qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
2,799,439 | <blockquote>
<p>Prove that if $p$ is a prime in $\Bbb Z$ that can be written in the form $a^2+b^2$ then $a+bi$ is irreducible in $\Bbb Z[i]$ .</p>
</blockquote>
<p>Let $a+bi=(c+di)(e+fi)\implies a-bi=(c-di)(e-fi)\implies a^2+b^2=(c^2+d^2)(e^2+f^2)\implies p|(c^2+d^2)(e^2+f^2)\implies p|c^2+d^2 $ or $p|e^2+f^2$
since $p$ is a prime.</p>
<p>How to show $e+fi $ or $c+di$ is a unit from here?</p>
| Jonathan Dunay | 538,622 | <p>Use different divides relations that you get from $a^2+b^2=(c^2+d^2)(e^2+f^2)$. You get that $c^2+d^2|p$ and $e^2+f^2|p$ which implies that one of $e+fi$ or $c+di$ them is a unit.</p>
|
3,485,441 | <p>I don't quite understand why Burnside's lemma
<span class="math-container">$$
|X/G|=\frac1{|G|}\sum_{g\in G} |X_g|
$$</span>
should be called a "lemma". By "lemma", we should mean there is something coming after it, presumably a theorem. However, I could not find a theorem which requires Burnside as a lemma. In every book I read, the author jumps into calculations using Burnside rather than further theorems.</p>
<p>Question: What are some important consequences of Burnside Lemma, and why is it called a "lemma"?</p>
| User7238 | 736,765 | <p>Suppose G acts on H where <span class="math-container">$|G|,|H|<\infty$</span>. For each <span class="math-container">$g\in G$</span>, consider the permutation character associated with the action. That is, <span class="math-container">$\chi(g)=|\{\alpha\in H:\alpha\cdot g=\alpha\}|$</span>.<br>
Then, <span class="math-container">$\sum_{g\in G}\chi(g)=\sum_{\alpha\in H}|G_{\alpha}|=n|G|$</span>, where n is the number of orbits of G on H (the result of Burside's Lemma).</p>
<p>Also, we have the Fundamental Counting Principle: if <span class="math-container">$O_{\alpha}$</span> is the orbit of <span class="math-container">$\alpha$</span>, then <span class="math-container">$|O_{\alpha}|=|G:G_{\alpha}|$</span>, and since G is finite, <span class="math-container">$|O_{\alpha}|$</span> divides <span class="math-container">$|G|$</span>. Then, as Chris stated, from here we get the class equation. </p>
<p>The number of orbits pops up all over Group Theory, so don't think of Burside's Lemma as a lemma that should have a theorem immediately following, but rather a "tool" that we can refer to very frequently for many theorems. </p>
|
1,748,001 | <p>I need to find a relation between $\sqrt{x+ia}$ and $\sqrt{\sqrt{x^2+a^2}+x}$</p>
<p>where $a>0$ $x\in \mathbb{R}$</p>
<p>Thank you</p>
| Ross Millikan | 1,827 | <p>When you multiply terms, you add exponents, so $x \cdot x^{\frac 23}=x^{\frac 53}$ They multiplied top and bottom of the first by $x$ to put the two over a common denominator.</p>
|
1,748,001 | <p>I need to find a relation between $\sqrt{x+ia}$ and $\sqrt{\sqrt{x^2+a^2}+x}$</p>
<p>where $a>0$ $x\in \mathbb{R}$</p>
<p>Thank you</p>
| ervx | 325,617 | <p>Recall that $x^{a}\cdot x^{b}=x^{a+b}$. Thus, to simplify your sum of fractions, we bring both of the original fractions to a common denominator using the aforementioned rule.</p>
<p>Note that $3x^{\frac{2}{3}}\cdot x=3x^{\frac{2}{3}+1}=3x^{\frac{5}{3}}$. Thus, </p>
<p>$$
\frac{1}{3x^{\frac{2}{3}}}+\frac{2}{3x^{\frac{5}{3}}}=\frac{1\cdot x}{3x^{\frac{2}{3}}\cdot x}+\frac{2}{3x^{\frac{5}{3}}}=\frac{x}{3x^{\frac{5}{3}}}+\frac{2}{3x^{\frac{5}{3}}}=\frac{x+2}{3x^{\frac{5}{3}}}.
$$</p>
|
1,251,914 | <p>I do not understand how to set up the following problem:</p>
<p>"Forces of 20 lb and 32 lb make an angle of 52 degrees with each other. find the magnitude of the resultant force."</p>
<p>An actually picture would really help.</p>
| Lee Mosher | 26,501 | <p>Multidimensional spaces occur naturally all around us. Take the possible positions of your arm, for example. </p>
<ol>
<li>You can rotate your shoulder with two degrees of freedom;</li>
<li>You can bend your elbow with one degree of freedom;</li>
<li>You can rotate your wrist with two degrees of freedom</li>
</ol>
<p>That's a total of five degrees of freedom for the possible positions of your arm.</p>
<p>In other words, the space of positions of your arm, a space with which you are personally familiar on an intimate level, is a 5-dimensional object.</p>
<p>And I'm ignoring the fact that twisting your forearm and flexing your hand bones and bending your finger bones adds at least another twenty or more dimensions.</p>
|
1,251,914 | <p>I do not understand how to set up the following problem:</p>
<p>"Forces of 20 lb and 32 lb make an angle of 52 degrees with each other. find the magnitude of the resultant force."</p>
<p>An actually picture would really help.</p>
| dtldarek | 26,306 | <p>There are text explanations, so I will post some pictures.</p>
<p>We will try to build some intuition by comparing to zero-, one-, two-, three- and four-dimensional tables/arrays.</p>
<p><strong>Introduction:</strong></p>
<p>What's zero-dimensional array you ask? It is a single cell, as single point is a zero-dimensional space.</p>
<p><img src="https://i.stack.imgur.com/Ry64U.png" alt="zero_and_one_dimension" /></p>
<p>Then there is one-dimensional array and it can be split into <span class="math-container">$n$</span> one-less-dimensional strucutres. Similarly a segment can be split into infinitely many points.</p>
<p>When we go into second dimension, we can split in different ways: either rows of columns first. Still, whatever way we do it, a 2D array is still a collection of <span class="math-container">$n$</span> 1D arrays, which can be split up into <span class="math-container">$n^2$</span> two-less-dimensional structures.</p>
<p><img src="https://i.stack.imgur.com/bBk7I.png" alt="two_dimensions" /></p>
<p>This pattern holds in three dimensions, where we can split and split and split our structures. In other words, a 3D array is a collection of <span class="math-container">$n$</span> 2D arrays, or a collection of <span class="math-container">$n^2$</span> 1D arrays, etc.</p>
<p><img src="https://i.stack.imgur.com/Q4y0F.png" alt="three_dimensions" /></p>
<p>That is, we can think of 3D array as 1D array of 2D arrays...</p>
<p><img src="https://i.stack.imgur.com/Vl182.png" alt="vector_of_matrices" /></p>
<p>... or 2D array of 1D arrays, etc.</p>
<p><img src="https://i.stack.imgur.com/SR8K0.png" alt="matrix_of_vectors" /></p>
<p><strong>The fourth dimension:</strong></p>
<p>Now, let's go up one dimension:</p>
<p><img src="https://i.stack.imgur.com/xGWLJ.png" alt="vector of 3D arrays" /></p>
<p>This is a 1D array of 3D array, or (1+3) = 4D structure. It's hard to imagine visually, but you could try scaling the cubes into the inside (and ignore that tiny little voice which says "such overlaps cannot happen in reality"). Another way to think about it is to mentally remove the inside of the cube and leave only the shell (6 faces), then such a thing can be scaled and combined with no overlaps. Trying with wireframes only it looks like this:</p>
<p><img src="https://i.stack.imgur.com/iVb1o.png" alt="scaling cube" /></p>
<p>Or we could move it to the side:</p>
<p><img src="https://i.stack.imgur.com/z2L0a.png" alt="moving cube" /></p>
<p>However, in 4D we can do yet something else, namely 2D array of 2D arrays:</p>
<p><img src="https://i.stack.imgur.com/wJVgr.png" alt="matrix of matrices" /></p>
<p>This is also (2+2) = 4D structure, and you can think of it as a product of two squares (with full insides). That would also be a bit hard to imagine, so to make it simple, take squares with no insides (only the edges). To further boost our intuition, observe that product of two circles gives you a torus. The wireframe for the 4D cube will look a bit alike, although remember that these are two very different objects.</p>
<p><img src="https://i.stack.imgur.com/iz1Gv.png" alt="from torus to hypercube" /></p>
<p>I hope this helps <span class="math-container">$\ddot\smile$</span></p>
|
1,251,914 | <p>I do not understand how to set up the following problem:</p>
<p>"Forces of 20 lb and 32 lb make an angle of 52 degrees with each other. find the magnitude of the resultant force."</p>
<p>An actually picture would really help.</p>
| N. Owad | 85,898 | <p>I am a firm believer in the idea that if something in math doesn't make sense to you, you just haven't found the proper way of thinking about it yet. And in some (or maybe most?) cases, there isn't anybody on earth who has found the right way. And this way is probably different for different people.</p>
<p>Now, to answer your question, I will tell you how I think of higher dimensional objects, as someone who prefers to visualize everything. Some of the other answers describe how from one $(x)$ to two $(x,y)$ dimensions we just "add a perpendicular line" to everything and call something in that new direction $y$. If that works that is fine. It makes sense for the transition from 2 to 3 dimensions also, for most people. When we go from 3 to 4, we have to get creative. Instead of a "adding a new line", I usually think about adding the color spectrum to each point in 3 space. So, at each point, we have $(x,y,z)$ just like in normal 3 space, and lets call it $c$, for color. To draw points in 4 space, now you need to place it in 3 space and also assign some color to it, to tell us where in the spectrum it is. </p>
<p>A line is the same way, but at each point in the line, we can pick a color. If the color varies continuously through the spectrum, the line is connected. But an abrupt change from red to green, for example, is like having a break in the line. In the picture below, the green line lives in 3 space, since it only uses one color, the rainbow line travels through the whole spectrum. Also, notice that the rainbow line in three dimensions crosses itself. But it doesnt actually intersect because it has two different colors.<img src="https://i.stack.imgur.com/1DQQ7.png" alt="A green line that does not intersect, and a Rainbow line intersects itself, but at a blue section and purple section, so it does not actually intersect."></p>
<p>So, this rainbow line lives in 4 dimensions right now, and the green line is only using 3 dimensions. This analogy still has drawbacks though. The three directional unit vectors make sense, but when you try to draw the new color one, you find yourself with "line" of colors, but in the 3 dimesions we are looking at, it is only at the origin, so it looks like a point to us. </p>
<p>One last example, and this might be a stretch: The 3-sphere, which lives in 4 dimensions. First a quick idea of how we make this. Just like in <em>flatland</em>, when the sphere (which in topology we call the 2-sphere) comes to visit, it shows up a a circle (or 1-sphere), picutred below. But it has a lot of circles of different size, depending on where the plane is intersecting the sphere. A second sphere is higher than the first, so it intersects a smaller circle in the plane.
<img src="https://i.stack.imgur.com/cgSyO.png" alt="A sphere intersecting a plane at the equator and one intersecting lower, so the circle is smaller than the equator."></p>
<p>So when the 3-sphere comes to 3 dimensions, we will only see it as a normal sphere (2-sphere). If that was confusing, I apologize. Here are some pictures where we get the normal unit sphere when we are green, which I am equating to $c=0$. but when we change colors, our spheres will get smaller and when the are red and violet, they will only be a point. That is because the color is "farther" from green, which we are thinking of as the 3 space we live in, and we need to satisfy the equation $x^2+y^2+z^2+c^2=1$. Well, I hope this helps.</p>
<p><img src="https://i.stack.imgur.com/jL8K3.png" alt="enter image description here"></p>
<p><img src="https://i.stack.imgur.com/xZHbV.png" alt="enter image description here"></p>
<p><img src="https://i.stack.imgur.com/AVT6j.png" alt="enter image description here"></p>
<p><img src="https://i.stack.imgur.com/erImG.png" alt="enter image description here"></p>
|
114,895 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/21282/show-that-every-n-can-be-written-uniquely-in-the-form-n-ab-with-a-squa">Show that every $n$ can be written uniquely in the form $n = ab$, with $a$ square-free and $b$ a perfect square</a> </p>
</blockquote>
<p>I am trying to prove that for every $n \ge 1$ there exist uniquely determined integers $a \gt 0$ and $b \gt 0$ such that $n = a^2b$ where $b$ is square-free.</p>
<p>The fact that such $a$ and $b$ exist is easy to prove.</p>
<p>From the fundamental theorem of arithmetic, $n$ can be uniquely represented as $p_1^{a_1} p_2^{a_2} \cdots p_s^{a_s}$ where $s$ is a positive integer. Thus</p>
<p>\begin{align*}
n & = \prod_{i=1}^s p_i^{a_i} \\\\
& = \prod_{i=1}^s p_i^{\left(2 \left\lfloor \frac{a_i}{2} \right\rfloor + a_i \bmod{2}\right)} \\\\
& = \prod_{i=1}^s p_i^{\left(2 \left\lfloor \frac{a_i}{2} \right\rfloor\right)} \cdot \prod_{i=1}^s p_i^{a_i \bmod{2}} \\\\
& = \left(\prod_{i=1}^s p_i^{\left\lfloor \frac{a_i}{2} \right\rfloor}\right)^2 \cdot \prod_{i=1}^s p_i^{a_i \bmod{2}}.
\end{align*}</p>
<p>Clearly, $\left(\prod_{i=1}^s p_i^{\left\lfloor \frac{a_i}{2} \right\rfloor}\right)^2$ is a perfect square and $\prod_{i=1}^s p_i^{a_i \bmod{2}}$ is square free. Hence, we have shown that such $a$ and $b$ exist.</p>
<p>Now, how do we show that such a pair of $a$ and $b$ is unique?</p>
<p>I know how to start proving such a theorem. Let us assume that $n = a^2b = a'^2b'$ such that $a' \ne a$ and $b' \ne b$. Now since this is not possible this should lead us to some contradiction. But, I'm unable to reach a contradiction from this assumption. Could you please help me?</p>
| Greg Martin | 16,078 | <p>For a proof using minimal machinery, I prefer:</p>
<p>Consider all representations of $n$ in the form $n=a^2b$, where $a$ and $b$ can be any positive integers. (These can be indexed by the set $A$ of positive integers $a$ such that $a^2\mid n$.) You can show that $A$ consists exactly of all divisors of a special integer $a_{max}\in A$, and then that in the representation $n=a^2b$ where $a\in A$, the corresponding $b$ is squarefree if and only if $a=a_{max}$. (Hint: if $a^2\mid n$ and $c^2\mid n$, then $(lcm[a,c])^2 \mid n$.)</p>
|
2,443,496 | <blockquote>
<p>Can someone point me in the right direction as to how to take the derivative of this function:
$$ f(x) = 2 \pi \sqrt{\frac{x^2}{c}} $$</p>
</blockquote>
<p>Thank you</p>
| valer | 403,899 | <p>$f(x) = 2\pi\sqrt\frac{x^2}{c}$</p>
<p>$f'(x) = 2\pi*\frac{1}{2\sqrt{\frac{x^2}{c}}}*({\frac{x^2}{c}})'=2\pi\frac{\frac{x}{c}}{\sqrt\frac{x^2}{c}}=2\pi*\frac{\sqrt\frac{x^2}{c}}{x}$</p>
|
3,251,754 | <p>Let <span class="math-container">$M$</span> be the set of all <span class="math-container">$m\times n$</span> matrices over real numbers.Which of the following statements is/are true??</p>
<ol>
<li>There exists <span class="math-container">$A\in M_{2\times 5}(\mathbb R)$</span> such that the dimension of the nullspace of <span class="math-container">$A $</span> is <span class="math-container">$2$</span>.</li>
<li>There exists <span class="math-container">$A\in M_{2\times 5}(\mathbb R)$</span> such that the dimension of the nullspace of <span class="math-container">$A $</span> is <span class="math-container">$0$</span>.</li>
<li>There exists <span class="math-container">$A\in M_{2\times 5}(\mathbb R)$</span> and <span class="math-container">$B\in M_{5\times 2}(\mathbb R)$</span> such that <span class="math-container">$AB$</span> is the <span class="math-container">$2\times 2$</span> identity matrix.</li>
<li>There exists <span class="math-container">$A\in M_{2\times 5}(\mathbb R)$</span> whose null space is <span class="math-container">$\{ (p,q,r,s,t)\in \mathbb R^5 | p=q, r=s=t\}$</span>.</li>
</ol>
<p>I am sure about the option <span class="math-container">$3$</span> definitely will not come. But I don't know about others..and then the dimension of the nullspace is <span class="math-container">$3$</span>??</p>
| 5xum | 112,884 | <p>Your are correct, and the proof is rather simple (not requiring the wall of text you wrote :)</p>
<p><span class="math-container">$$\begin{align}P(\neg B|C)&=\frac{P(\neg B \land C)}{P(C)} &\text{by definition}
\\&= \frac{P(C) - P(B\land C)}{P(C)} & \text{Because $B\land C$ and $\neg B\land C$ form a partition of $C$}
\\&=\frac{P(C)}{P(C)}-\frac{P(B\land C)}{P(C)}&\text{Algebraic manipulation}
\\&=1-P(B|C)&\text{by definition}\end{align}$$</span></p>
<hr>
<p><em>Note</em>: I assume here that <span class="math-container">$P(C)>0$</span>, i.e. that <span class="math-container">$C$</span> is not an impossible event. Things can get complicated quickly if we look at a more general solution.</p>
|
21,141 | <p>Is there a way to extract the arguments of a function? Consider the following example:</p>
<p>I have this sum</p>
<pre><code>g[1] + g[2] + g[3] + g[1]*g[3] + 3*g[1]*g[2] + 6*g[1]*g[2]*g[3]
</code></pre>
<p>Now, what I want to do is exctract the function arguments and apply them to another function <code>func</code> which takes the arguments as a list.</p>
<pre><code>func[{1}] + func[{2}] + func[{3}] + func[{1, 3}] + 3*func[{1, 2}] + 6*func[{1, 2, 3}]
</code></pre>
<p>I know there is <code>Extract[g[1]*g[3], Position[g[1]*g[3], _Integer]]</code> but that does not work if I have a multiplicative constant.</p>
<p>Is there a way to do this?</p>
| kglr | 125 | <pre><code>expr = g[1] + g[2] + g[3] + g[1]*g[3] + 3*g[1]*g[2] + 6*g[1]*g[2]*g[3];
expr /. Times[x___:1 , p__g] :> x func[{p}[[All, 1]]]
(* func[{1}] + func[{2}] + func[{3}] + 3 func[{1, 2}] + func[{1, 3}] + 6 func[{1, 2, 3}]*)
</code></pre>
|
3,181,502 | <p>We have <span class="math-container">$\tan(x)=\dfrac{\sin(x)}{\cos(x)}$</span>. I was wondering why <span class="math-container">$\tan(x+{\pi/2})=\tan(x)$</span>?</p>
<p>I wanted to Show </p>
<p><span class="math-container">$$\frac{\sin(x+\pi/2)}{\cos(x+\pi/2)}=\frac{\sin(x)}{\cos(x)}\iff\frac{\sin(x+\pi/2)\cos(x)}{\cos(x+\pi/2)\sin(x)}=1\iff\frac{\cos^2(x)}{-\sin^2(x)}=1$$</span></p>
<p>But this would imply that <span class="math-container">$\cos^2x+\sin^2x=0$</span>. But this is false</p>
<p>On the other hand we have that <span class="math-container">$$\tan(x+{\pi/2})=\tan(x)$$</span></p>
<p>don't we?</p>
| MPW | 113,214 | <p>No.</p>
<p><span class="math-container">$$\tan(x+\pi/2)=\frac{\sin(x+\pi/2)}{\cos(x+\pi/2)}$$</span>
<span class="math-container">$$=\frac{\sin x\cos \pi/2 + \cos x \sin \pi/2}{\cos x \cos \pi/2 - \sin x \sin \pi/2}$$</span>
<span class="math-container">$$=\frac{\cos x}{-\sin x}$$</span>
<span class="math-container">$$=-\cot x$$</span></p>
<p>which is generally not equal to <span class="math-container">$\tan x$</span>.</p>
<p>The period of <span class="math-container">$\tan x$</span> is <span class="math-container">$\pi$</span>. A similar line of reasoning as above shows that
<span class="math-container">$$\tan (x + \pi/2) = \frac{-\sin x}{-\cos x} = \tan x.$$</span></p>
|
2,588,968 | <p>I have the double integral</p>
<p>$$\int^{10}_0 \int^0_{-\sqrt{10y-y^2}} \sqrt{x^2+y^2} \,dx\,dy$$</p>
<p>And I am asked to evaluate this by changing to polar coordinates.</p>
| Community | -1 | <p>Set $x$ equal to the lower bound of your inner integral. This line defines a boundary of the region of integration. Plot it, along with the other boundaries $x=y=0$ and $y=10$ and see if you can express the region more naturally in polar coordinates. Make the necessary transformation and it should become clear how to proceed. </p>
|
3,831,073 | <p>Let <span class="math-container">$\alpha = \sqrt[3]{4+\sqrt{5}}$</span>. I would like to prove that <span class="math-container">$\left[ \mathbb{Q} \left( \alpha \right ) : \mathbb{Q} \right] = 6$</span>. We have <span class="math-container">$\alpha^3 = 4 + \sqrt{5}$</span>, and so <span class="math-container">$(\alpha^3 - 4)^2 = 5$</span>, hence <span class="math-container">$\alpha$</span> is a root of the polynomial <span class="math-container">$f(x)=x^6 - 8 x^3 + 11$</span>.
I tried to prove with various approaches that <span class="math-container">$f(x)$</span> is irreducible over <span class="math-container">$\mathbb{Q}$</span> without success, so I devised the following strategy.</p>
<p>Since <span class="math-container">$x^2 - 5$</span> is irreducible over <span class="math-container">$\mathbb{Q}$</span>, we have <span class="math-container">$\left[ \mathbb{Q} \left( \sqrt{5} \right ) : \mathbb{Q} \right] = 2$</span>. Now from <span class="math-container">$\alpha^3 = 4 + \sqrt{5}$</span> we get <span class="math-container">$\sqrt{5} \in \mathbb{Q} \left( \alpha \right )$</span>, so that <span class="math-container">$\mathbb{Q} \left( \sqrt{5} \right )$</span> is a subfield of <span class="math-container">$\mathbb{Q} \left( \alpha \right )$</span>, <span class="math-container">$\mathbb{Q} \left( \alpha \right )=\mathbb{Q}\left( \sqrt{5}\right) \left( \alpha \right)$</span>, and we have
<span class="math-container">\begin{equation}
\left[ \mathbb{Q} \left( \alpha \right ) : \mathbb{Q} \right] = \left[ \mathbb{Q} \left( \sqrt{5} \right)\left( \alpha \right ) : \mathbb{Q} \left (\sqrt{5} \right) \right] \left[ \mathbb{Q} \left( \sqrt{5} \right ) : \mathbb{Q} \right] .
\end{equation}</span>
Now <span class="math-container">$\alpha$</span> is a root of the polynomial <span class="math-container">$g(x) \in \mathbb{Q} \left (\sqrt{5} \right) [ x ]$</span> given by <span class="math-container">$g(x) = x^3 - 4 - \sqrt{5}$</span>. So to prove our thesis it is enough to prove that this polynomial is irreducible in <span class="math-container">$\mathbb{Q} \left (\sqrt{5} \right) [ x ]$</span>. Being <span class="math-container">$g(x)$</span> of third degree, if it were not irreducible, its factorization would have at least one linear factor, so that <span class="math-container">$g(x)$</span> would have some root in <span class="math-container">$\mathbb{Q} \left (\sqrt{5} \right)$</span>. Hence our problem boils down to show that there are no integers <span class="math-container">$m_0, m_1, n$</span>, with <span class="math-container">$n \neq 0$</span>, such that
<span class="math-container">\begin{equation}
\left( \frac{m_0}{n} + \frac{m_1}{n} \sqrt{5}\right)^3 = 4 + \sqrt{5},
\end{equation}</span>
which gives
<span class="math-container">\begin{equation}
m_0^3 + 5 \sqrt{5} m_1^3 +3 \sqrt{5} m_0^2 m_1 + 15 m_0 m_1^2 = 4 n^3 + \sqrt{5} n^3,
\end{equation}</span>
or
<span class="math-container">\begin{equation}
m_0^3 + 15 m_0 m_1^2 - 4 n^3 + \sqrt{5} \left( 5 m_1^3 +3 m_0^2 m_1 - n^3 \right)=0,
\end{equation}</span>
which implies, being <span class="math-container">$\sqrt{5}$</span> irrational,
<span class="math-container">\begin{cases}
m_0^3 + 15 m_0 m_1^2 - 4 n^3 = 0, \\ 5 m_1^3 +3 m_0^2 m_1 - n^3 = 0.
\end{cases}</span>
At this point I am stuck, because I do not know how to prove that this system admits the only integer solution <span class="math-container">$m_0 = m_1 = n = 0$</span>.</p>
<p>Any help is welcome!</p>
| Hagen von Eitzen | 39,174 | <p>As <span class="math-container">$\Bbb Q(\sqrt 5)$</span> is a subfield, its automorphism <span class="math-container">$\sqrt 5\to-\sqrt 5$</span> can be extended, which means that <span class="math-container">$\sqrt[3]{4-\sqrt 5}\in\Bbb Q(\sqrt[3]{4+\sqrt 5})$</span> and also
<span class="math-container">$$ \sqrt[3]{11}=\sqrt[3]{4-\sqrt 5}\sqrt[3]{4+\sqrt 5}\in\Bbb Q(\sqrt[3]{4+\sqrt 5}).$$</span> Now, <span class="math-container">$[\Bbb Q(\sqrt[3]{11}):\Bbb Q]$</span> is clearly <span class="math-container">$3$</span>. With a subfield of degree 2 and another of degree 3 over <span class="math-container">$\Bbb Q$</span>, ouf field must be <em>at least</em> of degree <span class="math-container">$6$</span>, hence exactly degree <span class="math-container">$6$</span>.</p>
|
2,544,864 | <p>I have been trying to prove the continuity of the function:
$f:\mathbb{R}\to \mathbb{R}, f(x) =x \sin(x) $ using the $\epsilon -\delta$ method. </p>
<p>The particular objective of posting this question is to understand <strong>the dependence of $\delta$ on $\epsilon$ and $x$</strong>. I know that $f(x) =x \sin(x) $ is not uniformly continuous, so $\delta$ depends on both. Here is my attempt:</p>
<p>We need to prove that $\forall \epsilon > 0 \: \exists\, \delta(\epsilon,x) >0$ such that $\lvert x - y \rvert < \delta \implies \lvert x \sin(x) - y \sin(y)\rvert < \epsilon$.</p>
<p>Let $x=2n\pi$ and $y=x-\frac{\delta}{2}$ so that $\lvert x - y \rvert < \delta$. </p>
<p>Then,
\begin{align}
\bigl\lvert x \sin(x) - y \sin(y)\bigr\rvert&=\biggl\lvert 2n\pi \sin(2n\pi) - (2n\pi-\frac{\delta}{2})\sin(2n\pi-\frac{\delta}{2})\biggr\rvert\\
&= \biggl\lvert (2n\pi-\frac{\delta}{2}) \: \sin(2n\pi-\frac{\delta}{2})\biggr\rvert
\end{align}
Now,
\begin{align}
\biggl\lvert (2n\pi-\frac{\delta}{2}) \sin(2n\pi-\frac{\delta}{2})\biggr\rvert \leq \biggl\lvert (2n\pi-\frac{\delta}{2}) \biggr\rvert \leq \epsilon
\end{align}
and hence, a $\delta $ chosen such as $4n\pi + 2\epsilon$ can be used. Since, this choice depends on $4n\pi$ which is $2x$ and $2\epsilon$, hence the function is continuous but not uniformly so.</p>
<p>Is my procedure correct? How can I prove it generally so $\forall x$?</p>
| Christian Blatter | 1,303 | <p>Forget about $2n\pi$ and stuff. The essential point is that $|\sin x-\sin y|\leq |x-y|$ for arbitrary $x$, $y\in{\mathbb R}$.</p>
<p>Let an $x\in{\mathbb R}$ be given and consider an increment $h$ with $|h|\leq1$ applied at $x$. Then
$$\eqalign{\bigl|(x+h)\sin(x+h)-x\sin x\bigr|&=\bigl|(x+h)(\sin(x+h)-\sin x)+h\sin x\bigr|\cr&\leq|h|\bigl(|x+h|+|\sin x|\bigr)\cr&\leq |h|\bigl(|x|+2\bigr)\ .\cr}$$
Given an $\epsilon>0$ we therefore can choose $$\delta:={\epsilon\over|x|+2}$$
in order to satisfy the standard $\epsilon/\delta$-requirement.</p>
<p>But be aware that the claim immediately follows from the continuity of the product $(x,y)\mapsto xy$ on ${\mathbb R}^2$ and standard facts about continuity.</p>
|
521,500 | <p>Today we proofed the (simple) Markov property for the Brownian motion. But I really don't get a crucial step in the proof.
The theorem states in particular that for $s\geq0$ fixed, the process $(C_t:=B_{t+s}-B_{s}, t\geq0)$ is independent from $\mathcal{F}_s=\sigma(B_u, 0\leq u\leq s)$.</p>
<p>The proof starts with the remark, that it suffices to show that $\forall n, 0\leq t_1<t_2\dots<t_n$ and $\forall m, 0\leq u_1<u_2\dots<u_m$ the two vectors $(C_{t_1},\dots,C_{t_n})$ and $(B_{u_1},\dots,B_{u_m})$ are independent. But I just cannot figure out why this is true?</p>
<p>Anyone got some advise? Thanks a lot!</p>
| Davide Giraudo | 9,849 | <p>If $D$ is a countable subset of $[0,s]$ and $S\in\mathcal B(\mathbb R^\infty)$, then $\{\omega,(B_u(\omega))_{u\in D}\in S)$ belongs to $\sigma(B_u,0\leqslant u\leqslant s)$. The collection of sets of this form is a $\sigma$-algebra. </p>
<p>So, after passing from countable intersections to finite ones, it's enough to prove that the process $(C_t,t\geqslant 0)$ is independent of $\{\omega,(B_u(\omega))_{u\in D}\in S)$ with $D$ finite ($|D|=d$) and $S\in\mathcal{B}(\mathbb R^d)$.</p>
<p>By a similar argument, we are reduced to check independence of sets of the form $\{\omega,(C_t(\omega)_{t\in D'}\in S'\}$, $S'\in\mathcal B(\mathbb R^{|D'|})$.</p>
|
9,085 | <p>So as the title says I am trying to make a list where each element is determined by a users choice of an element in a PopupMenu.</p>
<p>My first attempt:</p>
<pre><code>test = Table["A", {5}];
Table[PopupMenu[Dynamic[test[[n]]], {"A", "B", "C"}], {n, 5}]
</code></pre>
<p>Returned the following error</p>
<pre><code>Part::pspec: Part specification n is neither an integer nor a list of integers.
</code></pre>
<p>For some reason the dynamic(?) would not allow me to refer to specific elements in the list. I then tried to circumvent this issue by introducing an extra variable <em>temp</em>:</p>
<pre><code>Table[temp = n;PopupMenu[Dynamic[test[[temp]]], {"A", "B", "C"}], {n, 5}]
</code></pre>
<p>However, all this did was create 5 PopupMenus that all referred to the $5^{\text{th}}$ element of the list <em>test</em>. I tried to put a <code>Setting[]</code> around the <code>Dynamic[]</code>, but since that removes that effect of <code>Dynamic[]</code> nothing happened at all.</p>
<p>Any suggestions would be greatly appreciated.</p>
| b.gates.you.know.what | 134 | <p>This seems to work :</p>
<pre><code>test = Table["A", {5}];
PopupMenu[Dynamic[test[[#]]], {"A", "B", "C"}, "A"] & /@ Range[5] // Row
Dynamic @ test
(* {"C", "A", "B", "A", "C"} *)
</code></pre>
|
1,797,712 | <p>Let $G = \Bbb{Z}_{360} \oplus \Bbb{Z}_{150} \oplus \Bbb{Z}_{75} \oplus \Bbb{Z}_{3}$</p>
<p>a. How many elments of order 5 in $G$</p>
<p>b. How many elments of order 25 in $G$</p>
<p>c. How many elments of order 35 in $G$</p>
<p>d. How many subgroups of order 25 in $G$</p>
<p>I think I have done a,b,c correctly and got 124 elments of order 5, 3000 elements of order 25, and 0 elements from order 35,</p>
<p>But I'm not sure if that is correct, and how to approch d?</p>
| Quang Hoang | 91,708 | <p>Using the presentation theorem of finite abelian groups, one can restrict the problem to the $5$-primary part of $G$, which is
$$H=\mathbb Z_5\times \mathbb Z_{25}\times \mathbb{Z}_{25}.$$</p>
<p>$H$ has $5\times 25\times 25=3125$ elements. There should be $\color{red}{124}$ elements of order $5$, not $24$. </p>
<p>For the number of subgroup of order $25$, look at the two cases where the said subgroup is isomorphic to ether $\mathbb Z_5\times \mathbb Z_5$ or $\mathbb Z_{25}$.</p>
|
1,357,247 | <p>I'm sorry if my question is repeated.</p>
<p>Let $E$ be the set of all $x\in [0,1]$ whose decimal expansion contains only the digits $4$ and $7$. Is $E$ contable? Is $E$ dense in $[0,1]$? Is $E$ compact? Is $E$ perfect?</p>
<p><strong>Proof:</strong> It is easy to see that $E$ is uncountable. </p>
<p>Also $E$ is not dense in $[0,1]$. Because if we take point $2/3\in [0,1]$ then it's neighborhood with radius $1/3$ has an empty intersection with $E$ because minimal element in $E$ is $0,\overline{4}=0,444\cdots=4/9$.</p>
<p>The set $E$ is bounded in $[0,1]$ (also in $\mathbb{R}^1$). Moreover, $E$ is closed in $[0,1]$.</p>
<p><strong>Lemm:</strong> The set $E$ is closed in $[0,1]$.</p>
<p>We'll prove that $E^c$ is open in $[0,1]$. If $z\in E^c$ it means that decimal expansion of $z$ has minimal $j_0$ s.t. $a_{j_0}\notin \{4,7\}$. If $a_{j_0}\in \{1,2\}$ then we can take $\varepsilon=1/{10^{j_0}}$ and $N_{\varepsilon}(z)\subset E^c$ where $N_{\varepsilon}(z)=\{y\in [0,1]: d(y,z)<\varepsilon\}$. If $a_{j_0}\notin \{1,2,4,7\}$ then we can take $\varepsilon=1/{10^{j_0+1}}$ and $N_{\varepsilon(z)}\subset E^c$. It's easy to check these cases. Hence $E^c$ is open set. $\Box$</p>
<p>But $E$ is NOT closed in $\mathbb{R}^1$. We can take $z=0,\overline{9}\in E^c$ and for any $\varepsilon>0$ we have $N_{\varepsilon}(z)\nsubseteq E^c$.</p>
<p>By Heine-Borel theorem we got that $E$ is NOT compact in $\mathbb{R}^1$. But I have one question. Will it be compact in $[0,1]$? After all $[0,1]\subset \mathbb{R}^1$. Can anyone answer this question in detail?</p>
| ajotatxe | 132,456 | <p>$E$ is closed in $[0,1]$ and in $\Bbb R$. The number $z=0,\bar 9$ (which is $1$), is in $E^c$ (<em>to check whether $E$ is closed in $\Bbb R$, you must take the complementary respect to $\Bbb R$</em>) and the interval $(0.9,1.1)$ is contained in $E^c$.</p>
|
1,357,247 | <p>I'm sorry if my question is repeated.</p>
<p>Let $E$ be the set of all $x\in [0,1]$ whose decimal expansion contains only the digits $4$ and $7$. Is $E$ contable? Is $E$ dense in $[0,1]$? Is $E$ compact? Is $E$ perfect?</p>
<p><strong>Proof:</strong> It is easy to see that $E$ is uncountable. </p>
<p>Also $E$ is not dense in $[0,1]$. Because if we take point $2/3\in [0,1]$ then it's neighborhood with radius $1/3$ has an empty intersection with $E$ because minimal element in $E$ is $0,\overline{4}=0,444\cdots=4/9$.</p>
<p>The set $E$ is bounded in $[0,1]$ (also in $\mathbb{R}^1$). Moreover, $E$ is closed in $[0,1]$.</p>
<p><strong>Lemm:</strong> The set $E$ is closed in $[0,1]$.</p>
<p>We'll prove that $E^c$ is open in $[0,1]$. If $z\in E^c$ it means that decimal expansion of $z$ has minimal $j_0$ s.t. $a_{j_0}\notin \{4,7\}$. If $a_{j_0}\in \{1,2\}$ then we can take $\varepsilon=1/{10^{j_0}}$ and $N_{\varepsilon}(z)\subset E^c$ where $N_{\varepsilon}(z)=\{y\in [0,1]: d(y,z)<\varepsilon\}$. If $a_{j_0}\notin \{1,2,4,7\}$ then we can take $\varepsilon=1/{10^{j_0+1}}$ and $N_{\varepsilon(z)}\subset E^c$. It's easy to check these cases. Hence $E^c$ is open set. $\Box$</p>
<p>But $E$ is NOT closed in $\mathbb{R}^1$. We can take $z=0,\overline{9}\in E^c$ and for any $\varepsilon>0$ we have $N_{\varepsilon}(z)\nsubseteq E^c$.</p>
<p>By Heine-Borel theorem we got that $E$ is NOT compact in $\mathbb{R}^1$. But I have one question. Will it be compact in $[0,1]$? After all $[0,1]\subset \mathbb{R}^1$. Can anyone answer this question in detail?</p>
| nombre | 246,859 | <p>Compactness is an absolute notion -meaning that it is defined for a topological space regardless of the ambient space whence it may be a subset - contrary to the notion of being closed, for instance. </p>
<p>Now, if a space $X$ is compact, and if $F$ is a closed subset of $F$, then $F$ seen as a topological space (with the topology induced by that of $X$) is compact. </p>
<p>Here you have proven that $E$ is a closed subset of $[0;1]$ which is compact, therefore, $E$ with the topology induced from $[0;1]$ is compact. </p>
<p>It could have been that $E$ wouldn't be a closed subset of $\mathbb{R}$, then $E$ with the topology induced from that of $\mathbb{R}$ would not compact, but this topological space just wouldn't be the same as the former.</p>
<p>However, there is a mistake in your proof that $E$ isn't closed in $\mathbb{R}$, because $\frac{7}{9}$ is an upper bound of $E$ so $B(z,\frac{1}{9}) \subset E^c$. $(z = 0.\overline{9} = 1)$.
$E$ is actually closed in $\mathbb{R}$: it is closed in $[0;1]$, so there is a closed subset $F \subset \mathbb{R}$ such that $E = [0;1] \cap F$. Both $F$ and $[0;1]$ are closed in $\mathbb{R}$, so $E$ is too.</p>
|
131,206 | <p>According to the wiki of <a href="http://en.wikipedia.org/wiki/Kakutani_fixed-point_theorem">Kakutani's fixed-point theorem</a>, A set-valued mapping $\varphi$ from a topological space $X$ into a powerset $\wp(Y)$ called upper semi-continuous if for every open set $W \subseteq Y$, $\lbrace x| \varphi(x) \subseteq W \rbrace$ is an open set in $X$.</p>
<p>My question:</p>
<ol>
<li>What is the definition of continuity of a multi valued map $\varphi$?</li>
<li>What's the definition of open sets in $\wp(Y)$, in other words, what topology does $\wp(Y)$ have?</li>
</ol>
| Steven Landsburg | 10,503 | <p>$\phi$ is upper semicontinuous if, for every open $W\subset Y$, the set $\lbrace x | \phi(x)\subset W\rbrace $ is open in $X$.</p>
<p>$\phi$ is lower semicontinuous if, for every open $W\subset Y$, the set $\lbrace x | \phi(x)\cap W\neq \emptyset\rbrace$ is open in $X$.</p>
<p>$\phi$ is continuous if it is both upper semincontinuous and lower semicontinuous. </p>
|
2,761,509 | <p>I hope it's not a duplicate but I've been searching about this problem for some time on this site and I couldn't find anything. My problem is why a number $\in(-1,0)$ raised to $\infty$ is $0$. For example let's take
$$\lim_{n\to \infty} \left(\frac{-1}{2}\right)^n$$
Which is equivalent to
$$\left(\frac{-1}{2}\right)^\infty=(-1)^\infty\left(\frac{1}{2}\right)^\infty=0(-1)^\infty$$
But if a sequence converges all its subsequences converge to the same limit.
And $(-1)^{2n}$ is a subsequence of $(-1)^n$ that converges to $1$ when $(-1)^{2n + 1}$ is a subsequence of $(-1)^n$ that converges to $-1$. So $(-1)^\infty$ does not exist. It remains that$$\lim_{n\to \infty} \left(\frac{-1}{2}\right)^n=0(DNE)$$ </p>
| IDC | 422,221 | <p>Your assertion that $(\frac{-1}{2})^\infty=(-1)^\infty(\frac{1}{2})^\infty$ is wrong because the first limit on the right side does not exist. The limit of a product is only equal to the product of limits when all the limits exist.</p>
|
8,052 | <p>I wonder how you teachers walk the line between justifying mathematics because of
its many—and sometimes surprising—applications, and justifying it as the study
of one of the great intellectual and creative achievements of humankind?</p>
<p>I have quoted to my students G.H. Hardy's famous line,</p>
<blockquote>
<p>The Theory of Numbers has always been regarded as one of the most obviously useless branches of Pure Mathematics.</p>
</blockquote>
<p>and then contrasted this with the role number theory plays in contemporary
cryptography.
But I feel slightly guilty in doing so, because I believe that even without
the applications to cryptography that Hardy could not foresee—if in fact
number theory were completely "useless"—it
would nevertheless be well-worth studying for anyone.</p>
<p>One provocation is
Andrew Hacker's influential article in
the <em>NYTimes</em>,
<a href="http://www.nytimes.com/2012/07/29/opinion/sunday/is-algebra-necessary.html?_r=0" rel="noreferrer">Is Algebra Necessary?</a>
I believe your cultural education is not complete unless you understand
something of the achievements of mathematics, even if pure and useless.
But this is a difficult argument to make when you are teaching students how
to factor quadratic polynomials, e.g.,
<a href="https://matheducators.stackexchange.com/q/8020/511">The sum - product problem</a>.</p>
<p>So, to repeat, how do you walk this line?</p>
| MintChocolateIceCream | 5,141 | <p>Most of what you learn in school isn't directly useful. When I was in primary school I was taught the difference between warm and cold-blooded animals. I've never used that information; should I not have been taught it? Another example is that in any English-speaking country you still have to take English courses even after you speak the language fluently. In day-to-day life you'll never need to know about Shakespeare or whatever literature the curriculum exposes you to. But it's still important to develop your communication and reasoning skills, your ability to think critically and creatively, to expose you to things that are influential on our culture, to show you a field or a way of looking at life that might be interesting to you or that might influence your career choice, and so on. All of this is applicable to mathematics.</p>
|
8,052 | <p>I wonder how you teachers walk the line between justifying mathematics because of
its many—and sometimes surprising—applications, and justifying it as the study
of one of the great intellectual and creative achievements of humankind?</p>
<p>I have quoted to my students G.H. Hardy's famous line,</p>
<blockquote>
<p>The Theory of Numbers has always been regarded as one of the most obviously useless branches of Pure Mathematics.</p>
</blockquote>
<p>and then contrasted this with the role number theory plays in contemporary
cryptography.
But I feel slightly guilty in doing so, because I believe that even without
the applications to cryptography that Hardy could not foresee—if in fact
number theory were completely "useless"—it
would nevertheless be well-worth studying for anyone.</p>
<p>One provocation is
Andrew Hacker's influential article in
the <em>NYTimes</em>,
<a href="http://www.nytimes.com/2012/07/29/opinion/sunday/is-algebra-necessary.html?_r=0" rel="noreferrer">Is Algebra Necessary?</a>
I believe your cultural education is not complete unless you understand
something of the achievements of mathematics, even if pure and useless.
But this is a difficult argument to make when you are teaching students how
to factor quadratic polynomials, e.g.,
<a href="https://matheducators.stackexchange.com/q/8020/511">The sum - product problem</a>.</p>
<p>So, to repeat, how do you walk this line?</p>
| Kyle | 5,152 | <p>I'm no teacher, but my position is if you're only studying what has some useful application now, how are you ever going to discover the new applications that we don't know about? I think your Hardy example sums that up perfectly.</p>
<p>I know, that still probably somewhat misses the point of your question because it still implies that mathematics (or specific branches of it) are only worth learning for practical applications. I certainly don't think that's the case, but there are some people you are going to have a very difficult time convincing of that (much as there are some people that don't see the value in whatever human endeavor, from art to manned spaceflight). </p>
<p>I think at the very least you might be able to win them over somewhat with the prospect of discovery. The whole point of studying these things is because <strong>we don't know</strong> what they can be used for. And the only way to find out is to get them into the heads of as many people as possible so that when each of them goes out into the world, one of them stumbles onto something.</p>
|
8,052 | <p>I wonder how you teachers walk the line between justifying mathematics because of
its many—and sometimes surprising—applications, and justifying it as the study
of one of the great intellectual and creative achievements of humankind?</p>
<p>I have quoted to my students G.H. Hardy's famous line,</p>
<blockquote>
<p>The Theory of Numbers has always been regarded as one of the most obviously useless branches of Pure Mathematics.</p>
</blockquote>
<p>and then contrasted this with the role number theory plays in contemporary
cryptography.
But I feel slightly guilty in doing so, because I believe that even without
the applications to cryptography that Hardy could not foresee—if in fact
number theory were completely "useless"—it
would nevertheless be well-worth studying for anyone.</p>
<p>One provocation is
Andrew Hacker's influential article in
the <em>NYTimes</em>,
<a href="http://www.nytimes.com/2012/07/29/opinion/sunday/is-algebra-necessary.html?_r=0" rel="noreferrer">Is Algebra Necessary?</a>
I believe your cultural education is not complete unless you understand
something of the achievements of mathematics, even if pure and useless.
But this is a difficult argument to make when you are teaching students how
to factor quadratic polynomials, e.g.,
<a href="https://matheducators.stackexchange.com/q/8020/511">The sum - product problem</a>.</p>
<p>So, to repeat, how do you walk this line?</p>
| Community | -1 | <p>You are not learning math.</p>
<p>You are learning the very useful, highly transferable skills of abstraction and problem solving. It's not that math can solve many problems, it's the skill you learn in order to do math are useful in so many other problems.</p>
|
8,052 | <p>I wonder how you teachers walk the line between justifying mathematics because of
its many—and sometimes surprising—applications, and justifying it as the study
of one of the great intellectual and creative achievements of humankind?</p>
<p>I have quoted to my students G.H. Hardy's famous line,</p>
<blockquote>
<p>The Theory of Numbers has always been regarded as one of the most obviously useless branches of Pure Mathematics.</p>
</blockquote>
<p>and then contrasted this with the role number theory plays in contemporary
cryptography.
But I feel slightly guilty in doing so, because I believe that even without
the applications to cryptography that Hardy could not foresee—if in fact
number theory were completely "useless"—it
would nevertheless be well-worth studying for anyone.</p>
<p>One provocation is
Andrew Hacker's influential article in
the <em>NYTimes</em>,
<a href="http://www.nytimes.com/2012/07/29/opinion/sunday/is-algebra-necessary.html?_r=0" rel="noreferrer">Is Algebra Necessary?</a>
I believe your cultural education is not complete unless you understand
something of the achievements of mathematics, even if pure and useless.
But this is a difficult argument to make when you are teaching students how
to factor quadratic polynomials, e.g.,
<a href="https://matheducators.stackexchange.com/q/8020/511">The sum - product problem</a>.</p>
<p>So, to repeat, how do you walk this line?</p>
| CaptainObvious | 5,172 | <p>I'm not a teacher/professor but I'd like to give my input as a person who hated math for a big part of his life, especially at high school. </p>
<p>I'd like first to explain why I hated math that much and for so long. At high school we were basically receiving some input and had to spit out some output. I'm referring to the countless resolution of derivatives, integrals well design to apply some seen methods. At the end I didn't need my brain at all. Fortunately, It didn't stop me to go in a technical field (chemical engineering) but during all my years at the university I never reconsidered my feeling for the mathematics and I simply considered it as a necessary evil, some tool only needed to do some other cool stuffs. It's only lately that I've started to <strong>enjoy</strong> math by among other things reading about its history and doing some math puzzles.</p>
<p>Now this is what I'd have loved my teacher at high school to tell us. I'd have loved her to be bluntly honest. That there are some problems with the program, that it won't be necessary useful. That what is called math at their level is actually a tiny bit of what it is reality. Someone mentioned Lockhart's Lament, why not make them read it.</p>
<p>Of course, just that is not enough. I think it should go along with that other thought: humans don't do something because it's useful but because <strong>they enjoy it</strong>. Unfortunately we also <strong>have to</strong> do things we didn't choose to and this is they case with their curriculum. But people that are happy in their life are those who manage to find some enjoyment even in the things they have to do (are forced?). So encourage them to find a way to enjoy it (some might enjoy it because it's useful - but not much). Each of them will find something different. After all if you tell them that math is useful in chemistry you'll get the attention of only those interested in chemistry. </p>
<p>Of course you should give them some pointers. For instance numerical analysis was a big revelation for me,especially the methods for solving differential equations. For years I wonder how scientists who really had to solve these kind of equations in real life were doing that. I don't say to explain it, just mention it (sometime, just knowing something exist is an eye-opener).</p>
<p>Others things that reconciled me with math: </p>
<ul>
<li><a href="https://www.coursera.org/course/maththink">Coursera's course Introduction to mathematical thinking</a></li>
<li>The puzzles of Martin Gardner</li>
<li>Reading about the history of mathematics</li>
<li>Some blogs written by fun mathematicians like this <a href="http://mathwithbaddrawings.com/">one</a></li>
<li>Some resources explaining basic concept using intuition like this <a href="http://betterexplained.com/">one</a></li>
<li>Some Math/Computer programming challenges like the <a href="https://projecteuler.net/">project Euler</a></li>
</ul>
<p>tl;dr : I wouldn't try to show them it's useful, I'd show them it's in their best interesting to try to find <strong>enjoyment</strong> in it while being honest about the shortcoming of the educational system.</p>
|
8,052 | <p>I wonder how you teachers walk the line between justifying mathematics because of
its many—and sometimes surprising—applications, and justifying it as the study
of one of the great intellectual and creative achievements of humankind?</p>
<p>I have quoted to my students G.H. Hardy's famous line,</p>
<blockquote>
<p>The Theory of Numbers has always been regarded as one of the most obviously useless branches of Pure Mathematics.</p>
</blockquote>
<p>and then contrasted this with the role number theory plays in contemporary
cryptography.
But I feel slightly guilty in doing so, because I believe that even without
the applications to cryptography that Hardy could not foresee—if in fact
number theory were completely "useless"—it
would nevertheless be well-worth studying for anyone.</p>
<p>One provocation is
Andrew Hacker's influential article in
the <em>NYTimes</em>,
<a href="http://www.nytimes.com/2012/07/29/opinion/sunday/is-algebra-necessary.html?_r=0" rel="noreferrer">Is Algebra Necessary?</a>
I believe your cultural education is not complete unless you understand
something of the achievements of mathematics, even if pure and useless.
But this is a difficult argument to make when you are teaching students how
to factor quadratic polynomials, e.g.,
<a href="https://matheducators.stackexchange.com/q/8020/511">The sum - product problem</a>.</p>
<p>So, to repeat, how do you walk this line?</p>
| John | 6,433 | <p>My simplest answer to any student asking "Why study math?" is this</p>
<blockquote>
<p>Math gives you power to get what you want.</p>
</blockquote>
<p>This is not the most "politically correct" way of "selling math", but it's both true and often persuasive to students and adults who want more power (which ends up being a majority of students and adults).</p>
<p>The real question being asked here is this: "How do we sell math?" We are not just interested in selling math to students, but also in selling it to people in general. Some parts of math are easier to sell because the sales pitch is simple: you need to know how money works so that you can buy and sell things without feeling like you've made a "bad deal" or feeling like you've been "cheated". Most of the economic reasons for learning math are based on our inherent fear of loosing economic security or our inherent want for economic power.</p>
<p>Outside of this economic sphere there are deep social reasons for studying mathematics. Perhaps the most valuable is that of forethought. Mathematics, unlike history, demonstrates that by systematically applying a basic problem solving process like Polya's you can put together facts which show clearly and exactly the future consequences of present actions.</p>
<p>So far I've discussed the economic motivation, and the general problem solving motivation, now I move into the motivation from love. When someone asks a question like "Why study math?" we might hear, in the back of our curious mind, another question "Why do anything?" and this is sometimes what a pesky philosopher is apt to ask. I do not know how to answer the question "Why study x?" or "Why do x instead of y?", but I do know of at least one statement which provides a popularly accepted way of approaching these general problems:</p>
<blockquote>
<p>The good life is one inspired by love and guided by knowledge. - Bertrand Russell</p>
</blockquote>
<p>From this simple statement I say that we study math because some of us are inspired to love it: some people study math <em>because</em> they love it. Ask them why they love it and I'm sure they can tell you all sorts of silly facts that really get their blood pumping, but the truth is that they do just love it. Why do we love anything? Why do we do anything for love? It seems that there is often no use in questioning the love that inspires us, but only in using the knowledge that we have to guide that inspiration towards the good life.</p>
|
25,158 | <p>I'm trying to derive the LTE for CN applied to the linear heat equation;
$u_t = u_{xx}$.</p>
<p>The problem is that I end up with terms of the form $\frac{{\Delta t}^k}{{\Delta x}^2}$
when using a two dimensional Taylor expansion around $(x,t)$ for the term:</p>
<p>${\Delta x}^2 {\delta^2_x} = (u_{i+1}^{n+1} - 2 u_{i}^{n+1} + u_{i-1}^{n+1})$</p>
<p>What am I doing wrong?</p>
<p>(trying to prove LTE for the Crandall-Douglas scheme)</p>
| Community | -1 | <p>your taylor expansions should be around $\left(x_i,t_{n+1}\right)$</p>
|
412,482 | <p>I have a few (semi-)related questions regarding certain Hilbert space representations of locally compact groups that come up in the theory of automorphic forms. </p>
<p>Let $G$ be a unimodular locally compact Hausdorff group, $\Gamma$ a discrete (hence closed) subgroup of $G$, and $Z$ a closed subgroup of the center of $G$. The example of interest to me is:</p>
<p>(*) $G$ is the group of adelic points of a connected reductive $\mathbf{Q}$-group $\mathscr{G}$, $\Gamma=\mathscr{G}(\mathbf{Q})$, and $Z$ is the group of adelic points of the maximal split torus in the center of $\mathscr{G}$. </p>
<p>I would be happy to know the answer to the following question just in case (*).</p>
<p><strong>Is the group $Z\Gamma$ closed in $G$, and if so, is it unimodular?</strong></p>
<p>The answer to the first part of the question would be yes if one of $Z$, $\Gamma$ were compact, but this isn't so in many cases of interest to me (e.g. $G=\mathrm{GL}_n$). I suspect the answer to both parts of the question are "yes" because of the following definition which appears in the literature (I will assume $Z\Gamma$ is closed and unimodular to make the definition, although I'm hoping, at least in the case of interest mentioned above, that this is unnecessary):</p>
<p>For a unitary character $\omega$ of $Z$, let $L^2(Z\Gamma\setminus G,\omega)$ be the space of equivalence classes (the equivalence being almost every equality relative to the unique-up-to-scaling $G$-invariant measure on $\Gamma\setminus G$) of Borel measurable $f:\Gamma\setminus G\rightarrow\mathbf{C}$ with the following properties: for each $z\in Z$, $f(\Gamma zg)=\omega(z)f(\Gamma g)$ for almost every $\Gamma g\in\Gamma\setminus G$, $\vert f\vert$ is Borel measurable on $Z\Gamma\setminus G$ (this makes sense because of the unitary of $\omega$), and $\int_{Z\Gamma\setminus G}\vert f\vert^2<\infty$ (the integral being taken with respect to the unique-up-to-scaling $G$-invariant measure on $Z\Gamma\setminus G$). Assuming I've got this definition right, it will be a Hilbert space with the natural inner product.</p>
<p>My second question concerns the definition of the so-called cuspidal subspace $L_0^2(Z\Gamma/G,\omega)$, and I'm only interested in the case of (*) with $\mathscr{G}=\mathrm{GL}_n$, so $G=\mathrm{GL}_n(\mathbf{A}_\mathbf{Q})$, $\Gamma=\mathrm{GL}_n(\mathbf{Q})$, and $Z=\mathbf{A}_\mathbf{Q}^\times$ is the center of $G$. The (apparently) standard definition of $L_0^2(Z\Gamma\setminus G,\omega)$ is: the subspace of $L^2(Z\Gamma\setminus G,\omega)$ consisting of equivalence classes containing a function $f$ such that $\int_{\mathbf{A}_\mathbf{Q}/\mathbf{Q}}f\big(\big(\begin{smallmatrix}1&x\\0&1\end{smallmatrix}\big)g\big)dx=0$ for almost every $g\in G$ ($dx$ means Haar measure on $\mathbf{A}_\mathbf{Q}/\mathbf{Q}$). There are some things to check to make precise sense of this definition, but the important thing is that it should be a closed subspace of the full $L^2$-space. This is taken for granted in, e.g., Bump's book on automorphic representations and Godement's notes on Jacquet-Langlands. The closedness doesn't seem completely obvious to me, and I worry about it because some other authors, notably Lang in his book on $\mathrm{SL}_2$, define the cuspidal subspace as the Hilbert space completion of the space of bounded continuous functions on $\Gamma\setminus G$ with the appropriate properties. I would hope these definitions yield the same space but it isn't clear to me.</p>
<p><strong>Assuming the definition I've given of $L_0^2(Z\Gamma\setminus G,\omega)$ is technically correct (i.e. I haven't left out any conditions on the functions under consideration), is it closed in $L^2(Z\Gamma\setminus G,\omega)$?</strong></p>
| Bravo | 24,451 | <p>Consider <a href="http://en.wikipedia.org/wiki/Multivariate_normal_distribution#Conditional_distributions" rel="noreferrer">this theorem</a> on conditional expectation of multivariate Gaussian distribution. Let $x=X_u$ and $Y=\left[X_s \quad X_t\right]^\top$. Let $X_0=0$, so that $\mu_X=0$ and $\mu_Y=\left[0 \quad 0\right]^\top$. </p>
<p>\begin{align}
\Sigma_{XX}&=Cov(X_u,X_u)=\sigma^2u\\
\Sigma_{XY}&=E(XY^\top)=[E(X_uX_s)\quad E(X_uX_t)]=\sigma^2[s\quad u]\\
\Sigma_{YY}&=E(YY^\top)=\left[\begin{matrix}E(X_s^2) & E(X_sX_t) \\ E(X_tX_s) & E(X_t^2)\end{matrix}\right]=\sigma^2\left[\begin{matrix}s & s \\ s & t\end{matrix}\right]
\end{align}</p>
<p>Plugging the formula, the random variable $X_u\mid (X_s=x,X_t=y)$ is also Gaussian with </p>
<ul>
<li><strong>mean</strong>: $\Sigma_{XY}\Sigma_{YY}^{-1}\left[\begin{matrix}x \\ y\end{matrix}\right]=\frac{t-u}{t-s}x+\frac{u-s}{t-s}y$</li>
<li><strong>variance</strong>:$\Sigma_{XX}+\Sigma_{XY}\Sigma_{YY}^{-1}\Sigma_{YX}=\sigma^2u+\sigma^2[s\quad u]\left[\begin{matrix}s & s \\ s & t\end{matrix}\right]^{-1}\left[\begin{matrix}s \\ u \end{matrix}\right]=\sigma^2\frac{(t-u)(u-s)}{t-s}$</li>
</ul>
|
1,201,942 | <p>I apologize if this question gets down-voted ahead of time.</p>
<p>I've been working on the Collatz Conjecture all day with Python, because that is the language I'm most familiar with (I'm not a CS student, just majoring in math). Below is the function I'm using for your reference during my comments:</p>
<p>\begin{align}
T\left(n\right)&=\begin{cases}1 & \text{if}\;n=1\\
T\left(\frac{n}{2}\right) & \text{if}\;n\;\text{is even}\\
T\left(3n+1\right) & \text{if}\;n\;\text{is odd}\end{cases}\tag{1}
\end{align}</p>
<p>I've created graphs of the number of iterations vs. the integers up to $10^7$, one of which I've shown below for integers up to $2\cdot 10^6$:</p>
<p><img src="https://i.stack.imgur.com/8eZBA.png" alt="Collatz Conjecture"></p>
<p>I've read on <a href="http://mathworld.wolfram.com/CollatzProblem.html" rel="noreferrer">Wolfram's Mathworld</a> that the conjecture has been tested via computers for numbers up to $\approx 5.48\cdot 10^{18}$, which is quite impressive, although I was thinking ahead of time it would have been tested by this time for incredibly large integers given we've had computers for >50 years now.</p>
<p>Now to my simple question: would it be true that, given such an integer actually does exist that does not satisfy the conjecture, the smallest such number must be odd because if it is even there must exist a smaller such integer that does not satisfy the conjecture? </p>
<p>Further, suppose we did find such an integer. In my writings I've noticed that all integers within the sequences generated by this function eventually break down to a repeat of an earlier sequence. For instance, the following two sequences:</p>
<blockquote>
<p>$\color{red}{7}$-22-11-34-17-52-26-13-40-20-10-5-16-8-4-2-1</p>
<p>14-$\color{red}{7}$</p>
</blockquote>
<p>These are simple, but even more lengthy sequences of numbers I've found break down to prior sequences that have already been computed, and must if we are to end up at $1$ unless it is defining a new sequence to be called by a later one (I wonder what the frequency of this is). But if we find such a number and it is odd, then all other numbers contained within this sequence must also not satisfy the conjecture. But is this true? Or am I thinking of "not satisfying" the conjecture in the wrong way?</p>
<p>Thank you for your time,</p>
| hmakholm left over Monica | 14,366 | <p>Yes, you're right: If there are any counterexample, then the <em>smallest</em> counterexample must be odd. And all successors and predecessors of a counterexample are themselves counterexamples.</p>
<p>There are two conceivable kinds of counterexample.</p>
<p>The first is a finite cyclic sequence that differs from the trivial 1-4-2-1 cycle. Finding such a counterexample would immediately produce a <em>disproof</em> of the conjecture.</p>
<p>The second is a starting point from which the sequence continues indefinitely without ever hitting a cycle. Just coming <em>across</em> such a point would not directly yield a disproof of the conjecture, because one would need to prove that the sequence <em>does</em> indeed never join a cycle, and there's no known systematic way of findig such a proof.</p>
|
1,746,363 | <p>I got maybe easy problem. I am not sure if it is true that [$\mathbb Z_2[x]/f\mathbb Z_2[x]: \mathbb Z_2$]=deg $f$ where $f \in \mathbb Z_2[x]$ irreducible. Can anybody help me ? Thanks</p>
| Eman Yalpsid | 94,959 | <p>The rate of change at $x=\frac{5\pi}{6}$ is simply $h'(\frac{5\pi}{6}) = 12 \sin^2(\frac{5\pi}{6})\cos(\frac{5\pi}{6})$.</p>
<p>Your confusion probably arises from the fact that you are used to dealing with degrees instead of radians.</p>
<p>The period of $\sin, \cos$ is $2\pi$ or $360$ degrees. One radian is $\frac{360}{2\pi}$ degrees. Spend some time studying the unit circle,
<a href="https://i.stack.imgur.com/Eb4FS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Eb4FS.png" alt="unit circle"></a></p>
<p>and you'll get used to radians in no time. </p>
<p>In the parentheses you can also see the $\cos$ and $\sin$ values of the angles. Check out <a href="https://en.wikipedia.org/wiki/Unit_circle" rel="nofollow noreferrer">wikipedia</a> for more!</p>
|
1,578,783 | <p>A friend of mine found a book in which the author said that the dual space of $L^\infty$ is $L^1$, of course not with the norm topology but with the weak-* topology. Does anyone know where I can find this result? Thanks.</p>
| Community | -1 | <p>There is a general fact from duality of linear spaces (see Proposition 4.28 in Fabian-Habala-Hajek-Montesinos-Pelant-Zizler, Functional Analysis and Infinite-Dimensional Geometry): If we consider a linear subspace $F$ in the space of linear functionals on $E$, then the space of linear functionals on $E$ continuous in the corresponding weak topology on $E$ coincides with $F$.</p>
|
3,547,529 | <p>I did the following: I set <span class="math-container">$3^m+3^n+1=x^2$</span> where <span class="math-container">$x\in\Bbb{N}$</span> and assumed it was true for positive integer exponents and for all whole numbers x so that I can later on prove it's invalidity with contradiction. Since <span class="math-container">$3^m+3^n+1$</span> is odd we can write <span class="math-container">$3^m+3^n+1 = (2k+1)^2$</span> for <span class="math-container">$k\in\Bbb{N}$</span>. After a while I can't seem to prove it and I don't have any ideas on how else to approach this problem. </p>
<p>It would be very helpful if someone could give me a brief explanation on a possible proof since I'm currently practicing for a math competition. </p>
<p>Thanks in advance </p>
| Bart Michels | 43,288 | <p>It is clear that <span class="math-container">$A(y), B(y) \sim \pm (y/a)^{1/2n}$</span>, and the problem is to determine the lower order term in their expansion. Wlog suppose <span class="math-container">$B(y) < 0 < A(y)$</span>. By the mean value theorem,
<span class="math-container">$$(A(y) - (y/a)^{1/2n}) \cdot F'(\xi) \sim y - F((y/a)^{1/2n}) \,,$$</span>
for some <span class="math-container">$\xi \sim (y/a)^{1/2n}$</span>.
The LHS is <span class="math-container">$\sim 2an \cdot (y/a)^{(2n-1)/2n}$</span> times what we are looking for, and the RHS is <span class="math-container">$b(y/a)^{(2n-1)/2n} (1 + o(1))$</span>. We conclude that
<span class="math-container">$$A(y) - (y/a)^{1/2n} \sim \frac{b}{2an} \,.$$</span>
Replacing <span class="math-container">$F(x)$</span> by <span class="math-container">$F(-x)$</span>, we obtain
<span class="math-container">$$-B(y) - (y/a)^{1/2n} \sim \frac{-b}{2an} \,.$$</span>
The conclusion follows.</p>
|
2,571,909 | <p>$$\left|\frac{-10}{x-3}\right|>\:5$$</p>
<ul>
<li>Find the values that $x$ can take. </li>
</ul>
<p>I know that</p>
<p>$$\left|\frac{-10}{x-3}\right|>\:5$$
and
$$\left|\frac{-10}{x-3}\right|<\:-5$$</p>
| Community | -1 | <p>First note that $x=3$ cannot be a part of the solution. That in mind, we have: $$|\frac{-10}{x-3}|>5 \implies \frac{10}{5}> |x-3| \implies -2< x -3 < 2 \implies x \in (1,5) - \{3\}$$</p>
|
1,808,258 | <p>I was reading about orthogonal matricies and noticed that the $2 \times 2$ matrix
$$\begin{pmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{pmatrix} $$
is orthogonal for every value of $\theta$ and that every $2\times 2$ orthogonal matrix can be expressed in this form. I then wondered if this can be generalized to any smooth paramtrization of the unit circle. More precisely, let $x(t)$ and $y(t)$ be a smooth parametrization of the unit circle for all $t$ in some interval $I \subseteq \Bbb R$ such that $|\langle x(t),y(t) \rangle| = 1$ for all $t \in I$. Is the matrix
$$A= \begin{pmatrix}x(t)&y(t)\\x'(t) & y'(t) \end{pmatrix} $$
necessarily orthogonal?
At first I thought yes, but I'm having trouble proving it. Letting $v = \langle x(t), y(t) \rangle$, it suffices to show three things: 1) $|v| = 1$, 2) $v \cdot v' = 0$, and 3) $|v'| = 1$. (1) follows straight from how $x(t)$ and $y(t)$ were defined. (2) can be obtained by differentiating the equation $x(t)^2 + y(t)^2 = 1$:
\begin{align*}
&\frac{d}{dt} \Big[ x(t)^2 + y(t)^2 \Big]= 0 \\
&\implies 2x(t)x'(t) + 2y(t)y'(t)= 0 \\
&\implies v \cdot v' = 0.
\end{align*}
But I couldn't find a way to prove (3). Now I am unsure whether (3) is true at all. Is $A$ even orthogonal in the first place? Any hints would be much appreciated.</p>
| Travis Willse | 155,629 | <p>In general, (3) is not true. Since $\left\vert v \right\vert^2 = x(t)^2 + y(t)^2 = 1$ , we can write $$x(t) = \cos \theta(t), \quad y(t) = \sin \theta(t)$$
for some function $\theta(t)$. But computing gives $\left\vert\dot v\right\vert^2 = \dot\theta(t)^2$, so the condition $|\dot v|^2 = 1$ (which just says that $v$ is a <em>unit speed parameterization</em>) forces $\dot\theta(t) = \pm 1$, and this condition yields just the matrices of the given form (the <em>rotations</em>). Note that a composition of any rotation with a reflection is another reflection and hence orthogonal, but these reflections cannot be of the form in the question: Reflections have determinant $-1$ but matrices of the given form have determinant $(\cos \theta)(\cos \theta) - (\sin \theta)(-\sin \theta) = +1$.</p>
|
1,200,358 | <blockquote>
<p>Assume the $n$-th partial sum of a series $\sum_{n=1}^\infty a_n$ is the following:
$$S_n=\frac{8n-6}{4n+6}.$$
Find $a_n$ for $n > 1$.</p>
</blockquote>
<p>I'm really stuck on what to do here.</p>
| kobe | 190,421 | <p>Hint. Note $a_n = S_n - S_{n-1}$ for $n > 1$.</p>
|
446,326 | <blockquote>
<p>Let $Q$ be a $3\times3$ special orthogonal matrix. Show that $Q(u\times v)=Q(u)\times Q(v)$ for any vectors $u, v\in\mathbb R^3$.</p>
</blockquote>
<p>I have no idea how to start. I'm not sure if $Q(u)\cdot Q(V)=Q(u\cdot v)$ would helps. Please give me some help. Thanks.</p>
| Julien | 38,053 | <p>The cross-product $u\times v$ is the unique vector such that
$$
\det(u,v,w)=(u\times v)\cdot w\qquad \forall w
$$
where $\det(u,v,w)$ is the determinant of the $3\times 3$ matrix whose columns are $u,v,w$ in this order, that is the determinant of the linear map that sends the canonical basis to $(u,v,w)$. That's a common definition of the cross-product. See below if needed.</p>
<p>Recall that $Q$ special orthogonal means $Q^T=Q^{-1}$ and $\det Q=1$.</p>
<blockquote>
<p>We need to prove that $Q^T(Qu\times Qv)=u\times v$. So let us compute
$$
Q^T(Qu\times Qv)\cdot w=(Qu\times Qv)\cdot Qw=\det(Qu,Qv,Qw)=\det Q\det(u,v,w)=\det(u,v,w).
$$
By the uniqueness defining $u\times v$, this proves $Q^T(Qu\times Qv)=u\times v$, i.e. $Qu\times Qv=Q(u\times v)$.</p>
</blockquote>
<p><strong>Note:</strong> the same argument shows more generally that, as mentioned by lhf and wikipedia,
$$
M^T(Mu\times Mv)=(\det M) u\times v\quad\Rightarrow \quad (Mu\times Mv)=(\det M) M^{-T}(u\times v)
$$
for every invertible $3\times 3$ matrix $M$, where $M^{-T}=(M^{-1})^T=(M^T)^{-1}$. The formula on the left is true for every matrix $M$ and is just $0=0$ in the singular case, since we have $Mu\times Mv=0$ for every $u,v$ in this case.</p>
<hr>
<p>The fact that the identity $\det(u,v,w)=(u\times v)\cdot w$ is satisfied by every $u,v,w$ can be checked directly, by computations, from the determinant definition of $u\times v$. Another way to see it is to note that the map $(u,v,w)\longmapsto (u\times v)\cdot w$ is multilinear, anti-symmetric (or alternating), and sends the canonical basis to $1$, whatever definition of the cross-product you might have. So it must be the determinant. Uniqueness of $u\times v$ satisfying the identity follows from $(\mathbb{R}^{3})^\perp=\{0\}$, as $w_1\cdot w=w_2\cdot w$ for every $w$ implies $(w_1-w_2)\cdot w=0$ for every $w$, in particular for $w=w_1-w_2$, whence $\|w_1-w_2\|^2=0$.</p>
|
4,506,026 | <p>Consider the set of equations:
<span class="math-container">$$
\begin{cases}
x^2 &= -4y-10\\y^2 &= 6z-6\\z^2 &= 2x+2\\
\end{cases}$$</span></p>
<p>With <span class="math-container">$x,y,z$</span> being real numbers.</p>
<p>By adding the three equations, after simple manipulations, we easily obtain
<span class="math-container">$$
(x-1)^2+(y+2)^2+(z-3)^2=0
$$</span>
Which yields
<span class="math-container">$$
\begin{cases}
x&=1\\y&=-2\\z&=3\\
\end{cases}
$$</span>
However, plugging that in we find that the solution is invalid. How can one reason about this? In other words, <em>why</em> is it invalid? What causes it to be invalid?</p>
<p><a href="https://i.stack.imgur.com/EQMj6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EQMj6.jpg" alt="enter image description here" /></a></p>
| pierrecurie | 616,347 | <p>What you found is a necessary but not sufficient condition. This is what happens in general when you take linear combinations of your initial equations.</p>
<p>If you back substitute, you'll end up with a degree 8 polynomial with 8 complex roots. The manipulations you did aren't wrong - there are indeed 0 real roots.</p>
<p>Consider the simpler set of equations:
<span class="math-container">$$x^2=9 \\
y^2=16 \\
z^2=-25$$</span>
By inspection, the 8 roots are <span class="math-container">$(\pm3,\pm4,\pm5i)$</span>.</p>
<p>If you add the equations, you get
<span class="math-container">$$x^2+y^2+z^2=0$$</span>
Well, <span class="math-container">$(0,0,0)$</span> is clearly a solution of the new equation, but not the original set. OTOH, if you plug in the 8 true roots into this new derived equation, you will find that it's still true. So, as I said, necessary but not sufficient.</p>
|
3,182,802 | <p>Show that if <span class="math-container">$ \sigma $</span> is a solution to the equation <span class="math-container">$ x^2 + x + 1 = 0 $</span> then the following equality occurs:</p>
<p><span class="math-container">$$ (a +b\sigma + c\sigma^2)(a + b\sigma^2 + c\sigma) \geq 0 $$</span></p>
<p>I looked at the solution in my textbook and it says I should multiply the parentheses and take into account that <span class="math-container">$ \sigma + \sigma^2 + 1 = 0 $</span>.
I tried factoring the rest but I just can't seem to manage to solve it?</p>
<p>Maybe I messed up at multiplying the parentheses? Here's what I got:</p>
<p><span class="math-container">$$ a^2 + ab\sigma^2 + ac\sigma + ab\sigma + b^2\sigma^3 + bc\sigma^2 + ac\sigma^2 + bc\sigma^4 + c^2\sigma^3 $$</span> </p>
| Bernard | 202,857 | <p>I suppose <span class="math-container">$a,b,c$</span> are real numbers. Now observe <span class="math-container">$\sigma$</span> is a non-real cubic root of unity, so the other non-real root is <span class="math-container">$\bar \sigma=\sigma^2$</span>. Thus
<span class="math-container">\begin{align}
(a +b\sigma + c\sigma^2)(a + b\sigma^2 + c\sigma)&=(a +b\sigma + c\bar\sigma)(a + b\bar\sigma + c\sigma)=(a +b\sigma + c\bar\sigma)(\overline{a + b\sigma + c\bar\sigma})\\
&=\bigl|a +b\sigma + c\bar\sigma\bigr|^2.
\end{align}</span></p>
|
1,088,734 | <p>It's possible the integral bellow. What way I must to use for solve it.</p>
<p>$$\int \sin(x)x^2dx$$</p>
| heropup | 118,193 | <p>If $p$, $q$, $r$ are distinct positive integer roots of $P(x) = x^3 + ax^2 + bx - 26$, then we must have $$(x-p)(x-q)(x-r) = x^3 - (p+q+r)x^2 + (pq+qr+rp)x - pqr = P(x),$$ hence equating the constant coefficient, $$pqr = 26 = 2 \cdot 13.$$ Since the prime factorization of $26$ contains only $2$ and $13$, it immediately follows that there is only one unique solution (up to permutation) for this equation that satisfies the given conditions, namely $\{p, q, r\} = \{1, 2, 13\}$ in some order. The rest follows easily.</p>
|
582,283 | <p>$H$ is subgroup of $G$ with $H$ not equal $G$.</p>
<p>Be $S=G-H$. I am being asked to prove that $\langle S \rangle=G$.</p>
<p>Some tip to solve this? I think in $S_3$ is possible but I can´t prove.</p>
| egreg | 62,967 | <p>Hint. You have to use the hypothesis that $H\ne G$ and show that any element of $G$ can be written as a product of elements in $S$ or inverse thereof.</p>
<p>If $g\in S$, then of course $g\in\langle S\rangle$; if $g\in H$, pick an element $s\in S$; then $t=sg\notin H$ (why?); but $s^{-1}t=\dots$</p>
|
35,688 | <p>I'm looking for a fun (not too many tedious calculations) calculus one problem that uses the concept that, after subsitution, you have two integrals of diffent functions with different limits, but equal area. For example:</p>
<p><a href="http://www.wolframalpha.com/input/?i=int%20%28sin%28%28pi%5E2%29/x%29%29/%28x%5E2%29%20from%20pi%20to%20pi/2" rel="nofollow">Function one</a></p>
<p><a href="http://www.wolframalpha.com/input/?i=int%20%28sin%20x%29%2a%28-1/pi%5E2%29%20from%20pi%20to%202%20pi" rel="nofollow">Function two</a></p>
<p>Those two regions are the same. I'm thinking of a problem that asks the student to prove two regions are of equal area, and once they set up the integrals they can see it is a case of subsitution. </p>
<p>Do you know of such a problem?</p>
| André Nicolas | 6,312 | <p>The following is part of the standard machinery for proving the basic properties of logarithms, if one chooses to introduce them by using an integral.</p>
<p>For any number $c \ge 1$, define $L(c)$ by saying that $L(c)$ is the area under the curve $y=1/x$, above the $x$-axis, from $x=1$ to $x=c$.</p>
<p>Then $L(ab)$ is the area from $1$ to $ab$. This is the area from $1$ to $a$, <strong>plus</strong> the area from $a$ to $ab$.</p>
<p>But the area from $a$ to $ab$ is
$$\int_a^{ab}\frac{dx}{x}$$
Make the substitution $x=ua$. We find that
$$\int_a^{ab}\frac{dx}{x}=\int_1^b \frac{du}{u}$$</p>
<p>Conclusion: $L(ab)=L(a)+L(b)$.</p>
<p><strong>Edit:</strong> Here is another example, more usable. Let $a$ and $b$ be positive. Then the area of the top half of the ellipse $x^2/a^2+y^2/b^2=1$ is the same as the area of the top half of the circle with radius $\sqrt{ab}$.</p>
|
3,613,235 | <p>I know such integral: <span class="math-container">$\int_0^{\infty}\frac{\ln x}{e^x}\,dx=-\gamma$</span>. What about the integral <span class="math-container">$\int_0^{\infty}\frac{\ln x}{e^x+1}\,dx$</span>? </p>
<p>The answer seems very nice: <span class="math-container">$-\frac{1}{2}{\ln}^22$</span> but how it could be calculated? I tried integration by parts but the limit <span class="math-container">$\displaystyle{\lim_{x\to 0}\ln x\ln(1+e^{-x})}$</span> doesn't exist. Or I can also write the following equality
<span class="math-container">$$\int_0^{\infty}\frac{\ln x}{e^x+1}\,dx=\lim\limits_{t\to 0}\frac{d}{dt}\left(\int_0^{\infty}\frac{x^t}{e^x+1}\, dx\right)$$</span>
but I don't know what to do next. </p>
| Andronicus | 528,171 | <p>We can rewrite the sum as:</p>
<p><span class="math-container">$$\sum_{n=1}^{99} \frac{2}{\sqrt{n} + \sqrt{n+2}}\frac{\sqrt{n+2} - \sqrt{n}}{\sqrt{n+2} - \sqrt{n}}=
\sum_{n=1}^{99} \sqrt{n+2} - \sqrt{n}$$</span></p>
<p>It's a telescopic sum, where most of the terms are cancelled.</p>
<p>We can expand it:</p>
<p><span class="math-container">$$\sum_{n=1}^{99} \sqrt{n+2} - \sqrt{n}=
(\sqrt{3}-\sqrt{1})+(\sqrt{4}-\sqrt{2})+(\sqrt{5}-\sqrt{3})+\dots+(\sqrt{101}-\sqrt{99})=\sqrt{100}+\sqrt{101}-\sqrt{1}-\sqrt{2}$$</span></p>
|
382,293 | <p>Not sure where to go with this one. Clearly will have to use the axiom of choice at some point. I haven't been able to think of a good example for the set $A.$ Once we've got that, it'd be a matter of showing that a representation $($as a sum, $q+a)$ exists for each real number $($which should be the case by construction of $A)$, and then subsequently that this representation is unique.</p>
| Asaf Karagila | 622 | <p>If you already know the axiom of choice has to be involved (and it does), there is little hope to being able to write $A$ explicitly.</p>
<p>Here is a hint: consider the quotient $\Bbb{R/Q}$.</p>
|
1,649,907 | <p>Please kindly forgive me if my question is too naive, i'm just a <em>prospective</em> undergraduate who is simply and deeply fascinated by the world of numbers.</p>
<p>My question is: Suppose we want to prove that $f(x) > \frac{1}{a}$, and we <em>know</em> that $g(x) > a$, where $f,g$ and $a$ are all positive and $a$ is a nonzero real number.
<em>If we can show</em> that $f(x)g(x) > 1$, would that imply our required proof ?</p>
<p>EDIT: As demonstrated by various users in the solutions below, the answer is definitely <em>no</em>.
What about if we now want to prove the <em>reverse</em> inequality $f(x) \leq \frac{1}{a}$ given that $g(x) < a$, if we can show that $f(x)g(x)<1$, i guess our required result would follow ?</p>
| User1 | 305,659 | <p>Unless i'm missing something horribly wrong, it seems the <em>current</em> form of the question is straightforward (though this appears to contradict with fleablood's answer):</p>
<p>We know that $g(x)<a$, let $f(x)\leq t$. Therefore $f(x)g(x)<at$. But we also know that $f(x)g(x)<1$, hence we should have $t \leq \frac{1}{a}$, which yields $f(x)\leq t \leq \frac{1}{a}$, as required.</p>
|
2,315,739 | <p>I have an irregular quadrilateral.
I know the length of three sides (a, b and c) and the length of the two diagonals (e and f).
All angles are unknown
How do I calculate the length of the 4th side (d)?</p>
<p>Thank you for your help.
Regards,</p>
<p>Mo</p>
<p><a href="https://i.stack.imgur.com/Jtdzv.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jtdzv.jpg" alt="enter image description here"></a></p>
| Plus Twenty | 453,093 | <p>Hint : You could try re-arranging the Cosine Rule: $a^2 = c^2+b^2-2bc\cos A$ to try and find some of the angles of the triangles.</p>
|
3,394,277 | <p>I am new to this site and not familiar with how to type out math notation so I will do my best. I have a problem I am working on regarding the volume of a circle wrapped around a cylinder of variable radius. For the first part of the problem I had no issue creating a function to represent the cross sectional area. Using this function and I am not trying to integrate to find the volume for different values of cylindrical radius r. The first half of the integral was painless but I have been stuck on the second half for a while now and am looking for some help as I cannot find a solution anywhere. Here is the integral;</p>
<p><span class="math-container">$$\int_{-4}^4\sin(\sqrt{16-x^2})dx$$</span></p>
<p>This integral does include a couple other variable terms but they are treated as constants so there is little point in including them here as they will just complicate the problem. I am basically just looking for the technique used to integrate something like this because I am clueless. Thanks in advance.</p>
| Kavi Rama Murthy | 142,385 | <p>Note that <span class="math-container">$X$</span> is a non-decreasing function. For any non-decreasing function <span class="math-container">$f$</span> observe that <span class="math-container">$\{x: f (x) <a\}$</span> is an interval for any real number <span class="math-container">$a$</span>. Hence <span class="math-container">$X$</span> is measurable. </p>
|
2,473,780 | <p>So I have the limit $$\lim_{x\rightarrow \infty}\left(\frac{1}{2-\frac{3\ln{x}}{\sqrt{x}}}\right)=\frac{1}2,$$ I now want to motivate why $(3\ln{x}/\sqrt{x})\rightarrow0$ as $x\rightarrow\infty.$ I cam up with two possibilites:</p>
<ol>
<li><p>Algebraically it follows that $$\frac{3\ln{x}}{\sqrt{x}}=\frac{3\ln{x}}{\frac{x}{\sqrt{x}}}=\frac{3\sqrt{x}\ln{x}}{x}=3\sqrt{x}\cdot\frac{\ln{x}}{x},$$
Now since the last factor is a standard limit equal to zero as $x$ approaches infinity, the limit of the entire thing should be $0$. However, isn't it a problem because $\sqrt{x}\rightarrow\infty$ as $x\rightarrow \infty$ gives us the indeterminate value $\infty\cdot 0$?</p></li>
<li><p>One can, without having to do the arithmeticabove, directly motivate that the function $f_1:x\rightarrow \sqrt{x}$ increases faster than the function $f_2:x\rightarrow\ln{x}.$ Is this motivation sufficient? And, is the proof below correct?</p></li>
</ol>
<p>We have that $D(f_1)=\frac{1}{2\sqrt{x}}$ and $D(f_2)=\frac{1}{x}$. In order to compare these two derivateives, we have to look at the interval $(0,\infty).$ Since $D(f_1)\geq D(f_2)$ for $x\geq4$, it follows that $f_1>f_2, \ x>4.$</p>
| Siong Thye Goh | 306,553 | <p>When the numerator and denominator both go to $\infty$, it is in <a href="https://en.wikipedia.org/wiki/Indeterminate_form" rel="nofollow noreferrer">indeterminate form</a>, we can use <a href="https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule" rel="nofollow noreferrer">L'Hospital's rule</a>.</p>
<p>$$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{f'(x)}{g'(x)}$$</p>
<p>$$\lim_{x \to \infty} \frac{\ln x}{\sqrt{x}} = \lim_{x \to \infty} \frac{1/x}{1/(2\sqrt{x})}=\lim_{x \to \infty}\frac{2}{\sqrt{x}}=0$$</p>
|
362,881 | <p>I am going to try to explain this as easily as possible. I am working on a computer program that takes input from a joystick and controls a servo direction and speed. I have the direction working just fine now I am working on speed. To control the speed of rotation on the servo I need to send it so many pulses per second using PWM. The servo that I am using takes arguments for speed between 120-150. 120 is %100 speed and 150 is %0 or stopped. 135 is %50 speed. How would I convert percentage from 0-100 into a number between 120-150 including 1/10ths? I hope this makes sense if you need me to explain further please let me know. I really don't know what tag this falls under either.</p>
| ncmathsadist | 4,154 | <p>The $\delta_0$ functional is representable as a measure. You have
$$\delta_0(E) = \cases{1 & if $0 \in E$\cr 0 & otherwise }$$
Then the linear functional here is represented as
$$f \mapsto f(0) = \int_{R^d} f(\xi)d\delta_0(\xi). $$
It is not hard to show this relation will hold for any continuous function. Your result will follow right away.</p>
|
2,487,234 | <p>I want to prove that the following only have one solution, for $\zeta\in[0,1]$, at $\zeta =1$.</p>
<p>$$f(\zeta)=\frac{1}{1+\zeta}$$</p>
<p>$$g(\zeta) =
\frac{
(1-\zeta)(2-\zeta)\zeta
-
(2-\zeta)^2\log\left(2 - \zeta \right)
}
{\zeta(\zeta-4)(\zeta-1)^2}$$</p>
<p>These are plotted below. Note that $f(0)=1$, $\lim_{\zeta\rightarrow 0}g(\zeta)=\infty$ and $f(1)=1/2=g(1)$.</p>
<p>Both functions are convex and monotonically decreasing over this region, therefore if they intersect at $\zeta=1$ this is the only place they can? Can I shown they are both convex over $\zeta\in[0,1]$ by showing their second derivatives wrt $\zeta$ are +ve over this region? Or is this thinking too simplistic?</p>
<p><a href="https://i.stack.imgur.com/Jm6yn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jm6yn.png" alt="enter image description here"></a></p>
<p>Note, the limit as $g(\zeta)$ as $\zeta\rightarrow 1$ is proven below:
\begin{align}
\lim_{\zeta\rightarrow 1} g(\zeta)
%%$
&=
%%%
\lim_{\zeta\rightarrow 1}
\left(
\frac
{4\log(2 - \zeta) - 7\zeta - 2\zeta \log(2 - \zeta) + 3\zeta^2 + 4}
{4\zeta^3 - 18\zeta^2 + 18\zeta - 4}
\right)\\
%%%
&=
%%%
\lim_{\zeta\rightarrow 1}
\left(
\frac
{6\zeta - 2\log(2 - \zeta) - 9}
{12\zeta^2 - 36\zeta + 18}
\right)\\
%%%
&=
%%%
\frac
{6 - 9}
{12 - 36 + 18}\\
%%%
&=
%%%
\frac
{1}
{2}.
\end{align}</p>
<p>EDIT: So the second derivative of $g(\zeta)$ is pretty horrific. It does have a simplified form, but it is also pretty terrible.
<a href="http://www.wolframalpha.com/input/?i=(d%5E2%2Fdx%5E2+(+(1-x)(2-x)x+-+(2-x)%5E2%5Clog(2+-x)+)%2F(x(x-4)(x-1)%5E2))" rel="nofollow noreferrer">http://www.wolframalpha.com/input/?i=(d%5E2%2Fdx%5E2+(+(1-x)(2-x)x+-+(2-x)%5E2%5Clog(2+-x)+)%2F(x(x-4)(x-1)%5E2))</a></p>
<p>$$g'' =
\frac
{(x - 1) x (x (x (x (x (2 x - 7) - 13) + 124) - 224) + 64)
- 2 (x (x (x (x (3 (x - 12) x + 184) - 472) + 588) - 304) + 64) \log(2 - x))}
{(x - 4)^3 (x - 1)^4 x^3}$$</p>
<p>It is clear from the numerics that it is greater than $0$ in the region $\zeta\in[0,1]$.
<a href="http://www.wolframalpha.com/input/?i=(d%5E2%2Fdx%5E2+(+(1-x)(2-x)x+-+(2-x)%5E2%5Clog(2+-x)+)%2F(x(x-4)(x-1)%5E2))+%3D+0" rel="nofollow noreferrer">http://www.wolframalpha.com/input/?i=(d%5E2%2Fdx%5E2+(+(1-x)(2-x)x+-+(2-x)%5E2%5Clog(2+-x)+)%2F(x(x-4)(x-1)%5E2))+%3D+0</a></p>
| Daniel Pol | 481,873 | <p>Cutting places of this 2 functions are in zeros of equation $f(\zeta)-g(\zeta)=0$. I put this in form $f(\zeta)=g(\zeta)$. After this i put all powers together and try to separate $\log{(2-s)}$ in one side. I come to next expression :
$$
\log{(2-s)} = {{2(2s^3-5s^2+3s)} \over {(s-2)^2(s+1)}}
$$
Now i idea would to make a derivative in both sides ...</p>
<p>Will this help ? <br/>
Daniel</p>
|
1,519,952 | <p>Show that
$$S(n,k) = \sum_{m = k-1}^{n-1} {n-1 \choose m} S(m,k-1) $$</p>
<p>-I was having trouble with this proof in class and my professor suggested to look at it as another proof of the following theorem which states:</p>
<p>-For all $n\ge1$
$$B(n) = \sum_{k=0}^{n-1} {n-1 \choose k} B(k) $$
-Unfortunately I still do not understand how to solve this proof, I do see the similarities in structure although I am brand new to Stirling numbers and am unsure of how this would affect the proof. Any help is appreciated.</p>
| Brian M. Scott | 12,042 | <p>$S(n,k)$ is the number of partitions of the set $[n]=\{1,\ldots,n\}$ into exactly $k$ non-empty parts. Suppose that $\mathscr{P}$ is such a partition; then there must be a $P\in\mathscr{P}$ that contains the number $n$. </p>
<ul>
<li>The other $k-1$ parts must contain at least $k-1$ and at most $n-1$ elements of $[n]$; why? </li>
</ul>
<p>Let $m=|[n]\setminus P|$, the number of elements of $[n]$ that are <em>not</em> in the same part of $\mathscr{P}$ as $n$.</p>
<ul>
<li>Show that there are $\binom{n-1}m$ ways to choose the elements that are not in the same piece as $n$. </li>
<li>Show that there are $S(m,k-1)$ ways to divide them into exactly $k-1$ non-empty parts. </li>
</ul>
<p>Now combine the pieces to get the desired identity.</p>
|
2,375,736 | <p>If the Ratio of the roots of $ax^2+bx+c=0$ be equal to the ratio of the roots of $a_1x^2+b_1x+c_1=0$, then how one prove that $\frac{b^2}{b^2_1}=\frac{ac}{a_1 c_1}$?</p>
| David Quinn | 187,299 | <p>Let the roots of the first quadratic be $\alpha$ and $t\alpha$ and the roots of the second be $\beta$ and $t\beta$</p>
<p>Then $$\alpha(t+1)=\frac ca$$ and $$\alpha^2t=\frac ca$$</p>
<p>Then $$\frac {b^2}{a^2}=\alpha(t+1)^2=\frac{c}{at}(t+1)^2$$</p>
<p>In exactly the same way, $$\frac {b_1^2}{a_1^2}=\beta(t+1)^2=\frac{c_1}{a_1t}(t+1)^2$$</p>
<p>Now eliminate all the $t$ terms by division and the result follows immediately</p>
|
248,267 | <p>It is known that the transformation rule when you change coordinate frames of the Christoffel symbol is:</p>
<p>$$ \tilde \Gamma^{\mu}_{\nu\kappa} = {\partial \tilde x^\mu \over \partial x^\alpha} \left [ \Gamma^\alpha_{\beta \gamma}{\partial x^\beta \over \partial \tilde x^\nu}{\partial x^\gamma \over \partial \tilde x^\kappa} + {\partial ^2 x^\alpha \over \partial \tilde x^\nu \partial \tilde x^\kappa} \right ]$$</p>
<p>Is there any way to prove this rule using only the definition of the Christoffel via the metric tensor? That is, using:</p>
<p>$$ \Gamma^\mu _{\nu\kappa} = \frac{1}{2}g^{\mu\lambda}\left(g_{\lambda\kappa,\nu}+g_{\nu\lambda,\kappa}-g_{\nu\kappa,\lambda} \right)$$</p>
<p>All proofs have I've seen of the transformation law involve another method.</p>
| Andrews | 557,551 | <p>Let <span class="math-container">$(U, x^1, \ldots, x^n)$</span> be a chart of Riemannian manifold <span class="math-container">$M$</span>, smooth coordinate chang
<span class="math-container">$(U, g_{\mu\nu}, x^1, \ldots, x^n) \to (U, g'_{\mu'\nu'}, y^1, \ldots, y^n)$</span>.
Then <span class="math-container">$\partial^x_\mu =\frac{\partial}{\partial x^\mu}
= \frac{\partial y ^ {\mu'}}{\partial x^\mu} \frac{\partial}{\partial y^{\mu'}}
= \frac{\partial y ^ {\mu'}}{\partial x^\mu} \partial^y_{\mu'}$</span>,
<span class="math-container">$\partial^x_\nu = \frac{\partial y ^ {\nu'}}{\partial x^\nu} \partial^y_{\nu'}$</span>.</p>
<p>Inner product is invariant under coordinate change.</p>
<p>For <span class="math-container">$v, w\in T_p U, v = v^\mu \partial^x_\mu = v^\mu \frac{\partial y ^ {\mu'}}{\partial x^\mu} \partial^y_{\mu'},
w = w^\nu \partial^x_\nu = w^\nu \frac{\partial y ^ {\nu'}}{\partial x^\nu} \partial^y_{\nu'}$</span>.</p>
<p><span class="math-container">$\langle v, w\rangle = g_{\mu\nu}v^\mu w^\nu = g'_{\mu'\nu'}\frac{\partial y ^ {\mu'}}{\partial x^\mu}\frac{\partial y ^ {\nu'}}{\partial x^\nu} v^\mu w^\nu $</span>, thus
<span class="math-container">$g'_{\mu'\nu'} = \frac{\partial x^\mu}{\partial y ^ {\mu'}} \frac{\partial x^\nu}{\partial y ^ {\nu'}}g_{\mu\nu}$</span>. <span class="math-container">$g$</span> is a tensor.</p>
<p><span class="math-container">$\Gamma_{\mu \nu}^{\lambda}=\frac{1}{2} g^{\lambda \rho}(\partial_{\mu}g_{\nu \rho} +\partial_{\nu}g_{\rho \mu}-\partial_{\rho}g_{\mu \nu}),
{\Gamma'}_{\mu' \nu'}^{\lambda'}=\frac{1}{2} {g'}^{\lambda' \rho'}(\partial_{\mu'} {g'}_{\nu' \rho'} +\partial_{\nu'} {g'}_{\rho' \mu'}-\partial_{\rho'}{g'}_{\mu' \nu'})$</span>.</p>
<p><span class="math-container">$\partial_{\alpha '}g_{\beta ' \gamma '}'
=\frac{\partial x^{\alpha } }{\partial y^{\alpha '} }\frac{\partial x^{\beta } }{\partial y^{\beta '} }\frac{\partial x^{\gamma } }{\partial y^{\gamma'}} \partial_\alpha g_{\beta \gamma}
+ g_{\beta \gamma} (\frac{\partial x^{\gamma} }{\partial y^{\gamma'} }\frac{\partial^{2} x^{\beta } }{\partial y^{\alpha '}\partial y^{\beta '} }+ \frac{\partial x^{\gamma } }{\partial y^{\beta '} }\frac{\partial^{2} x^{\beta } }{\partial y^{\alpha '}\partial y^{\gamma '}})$</span>
<span class="math-container">\begin{align}\notag
&\partial_{\mu'} {g'}_{\nu' \rho'} + \partial_{\nu'} {g'}_{\rho' \mu'}-\partial_{\rho'}{g'}_{\mu' \nu'}\\
\notag&=\frac{\partial x^{\mu } }{\partial y^{\mu'} }\frac{\partial x^{\nu } }{\partial y^{\nu '} }\frac{\partial x^{\rho } }{\partial y^{\rho'}} \partial_\mu g_{\nu \rho}
+ g_{\nu \rho} (\frac{\partial x^{\rho} }{\partial y^{\rho'} }\frac{\partial^{2} x^{\nu } }{\partial y^{\mu '}\partial y^{\nu '} }+ \frac{\partial x^{\rho } }{\partial y^{\nu '} }\frac{\partial^{2} x^{\nu } }{\partial y^{\mu '}\partial y^{\rho '}})\\
\notag&+ \frac{\partial x^{\nu } }{\partial y^{\nu '} }\frac{\partial x^{\rho } }{\partial y^{\rho '} }\frac{\partial x^{\mu } }{\partial y^{\mu'}} \partial_\nu g_{\rho \mu}
+ g_{\rho \mu} (\frac{\partial x^{\mu} }{\partial y^{\mu'} }\frac{\partial^{2} x^{\rho } }{\partial y^{\nu '}\partial y^{\rho '} }+ \frac{\partial x^{\mu} }{\partial y^{\rho '} }\frac{\partial^{2} x^{\rho } }{\partial y^{\nu '}\partial y^{\mu '}})\\
\notag&-\frac{\partial x^{\rho } }{\partial y^{\rho '} }\frac{\partial x^{\mu } }{\partial y^{\mu '} }\frac{\partial x^{\nu } }{\partial y^{\nu'}} \partial_\rho g_{\mu \nu}
- g_{\mu\nu} (\frac{\partial x^{\nu} }{\partial y^{\nu'} }\frac{\partial^{2} x^{\mu } }{\partial y^{\rho'}\partial y^{\mu'} }+ \frac{\partial x^{\nu} }{\partial y^{\mu'} }\frac{\partial^{2} x^{\mu} }{\partial y^{\rho '}\partial y^{\nu'}})\\
\notag&=\frac{\partial x^{\mu } }{\partial y^{\mu'} }\frac{\partial x^{\nu } }{\partial y^{\nu '} }\frac{\partial x^{\rho } }{\partial y^{\rho'}}(\partial_\mu g_{\nu \rho}+\partial_\nu g_{\rho \mu}-\partial_\rho g_{\mu \nu})+g_{\nu \rho} (\frac{\partial x^{\rho} }{\partial y^{\rho'} }\frac{\partial^{2} x^{\nu } }{\partial y^{\mu '}\partial y^{\nu '} }+ \frac{\partial x^{\rho } }{\partial y^{\nu '} }\frac{\partial^{2} x^{\nu } }{\partial y^{\mu '}\partial y^{\rho '}}\\
\notag&\quad+\frac{\partial x^{\rho} }{\partial y^{\mu'} }\frac{\partial^{2} x^{\nu } }{\partial y^{\nu '}\partial y^{\rho '} }+ \frac{\partial x^{\rho} }{\partial y^{\rho '} }\frac{\partial^{2} x^{\nu } }{\partial y^{\nu '}\partial y^{\mu '}}
-\frac{\partial x^{\rho} }{\partial y^{\nu'} }\frac{\partial^{2} x^{\nu } }{\partial y^{\rho'}\partial y^{\mu'} }- \frac{\partial x^{\rho} }{\partial y^{\mu'} }\frac{\partial^{2} x^{\nu} }{\partial y^{\rho '}\partial y^{\nu'}})\\
\notag&=\frac{\partial x^{\mu } }{\partial y^{\mu'} }\frac{\partial x^{\nu } }{\partial y^{\nu '} }\frac{\partial x^{\rho } }{\partial y^{\rho'}}(\partial_\mu g_{\nu \rho}+\partial_\nu g_{\rho \mu}-\partial_\rho g_{\mu \nu})
+2 g_{\nu \rho} \frac{\partial x^{\rho} }{\partial y^{\rho'} }\frac{\partial^{2} x^{\nu } }{\partial y^{\mu '}\partial y^{\nu '} }
\end{align}</span>
<span class="math-container">\begin{align}\notag
{\Gamma'}_{\mu' \nu'}^{\lambda'}
&=\frac{1}{2} {g'}^{\lambda' \rho'}(\partial_{\mu'} {g'}_{\nu' \rho'} +\partial_{\nu'} {g'}_{\rho' \mu'}-\partial_{\rho'}{g'}_{\mu' \nu'})\\
\notag&=\frac{1}{2} (\frac{\partial y ^ {\lambda'}}{\partial x^\lambda} \frac{\partial y ^ {\rho'}}{\partial x^\beta}g^{\lambda\beta})(\frac{\partial x^{\mu } }{\partial y^{\mu'} }\frac{\partial x^{\nu } }{\partial y^{\nu '} }\frac{\partial x^{\rho } }{\partial y^{\rho'}}(\partial_\mu g_{\nu \rho}+\partial_\nu g_{\rho \mu}-\partial_\rho g_{\mu \nu})
+2 g_{\nu \rho} \frac{\partial x^{\rho} }{\partial y^{\rho'} }\frac{\partial^{2} x^{\nu } }{\partial y^{\mu '}\partial y^{\nu '} })\\
\notag&=\frac{1}{2}g^{\lambda\beta}\delta_\beta^\rho \frac{\partial y ^ {\lambda'}}{\partial x^\lambda}\frac{\partial x^{\mu } }{\partial y^{\mu'} }\frac{\partial x^{\nu } }{\partial y^{\nu '}} (\partial_\mu g_{\nu \rho}+\partial_\nu g_{\rho \mu}-\partial_\rho g_{\mu \nu})+
g_{\nu \rho}g^{\lambda\beta}\delta_\beta^\rho \frac{\partial y ^ {\lambda'}}{\partial x^\lambda} \frac{\partial^{2} x^{\nu } }{\partial y^{\mu '}\partial y^{\nu '}}\\
\notag&=\frac{\partial y ^ {\lambda'}}{\partial x^\lambda}\frac{\partial x^{\mu } }{\partial y^{\mu'} }\frac{\partial x^{\nu } }{\partial y^{\nu '}} \frac{1}{2}g^{\lambda\rho}(\partial_\mu g_{\nu \rho}+\partial_\nu g_{\rho \mu}-\partial_\rho g_{\mu \nu})+\frac{\partial y ^ {\lambda'}}{\partial x^\lambda} \frac{\partial^{2} x^{\lambda } }{\partial y^{\mu '}\partial y^{\nu '}}\\
\notag&=\frac{\partial y ^ {\lambda'}}{\partial x^\lambda}\frac{\partial x^{\mu } }{\partial y^{\mu'} }\frac{\partial x^{\nu } }{\partial y^{\nu '}}\Gamma^\lambda_{\mu\nu}+\frac{\partial y ^ {\lambda'}}{\partial x^\lambda} \frac{\partial^{2} x^{\lambda } }{\partial y^{\mu '}\partial y^{\nu '}}
\end{align}</span>
Thus Christoffel symbol is not a tensor.</p>
|
3,807,900 | <p>I have just met this exercise in functional analysis, asking us to determine if these two subspaces of the Hilbert space <span class="math-container">$\ell^2$</span> of square-summable complex sequences are closed:</p>
<blockquote>
<ol>
<li>The set of all sequences <span class="math-container">$\{x_n\}_{n=1}^{\infty}$</span> satisfying
<span class="math-container">$$\sum_{n=1}^{\infty} \frac{1}{n} x_n = 0 $$</span></li>
</ol>
</blockquote>
<blockquote>
<ol start="2">
<li>The set of all sequences <span class="math-container">$\{x_n\}_{n=1}^{\infty}$</span> satisfying
<span class="math-container">$$\sum_{n=1}^{\infty} x_n = 0 $$</span></li>
</ol>
</blockquote>
<p>I know what I am supposed to do: to prove the subspace is closed I need to consider a general Cauchy sequence in the subspace and show its limit is also in the subspace and to prove it is not closed I only need to find one Cauchy sequence in the subspace whose limit is not in it. However, these two subspaces have me stuck, I do not know if they are closed or not so I have no idea on this. I thank all helpers.</p>
| Brian Moehring | 694,754 | <p>For (1), as has been noted by past answers, we can use Cauchy-Schwarz to see that <span class="math-container">$f(x)= \sum_{n=1}^\infty \frac{1}{n}x_n$</span> is continuous <span class="math-container">$\ell^2 \to \mathbb{C}$</span>, so the set <span class="math-container">$\{x \in \ell^2 \mid f(x) = 0\} = f^{-1}(0)$</span> is closed.</p>
<p>For (2), consider the sequence <span class="math-container">$x(k) \in \ell^2$</span> of the form
<span class="math-container">$$x(k)_n = \begin{cases} 1 & n=1 \\ -\frac{1}{k} & 2 \leq n \leq k+1 \\ 0 & n\geq k+2 \end{cases}$$</span>
Then for all <span class="math-container">$k\geq 1$</span>, we have <span class="math-container">$\sum_{n=1}^\infty x(k)_n = 0$</span> and <span class="math-container">$\sum_{n=2}^\infty x(k)_n^2 = \frac{1}{k} \to 0,$</span> so <span class="math-container">$x(k) \to (1,0,0,0,\ldots)$</span> in <span class="math-container">$\ell^2$</span>, but <span class="math-container">$(1,0,0,0,\ldots) \not\in \{x \in \ell^2 \mid \sum_{n=1}^\infty x_n = 0\}$</span>, so the subspace is not closed.</p>
|
176,488 | <p><strong>Summary:</strong> My question, in a nutshell, is how we should intuitively imagine a generic real number (as opposed to a random one), and whether we can construct numbers which empirically behave like generic numbers in the same way that $e$ or $\pi$ behave empirically like random ones. I hope this is not too vague, informal or philosophical for MO. Let me explain what I'm asking in greater detail.</p>
<p><strong>Background:</strong> First recall the classical "duality" between (Lebesgue) measure and (topological) category: a subset of $\mathbb{R}$ is said to be "negligible" iff it is of Lebesgue measure zero, and "meager" iff it is contained in a countable union of nowhere dense closed sets (i.e., closed sets with empty interior). By Lebesgue measure theory, resp. by the Baire category theorem, a negligible, resp. meager set has empty interior. Both are "small" in a certain sense, but in an incompatible way since $\mathbb{R}$ is the union of a negligible and a meager set (as a simple example is given below). There is also a classical theorem by Erdős (refining an earlier result by Sierpiński) showing that, under the Continuum Hypothesis, there is an involution of $\mathbb{R}$ to itself which takes negligible subsets to meager ones and conversely.</p>
<p>[What follows is badly written: jump to "edit/clarification" below for an attempt at saying things more clearly.]</p>
<p>There are various meanings of the word "random", but the general flavor is that a real number is random iff it does not belong to a negligible Borel set which can somehow be described or coded in a simple way (e.g., random over a transitive model of set theory means that it does not belong to a negligible Borel set coded by a sequence in that model; but there are some weaker meanings of "random" where we forbid belonging only to negligible Borel sets of simpler description, e.g., those which can be described by a code computable by a Turing machine). For example, a random number will be normal in every base, because the set of real numbers which are not normal in every base is a simply described Borel set which is negligible.</p>
<p>Now a real number like $e$ or $\pi$ is not random even in the weakest sense, because it is, well, equal to $e$ or $\pi$, and that is not random (it belongs to the Borel set $\{e\}$ or $\{\pi\}$ which is negligible and certainly computable). It does seem to be, however, "empirically random", in a sense that we (or at least, I) don't know how to make precise, but the idea being that it won't belong to any simply defined negligible Borel set which hasn't been specifically constructed to contain it. For example, it is conjectured that $e$ and $\pi$ are normal in every base: we expect their decimals to pass statistical tests of randomness. The same holds for a huge number of "naturally defined" real numbers (and not just real numbers: see <a href="https://mathoverflow.net/questions/156301/distribution-of-digits-of-pq-adic-idempotents-aka-automorphic-numbers">this question</a> for another case). Philosophically, it is also generally expected that the real number whose binary expansion is obtained by flipping an unbiased coin (or, better, taking some physical source of randomness) will be random in a strong sense. So, anyway, we have a good intuition of what a random number feels like.</p>
<p>The dual notion of a "generic" number, however, is more obscure: a real number is generic iff it does not belong to a meager Borel set which can be somehow described or coded in a simple way (e.g., belonging to a transitive model of set theory, or Turing-computable, or something like this).</p>
<p>Here is an example of something we can say about generic numbers: call a real number (between $0$ and $1$, say) an "oft-repeater in base $b$" iff its expansion in base $b$ repeats an infinite number of times <em>all</em> the digits up to that point. In other words, there exist arbitrarily large $n$ such that the digits $c_n$ to $c_{2n-1}$ are equal to $c_0$ to $c_{n-1}$. It is easy to see that the set of oft-repeaters in base $b$ (and therefore, in any base) is negligible but comeager (=contained in a countable intersection of open dense subsets): so a random real number is <em>not</em> an oft-repeater in any base, but a generic real number <em>is</em> an oft-repeater in every base.</p>
<p>We don't know this either way, but I don't think anyone would seriously conjecture that $e$ or $\pi$ is an oft-repeater in any base: clearly we expect them to be empirically random and <em>not</em> empirically generic. Or in other words, we expect measure theory to be a better predictor of what $e$ and $\pi$ behave like than category. An "empirically generic" real number, however, would be (among many other things) an oft-repeater in every base (and it would <em>not</em> be normal in any base: for example, in a generic number, there are arbitrarily large $n$ such that all digits $c_n$ to $c_{n^2}$ are zero — the set of normal reals is meager).</p>
<p><strong>Questions:</strong> So, my questions are something like this:</p>
<ul>
<li><p>Are there "naturally defined" real numbers which are "empirically generic" rather than "empirically random"? Or at least, can we give some examples of (non-"naturally defined") such numbers?</p></li>
<li><p>Is there some kind of process (physical or idealized), analog to throwing a coin, that would produce a (somewhat!) generic real number?</p></li>
<li><p>Is there a philosophical argument explaining why measure theory predicts better than category how the numbers naturally encountered in mathematics behave? Why should we expect $e$ and $\pi$ to behave more randomly than generically (when, in fact, they are neither)?</p></li>
<li><p>How can one intuitively visualize a generic random number? (I think I can picture a random one, and the idea of it being normal makes sense, but the fact that a generic random number has infinitely large $n$ such that all digits $c_n$ to $c_{n^2}$ are zero seems very difficult to imagine.)</p></li>
<li><p>How would one even test empirically if a given real number is generic? (Assume you have a true generic oracle and a fake one: how would you proceed to detect which is the true one? Can we have a "genericity test" like we have randomness tests?)</p></li>
</ul>
<p>Perhaps the idea that genericity should behave symmetrically to randomness is naïve: please don't hesitate to tell me why this is naïve!</p>
<hr>
<p><strong>Edit/Clarification:</strong> The above discussion was probably too messy or informal. Let me try to give a clearer restatement:</p>
<p><strong>Definition:</strong> If $\mathscr{T}$ is a set of Turing degrees, a real number is said to be $\mathscr{T}$-random, resp. $\mathscr{T}$-generic, iff it does not belong to any negligible, resp. meager, Borel set which can be coded by a sequence whose Turing degree is in $\mathscr{T}$ ("coding" of Borel sets being done, say, as in Jech's <em>Set Theory</em>).</p>
<p>If $\mathscr{T}$ is the set of degrees belonging to some transitive model $\mathfrak{M}$ of ZFC, the reals in question are said to be random over $\mathfrak{M}$ resp. Cohen/generic over $\mathfrak{M}$ (cf. Jech, definition 26.3 and lemma 26.4 in the Third Millennium edition). If $\mathscr{T}$ is simply the degree $\mathbf{0}$ of Turing computability, I think we get a definition equivalent to Martin-Löf random numbers, and something analogous for "generic": of course, this is a much weaker property than being random, resp. generic, over a model of ZFC. We could perhaps define even weaker versions of "random", resp. "generic", by replacing $\mathscr{T}$ by a set of finer degrees, maybe primitive recursive degrees (but if the degrees are too fine, then the definition will become too sensitive on how Borel sets are coded and probably not the right way to proceed).</p>
<p><strong>Question number 1:</strong> While I think I have an intuitive grasp of how a random number real behaves (irrespective of what $\mathscr{T}$ is), e.g., by imagining an coin being tossed an infinite number of times, the corresponding "generic" notion is much more obscure. Is there some way to picture it intuitively?</p>
<p>Now there is the matter of numbers like $e$ and $\pi$. Of course <strong>these numbers are not random (nor are they generic)</strong> in the sense of the above definition, or even of any reasonable weakening I can imagine.</p>
<p>Nevertheless, $e$ and $\pi$ behave <em>in certain ways</em> like random real numbers, and I claim that they behave "more like random reals than like generic reals". For example, if we are to make a conjecture as to the lim.sup. and lim.inf. of the sequence $\frac{1}{n}\sum_{k=0}^{n-1} c_k$ where $c_k$ denotes the $k$-th binary digit of $\pi$, and if Pr. Eugsebel predicts "I conjecture that the limit is $\frac{1}{2}$, because the set of real numbers for which this is the case is of full measure", while Pr. Eriab predicts "I conjecture that the lim.inf. is $0$ and the lim.sup. is $1$, because the set of real numbers for which this is the case is comeager", then experimentally, it appears that Pr. Eugsebel is right and Pr. Eriab is wrong: measure theory seems to predict the <em>empirical</em> behavior of the decimals of $\pi$ correctly, and category does not; alternatively, $\pi$ behaves <em>empirically</em> like a random number (even though it is not at all random!), in this limited respect, and it <em>does not</em> behave empirically like a generic number.</p>
<p><strong>Informal definition</strong> (which probably cannot be made rigorous): Say that a real number is "empirically random" when it behaves like a random real number for this kind of simple tests. (Perhaps "pseudorandom" would be a better term for this.) For example, an "empirically random" real number should, at least, be normal in any base (note that the set of real numbers that are normal in any base is of full measure). Analogously, we want to define a number to be "empirically generic" (or "pseudogeneric") when it behaves like a generic real number. For example, an "empirically generic" real number should at least be an "oft-repeater" in any base (meaning that there exist arbitrarily large $n$ such that the digits $c_n$ to $c_{2n-1}$ are equal to $c_0$ to $c_{n-1}$); also, the lim.sup. and lim.inf. of the sequence $\frac{1}{n}\sum_{k=0}^{n-1} c_k$ where $c_k$ denotes the $k$-th binary digit of the number should be $0$ and $1$ (note that the set of real numbers satisfying these criteria comeager).</p>
<p><strong>Question number 2:</strong> Whereas $e$, $\pi$ and many others can reasonably be conjectured to be "empirically random", is there, dually, any real number that has been explicitly defined in mathematics that one can reasonably expect to be "empirically generic"? Or could one be defined?</p>
<p>(By "explicitly", I mean to forbid something like "take some number outside of the union of all meager Borel sets with a computable code": this would indeed define a generic real number, not just an "empirically generic" one, but this is not explicit by any means.)</p>
<p>The underlying philosophical question is something like this: "How come is it that randomness appears to be a much more natural notion than genericity?" But I don't really expect anyone to have an answer to that.</p>
| Robert Israel | 13,650 | <p>You can't really have a "given" empirically-generic number, because any explicit description of $x$ can be turned into an explicit description of the meagre closed set $\{x\}$ to which it belongs.</p>
|
2,003,916 | <p>Probably a very simple question:</p>
<p>Suppose a hospital orders defibrillators from a manufacturer. It is well known that defibrillations are often not effective, even when the defibrillators themselves are working properly. Suppose research shows that only 15% percent of defibrillations are effective. Over the next few months the hospital performs about 3000 defibrillations. What is the probability that none of these are successful?</p>
| Bill Dubuque | 242 | <p>It's a special case of the following homogeneous generalization of the Euler-Fermat theorem.</p>
<p><strong>Theorem</strong> <span class="math-container">$ $</span> Suppose that <span class="math-container">$\rm\ n\in \mathbb N\ $</span> has the prime factorization <span class="math-container">$\rm\:n = p_1^{e_{\:1}}\cdots\:p_k^{e_k}\ $</span> and suppose that for all <span class="math-container">$\rm\,i,\,$</span> <span class="math-container">$\rm\ e\ge e_i\ $</span> and <span class="math-container">$\rm\ \phi(p_i^{e_{\:i}})\mid f.\ $</span> Then <span class="math-container">$\rm\ n\mid (ab)^e\,(a^f-b^f)\ $</span> for all <span class="math-container">$\rm\: a,b\in \mathbb Z.$</span></p>
<p><strong>Proof</strong> <span class="math-container">$\ $</span> Notice that if <span class="math-container">$\rm\ p_i\mid ab\ $</span> then <span class="math-container">$\rm\:p_i^{e_{\:i}}\ |\ (ab)^e\ $</span> by <span class="math-container">$\rm\ e_i \le e.\: $</span> Else <span class="math-container">$\rm\:a\:$</span> and <span class="math-container">$\rm\:b\:$</span> are coprime to <span class="math-container">$\rm\: p_i\:$</span> so by Euler's phi theorem, <span class="math-container">$\rm\ mod\ q = p_i^{e_{\:i}}\!: \ a^{\phi(q)}\equiv 1\equiv b^{\phi(q)} \Rightarrow\ a^f\equiv 1\equiv b^f\: $</span> by <span class="math-container">$\rm\: \phi(q)\mid f.\ $</span> Thus since all <span class="math-container">$\rm\ p_i^{e_{\:i}}\ |\ (ab)^e\ (a^f - b^f\:)\ $</span> <a href="https://math.stackexchange.com/a/3841440/242">so too does their lcm = product</a> = <span class="math-container">$\rm\: n.$</span></p>
<p><strong>Corollary</strong> <span class="math-container">$\, \rm[e_i=1]\quad\ \ \ n\mid ab(a^f\,-\,b^f)\ $</span> if <span class="math-container">$\,n\,$</span> is squarefree and prime <span class="math-container">$\,\rm p\mid n\,\Rightarrow\, p\!-\!1\mid f$</span></p>
<p><strong>Corollary</strong> <span class="math-container">$\,\rm[OP]\quad 2\cdot 3\cdot p\mid ab(a^{p-1}\! - b^{p-1})\: $</span> for all <span class="math-container">$\rm\:a,b\in \mathbb Z,\:$</span> prime <span class="math-container">$\rm\:p>3$</span></p>
<p><strong>Corollary <span class="math-container">$3\ $</span></strong> <span class="math-container">$ $</span> If we restrict the theorem to all <span class="math-container">$\,a,b\,$</span> such that <span class="math-container">$\,p_1\mid a\!\iff\! p_1\mid b\,$</span> then we can relax <span class="math-container">$\,e\ge e_1$</span> to <span class="math-container">$\,2e+f\ge e_1$</span> since then we get <span class="math-container">$\,p_1^{2e+f}\mid (ab)^e(a^f-b^f)$</span> in the case one (so both) of <span class="math-container">$\,a,b\,$</span> are divisible by <span class="math-container">$\,p.\,$</span> Instead of Euler <span class="math-container">$\phi$</span> we we can use <a href="https://en.wikipedia.org/wiki/Carmichael_function" rel="nofollow noreferrer">Carmichael <span class="math-container">$\,\lambda$</span></a> (= half of <span class="math-container">$\phi$</span> for <span class="math-container">$2^e\ge 8),\,$</span> <a href="https://math.stackexchange.com/q/4054036/242">e.g. if <span class="math-container">$e=1,f = 2, n = 2^3\cdot 3\,$</span></a> then <span class="math-container">$\,24\mid ab(a^2-b^2)\,$</span> when <span class="math-container">$\,2\mid a\!\iff\! 2\mid b\,$</span> since <span class="math-container">$\,4=2e+f \ge e_1 = 3,\,$</span> and <span class="math-container">$\lambda(2^3)=2\mid f$</span>.</p>
|
3,811,498 | <p>Solve, by bringing the equation to Bernoulli form:</p>
<p><span class="math-container">$$
y’ = \frac{2-xy^3}{3x^2y^2}
$$</span></p>
<hr />
<p>Therefore we want to bring it to a form like:</p>
<p><span class="math-container">$$
y’ + p(x)y = q(x)y^n
$$</span></p>
<p>So working with the equation i get:</p>
<p><span class="math-container">$$
y’ - \frac{2}{3x^2y^2} = -\frac{xy^3}{3x^2y^2} = -\frac{y}{3x}
$$</span></p>
<p>I don’t see how to get to Bernoulli equation from here...</p>
| Lutz Lehmann | 115,115 | <p>You selected the wrong term. Exchange the position of terms formerly on the right side to get
<span class="math-container">$$
y'+\frac1{3x}y = \frac2{3x^2}y^{-2}.
$$</span></p>
<p>With some contemplation of the formula you could also directly detect that
<span class="math-container">$$
(xy^3)'=3xy^2y'+y^3=\frac2x.
$$</span></p>
|
4,539,043 | <p>I have tried to prove this statement by utilizing the proof by cases method. My cases are (1)<span class="math-container">$x=6$</span>, (2)<span class="math-container">$x>6$</span> and (3)<span class="math-container">$x<6$</span>.</p>
<p>For (3) for some reason it's not true</p>
<p>Case (1):
For <span class="math-container">$x>6$</span>, I know that <span class="math-container">$x^2+x>11$</span> is true</p>
<p>Case (2):
For <span class="math-container">$x=6$</span>, clearly <span class="math-container">$36>5$</span></p>
<p>Case (3):</p>
<p>For <span class="math-container">$x<6$</span>, <span class="math-container">$x^2+x>11$</span> it is not true. (e.g 1,2,3...)</p>
<p>Thanks</p>
| Bob Dobbs | 221,315 | <p>For your case <span class="math-container">$(3)$</span>, notice that if <span class="math-container">$x<6$</span> then <span class="math-container">$|x-6|=6-x$</span>. Then, your inequality becomes
<span class="math-container">$$x^2+6-x>5$$</span>
<span class="math-container">$$x^2-x+1>0$$</span>
<span class="math-container">$$(x-\frac{1}{2})^2+\frac{3}{4}>0$$</span>
which is true.</p>
|
876,310 | <p>So I <em>think</em> I understand what differentials are, but let me know if I'm wrong.</p>
<p>So let's take $y=f(x)$ such that $f: [a,b] \subset \Bbb R \to \Bbb R$. Instead of defining the derivative of $f$ in terms of the differentials $\text{dy}$ and $\text{dx}$, we take the derivative $f'(x)$ as our "primitive". Then to define the differentials we do as follows:</p>
<p>We find some $x_0 \in [a,b]$ where there is some neighborhood of $x_0$, $N(x_0)$, such that all $f(x)$ in $\{f(x) \in \Bbb R \mid x \in N(x_0)\}$ are differentiable. Then we choose another point in $N(x_0)$, let's call it $x_1$, such that $x_1 \ne x_0$. Then let $dx = \Delta x = x_1 - x_0$. Now this $\Delta x$ doesn't actually have to be very small like we're taught in Calculus 1 (in particular it's not infinitesimal, it's finite). In fact, as long as $f(x)$ is differentiable for all $x \in [-10^{10}, 10^{10}]$ we could choose $x_0 = -10^{10}$ and $x_1 = 10^{10}$.</p>
<p>Then we know that $\Delta y = f'(x_0) \Delta x + \epsilon(\Delta x)$, where $\epsilon(\Delta x)$ is some nonlinear function of $\Delta x$. If $f(x)$ is smooth, we know that $\epsilon(\Delta x)$ is equal to the sum of powers of $\Delta x$ with some coefficients, by Taylor's theorem. But of course, $\epsilon(\Delta x)$ won't be so easy to describe if $f(x)$ is only once differentiable. So we define $dy$ as $dy = f'(x_0) dx$: that is, $dy$ is the <em>linear part</em> of $\Delta y$. This has the very useful property that $\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \frac{dy}{dx} = f'(x_0)$. This is then <em>not a definition</em> of the derivative, but a consequence of our definitions.</p>
<p>It can be seen from this $dy$ really depends on what we choose as $dx$, but $f'$ is independent of both. </p>
<p>This definition can be extended to functions of multiple variables, like $z = f(x, y)$ as well, by letting $\Delta x = dx,\ \Delta y=dy$ and defining $dz$ as $dz = \frac{\partial f(x_0, y_0)}{\partial x}dx + \frac{\partial f(x_0, y_0)}{\partial y} dy$. So $dz$ is the linear part of $\Delta z$. Does all of the above look correct?</p>
<p>If so, then where I'm having a problem is: <br>1) how then <em>do</em> we define the derivative of $f(x)$ if not by $f'(x_0) = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$? <br>2) how do we apply this definition of $dx$ to $\int_a^b f(x)dx$? It seems like the inherit arbitrariness of $dx$ is really going to get in the way of a good definition of the integral.</p>
| user121955 | 121,955 | <p>Differentials are infinitely small changes in x or y. For instance, the concept of the integral is the sum of the areas of an infinite number of rectangles under a curve. The height of each is f(x) and the width is dx.</p>
|
1,794,072 | <p>My attempt :</p>
<p>If $n$ is odd, then the square must be 2 (mod 3), which is not possible.</p>
<p>Hence $n =2m$</p>
<p>$2^{2m}+3^{2m}=(2^m+a)^2$</p>
<p>$a^2+2^{m+1}a=3^{2m}$</p>
<p>$a (a+2^{m+1})=3^{2m} $</p>
<p>By fundamental theorem of arithmetic, </p>
<p>$a=3^x $</p>
<p>$3^x +2^{m+1}=3^y $</p>
<p>$2^{m+1}=3^x (3^{y-x}-1) $</p>
<p>Which is not possible by Fundamental theorem of Arithmetic </p>
<p>Is there a better method? I even have a nagging feeling this is wrong as only once is the equality of powers of 2 and 3 $(n) $are used.</p>
| S.C.B. | 310,930 | <p>Your method is slightly wrong, as you have to deal with the $x=0$ case. However, the remaining diophantine equation is simple. </p>
<p>Here, I provide you with an alternative approach, though admittedly yours seems better. </p>
<p>For the case $m \equiv 0 \pmod 2$, note that $$2^{2m}+3^{2m}=n^2$$ implies that $2^m=2ab$, and $3^m=a^2-b^2$ where $\gcd(a,b)=1, a \not \equiv b \pmod 2$. This implies that $a=2^{m-1}, b=1$. So all that remains is finding $m,y$ such that $3^{m}=2^{2m-2}-1$. </p>
|
1,794,072 | <p>My attempt :</p>
<p>If $n$ is odd, then the square must be 2 (mod 3), which is not possible.</p>
<p>Hence $n =2m$</p>
<p>$2^{2m}+3^{2m}=(2^m+a)^2$</p>
<p>$a^2+2^{m+1}a=3^{2m}$</p>
<p>$a (a+2^{m+1})=3^{2m} $</p>
<p>By fundamental theorem of arithmetic, </p>
<p>$a=3^x $</p>
<p>$3^x +2^{m+1}=3^y $</p>
<p>$2^{m+1}=3^x (3^{y-x}-1) $</p>
<p>Which is not possible by Fundamental theorem of Arithmetic </p>
<p>Is there a better method? I even have a nagging feeling this is wrong as only once is the equality of powers of 2 and 3 $(n) $are used.</p>
| Ant | 66,711 | <p>This is more convoluted than your way, but I already wrote it so I'll just leave it here :-) </p>
<p>$4^m = (a-3^m)(a+3^m) \implies a-3^m = 2^p, a+3^m = 2^q$ with $p+q = 2m$. Note that $q \ge p$</p>
<p>Then $a = 2^p + 3^m = 2^q - 3^m \iff 2^p (2^{q-p}-1) = 2\cdot 3^m$</p>
<p>But then it must be $p=1$, $q = 2m-1$, and we have </p>
<p>$$2^{2(m-1)}-1 = (2^{m-1}-1)(2^{m-1}+1) = 3^m$$</p>
<p>Since only one of the factors in the LHS can be divisible by $3$, it means the other one must be $1$, hence we get $2^{m-1}-1 = 1 \implies m = 2$</p>
<p>But in this case $2^{m-1}+1 = 3 \neq 3^2 $, so this is impossible.</p>
|
105,750 | <p>Given a <code>ContourPlot</code> with a set of contours, say, this:</p>
<p><a href="https://i.stack.imgur.com/cKoyo.jpg"><img src="https://i.stack.imgur.com/cKoyo.jpg" alt="enter image description here"></a></p>
<p>is it possible to get the contours separating domains with the different colors in the form of lists? </p>
<p>For example, how to extract the boundaries of the blue domain in the image above?
Or just for the sake of trial, from such a simple example:</p>
<pre><code> ContourPlot[x*Exp[-x^2 - y^2], {x, 0, 3}, {y, -3, 3},
PlotRange -> {0, 0.5}, ColorFunction -> "Rainbow"]
</code></pre>
<p><a href="https://i.stack.imgur.com/Beuzu.jpg"><img src="https://i.stack.imgur.com/Beuzu.jpg" alt="enter image description here"></a></p>
<p>The same task, let us find the lists corresponding to the blue domain boundaries.</p>
<p>To make it clear, I am not asking of how to get the lines from the function behind. This I understand. I ask of how to extract the contour lines that are generated by Mma.</p>
<p>Let us put this question another way around. Is it possible to define the areas with the same color as separate geometric regions in the sense of the computation geometry, and then work with these domains separately?</p>
| Jason B. | 9,490 | <p>I don't know how to do this in an automated way, but here is something at least:</p>
<p>Make your plot, extract the lines, convert them to regions, and then take the <code>RegionDifference</code> between them</p>
<pre><code>plot = ContourPlot[x*Exp[-x^2 - y^2], {x, 0, 3}, {y, -3, 3},
PlotRange -> {0, 0.5}, ColorFunction -> "Rainbow"]
points = Cases[Normal@plot, Line[pts__] -> pts, Infinity];
regions = BoundaryMeshRegion[#, Line[Range[Length@#]]] & /@ points;
regiondiffs = RegionDifference[#2, #1] & @@@ Partition[regions, 2, 1];
PrependTo[regiondiffs, regions[[1]]]
</code></pre>
<p><a href="https://i.stack.imgur.com/2L64D.png" rel="noreferrer"><img src="https://i.stack.imgur.com/2L64D.png" alt="enter image description here"></a></p>
<pre><code>Show[Table[
RegionPlot[regiondiffs[[n]], PlotStyle -> Hue[n/8]], {n, 8, 1, -1}]]
</code></pre>
<p><a href="https://i.stack.imgur.com/ViSed.png" rel="noreferrer"><img src="https://i.stack.imgur.com/ViSed.png" alt="enter image description here"></a></p>
<p><strong>Edit</strong>
@AlexeiBoulbitch - I had tried to extract the points without <code>Normal</code> at first, and the result was a list of integers rather than <code>{x,y}</code> coordinates. And Szabolcs had just <a href="https://mathematica.stackexchange.com/a/105653/9490">shown me</a> that <code>Normal</code> is useful when extracting regions from contour plots so I tried it. </p>
<p>What <code>Normal</code> is doing is removing the <code>GraphicsComplex</code> head:</p>
<blockquote>
<p>GraphicsComplex[{$pt_1$, $pt_2$ ,...},$data$]
represents a graphics complex in which coordinates given as integers i
in graphics primitives in data are taken to be $pt_i$. >></p>
</blockquote>
<p>So after using <code>Normal</code>, you get the actual coordinates for the lines.</p>
|
718,748 | <p>How can I calculate the volume of the solid under the surface $z = 6x + 4y + 7$ and above the plane $z = 0$ over a given rectangle $R = \{ (x, y): -4 \leq x \leq 1, 1 \leq y \leq 4 \}$?</p>
<p>I know I have to integrate some function, but since the surface takes both positive and negative values I don't know what to integrate.</p>
| gt6989b | 16,192 | <p><strong>Hint</strong></p>
<ul>
<li>Find the equation of the curve of the surface where you have $z=0$ over your region (that will be a curve in $x,y$).</li>
<li>Now draw the region with your curve there and indicate where you have $z>0$ and where $z<0$.</li>
<li>Set up the double integral just over the piece where $z > 0$.</li>
<li>Evaluate</li>
</ul>
|
718,748 | <p>How can I calculate the volume of the solid under the surface $z = 6x + 4y + 7$ and above the plane $z = 0$ over a given rectangle $R = \{ (x, y): -4 \leq x \leq 1, 1 \leq y \leq 4 \}$?</p>
<p>I know I have to integrate some function, but since the surface takes both positive and negative values I don't know what to integrate.</p>
| kokocijo | 136,606 | <p><strong>Hint</strong></p>
<ul>
<li>Graph the $x$, $y$, and $z$ intercepts of the given surface (plane) to see where in the $xy$ plane you have $z>0$.</li>
<li>Adjust the rectangle you are integrating over to only cover area where $z>0$. It may help to draw the rectangle in the $xy$ plane. (It looks like in this question $y$ needs no adjustment, but $x$ needs to be within a smaller interval than $[-4,1]$.)</li>
<li>Write down the double integral of $z$ integrating over $x$ and $y$ as specified by your found limits. Integrate.</li>
</ul>
|
909,734 | <p>I have answered this question to the best of my knowledge but somehow I feel as if I am missing something? Can I further prove this statement or add anything to it? </p>
<p>Question: </p>
<p>Let $m \in \mathbb N$. Prove that the congruence modulo $m$ relation on $\mathbb Z$ is transitive. </p>
<p>My attempt:</p>
<p>Let $a\equiv b \pmod{m}$ and $b\equiv c \pmod{m}$.</p>
<p>Then $a-b \equiv 0 \pmod{m}$ and $b-c\equiv 0 \pmod{m}$.</p>
<p>Adding, $a-c\equiv 0 \pmod{m}$, so $a\equiv c\pmod{m}$.</p>
| Bill Cook | 16,423 | <p>In some sense, in your proof, you are assuming what you are trying to prove. </p>
<p>When showing a basic property like the transitivity of a congruence relation, it's important to keep in mind what you already "know". </p>
<p>For example: It seems obvious that $a \equiv b$ (mod $m$) implies that $b \equiv a$ (mod $m$). But at the very beginning, we shouldn't assume this (until it's proven). </p>
<p>To prove something like this you need to either use propositions/theorems you've already established or go back to the definition. </p>
<p>For a homework problem like this I imagine you don't really have any theorems about congruences to work with. This means you're stuck going back to the definition.</p>
<p>Question: How did you define congruences? There are many equivalent ways. </p>
<p>Let's say you defined $a \equiv b$ (mod $m$) to mean there is some integer $k$ such that $a=b+km$. If so we can use what we know about integer arithmetic to rewrite this as: $b=a-km$. Notice that $-k$ is also an integer (since $k$ is) so $b$ and $a$ differ by an integer and we've established that $b \equiv a$ (mod $m$).</p>
<p>Now back to your problem...</p>
<p>Assume $a \equiv b$ (mod $m$) and $b \equiv c$ (mod $m$). So, for example, there exists some $k$ such that $a=b+km$. We can say something about $b$ and $c$ as well. Then combine these two facts and get $a = c + ???m$, make sure ??? is an integer and we'll have established that $a \equiv c$ (mod $m$). Done! :)</p>
|
875,458 | <p>Is it possible to draw a triangle, if the length of its medians $(m_1, m_2, m_3)$ are given only?</p>
<p>Someone asked me this question, but I can not see it. Is it really possible?</p>
<p><strong>UPDATE</strong></p>
<p>Apart from the algebraic solution given by <em>JimmyK4542</em>, can anyone give me a direct construction? I mean, it should sound like:</p>
<blockquote>
<p>Draw a line segment sufficiently long. Cut the length of $m_1$ from it. Then $\ldots$</p>
</blockquote>
| JimmyK4542 | 155,509 | <p>The formulas for the lengths of the <a href="http://en.wikipedia.org/wiki/Median_(geometry)" rel="nofollow">medians</a> of a triangle given the sidelengths are: </p>
<p>$m_a^2 = \dfrac{2b^2+2c^2-a^2}{4}$</p>
<p>$m_b^2 = \dfrac{2c^2+2a^2-b^2}{4}$</p>
<p>$m_c^2 = \dfrac{2a^2+2b^2-c^2}{4}$</p>
<p>Solving for $a,b,c$ in terms of $m_a,m_b,m_c$ gives: </p>
<p>$a^2 = \dfrac{8m_b^2+8m_c^2-4m_a^2}{9}$</p>
<p>$b^2 = \dfrac{8m_c^2+8m_a^2-4m_b^2}{9}$</p>
<p>$c^2 = \dfrac{8m_a^2+8m_b^2-4m_c^2}{9}$</p>
<p>This gives you the lengths of the three sides of the triangle.</p>
|
619,477 | <blockquote>
<p>Alice opened her grade report and exclaimed, "I can't believe Professor Jones flunked me in Probability." "You were in that course?" said Bob.
"That's funny, i was in it too, and i don't remember ever seeing you there."
"Well," admitted Alice sheepishly, "I guess i did skip class a lot." "Yeah, me too" said Bob. Prove that either Alice or Bob missed at least half of the classes. </p>
</blockquote>
<p>Proof:</p>
<p>Let $A$ be the set of lectures Alice attended and missed, let's assume she attended them in no particular order, similarly for Bob $B$ is the set of all lectures Bob attended and missed in random order.
Let $f$ be one-to-one and onto, we define $f:A\to B$ to be the mapping that matches the lectures Alice attended to the lectures that Bob missed and the lectures that Alice missed to the ones that Bob attended. If we consolidate the contiguous entries in the sets $A$ and $B$ into two groups, the group of lectures that Alice attended and the group of lectures she didn't attend and similarly for $B$ then the function $f$ can only be one-to-one and onto if both Alice and Bob attended the same number of lectures they missed.</p>
<p>I understand i've shown that Bob and Alice missed half the classes, how do i show that they could've missed more with this method?? </p>
| nadia-liza | 113,971 | <p>Let $n$ be number all lectures.</p>
<p>Let $k$ be number of the lectures that Bob missed. if $k<n/2$ then $-k>-n/2$</p>
<p>Let $m$ be number of the lectures that Alice missed.Then $m>=n-k>n/2$</p>
|
619,477 | <blockquote>
<p>Alice opened her grade report and exclaimed, "I can't believe Professor Jones flunked me in Probability." "You were in that course?" said Bob.
"That's funny, i was in it too, and i don't remember ever seeing you there."
"Well," admitted Alice sheepishly, "I guess i did skip class a lot." "Yeah, me too" said Bob. Prove that either Alice or Bob missed at least half of the classes. </p>
</blockquote>
<p>Proof:</p>
<p>Let $A$ be the set of lectures Alice attended and missed, let's assume she attended them in no particular order, similarly for Bob $B$ is the set of all lectures Bob attended and missed in random order.
Let $f$ be one-to-one and onto, we define $f:A\to B$ to be the mapping that matches the lectures Alice attended to the lectures that Bob missed and the lectures that Alice missed to the ones that Bob attended. If we consolidate the contiguous entries in the sets $A$ and $B$ into two groups, the group of lectures that Alice attended and the group of lectures she didn't attend and similarly for $B$ then the function $f$ can only be one-to-one and onto if both Alice and Bob attended the same number of lectures they missed.</p>
<p>I understand i've shown that Bob and Alice missed half the classes, how do i show that they could've missed more with this method?? </p>
| Martin | 34,537 | <p>Suppose, to the contrary, that neither Alice nor Bob missed at least half of the classes. Hence, both must have attended more than half of the classes. </p>
<p>Then clearly, out of the $n \geq 1$ classes total, it cannot be the case that Alice and Bob were never in the same class. </p>
<p>Why? <em>More</em> than half of the classes were attended by Alice. Hence, in order for Bob to have never attended a class that Alice has attended, he can at most have attended <em>less</em> than half of classes. But this in contradiction to the assumption that Bob has attended <em>more</em> than half of the classes.</p>
|
1,957,304 | <p>I'm proving the compact-to-Hausdorff lemma (probably not a universal name for it) which is stated as:</p>
<blockquote>
<p>If $X$ is compact, $Y$ Hausdorff, $f:X \rightarrow Y$ a continuous bijection, then $f$ is a homeomorphism.</p>
</blockquote>
<p>However, the following line has popped up in a proof of it:</p>
<blockquote>
<p>If $f: X \rightarrow Y$ is a bijection, then $f(U) = Y\setminus f(X \setminus U)$ (where $U$ is open in $X$)($\star$).</p>
</blockquote>
<p>I know that if $f$ is a bijection, then $f(X\setminus U) = f(X) \setminus f(U)$. Using this, I've tried to draw a little picture to try to see that $f(U) = Y\setminus f(X \setminus U)$, but it hasn't actually helped.</p>
<p>What's a proof of ($\star$)?</p>
| Moritz | 215,955 | <p>My pleasure: </p>
<p>How about $f(U) \cup f(X\setminus U) = f(U\cup (X\setminus U)) = f(X) = Y$? Since $f(U)$ and $f(X\setminus U)$ are disjoint, you get $f(U) = (f(U) \cup f(X\setminus U))\setminus f(X\setminus U) = Y\setminus f(X\setminus U)$.</p>
|
197,730 | <blockquote>
<p>Prove that the states of the 8-puzzle are divided into two disjoint sets such that any
state in one of the sets is reachable from any other state in that set, but not from any state in the other set. To do so, you can use the following fact: think of the board as a one-dimensional array, arranged in row-major order. Define an inversion as any pair of contiguous tiles (other than the blank tile) in this arrangement such that the higher
numbered tile precedes the lower numbered tile. Let N be the sum of the total number of
inversions plus the number of the row in which the blank tile appears. Then (N mod 2) is
invariant under any legal move in the puzzle.</p>
</blockquote>
<p>I know how to show that any state in one set is not reachable from another set, due to the invariant, but I'm trying to show that the union of the two disjoint sets encompass the entire state space. One thing I've tried is calculating the total possible arrangements (9!), and then the number of possible arrangements in each of the disjoint sets, but I haven't thought of a good way to calculate the latter.</p>
| Brian M. Scott | 12,042 | <p>You’re working too hard: every state has an associated value of $N$, and that value is either even or odd.</p>
|
66,463 | <p>Hi,</p>
<p>Let $\Gamma$ be a free subgroup of rank 2 in $\mathbb{G}_m^2(\mathbb{Q})$. For all but finitely many primes p we can reduce $\Gamma$ modulo p. Let $S$ be the of primes for which $\Gamma$ does not reduce modulo p, and for any $p$ not in $S$, let $\gamma_p$ be the size of $\Gamma \mod p$. My question is what is known about the function</p>
<blockquote>
<p>$f(x)= \sum_{p\not\in S,\ p\leq x}\frac{\log p }{\gamma_p}$</p>
</blockquote>
<p>In particular what is the asymptotic behavior of $f$? Is the corresponding infinite series convergent whenever $\Gamma$ is <em>not</em> contained in an algebraic subgroup of $\mathbb{G}_m^2$? Do you know of any references that might be relevant to those questions?</p>
<p>Thanks in advance,</p>
| Tzanko Matev | 421 | <p>I would just like to give a small update for the question. In my thesis <a href="https://epub.uni-bayreuth.de/1721/1/thesis.pdf" rel="nofollow">https://epub.uni-bayreuth.de/1721/1/thesis.pdf</a> I showed that the group $\Gamma\ \mod{p}$ has two generators for almost all primes p. So I would conjecture that on average $\gamma_p \sim p^2$ which would imply that the sum above indeed converges.</p>
|
3,578,357 | <p>The problem is like this : How do you solve <span class="math-container">$$ \lim _{x\to 0}\:\:\frac{x^m-sin^n(x)}{x^{n+2}} $$</span> for different values of <span class="math-container">$ n \in \Bbb N $</span>
Now, what i've started doing is to add <span class="math-container">$$ \lim _{x\to 0}\:\:\frac{x^m-x^n+x^n-sin^n(x)}{x^{n+2}} $$</span> then i split the limit into two limits like this <span class="math-container">$$ \lim _{x\to 0}\:\:\frac{x^m-x^n}{x^{n+2}} + \lim _{x\to 0}\:\:\frac{x^n-sin^n(x)}{x^{n+2}} $$</span> and i was thinking for the second limit to apply the formula : <span class="math-container">$(a-b)^n$</span> . The problem is that i don't know what to do with the first limit which has <span class="math-container">$x^m$</span>, at first i thought that i was a mistake in my textbook, but i am not sure . </p>
| Fred | 380,717 | <p>Suppose that there is <span class="math-container">$x_0 \in \mathbb R$</span> such that <span class="math-container">$f(x_0) \ne 0.$</span> Then there is an intervall <span class="math-container">$I$</span> such that <span class="math-container">$x_0 \in I$</span> and <span class="math-container">$f(x) \ne 0$</span> for all <span class="math-container">$x \in I.$</span> Now define <span class="math-container">$g:=1/f$</span> on <span class="math-container">$I$</span> . Then it is easy to see that <span class="math-container">$g'=-1$</span> on <span class="math-container">$I$</span>. Hence there is <span class="math-container">$C$</span> such that</p>
<p><span class="math-container">$f(x)=\frac{1}{C-x}$</span> for all <span class="math-container">$x \in I.$</span> This contradicts the facts, that the domain of <span class="math-container">$f$</span> is all of <span class="math-container">$ \mathbb R$</span> and <span class="math-container">$f$</span> is continuous at <span class="math-container">$C$</span>.</p>
<p>That <span class="math-container">$f$</span> is not injective is not needed.</p>
|
3,578,357 | <p>The problem is like this : How do you solve <span class="math-container">$$ \lim _{x\to 0}\:\:\frac{x^m-sin^n(x)}{x^{n+2}} $$</span> for different values of <span class="math-container">$ n \in \Bbb N $</span>
Now, what i've started doing is to add <span class="math-container">$$ \lim _{x\to 0}\:\:\frac{x^m-x^n+x^n-sin^n(x)}{x^{n+2}} $$</span> then i split the limit into two limits like this <span class="math-container">$$ \lim _{x\to 0}\:\:\frac{x^m-x^n}{x^{n+2}} + \lim _{x\to 0}\:\:\frac{x^n-sin^n(x)}{x^{n+2}} $$</span> and i was thinking for the second limit to apply the formula : <span class="math-container">$(a-b)^n$</span> . The problem is that i don't know what to do with the first limit which has <span class="math-container">$x^m$</span>, at first i thought that i was a mistake in my textbook, but i am not sure . </p>
| Peter Szilas | 408,605 | <p>Partial answer.</p>
<p>FTC:</p>
<p>Assume there exist <span class="math-container">$x_2 >x_1$</span> s.t. <span class="math-container">$f(x_2)=f(x_1)$</span>.</p>
<p>Then <span class="math-container">$f(x_2)-f(x_1)=\displaystyle{\int_{x_1}^{x_2}}f'(x)dx=$</span></p>
<p><span class="math-container">$\displaystyle{\int_{x_1}^{x_2}}f^2(x)dx =0$</span>.</p>
<p>Since <span class="math-container">$f$</span> is continuos this implies <span class="math-container">$f(x)\equiv 0$</span> for <span class="math-container">$x \in [x_1,x_2]$</span>.</p>
<p>What about <span class="math-container">$\mathbb{R}$</span> \ <span class="math-container">$[x_1,x_2]$</span>?</p>
|
268,360 | <p>Why is $\log_xy=\frac{\log_zy}{\log_zx}$? Can we prove this using the laws of exponents?</p>
| latiff | 54,820 | <p>By definition,
log<sub>z</sub> x<sup>log<sub>x</sub>y</sup> = log<sub>z </sub>y, also by definition we have, log<sub>z</sub> x<sup>log<sub>x</sub>y</sup> = log<sub>x</sub>y*log<sub>z</sub>x. So log<sub>z</sub>y = log<sub>x</sub>y*log<sub>z</sub>x. With division this gives us the result: log<sub>z</sub>y/log<sub>z</sub>x = log<sub>x</sub>y</p>
|
882,590 | <p><img src="https://i.stack.imgur.com/BMapu.png" alt="enter image description here"></p>
<p>long method: Determine an equation for each and solve using average value formula</p>
<p>alternative methods? </p>
<p>How could you prove the average value to be C over an interval [a,b] if you are given a graph.... looking for most efficient/unique methods. </p>
| Petite Etincelle | 100,564 | <p>Graph A:
The part for $x\in[0,2)$ compensates with the part for $x\in (2,4]$, giving an average of 2 over the interval $[0,4]$. But the part for $x\in(4,6]$ is above the line $y=2$. So A is not good.</p>
<p>GraphB:
Define a new function $g$ whose graph is composed of one straight line between $(0,0)$ and $(3,4)$ and the other straight line between $(3,4)$ and $(6,0)$. Similarly as in case A, we can see average of the function $g$ is equal to 2. But the graph of $f$ is always above $g$. So B is not good.</p>
<p>Similarly, we find average in graph C is 2 and average in Graph D is 4</p>
|
99,237 | <p>If we have a directed graph $G = (V,E)$ and we want to find if there is such node $s \in V$ that we can reach all other nodes of $G$</p>
<p>What is a good algorithm to solve this problem and what is its execution time?</p>
| Gerry Myerson | 8,269 | <p>Perhaps the section about algorithms at the wikipedia article on <a href="http://en.wikipedia.org/wiki/Reachability" rel="nofollow">reachability</a> has what you need. Why not have a look and let us know? </p>
<p>You can also search for "Warshall's algorithm," although this may do more than you need. </p>
|
1,615,883 | <p>A cubic polynomial with real coefficients, $a x^3 + b x^2 + c x + d$, has either three real roots, or one real root and a pair of complex conjugate ones. If the latter happens, what is the explicit formula for this real solution, and what conditions can be placed on $a,b,c$ and $d$ to guarantee that the real root is positive?</p>
| Asinomás | 33,907 | <p>No matter where we place the first book, the probability each of the other two is on the same shelf as the first book is $\frac{1}{5}$. I agree, the probability is $\frac{1}{25}$.</p>
|
11,266 | <p>I have a list of time durations, which are strings of the form: <code>"hh:mm:ss"</code>. Here's a sample for you to play with:</p>
<pre><code>durations = {"00:09:54", "00:31:24", "00:40:07", "00:11:58", "00:13:51", "01:02:32"}
</code></pre>
<p>I want to convert all of these into numbers in seconds, so that I can actually do useful things with this data!</p>
<p>How would I go about doing this? There must be some way to get <em>Mathematica</em> to extract the relevant things from each string.</p>
<p>Thanks.</p>
<p>Edit: I'm surprised that there doesn't seem to be a question related to this already - but it may be that I was using the wrong search terms.</p>
| VLC | 685 | <p>A method based on <code>AbsoluteTime</code>:</p>
<pre><code>AbsoluteTime /@ durations - AbsoluteTime["00:00:00"]
(* {594, 1884, 2407, 718, 831, 3752} *)
</code></pre>
|
12,114 | <p>I retired after 25 years of teaching and moved to Israel a year ago. My Hebrew is okay, but before moving here, I had no experience talking about math in Hebrew. I have been learning Hebrew math vocabulary by reading math textbooks and taking an online math course in Hebrew. </p>
<p>I recently started volunteering in an after school program to help Hebrew-speaking students prepare for tests in mathematics. These tests (called bagruyot) are standardized tests taken at the end of high school and used to determine entrance to college. I have no trouble understanding the problems on the practice exams, but I see that it is difficult for me to teach in Hebrew.</p>
<p>Currently my strategies (when I can't explain clearly in Hebrew) include: acting out problems with students, writing out solutions in math symbols, and finding similar problems with solutions in the textbooks This is not how I taught in my native language!</p>
<p>I am interested in any tips to becoming a better teacher in a language that is foreign to me but native to my students.</p>
| kjetil b halvorsen | 122 | <p>Well, I was in just such a situation when I had to learn Spanish and learn to teach in Spanish simultaneously. What I did was sitting every evening with the books and dictionaries and translating, and writing down, word for word, what I had to say next day. Timeconsuming, yes, but it worked. It did help to have some students in class which just knew some english and could help. </p>
|
2,555,399 | <p>The question is to find out the coefficient of $x^3$ in the expansion of $(1-2x+3x^2-4x^3)^{1/2}$</p>
<p>I tried using multinomial theorem but here the exponent is a fraction and I couldn't get how to proceed.Any ideas?</p>
| nonuser | 463,553 | <p>Say $$(1-2x+3x^2-4x^3)^{1/2} =a+bx+cx^2+dx^3...$$
then </p>
<p>$$1-2x+3x^2-4x^3 =(a+bx+cx^2+dx^3...)^2$$
but $$(a+bx+cx^2+dx^3...)^2 = a^2+2abx+(2ac+b^2)x^2+2(ad+bc)x^3+...$$</p>
<p>So $a=\pm 1$. </p>
<p>If $a=1$ then $b=-1$ and $c=1$ and $d=-1$</p>
|
4,034,709 | <p>What will be the operator norm of the matrix <span class="math-container">$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix},$</span> where <span class="math-container">$a,b,c,d \in \Bbb C\ $</span>?</p>
<p>According to the definition of the operator norm it turns out that <span class="math-container">$$\|A\|^2 = \sup \left \{ \left (|a|^2 + |c|^2 \right ) |z|^2 + \left (|b|^2 + |d|^2 \right ) |w|^2 + 2\ \mathfrak R\ (a \overline b z \overline w) + 2\ \mathfrak R\ (c \overline d z \overline w) \ \bigg |\ |z|^2 + |w|^2 = 1,\ z,w \in \Bbb C \right \}.$$</span></p>
<p>Is there any way to simplify the above expression?
Any help will be highly appreciated.</p>
<p>Thanks for your time.</p>
| donaastor | 251,847 | <p>Yes, there is. Please, keep in mind that I do not know the definition of operator norm, but if you did that part of job correctly, then there is a simplification.</p>
<p>Firstly, focus on the last half of your expression, the one with real parts. That sum is equal to <span class="math-container">$\Re(C_3\cdot z\overline{w})$</span>, where <span class="math-container">$C_3=2(a\overline{b}+c\overline{d})$</span>. Rotating <span class="math-container">$z$</span> or <span class="math-container">$w$</span> around the zero would affect only this summand, so you are free to multiply <span class="math-container">$z$</span> and/or <span class="math-container">$w$</span> with appropriate constant <span class="math-container">$e^{i\varphi}$</span> to achieve <span class="math-container">$\Re(C_3\cdot z\overline{w})=|C_3|\cdot|zw|$</span>.</p>
<p>Now, your expression becomes <span class="math-container">$$\sup\{C_1\cdot|z|^2+C_2\cdot|w|^2+|C_3|\cdot|zw|\mid|z|^2+|w|^2=1,z,w\in\mathbb{C}\}.$$</span>
By agreeing that <span class="math-container">$C_3$</span> replaces the <span class="math-container">$|C_3|$</span> from now on, your problem reduces to problem of finding the maximum of <span class="math-container">$$C_1\cdot a^2+C_2\cdot b^2+C_3\cdot ab,\text{ where }a^2+b^2=1,$$</span>where all numbers involved are real and positive. Now you can apply methods like Lagrangian multipliers to find the maximum and then substitute the expressions behind the constants to get what you want. But it will still be an ugly expression.</p>
|
867,209 | <p>I tried to do the implication part. Please, see what I need to do to fix it.</p>
<p>claim: $n|a – b → n|a^2 – b^2$.</p>
<p>claim: $nk = a – b$ for some $k \in \mathbb Z \to nk' = a^2 – b^2$ for some $k' \in Z$.</p>
<p>$(a + 1)^2 – (b +1)^2$</p>
<p>$= a^2 + 2a + 1 -(b^2 +2b + 1)$</p>
<p>$= a^2 + 2a -b^2 -2b$</p>
<p>$= a^2 – b^2 + 2(a – b)$</p>
<p>$= (a + b)(a – b) + 2(a – b)$</p>
<p>$= (a – b)[(a + b) + 2]$</p>
<p>$= nk[(a + b) + 2]$</p>
<p>$= n[(a + b) + 2k]$</p>
| Adam | 82,101 | <p>Any set $$R \subseteq A \times A = \{(x,y) \mid x \in A ,\,y \in A\} $$ is a relation on $A $. Since $$\{(b,c), (b,d)\} \subseteq A \times A$$ holds, it is indeed a relation on $ A$. </p>
|
3,598,476 | <p>I have proven that if <span class="math-container">$|x|<\varepsilon,\forall\varepsilon>0$</span>, then <span class="math-container">$x=0$</span>. Further I have proven that ,<span class="math-container">$L=\displaystyle\lim_{n\to\infty}\frac{1}{n} = 0$</span> so that by definition <span class="math-container">$(\forall\varepsilon >0)(\exists N>0)(\forall n>N)\left|\frac{1}{n}-L\right|<\varepsilon$</span>, but <span class="math-container">$\frac{1}{n}\ne 0, \forall n\in\mathbb N$</span>.</p>
<p>Is there a contradiction or have I made an error in analysis?</p>
| Kavi Rama Murthy | 142,385 | <p>There is no contradiction. You are getting <span class="math-container">$|\frac 1n |<\epsilon$</span> only under the extra condition <span class="math-container">$n >N$</span>. If you had <span class="math-container">$|\frac 1n |<\epsilon$</span> without any precondition you could say <span class="math-container">$\frac 1 n=0$</span> and get a contradiction, but that is not the case. </p>
<p>You should note that as <span class="math-container">$\epsilon$</span> becomes smaller and smaller the integer <span class="math-container">$N$</span> becomes larger and larger and you cannot have any <span class="math-container">$n$</span> which exceeds <span class="math-container">$N$</span> for all these values of <span class="math-container">$N$</span>.</p>
|
588,802 | <p>The problem is: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$</p>
<p>The first thing I did was use the divergence test which didn't help since the result of the limit was 0.</p>
<p>If I multiply it through, the result is $\sum_{n=1}^{\infty} \frac{1}{n^2+3n}$</p>
<p>I'm wondering if I can consider this as a p-series and simply use the largest power. In this case the power would be 2 which would mean it converges. If this is the correct way to go about this, how do I find where it converges to.</p>
| Alec Teal | 66,223 | <p>Bound it above! Note $n(n+3)=n^2+3n>n^2$</p>
<p>so $\frac{1}{n(n+3)}<\frac{1}{n^2}$</p>
<p>Each term is clearly > 0 btw.</p>
<p>So! $\sum\frac{1}{n(n+3)}<\sum\frac{1}{n^2}$ which you ought to know (but can trivially show) converges.</p>
<p>Finally a question I can answer here!</p>
|
471,153 | <p>Let $k$ be a number field. Define a prime of $k$ to be an equivalence class of absolute values on $k$. If $\sigma:k\hookrightarrow \mathbb{C}$ is an embedding of $k$ into the complex numbers then we know that $|a|=|\sigma(a)|$ defines an absolute value on $k$. </p>
<p>My question is: </p>
<p>Why is there exactly <em>one</em> prime for each real embedding and for each pair of complex embeddings of $k$? </p>
<p>How can we show this?</p>
| anon | 90,978 | <p>Fix a generator $a \in k$, so that $k = \mathbb Q(a)$. Let $f \in \mathbb Q[X]$ be its minimal polynomial, and let
$$ \alpha_1, \dots, \alpha_n \in \mathbb C $$
be the complex roots of $f$; then any embedding $\sigma \colon k \to \mathbb C$ is determined by where it sends $a$, which is one of the $\alpha_i$'s. </p>
<p>Suppose we have two embeddings $\sigma_1, \sigma_2$, which map $a$ to $\alpha_1$ and $\alpha_2$ respectively, such that $|\sigma_1(t) | = |\sigma_2(t)|$ for all $t \in k$. </p>
<p>Then in particular for all $x, y \in \mathbb Q$, we have
$$ |x + y \alpha_1 |^2 = |x + y \alpha_2|^2,$$
which holds if and only if
$$Re(\alpha_1) = Re(\alpha_2) \qquad \text{ and} \qquad Im(\alpha_1) = \pm Im(\alpha_2). $$</p>
<p>This means that either $\alpha_1 = \alpha_2$ or that $\alpha_1 = \overline{\alpha_2}$; in the first case the embeddings are the same, in the second they are conjugate. </p>
|
471,153 | <p>Let $k$ be a number field. Define a prime of $k$ to be an equivalence class of absolute values on $k$. If $\sigma:k\hookrightarrow \mathbb{C}$ is an embedding of $k$ into the complex numbers then we know that $|a|=|\sigma(a)|$ defines an absolute value on $k$. </p>
<p>My question is: </p>
<p>Why is there exactly <em>one</em> prime for each real embedding and for each pair of complex embeddings of $k$? </p>
<p>How can we show this?</p>
| walcher | 89,844 | <p>There is the so called "Extension Theorem".<br><br>
<strong>Theorem:</strong> Let $\mid \;\mid_v$ be an absolute value on a field $K$ and $L$ an algebraic extension. Denote by $K_v$ the completion of $K$ w.r.t $\mid \;\mid_v$ and by $\bar K_v$ its algebraic closure. Note that $\mid\;\mid_v$ extends uniquely to $\bar K_v$, because $K$ is dense in $K_v$, which is complete w.r.t. $\mid\;\mid_v$. Then the following hold:<br>(1) Every extension of $\mid\;\mid_v$ to $L$ is of the form $|x|_{\tau v}=|\tau x|_v$ for some $K$-embedding $\tau :L\to \bar K_v$<br>
(2) Two extensions $\mid\;\mid_{\tau v}$ and $\mid\;\mid_{\tau'v}$ are equivalent if and only if $\tau$ and $\tau'$ are conjugate over $K_v$, i.e. there is a $\sigma \in Gal(\bar K_v/K_v)$ such that $\sigma \circ \tau =\tau'$.<br><br></p>
<p>Apply this to $K=\Bbb Q, L=k$ with the usual absolute value, then $K_v=\Bbb R, \bar K_v=\Bbb C$. Therefore two embeddings $\tau, \tau'$ are conjugate if and only if they are equal or complex conjugates of each other, because $Gal(\Bbb C/\Bbb R)=\{id, \sigma\}$, where $\sigma$ is complex conjugation. Hence, every infinite prime of a number field is either a pair of complex conjugate embeddings or a real embedding (which is its own complex conjugate).</p>
|
2,369,431 | <p>I took a course some years ago and in it was a treatment of how to associate a field to an abstract geometry. </p>
<p>I would very much appreciate some reading on this, as I have been unsuccessful on where to find any resources on such ideas, and I have since lost my notes!</p>
<blockquote>
<p>Is there a book which treats, thoroughly, the connection between an abstract geometry and a field?</p>
</blockquote>
<p>In particular:</p>
<blockquote>
<p>Is there a way to consider a discrete geometry, like projective Steiner triple systems, and associate to these objects a field?</p>
</blockquote>
<p>Any information would be extremely useful, introductory or extremely advanced. I would be particularly interested in anyones knowledge of resources which have connections to class field theory.</p>
<p>Thank you for your time and consideration.</p>
| xxxxxxxxx | 252,194 | <p>For the most part, you can start with a geometric object such as a projective plane, and coordinatize. This leads to a structure called a planar ternary ring. The ring you get is not unique up to isomorphism, but rather up to a relation known as <em>isotopism</em>.</p>
<p>I'm not sure about other objects, you can also do this with biplanes though at least.</p>
<p>I don't have references on hand, but suggest searching for "planar ternary ring" along with the geometric structure you are interested in.</p>
|
151,425 | <p>I've considered the following spectral problems for a long time, I did not kow how to tackle them. Maybe they needs some skills with inequalities.</p>
<p>For the first, suppose $T:L^{2}[0,1]\rightarrow L^{2}[0,1]$ is defined by
$$Tf(x)=\int_{0}^{x} \! f(t) \, dt$$</p>
<p>How can I calculate:</p>
<ul>
<li>the radius of the spectrum of $T$?</li>
<li>$T^{*}T$?</li>
<li>the norm of $T$?</li>
</ul>
<p>I guess $r(T)$ should be $0$. but I did know how to prove it. My idea is use Fourier to tackle it, however it does not seem to work.</p>
<p>The other problem may be very similar to this one. Let $T:C[0,1]\rightarrow C[0,1]$ be defined by
$$Tf(x)=\int_{0}^{1-x}f(t)dt$$</p>
<p>It is obvious that $T$ is compact and I guess its radius of spectrum is zero, but I do not know how to prove it.</p>
<p>Any references and advice will be much appreciated.</p>
| yaoxiao | 21,273 | <p>thanks PZZ's answer for the first problem. Now I know why I did realize why I am failed in the second problem.</p>
<p>$$Tf(x)=\int_{0}^{1-x}f(t)dt$$</p>
<p>It is trival that this operator is is linaer compact operator according to Arzela-Ascoli theorem, so we get $\sigma(T)/{0} \subset \sigma_{p}(T)$, use simple computation we can get that $0\notin \sigma_{p}(T)$, for C[0,1] is a infinite dimensional space, we can get $0\in \sigma(T)$. $\forall \lambda \in \sigma_{p}(T)$, and $\lambda \neq 0$, we can get that </p>
<p>$$\int_{0}^{1-x}f(t)dt=\lambda f(x)$$</p>
<p>which implies that $f \in C^{\infty}[0,1]$, then we can get<br>
$$-f(1-x)=\lambda f'(x)$$
notice that
$$f(1)=0$$
further we get
$$\lambda^{2}f''(x)=-f(x)$$, solve this second order of differential equation, we can calulated every $\lambda_{n}=\frac{2\pi}{2n+1}, \qquad n\in Z$.</p>
|
4,187,238 | <p>How many points are common to the graphs of the two equations <span class="math-container">$(x-y+2)(3x+y-4)=0$</span> and <span class="math-container">$(x+y-2)(2x-5y+7)=0$</span>?</p>
<p><span class="math-container">\begin{align*}
(x-y+2)(3x+y-4) &= 0\tag{1}\\
(x+y-2)(2x-5y+7) &= 0\tag{2}
\end{align*}</span></p>
<p>In equation <span class="math-container">$1$</span>, it is not possible for both <span class="math-container">$x-y+2$</span> and <span class="math-container">$3x+y-4$</span> to be nonzero. Like-wise in equation <span class="math-container">$2$</span>, it is not possible for both <span class="math-container">$x+y-2$</span> and <span class="math-container">$2x-5y+7$</span> to be nonzero.
Therefore the LHS of either equation must involve a product of <span class="math-container">$0$</span> and some number.
This means there are infinitely many solutions to both equations.</p>
<p>However this doesn't shead any light on which points are common between the two equations. I'm confused. It seems I've made an error in my judgement. What should I reconsider?</p>
| lab bhattacharjee | 33,337 | <p>If <span class="math-container">$x-y+2=0,y=?$</span></p>
<p>Replace the value of<span class="math-container">$y$</span> in terms of <span class="math-container">$x,$</span> in the second equation to find</p>
<p><span class="math-container">$$0=(x+x+2-2)(3x+x+2-4)=2x(4x-2)$$</span></p>
<p><span class="math-container">$$x=?,?$$</span></p>
<p>Had there been <span class="math-container">$n$</span> factors, we would have <span class="math-container">$n$</span> degree equation in <span class="math-container">$x$</span></p>
<p>Similarly for <span class="math-container">$3x+y-4=0\implies y=4-3x$</span></p>
<p>Can you take it from here?</p>
|
1,990,804 | <p>I know that we can define the exponential by a function $f: \mathbb{N}^2 \rightarrow \mathbb{N}$ by letting:</p>
<p>$f(m,o) = 1$ and $f(m,n+1) = f_x(m^n,m)$ where $f_x$ is the multiplication function, which we know is recursive.</p>
<p>I would then let $g = s(z)$ and hence $g(n) = 1$ for each $n \in \mathbb{N}$ , and let $h = f_x$ </p>
<p>However I am unsure where to go from here in order to complete a recursive definition for the exponential function. Could somebody please guide me into the approach to finish this proof? </p>
| Noah Schweber | 28,111 | <p>Your usual sources of examples - $\mathbb{C}$ and its many variations - won't help here: this operation is <em>non-associative</em>. For example, $$a*(a*b)=a*a=b\quad\mbox{but}\quad (a*a)*b=b*b=a.$$ (In particular, I don't understand your claim that you get semigroups out of this.) Indeed it's not even <a href="https://en.wikipedia.org/wiki/Power_associativity" rel="nofollow">power associative</a>, since we can replace "$b$" with "$(a*a)$" in the above example.</p>
<p>So this isn't even a semigroup. It also isn't a loop, since it doesn't have an identity element. And it's neither left nor right cancellative: $a*b=b*b=b*a$. So it's not a <a href="https://en.wikipedia.org/wiki/Quasigroup#Loop" rel="nofollow">quasigroup</a> either.</p>
|
617,389 | <p>How to sketch $y = \frac1{\sqrt{x-1}}$</p>
<p>My way:(which does not work here)</p>
<p>I normally solve these problems by squaring and converting them to equations of 2 degree curves(such as parabola, hyperbola, etc.) which I can easily plot. But this seems to go 3 degree as $xy^2$ term is coming.</p>
<p>Please help me to solve this.</p>
<p>Note: Please don't say to use a graph plotter and see for myself since in the exam if this question comes I won't have the graph plotter with me. </p>
| mathlove | 78,967 | <p>HINT : What is the domain of $x$? What happens if you make $x$ larger, larger, to infinity? What happens if you make $x$ closer to $1$ from the right side?</p>
|
28,811 | <p>There are lots of statements that have been conditionally proved on the assumption that the Riemann Hypothesis is true.</p>
<p>What other conjectures have a large number of proven consequences?</p>
| Jim Humphreys | 4,231 | <p>What is usually referred to as <em>Lusztig's Conjecture</em> in the modular representation theory of semisimple algebraic groups has been enormously influential, as seen in Jantzen's treatise <em>Representations of Algebraic Groups</em>. It is actually a series of closely related conjectures, from 1979 on, inspired by the (soon proved) Kazhdan-Lusztig Conjecture (1979) on the formal characters of the usually infinite dimensional simple highest weight modules for a complex semisimple Lie algebra: such a character can be written as a $\mathbb{Z}$-linear combination of the known formal characters of Verma modules whose coefficients are values at 1 of Kazhdan-Lusztig polynomials for the Iwahori-Hecke algebra of the Weyl group $W$. The original characteristic $p$ conjecture has a similar flavor, but with the affine Weyl group (whose translations are multiplied by $p$) replacing $W$ and with the essential proviso that $p$ be not too small. It is expected that the Coxeter number of $W$ will be a suitable lower bound, but so far the partial proofs by Andersen-Jantzen-Soergel, Fiebig, and Bezrukavnikov-Mirkovic do not achieve a reasonable bound. </p>
<p>If proved, the conjecture would combine with older results of Curtis and Steinberg to yield all modular irreducible characters of finite groups of Lie type in the defining characteristic (but still with the lower bound on $p$),
as well as the formal characters and dimensions of all restricted representations of the Lie algebra of the given semisimple group. Andersen and others have formulated further consequences, in terms of the structure of Weyl modules, the extensions and cohomology of simple or Weyl modules, etc. (Adapted to general linear groups, there are also implications for modular characters of symmetric groups.) The later conjectures of Lusztig, proved for large enough $p$ in a preprint by Bezrukavnikov and Mirkovic, go further with the non-restricted Lie algebra representations as well in a unified geometric setting which promises further applications. </p>
<p>ADDED: I should point out that many special cases of the more general results which would follow from Lusztig's Conjecture have in fact been verified, but usually by computational or somewhat ad hoc methods. Plus the existing proofs of the conjecture itself for "large enough" primes, which don't seem improvable without new methods.</p>
|
118,406 | <p>I have a single flat directory with over a million files. I just wanted to take a sample of the first few files but <code>FileNames</code> doesn't include a "only the first n" option, and so it took over a minute:</p>
<p><a href="https://i.stack.imgur.com/s5cBS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/s5cBS.png" alt="enter image description here"></a></p>
<p>Is there a faster way?</p>
| MarcoB | 27,951 | <p>Using OS shells commands seems to be much faster, although their output will need some massaging to obtain only the file names. </p>
<p>For instance, the following works quite well on my system (Win7-64):</p>
<pre><code>Import["!dir /b /a-d C:\\Windows\\*", "Text"] // StringSplit[#, "\n"] &
</code></pre>
<p>This command takes 0.11s to execute; for comparison the corresponding FileNames expression (<code>FileNames["*", "C:\\Windows\\"]</code>) takes 0.33s. The difference is even more pronounced with larger/deeper directories.</p>
<p>In the dir command above, <code>/a-d</code> selects files that are not directories; <code>/b</code> produces minimal output which is easier to parse. If you want to traverse subdirectories, you can use the <code>/s</code> option.</p>
|
889,719 | <p>Example $5.9$ on page $103$ of John Lee's Smooth Manifolds says the following:</p>
<p>The intersection of $S^n$ with the open subset $\{x:x^i>0\}$ is the graph of the smooth function
$$
x^i=f(x^1,\dots,x^{i-1},x^{i+1},\dots,x^{n+1})
$$
where $f\colon B^n\to\mathbb{R}$ is $f(u)=\sqrt{1-|u|^2}$. The intersection of $S^n$ with $\{x:x^i<0\}$ is the graph of $-f$. Since every point in $S^n$ is in one of these sets, $S^n$ satisfies the local $n$-slice condition, this is an embedded submanifold.</p>
<p>The terminology is that if $M$ is a smooth manifold, and $S\subset M$ a subset, then $S$ satisfies the local $k$-slice condition if each point of $S$ is contained in the domain of a smooth chart $(U,\varphi)$ for $M$ such that $S\cap U$ is a single $k$-slice in $U$. </p>
<p>I don't see how this makes $S^n$ satisfy the local $n$-slice condition. Presumably the chart on $\mathbb{R}^{m+1}$ is $(U=\{x:x^i>0\},\mathrm{id})$, so that $S^n\cap\{x:x^i>0\}$ is an $n$-slice of $U$? But this doesn't seem right since $\mathrm{id}(S^n\cap U)$ is a hemisphere of $S^n$, but that's not a $n$-slice in the corresponding half-place $\mathrm{id}(U)$?</p>
| PhoemueX | 151,552 | <p>The chart is</p>
<p>$$
\{(x_1, \dots , x_{n+1}) \mid (x_1, \dots x_{i-1}, x_{i+1}, \dots x_{n+1}) \in B^n \text{ and } x_i > 0\} \to \Bbb{R}^{n+1}, (x_1, \dots,x_{n+1}) \mapsto (x_1, \dots, x_{i-1},x_i - f(x_1, \dots x_{i-1}, x_{i+1}, \dots x_{n+1}), x_{i+1}, \dots x_{n+1}) .
$$</p>
|
1,032,714 | <p>'Let $X$ be a topological space and let $(U_i)_{i \in I}$ be a cover of $X$ by connected subspaces $U_i$. Supposed for all $i,j \in I$ there exists some $n \geq 0$ and $k_0,...,k_n \in I$ such that $k_0 = i, k_n = j$ and $$U_{k_0} \cap U_{k_1} \neq \emptyset, U_{k_1} \cap U_{k_2} \neq \emptyset, ..., U_{k_{n-1}} \cap U_{n} \neq \emptyset$$
Show that $X$ is connected.</p>
<p>My intuition tells me the proof is along the lines of since each $U_i$ is connected, there are no subsets $A,B \subseteq U_i$ for $i \in I$ such that $A$ and $B$ are disjoint (unless they are the empty set.) Also, since the intersection of $U_{k_i} \cap X/U_{k_i}$ is non empty (as there always exists at least some other subcover it intersects with) there are no two nonempty subsets $A,B \subset X$ such that $A \cap B = X$. Thus $X$ is connected.</p>
<p>I was looking for some help formulating this into a nice mathematical proof. Thanks.</p>
| user178543 | 178,543 | <p>If $\omega \in B$, then $\{\omega\} \cap B = \{\omega\}$.</p>
|
1,456,407 | <p><a href="https://i.stack.imgur.com/oy6T7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oy6T7.jpg" alt="enter image description here"></a></p>
<p>We need to find the area of the shaded region , where curves are in polar forms as $r = 2 \sin\theta$ and $r=1$.</p>
<p>I formulated the double integral as follows : </p>
<p>We find the area in the first quadrant and then multiply it by $2$ , </p>
<p>Area of the circle $r=1$ in the first quadrant is $\frac{\pi}{4}$ , we need to subtract the area of the curve $r = 2\sin\theta$ from this , thus , area is given by : </p>
<p>$[\dfrac{\pi}{4} - \int^{\dfrac{\pi}{6}}_{0}\int^{2\sin\theta}_{0}r.dr.d\theta]\times2$ </p>
<p>Is this correct ?
The solution says , " first consider $0< \theta < \dfrac{\pi}{6}$ and then $\dfrac{\pi}{6}< \theta < \dfrac{\pi}{2}$ etc etc..... "</p>
| Yiyuan Lee | 104,919 | <p>Note that $$\begin{align}42^n - 1 &\equiv 1 - 1 \\&\equiv 0 \pmod{41}\end{align}$$</p>
<p>so the only way for $42^n - 1$ to be a prime is for $n$ to be $1$.</p>
<p>In general, for $a^n - 1$ to be a prime, where $a, n \in\mathbb{Z}^+$, either <a href="https://en.wikipedia.org/wiki/Mersenne_prime">$a = 2$</a> or $n = 1$.</p>
<p>(Not sure if this counts as indirect, but you could turn it into some form of contradiction)</p>
|
467,609 | <blockquote>
<p>Find the value of
$$\int _0 ^ \pi \dfrac{x}{1+\sin^2(x)} dx $$</p>
</blockquote>
<p>I have tried using $\int_a ^bf(x) dx=\int_a^b f(a+b-x)dx$</p>
<p>$\displaystyle \int _0 ^ \pi \dfrac{x}{1+\sin^2(x)} dx=\int _0 ^ \pi \dfrac{\pi-x}{1+\sin^2(x)} dx=I$</p>
<p>I couldn't go any further with that!</p>
| Suraj M S | 85,213 | <p>$I$ =$\int _0 ^ \pi \dfrac{x}{1+\sin^2(x)} dx$</p>
<p>also $\int _0 ^ \pi \dfrac{\pi-x}{1+\sin^2(x)} dx=I$</p>
<p>add both to get,
$2I$ = $\int _0 ^ \pi \dfrac{\pi}{1+\sin^2(x)} dx$</p>
<p>AND FURTHER SOLVE IT.</p>
|
1,269,738 | <p>I'm looking for problems that due to modern developments in mathematics would nowadays be reduced to a rote computation or at least an exercise in a textbook, but that past mathematicians (even famous and great ones such as Gauss or Riemann) would've had a difficult time with. </p>
<p>Some examples that come to mind are <em><a href="http://en.wikipedia.org/wiki/Group_testing">group testing problems</a></em>, which would be difficult to solve without a notion of error-correcting codes, and -- for even earlier mathematicians -- calculus questions such as calculating the area of some $n$-dimensional body.</p>
<p>The questions have to be understandable to older mathematicians and elementary in some sense. That is, past mathematicians should be able to appreciate them just as well as we can. </p>
| Oscar Cunningham | 1,149 | <p>That there exist transcendental numbers. This was first shown by Liouville, who proved that Liouville's number:
<span class="math-container">$$\sum_{i=0}^\infty10^{-i!}$$</span>
is transcendental.</p>
<p>The "modern" proof would be due to Cantor:</p>
<blockquote>
<p>There are countably many algebraic numbers and uncountably many reals. Therefore there exists a transcendental number.</p>
</blockquote>
<p>Proving that Liouville's number is transcendental isn't so hard, but compared to the above it seems quite torturous.</p>
|
25,917 | <p>$\sqrt{2+\sqrt{2+\sqrt{2+\dots}}}$ </p>
<p>$\dots\sqrt{2+\sqrt{2+\sqrt{2}}}$</p>
<p>Why they are different?</p>
| Arturo Magidin | 742 | <p>To prove something by strong induction, you have to prove that</p>
<blockquote>
<p>If all natural numbers strictly less than $N$ have the property, then $N$ has the property.</p>
</blockquote>
<p>is true for all $N$. </p>
<p>So: our induction hypothesis is going to be:</p>
<blockquote>
<p>Every natural number $k$ that is strictly less than $n$ can be written as a product of a power of $2$ and an odd number.</p>
</blockquote>
<p>And we want to prove that from this hypothesis, we can conclude that $n$ itself can be written as the product of a power of $2$ and an odd number.</p>
<p>Well, we have two cases: either $n$ is odd, or $n$ is even. If we can prove the result holds in both cases, we'll be done.</p>
<p><strong>Case 1:</strong> $n$ is odd. Then we can write $n=2^0\times n$, and we are done. So in Case 1, the result holds for $n$.</p>
<p><strong>Case 2:</strong> If $n$ is even, then we can write $n=2k$ for some natural number $k$. But then $k\lt n$, so we can apply the induction hypothesis to $k$. We conclude that <code>...and you should finish this part...</code></p>
<p>So we conclude that the result holds for all natural numbers by strong induction.</p>
|
576,553 | <p>Please, forgive me if this is an elementary question, as well as my the sloppy phrasing and notation.</p>
<p>Suppose we have two discrete probability distributions $p = {\lbrace p_i \rbrace}$ and $q={\lbrace q_i \rbrace}$, $i=1,\dots,n$, where $p_i=P(p=p_i)$ and $q_i=P(q=q_i)$. Let's represent them as vectors $\boldsymbol{p} = [p_i], \boldsymbol{q}= [q_i] \in \mathbb{R}^n$.</p>
<p>If we take the two p-norms $||\cdot||_a$ and $||\cdot||_b$, excluding 1-norm and max-norm then if $||\boldsymbol{p}||_a>||\boldsymbol{q}||_a$ is it the case that also $||\boldsymbol{p}||_b>||\boldsymbol{q}||_b$ holds? In other words, will all the p-norms induce the same 'ranking' of $\boldsymbol{p}$ and $\boldsymbol{q}$?</p>
<p>Would anything change if at least one of the $||\cdot||_a$ and $||\cdot||_b$ were p-quasinorms i.e., $a,b\in(0,1)$ instead?</p>
| Nick | 110,656 | <p>Nate, first of all thank you for your time. However, I am afraid I am not persuaded by your example.</p>
<p>We have $||\boldsymbol{q}||_a = (q_{2}^a)^{1/a} = q_2 = 1, \forall a$.</p>
<p>On the other hand, we will have $||\boldsymbol{p}||_1 = (1-x) + x = 1 = ||\boldsymbol{q}||_1, \forall x$. The 1-norm case is trivial, since p and q are probability distributions, hence by definition $||\boldsymbol{p}||_1 = ||\boldsymbol{q}||_1=1$, hence I ignore it in my question.</p>
<p>However if $a>1$ then $||\boldsymbol{p}||_a < 1 = ||\boldsymbol{q}||_a, \forall x \in (0,1)$ and we do get exactly $||\boldsymbol{p}||_a = 1 = ||\boldsymbol{q}||_a$ for $x \in \lbrace 0,1 \rbrace$, when we actually have $p\equiv q$ if we consider the permutations [1,0,0], [0,1,0], [0,0,1] to correspond to different representations of the same probability distribution (I know I did not mention this in my original question...).</p>
<p>Am I missing something?</p>
|
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