|
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