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 |
|---|---|---|---|---|
3,599,893 | <p>I had this idea to build a model of Earth in Minecraft. In this game, everything is built on a 2D plane of infinite length and width. But, I wanted to make a world such that someone exploring it could think that they could possibly be walking on a very large sphere. (Stretching or shrinking of different places is OK.) </p>
<p>What I first thought about doing was building a finite rectangular model of the world as like a mercator projection, and tessellating this model infinitely throughout the plane. </p>
<p><a href="https://i.stack.imgur.com/bzdjA.png" rel="noreferrer"><img src="https://i.stack.imgur.com/bzdjA.png" alt="enter image description here"></a></p>
<p>Someone starting in the US could swim eastwards in a straight line across the Atlantic, walk across Africa and Asia, continue through the Pacific and return to the US. This would certainly create a sense of 3D-ness. However, if you travel north from the North Pole, you would wind up immediately at the South Pole. That wouldn't be right.</p>
<p>After thinking about it, I hypothesized that an explorer of this model might conclude that they were walking on a donut-shaped world, since that would be the shape of a map where the left was looped around to the right (making a cylinder), and then the top was looped to the bottom. For some reason, by simply tessellating the map, I was creating a hole in the world.</p>
<p>Anyway, to solve this issue, I thought about where one ends up after travelling north from various parts of the world. Going north from Canada, and continuing to go in that direction, you end up in Russia and you face south. The opposite is true as well: going north from Russia, you end up in Canada pointing south. Thus, I started to modify the tessellation to properly connect opposing parts of Earth at the poles. </p>
<p>When going north of a map of Earth, the next (duplicate) map would have to be rotated 180 degrees to reflect the fact that one facing south after traversing the north pole. This was OK. However, to properly connect everything, the map also had to be <em>flipped</em> about the vertical axis. On a globe, if Alice starts east of Bob and they together walk North and cross the North Pole, Alice still remains east of Bob. So, going north from a map, the next map must be flipped to preserve the east/west directions that would have been otherwise rotated into the wrong direction.</p>
<p><a href="https://i.stack.imgur.com/U5n9t.png" rel="noreferrer"><img src="https://i.stack.imgur.com/U5n9t.png" alt="enter image description here"></a></p>
<p>Now the situation is hopeless. After an explorer walks across the North Pole in this Minecraft world, he finds himself in a mirrored world. If the world were completely flat, it would feel as if walking North will take you from the outside of a 3D object to its inside.</p>
<p>Although I now think that it is impossible to trick an explorer walking on infinite plane into thinking he is on a sphere-like world, a part of me remains unconvinced. Is it really impossible? Also, how come a naive tessellation introduces a hole? And finally, if an explorer were to roam the world where crossing a pole flips everything, what would he conclude the shape of the world to be?</p>
| Magma | 599,209 | <p>Unlike the other answers to this question, I claim that it's possible to trick an explorer on an infinite plane into thinking he's on a sphere. In fact, I'm about to trick you, by providing just a screenshot of <a href="https://github.com/Ralith/hypermine" rel="noreferrer">a work-in-progress video game project</a> I'm currently working on, together with a bunch of other excellent internet strangers.</p>
<p>The previous answers have all shown that it isn't really possible to do what you want to satisfaction in euclidean space. So instead let's do it in <a href="https://en.wikipedia.org/wiki/Hyperbolic_space" rel="noreferrer">hyperbolic space</a>. In hyperbolic space, planes are hyperbolic planes, they have negative curvature, and interesting surfaces with zero curvature are spheres of infinite radius called horospheres (in hyperbolic space, this is not the same as a plane!) Horospheres work intrinsically just like euclidean space, but they look curved when embedded into hyperbolic space.</p>
<p>Here's a screenshot of a horosphere with some terrain:</p>
<p><a href="https://i.stack.imgur.com/h1qdw.png" rel="noreferrer"><img src="https://i.stack.imgur.com/h1qdw.png" alt="Horosphere"></a></p>
<p>Looks pretty round, right? Well, it is indeed round. However, it's not spherical, even if you ignore the altitude variations. It's a horosphere. The surface of this "planet" is actually Euclidean. You can draw a square grid onto the surface without distorting it.</p>
<p>To further confuse you, here's a different view, this time with inverted terrain (air is now <em>inside</em> the horosphere instead of outside of it).</p>
<p><a href="https://i.stack.imgur.com/vMiDM.png" rel="noreferrer"><img src="https://i.stack.imgur.com/vMiDM.png" alt="Inverted horosphere"></a></p>
<p>Looks much bigger on the inside, I know, but it's the exact same horosphere as before! From this perspective it should be easier to tell that the horosphere is infinite, but seen in action this can still convince explorers from the Euclidean world that they're inside a relatively small sphere if they don't happen to look up.</p>
<p>It's also possible to achieve a very different effect: If you place terrain along a hyperbolic plane, it will look perfectly flat at ground level:</p>
<p><a href="https://i.stack.imgur.com/pAHLw.png" rel="noreferrer"><img src="https://i.stack.imgur.com/pAHLw.png" alt="Hyperbolic plane from ground level"></a></p>
<p>However, if you look at it from just a few blocks up, it begins to seem like a pretty small planet:</p>
<p><a href="https://i.stack.imgur.com/1P4IB.png" rel="noreferrer"><img src="https://i.stack.imgur.com/1P4IB.png" alt="Hyperbolic plane from above"></a></p>
<p>None of these screenshots involve cheap rendering tricks, this is exactly what hyperbolic space actually looks like.</p>
|
3,223,618 | <p>I have this system of linear equations with parameter:</p>
<p><span class="math-container">$ ax + 4y + z =0 $</span></p>
<p><span class="math-container">$2y + 3z = 1$</span> </p>
<p><span class="math-container">$3x -cz=-2$</span></p>
<p>What I did was to put those equations into a matrix and transform that matrix it into a triangular matrix. Then I got these results on the right side:
<span class="math-container">$ \frac{-c+10}{ac-15}$</span>, <span class="math-container">$\frac{3+(c-6)a}{2ac-30}$</span>, <span class="math-container">$\frac{2a-6}{ac-15}$</span>. However, I don't know what to do now. </p>
<p>Thanks for any help!</p>
| Dr. Sonnhard Graubner | 175,066 | <p>From the first equation we get
<span class="math-container">$$z=-ax-4y$$</span> so we get with the third equation
<span class="math-container">$$-3ax-10y=1$$</span> and with the last one:
<span class="math-container">$$3x+acx+4yc=-2$$</span></p>
<p>From the second equation above <span class="math-container">$$y=-\frac{3}{10}ax-\frac{1}{10}$$</span> so we obtain
finally</p>
<p><span class="math-container">$$(15-ac)x=2c-1$$</span>
Can you finish?The solution is given by <span class="math-container">$$\left\{ x=-2\,{\frac {c-5}{ac-15}},y=1/2\,{\frac {ac-6\,a+3}{ac-15}},
z=2\,{\frac {a-3}{ac-15}} \right\}
$$</span>
It remains to consider the case <span class="math-container">$$ac-15=0$$</span></p>
|
636,246 | <blockquote>
<p>Let $g(x)=x^2\sin(1/x)$, if $x \neq 0$ and $g(0)=0$. If $\{r_i\}$ is the numeration of all rational numbers in $[0,1]$, define
$$
f(x)=\sum_{n=1}^\infty \frac{g(x-r_n)}{n^2}
$$
Show that $f:[0,1] \rightarrow R$ is differentiable in each point over [0,1] but $f'(x)$ is discontinuous over each $r_n$. Is possible that the set of all discontinuous points of $f'$ is precisely $\{r_n\}$?</p>
</blockquote>
<p>I'm not seeing how this function is working. I could not even derive it. I need to fix some $ n $ to work? And to see the discontinuity of $ f '(x) $ after that? Can anyone give me any tips? I am not knowing how to work with this exercise and do not know where to start.</p>
| Cameron Williams | 22,551 | <p>First a stylistic comment: you should use the word "differentiable" in place of "derivable." Second: you should show that $f$ is well-defined on $[0,1]$ so that you can take its derivative (this is easy). We want to consider the difference quotient $\frac{f(x)-f(y)}{x-y}$ and what happens as $x\rightarrow y$ (I will leave it to you to argue why you can interchange sum and limit.. Weierstrass M-test and uniform convergence are your friends). I'll help out with part of the solution.</p>
<p>So we want to evaluate the limit of the difference quotient. To do so, we consider two cases: when $y$ is irrational and when $y$ is rational.</p>
<p><strong>Case 1</strong>: $y$ <em>is irrational.</em></p>
<p>What we have is </p>
<p>$$\lim_{x\rightarrow y}\frac{f(x)-f(y)}{x-y} = \sum_{n=0}^{\infty}\frac{1}{n^2}\lim_{x\rightarrow y}\frac{g(x-r_n)-g(y-r_n)}{x-y}.$$</p>
<p>The important part here is the limit, so I'll focus on that. The limit looks eerily close to a difference quotient (and after a clever introduction of $0$, we see that it is):</p>
<p>$$\lim_{x\rightarrow y}\frac{g(x-r_n)-g(y-r_n)}{x-y} = \lim_{x\rightarrow y}\frac{g(x-r_n)-g(y-r_n)}{(x-r_n)-(y-r_n)}.$$</p>
<p>This is just the derivative of $g$ evaluated at $y-r_n$! Which is equal to $2(y-r_n)\sin\left(\frac{1}{y-r_n}\right)-\cos\left(\frac{1}{y-r_n}\right)$. This is well-defined for all $r_n$ since $y$ is irrational. Since this function is bounded for all $r_n$ and $y$ by the value of $4$, the series converges and so $f'(y)$ is well-defined if $y$ is irrational.</p>
<p><strong>Case 2</strong>: $y$ <em>is rational.</em></p>
<p>If $y$ is a rational in $[0,1]$, then $y=r_k$ for some $k$. Then our difference quotient is</p>
<p>$$\lim_{x\rightarrow r_k}\frac{f(x)-f(r_k)}{x-r_k} = \lim_{x\rightarrow r_k}\left(\frac{1}{k^2}\frac{g(x-r_k)-g(r_k-r_k)}{x-r_k}+\sum_{n\neq k}\frac{1}{n^2}\frac{g(x-r_n)-g(r_k-r_n)}{x-r_k}\right).$$</p>
<p>We could not haphazardly apply the trick from above in this case because we required that $y$ be irrational above (else the denominator in the trigonometric functions will be ill-defined) which is why we split off the term in the series corresponding to $y$ in this case. Notice that the remainder of the series is now susceptible to the trick we did above and the term we pulled out is easy to handle. This gives us:</p>
<p>$$\lim_{x\rightarrow r_k} \frac{f(x)-f(r_k)}{x-r_k} = \lim_{x\rightarrow r_k}\left(\frac{1}{k^2}\frac{g(x-r_k)}{x-r_k} + \sum_{n\neq k}\frac{1}{n^2}\frac{g(x-r_n)-g(r_k-r_n)}{(x-r_n)-(r_k-r_n)}\right).$$</p>
<p>The first part is simply $\frac{1}{k^2}g'(0)$ and the second part is exactly like above.</p>
<p>Do you see how this is also well-defined making $f'$ differentiable everywhere? Can you take it from here?</p>
|
1,545,583 | <p>Suppose that $K$ is an infinite compact metric space. Define $c_0=\{ (x_n)_{n \in \mathbb{N}}| \lim_n{\| x_n \|}=0 \}$.</p>
<p>Is it true that $c_0$ complemented in $C(K)$, the set of continuous functions on $K$?</p>
<p>It seems true based on this <a href="http://www.ams.org/journals/proc/2006-134-04/S0002-9939-05-08301-2/S0002-9939-05-08301-2.pdf" rel="nofollow">paper</a>(first sentence in the proof of Theorem $5$).</p>
<p>Can anyone prove it?</p>
| Jochen | 38,982 | <p>Let $t_n \in K$ be a convergent sequence of distinct points with limit $t_\infty$ also distinct from all $t_n$, consider
$C_0(K)=\lbrace f\in C(K): f(t_\infty)=0\rbrace$ and consider $P:C_0(K)\to c_0$, $f\mapsto (f(t_n))_{n\in\mathbb N}$. Conversely, choose peak functions $\varphi_n \in C(K)$ with disjoint supports and $\varphi_n(t_k)=\delta_{n,k}$ and define $I:c_0\to C_0(K)$, $(x_n)_{n\in\mathbb N} \mapsto \sum_{n=1}^\infty x_n \varphi_n$ (this is only formally a series, for each $t\in K$ at most one term of $\sum_{n=1}^\infty x_n \varphi_n(t)$ does not vanish). This should establish $c_0$ as a complemented subspace of $C_0(K)$. It remains to see that $C(K)\to C_0(K)$, $f\mapsto f-f(t_\infty)$ is a projector.</p>
|
1,545,583 | <p>Suppose that $K$ is an infinite compact metric space. Define $c_0=\{ (x_n)_{n \in \mathbb{N}}| \lim_n{\| x_n \|}=0 \}$.</p>
<p>Is it true that $c_0$ complemented in $C(K)$, the set of continuous functions on $K$?</p>
<p>It seems true based on this <a href="http://www.ams.org/journals/proc/2006-134-04/S0002-9939-05-08301-2/S0002-9939-05-08301-2.pdf" rel="nofollow">paper</a>(first sentence in the proof of Theorem $5$).</p>
<p>Can anyone prove it?</p>
| Tomasz Kania | 17,929 | <p>Even more is true. Every copy of $c_0$ in $C(K)$ for $K$ compact, metric is complemented by a projection of norm at most 2. Indeed, $C(K)$ is in this case separable (as $K$ is second-countable we may use the <a href="https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem" rel="nofollow">Stone–Weierstrass theorem</a> to get the claim) and then we may apply <a href="http://matematicas.unex.es/~fcabello/files/printable/21.pdf" rel="nofollow">Sobczyk's theorem</a>.</p>
|
10,942 | <p>I heard about it sometime somewhere and want to read about it now, but I can't recall what the name is:</p>
<p>Start with $a_1 = \ldots =a_n=1$. Choose a number between 1 and $n$ with probability $a_i/(a_1+ \ldots + a_n)$ to choose $i$. If $i_0$ is the number chosen, increase $a_i$ by 1 and now choose another number and so on indefinitely.</p>
| Shai Covo | 2,810 | <p>For this process and a substantial generalization, see <a href="http://www.combinatorics.net/aoc/toc/v7n2/Chung/7_2_141.pdf" rel="nofollow">this</a>.
The case $p=0$ and $\gamma = 1$ corresponds to the process you described.</p>
|
302,061 | <p>Can you say how to find number of non-abelian groups of order n?</p>
<p>Suppose n is 24 ,then from structure theorem of finite abelian group we know that there are 3 abelian groups.But what can you say about the number of non-abelian groups of order 24?</p>
<p>The following link is a list of number of groups of order n:
<a href="http://oeis.org/wiki/Number_of_groups_of_order_n" rel="nofollow">http://oeis.org/wiki/Number_of_groups_of_order_n</a>
.But here also they did not mention anything how to find number of non-abelian groups of order n.</p>
| Andreas Caranti | 58,401 | <p>The point is, the numbers grow very fast, particularly for prime powers. Look at <a href="http://groupprops.subwiki.org/wiki/Groups_of_order_2%5En" rel="nofollow">this list</a> for groups of order $2^{k}$, for $k \le 10$. (I will try and provide a better reference later.)</p>
|
3,242,363 | <blockquote>
<p>Why does this function, <span class="math-container">$$\tan\left(x ^ {1/x}\right)$$</span>
have a maximum value at <span class="math-container">$x=e$</span>?</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/pqE0Q.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pqE0Q.png" alt="Graph"></a></p>
| Vineet | 196,541 | <p>This is going to be clumsy.</p>
<p><span class="math-container">$a=(x+b)\cdot0.029+0.3 ~~~~~~~~~(i)\\
b=(x+a)\cdot0.015 ~~~~~~~~~~~~~~~~~~(ii)\\
\frac{b}{0.015}=x+a\Rightarrow \boxed{a = \frac{b}{0.015}-x}$</span></p>
<p>Putting this in equation <span class="math-container">$(i)$</span>,</p>
<p><span class="math-container">$\frac{b}{0.015}-x -0.3=(x+b)\cdot0.029$</span></p>
<p><span class="math-container">$\frac{b}{0.015}-x -0.3=b\cdot0.029+x\cdot0.029$</span></p>
<p><span class="math-container">$x\cdot0.029+x=\frac{b}{0.015}-b\cdot0.029-0.3$</span></p>
<p><span class="math-container">$x\cdot1.029=\frac{b}{0.015}-b\cdot0.029-0.3$</span></p>
<p><span class="math-container">$x\cdot1.029+0.3=b(\frac{1}{0.015}-0.029)$</span></p>
<p><span class="math-container">$x\cdot1.029+0.3=66.6376\cdot b$</span></p>
<p><span class="math-container">$\therefore \boxed{b = x\cdot\frac{1.029}{66.6376}+\frac{0.3}{66.6376}}$</span></p>
<p>We know, <span class="math-container">$y=x+a+b$</span></p>
<p><span class="math-container">$y=x+\frac{b}{0.015}-x+x\cdot\frac{1.029}{66.6376}+\frac{0.3}{66.6376}$</span></p>
<p><span class="math-container">$y=66.66b+0.004+.0154x$</span></p>
<p>substituting, the value of <span class="math-container">$b$</span> will give us a linear equation in <span class="math-container">$x$</span> something like <span class="math-container">$ax+b = y$</span>. I am not sure how you gonna interpret this.</p>
|
2,012,318 | <p>Find the volume of:</p>
<p>$V=[(x,y,z): 0 \leqslant z \leqslant 4 - \sqrt{x^2+y^2}, 2x \leqslant x^2+ y^2 \leqslant 4x] $</p>
<p>I should somehow construct triple integral here in order to solve this, which means that i have to find limits of integration for three variables, but i am just not quite sure how, i assume that i should first integrate for $z$ since we could say that limits for $z$ are already given, but what i am supposed to do with other two variables. When i find limits, what function i am going to integrate, is it going to be just $ \iiint dzdydx $?</p>
| Eugene Zhang | 215,082 | <p>\begin{align}
(x,y)\in ((A \times B) -(A\times C))&\iff (x,y)\in (A \times B) \land (x,y) \not \in (A \times C)
\\
&\iff(x\in A \land y\in B) \land (x\not \in A \lor y \not \in C)
\\
&\iff (x \in A \land y \in B \land x\not \in A) \lor (x \in A \land y \in B \land y\not \in C)
\\
&\iff 0\lor (x \in A \land y \in B \land y\not \in C)
\\
&\iff x \in A \land y \in B \land y\not \in C
\\
&
\iff x \in A \land y \in (B - C)
\\
&\iff (x,y)\in A\times(B-C)
\end{align}
Hence
$$
(A \times B) -(A\times C)=A\times(B-C)
$$</p>
|
2,994,970 | <p>As far that i have known, i understand the notion "a function on the circle" by each one of the followings (both equivalent):</p>
<ol>
<li>A function is defined on <span class="math-container">$\mathbb{R}$</span> that is <span class="math-container">$2\pi-$</span>periodic.</li>
<li>A function that is defined on <span class="math-container">$[a,b]$</span> with <span class="math-container">$b-a=2\pi$</span> and <span class="math-container">$f(a)=f(b).$</span> So we can extend this function to get a <span class="math-container">$2\pi-$</span>periodic function.</li>
</ol>
<p>And my problem is: i am confused every time the author (of the book i have been reading) uses the notion "integrable on the circle". So, by "a function that is integrable on the circle", do i have to get the meaning in which way:</p>
<ol>
<li>A function that is integrable on all the interval of length <span class="math-container">$2\pi.$</span> Just like this <a href="http://math.uchicago.edu/~may/REU2017/REUPapers/Xue.pdf" rel="nofollow noreferrer">http://math.uchicago.edu/~may/REU2017/REUPapers/Xue.pdf</a> (page 1)</li>
<li>A function that is needed firstly be a function on the circle and then, it is integrable on some interval of length <span class="math-container">$2\pi$</span> (because the integral gets the same value over any interval of length <span class="math-container">$2\pi$</span>) <a href="https://i.stack.imgur.com/w7dKa.png" rel="nofollow noreferrer">like this</a></li>
</ol>
<p>I ask the question because of <a href="https://i.stack.imgur.com/2n0qT.png" rel="nofollow noreferrer">the passage 1</a> and <a href="https://i.stack.imgur.com/3MZH4.png" rel="nofollow noreferrer">the passage 2</a>. So do i need a function getting the same value at the end-points of every interval of lenth <span class="math-container">$2\pi?$</span> By the way, the author approachs the integral in the Riemann sense, if it helps.</p>
| edm | 356,114 | <p>The whole theory of Fourier series is built to study periodic functions. The correct interpretation of "integrable function on the circle" is 2.</p>
|
231,479 | <p>Is there a function that can create hexagonal grid?</p>
<p>We have square grid graph, where we can specify <code>m*n</code> dimensions:</p>
<pre><code>GridGraph[{m, n}]
</code></pre>
<p>We have triangular grid graph (which works only for argument <code>n</code> up to 10 - for unknown reason):</p>
<pre><code>GraphData[{"TriangularGrid", n}, "Graph"]
</code></pre>
<p>I can not find a function that would generate a hexagonal grid graph. I would like it like it is with <code>GridGraph</code> something like <code>HexagonalGridGraph[{m,n,o}]</code> where <code>m,n,o</code> are dimensions <code>m*n*o</code> of planar graph - or other way said - "lengths" of the sides of the graph.</p>
<p>I can make my own code, I am asking just in case there already exist implemented function.</p>
<p><strong>UPDATE:</strong></p>
<p>What I mean by <code>m*n*o</code> hexagonal grid is for example this <code>3*5*7</code> hexagonal grid:</p>
<p><a href="https://i.stack.imgur.com/r8yTS.png" rel="noreferrer"><img src="https://i.stack.imgur.com/r8yTS.png" alt="enter image description here" /></a></p>
<p>My code for producing it is very long and cumbersome so I will not upload it unless I can make it simpler.</p>
| Daniel Huber | 46,318 | <p>You can make a hexagonal grid using only MMA built in functions. You may adapt the code to your liking:</p>
<pre><code>c3 = Cos[30 Degree]; s3 = Sin[30 Degree];
del1 = {Sqrt[c3^2 + Sqrt[(1 + s3^2)^2 + c3^3]], c3} // N;
del2 = {-Sqrt[c3^2 + Sqrt[(1 + s3^2)^2 + c3^3]], c3} // N;
del3 = {0, 2 c3};
trans[del_] := Map[(del + #) &, hex, {2}];
n = 3;
grid = Flatten[
Table[trans[i1 del1 + i2 del2 + i3 del3], {i1, n}, {i2, n}, {i3,
n}], 2];
Graphics[Line /@ grid]
</code></pre>
<p><a href="https://i.stack.imgur.com/Ld0u3.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Ld0u3.png" alt="enter image description here" /></a></p>
|
231,479 | <p>Is there a function that can create hexagonal grid?</p>
<p>We have square grid graph, where we can specify <code>m*n</code> dimensions:</p>
<pre><code>GridGraph[{m, n}]
</code></pre>
<p>We have triangular grid graph (which works only for argument <code>n</code> up to 10 - for unknown reason):</p>
<pre><code>GraphData[{"TriangularGrid", n}, "Graph"]
</code></pre>
<p>I can not find a function that would generate a hexagonal grid graph. I would like it like it is with <code>GridGraph</code> something like <code>HexagonalGridGraph[{m,n,o}]</code> where <code>m,n,o</code> are dimensions <code>m*n*o</code> of planar graph - or other way said - "lengths" of the sides of the graph.</p>
<p>I can make my own code, I am asking just in case there already exist implemented function.</p>
<p><strong>UPDATE:</strong></p>
<p>What I mean by <code>m*n*o</code> hexagonal grid is for example this <code>3*5*7</code> hexagonal grid:</p>
<p><a href="https://i.stack.imgur.com/r8yTS.png" rel="noreferrer"><img src="https://i.stack.imgur.com/r8yTS.png" alt="enter image description here" /></a></p>
<p>My code for producing it is very long and cumbersome so I will not upload it unless I can make it simpler.</p>
| kglr | 125 | <p>We can generate the vertex coordinates using a slightly modified version of azerbajdan's <code>cells</code> and use them with <a href="https://reference.wolfram.com/language/ref/NearestNeighborGraph.html" rel="noreferrer"><code>NearestNeighborGraph</code></a>:</p>
<pre><code>ClearAll[vCoords]
vCoords = DeleteDuplicates @ Flatten[
Table[CirclePoints[{(2 j + k + l - 4) Sqrt[3] , 3 k - 2 - 3 l}, {2, π/2}, 6],
{j, #}, {k, #2}, {l, #3}], 3] &;
ClearAll[hexGridGraph]
hexGridGraph = Module[{v = vCoords @@ #},
NearestNeighborGraph[v, ##2, VertexCoordinates -> v]] &;
</code></pre>
<p><em><strong>Examples:</strong></em></p>
<pre><code>hexGridGraph[{3, 5, 7},
VertexLabels -> Placed["Index", Center],
VertexSize -> .7,
VertexStyle -> White,
VertexLabelStyle -> 8,
ImageSize -> 400]
</code></pre>
<p><a href="https://i.stack.imgur.com/Vjoif.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Vjoif.png" alt="enter image description here" /></a></p>
<pre><code>args = Sort /@ Tuples[Range[4], {3}] // Union;
hexGridGraph[#, PlotLabel -> #, ImageSize -> {100, 100}] & /@ args //
Multicolumn[#, 5] &
</code></pre>
<p><a href="https://i.stack.imgur.com/hPa1t.png" rel="noreferrer"><img src="https://i.stack.imgur.com/hPa1t.png" alt="enter image description here" /></a></p>
|
1,028,720 | <p>I was wondering the following:</p>
<blockquote>
<p><strong>Background Question:</strong> Does there exist a Banach space $X$ which contains a copy $X_0 \subset X$ of itself that is not complemented?</p>
</blockquote>
<p>By "$X_0$ is a copy of $X$", I mean $X_0 \cong X$ via an invertible, bounded, linear map. Some googling turned up that the answer to the above question is "yes". This <a href="https://mathoverflow.net/questions/140557/quotients-of-linfty">thread</a> points to a <a href="http://archive.numdam.org/ARCHIVE/CM/CM_1981__43_1/CM_1981__43_1_133_0/CM_1981__43_1_133_0.pdf" rel="nofollow noreferrer">paper</a> containing an example of an uncomplemented copy of $\ell^1$ in $\ell^1$. Since the proof seems to depend on some pretty fiddly analysis, and since answering the above question is only incidental to the aims of that paper, I was wondering whether there might be a simpler example of this phenomenon. In particular, I thought that the following example might work:</p>
<blockquote>
<p><strong>Example(?):</strong> The most famous, and probably the most elementary, example of an uncomplemented subspace is the canonical copy of $c_0$ in $\ell_\infty$. With this in mind, I thought maybe one could use the following:
\begin{align*}
X = c_0 \oplus \ell_\infty \oplus \ell_\infty \oplus \ell_\infty \oplus \ldots && X_0 = \{0\} \oplus c_0 \oplus \ell_\infty \oplus \ell_\infty \oplus \ell_\infty \oplus \ldots
\end{align*}
The hope is that, if $F \subset X$ was a complement for $X_0$, then this would imply $\pi_2(F) \subset \ell_\infty$ was a complement for $c_0 \subset \ell_\infty$ (a contradiction). Here, $\pi_2$ is projection onto the second factor $X \to \ell_\infty$. </p>
</blockquote>
<p>Towards completing the argument, I have already asked <a href="https://math.stackexchange.com/questions/1027012/if-e-is-not-complemented-in-x-is-e-oplus-0-not-complemented-in-x-o">this question</a>, but it hasn't attracted much attention. So, I am asking this (hopefully) better motivated question in addition. So, to reiterate, my question is:</p>
<blockquote>
<p><strong>Question:</strong> Is $\{0\} \oplus c_0 \oplus \ell_\infty \oplus \ell_\infty \oplus \ell_\infty \oplus \ldots$ complemented in $c_0 \oplus \ell_\infty \oplus \ell_\infty \oplus \ell_\infty\oplus \ldots$?</p>
</blockquote>
| Mike F | 6,608 | <p>Tomek's answer is very comprehensive. However, I thought I would add a "lower level" explanation.</p>
<blockquote>
<p><strong>Proposition:</strong> $\{0\} \oplus c_0 \oplus \ell_\infty \oplus \ell_\infty \oplus \ell_\infty \oplus \ldots$ is not complemented in $c_0 \oplus \ell_\infty \oplus \ell_\infty \oplus \ell_\infty\oplus \ldots$.</p>
</blockquote>
<p>The idea is to use the following simple lemma: </p>
<blockquote>
<p><strong>Lemma:</strong> Let $X \subset Y \subset Z$ be nested Banach spaces.</p>
<ol>
<li>If $X$ is complemented in $Z$, then $X$ is complemented in $Y$.</li>
<li>If $Y$ is complemented in $Z$, then $Y/X$ is complemented in $Z/X$.</li>
</ol>
</blockquote>
<p>Now we prove the proposition as follows. </p>
<ul>
<li>Suppose that $\{0\} \oplus c_0 \oplus \ell_\infty \oplus \ell_\infty \oplus \ell_\infty \oplus \ldots$ is complemented in $c_0 \oplus \ell_\infty \oplus \ell_\infty \oplus \ell_\infty\oplus \ldots$. </li>
<li>Using (2) to quotient out $\{0\} \oplus \{0\}\oplus \ell_\infty \oplus \ell_\infty \oplus \ldots$, we get that $\{0\} \oplus c_0$ is complemented in $c_0 \oplus \ell_\infty$.</li>
<li>Using (1), with the intermediate subspace being $\{0\} \oplus \ell_\infty$, we get that $\{0\} \oplus c_0$ is complemented in $\{0\} \oplus \ell_\infty$.</li>
<li>The latter was equivalent to "the usual copy of $c_0$ is complemented in $\ell_\infty$", which contradicts Phillip's Lemma.</li>
</ul>
|
4,236,077 | <p>Suppose that <span class="math-container">$A\subset B$</span> and <span class="math-container">$A\subset C$</span>. Why does this imply <span class="math-container">$A\subset B\cup C?$</span></p>
<p>If <span class="math-container">$x\in A$</span>, then since <span class="math-container">$A\subset B$</span> and <span class="math-container">$A\subset C$</span>, we know <span class="math-container">$x\in B$</span> and <span class="math-container">$x\in C.$</span> This implies <span class="math-container">$x\in B \cap C.$</span></p>
<p>But it is not generally true that <span class="math-container">$B\cap C\subset B\cup C $</span> since <span class="math-container">$B$</span> and <span class="math-container">$C$</span> may be disjoint.</p>
<p>So why does <span class="math-container">$A\subset B$</span> and <span class="math-container">$A\subset C$</span> imply <span class="math-container">$A\subset B\cup C?$</span></p>
| Shaun | 104,041 | <p>This is trivial if <span class="math-container">$A=\varnothing$</span>.</p>
<p>Let <span class="math-container">$x\in A$</span>. If <span class="math-container">$A\subset B$</span>, then <span class="math-container">$x\in A$</span> implies <span class="math-container">$x\in B$</span>. If <span class="math-container">$x\in B$</span>, then certainly <em>either</em> <span class="math-container">$x\in B$</span> or <span class="math-container">$x\in C$</span>. Thus <span class="math-container">$x\in B\cup C$</span>. Thus <span class="math-container">$A\subset B\cup C$</span>.</p>
|
4,236,077 | <p>Suppose that <span class="math-container">$A\subset B$</span> and <span class="math-container">$A\subset C$</span>. Why does this imply <span class="math-container">$A\subset B\cup C?$</span></p>
<p>If <span class="math-container">$x\in A$</span>, then since <span class="math-container">$A\subset B$</span> and <span class="math-container">$A\subset C$</span>, we know <span class="math-container">$x\in B$</span> and <span class="math-container">$x\in C.$</span> This implies <span class="math-container">$x\in B \cap C.$</span></p>
<p>But it is not generally true that <span class="math-container">$B\cap C\subset B\cup C $</span> since <span class="math-container">$B$</span> and <span class="math-container">$C$</span> may be disjoint.</p>
<p>So why does <span class="math-container">$A\subset B$</span> and <span class="math-container">$A\subset C$</span> imply <span class="math-container">$A\subset B\cup C?$</span></p>
| ryang | 21,813 | <p>Continuing from where you left off:
<span class="math-container">\begin{aligned}x&\in B\cap C\\&\subset B\cup C.\end{aligned}</span></p>
<p>Therefore, since <span class="math-container">$x\in A\implies x\in B\cup C,$</span> <span class="math-container">$$A\subset B\cup C.$$</span></p>
|
210,110 | <p>A good approximation of $(1+x)^n$ is $1+xn$ when $|x|n << 1$. Does this approximation have a name? Any leads on estimating the error of the approximation?</p>
| Christian Fries | 75,356 | <p>Wikipedia calls it <a href="https://en.wikipedia.org/wiki/Binomial_approximation" rel="nofollow noreferrer">Binomial Approximation</a>.</p>
|
117,285 | <p>Let $R \subseteq A \times A$ and $S \subseteq A \times A$ be two arbitary equivalence relations.
Prove or disprove that $R \cup S$ is an equivalence relation.</p>
<p>Reflexivity: Let $(x,x) \in R$ or $(x,x) \cup S \rightarrow (x,x) \in R \cup S$</p>
<p>Now I still have to prove or disprove that $R \cup S$ is symmetric and transitive. How can I do that?</p>
<p>My guess for symmetry is: R and S are equivalence relations, which means that $(x,y)\ and\ (y,x) \in R,S$ For each (x,y) in R and S there is an (y,x) in R and S so that (x,y) ~ (y,x). Is that correct?</p>
<p>Transitivity: ?</p>
| dtldarek | 26,306 | <p>Transitivity fails. Let $A = \{1,2,3\}$, $R = \{(1,1), (2,2), (3,3), (1,2), (2,1)\}$ and $S = \{(1,1), (2,2), (3,3), (2,3), (3,2)\}$, then $R \cup S$ contains both $(1,2)$ and $(2,3)$, but not $(1,3)$.</p>
<p>The symmetry could be worded better, but is alright. The important thing is that if $(x,y) \in R \cup S$ then $(x,y) \in R$ or $(x,y) \in S$, but both $R$ and $S$ are symmetrical so $(y,x)$ must be contained in at least one of them (here I use "or" instead of "and" you have used).</p>
|
4,074,630 | <p>Let <span class="math-container">$f: [a,b] \to [0,\infty)$</span> and <span class="math-container">$f$</span> is Riemann Integrable on every subinterval <span class="math-container">$[a + \epsilon,b]$</span> for <span class="math-container">$\epsilon > 0$</span>. Suppose that the improper Riemann integral exists. That is
<span class="math-container">$$I = \lim_{\epsilon \to 0} \int_{a + \epsilon}^{b} f(x) dx < \infty$$</span>
exists. Prove that <span class="math-container">$f$</span> is Lebesgue integrable on <span class="math-container">$[a,b]$</span> and that <span class="math-container">$\int_{[a,b]} f(x) dx = I$</span>.</p>
<p>I found another posting of this problem written in the same way, but it used dominated convergence theorem, which I have not covered. The others I saw were written slightly differently and/or weren't making too much sense.</p>
<p>Also, maybe I'm not seeing something clearly, but here's something I thought:</p>
<p>The improper integral I is finite (it exists). If the improper integral exists, isn't it equal to the "proper" integral? That would mean the "normal" proof of showing that every Riemann Integrable function is Lebesgue integrable would apply. However, I have a feeling this is not the way to prove it or else this question wouldn't be asked.</p>
<p>Thanks for the help!</p>
| RRL | 148,510 | <p>Since <span class="math-container">$f$</span> is nonnegative, the Lebesgue integral must exist (but may be infinite). With the existence of the improper Riemann integral we can show that the Lebesgue integral is finite and <span class="math-container">$f$</span> is "Lebesgue integrable" on <span class="math-container">$[a,b]$</span>.</p>
<p>For all sufficiently large <span class="math-container">$n$</span> we have <span class="math-container">$[a+1/n,b] \subset [a,b]$</span>. Since <span class="math-container">$f \chi_{[a+1/n,b]} \nearrow f$</span> as <span class="math-container">$n \to \infty$</span>, it follows by the monotone convergence theorem that</p>
<p><span class="math-container">$$\int_{[a,b]} f = \int_{[a,b]} \lim_{n \to \infty}f \chi_{[a+1/n,b]} = \lim_{n \to \infty}\int_{[a,b]} f \chi_{[a+1/n,b]} = \lim_{n \to \infty}\int_{a+ 1/n}^b f(x) \, dx = \int_a^bf(x) \, dx < +\infty$$</span></p>
<p>Here we have applied the equivalence of the Lebesgue and Riemann integrals on the interval <span class="math-container">$[a+1/n,b]$</span></p>
|
293,937 | <p>Let me given with an obvious example. Let $\Omega\subset{\mathbb R}^n$ be an open domain. If $f,g\in L^1(\Omega)$ and $f,g\ge0$, then $\sqrt{fg}\,\in L^1(\Omega)$.</p>
<p>Now let me replace the absolutely continuous measures $f(x)dx$ and $g(x)dx$, by a pair $\lambda,\mu$ of non-negative bounded measures on $\Omega$.</p>
<blockquote>
<p>Is there a natural way to define a geometric mean $\sqrt{\lambda\mu}$ as a non-negative measure ?</p>
</blockquote>
<p>Here is an incomplete attempt: remark that if $a,b\in[0,\infty)$, then $\sqrt{ab}=\inf_{p,q>0}\frac{ap+bq}{2\sqrt{pq}}\,$. This suggests to define
$$\langle\sqrt{\lambda\mu},\phi\rangle:=\inf_h\{\langle\lambda,h\phi\rangle+\langle\mu,\frac\phi{h}\rangle\}\qquad\forall\phi\in C^+_K(\Omega)$$
where the infimum runs over function $h\in C(\overline{\Omega})$ that are strictly positive. This definition has several nice features: - it yields the correct function when applied to $L^1$-functions, - as a function of $\phi$, it is positive and homogeneous of degree one. However, it is not clear whether this defines a linear form.</p>
| Fedor Petrov | 4,312 | <p>Any two measures are a.c. with respect to their sum, and you may take the geometric mean of the densities. This is the same definition as yours with infimum. </p>
|
293,937 | <p>Let me given with an obvious example. Let $\Omega\subset{\mathbb R}^n$ be an open domain. If $f,g\in L^1(\Omega)$ and $f,g\ge0$, then $\sqrt{fg}\,\in L^1(\Omega)$.</p>
<p>Now let me replace the absolutely continuous measures $f(x)dx$ and $g(x)dx$, by a pair $\lambda,\mu$ of non-negative bounded measures on $\Omega$.</p>
<blockquote>
<p>Is there a natural way to define a geometric mean $\sqrt{\lambda\mu}$ as a non-negative measure ?</p>
</blockquote>
<p>Here is an incomplete attempt: remark that if $a,b\in[0,\infty)$, then $\sqrt{ab}=\inf_{p,q>0}\frac{ap+bq}{2\sqrt{pq}}\,$. This suggests to define
$$\langle\sqrt{\lambda\mu},\phi\rangle:=\inf_h\{\langle\lambda,h\phi\rangle+\langle\mu,\frac\phi{h}\rangle\}\qquad\forall\phi\in C^+_K(\Omega)$$
where the infimum runs over function $h\in C(\overline{\Omega})$ that are strictly positive. This definition has several nice features: - it yields the correct function when applied to $L^1$-functions, - as a function of $\phi$, it is positive and homogeneous of degree one. However, it is not clear whether this defines a linear form.</p>
| R W | 8,588 | <p>This is what is called "Hellinger integral" and appears in the definition of the <a href="https://en.wikipedia.org/wiki/Hellinger_distance" rel="nofollow noreferrer">Hellinger distance</a>.</p>
|
1,913,835 | <p>I'm having a difficult time explaining/understanding a (seemingly) simple argument of an algorithm that I know I can use to determine if a directed graph <strong>G</strong> is strongly connected.</p>
<p>The algorithm that I know (does this have a name?) goes like this:</p>
<pre><code>Use BFS (breadth-first-search) on G staring from some node S
IF every node is found
Construct G^ (G with reversed Edges, G transpose?)
Use BFS on G^ starting from the same node S
IF every node is found
G is strongly connected
ELSE
G is not strongly connected
ELSE
G is not strongly connected
End
</code></pre>
<p>So the first run of BFS ensures that the node S can reach every other node on G, which makes sense. If it cant, then clearly G is not strongly connected.</p>
<p>The second run of BFS on G^, as I understand, will show that any node on the graph can also make it to S in G. <strong>This is the part I can't fully explain to myself.</strong> </p>
<p>The conclusion of the algorithm I understand, if S can make it to every node and every node can make it to S, then any two nodes will always be able to make it to each other through S.</p>
<p>To reiterate my question, I'm looking for an explanation on why determining S can make it to every node in G^ shows that every node can make it to S in G. I've tried doing a proof by contradiction, but I'm having a hard time wording it out. Any explanation will help me. Thank you!</p>
| angryavian | 43,949 | <p>I think the analysis is simple if the starting note $S$ for the second run of BFS is the same as the starting node of the first run.</p>
<p>The first run tells you that every node can be reached from $S$; the second run tells you every node can reach $S$. Combining these two facts shows strong connectivity: you can start from any node, go to $S$, and then go to any other node.</p>
|
19,356 | <p>So I was wondering: are there any general differences in the nature of "what every mathematician should know" over the last 50-60 years? I'm not just talking of small changes where new results are added on to old ones, but fundamental shifts in the nature of the knowledge and skills that people are expected to acquire during or before graduate school.</p>
<p>To give an example (which others may disagree with), one secular (here, secular means "trend over time") change seems to be that mathematicians today are expected to feel a lot more comfortable with picking up a new abstraction, or a new abstract formulation of an existing idea, even if the process of abstraction lies outside that person's domain of expertise. For example, even somebody who knows little of category theory would not be expected to bolt if confronted with an interpretation of a subject in his/her field in terms of some new categories, replete with objects, morphisms, functors, and natural transformations. Similarly, people would not blink much at a new algebraic structure that behaves like groups or rings but is a little different.</p>
<p>My sense would be that the expectations and abilities in this regard have improved over the last 50-60 years, partly because of the development of "abstract nonsense" subjects including category theory, first-order logic, model theory, universal algebra etc., and partly because of the increasing level of abstraction and the need for connecting frameworks and ideas even in the rest of mathematics. I don't really know much about how mathematics was taught thirty years ago, but I surmised the above by comparing highly accomplished professional mathematicians who probably went to graduate school thirty years ago against today's graduate students.</p>
<p>Some other guesses:</p>
<ol>
<li>Today, people are expected to have a lot more of a quick idea of a larger number of subjects, and less of an in-depth understanding of "Big Proofs" in areas outside their subdomain of expertise. Basically, the Great Books or Great Proofs approach to learning may be declining. The rapid increase in availability of books, journals, and information via the Internet (along with the existence of tools such as Math Overflow) may be making it more profitable to know a bit of everything rather than master big theorems outside one's area of specialization.</li>
<li>Also, probably a thorough grasp of multiple languages may be becoming less necessary, particularly for people who are using English as their primary research language. Two reasons: first, a lot of materials earlier available only in non-English languages are now available as English translations, and second, translation tools are much more widely available and easy-to-use, reducing the gains from mastery of multiple languages.</li>
</ol>
<p>These are all just conjectures. Contradictory information and ideas about other possible secular trends would be much appreciated.</p>
<p>NOTE: This might be too soft for Math Overflow! Moderators, please feel free to close it if so.</p>
| Georges Elencwajg | 450 | <p>Many, many things have changed in the last 60 years. A mathematician of the fifties (in Europe) was required to know descriptive geometry, rational mechanics, maybe some astronomy, and a lot of physics. He (yes!) was supposed to know how to calculate rather difficult primitives and have many tricks at his fingertips for checking the convergence of a series. Masterful use of logarithms tables and slide-rules went without saying. Nomography, the graphical representation of mathematical relationships (I guess even the word is forgotten), was a popular option, etc...</p>
|
4,614,334 | <p>I'm trying to resolve this recurrence equation <span class="math-container">$T(n)=4T(\frac{n}{2})+cn$</span>.</p>
<p>The solution I discovered online is <span class="math-container">$T(n)=\theta(n^2)$</span>.</p>
<p>The steps I follow was these: <br/>
a) Create the tree of recurrence as this:</p>
<pre><code> cn cost cn
/ / \ \
c(n/2) c(n/2) c(n/2) c(n/2) cost 2cn
//\\ //\\ //\\ //\\
c(n/2)^2 c(n/2)^2 c(n/2)^2 ... c(n/2)^2 cost 2^2cn
... ... ... ... ...
T(1) T(1) ... T(1) T(1) cost 2^icn
</code></pre>
<p>With <span class="math-container">$\log_2n$</span> expected level. So the longest path and calculation shall be <span class="math-container">$2^i\log_2cn$</span>, now I think I can ignore the <span class="math-container">$2^i$</span> and got an <span class="math-container">$O(n\log n)$</span>. That differs from the solution.</p>
<p>I tried to follow with the substitution method: <br/>
b) let's assume O(n\log n), <span class="math-container">$f(n) = dn\log_2n$</span> with d a constant greater than zero be more of our function, with the constant c set as the same value <span class="math-container">$d$</span>.</p>
<p><span class="math-container">$T(n) <= 4d\frac{n}{2}\log_2n+cn$</span> <br/>
<span class="math-container">$T(n) <= 2dn\log_2n+cn$</span> <br/>
<span class="math-container">$dn\log_2n <= 2dn\log_2n+cn$</span> <br/>
<span class="math-container">$-dn\log_2n <= cn$</span> <br/>
true for all <span class="math-container">$d>=\frac{c}{\log_2n}$</span></p>
<p>Where am I wrong?</p>
<p>Source of the solutions:
a) That says shall be <span class="math-container">$n^2+2cn$</span>: <a href="https://github.com/gzc/CLRS/blob/master/C04-Recurrences/4.2.md" rel="nofollow noreferrer">https://github.com/gzc/CLRS/blob/master/C04-Recurrences/4.2.md</a>, point 4.2-3
b) WolframAlpha, that say is, is a <span class="math-container">$\theta$</span> of <span class="math-container">$n^2$</span> <a href="https://www.wolframalpha.com/input?i=g%28n%29%3D4g%28n%2F2%29" rel="nofollow noreferrer">https://www.wolframalpha.com/input?i=g%28n%29%3D4g%28n%2F2%29</a></p>
| sibillalazzerini | 1,137,888 | <p><span class="math-container">$T(n)=4T(n/2)+cn$</span></p>
<p><span class="math-container">$ =4[4T(n/4)+cn/2]+cn$</span></p>
<p><span class="math-container">$ =16T(n/4)+3cn$</span></p>
<p><span class="math-container">$ =16[4T(n/8)+cn/4]+3cn$</span></p>
<p><span class="math-container">$ =64T(n/8)+7cn$</span></p>
<p><span class="math-container">$ =64[4T(n/16)+cn/8]+7cn$</span></p>
<p><span class="math-container">$ =256T(n/16)+15cn$</span></p>
<p>......................</p>
<p>......................</p>
<p>After k substitutions:</p>
<p><span class="math-container">$ =4^{k+1}T(n/2^{k+1})+(2^{k+1}-1)cn$</span></p>
<p>Putting <span class="math-container">$k=log(n)$</span> as after <span class="math-container">$log(n)$</span> steps the <span class="math-container">$T(n/2^k)$</span> converges to <span class="math-container">$T(1)$</span>,</p>
<p><span class="math-container">$ =4n^2T(1/2)+(2n-1)cn$</span></p>
<p><span class="math-container">$ =n^2.c' + n(2n-1)c$</span></p>
<p><span class="math-container">$ = ~ O(n^2)$</span></p>
|
4,614,334 | <p>I'm trying to resolve this recurrence equation <span class="math-container">$T(n)=4T(\frac{n}{2})+cn$</span>.</p>
<p>The solution I discovered online is <span class="math-container">$T(n)=\theta(n^2)$</span>.</p>
<p>The steps I follow was these: <br/>
a) Create the tree of recurrence as this:</p>
<pre><code> cn cost cn
/ / \ \
c(n/2) c(n/2) c(n/2) c(n/2) cost 2cn
//\\ //\\ //\\ //\\
c(n/2)^2 c(n/2)^2 c(n/2)^2 ... c(n/2)^2 cost 2^2cn
... ... ... ... ...
T(1) T(1) ... T(1) T(1) cost 2^icn
</code></pre>
<p>With <span class="math-container">$\log_2n$</span> expected level. So the longest path and calculation shall be <span class="math-container">$2^i\log_2cn$</span>, now I think I can ignore the <span class="math-container">$2^i$</span> and got an <span class="math-container">$O(n\log n)$</span>. That differs from the solution.</p>
<p>I tried to follow with the substitution method: <br/>
b) let's assume O(n\log n), <span class="math-container">$f(n) = dn\log_2n$</span> with d a constant greater than zero be more of our function, with the constant c set as the same value <span class="math-container">$d$</span>.</p>
<p><span class="math-container">$T(n) <= 4d\frac{n}{2}\log_2n+cn$</span> <br/>
<span class="math-container">$T(n) <= 2dn\log_2n+cn$</span> <br/>
<span class="math-container">$dn\log_2n <= 2dn\log_2n+cn$</span> <br/>
<span class="math-container">$-dn\log_2n <= cn$</span> <br/>
true for all <span class="math-container">$d>=\frac{c}{\log_2n}$</span></p>
<p>Where am I wrong?</p>
<p>Source of the solutions:
a) That says shall be <span class="math-container">$n^2+2cn$</span>: <a href="https://github.com/gzc/CLRS/blob/master/C04-Recurrences/4.2.md" rel="nofollow noreferrer">https://github.com/gzc/CLRS/blob/master/C04-Recurrences/4.2.md</a>, point 4.2-3
b) WolframAlpha, that say is, is a <span class="math-container">$\theta$</span> of <span class="math-container">$n^2$</span> <a href="https://www.wolframalpha.com/input?i=g%28n%29%3D4g%28n%2F2%29" rel="nofollow noreferrer">https://www.wolframalpha.com/input?i=g%28n%29%3D4g%28n%2F2%29</a></p>
| zwim | 399,263 | <p>Rewrite <span class="math-container">$T(2n)=4T(n)+2cn$</span> and set <span class="math-container">$U(n)=\dfrac{4T(n)}{n}+\alpha \dfrac{2cn}n$</span></p>
<p>(this is motivated by this -> <a href="https://math.stackexchange.com/a/3002925/399263">https://math.stackexchange.com/a/3002925/399263</a>)</p>
<p>After substitution the <span class="math-container">$n$</span> gets to simplify and you have :</p>
<p><span class="math-container">$U(2n)=\dfrac{4T(2n)}{2n}+2\alpha c=\dfrac{16T(n)+8cn}{2n}+2\alpha c=2U(n)+(4c-2\alpha c)$</span></p>
<p>We see that setting <span class="math-container">$\alpha=2$</span> gets rid of the constant term and simplifies the relation to <span class="math-container">$$U(2n)=2U(n)$$</span></p>
<p>Which solves to <span class="math-container">$U(n)=n\,U(1)$</span></p>
<p>( <em>side note: <span class="math-container">$n=2^p$</span> and <span class="math-container">$V(p)=U(n)$</span> gives <span class="math-container">$V(p+1)=2V(p)\iff U(n)=V(p)=2^pV(0)=nV(0)=nU(1)$</span></em> )</p>
<p>Reporting in initial term gives <span class="math-container">$$T(n)=\frac 14n^2\,U(1)-cn=n^2\,T(1)+n(n-1)c=\mathcal O(n^2)$$</span></p>
|
2,322,678 | <p>We have $n$ different elements $(a_1,...,a_n)$ that are all the elements of $K$ and $\in$ finite field $K$.
I want to prove, that $\prod_{i=1}^{n} (X - a_i) + 1 \in K[X]$ doesn't have roots</p>
<p>I know, that if $a_i$ is a root of polynomial $p \in K[X]$ , then exists $f \in K[X]$ such that $p = (x - a_i)f$</p>
<p>How can we use this fact in order to prove the statement?</p>
| SvanN | 446,362 | <p>We know that $\lim_{x \to \infty} \frac{\sin{x}}{x} = 0$. Because the sine oscillates between being positive and negative, we can divide the integral up into 'chunks', on each of which the function is either wholly positive or wholly negative. If the integral exists, then this will be equal to the series of these 'chunks'. Then, by the alternating series theorem, the series (which equals the integral should it exist) converges - the terms go to zero monotonically, for sine is bounded and $\frac{1}{x}$ goes to zero monotonically.</p>
|
2,322,678 | <p>We have $n$ different elements $(a_1,...,a_n)$ that are all the elements of $K$ and $\in$ finite field $K$.
I want to prove, that $\prod_{i=1}^{n} (X - a_i) + 1 \in K[X]$ doesn't have roots</p>
<p>I know, that if $a_i$ is a root of polynomial $p \in K[X]$ , then exists $f \in K[X]$ such that $p = (x - a_i)f$</p>
<p>How can we use this fact in order to prove the statement?</p>
| Angina Seng | 436,618 | <p>Integration by parts gives
$$\int_0^N\frac{\sin x}{x}\,dx
=\left[\frac{1-\cos x}{x}\right]_0^N+\int_0^N\frac{1-\cos x}{x^2}\,dx
=\frac{1-\cos N}{N}+\int_0^N\frac{1-\cos x}{x^2}\,dx.$$
So
$$\lim_{N\to\infty}\int_0^N\frac{\sin x}{x}\,dx
=\lim_{N\to\infty}\int_0^N\frac{1-\cos x}{x^2}\,dx.$$
This last is a convergent integral.</p>
|
2,322,678 | <p>We have $n$ different elements $(a_1,...,a_n)$ that are all the elements of $K$ and $\in$ finite field $K$.
I want to prove, that $\prod_{i=1}^{n} (X - a_i) + 1 \in K[X]$ doesn't have roots</p>
<p>I know, that if $a_i$ is a root of polynomial $p \in K[X]$ , then exists $f \in K[X]$ such that $p = (x - a_i)f$</p>
<p>How can we use this fact in order to prove the statement?</p>
| DonAntonio | 31,254 | <p>About your last question in your post:</p>
<p>$$\lim_{x\to\infty}-\frac{\cos c}c=-\lim_{c\to\infty}\,\frac1c\cdot\cos c=0$$</p>
<p>since the last is the limit of a function whose limit zero times a <strong>bounded</strong> one.</p>
|
2,595,418 | <p>Let there be a graph $G$ and it's complement $G'$ , if the degree of a vertex in $G$ is added with degree of the corresponding vertex of $G'$ , the sum will be $(n-1)$; where $n$ is the number of vertices. How to prove this ?</p>
| Community | -1 | <p><strong>Hint:</strong></p>
<p>The union of $G$ and $\overline G$ contains an edge between each pair of the $n$ vertices. Hence, the union is $K_n$ which is the <a href="https://en.m.wikipedia.org/wiki/Complete_graph" rel="nofollow noreferrer">complete graph</a> of $n$ vertices. </p>
<p>What is the degree of each vertex of $K_n$? </p>
|
14,140 | <p>One of the most annoying "features" of <em>Mathematica</em> is that the <code>Plot</code> family does extrapolation on <code>InterpolatingFunction</code>s without any warning. I'm sure it was discussed to hell previously, but I cannot seem to find any reference. While I know how to simply overcome the problem by defining a global variable for the domain of the interpolation, from time to time I forget to do this and then I spend days figuring out where the numerical error originates. This could be avoided if <code>Plot</code> was to give a warning.</p>
<p>Consider the following example. An ODE system is defined and integrated for two different time ranges:</p>
<pre><code>odes = {
a'[t] == -a[t] - .2 a[t]^2 + 2. b[t],
b'[t] == a[t] + .1 a[t]^2 - 1.1 b[t], a[0] == 1, b[0] == 1
};
sol100 = First@NDSolve[odes, {a, b}, {t, 0, 100}];
sol500 = First@NDSolve[odes, {a, b}, {t, 0, 500}];
</code></pre>
<p>Now querying the function value for a point outside of the range correctly gives a warning:</p>
<pre><code>(a /. sol100)[500]
</code></pre>
<blockquote>
<pre><code>InterpolatingFunction::dmval: Input value {500} lies outside
the range of data in the interpolating function. Extrapolation will be used. >>
651.034
</code></pre>
</blockquote>
<p>The same is not done when we use the function in <code>Plot</code>:</p>
<pre><code>Show[
Plot[{a[t], b[t]} /. sol100, {t, 0, 400}, PlotStyle -> {Thick, Red}],
Plot[{a[t], b[t]} /. sol500, {t, 0, 400}, PlotStyle -> {Thick, Blue}]
]
</code></pre>
<p><img src="https://i.stack.imgur.com/X7xaL.png" alt="Mathematica graphics"></p>
<p>I've tried to force a warning, with no avail. The following example won't give a warning.</p>
<pre><code>On[InterpolatingFunction::dmval]
Check[Plot[{a[t], b[t]} /. sol100, {t, 0, 500}], "Error",
InterpolatingFunction::dmval]
</code></pre>
<p>Interestingly, one can be sure that <code>InterpolatingFunction::dmval</code> is NOT turned off at all inside the <code>Plot</code> family. In the following example, <code>LogLinearPlot</code> is able to drop a warning about sampling from below the domain (that can be ignored being unrelated, see <a href="https://mathematica.stackexchange.com/q/5986/89">this post</a>, also it seems to be fixed in v9), but it does not give the same warning when sampling from <strong>above</strong> (> 100)! </p>
<pre><code>LogLinearPlot[{a[t], b[t]} /. sol100, {t, 0.1, 500}]
</code></pre>
<blockquote>
<pre><code>InterpolatingFunction::dmval: Input value {-2.30241} lies outside the
range of data in the interpolating function. Extrapolation will be used. >>
</code></pre>
</blockquote>
<p>It is even more disturbing to see that <code>Plot</code> checks the lower boundary but not the upper (thanks to <a href="https://mathematica.stackexchange.com/users/50/j-m">J.M.</a> for the comment):</p>
<pre><code>Plot[{a[t], b[t]} /. sol100, {t, -1, 500}]
</code></pre>
<blockquote>
<pre><code>InterpolatingFunction::dmval: Input value {-0.989765} lies outside the
range of data in the interpolating function. Extrapolation will be used. >>
</code></pre>
</blockquote>
<p>As <a href="https://mathematica.stackexchange.com/users/312/oleksandr-r">Oleksandr</a> has pointed out, it is not about lower vs. upper boundaries but first point vs. the rest. </p>
<pre><code>Plot[{a[t], b[t]} /. sol100, {t, 101, 500}]
</code></pre>
<blockquote>
<pre><code>InterpolatingFunction::dmval: Input value {101.008} lies outside the
range of data in the interpolating function. Extrapolation will be used. >>
</code></pre>
</blockquote>
<h2><strong>Questions</strong></h2>
<ol>
<li>Why <code>Plot</code> does not give a warning when extrapolating an <code>InterpolatingFunction</code>? Is there some higher-level consideration that justifies this behaviour, or is it a bug?</li>
<li>How can one force <code>Plot</code> to give a warning? Is there any workaround that forces <code>InterpolatingFunction::dmval</code> not to be attenuated inside <code>Plot</code>?</li>
</ol>
| Rojo | 109 | <p>As you can see with </p>
<pre><code>if = Interpolation[Range[10]];
With[{mess = $Messages},
Plot[
Block[{$Messages = mess}, if[x]], {x, 5, 20}, Evaluated -> False]
]~Quiet~Message::msgl
</code></pre>
<p>the problem is that <code>$Messages</code> gets blocked somehow. I have so far no good recommendations as to how to fix it properly.</p>
|
3,587,891 | <blockquote>
<p>Suppose <span class="math-container">$f(x, y, z): \mathbb{R}^{3} \rightarrow \mathbb{R}$</span> is a <span class="math-container">$C^{2}$</span> harmonic function, that is, it satisfies <span class="math-container">$f_{x x}+f_{y y}+f_{z z}=0 .$</span> Let <span class="math-container">$E \subset \mathbb{R}^{3}$</span> be a region to which Gauss's Theorem can be applied. I want to show that
<span class="math-container">$$
\iint_{\partial E} f \nabla f \cdot \vec{n} d \sigma=\iiint_{E}|\nabla f|^{2}\mathrm{d}(x,y,z)
$$</span>
where, as usual, <span class="math-container">$\nabla f=\left(f_{x}, f_{y}, f_{z}\right)$</span> is the gradient.</p>
</blockquote>
<p>Can you help, how can I show?</p>
<h1>StayAtHome</h1>
| Henry Swanson | 55,540 | <p>Yeah, that's the approach you're supposed to take, I think.</p>
<p>Picking up where you left off, you want to show that, given two sections <span class="math-container">$f_i \in \mathscr F(U_i)$</span>, <span class="math-container">$f_j \in \mathscr F(U_j)$</span>, that they agree when restricted to <span class="math-container">$U_i \cap U_j$</span>.</p>
<p>At any point <span class="math-container">$p \in U_i \cap U_j$</span>, both <span class="math-container">$(f_i)\mid_{U_i \cap U_j}$</span> and <span class="math-container">$(f_j)\mid_{U_i \cap U_j}$</span> have the stalk <span class="math-container">$s_p$</span>. But by 2 4.A, two sections being everywhere stalk-wise equal means that they are the same section.</p>
|
3,046,083 | <p>Is it true that the intersection of the closures of sets <span class="math-container">$A$</span> and <span class="math-container">$B$</span> is equal to the closure of their intersection?
<span class="math-container">$ cl(A)\cap{cl(B)}=cl(A\cap{B})$</span> ?</p>
| bubba | 31,744 | <p>No, it's not true. Look at <a href="https://en.wikipedia.org/wiki/Closure_(topology)#Facts_about_closures" rel="nofollow noreferrer">this page</a>, or <a href="https://math.stackexchange.com/questions/1851554/closure-of-intersection-is-subset-of-intersection-of-closures?rq=1">this question</a>.</p>
<p>A simple counterexample stolen from the question cited above is as follows:</p>
<p>Take <span class="math-container">$A = (0,1)$</span> and <span class="math-container">$B = (1,2)$</span>. Then we have
<span class="math-container">$$\operatorname{cl}(A) \cap \operatorname{cl}(B) = [0,1] \cap [1,2] = \lbrace 1 \rbrace $$</span>
but
<span class="math-container">$$\operatorname{cl}(A \cap B) = \operatorname{cl}(\emptyset) = \emptyset$$</span></p>
|
3,779,785 | <p>So I have this problem, <span class="math-container">$W=3^n -n -1$</span>. How to find all <span class="math-container">$n$</span> so <span class="math-container">$W$</span> can be divided by <span class="math-container">$5$</span>.</p>
<p><em>what I tried:</em>
I found all the remainders of <span class="math-container">$3^n$</span> divided by <span class="math-container">$5$</span>, they are: <span class="math-container">$1,3,4,2$</span>.</p>
<p>if there isn't (<span class="math-container">$-n$</span>), it's easy but with it I can't continue.</p>
<p>this is the solutions of the problem: <span class="math-container">$n=20k+11$</span>, <span class="math-container">$n=20k+18$</span>, <span class="math-container">$n=20k+17$</span>, <span class="math-container">$n=20k$</span>. but I don't know how to find them</p>
<p>thanks</p>
| lhf | 589 | <p>Let <span class="math-container">$W_n=3^n -n -1$</span>.
Then <span class="math-container">$W_n \bmod 5$</span> is periodic of period <span class="math-container">$20$</span>:
<span class="math-container">$$
0,1,1,3,1,2,2,4,2,3,3,0,3,4,4,1,4,0,0,2,0,1,1,3,\dots
$$</span></p>
<p>This follows by induction because <span class="math-container">$W_n= 3^n -n1^n -1^n$</span> and so <span class="math-container">$W_{n+3}=5W_{n+2}-7W_{n+1}+3W_n$</span> because <span class="math-container">$(x-3)(x-1)^2=x^3 - 5 x^2 + 7 x - 3$</span>.</p>
|
3,773,856 | <p>I'm having trouble with part of a question on Cardano's method for solving cubic polynomial equations. This is a multi-part question, and I have been able to answer most of it. But I am having trouble with the last part. I think I'll just post here the part of the question that I'm having trouble with.</p>
<p>We have the depressed cubic equation :
<span class="math-container">\begin{equation}
f(t) = t^{3} + pt + q = 0
\end{equation}</span>
We also have what I believe is the negative of the discriminant :
<span class="math-container">\begin{equation}
D = 27 q^{2} + 4p^{3}
\end{equation}</span>
We assume <span class="math-container">$p$</span> and <span class="math-container">$q$</span> are both real and <span class="math-container">$D < 0$</span>. We also have the following polynomial in two variables (<span class="math-container">$u$</span> and <span class="math-container">$v$</span>) that results from a variable transformation <span class="math-container">$t = u+v$</span> :
<span class="math-container">\begin{equation}
u^{3} + v^{3} + (3uv + p)(u+v) + q = 0
\end{equation}</span>
You also have the quadratic polynomial equation :
<span class="math-container">\begin{equation}
x^{2} + qx - \frac{p^{3}}{27} = 0
\end{equation}</span>
The solutions to the 2-variable polynomial equation satisfy the following constraints :
<span class="math-container">\begin{equation}
u^{3} + v^{3} = -q
\end{equation}</span>
<span class="math-container">\begin{equation}
uv = -\frac{p}{3}
\end{equation}</span>
The first section of this part of the larger question asks to prove that the solutions of the quadratic equation are non-real complex conjugates. Here the solutions to the quadratic are equal to <span class="math-container">$u^{3}$</span> and <span class="math-container">$v^{3}$</span> (this relationship between the quadratic polynomial and the polynomial in two variables was proven in an earlier part of the question). I was able to do this part. The second part of this sub-question is what I'm having trouble with.</p>
<p>The question says, let :
<span class="math-container">\begin{equation}
u = r\cos(\theta) + ir\sin(\theta)
\end{equation}</span>
<span class="math-container">\begin{equation}
v = r\cos(\theta) - ir\sin(\theta)
\end{equation}</span>
The question then asks the reader to prove that the depressed cubic equation has three real roots :
<span class="math-container">\begin{equation}
2r\cos(\theta) \text{ , } 2r\cos\left( \theta + \frac{2\pi}{3} \right) \text{ , } 2r\cos\left( \theta + \frac{4\pi}{3} \right)
\end{equation}</span>
In an earlier part of the question they had the reader prove that given :
<span class="math-container">\begin{equation}
\omega = \frac{-1 + i\sqrt{3}}{2}
\end{equation}</span>
s.t. :
<span class="math-container">\begin{equation}
\omega^{2} = \frac{-1 - i\sqrt{3}}{2}
\end{equation}</span>
and :
<span class="math-container">\begin{equation}
\omega^{3} = 1
\end{equation}</span>
that if <span class="math-container">$(u,v)$</span> is a root of the polynomial in two variables then so are :
<span class="math-container">$(u\omega,v\omega^{2})$</span> and <span class="math-container">$(u\omega^{2},v\omega)$</span>. I think that the part of the question I'm having trouble with is similar. I suspect that :
<span class="math-container">\begin{equation}
2r \cos\left( \theta + \frac{2\pi}{3} \right) = u\omega + v\omega^{2} \text{ or } u\omega^{2} + v\omega \tag{1}
\end{equation}</span>
and :
<span class="math-container">\begin{equation}
2r \cos\left( \theta + \frac{4\pi}{3} \right) = u\omega + v\omega^{2} \text{ or } u\omega^{2} + v\omega \tag{2}
\end{equation}</span>
I have derived that :
<span class="math-container">\begin{equation}
\omega = \cos(\phi) + i\sin(\phi)
\end{equation}</span>
where <span class="math-container">$\phi = \frac{2\pi}{3}$</span>. Also :
<span class="math-container">\begin{equation}
\omega^{2} = \cos(2\phi) + i \sin(2\phi)
\end{equation}</span>
So that the goal of the question may be to prove equations <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span>. I have tried to do this but haven't been able to.</p>
<p>Am I approaching this question in the correct way ? If I am approaching it the right way can someone show me how to use trigonometric identities to prove equations #1 and #2 ?</p>
| Community | -1 | <p><strong>An alternative method:</strong></p>
<p>We can try to turn
<span class="math-container">$$t^3+pt+q=0$$</span>
into
<span class="math-container">$$4\cos^3\theta-3\cos\theta=a$$</span> by a change of variable: we set</p>
<p><span class="math-container">$$t=\lambda \cos\theta$$</span> and solve</p>
<p><span class="math-container">$$-\frac3{4\lambda^2}=p,$$</span> or</p>
<p><span class="math-container">$$\lambda=\sqrt{-\frac3{4p}}.$$</span> This establishes</p>
<p><span class="math-container">$$4\cos^3\theta-3\cos\theta=-4q\lambda^3.$$</span></p>
<p>But the LHS is just</p>
<p><span class="math-container">$$\cos3\theta.$$</span></p>
|
3,795,234 | <p>(disclaimer: I am not well versed in mathematics so please excuse my poor notation / explanation)</p>
<p>Given a hexagon grid that defines it's "neighbours" via offsets on the axis' <span class="math-container">$q$</span> & <span class="math-container">$r$</span> like this :
<a href="https://i.stack.imgur.com/AWkKH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AWkKH.png" alt="Image credit to: https://www.redblobgames.com/grids/hexagons/#coordinates" /></a></p>
<p>(the hexagon containing just <span class="math-container">$q$</span> & <span class="math-container">$r$</span> is the <span class="math-container">$(0,0)$</span> node)</p>
<p>I am looking for a function <span class="math-container">$f$</span> that calculates a unique natural number (like an identifier) for a given <span class="math-container">$(q,r)$</span> offset and a given max neighbour distance <span class="math-container">$r$</span>. The generated natural numbers should be between 1 and the neighbour count in regards to a maximum distance <span class="math-container">$d$</span><sup>[2]</sup> while also being "gapless": for <span class="math-container">$d = 1$</span> the numbers 1 to 6 should be assigned assigned to the neighbours while <span class="math-container">$d = 2$</span> implies the numbers from 1 to 18 are assigned.</p>
<p>I know that the value pairs of the <span class="math-container">$(q,r)$</span> offsets are unique but struggle to find a way to map them onto the wanted value range. A classic square grid based algorithm like <span class="math-container">$row*columncount + column$</span> is a good starter but of course has gaps in a hexagon based grid.</p>
<p>[2]: <span class="math-container">$3(d^2+d)$</span>,d > 0</p>
| Somos | 438,089 | <p>Given hexagon integer coordinates <span class="math-container">$\,q\,$</span> and <span class="math-container">$\,r,\,$</span> define <span class="math-container">$\,s:=-q-r\,$</span>
which implies <span class="math-container">$\,0=q+r+s.\,$</span> Two hexagons are "neighbors" exactly when one
of the <span class="math-container">$\,q,r,s\,$</span> coordinates is the same and one of the other coordinates
differs by <span class="math-container">$\,1\,$</span> in one direction and the third coordinate differs by
<span class="math-container">$\,1\,$</span> in the opposite direction.</p>
<p>The hexagons whose coordinates satisfy the conditions
<span class="math-container">$\,q\ge 0\,$</span> and <span class="math-container">$\,s\ge 1\,$</span> form a wedge shaped region.
The function <span class="math-container">$\,f(q,r,s):=3r^2+2r-s+1\,$</span> defines a numbering of the
hexagons in the region yielding <span class="math-container">$\,1,7,8,19,20,21,\dots.$</span></p>
<p>For <span class="math-container">$\,r\le -1,$</span> and <span class="math-container">$\,s\le 0\,$</span> the function <span class="math-container">$\,f(q,r,s) :=
3q^2-2q-s+1\,$</span> yielding <span class="math-container">$\,2,9,10,22,23,24,\dots.$</span></p>
<p>For <span class="math-container">$\,q\ge 1\,$</span> and <span class="math-container">$\,r\ge 0\,$</span> the function <span class="math-container">$\,f(q,r,s) := 3s^2+s+r+1\,$</span> yielding <span class="math-container">$\,3,11,12,25,26,27,\dots.$</span></p>
<p>For <span class="math-container">$\,q\le 0\,$</span> and <span class="math-container">$\,s\le -1\,$</span> the <span class="math-container">$\,f:=3r^2+r+s+1.\,$</span></p>
<p>For <span class="math-container">$\,r\ge 1\,$</span> and <span class="math-container">$\,s\ge 0\,$</span> the <span class="math-container">$\, f:=3q^2-2q-r+1.\,$</span></p>
<p>For <span class="math-container">$\,q\le -1\,$</span> and <span class="math-container">$\,r\le 0\,$</span> the <span class="math-container">$\, f:=3s^2+2s-r+1.$</span></p>
<p>The numbering I used starts with <span class="math-container">$0$</span> at the origin and the numbers
go clockwise around each hexagonal ring of hexagons. There are
other choices for numbering and some may be better than others.</p>
|
3,012,090 | <p>Let <span class="math-container">$x>0$</span>. I have to prove that</p>
<p><span class="math-container">$$
\int_{0}^{\infty}\frac{\cos x}{x^p}dx=\frac{\pi}{2\Gamma(p)\cos(p\frac{\pi}{2})}\tag{1}
$$</span></p>
<p>by converting the integral on the left side to a double integral using the expression below:</p>
<p><span class="math-container">$$
\frac{1}{x^p}=\frac{1}{\Gamma(p)}\int_{0}^{\infty}e^{-xt}t^{p-1}dt\tag{2}
$$</span></p>
<p>By plugging <span class="math-container">$(2)$</span> into <span class="math-container">$(1)$</span> I get the following double integral:</p>
<p><span class="math-container">$$
\frac{1}{\Gamma(p)}\int_{0}^{\infty}\int_{0}^{\infty}e^{-xt}t^{p-1}\cos xdtdx\tag{3}
$$</span></p>
<p>However, I unable to proceed any further as I am unclear as to what method should I use in order to compute this integral. I thought that an appropriate change of variables could transform it into a product of two gamma functions but I cannot see how that would work. Any help would be greatly appreciated.</p>
| Yadati Kiran | 490,720 | <p>Hint: <span class="math-container">$\displaystyle\int_{0}^{\infty}\frac{\cos x}{x^p}dx= \text{Real part of}\:\int_{0}^{\infty}\frac{e^{iz}}{z^p}dz$</span> and use residue theorem. This has a pole of order <span class="math-container">$p$</span> hence the term <span class="math-container">$\Gamma (p)$</span> in the denominator.</p>
|
3,012,090 | <p>Let <span class="math-container">$x>0$</span>. I have to prove that</p>
<p><span class="math-container">$$
\int_{0}^{\infty}\frac{\cos x}{x^p}dx=\frac{\pi}{2\Gamma(p)\cos(p\frac{\pi}{2})}\tag{1}
$$</span></p>
<p>by converting the integral on the left side to a double integral using the expression below:</p>
<p><span class="math-container">$$
\frac{1}{x^p}=\frac{1}{\Gamma(p)}\int_{0}^{\infty}e^{-xt}t^{p-1}dt\tag{2}
$$</span></p>
<p>By plugging <span class="math-container">$(2)$</span> into <span class="math-container">$(1)$</span> I get the following double integral:</p>
<p><span class="math-container">$$
\frac{1}{\Gamma(p)}\int_{0}^{\infty}\int_{0}^{\infty}e^{-xt}t^{p-1}\cos xdtdx\tag{3}
$$</span></p>
<p>However, I unable to proceed any further as I am unclear as to what method should I use in order to compute this integral. I thought that an appropriate change of variables could transform it into a product of two gamma functions but I cannot see how that would work. Any help would be greatly appreciated.</p>
| omegadot | 128,913 | <p>So let us follow your initial line of thought and convert the integral to a double integral. As you correctly observe, as
<span class="math-container">$$\frac{1}{x^p} = \frac{1}{\Gamma (p)} \int_0^\infty e^{-xt} t^{p - 1} \, dt,$$</span>
which, by the way, is just the Laplace transform for the function <span class="math-container">$x^{p -1}$</span>, as a double integral your integral can be rewritten as
<span class="math-container">$$\int_0^\infty \frac{\cos x}{x^p} \, dx = \frac{1}{\Gamma (p)} \int_0^\infty \int_0^\infty e^{-xt} \cos x t^{p - 1} \, dt \, dx,$$</span>
or
<span class="math-container">$$\int_0^\infty \frac{\cos x}{x^p} \, dx = \frac{1}{\Gamma (p)} \int_0^\infty t^{p - 1} \int_0^\infty e^{-xt} \cos x \, dx \, dt,$$</span>
after changing the order of integration.</p>
<p>The inner <span class="math-container">$x$</span>-integral can be readily found. Either using integration by parts twice, or recognising the integral as the Laplace transform for the function <span class="math-container">$\cos x$</span>, as
<span class="math-container">$$\int_0^\infty e^{-xt} \cos x \, dx = \frac{t}{1 + t^2},$$</span>
we have
<span class="math-container">$$\int_0^\infty \frac{\cos x}{x^p} \, dx = \frac{1}{\Gamma (p)} \int_0^\infty \frac{t^p}{1 + t^2} \, dt.$$</span>
Enforcing a substitution of <span class="math-container">$t \mapsto \sqrt{t}$</span> leads to
<span class="math-container">\begin{align}
\int_0^\infty \frac{\cos x}{x^p} \, dx &= \frac{1}{2 \Gamma (p)} \int_0^\infty \frac{t^{\frac{p}{2} - \frac{1}{2}}}{1 + t} \, dt = \frac{1}{2 \Gamma (p)} \int_0^\infty \frac{t^{\frac{p + 1}{2} - 1}}{(1 + t)^{\frac{p + 1}{2} + \frac{1 - p}{2}}}.
\end{align}</span>
As this is exactly of the form of the <a href="https://en.wikipedia.org/wiki/Beta_function#Properties" rel="nofollow noreferrer">Beta function</a> (see the second of the integral representations in the link) we have
<span class="math-container">\begin{align}
\int_0^\infty \frac{\cos x}{x^p} \, dx &= \frac{1}{2 \Gamma (p)} \text{B} \left (\frac{p + 1}{2}, \frac{1 - p}{2} \right )\\
&= \frac{1}{2 \Gamma (p)} \Gamma \left (\frac{p}{2} + \frac{1}{2} \right ) \Gamma \left (\frac{1}{2} - \frac{p}{2} \right ) \\
&= \frac{1}{2 \Gamma (p)} \Gamma \left (\frac{p}{2} + \frac{1}{2} \right ) \Gamma \left [1 - \left (\frac{p}{2} + \frac{1}{2} \right ) \right ] \\
&= \frac{1}{2 \Gamma (p)} \frac{\pi}{\sin (p + 1)\pi/2} \qquad (*)\\
&= \frac{\pi}{2 \Gamma (p) \cos \left (\frac{\pi p}{2} \right )},
\end{align}</span>
as required. Note <a href="https://en.wikipedia.org/wiki/Gamma_function#General" rel="nofollow noreferrer">Euler's reflection formula</a> was used in (<span class="math-container">$*$</span>). </p>
|
2,837,934 | <p>A cyclist gets left behind by $500$ meters every $minute$ by motorcyclist, because of that he takes $2$ $hour$ and $42$ $minute$ more than motorcyclist to cover $52$ $km$.
Find both of their speed.</p>
<p>My approach: $v_2-v_1=30km/h$ (converted 500 meter per minute to km/h)</p>
<p>$v_2=52/t$<br>
$v_1=52/(t+2.42)$</p>
<p>then I plug them in the first formula, seems wrong. How can I solve this?</p>
| fleablood | 280,126 | <p>$500 \frac {meters}{minute} * \frac {km}{1000 meters}*\frac {60 minutes}{hr}= 30 \frac {km}{hr}$</p>
<p>$2 hr 42 minutes = 2\frac {42}{60} hr = 2\frac {7}{10} = 2.7 hr$.</p>
<p>$v_2 - v_1 = 30\frac {km}{hr}$ </p>
<p>$v_2 = \frac {52}{t}$</p>
<p>$v_1 = \frac {52}{t + 2.7}$ so </p>
<p>$\frac {52}{t}-\frac {52}{t + 2.7} = 30$</p>
<p>$52(t+2.7) - 52t = 30t(t+2.7)$</p>
<p>$30t^2 + 81t - 140.4 = 0$.</p>
<p>$t = \frac {-81 \pm \sqrt{81^2 +4*30*140.4}}{2*30}$ (but we know it is positive)</p>
<p>$= \frac {-81 +\sqrt {6561 + 16848}}{60}=\frac {-81 + \sqrt{23409}}{60} =\frac {-81 + 153}{60} = {72}{60} =\frac {12}{10} = 1.2$ (which is an $1 hr 12 minutes$ BTW-- which is how long the car took.)</p>
<p>So $v_2 = \frac {52 km}{1.2 hr} = 43 \frac 13 \frac {km}{hr}$ and </p>
<p>An $v_1 = \frac {52 km}{1.2 + 2.7 hr} = \frac {52}{3.9}\frac {km}{hr} = 13\frac 13 \frac {km}{hr}$</p>
<p>(And that confirms that $v_2 - v_1 = 43\frac 13\frac {km}{hr} - 13\frac 13\frac {km}{hr} = 30\frac {km}{hr}$)</p>
|
2,375,529 | <p>Let $H$ be a Hilbert space and let $T\in \mathcal{B}(H)$ such that $T$ is self-adjoint. I want to show that if $T$ is non-zero, then $T^n\neq 0$ for all $n\in \mathbb{N}$.</p>
<p>Suppose $n$ be the least positive integer such that $T^n=0$. Then for all $x,y\in H$, we have $\langle T^nx,y\rangle=0\implies \langle T^{n-2}x,T^2y\rangle=0$. Herefrom can I show that $T^{n-1}=0$? If it is possible, then I am done. Please suggest.</p>
| Angina Seng | 436,618 | <p>If $n$ is even then $0=\left<T^n x,x\right>=\left<T^{n/2} x,T^{n/2}x\right>$
so $T^{n/2}x=0$. What if $n$ is odd?</p>
|
1,671,111 | <p>I'm looking for an elegant way to show that, among <em>non-negative</em> numbers,
$$
\max \{a_1 + b_1, \dots, a_n + b_n\} \leq \max \{a_1, \dots, a_n\} + \max \{b_1, \dots, b_n\}
$$</p>
<p>I can show that $\max \{a+b, c+d\} \leq \max \{a,c\} + \max \{b,d\}$ by exhaustively checking all possibilities of orderings among $a,c$ and $b,d$.</p>
<p>But, I feel like there should be a more intuitive/efficient way to show this property for arbitrary sums like the one above.</p>
| marty cohen | 13,079 | <p>More than this is true.</p>
<p>Let $P$ be a permutation of
$[1, 2, ..., n]$.</p>
<p>Then
$\max \{a_1 + b_{P(1)}, \dots, a_n + b_{P(n)}\}
\leq \max \{a_1, \dots, a_n\} + \max \{b_1, \dots, b_n\}
$.</p>
<p>This is proved
in the same way
as carmichael561's proof:</p>
<p>For all $i$ from
$1$ to $n$,
$a_i \le \max \{a_1, \dots, a_n\} $
and
$b_{P(i)} \le \max \{b_1, \dots, b_n\}
$
so
$a_i+b_{P(i)}
\le \max \{a_1, \dots, a_n\} + \max \{b_1, \dots, b_n\}
$.</p>
<p>The similar,
but reversed inequality
holds
for $\min$.</p>
|
3,075,263 | <p>Let <span class="math-container">$A$</span> be a positive semi-definite matrix. How to show that Frobenius norm is less than trace of the matrix? Formally,
<span class="math-container">$$\sqrt{\text{Tr}(A^2)} \leq \text{Tr}(A)$$</span>
Also, show when <span class="math-container">$A$</span> is an <span class="math-container">$n \times m$</span> the following is true
<span class="math-container">$$\sqrt{\text{Tr}(A^TA)} \leq \|A\|_*$$</span>
where <span class="math-container">$\|\cdot\|_*$</span> is nuclear norm which is the summation of the singular values.</p>
| angryavian | 43,949 | <p>Let <span class="math-container">$\sigma_1, \ldots, \sigma_r$</span> be the singular values of <span class="math-container">$A$</span>.
Then
<span class="math-container">$$\sqrt{\text{Tr}(A^\top A)} = \sqrt{\sum_i \sigma^2_r} \le \sum_i |\sigma_r| = \|A\|_*.$$</span></p>
|
2,629,408 | <p>How to evaluate this given expression?
$$\int\frac{du}{\sqrt{9e^{-2u}-1}}$$
I got so many tries but I'm not sure of my answer because somebody said that it was wrong, they told me that I used a wrong formula applied!
That's why I ask a support here I want correct explanation and answer of this given!</p>
<p>Thanks!</p>
| Claude Leibovici | 82,404 | <p>Considering $$I=\int\frac{du}{\sqrt{9e^{-2u}-1}}$$ use
$$\sqrt{9e^{-2u}-1}=t\implies u=-\frac{1}{2} \log \left(\frac{1}{9} \left(t^2+1\right)\right)\implies du=-\frac{t}{t^2+1}\,dt$$ This makes
$$I=-\int\frac{dt}{t^2+1}$$</p>
|
3,454,095 | <p>Minimize <span class="math-container">$\;\;\displaystyle \frac{(x^2+1)(y^2+1)(z^2+1)}{ (x+y+z)^2}$</span>, if <span class="math-container">$x,y,z>0$</span>.
By setting gradient to zero I found <span class="math-container">$x=y=z=\frac{1}{\displaystyle\sqrt{2}}$</span>, which could minimize the function.</p>
<blockquote>
<p>Question from Jalil Hajimir</p>
</blockquote>
| Michael Rozenberg | 190,319 | <p>Let <span class="math-container">$x=\frac{a}{\sqrt2},$</span> <span class="math-container">$y=\frac{b}{\sqrt2}$</span> and <span class="math-container">$z=\frac{c}{\sqrt2}.$</span></p>
<p>Thus, since we can assume that <span class="math-container">$(a^2-1)(b^2-1)\geq0,$</span> by C-S we obtain:
<span class="math-container">$$\frac{(x^2+1)(y^2+1)(z^2+1)}{(x+y+z)^2}=\frac{(a^2+2)(b^2+2)(c^2+2)}{4(a+b+c)^2}\geq$$</span>
<span class="math-container">$$\geq\frac{3(a^2+b^2+1)(1+1+c^2)}{4(a+b+c)^2}\geq\frac{3(a+b+c)^2}{4(a+b+c)^2}=\frac{3}{4}.$$</span>
The equality occurs for <span class="math-container">$x=y=z=\frac{1}{\sqrt2},$</span> which says that we got a minimal value.</p>
|
2,078,535 | <p>I'm kinda new with this and find hard to solve the problems related to LA although I can visually and conceptualize stuff easily. Please help me to find $\{(v_1, v_2, v_3) \in \Bbb R^3 \mid 5v_1 - 3v_2 + 2v_3 = 0\}$.</p>
| hamam_Abdallah | 369,188 | <p>A vector $u=(v_1,v_2,v_3)$ which satisfies the condition $: 5v_1-3v_2+2v_3=0$ can be written as</p>
<p>$$u=(v_1,v_2,\frac{1}{2}(-5v_1+3v_2))$$
$$=v_1(1,0,-\frac{5}{2})+v_2(0,1,\frac{3}{2})$$</p>
<p>Put $ u_1=(1,0,-\frac{5}{2})$ and
$u_2=(0,1,\frac{3}{2})$. </p>
<p>$u_1$ and $u_2$ are independent so your space is span$(u_1,u_2)$.</p>
|
151,956 | <p>I'm looking for a general method to evaluate expressions of the form</p>
<p>$$\frac{\mathrm{d}(u^v)}{\mathrm{d}u}\text{ and }\frac{\mathrm{d}(u^v)}{\mathrm{d}v}\;.$$</p>
<p>I know that the answers to these are, respectively, $u^{v-1}v$ and $u^v\mathrm{ln}u$, but am unsure of how to obtain them, and how the chain rule applies here.</p>
<p>I'd be very grateful of any enlightenment.</p>
<p>With very many thanks,</p>
<p>Froskoy.</p>
| Brian M. Scott | 12,042 | <p>If $v$ is a constant, $\frac{d}{du}u^v=vu^{v-1}$, but the chain rule is not required. If $v$ is a function of $u$, then your formula is simply wrong: $u^v=\left(e^{\ln u}\right)^v=e^{v\ln u}$, so</p>
<p>$$\frac{d}{du}u^v=\frac{d}{du}e^{v\ln u}=e^{v\ln u}\left(\frac{v}u+\ln u\frac{dv}{du}\right)=u^v\left(\frac{v}u+\ln u\frac{dv}{du}\right)\;,$$ which can be written $$vu^{v-1}+u^n\ln u\frac{dv}{du}\;,$$ if you really wish.</p>
<p>Similarly, if $u$ is constant, $\frac{d}{dv}u^v=u^v\ln u$, no chain rule being required, but if $u$ is a function of $v$, then $$\frac{d}{dv}u^v=v^u\left(\frac{u}v+\ln v\frac{du}{dv}\right)\;.$$</p>
|
23,268 | <p>I'm the sort of mathematician who works really well with elements. I really enjoy point-set topology, and category theory tends to drive me crazy. When I was given a bunch of exercises on subjects like limits, colimits, and adjoint functors, I was able to do them, although I am sure my proofs were far longer and more laborious than they should have been. However, I felt like most of the understanding I gained from these exercises was gone within a week. I have a copy of MacLane's "Categories for the Working Mathematician," but whenever I pick it up, I can never seem to get through more than two or three pages (except in the introduction on foundations).</p>
<p>A couple months ago, I was trying to use the statements found in Hartshorne about glueing schemes and morphisms and realized that these statements were inadequate for my purposes. Looking more closely, I realized that Hartshorne's hypotheses are "wrong," in roughly the same way that it is "wrong" to require, in the definition of a basis for a topology that it be closed under finite intersections. (This would, for instance, exclude the set of open balls from being a basis for $\mathbb{R}^n$.) Working through it a bit more, I realized that the "right" statement was most easily expressed by saying that a certain kind of diagram in the category of schemes has a colimit. At this point, the notion of "colimit" began to seem much more manageable: a colimit is a way of gluing objects (and morphisms).</p>
<p>However, I cannot think of any similar intuition for the notion of "limit." Even in the case of a fibre product, a limit can be anything from an intersection to a product, and I find it intimidating to try to think of these two very different things as a special cases of the same construction. I understand how to show that they are; it just does not make intuitive sense, somehow.</p>
<p>For another example, I think (and correct me if I am wrong) that <strike>the sheaf condition on a presheaf can be expressed as stating that the contravariant functor takes colimits to limits</strike>. [This is not correct as stated. See Martin Brandenburg's answer below for an explanation of why not, as well as what the correct statement is.] It seems like a statement this simple should make everything clearer, but I find it much easier to understand the definition in terms of compatible local sections gluing together. I can (I think) prove that they are the same, but by the time I get to one end of the proof, I've lost track of the other end intuitively.</p>
<p>Thus, my question is this: Is there a nice, preferably geometric intuition for the notion of limit? If anyone can recommend a book on category theory that they think would appeal to someone like me, that would also be appreciated.</p>
| Buschi Sergio | 6,262 | <p>Of course your "intuition" request can be only about Set-based category where limits and colimits are based on Set analogue.</p>
<p>ABout Colimit you can think as a "amalgamated" union like glueing for a descent data (see also Boubaky- Topology (I vol.)).</p>
<p>ABout limits, is different, limits belong to the prodoct (discrete limit), and for a "geometric" representation the dimention grow...</p>
<p>Anyway an elements of $Lim_{i\in I}X_i$ can be view as a coherent chain (of the some shape of $I$) of elements: just one $x_i$ for any space $X_i$, these are connected togheter by links i.e. diagram morphism maps, and the coherence means that these elements are mapped togheter by these maps.</p>
<p>I hope this can help you. </p>
|
23,268 | <p>I'm the sort of mathematician who works really well with elements. I really enjoy point-set topology, and category theory tends to drive me crazy. When I was given a bunch of exercises on subjects like limits, colimits, and adjoint functors, I was able to do them, although I am sure my proofs were far longer and more laborious than they should have been. However, I felt like most of the understanding I gained from these exercises was gone within a week. I have a copy of MacLane's "Categories for the Working Mathematician," but whenever I pick it up, I can never seem to get through more than two or three pages (except in the introduction on foundations).</p>
<p>A couple months ago, I was trying to use the statements found in Hartshorne about glueing schemes and morphisms and realized that these statements were inadequate for my purposes. Looking more closely, I realized that Hartshorne's hypotheses are "wrong," in roughly the same way that it is "wrong" to require, in the definition of a basis for a topology that it be closed under finite intersections. (This would, for instance, exclude the set of open balls from being a basis for $\mathbb{R}^n$.) Working through it a bit more, I realized that the "right" statement was most easily expressed by saying that a certain kind of diagram in the category of schemes has a colimit. At this point, the notion of "colimit" began to seem much more manageable: a colimit is a way of gluing objects (and morphisms).</p>
<p>However, I cannot think of any similar intuition for the notion of "limit." Even in the case of a fibre product, a limit can be anything from an intersection to a product, and I find it intimidating to try to think of these two very different things as a special cases of the same construction. I understand how to show that they are; it just does not make intuitive sense, somehow.</p>
<p>For another example, I think (and correct me if I am wrong) that <strike>the sheaf condition on a presheaf can be expressed as stating that the contravariant functor takes colimits to limits</strike>. [This is not correct as stated. See Martin Brandenburg's answer below for an explanation of why not, as well as what the correct statement is.] It seems like a statement this simple should make everything clearer, but I find it much easier to understand the definition in terms of compatible local sections gluing together. I can (I think) prove that they are the same, but by the time I get to one end of the proof, I've lost track of the other end intuitively.</p>
<p>Thus, my question is this: Is there a nice, preferably geometric intuition for the notion of limit? If anyone can recommend a book on category theory that they think would appeal to someone like me, that would also be appreciated.</p>
| Anton Fetisov | 10,605 | <p>There's an enlightening example of limit coming from topology. Arguably it was one of the motivating examples for the notion of categorical limit. In general topology it is known as limit over a filter of subsets.</p>
<p>Consider a category $Ouv_X$ of open subsets for a topological space $X$, morphisms being the obvious inclusions (I consider it as a subcategory of $Set$). Assume that $f: X \to Y$ is a (not necessarily continuous) mapping. You have a pair of functors associated with it: $f^{-1}$ (a preimage) and $f_*$.
$$f^{-1}: Ouv_Y \to Ouv_X$$
$$f_*: Ouv_X \to Ouv_Y,~ f_*(U) = \mathrm{Lim}_{V\supset f(U)} V$$</p>
<p>A category $Ouv$ is a preorder: you have at most one morphism between any two objects. What is a limit in a preorder? It is just the infinum! So the right-hand side in the equation defining $f_*$ just means "the smallest open subset containing the image of U". Now consider a full subcategory $\tau_p \subset Ouv_X$
$$\tau_p = \{ U\subset X \vert p\in U \}$$</p>
<p>This is a set of open subsets containing a given point $p\in X$. If X is hausdorff, then the limit of $\tau_p$, considered as a diagram of sets, is just the point $p$. Now consider $\mathrm{Lim}\;f_* \circ \tau_p$. In general it can be anything, but if $f$ is continuous, then it is easy to show directly that
$$\mathrm{Lim}\;f_* \circ \tau_p = f(p) = f(\mathrm{Lim}\; \tau_p)$$</p>
<p>That's a familiar identity $f(\lim x_n) = \lim f(x_n)$: a continuous function preserves limits! This limit-preservation property can also be figured out more abstractly, 'cause iff $f$ is continuous, then $f^{-1} \rightleftharpoons f_*$ is a pair of adjoint functors, and a right adjoint functor always preserves limits.</p>
|
96,369 | <p>Is there a mathematical term which would include both liminf and limsup? (In a similar way we talk about extrema to describe both maxima and minima?)</p>
<p>The only thing I was able to find was that some authors use the name extreme limits; see google books: <a href="http://www.google.com/search?tbm=bks&tbo=1&q=%22extreme+limits%22+%22lim+sup%22" rel="nofollow">"extreme limits" "lim sup"</a>. It is used e.g. in Thomson, Bruckner, Bruckner: Elementary real analysis, <a href="http://books.google.com/books?id=6l_E9OTFaK0C&pg=PA70" rel="nofollow">p.70</a></p>
<p>EDIT: I've checked a few more searches to see to which extent the
above term is widespread. Searching for
<a href="http://www.google.com/search?tbm=bks&tbo=1&q=%22extreme+limits%22" rel="nofollow">"extreme limits"</a> only
gives a lot of non-mathematical result, so these are basically
attempts to filter out results that are interesting for this
question.</p>
<ul>
<li>Google Books: <a href="http://www.google.com/search?tbm=bks&tbo=1&q=%22extreme+limits%22+sequence" rel="nofollow">"extreme limits" sequence</a></li>
<li>Google Books: <a href="http://www.google.com/search?tbm=bks&tbo=1&q=%22extreme+limits%22+function" rel="nofollow">"extreme limits" function</a></li>
<li>Google books: <a href="http://www.google.com/search?tbm=bks&q=%22Subject%3AMathematics%22+%22extreme+limits%22" rel="nofollow">"Subject:Mathematics" "extreme limits"</a></li>
<li>Google Scholar - searching in Engineering, Computer Science, and
Mathematics: <a href="http://scholar.google.com/scholar?as_subj=eng&q=%22extreme%20limits%22" rel="nofollow">"extreme limits"</a></li>
</ul>
<p>NOTE: This question arose more or less out of curiosity. When I discussed limit superior and limit inferior with some other MSE users, we found out that none of us was familiar with a name which would describe these two things. (Of course, it might be the case that such term is not needed very much, if we are only able to find something which is used frequently.)</p>
| leslie townes | 18,076 | <p>For what it is worth I do not think there is any well-established terminology for what you want. I have never heard of any, anyway.</p>
<p>I think it is worthwhile to change your question to a slightly broader question: "what kinds of common generalizations could one make of these ideas, so that they become different instances of the same general thing, and what terminology might one use for the generalization?"</p>
<p>A question like has more than one right answer. Depending on what properties you regard as essential to any abstraction, and what your perspective is, you will get different results. (Even something more fundamental than all of these limitlike operations, namely the completeness property of the set of real numbers, takes on very different "flavors" depending on your point of view: is it an order-theoretic property, or a metric property, or what?)</p>
<p>Here is one view of the issue based on a "dynamical" sort of perspective: you might justifiably call each of <span class="math-container">$\liminf$</span>, <span class="math-container">$\limsup$</span>, and <span class="math-container">$\lim$</span> "asymptotic properties" of a sequence. Let me formalize a definition here.</p>
<p>Consider the map <span class="math-container">$\sigma$</span> on the set of all sequences of real numbers defined by sending the sequence <span class="math-container">$(x_n)_{n=1}^{\infty}$</span> to <span class="math-container">$(x_{n+1})_{n=1}^{\infty}$</span> (what many call "the shift").</p>
<p>The natural domains of <span class="math-container">$\liminf$</span>, <span class="math-container">$\limsup$</span>, and <span class="math-container">$\lim$</span> are all invariant under <span class="math-container">$\sigma$</span> (by "natural domain" I just mean "the largest set of sequences on which the operation can be defined and gives a real result", which in the cases of <span class="math-container">$\liminf$</span>, <span class="math-container">$\limsup$</span>, and <span class="math-container">$\lim$</span> are the sets of bounded below, bounded above, and convergent sequences, respectively). By this I mean that if a sequence <span class="math-container">$x$</span> is in the domain of any one of these operations, then <span class="math-container">$\sigma(x)$</span> is also in that domain. And of course each of the operations <span class="math-container">$\liminf$</span>, <span class="math-container">$\limsup$</span>, <span class="math-container">$\lim$</span> is invariant under <span class="math-container">$\sigma$</span>, by which I mean that the result of applying the operation to a sequence <span class="math-container">$x$</span> in its domain is the same as the result of applying the operation to <span class="math-container">$\sigma(x)$</span>.</p>
<p>Here's a first approximation to a general definition:</p>
<p>Suppose that <span class="math-container">$\mathcal{D}$</span> is a subset of the set of all real sequences with the property that <span class="math-container">$\sigma(\mathcal{D}) \subseteq \mathcal{D}$</span>. Then a function <span class="math-container">$F: \mathcal{D} \to \mathbb{R}$</span> is called an asymptotic property of sequences in <span class="math-container">$\mathcal{D}$</span> if <span class="math-container">$F(\sigma(x)) = F(x)$</span> holds for all <span class="math-container">$x$</span> in <span class="math-container">$\mathcal{D}$</span>.</p>
<p>By this definition <span class="math-container">$\liminf$</span> and <span class="math-container">$\limsup$</span> and <span class="math-container">$\lim$</span> are all asymptotic properties of sequences in their domains, whereas e.g. <span class="math-container">$\inf$</span> or <span class="math-container">$\sup$</span> or "evaluate at the 200th term", in <em>their</em> natural domains, are not.</p>
<p>The intuitive reason for choosing the word "asymptotic" is hopefully clear: if <span class="math-container">$F$</span> has the property given in the definition, then although the number <span class="math-container">$F(x)$</span> depends on <span class="math-container">$x$</span>, it does not depend on any finite number of terms taken from <span class="math-container">$x$</span>. So in a vague sense it depends only on what is happening "way out there".</p>
<p>(To make this notion slightly more reasonable, you might require that the domain <span class="math-container">$\mathcal{D}$</span> be more than just a shift-invariant set of sequences. To avoid trivialities it seems natural to require that <span class="math-container">$\mathcal{D}$</span> contain, or even <em>properly</em> contain, the set of constant sequences, and also to require that <span class="math-container">$F$</span> coincide with "evaluation at the first entry" on the set of constant sequences.)</p>
<p>More structured versions of this "asymptotic property" concept do exist and have names. For example: require <span class="math-container">$\mathcal{D}$</span> to be the set of <em>all</em> bounded sequences, regarded as a topological vector space with the componentwise vector space operations and the topology coming from the supremum norm, and require <span class="math-container">$F$</span> to be a <em>linear</em> map that is <em>continuous</em> with respect to this topology, and require additionally that <span class="math-container">$F$</span> coincide with <span class="math-container">$\lim$</span> on the set of convergent sequences (and perhaps also add a positivity condition), and you have what is called a <a href="https://en.wikipedia.org/wiki/Banach_limit" rel="nofollow noreferrer">Banach limit</a>. But <span class="math-container">$\limsup$</span> and <span class="math-container">$\liminf$</span>, not being linear, are not instances of this concept.</p>
|
96,369 | <p>Is there a mathematical term which would include both liminf and limsup? (In a similar way we talk about extrema to describe both maxima and minima?)</p>
<p>The only thing I was able to find was that some authors use the name extreme limits; see google books: <a href="http://www.google.com/search?tbm=bks&tbo=1&q=%22extreme+limits%22+%22lim+sup%22" rel="nofollow">"extreme limits" "lim sup"</a>. It is used e.g. in Thomson, Bruckner, Bruckner: Elementary real analysis, <a href="http://books.google.com/books?id=6l_E9OTFaK0C&pg=PA70" rel="nofollow">p.70</a></p>
<p>EDIT: I've checked a few more searches to see to which extent the
above term is widespread. Searching for
<a href="http://www.google.com/search?tbm=bks&tbo=1&q=%22extreme+limits%22" rel="nofollow">"extreme limits"</a> only
gives a lot of non-mathematical result, so these are basically
attempts to filter out results that are interesting for this
question.</p>
<ul>
<li>Google Books: <a href="http://www.google.com/search?tbm=bks&tbo=1&q=%22extreme+limits%22+sequence" rel="nofollow">"extreme limits" sequence</a></li>
<li>Google Books: <a href="http://www.google.com/search?tbm=bks&tbo=1&q=%22extreme+limits%22+function" rel="nofollow">"extreme limits" function</a></li>
<li>Google books: <a href="http://www.google.com/search?tbm=bks&q=%22Subject%3AMathematics%22+%22extreme+limits%22" rel="nofollow">"Subject:Mathematics" "extreme limits"</a></li>
<li>Google Scholar - searching in Engineering, Computer Science, and
Mathematics: <a href="http://scholar.google.com/scholar?as_subj=eng&q=%22extreme%20limits%22" rel="nofollow">"extreme limits"</a></li>
</ul>
<p>NOTE: This question arose more or less out of curiosity. When I discussed limit superior and limit inferior with some other MSE users, we found out that none of us was familiar with a name which would describe these two things. (Of course, it might be the case that such term is not needed very much, if we are only able to find something which is used frequently.)</p>
| Martin Sleziak | 8,297 | <p>It seems that the term <em>extreme limits</em> is used, although not very frequently.</p>
<p>Some people call limit superior and limit inferior <em>inner and <a href="https://en.wikipedia.org/wiki/Outer_limit" rel="nofollow noreferrer">outer limits</a></em>, see e.g. this spikedmath comic <a href="http://spikedmath.com/479.html" rel="nofollow noreferrer">strip</a> (<a href="https://web.archive.org/web/20201123085847/http://spikedmath.com/479.html" rel="nofollow noreferrer">Wayback Machine</a>). This is not a single-word name, but it was interesting for me, since I did not know it before.</p>
<p>Some examples of usage of this usage terms can be found here (although many non-mathematical uses appear there, too):</p>
<p>Extreme limits</p>
<ul>
<li>Google Books: <a href="https://www.google.com/search?tbm=bks&tbo=1&q=%22extreme+limits%22+sequence" rel="nofollow noreferrer">"extreme limits" sequence</a></li>
<li>Google Books: <a href="https://www.google.com/search?tbm=bks&tbo=1&q=%22extreme+limits%22+function" rel="nofollow noreferrer">"extreme limits" function</a></li>
<li>Google Scholar - searching in Engineering, Computer Science, and
Mathematics: <a href="https://scholar.google.com/scholar?as_subj=eng&q=%22extreme%20limits%22" rel="nofollow noreferrer">"extreme limits"</a></li>
</ul>
<p>Inner and outer</p>
<ul>
<li>Google Books: <a href="https://www.google.com/search?tbm=bks&tbo=1&q=inner+%22outer+limits%22+sequence" rel="nofollow noreferrer">inner "outer limits" sequence</a></li>
<li>Google Books: <a href="https://www.google.com/search?tbm=bks&tbo=1&q=inner+%22outer+limits%22+function" rel="nofollow noreferrer">inner "outer limits" function</a></li>
<li>Google Scholar - searching in Engineering, Computer Science, and Mathematics: <a href="https://scholar.google.com/scholar?as_subj=eng&q=inner+%22outer%20limits%22" rel="nofollow noreferrer">inner "outer limits"</a></li>
</ul>
|
181,000 | <p>Let $p$ be a prime number greater than or equal to 11. Are there any cospectral non-isomorphic graphs with circulant graphs on $p$ vertices.
Which circulant graphs over prime number of vertices greater than or equal to 11 are determined by the spectrum?</p>
| Brendan McKay | 9,025 | <p>No, and you don't need $p\ge 11$ either.</p>
<p>B Elspas, J Turner
Graphs with circulant adjacency matrices,
J. Combinatorial Theory, 9 (1970), pp. 297–307.</p>
<p>This paper shows that no two non-isomorphic circulant graphs on a prime number of vertices have the same spectrum. It doesn't say anything about circulant graphs cospectral with non-circulant graphs.</p>
|
181,000 | <p>Let $p$ be a prime number greater than or equal to 11. Are there any cospectral non-isomorphic graphs with circulant graphs on $p$ vertices.
Which circulant graphs over prime number of vertices greater than or equal to 11 are determined by the spectrum?</p>
| Shahrooz | 19,885 | <p>We know that all groups with prime order is CI-group. So, if two circulant graph with $p$ vertices be cospectral, then they are isomorphic. So, as Dear Brendan said, the answer is no.</p>
<p>But about your comment, the answer is yes. Just look at strongly regular graphs of order $29$. In this order we have Paley graph that is circulant, but also we have exactly $40$ other graphs that are cospectral with this graph and some of them are not circulant. You can see the page $856$ of book: $Handbook\; of\; combinatorial\; design$.</p>
<p>For $p=11$ the asnwer is yes. All circulant graphs with $11$ vertices are DS. Also, All circulant graphs with $13$ vertices are DS. But, as I said for $p=29$ there are at least $40$ non-isomorphic graphs cospectral with Paley graph. So, the answer to this question is not known in general and I think it is very hard by some evidences.</p>
|
295,597 | <p>I'm trying to solve this simple integral:</p>
<p>$$\frac12 \int \frac{x^2}{\sqrt{x + 1}} dx$$</p>
<p>Here's what I have done so far:</p>
<ol>
<li><p>$\displaystyle t = \sqrt{x + 1} \Leftrightarrow x = t^2 - 1 \Rightarrow dx = 2t dt$</p></li>
<li><p>$\displaystyle \frac12 \int \frac{x^2}{\sqrt{x + 1}} dx = \int \frac{t (t^2 - 1)^2}t dt$</p></li>
<li><p>$\displaystyle \int (t^2 - 1)^2 dt = \frac15 t^5 - \frac23 t^3 + t + C$</p></li>
<li><p>$\displaystyle \frac15 t^5 - \frac23 t^3 + t + C = \frac15 \sqrt{(x + 1)^5} - \frac23 \sqrt{(x + 1)^3} + \sqrt{x + 1} + C$</p></li>
</ol>
<p>WolframAlpha tells me steps 1 and 3 are right so the mistake must be somewhere in steps 2 and 4, but I really can't see it.</p>
| Tapu | 17,142 | <p>No, everything is fine. Now simplify what you have got after integration. Note that $\sqrt{a^5}=\sqrt{a}.a^2$ etc.</p>
|
2,900,372 | <p>Let $R=\mathbb{Z}[\sqrt{-2}]$</p>
<blockquote>
<p>1) Is it true that for every free $R-$module $M$ of rank=$n$ that $M$ is a free $\mathbb{Z}-$module of rank=$2n$?</p>
<p>2) Find two non-isomorphic $R$-modules with $19$ elements each.</p>
</blockquote>
<p>For (1) I tried to begin with a basis $\{m_1,...,m_n\}$ of $M$ as a $R-$module and construct a basis $\{m_1',...,m_{2n}'\}$ of it as a $\mathbb{Z}-$module.</p>
<p>For (2) I have no idea.</p>
<p>Any hints?? </p>
| Sarvesh Ravichandran Iyer | 316,409 | <p>Let $M = [m_1,...,m_n]$. Then, every element of $M$ can be written uniquely as $\sum a_im_i$ where $a_i \in \mathbb Z[\sqrt -2] = b_i + c_i \sqrt{-2}$.</p>
<p>Clearly, $M$ is then generated by $[m_i, m_i\sqrt{-2}]$ over $\mathbb Z$.The question is : is this set linearly independent over $\mathbb Z$?</p>
<p>Well, let $\sum c_im_i + \sum d_i m_i\sqrt{-2} = 0$, where $c_i,d_i$ are integers. Then, $\sum(c_i + d_i \sqrt {-2}) m_i = 0$, so by the fact that $m_i$ are linearly independent over $\mathbb Z[\sqrt{-2}]$, we get that $c_i,d_i = 0$ for all $i$, as $c_i + d_i \sqrt{-2} = 0$ for all $i$.</p>
<hr>
<p>The second question is already answered above , but I wish to point out something important, and be more elaborate. This will result in an expanded answer.</p>
<p>The point is, $\frac{R}{\langle \alpha\rangle} \not \equiv \frac{R}{\langle\beta\rangle}$ <strong>as $R$- modules</strong> is important, because they are indeed isomorphic as <em>additive groups</em> (modules are abelian groups under addition) , having the same cardinality $19$ which is prime, so both are cyclic and we may map a generator to another generator of the additive abelian group.</p>
<p>What is the difference? An $R$-module homomorphism $\phi : \frac{R}{\langle \alpha\rangle} \to \frac{R}{\langle \beta\rangle}$, in addition to the fact that it is additive, also must satisfy $f(rx) = rf(x)$ for all $r \in \mathbb Z[\sqrt{-2}]$ and $x \in \frac{R}{\langle \alpha\rangle}$. </p>
<hr>
<p>This very key fact makes a great difference. For example, let us consider $f(1 + \langle \alpha\rangle)$. This element satisfies the fact that $(\alpha)(1 + \langle\alpha\rangle) = 0 + \langle \alpha \rangle$. Hence, it follows that $\alpha \times f(1 + \langle \alpha \rangle) = 0 + \langle \beta\rangle$. That is, if $f(1 + \langle \alpha\rangle) = c + \langle \beta\rangle$, then $\beta | \alpha c$, as $\alpha c -0 \in \langle \beta\rangle$.</p>
<p>The catch? Well, $\beta$ is co-prime to $\alpha$, because $a$ and $b$ have the same norm which is $19$ , a prime number, and they are not associates, so no element of non-unit norm can divide both. So $\beta | c$ (note that we are in a unique factorization domain), and consequently, we get $f(1+\langle\alpha\rangle) = 0$. If one generator maps to zero, the whole group maps to zero. Consequently, every module homomorphism between the two modules is the zero map, and therefore none is an isomorphism.</p>
|
56,134 | <p>As mentioned,
I wish to read the first line of a file, and if needed, overwrite it with a new string.
The aim is to have a CSV with a list of possible elements.
Example:</p>
<pre><code>Adding the elements: 5 A, 6 B, and 7 C to a blank CSV:
A B C
5 6 7
Adding 4 A, 9 D:
A B C D
5 6 7
4 0 0 9
Adding 2 B, 7 E
A B C D E
5 6 7
4 0 0 9
0 2 0 0 7
</code></pre>
<p>Rewriting the file is okay for small files, but the files are expected to grow to sizes of hundreds of MB, so repeatedly overwriting the entire file is not practiceable, since the entries are preferably done quickly. </p>
<p>How would one go about overwriting only the first line of the file? Reading the first line is possible from <a href="https://mathematica.stackexchange.com/questions/5179/how-to-read-data-file-quickly#_=_">https://mathematica.stackexchange.com/questions/5179/how-to-read-data-file-quickly#<em>=</em></a>, but what about writing?</p>
| Nick Lariviere | 18,824 | <p>Low-level file operators like Write won't work here because OutputStreams (such as you get with OpenWrite and OpenAppend) can't have their StreamPosition set before the end of the file. In general overwriting characters in an existing file isn't terribly trivial; you can use c-functions like<code>file_ptr = fopen(file, "rb+")</code> to do this, but you're overwriting character-by-character, so if you remove a character or add one it can throw things off.</p>
<p>Generally you're better off creating a copy of the data, along the lines of:</p>
<pre><code> replaceNthRecord[file_String, n_Integer, replaceWith_] := Module[
{readStream = OpenRead[file], writeStream = OpenWrite[file <> "temp"], temp},
Do[Write[writeStream, Read[readStream]], {n - 1}];
Write[writeStream, replaceWith]; Read[readStream];
While[temp =!= EndOfFile,
temp = Read[readStream];
If[UnsameQ[temp, EndOfFile], Write[writeStream, temp]];
]; Close /@ {readStream, writeStream}
]
Close[stream]
</code></pre>
<blockquote>
<p>"file.dat"</p>
</blockquote>
<pre><code>replaceNthRecord["file.dat", 3, 99]
</code></pre>
<blockquote>
<p>{"file.dat", "file.dattemp"}</p>
</blockquote>
<p>Make sure that "file.dattemp" is showing what you want, and then delete the original file and replace it with the temp one:</p>
<pre><code>ReadList["file.dattemp"]
</code></pre>
<blockquote>
<p>{1, 2, 99, 4, 5, 6, 7, 8, 9, 10}</p>
</blockquote>
<p>(then just add this to the end of the function:</p>
<pre><code>CopyFile[file <> "temp", file, OverwriteTarget -> True];
DeleteFile[file <> "temp"]
</code></pre>
<p>Note that using <code>ReadList</code> appropriately here will be more efficient than <code>Do[Write[_,Read[_]]]</code> but you'll need to do a bit more work to figure out what an appropriate batch size is to do this in and make sure you stop in the right place.</p>
|
419,625 | <p>Please help me, I have two functions:</p>
<blockquote>
<pre><code>a := x^3+x^2-x;
b := 20*sin(x^2)-5;
</code></pre>
</blockquote>
<p>and I would like to change a background color and fill the areas between two curves. I filled areas but I dont know how can I change the background, any idea?</p>
<blockquote>
<pre><code>plots[display]([plottools[transform]((x, y)->
[x, y+x^3+x^2+x])(plot(20*sin(x^2)-5-
x^3-x^2-x, x = o .. s, filled = true, color = red)),
plot([x^3+x^2+x, 20*sin(x^2)-5],
x = -3 .. 3, y = -30 .. 30, color = black)]);
</code></pre>
</blockquote>
| acer | 12,448 | <p>One way is to add a colored rectangle to the displayed items. Note that the filled red region now gets transparency=0.0 so that the green doesn't bleed through it.</p>
<pre><code>a := x^3+x^2-x:
b := 20*sin(x^2)-5:
plots[display](
plottools[transform]((x, y)-> [x, y+x^3+x^2+x])
(plot(20*sin(x^2)-5-x^3-x^2-x,
x = 0 .. 3, filled = true, color = red,
transparency=0.0)),
plot([x^3+x^2+x, 20*sin(x^2)-5],
x = -3 .. 3, y = -30 .. 30, color = black),
plottools:-rectangle([-3,-30],[3,40],colour=green));
</code></pre>
<p><img src="https://i.stack.imgur.com/svUOX.gif" alt="enter image description here"></p>
<p>And here is a version for those who will (often ill-advisedly) want to copy blocks of plaintext 1D Maple Notation code and paste it into a 2D Math input region.</p>
<pre><code>H := plottools[transform](proc(x, y) [x, y+x^3+x^2+x]; end proc):
plots[display](H(plot(20*sin(x^2)-5-x^3-x^2-x,
x = 0 .. 3, filled = true, color = red,
transparency=0.0)),
plot([x^3+x^2+x, 20*sin(x^2)-5],
x = -3 .. 3, y = -30 .. 30, color = black),
plottools:-rectangle([-3,-30],[3,40],colour=green));
</code></pre>
<p>The 2D Math parser doesn't like the first version above since the formatted code involves extra spaces between the <code>transform</code> result and its bracketed arguments. A space in such a function call gets parsed in 2D Math mode as an implicit multiplication. All these parsing issues have nothing in particular to do with this solution, or with background colors on plots, or with <code>plots[display]</code>.</p>
|
2,389,782 | <p>Other than <a href="http://rads.stackoverflow.com/amzn/click/3885380064" rel="nofollow noreferrer">Engelking General Topology</a>, I also come across other graduate general topology text such as <a href="http://rads.stackoverflow.com/amzn/click/0697068897" rel="nofollow noreferrer">Dugundji</a> and <a href="http://www.springer.com/gp/book/9780387901251" rel="nofollow noreferrer">Kelley</a>, which I also find them interesting. </p>
<p>However, I find Engelking's book more comprehensive compare to the two graduate texts. </p>
<blockquote>
<p>Question: Does there exist a general topology book which is more
comprehensive than Engelking? If yes, what is the title?</p>
</blockquote>
| Alex Ravsky | 71,850 | <p>As far as I know, no. There are big general topology books “Topology” by Kuratowski (available at LibGen) and “Handbook of set-theoretic topology” (eds. Kunen and Vaughan) (also available in the Internet), but they consider selected topics. </p>
|
2,389,782 | <p>Other than <a href="http://rads.stackoverflow.com/amzn/click/3885380064" rel="nofollow noreferrer">Engelking General Topology</a>, I also come across other graduate general topology text such as <a href="http://rads.stackoverflow.com/amzn/click/0697068897" rel="nofollow noreferrer">Dugundji</a> and <a href="http://www.springer.com/gp/book/9780387901251" rel="nofollow noreferrer">Kelley</a>, which I also find them interesting. </p>
<p>However, I find Engelking's book more comprehensive compare to the two graduate texts. </p>
<blockquote>
<p>Question: Does there exist a general topology book which is more
comprehensive than Engelking? If yes, what is the title?</p>
</blockquote>
| William Elliot | 426,203 | <p>I like <em>General Topology</em> by Willard for reference of difficult theorems.<br>
It is point set topology orientated instead of analysis orientated.</p>
|
3,965,834 | <p>Does this sum converge or diverge?</p>
<p><span class="math-container">$$ \sum_{n=0}^{\infty}\frac{\sin(n)\cdot(n^2+3)}{2^n} $$</span></p>
<p>To solve this I would use <span class="math-container">$$ \sin(z) = \sum \limits_{n=0}^{\infty}(-1)^n\frac{z^{2n+1}}{(2n+1)!} $$</span></p>
<p>and make it to <span class="math-container">$$\sum \limits_{n=0}^{\infty}\sin(n)\cdot\frac{(n^2+3)}{2^n} = \sum \limits_{n=0}^{\infty}(-1)^n\frac{n^{2n+1}}{(2n+1)!} \cdot \sum \limits_{n=0}^{\infty}\frac{(n^2+3)}{2^n} $$</span></p>
<p>and since <span class="math-container">$$\sum \limits_{n=0}^{\infty}\frac{(n^2+3)}{2^n} \text{ and } \sum \limits_{n=0}^{\infty}(-1)^n\frac{n^{2n+1}}{(2n+1)!} $$</span></p>
<p>converges <span class="math-container">$$ \sum \limits_{n=0}^{\infty}\frac{\sin(n)\cdot(n^2+3)}{2^n} $$</span>
would also converge.</p>
<p>Is my assumption true? I'm also a bit scared to use it since I've got the sin(z) equation from a source outside the stuff that my professor gave us</p>
| CHAMSI | 758,100 | <p>Notice that for whatever <span class="math-container">$ p\geq 2 $</span>, <span class="math-container">$ \lim\limits_{n\to +\infty}{\left|\frac{n^{p}\sin{n}\left(n^{2}+3\right)}{2^{n}}\right|}=0 $</span>.</p>
<p>Thus <span class="math-container">$ \left|\frac{\sin{n}\left(n^{2}+3\right)}{2^{n}}\right|=\underset{\overset{n\to +\infty}{}}{\scriptsize\mathcal{O}}\left(\frac{1}{n^{p}}\right) $</span>. Since <span class="math-container">$ \sum\limits_{n\geq 1}{\frac{1}{n^{p}}} $</span> converges for any <span class="math-container">$ p\geq 2 $</span>, the series <span class="math-container">$ \sum\limits_{n\geq 1}{\left|\frac{\sin{n}\left(n^{2}+3\right)}{2^{n}}\right|} $</span> does also converge.</p>
<p>Our series does, hence, converge absolutely, which means it does converge.</p>
|
4,201,477 | <blockquote>
<p>Integrate <span class="math-container">$$\int \frac{\cos 2x}{(\sin x+\cos x)^2}\mathrm dx$$</span></p>
</blockquote>
<p>I was integrating my own way.</p>
<p><span class="math-container">$$\int \frac{\cos 2x}{\sin^2x+2\sin x\cos x+cos^2}\mathrm dx$$</span>
<span class="math-container">$$\int \cot 2x \mathrm dx$$</span>
<span class="math-container">$$\frac{1}{2}\ln |\sin2x|+c$$</span></p>
<p>I guess I didn't do any mistake. But, my book had derived something else.</p>
<p><span class="math-container">$$\int \frac{\cos 2x \mathrm dx}{1+\sin 2x}$$</span> By taking <span class="math-container">$1+\sin 2x=z$</span> By differentiating the value,
<span class="math-container">$\cos 2x \mathrm dx=\frac{1}{2}\mathrm dz$</span>
Continue to main equation,
<span class="math-container">$$\frac{1}{2}\int \frac{1}{z}\mathrm dz$$</span>
<span class="math-container">$$\frac{1}{2}ln|z|+c=\frac{1}{2}\ln |1+\sin 2x|+c$$</span></p>
<p>Why both answers are different? I don't think it's possible to derive one to another.</p>
| Henry Lee | 541,220 | <p>Here is another alternative:
<span class="math-container">$$\frac{\cos 2x}{(\cos x+\sin x)^2}=\frac{\cos^2 x-\sin^2x}{(\cos x+\sin x)^2}\\
=\frac{(\cos x+\sin x)(\cos x-\sin x)}{(\cos x+\sin x)^2}=\frac{\cos x-\sin x}{\cos x+\sin x}\\
=\frac{(\cos x+\sin x)'}{(\cos x+\sin x)}$$</span>
now sub <span class="math-container">$u=\cos x+\sin x$</span></p>
|
2,146,457 | <p>Simplify the ring $\mathbb Z[\sqrt{-13}]/(2)$. I have so far:</p>
<p>$$\mathbb Z[\sqrt{-13}]/(2) \cong \mathbb Z[x]/(2, x^2 + 13) \cong \mathbb Z_2[x] / (x^2 + 1)$$</p>
<p>Now how do I simplify it further? I know that $x^2 + 1 = (x + 1)^2$ in $\mathbb Z_2[x]$, is this useful?</p>
| user416819 | 416,819 | <p>Make the substitution $y = x + 1$ to get an isomorphism $(Z/2Z)[x] \cong (Z/2Z)[y]$. The answer is $(Z/2Z)[y]/(y^2)$.</p>
|
1,130,487 | <p>Jessica is playing a game where there are 4 blue markers and 6 red markers in a box. She is going to pick 3 markers without replacement.
If she picks all 3 red markers, she will win a total of 500 dollars. If the first marker she picks is red but not all 3 markers are red, she will win a total of 100 dollars. Under any other outcome, she will win 0 dollars. </p>
<p><strong>Solution</strong>
The probability of Jessica picking 3 consecutive red markers is: $\left(\frac16\right)$</p>
<p>The probability of Jessica's first marker being red, but not picking 3 consecutive red markers is:<br/>$\left(\frac35\right)-\left(\frac16\right)=\left(\frac{13}{30}\right)$
<br/>
So i am bit stuck here<br/></p>
<p><strong>what i think</strong> is it shouldn't be that complex it should be as simple as
chance of Jessica's first marker being red=chance of getting red 1 time
i.e P(First marker being red)=$\left(\frac{6}{10}\right)$
can any explain me the probability of Jessica's first marker being red=$\left(\frac{13}{30}\right)$?</p>
| Emanuele Paolini | 59,304 | <p>Nothing is wrong. You have found two different antiderivatives of the same function.
You know that antiderivatives differ only by a constant on each interval where they are defined... you can check that this is the case in your computation.</p>
|
1,393,869 | <p>Given a cubic polynomial with real coefficients of the form $f(x) = Ax^3 + Bx^2 + Cx + D$ $(A \neq 0)$ I am trying to determine what the necessary conditions of the coefficients are so that $f(x)$ has exactly three distinct real roots. I am wondering if there is a way to change variables to simplify this problem and am looking for some clever ideas on this matter or on other ways to obtain these conditions.</p>
| Travis Willse | 155,629 | <p>Suppose that (including multiplicity) the roots of <span class="math-container">$$f(x) = A x^3 + B x^2 + C x + D,$$</span> <span class="math-container">$A \neq 0$</span>, are <span class="math-container">$r_1, r_2, r_3$</span>. Consider the quantity
<span class="math-container">$$\Delta(f) := A^4 (r_3 - r_2)^2 (r_1 - r_3)^2 (r_2 - r_1)^2,$$</span>
called the <a href="https://en.wikipedia.org/wiki/Discriminant#Discriminant_of_a_polynomial" rel="nofollow noreferrer"><em>(polynomial) discriminant</em></a> of <span class="math-container">$f$</span>.</p>
<p>If <span class="math-container">$r_1, r_2, r_3$</span> are all real and pairwise distinct, then we see that <span class="math-container">$\Delta(f) > 0$</span>. On the other hand, if <span class="math-container">$f$</span> has a repeated root, then <span class="math-container">$\Delta(f) = 0$</span>, and if <span class="math-container">$f$</span> has one real root and two nonreal (necessarily conjugate) roots, substituting gives <span class="math-container">$\Delta(f) < 0$</span>. (The coefficient <span class="math-container">$A^4$</span> is unnecessary for <span class="math-container">$\Delta$</span> to enjoy these properties, but among other things, its inclusion makes the below formula nicer.)</p>
<p>These three cases exhaust all of the possibilities, so we conclude:</p>
<p><span class="math-container">$$\color{#bf0000}{\fbox{A cubic polynomial $f$ has three distinct, real roots iff $\Delta(f) > 0$.}}$$</span></p>
<p>The above definition of <span class="math-container">$\Delta$</span> is not immediately practical, since explicit formulas for the roots of a general cubic polynomial are unwieldy. On the other hand, with some work (say, by expanding and using <a href="https://en.wikipedia.org/wiki/Newton%27s_identities" rel="nofollow noreferrer">Newton's Identities</a> and <a href="https://en.wikipedia.org/wiki/Vieta%27s_formulas" rel="nofollow noreferrer">Vieta's Formulas</a>) we can write <span class="math-container">$\Delta(f)$</span> as a homogeneous quartic expression in the coefficients <span class="math-container">$A, B, C, D$</span>:
<span class="math-container">$$\Delta(f) = -27 A^2 D^2 + 18 ABCD - 4 A C^3 - 4 B^3 D + B^2 C^2.$$</span> This formula gives a computationally practical answer to the question:</p>
<p><span class="math-container">$$
\bbox[0.5mm,border: 1px solid #bf0000]{
\color{#bf0000}{
\begin{array}{c}
\textrm{A cubic polynomial} \\
f(x) = A x^3 + B x^2 + C x + D \\
\textrm{has three distinct, real roots iff} \\
-27 A^2 D^2 + 18 ABCD - 4 A C^3 - 4 B^3 D + B^2 C^2 > 0 \textrm{.}
\end{array}
}
}
$$</span></p>
<p>It's apparent that one can generalize the notion of discriminant to polynomials <span class="math-container">$p$</span> of any degree <span class="math-container">$> 1$</span>, producing an expression homogeneous of degree <span class="math-container">$2(\deg p - 1)$</span> in the polynomial coefficients. In each case, up to a constant that depends on the degree and the leading coefficient of <span class="math-container">$f$</span>, <span class="math-container">$\Delta(f)$</span> is equal to the <a href="https://en.wikipedia.org/wiki/Resultant" rel="nofollow noreferrer">resultant</a> <span class="math-container">$R(f, f')$</span> of <span class="math-container">$f$</span> and its derivative.</p>
<p>By making a suitable real, affine change of variables <span class="math-container">$x \rightsquigarrow y = a x + b$</span>, by the way, one can transform any given real cubic polynomial to the so-called <em>depressed</em> form
<span class="math-container">$$\tilde{f}(y) = y^3 + P y + Q .$$</span>
such transformations do not change the number of real roots or the multiplicities of roots. For a cubic polynomial in this reduced form the discriminant takes the simpler and well-known form
<span class="math-container">$$\Delta(\tilde f) = -4 P^3 - 27 Q^2.$$</span></p>
|
759,810 | <p>I'm trying to find the last two digits of ${2012}^{2012}$. I know you can use (mod 100) to find them, but I'm not quite sure how to apply this. Can someone please explain it?</p>
| Sandeep Silwal | 138,892 | <p>$\phi(25) = 20$ so $2012^{2012} \equiv 2012^{12} \pmod {25}$.</p>
<p>Then $2012^{12} \equiv 12^{12} \equiv 144^6 \equiv 36^3 \equiv 6 \pmod {25}$. </p>
<p>Then solving the system $N \equiv 6 \pmod {25}, \equiv 0 \pmod 4$ gives $n \equiv 56 \pmod {100}$. </p>
|
3,982,937 | <p>To avoid typos, please see my screen captures below, and the red underline. The question says <span class="math-container">$h \rightarrow 0$</span>, thus why <span class="math-container">$|h|$</span> in the solution? Mustn't that <span class="math-container">$|h|$</span> be <span class="math-container">$h$</span>?</p>
<p><img src="https://i.stack.imgur.com/rweRh.jpg" alt="enter image description here" /></p>
<p>Spivak, <em>Calculus</em> 2008 4 edn. <a href="https://mathpop.com/" rel="nofollow noreferrer">His website's errata</a> lists no errata for these pages.</p>
| Ben Grossmann | 81,360 | <p>Note that the definition of a limit involves the inequality <span class="math-container">$0 < |x - a| < \delta$</span>. Substituting <span class="math-container">$a = 0$</span> and <span class="math-container">$x = h$</span> yields
<span class="math-container">$$
0 < |h| < \delta,
$$</span>
which is the inequality that you've underlined.</p>
|
3,497,679 | <p>I was given this function:
<span class="math-container">$$
f(x)=
\begin{cases}
x+x^2, & x\in\Bbb Q\\
x, & x\notin \Bbb Q
\end{cases}
$$</span>
I first proved that it is continuous at <span class="math-container">$x=0$</span>.</p>
<p>Now I need to prove that that for every <span class="math-container">$x_0 \in \mathbb R\setminus\{0\}$</span> the limit <span class="math-container">$\lim \limits_{x \to x_0}f(x)$</span> does not exist.</p>
<p>I know that I need to start by assuming that the limit does exist but I don't know how to reach a contradiction.</p>
| davidlowryduda | 9,754 | <blockquote>
<p>I know that I need to start by assuming that the limit does exist but I don't know how to reach a contradiction.</p>
</blockquote>
<p>That's not necessary. It is sufficient to simply prove that the limit doesn't exist.</p>
<p>And the easiest way to do that is to consider a limit along rational values tending to <span class="math-container">$x_0$</span> and compare it with the limit along irrational values tending to <span class="math-container">$x_0$</span>. That these are different amounts to saying that <span class="math-container">$x^2 + x \neq x$</span> when <span class="math-container">$x \neq 0$</span>.</p>
|
3,497,679 | <p>I was given this function:
<span class="math-container">$$
f(x)=
\begin{cases}
x+x^2, & x\in\Bbb Q\\
x, & x\notin \Bbb Q
\end{cases}
$$</span>
I first proved that it is continuous at <span class="math-container">$x=0$</span>.</p>
<p>Now I need to prove that that for every <span class="math-container">$x_0 \in \mathbb R\setminus\{0\}$</span> the limit <span class="math-container">$\lim \limits_{x \to x_0}f(x)$</span> does not exist.</p>
<p>I know that I need to start by assuming that the limit does exist but I don't know how to reach a contradiction.</p>
| pre-kidney | 34,662 | <p>If <span class="math-container">$\lim_{x\to x_0}f(x)$</span> exists, then for any sequence of numbers <span class="math-container">$x_1,x_2,\ldots$</span> converging to <span class="math-container">$x_0$</span>, we must have that <span class="math-container">$f(x_1),f(x_2),\ldots$</span> converges to <span class="math-container">$f(x_0)$</span>. In particular we can consider two sequences converging to <span class="math-container">$x_0$</span>: the first uses only rational numbers, and the second uses only irrational numbers. (This is possible since both <span class="math-container">$\mathbb Q$</span> and <span class="math-container">$\mathbb R\setminus \mathbb Q$</span> are <a href="https://en.wikipedia.org/wiki/Rational_number#Properties" rel="nofollow noreferrer">dense</a>.) Then, since both the function <span class="math-container">$x\mapsto x^2+x$</span> and the identity function are continuous, the limit of the rational sequence will be <span class="math-container">$x_0^2+x_0$</span> and that of the irrational sequence will be <span class="math-container">$x_0$</span>. However if <span class="math-container">$x_0\not=0$</span> these quantities differ, forming a contradiction.</p>
|
3,497,679 | <p>I was given this function:
<span class="math-container">$$
f(x)=
\begin{cases}
x+x^2, & x\in\Bbb Q\\
x, & x\notin \Bbb Q
\end{cases}
$$</span>
I first proved that it is continuous at <span class="math-container">$x=0$</span>.</p>
<p>Now I need to prove that that for every <span class="math-container">$x_0 \in \mathbb R\setminus\{0\}$</span> the limit <span class="math-container">$\lim \limits_{x \to x_0}f(x)$</span> does not exist.</p>
<p>I know that I need to start by assuming that the limit does exist but I don't know how to reach a contradiction.</p>
| Michael Hardy | 11,667 | <p>Suppose <span class="math-container">$x_0\ne0$</span> and there is a limit:
<span class="math-container">$$
L = \lim_{x\,\to\, x_0} f(x).
$$</span>
Let <span class="math-container">$\varepsilon = x_0^2/4.$</span> Then:</p>
<ul>
<li>There is some number <span class="math-container">$\delta_1>0$</span> such that for all values of <span class="math-container">$x$</span> between <span class="math-container">$x_0\pm\delta_1$</span> we have <span class="math-container">$|f(x)-L|<\varepsilon.$</span></li>
<li>There is some number <span class="math-container">$\delta_2>0$</span> such that for all values of <span class="math-container">$x$</span> between <span class="math-container">$x_0\pm\delta_2,$</span> we have <span class="math-container">$|(x+x^2)-(x_0+x_0^2)|<\varepsilon.$</span></li>
</ul>
<p>Let <span class="math-container">$\delta=\min\{\delta_1,\delta_2,\varepsilon\}.$</span></p>
<p>A crucial fact is that no matter how small <span class="math-container">$\delta$</span> is, there are both rational and irrational numbers between <span class="math-container">$x_0\pm\delta.$</span> So let <span class="math-container">$x_1$</span> be a rational number in that interval and let <span class="math-container">$x_2$</span> be an irrational number in that interval.
<span class="math-container">\begin{align}
& |f(x_1)-L|< \varepsilon, \\[8pt]
\text{which means } & |(x_1+x_1^2) - L| < \varepsilon. \tag 1 \\[8pt]
& |(x_1+x_1^2)-(x_0+x_0^2)| < \varepsilon. \tag 2 \\[8pt]
\text{The triangle } & \text{inequality applied to $(1)$ and $(2)$ yields:} \\
& |L-(x_0+x_0^2)|<2\varepsilon. \tag 3 \\[8pt]
\text{We also have } & |f(x_2)-L|< \varepsilon, \\[8pt]
\text{which means } & |x_2-L|<\varepsilon. \tag 4 \\[8pt]
\text{And we have } & |x_0-x_2| < \delta\le\varepsilon. \tag 5 \\[8pt]
\text{The triangle } & \text{inequality applied to $(4)$ and $(5)$ yields:} \\
& |x_0-L|<2\varepsilon. \tag 6 \\[8pt]
\text{Now the triangle } & \text{inequality applied to $(3)$ and $(6)$ gives us:} \\
& |x_0 - (x_0+x_0^2)|<4\varepsilon. \\[10pt]
\text{So } & x_0^2 < 4\varepsilon \\
\text{and we have } & \text{a contradiction.}
\end{align}</span></p>
|
2,202,339 | <p>It's easy to prove $x^2+1$ is never divisible by $4k+3$ primes. I know a non-constructive proof for existing $x$ so that $p|x^2+1$ for $4k+1$ primes. is there any constructive one?</p>
| aras | 277,932 | <p>First note that</p>
<p>$(p-1)! = -1 \mod p$</p>
<p>by Wilson's Theorem. Since $p = 4k+ 1$, we can write this product as</p>
<p>$1\cdot 2 \cdots (\frac{p-1}{2}) (\frac{p+1}{2}) \cdots (p-1) \equiv -1 \mod p$</p>
<p>Note that this can be written as</p>
<p>$1 \cdot 2 \cdots (\frac{p-1}{2}) \cdot (p-\frac{p-1}{2}) \cdots (p-1) \equiv -1 \mod p$</p>
<p>so</p>
<p>$1 \cdot 2 \cdots (\frac{p-1}{2}) \cdot (-\frac{p-1}{2}) \cdots (-1) \equiv -1 \mod p$</p>
<p>so</p>
<p>$(\frac{p-1}{2})^2 \cdot (-1)^{\frac{p-1}{2}} \equiv -1 \mod p$</p>
<p>Since $p \equiv 1 \mod 4$, $\frac{p-1}{2}$ is even. Thus we are left with</p>
<p>$(\frac{p-1}{2})^2 \equiv -1 \mod 2$.</p>
|
3,020,365 | <p>Let <span class="math-container">$A=\{t\sin(\frac{1}{t})\ |\ t\in (0,\frac{2}{\pi})\}$</span>.</p>
<p>Then </p>
<ol>
<li><p><span class="math-container">$\sup (A)<\frac{2}{\pi}+\frac{1}{n\pi}$</span> for all <span class="math-container">$n\ge 1$</span>.</p></li>
<li><p><span class="math-container">$\inf (A)> \frac{-2}{3\pi}-\frac{1}{n\pi}$</span> for all <span class="math-container">$n\ge 1$</span>.</p></li>
<li><span class="math-container">$\sup (A)=1$</span></li>
<li><span class="math-container">$\inf (A)=-1$</span></li>
</ol>
<p>My answer was options <span class="math-container">$1$</span> and <span class="math-container">$2$</span> which matched with prelim answer key provided by organization, but the final answer key changed the answer to option <span class="math-container">$1$</span> only. Why is option <span class="math-container">$2$</span> not correct?</p>
<p><strong>My attempt-</strong> </p>
<p>As <span class="math-container">$-t\le t\sin(1/t)\le t$</span>. So <span class="math-container">$t\sin(1/t)$</span> will always be less than <span class="math-container">$\frac{2}{\pi}$</span> which is strictly less than <span class="math-container">$1$</span>. So option <span class="math-container">$3$</span> is false and option <span class="math-container">$1$</span> is true. Also value of <span class="math-container">$t\sin(1/t)$</span> is always greater than or equal to <span class="math-container">$-2/\pi=-0.63$</span>, so infimum cannot be <span class="math-container">$-1$</span>.</p>
<p>Now at <span class="math-container">$t=2/3\pi$</span> , function has value <span class="math-container">$\frac{-2}{3\pi}$</span>. So if option <span class="math-container">$2$</span> is false we must have some value of <span class="math-container">$t$</span> in <span class="math-container">$A$</span> for which the function attains value strictly less than <span class="math-container">$\frac{-2}{3\pi}$</span>. Right? </p>
<p>Can you please help me with option <span class="math-container">$2$</span> now. Thanks in advance. </p>
| Mostafa Ayaz | 518,023 | <p>If the option <span class="math-container">$2$</span> is true, it means that <span class="math-container">$$\inf A\ge -{2\over 3\pi}$$</span> since <span class="math-container">$\inf A> -{2\over 3\pi}-{1\over n\pi}$</span> is true for all <span class="math-container">$n>1$</span>. Since <span class="math-container">$$\large t\sin {1\over t}\Big|_{t={2\over 3\pi}\in A}=-{2\over 3\pi}$$</span> then we must have <span class="math-container">$\inf A= -{2\over 3\pi}$</span> and <span class="math-container">$$\large {d\over dt}t\sin {1\over t}\Big|_{t={2\over 3\pi}\in A}=0$$</span> because the infimum is equal to minimum now (since it is attained in <span class="math-container">$t={2\over 3\pi}\in A$</span>) and as the function is differentiable in <span class="math-container">$A$</span>. But this doesn't hold since <span class="math-container">$$\large {d\over dt}t\sin {1\over t}\Big|_{t={2\over 3\pi}\in A}=\large \sin {1\over t}-{1\over t}\cos {1\over t}\Big|_{t={2\over 3\pi}}=-1\ne 0$$</span>therefore option <span class="math-container">$2$</span> is incorrect.</p>
|
16,749 | <p>I wanted to remove the <code>Ticks</code> in my coding but i can't. Here when i try to remove the <code>Ticks</code> the number also gone. I need numbers without <code>Ticks</code>, <code>Ticks</code> and <code>GridLines</code> should be automatic and don't use<code>PlotRange</code> .</p>
<pre><code>BarChart[{{1, 2, 3}, {4, 5, 6}}, ImageSize -> 400, BarOrigin -> Left,
ChartLayout -> "Stacked", ImageSize -> {500, 300},
GridLines -> {Automatic, None}, Ticks -> {{Automatic}, None}
,
LabelStyle -> Directive[Opacity[1]],
TicksStyle -> Directive[Opacity[.3]]
,
Axes -> {True, False}, AxesStyle -> Opacity[.0],
ChartStyle -> {RGBColor[.06, .29, .66], RGBColor[.01, .56, .61],
RGBColor[1, .58, 0]},
ChartBaseStyle -> EdgeForm[GrayLevel[.6]]]
</code></pre>
<p><img src="https://i.stack.imgur.com/s28b2.png" alt="enter image description here"></p>
| Mr.Wizard | 121 | <p>Simply use option:</p>
<pre><code>TicksStyle -> Directive[20, Opacity[0], FontOpacity -> 0.3]
</code></pre>
<p><img src="https://i.stack.imgur.com/yuedn.png" alt="Mathematica graphics"></p>
<p>Credit to <a href="https://mathematica.stackexchange.com/users/5/rm-rf">R.M</a> for showing me <code>FontOpacity</code>.</p>
|
425,460 | <p>While browsing through several pages of nlab(mainly on n-Categories), I encountered the notion "foo" several times. However, there seems to be article on nlab about this notion. Is this some kind of category theorist slang? Please explain to me what this term means.</p>
| Haizum Skallah | 86,428 | <p>1.1 More than the sum of their parts.</p>
<p>We motivate this first chapter by noticing that while many real-world structures are compositional, the results of observing them are often not. The reason is that observation is inherently “lossy”: in order to extract information from something, one must drop the details. For example, one stores a real number by rounding it to some precision. But if the details are actually relevant in a given system operation, then the observed result of that operation will not be as expected. This is clear in the case of roundoff error, but it also shows up in non-numerical domains: observing a complex system is rarely enough to predict its behavior because the observation is lossy.</p>
<p>Acentral themein category theory is the study of structures and structure-preserving maps. A map f : X ! Y is a kind of observation of object X via a specified relationship it has with another object, Y. For example, think of X as the subject of an experiment and Y as a meter connected to X, which allows us to extract certain features of X by looking at the reaction of Y.</p>
<p>Asking which aspects of X one wants to preserve under the observation f becomes the question “what category are you working in?.” As an example, there are many functions f from R to R, and we can think of them as observations: rather than view x “directly”, we only observe f(x). Out of all the functions f : R -> R, only some of them preserve the order of numbers, only some of them preserve the distance between numbers, only some of them preserve the sum of numbers, etc. Let’s check in with an exercise; a solution can be found in Chapter 1.
Exercise 1.1. Some terminology: a function f : R -> R is said to be
(a) order-preserving ...
(b) metric-preserving ...
(c) addition-preserving ...</p>
<p><strong>For each of the three properties defined above—call it foo—find an f that is foo-preserving and an example of an f that is not foo-preserving.</strong></p>
<p>From: Seven Sketches in Compositionality: An Invitation to Applied Category Theory by Brendan Fong and David I. Spivak pp. 1-2 (2018)</p>
|
1,749,128 | <p>How to compute taylor series $f(x)=\frac{1}{1-x}$ about $a=3$? It should be associated with the geometric series. Setting $t=x-3,\ x=t+3$, then I don't know how to continue, could someone clarify the procedure?</p>
| Doug Spoonwood | 11,300 | <p>According to Wikipedia, Alfred Tarski came up with a set of axioms for a substantial fragment of Euclidean Geometry. The axioms require <a href="https://en.wikipedia.org/wiki/Tarski's_axioms" rel="nofollow">no set theory</a>. The axioms are first-order, stand as precise, and have gotten explored in an automated <a href="http://www.michaelbeeson.com/research/papers/TarskiFormalization.pdf" rel="nofollow">theorem</a> <a href="http://www.automatedreasoning.net/docs_and_pdfs/an_amazing_approach%20to_plane_geometry.pdf" rel="nofollow">proving</a> context.</p>
<p>So, yes, mathematical objects can exist without set theory.</p>
<p>A specific function or a specific relation could just be something which satisfies some axioms.</p>
<p>For instance, if we have the axioms </p>
<ol>
<li>C(x, C(y, x)) </li>
</ol>
<p>and </p>
<ol start="2">
<li>C(C(x, C(y, z)), C(C(x, y), C(x, z)))</li>
</ol>
<p>A specific function C could be anything which makes the above two formulas true. The same can hold for a specific relation.</p>
|
2,957,315 | <p><span class="math-container">$j_{1,1}$</span> denotes the first zero of the first Bessel function of the first kind. (That's a lot of firsts!) It's approximately equal to <span class="math-container">$3.83$</span>. My question is, is there any closed form expression for its value? Even a infinite series or infinite product that yields it would be good.</p>
<p>I ask because this value is used in physics, in the context of diffraction of light through a circular aperture, and students often make the mistake of thinking that the number just pops out of nowhere.</p>
| Claude Leibovici | 82,404 | <p>If you want an <em>approximation</em> of the first root without invoking at any time the Bessel functions, what you could consider is an <span class="math-container">$[2m+1,2n]$</span> Padé approximation of <span class="math-container">$J_1(x)$</span> built at <span class="math-container">$x=0$</span>. It will be looking like
<span class="math-container">$$J_1(x)\approx\frac{x \left(\frac12+\sum_{j=1}^m a_j \,x^{2j}\right)}{1+\sum_{k=1}^n b_k\, x^{2k} }$$</span> The coefficients are "easy" to compute since, using the infinite series given in Geremia's answer, we know exactly all required derivatives of <span class="math-container">$J_1(x)$</span> at <span class="math-container">$x=0$</span>. These approximations are equivalent to Taylor expansions to <span class="math-container">$O\left(x^{2(m+n+1})\right)$</span>.</p>
<p>In order to keep it simple (solution of a quadratic equation in <span class="math-container">$x^2$</span>), select <span class="math-container">$m=2$</span>. The following table gives the equation to be solved, its first root and its decimal representation.
<span class="math-container">$$\left(
\begin{array}{cccc}
n & \text{equation} & \text{root} & \text{value} \\
1 & x^4-40 x^2+384=0 & 4 & 4.00000 \\
2 & 11 x^4-552 x^2+5760=0 & \sqrt{\frac{12}{11} \left(23-\sqrt{89}\right)} &
3.84698 \\
3 & 23 x^4-1296 x^2+14080=0 & \sqrt{\frac{8}{23} \left(81-\sqrt{1501}\right)} &
3.83382 \\
4 & 97 x^4-5810 x^2+64400=0 & \sqrt{\frac{5}{97} \left(581-\sqrt{87689}\right)} &
3.83202
\end{array}
\right)$$</span> while the "exact" value is <span class="math-container">$3.83171$</span>.</p>
<p>Just for the fun of it, plot on the same graph, for <span class="math-container">$0 \leq x \leq 4$</span>, <span class="math-container">$J_1(x)$</span> and <span class="math-container">$\frac{x(11 x^4-552 x^2+5760) }{ 4 x^4+336 x^2+11520}$</span></p>
|
259,734 | <p>Basically, what the title says. </p>
<p>Presumably, one could use the fact that monoidal categories (resp. strict monoidal categories) are one-object bicategories (resp. 2-categories) and use the Lack model structure on those, but I am unsure if this would work or not.</p>
| Karol Szumiło | 12,547 | <p>I'm not sure about the case of general monoidal categories. (Although I seem to recall a remark that there is no such structure since the category of monoidal categories is not cocomplete and a suitable replacement would be the category of multicategories. Perhaps somebody can confirm this.)</p>
<p>However, the case of strict monoidal categories is known since they are simply monoids internal to the category of categories. It is a theorem of Schwede and Shipley (Theorem 4.1 in <em>Algebras and Modules in Monoidal Model Categories</em>) that if a monoidal model category satisfies so called <em>monoid axiom</em>, then the category of monoids in it inherits a model structure. The monoid axiom is a technical condition which is, in particular, automatically satisfied if all objects are cofibrant as is the case in the canonical model structure on the category of categories.</p>
|
3,423,225 | <p>I know that the angle <span class="math-container">$\theta$</span> of a right-angled triangle, centered at the origin, is defined as the radian measure of its intersection point with the unit circle, and that <span class="math-container">$\cos(\theta)$</span> and <span class="math-container">$\sin(\theta)$</span> are defined to be the x- and y-cordinates of that intersection point:</p>
<p><a href="https://i.stack.imgur.com/axHKG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/axHKG.png" alt="enter image description here"></a></p>
<p>However, for larger right-angled triangles, we compute <span class="math-container">$\cos(\theta)=A/C$</span> and <span class="math-container">$\sin(\theta)=B/C$</span> by dividing the respective side length with the hypotenuse, respectively. But, for this to work, right-angled triangles with a common angle <span class="math-container">$\theta$</span> must have proportional side lengths.</p>
<p>How do I see that right-angled triangles with a common angle <span class="math-container">$\theta$</span> have proportional side lengths? </p>
| Dr. Richard Klitzing | 518,676 | <p>It simply is the <strong>theorem on intersecting lines</strong> (aka on confocal rays, or aka about similar triangles), which provides
<span class="math-container">$$\frac{A}{\cos(\theta)}=\frac{C}1=\frac{B}{\sin(\theta)}$$</span>
Solving the right or left equation then provides the searched for relations.</p>
<p>--- rk</p>
|
3,802,269 | <p>Evaluate <span class="math-container">$\lim_{x\rightarrow \infty} x\int_{0}^{x}e^{t^2-x^2}dt$</span></p>
<p>My approach:</p>
<p><span class="math-container">$$
\lim_{x\rightarrow \infty} x\int_{0}^{x}e^{t^2-x^2}dt = \lim_{x\rightarrow \infty} \frac{\int_{0}^{x}e^{t^2}dt}{x^{-1}e^{x^2}}
$$</span></p>
<p>Both the numerator and denominator <span class="math-container">$\rightarrow \infty$</span> as <span class="math-container">$x\rightarrow \infty$</span>. Apply L'Hopital's rule and FTC:</p>
<p><span class="math-container">$$
\lim_{x\rightarrow \infty} x\int_{0}^{x}e^{t^2-x^2}dt = \lim_{x\rightarrow \infty} \dfrac{e^{x^2}}{2e^{x^2}-x^{-2}e^{x^2}}=\frac{1}{2}
$$</span></p>
<p>I am looking for a verification of my result. Thank you!</p>
| Eric Towers | 123,905 | <p><span class="math-container">\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x} x^{-1} \mathrm{e}^{x^2}
&= \frac{\mathrm{d}}{\mathrm{d}x}(x^{-1})\mathrm{e}^{x^2} + x^{-1} \frac{\mathrm{d}}{\mathrm{d}x}\mathrm{e}^{x^2} \\
&= -x^{-2}\mathrm{e}^{x^2} + x^{-1}2x\mathrm{e}^{x^2} \text{,}
\end{align*}</span>
which is not what you have in your denominator...</p>
|
3,802,269 | <p>Evaluate <span class="math-container">$\lim_{x\rightarrow \infty} x\int_{0}^{x}e^{t^2-x^2}dt$</span></p>
<p>My approach:</p>
<p><span class="math-container">$$
\lim_{x\rightarrow \infty} x\int_{0}^{x}e^{t^2-x^2}dt = \lim_{x\rightarrow \infty} \frac{\int_{0}^{x}e^{t^2}dt}{x^{-1}e^{x^2}}
$$</span></p>
<p>Both the numerator and denominator <span class="math-container">$\rightarrow \infty$</span> as <span class="math-container">$x\rightarrow \infty$</span>. Apply L'Hopital's rule and FTC:</p>
<p><span class="math-container">$$
\lim_{x\rightarrow \infty} x\int_{0}^{x}e^{t^2-x^2}dt = \lim_{x\rightarrow \infty} \dfrac{e^{x^2}}{2e^{x^2}-x^{-2}e^{x^2}}=\frac{1}{2}
$$</span></p>
<p>I am looking for a verification of my result. Thank you!</p>
| Mark Viola | 218,419 | <p>The numerator after applying L'Hospital's Rule in the OP is incorrect as it is written. Note that <span class="math-container">$$\frac{d}{dx} \int_0^x e^{t^2}\,dt=e^{x^2}\ne e^{x^2}-1$$</span></p>
<hr />
<p>Applying L'Hospital's Rule reveals that</p>
<p><span class="math-container">$$\begin{align}
\lim_{x\to \infty }x\int_0^x e^{t^2-x^2}\,dt&=\lim_{x\to \infty }\frac{\int_0^x e^{t^2}\,dt}{\frac{e^{x^2}}x}\\\\
&=\lim_{x\to \infty }\frac{e^{x^2}}{2e^{x^2}-\frac{e^{x^2}}{x^2}}\\\\
&=\lim_{x\to \infty }\frac{1}{2-\frac{1}{x^2}}\\\\
&=\frac12
\end{align}$$</span></p>
<p>and we are done!</p>
<hr />
<p>I thought it might be instructive to present an approach that does not rely on L'Hospital's Rule. To that end, we now proceed.</p>
<p>Enforcing the substitution <span class="math-container">$t\mapsto \sqrt{x^2-t}$</span>, we have</p>
<p><span class="math-container">$$\begin{align}
\lim_{x\to\infty} x\int_0^x e^{t^2-x^2}\,dt&=\lim_{x\to\infty} x\int_0^{x^2} \frac{e^{-t}}{2\sqrt{x^2-t}}\,dt\\\\
&=\frac12 \lim_{x\to\infty}\int_0^{x^2} \frac{e^{-t}}{\sqrt{1-\frac{t}{x^2}}}\,dt\\\\
&=\frac12\int_0^\infty e^{-t}\,dt\\\\
&=\frac12
\end{align}$$</span></p>
<p>as expected!</p>
|
3,056,121 | <p>I'm trying to find a function with infinitely many local minimum points where x <span class="math-container">$\in$</span> [0,1] and f has only 1 root. No interval should exist where the function is constant.</p>
| Zacky | 515,527 | <p>For the first part of the question we can substitute <span class="math-container">$x=\frac{1}{t}$</span> in order to get: <span class="math-container">$$I=\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x=\int^0_\infty \frac{\ln\left(\frac{1}{t}\right)}{t^2\left(1+\frac{1}{t^2}\right)^3}\frac{-dt}{t^2}=\int_0^\infty \frac{t^2 \ln\left(\frac{1}{t}\right)}{(1+t^2)^3}dt=-I$$</span>
So we just saw that <span class="math-container">$I=-I\Rightarrow I=0$</span></p>
<hr>
<blockquote>
<p>I am currently really interested in proving this
<span class="math-container">$$ \int_0^\infty \frac{\ln(x)}{(1+x^2)^2} {\rm d}x = \int_0^\infty \frac{\ln(x)}{(1+x^2)^3} {\rm d}x = -\frac{\pi}{4} $$</span></p>
</blockquote>
<p>Well now that we showed that both integrals are equal it's enough to compute only one of it.
<span class="math-container">$$\Omega=\int_0^\infty \frac{\ln x}{(1+x^2)^2}dx\overset{x=\tan t}=\int_0^\frac{\pi}{2} \ln(\tan t)\cos^2 t\,\mathrm dt =\frac12\int_0^\frac{\pi}{2} \ln(\tan t) (1+\cos(2t))dt$$</span>
Now we will split into two integrals and show that the first one vanishes using the following property of the definite integrals:
<span class="math-container">$$\int_a^b f(x)dx=\int_a^b f(a+b-x)dx$$</span>
<span class="math-container">$$J=\int_0^\frac{\pi}{2}\ln(\tan t)dt=\int_0^\frac{\pi}{2} \ln(\cot t)dt=-\int_0^\frac{\pi}{2} \ln (\tan t)dt=-J\Rightarrow J=0$$</span>
<span class="math-container">$$\Rightarrow \Omega=\frac12 \int_0^\frac{\pi}{2}\ln(\tan t)\cos (2t)\mathrm dt=\frac12 \int_0^\frac{\pi}{2} \ln(\tan t)\left(\frac12 \sin(2t)\right)' \mathrm dt=$$</span>
<span class="math-container">$$=\frac14 \underbrace{\ln(\tan t)\sin(2t)\bigg|_0^\frac{\pi}{2}}_{=0}-\frac14\int_0^\frac{\pi}{2} \frac{\sec^2 t}{\tan t}\sin(2t)\mathrm dt=-\frac12 \int_0^\frac{\pi}{2} dt=-\frac{\pi}{4}$$</span></p>
|
3,056,121 | <p>I'm trying to find a function with infinitely many local minimum points where x <span class="math-container">$\in$</span> [0,1] and f has only 1 root. No interval should exist where the function is constant.</p>
| Jia Ming جيا ميڠ | 383,520 | <p>I found a way to directly evaluate both of those integrals!</p>
<p>We first use the well known result from <a href="https://math.stackexchange.com/questions/290200/int-0-infty-frac-ln-xx2a2-mathrmdx-evaluate-integral"><span class="math-container">$\int_0^{\infty}\frac{\ln x}{x^2+a^2}\mathrm{d}x$</span> Evaluate Integral</a>
<span class="math-container">$$
\int_0^\infty \frac{\ln(x)}{x^2 +\alpha^2} {\rm d}x = \frac{\pi}{2\alpha} \ln(\alpha)
$$</span>
and use differentiation under the integral sign.</p>
<p><span class="math-container">$$
\int_0^\infty \frac{-2\alpha\ln(x)}{(x^2 +\alpha^2)^2} {\rm d}x
= \frac{\pi}{2} \frac{1-\ln(\alpha)}{\alpha^2} \\
\implies \int_0^\infty \frac{\ln(x)}{(x^2 +\alpha^2)^2} {\rm d}x
= \frac{\pi}{4\alpha^3} (\ln(\alpha)-1)
$$</span></p>
<p>And again, to get
<span class="math-container">$$
\int_0^\infty \frac{\ln(x)}{(x^2 +\alpha^2)^3} {\rm d}x
= \frac{\pi}{16\alpha^5}(3\ln(\alpha) - 4)
$$</span></p>
<blockquote>
<p>And so setting <span class="math-container">$\alpha = 1$</span> gives us the immediate result
<span class="math-container">$$ \int_0^\infty \frac{\ln(x)}{(1+x^2)^2} {\rm d}x = \int_0^\infty \frac{\ln(x)}{(1+x^2)^3} {\rm d}x = -\frac{\pi}{4} $$</span></p>
</blockquote>
<p>Another method to evaluate this is to use the integral from the 2015 MIT Integration Bee, and is also how I first came across this integral.</p>
<p>From this result (solved by using the substitution <span class="math-container">$u=\frac1x$</span>)
<span class="math-container">$$ \int_0^\infty \frac{1}{(1+x^2)(1+x^\alpha)} {\rm d}x = \frac\pi4 $$</span></p>
<p>we will get, by differentiating with respect to <span class="math-container">$\alpha$</span>,</p>
<blockquote>
<p><span class="math-container">$$ \int_0^\infty \frac{x^\alpha \ln(x)}{(1+x^2)(1+x^\alpha)^2} {\rm d}x = 0 $$</span></p>
</blockquote>
<p>And finally setting <span class="math-container">$\alpha = 2$</span>.</p>
|
138,921 | <p>If $A\colon H\to S$ is a bounded operator on a Hilbert space $H$, and $S\subset H$. It is known that $\operatorname{trace}(A)=\sum_{n} \langle Af_n,f_n\rangle$ for any orthonormal basis $\{f_{n}\}$. Is there a relation between $\operatorname{trace}(A)$, $\operatorname{rank}(A)$, and dimension of $\operatorname{range}(S)$?</p>
<p><strong>Edit:</strong> What if $A$ is a composition of two orthogonal projections $A_{1}:H\to S_{1}$,$A_{2}:H\to S_{2}$, such that $A=A_{1}oA_{2}$, for $S_{1},S_{2}\subset H$. I need to show that $\operatorname{trace}(A)\leq \operatorname{rank}(A)\leq \dim(S_{2})$</p>
| Shuhao Cao | 7,200 | <p>We could simply apply the chain rule, to avoid some confusions we let $ C(x,t) = C(x^* + vt,t^*) = C^*(x^*,t^*)$:
$$
\frac{\partial C}{\partial x} = \frac{\partial C^*}{\partial x^{\phantom{*}}}= \frac{\partial C^*}{\partial x^*} \frac{\partial x^*}{\partial x^{\phantom{*}}} + \frac{\partial C^*}{\partial t^*} \frac{\partial t^*}{\partial x^{\phantom{*}}} = \frac{\partial C}{\partial x^*}
$$
remember here in chain rule, the partial derivative is being taken wrt the first and second variable if not to confuse this wrt the total derivative, similary we could have $\displaystyle \frac{\partial^2 C}{\partial x^2} = \frac{\partial^2 C^*}{\partial {x^*}^2} $,
$$
\frac{\partial C}{\partial t} = \frac{\partial C^*}{\partial t} = \frac{\partial C^*}{\partial x^*} \frac{\partial x^*}{\partial t^{\phantom{*}}} + \frac{\partial C^*}{\partial t^*} \frac{\partial t^*}{\partial t^{\phantom{*}}} = -v\frac{\partial C^*}{\partial x^*} + \frac{\partial C^*}{\partial t^*}
$$
Plugging back to the original equation you will see the convection term is gone if we have done this velocity cone rescaling, you could think the original equation like a diffusion on a car with velocity $v$ measured by a standing person, after the change of variable it is just a pure diffusion measured on a car:
$$
\frac{\partial C^*}{\partial t^*} = D\frac{\partial^2 C^*}{\partial {x^*}^2}
$$
and the initial condition changes to $C^*(x^*,0) = C(x^*+vt^*,t^*)\Big\vert_{t^*=0}= f(x^*)$, the boundary condition remains the same.</p>
|
3,757,864 | <p>Given a diagram like this,
<a href="https://i.stack.imgur.com/Xwum0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Xwum0.png" alt="enter image description here" /></a></p>
<p>Where <span class="math-container">$O$</span> is the center and <span class="math-container">$OA = \sqrt{50}$</span>, <span class="math-container">$AB = 6$</span>, and <span class="math-container">$BC = 2$</span>. The question was to find the length of <span class="math-container">$OB$</span>. <span class="math-container">$\angle ABC = 90^o$</span></p>
<p>What I've done is so far:</p>
<p>I made the triangle <span class="math-container">$ABC$</span> and named <span class="math-container">$\angle BAC = \alpha$</span> . By trigonometry, I have the values for <span class="math-container">$\sin{\alpha}$</span> and <span class="math-container">$\cos{\alpha}$</span>. I get <span class="math-container">$\cos{\alpha}=\frac{6}{\sqrt{40}}$</span>.</p>
<p>Then I made the triangle <span class="math-container">$OCA$</span> and named <span class="math-container">$\angle OAB = \beta$</span> so <span class="math-container">$\angle OAC = \alpha + \beta$</span>. By using the cosinus rule, I have <span class="math-container">$\cos(\alpha + \beta) = \frac{1}{\sqrt{5}}$</span>.</p>
<p>Using the formula, <span class="math-container">$\cos(\alpha + \beta) = \cos{\alpha}.\cos{\beta} - \sin{\alpha}.\sin{\beta}$</span> and making <span class="math-container">$\sin{\beta} = \sqrt{1 -\cos^2{\beta}}$</span> I finally get that <span class="math-container">$\cos{\angle OAB} = \frac{1}{\sqrt{2}}$</span>.</p>
<p>Finally, by using the cosinus rule on the triangle <span class="math-container">$AOB$</span> I get <span class="math-container">$OB = \sqrt{26}$</span>.</p>
<p>My only problem is this takes me way too long! I am interested in a quicker way to do this (i.e. I now know that <span class="math-container">$\angle OAB = 45^o$</span> from trigonometry, but is there a quicker way to recognize it?)</p>
| g.kov | 122,782 | <p><a href="https://i.stack.imgur.com/IVsYw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IVsYw.png" alt="enter image description here" /></a></p>
<p>A slight variation of the solution</p>
<p>Note that <span class="math-container">$R$</span> is circumradius of <span class="math-container">$\triangle ADC$</span>,</p>
<p><span class="math-container">\begin{align}
|CD|&=2R\sin\alpha=2\sqrt5
,\\
|BD|&=\sqrt{|CD|^2-a^2}=4
,\\
|AD|&=c+BD=10
.
\end{align}</span></p>
<p>By the <a href="https://en.wikipedia.org/wiki/Stewart%27s_theorem" rel="nofollow noreferrer">Stewart’s Theorem</a>
for <span class="math-container">$\triangle AOD$</span>,</p>
<p><span class="math-container">\begin{align}
|OD|^2\cdot c+|OA|^2\cdot |BD|
-|AD|\cdot(|OB|^2+c\cdot |BD|)
&=0
,\\
|AD|\cdot (
R^2
-|OB|^2-c\cdot |BD|)
&=0
,
\end{align}</span></p>
<p><span class="math-container">\begin{align}
|OB|^2&=
R^2-c\cdot |BD|
\\
&=50-6\cdot4
=26
.
\end{align}</span></p>
|
68,386 | <p>I'm looking for a theorem of the form </p>
<blockquote>
<p>If $R$ is a nice ring and $v$ is a reasonable element in $R$ then Kr.Dim$(R[\frac{1}{v}])$ must be either Kr.Dim$(R)$ or Kr.Dim$(R)-1$.</p>
</blockquote>
<p>My attempts to do this purely algebraically are not working, so I started looking into methods from algebraic geometry. I thought that Grothendieck's Vanishing Theorem might help (i.e. if dim$(X)=n$ then $H^i(X,\mathcal{F})=0$ for any sheaf of abelian groups $\mathcal{F}$ and any $i>n$) but the problem is that the converse for this theorem fails, so I can't conclude anything about dimension. Perhaps this theorem could give some sort of test for when dimension drops, but I'm hoping for a better answer.</p>
<p>We'll definitely need some hypotheses. For the application I have in mind we can assume $R$ is commutative and is finitely generated over some base ring (e.g. $\mathbb{Z}_{(2)}$), but we should not assume it's an integral domain. If necessary we can assume it's Noetherian and local, but I'd rather avoid this. As for $v$, it's not in the base ring and it has only a few relations with other elements in $R$, none of which are in the base ring. If we can't get the theorem above, perhaps we can figure out something to help me get closer:</p>
<blockquote>
<p>Are there any conditions on $v$ such that the dimension would drop by more than 1 after inverting $v$?</p>
</blockquote>
<p>One thing I know: to have any hope of dimension dropping by $1$ I need to be inverting a maximal irreducible component. I'm curious as to the algebraic condition this puts on $v$. </p>
| Hailong Dao | 2,083 | <p>The dimension of $R[1/v]$ is the biggest height of some prime ideal $P$ such that $v\notin P$. So, let $I_{d-1}$ be the intersection of all primes of height at least $d-1$ ($d= \dim R$), then</p>
<blockquote>
<p>$\dim R[1/v] \geq d-1$ if and only if $v\notin I_{d-1}$. </p>
</blockquote>
<p>Under a mild condition (all maximal ideals has height at least $d-1$), then $I_{d-1}$ is inside the Jacobson radical of $R$. So in this case $v$ is not inside the Jacobson radical would suffice. </p>
<p>EDIT (in response to the comment by the OP): Now let $I_d$ be the intersection of all primes with height $d$. Then $\dim R[1/v] \leq d-1$ iff $v\in I_d$. So </p>
<blockquote>
<p>$\dim R[1/v] =d-1$ iff $v \in I_d- I_{d-1}$.</p>
</blockquote>
<p>Here is a short calculation to show that in Fernando's example above, $I_1=(x)$. Any prime that does not contain $x$ would have to contain $(x-1, y, z)$, but this is a maximal ideal of height $0$. So $x \in I_1$. On the other hand $R/(x)$ is $k[y,z]$ so the intersection of all height one there is $0$. Thus $I_1=(x)$. This gives you precisely what elements satisfies your condition. </p>
|
3,464,247 | <p>I'm wondering if my reasoning is justified when determining if a vector is in the span of of a set of vectors.</p>
<p><span class="math-container">$$T = \{(1, 1, 0), (-1, 3, 1)\}$$</span></p>
<p>For which <span class="math-container">$a$</span> is <span class="math-container">$(a^2, a+2, 2) \in span(T)$</span></p>
<p>I've come to the conclusion that there DNE an "<span class="math-container">$a$</span>" which could make the vector be in the span of set <span class="math-container">$T$</span>. However I'm not sure if my reasoning is justified. </p>
<p>For vector <span class="math-container">$V$</span> to be in the span of set <span class="math-container">$T$</span>,it would have to be a linear combination of the two vectors in set <span class="math-container">$T$</span>, a scalar of the vectors in set <span class="math-container">$T$</span>, or combination of both. </p>
<p>In order to satisfy the <span class="math-container">$2$</span> in the third entry of vector <span class="math-container">$V$</span>, we must have <span class="math-container">$2(-1,3,1)$</span>. While we have our <span class="math-container">$2$</span> in the third entry, it does not fit the 1st and 2nd entries of vector <span class="math-container">$V$</span>, so I would need add/subtract a scalar of the first vector <span class="math-container">$(1,1,0)$</span>. But I've realized that it can never fit the requirements of vector <span class="math-container">$V$</span> so I've come to the conclusion that an "<span class="math-container">$a$</span>" that would make vector <span class="math-container">$V$</span> to be in the span of set <span class="math-container">$T$</span> DNE.</p>
| José Carlos Santos | 446,262 | <p>Actually, the answer depends upen the field that you are working with. In fact,<span class="math-container">\begin{align}\begin{bmatrix}a^2\\a+2\\2\end{bmatrix}\in\operatorname{span}\left\{\begin{bmatrix}1\\1\\0\end{bmatrix},\begin{bmatrix}-1\\3\\1\end{bmatrix}\right\}&\iff\begin{vmatrix}1&-1&a^2\\1&3&a+2\\0&1&2\end{vmatrix}=0\\&\iff a^2-a+6=0.\end{align}</span>Over <span class="math-container">$\mathbb Q$</span>, for instance, this equation has no solutions, whereas of <span class="math-container">$\mathbb C$</span>, for instance, it has.</p>
|
3,290,514 | <p>I need to make the navigation and guidance of a vehicle (a quadcopter) in a platform. This platform can be seen like this:</p>
<p><a href="https://i.stack.imgur.com/jeJ34.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jeJ34.png" alt="enter image description here"></a></p>
<p>where the blue dots are the center of each square, and the <span class="math-container">$x$</span> distances are all the same, and the <span class="math-container">$y$</span> distances are all the same.</p>
<p>I need the distance between each blue dot to the center (the blue dot of the <span class="math-container">$(2;2)$</span>), but that distance depends on the <span class="math-container">$yaw$</span> angle. For example, if <span class="math-container">$yaw=0^\circ$</span>, the situation is like this:</p>
<p><a href="https://i.stack.imgur.com/QGwlE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QGwlE.png" alt="enter image description here"></a></p>
<p>and the distances are:</p>
<p><span class="math-container">$$d_{1;1} = (-d_x; -d_y)$$</span>
<span class="math-container">$$d_{1;2} = (-d_x; 0)$$</span>
<span class="math-container">$$d_{1;3} = (-d_x; d_y)$$</span></p>
<p><span class="math-container">$$d_{2;1} = (0; -d_y)$$</span>
<span class="math-container">$$d_{2;2} = (0; 0)$$</span>
<span class="math-container">$$d_{2;3} = (0; d_y)$$</span></p>
<p><span class="math-container">$$d_{3;1} = (d_x; -d_y)$$</span>
<span class="math-container">$$d_{3;2} = (d_x; 0)$$</span>
<span class="math-container">$$d_{3;3} = (d_x; d_y)$$</span></p>
<p>If the situation is with <span class="math-container">$yaw=180^\circ$</span>:</p>
<p><a href="https://i.stack.imgur.com/Y0A6P.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Y0A6P.png" alt="enter image description here"></a></p>
<p>the distances are the same but with the opposite sign, i.e,</p>
<p><span class="math-container">$$d_{1;1} = (d_x; d_y)$$</span>
<span class="math-container">$$d_{1;2} = (d_x; 0)$$</span>
<span class="math-container">$$d_{1;3} = (d_x; -d_y)$$</span></p>
<p><span class="math-container">$$d_{2;1} = (0; d_y)$$</span>
<span class="math-container">$$d_{2;2} = (0; 0)$$</span>
<span class="math-container">$$d_{2;3} = (0; -d_y)$$</span></p>
<p><span class="math-container">$$d_{3;1} = (-d_x; d_y)$$</span>
<span class="math-container">$$d_{3;2} = (-d_x; 0)$$</span>
<span class="math-container">$$d_{3;3} = (-d_x; -d_y)$$</span></p>
<p>If <span class="math-container">$yaw=90^\circ$</span>, the situation is like this:</p>
<p><a href="https://i.stack.imgur.com/B6a8b.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B6a8b.png" alt="enter image description here"></a></p>
<p>and the distances (see the difference between <span class="math-container">$d_x$</span> and <span class="math-container">$d_y$</span>) would be:</p>
<p><span class="math-container">$$d_{1;1} = (-d_y; d_x)$$</span>
<span class="math-container">$$d_{1;2} = (-d_y; 0)$$</span>
<span class="math-container">$$d_{1;3} = (-d_y; d_x)$$</span></p>
<p><span class="math-container">$$d_{2;1} = (0; -d_x)$$</span>
<span class="math-container">$$d_{2;2} = (0; 0)$$</span>
<span class="math-container">$$d_{2;3} = (0; d_x)$$</span></p>
<p><span class="math-container">$$d_{3;1} = (d_y; -d_x)$$</span>
<span class="math-container">$$d_{3;2} = (d_y; 0)$$</span>
<span class="math-container">$$d_{3;3} = (d_y; d_x)$$</span></p>
<p>If <span class="math-container">$yaw = -90^\circ$</span>:</p>
<p><a href="https://i.stack.imgur.com/6Zk2f.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6Zk2f.png" alt="enter image description here"></a></p>
<p>the distances would be:</p>
<p><span class="math-container">$$d_{1;1} = (d_y; d_x)$$</span>
<span class="math-container">$$d_{1;2} = (d_y; 0)$$</span>
<span class="math-container">$$d_{1;3} = (d_y; -d_x)$$</span></p>
<p><span class="math-container">$$d_{2;1} = (0; d_x)$$</span>
<span class="math-container">$$d_{2;2} = (0; 0)$$</span>
<span class="math-container">$$d_{2;3} = (0; -d_x)$$</span></p>
<p><span class="math-container">$$d_{3;1} = (-d_y; d_x)$$</span>
<span class="math-container">$$d_{3;2} = (-d_y; 0)$$</span>
<span class="math-container">$$d_{3;3} = (-d_y; -d_x)$$</span></p>
<p>I need to write a matrix that uses the information of the <span class="math-container">$yaw$</span> angle and returns the distances from each angle (not just 0, 90, -90 and 180, but also 1, 2, 3, ...)</p>
<p>I tried to write it but I couldn't find the solution.</p>
<p>Thank you very much. I really need this help</p>
<p>Edit: please note that the coordinate frame moves with the quadcopter, like in this image:</p>
<p><a href="https://i.stack.imgur.com/Vge0h.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Vge0h.png" alt="enter image description here"></a></p>
<p>Edit 2: for example, if <span class="math-container">$yaw=45^\circ$</span>, then the distance from <span class="math-container">$(3;3)$</span> to <span class="math-container">$(2;2)$</span> is <span class="math-container">$\sqrt{d_x^2+d_y^2}$</span> in <span class="math-container">$x$</span> and <span class="math-container">$0$</span> in <span class="math-container">$y$</span>.</p>
| Robert Israel | 8,508 | <p>One way to fit points to a circle is to do linear regression on the equation in the form <span class="math-container">$x^2 + a x + y^2 + b y + c = 0$</span>. Writing this as
<span class="math-container">$(x+a/2)^2 + (y+b/2)^2 = (a^2 + b^2)/4 - c$</span>, this corresponds to a circle of centre <span class="math-container">$(-a/2, -b/2)$</span> and radius <span class="math-container">$\sqrt{(a^2+b^2)/4 - c}$</span>.</p>
|
2,785,698 | <p>Please help me go over this problem; I am a bit confused.</p>
<p>Find ${\displaystyle \frac{\mathrm d}{\mathrm dt} \int_2^{x^2}e^{x^3}\mathrm dx}$.</p>
| tien lee | 557,074 | <p>The Fundamental Theorem of Calculus states that if
$$g(x) = \int_{a}^{f(x)} h(t)~{\rm d}t$$
where $a$ is any constant, then
$$g'(x) = h(f(x)) \cdot f'(x)$$
Using this with the integral, $g(x) = g(x)$, $f(x) = x^2$, and $h(x) = e^{x^3}$.</p>
<p>So, $$\frac{\mathrm d}{\mathrm dx} \int_2^{x^2}e^{x^3}\mathrm dx=(2x)e^{x^6}$$</p>
|
4,328,117 | <p>I have this lambda expression</p>
<p><span class="math-container">$$(\lambda xyz.xy(zx)) \;1\; 2\; 3$$</span></p>
<p>or</p>
<p><span class="math-container">$$(\lambda x. (\lambda y. (\lambda z.xy(zx))))\;1\;2\;3$$</span>
<span class="math-container">$$(\lambda y. (\lambda z.1y(z1))))\;2\;3$$</span>
<span class="math-container">$$(\lambda z.12(z1))))\;3$$</span>
<span class="math-container">$$1\;2\;(3\;1)$$</span></p>
<p>What exactly is the point of the inner parentheses in the original expression -- which I dutifully carried along to the end? What's the difference between <span class="math-container">$1\;2\;(3\;1)$</span> and <span class="math-container">$1\;2\;3\;1$</span>? The outer is to distinguish between the expression and the input, I'm assuming, but the inner <span class="math-container">$\ldots(zx)\ldots$</span> isn't clear to me. Parentheses in math indicate "belonging together," but I don't understand this "belonging together." Even if I wanted to have two <span class="math-container">$x$</span> variables in the body, I'd just have them without needing parentheses. I checked out <a href="https://math.stackexchange.com/questions/50401/use-of-parenthesis-in-lambda-calculus">this</a>, but I don't think it's the same parentheses issue.</p>
| Marc van Leeuwen | 18,880 | <p>Syntactically, function application is a two-operand non-associative operation, where the latter means that <span class="math-container">$f(g\,x)$</span> has to be distinguished from <span class="math-container">$(f\,g) x$</span> (the function application operation conventionally has an empty operator symbol, in other words the two operands are just juxtaposed; for the current discussion it would however make no difference if a visible operator symbol were used). The basic solution to avoid ambiguity for formulas that contain more than one such operation is to require parentheses around every non atomic sub-expression that is itself operand of another application, as in the two examples I cited. One can reduce the number of required parentheses somewhat by stipulating that in one of the two operand positions the parentheses can be omitted, and indeed in <span class="math-container">$\lambda$</span>-calculus such omission is allowed for the left operand, so <span class="math-container">$(f\,g)x$</span> can be written simply as <span class="math-container">$f\,g\,x$</span> (the operation is then said to associate to the left). But of course the parentheses must by retained in the case of a right operand, to be able to distinguish <span class="math-container">$f(g\,x)$</span> from <span class="math-container">$f\,g\,x$</span>.</p>
<p>Since <span class="math-container">$\lambda$</span>-calculus also has <span class="math-container">$\lambda$</span>-expressions, it may be useful to state how these enter into the syntactic game. By convention <span class="math-container">$\lambda$</span>-abstraction binds weaker than function application, which means that a <span class="math-container">$\lambda$</span>-expression that is an operand of a function application, either left or right, always needs to be parenthesised; on the other hand no additional parentheses are needed to hold a function application that is the "body" of the <span class="math-container">$\lambda$</span>-expression together. A new <span class="math-container">$\lambda$</span>-expression that occurs directly as body of another one also requires no additional parentheses, since in this case there is actually no ambiguity on how to parse something like <span class="math-container">$\lambda x.\lambda y.x$</span> in the first place.</p>
|
463,239 | <p>Integrate $$\int{x^2(8x^3+27)^{2/3}}dx$$</p>
<p>I'm just wondering, what should I make $u$ equal to?</p>
<p>I tried to make $u=8x^3$, but it's not working. </p>
<p>Can I see a detailed answer?</p>
| Keith Afas | 85,076 | <p>You would use $u,du$ substitution</p>
<p>$$\int{x^2(8x^3+27)^{2/3}}dx$$
$$=\int{x^2(u)^{2/3}}dx$$
$$u=8x^3+27, du=24x^2 \implies \frac{du}{24}=x^2dx$$
$$\int{x^2(u)^{2/3}}dx$$
$$=\int{(u)^{2/3}x^2}dx$$
$$=\int{(u)^{2/3}}\frac{du}{24}$$
$$=\frac{1}{24}\int{(u)^{2/3}}du$$
$$=\frac{1}{24}\int{(u)^{2/3}}du$$
$$=\frac{1}{24}\left(\frac{(u)^{2/3+1}}{2/3+1}\right)+C$$
$$=\left(\frac{(u)^{5/3}}{24(5/3)}\right)+C$$
$$=\frac{u^{5/3}}{40}+C$$</p>
|
1,380,697 | <p>I am currently an undergraduate and thinking about applying to graduate school for math. The problem is that I don't know what field I want to go. Taking graduate classes even more confuse me because the more I learn the less I know what specifically I want to do. My question is to where to find an information about different fields of mathematics? Maybe you can recommend me some good journals about math with overview of top areas of math or popular fields. I already spoke with my professors, asked graduate students about their history but I think that my knowledge about math in a broad sense grows slower than I want it. </p>
<p>Maybe there is some good website with people chatting about different fields of research? What about conferences: is there a conference available for undergrad about top-trands in mathematics? </p>
<p>All sources and all answers are welcome. </p>
<p>I am mostly interested in pure math, but I also like applied math. </p>
| Thomas | 26,188 | <p>It is hard to get a good overview of all of mathematics. The best really is to <strong>take a good broad variety of classes</strong>. This you usually do (to some extent) the first couple of years in graduate school. Here you will learn the basic language of the main areas of mathematics. Here you will be get to <strong>meet various professors</strong>.</p>
<p>For now, then, I wouldn't worry too much about what field you want to go into. If you don't already have an interest in an area, then it is hard to say much that can push you in a certain direction. You might be thinking that you want to find an area so that you can go to a school that specializes in that area. Again, <strong>don't worry too much about it</strong>.</p>
<p>One thing that you probably can decide on now is whether you want to do applied mathematics or pure mathematics. Some departments are very strong in one and not the other. It sounds like you are more into pure mathematics.</p>
<p>If you are wanting to do a Ph.D., then I would also point out the importance of having a good advisor. Some are highly motivated and work hard on their own, but others need good guidance. If you take an area like algebra, then you have to remember that this area is huge. There are many subareas of algebra and what you will end up doing research in will (usually) depend heavily on what your advisor does. I say this to point out that the <strong>choice of advisor for many is almost as important as picking a good area</strong>. (Again, others already know what they want to do and just need an environment where they can work in peace.)</p>
<p>About conferences. Ask your university/department about funding to go to <strong>MathFest</strong> or the <strong>Joint Meetings.</strong> Do you have a Mathclub at your school that might be able to help you? Next week happens to be MathFest and here many undergraduate students will present work they have done over the summer (or past semesters). These talks are a great introduction to various areas. The level is low enough that an advanced undergraduate student should be able to learn something. There are also student activities at the Join Meetings in January. <strong>Talk to your department about this now!</strong></p>
<p>Another great suggestion is to <strong>attend an REU</strong> (Reseach Experience for Undergraduates). These are usually summer programs where you get to go into depth on a specific topic. Usually there isn't much of a requirement for background). Google this and start to talk to your department about how to apply.</p>
<p>Lastly, try to talk to as many people as possible about what they do. It sounds like you have already been doing this, but ask each professor in your department what they work on. Talk to graduate students about what they do. </p>
|
3,681,916 | <p>Given a matrix A, Does
<span class="math-container">$$\lim_{n \to \infty}A^n=0$$</span>
Imply that(by lower then I mean that every number in the lower matrix is closer or the same distance to 0 then it’s counterpart in the bigger one)
<span class="math-container">$$\text{if }a>b>0 \text{ then }A^a<A^b$$</span>
In R it’s true, a Number who is in between -1 and 1 is always getting closer to 0 and never goes up and back down.<br>
I’m unsure of it’s truth for matrices especially with my definition.
Can any of you provide hints/proves for it?<br>
Maybe counter example?<br>
Thanks in advance.</p>
| Clement Yung | 620,517 | <p>This is false. Consider:
<span class="math-container">$$ A :=
\begin{pmatrix}
0 & 0.5 \\
0.5 & 0
\end{pmatrix}
$$</span>
Then the sequence of matrix is as follows:
<span class="math-container">\begin{align*}
A^2 =
\begin{pmatrix}
0.25 & 0 \\
0 & 0.25
\end{pmatrix}, \;
A^3 =
\begin{pmatrix}
0 & 0.125 \\
0.125 & 0
\end{pmatrix}, \;
A^4 =
\begin{pmatrix}
0.0625 & 0 \\
0 & 0.0625
\end{pmatrix}, \;
\dots
\end{align*}</span>
which tends to <span class="math-container">$0$</span> (under any norm), but <span class="math-container">$A^2 \not< A$</span> (under your definition of <span class="math-container">$<$</span>).</p>
|
1,910,109 | <p>$$\int \frac{1}{\sqrt{x} (1 - 3\sqrt{x})}$$</p>
<p>I tried with the substitution $u = 1-3\sqrt{x}$</p>
<p>I am confused with how to finish this problem I know I am supposed to substitute $u$ and $\text{d}u$ in but I am not sure how to finish it.</p>
| Enrico M. | 266,764 | <p>Let's set the problem, then you may proceed by yourself.</p>
<p>Performing the substitution</p>
<p>$$u = 1 - 3\sqrt{x} ~~~ \text{and so} ~~~ \sqrt{x} = -\frac{u-1}{3}$$</p>
<p>Hence</p>
<p>$$\text{d}u = -\frac{3}{2\sqrt{x}} = -\frac{3}{2}\frac{-3}{u-1}\ \text{d}x$$</p>
<p>$$\text{d}u = \frac{9}{2(u-1)}\ \text{d}x ~~~~~ \to ~~~~~ \text{d}x = \frac{2(u-1)}{9}\ \text{d}u$$</p>
<p>Then your integral becomes</p>
<p>$$\int -\frac{1}{\frac{u-1}{3}u}\frac{2(u-1)}{9} = -\int \frac{2}{3u}\ \text{d}u$$</p>
<p>Now you can proceed for sure.</p>
|
3,931,831 | <p>For the scenario given below, I am confused about if the samples are dependent or independent since the scenario does not mention anything about the samples being paired/related or vice versa.</p>
<p>I am aware if terms such as paired, repeated measurements, within-subject effects, matched pairs, and pretest/posttest are instructed in scenarios then it indicates that the samples are dependent and the opposite applies to independent samples, but I am clueless for the given scenario. Any help would be appreciated.</p>
<p><em>Alice and Bob work evening shifts in a supermarket. Alice has complained to
the manager that she works, on average, much more than Bob. The manager claims that on
average they both work the same amount of time, i.e. the competing claim is that the average
working hours are different. After a short discussion between the manager and Alice, the manager
randomly selected 50 evenings when Alice and Bob both worked.</em></p>
| PierreCarre | 639,238 | <p>In this case, it is really not a great idea to use Lagrange multipliers. We can write <span class="math-container">$x$</span> in terms of <span class="math-container">$y$</span> (or vice-versa) using the restriction and reduce this question to a one variable optimisation problem. Substituting <span class="math-container">$x = 3 - 2y$</span>, the problem reduces to finding the extrema of <span class="math-container">$f(y)=-7 y^3+36 y^2-51 y+18$</span>. The critical points are <span class="math-container">$y = \frac{17}{7}$</span> and <span class="math-container">$y = 1$</span> and, considering that <span class="math-container">$f''(1)>0$</span> and <span class="math-container">$f''(\frac{17}{7})<0$</span>, they are a local minimum and maximum, respectively.</p>
<p>It is also worth noticing that, since <span class="math-container">$\displaystyle \lim_{y \to \pm \infty} f(y) = \mp \infty$</span>, there will not be any global extrema.</p>
|
1,832,812 | <blockquote>
<p>Let $M$ be a nonempty subset of $\mathbb Z^+$ such that for every element $x$ in $M,$ the numbers $4x$ and $\lfloor \sqrt x \rfloor$ also belong to $M.$ Prove that $M = \mathbb Z^+$.</p>
</blockquote>
<p>Suppose $a \in M$. Then so are $4ak$ and $\lfloor \sqrt{4ak} \rfloor$ for every positive integer multiple of $k$. Also we could take a multiple of $4$, then do the square root and floor it etc. in many different combinations. How do we prove that $M$ is all of $\mathbb{Z}^+$?</p>
| Ethan Alwaise | 221,420 | <p>Use the fact that $\lfloor \sqrt{x} \rfloor \in M$ for every $x \in M$ to show that $1 \in M$, hence $M$ contains every multiple of $4$. To finish, use the fact that there is always a multiple of $4$ between any two consecutive squares. </p>
|
1,201,955 | <blockquote>
<p><em>Question</em>: If $f(n)$ is $O(g(n))$ and $g(n)$ is $O(f(n))$, is $f(n) = g(n)$?</p>
</blockquote>
<p>I'm studying for a discrete mathematics test, and I'm not sure if this is true or false. Since Big-OH ignores constant multiples, wouldn't $f(n) = c\, g(n)$?</p>
| marty cohen | 13,079 | <p>Consider
$f(n) = an^2$
and
$g(n) = bn^2$
where $a$ and $b$ are
distinct positive reals.</p>
<p>Both
$f(n) = O(g(n))$
and
$g(n) = O(f(n))$
are true,
but $f(n) \ne g(n)$
for $n > 0$.</p>
<p>This is often written as
$f(n) = \Theta(g(n))$.</p>
|
2,170,832 | <p>What is $\mathbb{Z}^*_p$ ?</p>
<p>I think $\mathbb{Z}_p$ is the ring of p-adic integers but not sure what the * represents.</p>
| Ege Erdil | 326,053 | <p>It's the group of $ p $-adic units.</p>
|
2,170,832 | <p>What is $\mathbb{Z}^*_p$ ?</p>
<p>I think $\mathbb{Z}_p$ is the ring of p-adic integers but not sure what the * represents.</p>
| Dietrich Burde | 83,966 | <p>The star usually refers to the group of units of a ring $R$. Another notation is $U(R)$. </p>
|
456,826 | <p>I need to find the derivative of this function. I know I need to separate the integrals into two and use the chain rule but I am stuck.</p>
<p>$$y=\int_\sqrt{x}^{x^3}\sqrt{t}\sin t~dt~.$$</p>
<p>Thanks in advance</p>
| Community | -1 | <p><strong>Hint</strong></p>
<p>By the chain rule we prove easly:</p>
<p>If
$$F(x)=\int_{u(x)}^{v(x)}f(t)dt$$
then
$$F'(x)=f(v(x))v'(x)-f(u(x))u'(x)$$</p>
|
3,989,878 | <p>I can't solve this problem. I tried to find <span class="math-container">$\tan x$</span> directly by solving cubic equations but I failed.</p>
<p>The problem is to find <span class="math-container">$\tan x\cot 2x$</span> given that
<span class="math-container">$$\tan x+ \tan 2x=\frac{2}{\sqrt{3}}, \>\>\>\>\>0<x<\pi/4$$</span></p>
<p>How am I supposed to solve this problem?</p>
| Raffaele | 83,382 | <p><span class="math-container">$$\frac{2 \tan x}{1-\tan ^2 x}+\tan x=^*\frac{2}{\sqrt{3}}$$</span>.
<span class="math-container">$$\tan x=-1.45424, \tan x=0.35178, \tan x =2.25716$$</span>
<span class="math-container">$L=\tan x \cot 2x$</span> can be written<span class="math-container">$^{**}$</span> as <span class="math-container">$\frac{1}{2} \left(1-\tan ^2 x\right)$</span> thus possible values are <span class="math-container">$$L_1=-0.557406,L_2=0.438125,L_3=-2.04739$$</span></p>
<hr />
<p><span class="math-container">$$(^*)\quad \tan x=u \to u^3-\frac{2 u^2}{\sqrt{3}}-3 u+\frac{2}{\sqrt{3}}=0$$</span>
see solutions <a href="https://www.wolframalpha.com/input/?i=solve%202%2FSqrt%5B3%5D%20-%203%20u%20-%20%282%20u%5E2%29%2FSqrt%5B3%5D%20%2B%20u%5E3%20%3D%3D%200" rel="nofollow noreferrer">here</a>. (press <em>Approximate forms</em> button)
<span class="math-container">$$(^{**})\quad\tan x \cot 2x=\frac{\sin x\left(\cos ^2 x-\sin ^2 x\right)}{\cos x(2 \sin x \cos x)}=\frac{\cos ^2 x-\sin ^2 x}{2\cos^2 x}=\frac{1}{2}\left(1-\tan^2 x\right)$$</span></p>
|
2,400,654 | <p>I am told that the statement "any closed set has a point on its boundary" is false, yet I don't know how to disprove it. In fact, I think it is true. </p>
<p>Suppose we have [a,b], a closed set. Then, the boundary would be {a,b}, both of which are the elements of the set. So, there we have a closed set that has point(s) on its boundary.</p>
<p>I think one reason why I was told that the statement is false can be related to the empty set since</p>
<ul>
<li>Empty set is also closed (clopen, but still)</li>
<li>It does not have a point on its boundary</li>
<li>The statement goes "any closed set..", hence empty set is not
excluded.</li>
</ul>
<p>Supposing that the statement is false due to the above-mentioned reasoning (fact that empty set was not excluded), would "any nonempty closed set has a point on its boundary" be correct?</p>
<p>Thanks.</p>
| Michael Hardy | 11,667 | <p>On the "pro" side I would mention two facts:</p>
<ul>
<li><p>For $\alpha>0,$ let $$f_\alpha(x) = \frac 1 {\Gamma(\alpha)} x^{\alpha-1} e^{-x}.$$ Then for $\alpha,\beta>0$ we have the convolution $$f_\alpha * f_\beta = f_{\alpha+\beta}.$$ (And recall that the convolution of two probability density functions is the density of the sum of two independent random variables.)</p></li>
<li><p>For $A\subseteq[0,1],$ suppose $$\Pr(X\in A) = \frac 1 {\operatorname{B}(\alpha,\beta)} \int_A x^{\alpha-1} (1-x)^{\beta-1} \, dx.$$ Then $$\operatorname{E}(X) = \frac \alpha {\alpha+\beta}.$$</p></li>
</ul>
<p>(I wonder if the usual convention can be made less painful to those who don't like it by writing it like this:
$$
\Gamma(\alpha) = \int_0^\infty x^\alpha e^{-x} \, \frac{dx} x.
$$</p>
|
3,085,591 | <p>I'm trying to prepare for an exam and came across the following question:</p>
<blockquote>
<p>Given <span class="math-container">$n \times n$</span> matrix <em>A</em>, let <em>U</em> represent row space and <em>W</em>
represent column space. </p>
<p>A) Prove: <span class="math-container">$W \subseteq$</span> <span class="math-container">$U^{\perp} $</span> if and only if <span class="math-container">$A^2=0$</span></p>
<p>B) Prove that if <span class="math-container">$dim(U)>n/2$</span>, then <span class="math-container">$A^2 \ne 0$</span>.</p>
</blockquote>
<p>I was able to do part A pretty easily as follows: </p>
<p><span class="math-container">$\Rightarrow$</span>: Let rows of <em>A</em> be {<span class="math-container">$ \vec{r}_1, \vec{r}_2 ,...,\vec{r}_n $</span>} and columns be {<span class="math-container">$ \vec{c}_1, \vec{c}_2 ,...,\vec{c}_n $</span>}. Assume <span class="math-container">$W \subseteq$</span> <span class="math-container">$U^{\perp} $</span>. Entry <span class="math-container">$a_{ij}$</span> (in row <em>i</em>, column <em>j</em> of matrix <span class="math-container">$A^2=AA$</span>) equals the dot product of <span class="math-container">$\vec{r}_i \cdot \vec{c}_j $</span>, and by assumption <span class="math-container">$\vec{c}_j$</span> is orthogonal to <span class="math-container">$ \vec{r}_1, \vec{r}_2 ,...,\vec{r}_n $</span>, so it follows that <span class="math-container">$a_{ij}=0$</span> for all <em>i,j</em>. Thus <span class="math-container">$A^2=0$</span></p>
<p><span class="math-container">$\Leftarrow$</span>: Assume <span class="math-container">$A^2=0$</span>. We can represent the columns of <span class="math-container">$A^2$</span> as </p>
<p><span class="math-container">$$ \pmatrix{\vec{r}_1 \\ \vec{r}_2 \\ \vdots \\ \vec{r}_n} \cdot \vec{c}_1 , \pmatrix{\vec{r}_1 \\ \vec{r}_2 \\ \vdots \\ \vec{r}_n} \cdot \vec{c}_2 , ... , \pmatrix{\vec{r}_1 \\ \vec{r}_2 \\ \vdots \\ \vec{r}_n} \cdot \vec{c}_n ,$$</span>
But by assumption, <span class="math-container">$A^2=0$</span>, meaning all columns of <span class="math-container">$A^2=0$</span> and all the above dot products are zero. Hence, <span class="math-container">$\vec{c}_i$</span> is orthogonal to all vectors in {<span class="math-container">$\vec{r}_1, \vec{r}_2 ,...,\vec{r}_n $</span>}, and thus all linear combination of those vectors. Since this applies to all <span class="math-container">$\vec{c}_i$</span>, then <span class="math-container">$W \subseteq$</span> <span class="math-container">$U^{\perp} $</span>.</p>
<p>As for Part B, I'm having trouble and not sure exactly where to begin. Presumably, I would use the results from part A, meaning that if <span class="math-container">$A^2 \ne 0$</span>, then <span class="math-container">$W \not\subseteq$</span> <span class="math-container">$U^{\perp} $</span>, that is, there is some vector in column space that is not orthogonal to every vector in row space.</p>
<p>Some properties I'm aware of that may be useful but not sure if or how:</p>
<ul>
<li><span class="math-container">$dim(U)=dim(W)$</span> </li>
<li>The orthogonal complement of the row space is null space</li>
<li>the dimension of the row space and nullspace equals <span class="math-container">$n$</span></li>
<li>dimension of column space of <em>A</em> equals dimension of row space of <span class="math-container">$A^T$</span></li>
</ul>
<p>Anyway, any tips would be greatly appreciated!</p>
| jmerry | 619,637 | <p>Hint: compare the dimensions of <span class="math-container">$W$</span> and <span class="math-container">$U^{\perp}$</span>, given that <span class="math-container">$\dim(U) >\frac n2$</span>. We can't include a vector space inside one of smaller dimension...</p>
|
3,292,918 | <p>Let <span class="math-container">$X$</span> be a Banach space, and denote by <span class="math-container">$B_r (x)$</span> the closed ball of radius <span class="math-container">$r > 0$</span>
around <span class="math-container">$x \in X$</span>. Furthermore, let <span class="math-container">$A \subset X$</span> be compact and <span class="math-container">$N \in \Bbb{N}$</span>.
I am interested in "optimally" covering <span class="math-container">$A$</span> by <span class="math-container">$N$</span> balls, i.e,. with as small radius as possible.
More precisely, define
<span class="math-container">$$
r_0
:= \inf \Big\{
r > 0
\colon
\exists \, x_1,\dots,x_N \in X \text{ such that } A \subset \bigcup_{i=1}^N B_r (x_i)
\Big\},
$$</span>
and assume that <span class="math-container">$r_0 > 0$</span>.</p>
<blockquote>
<p>I would like to know whether there necessarily exist <span class="math-container">$x_1,\dots,x_N \in X$</span>
such that <span class="math-container">$A \subset \bigcup_{i=1}^N B_{r_0} (x_i)$</span>.
In other words, I would like to know if the infimum above is actually a minimum.</p>
</blockquote>
<p><strong>Note</strong> I am taking the <span class="math-container">$x_i$</span> from the "surrounding" space <span class="math-container">$X$</span>, not from the compact set <span class="math-container">$A$</span>.</p>
<p>I can prove the claim in case that <span class="math-container">$X$</span> is reflexive (even only assuming that <span class="math-container">$A$</span> is bounded),
but I am not sure whether it is true for more general Banach spaces.
I will give my proof for the reflexive case below, in case one can either generalize it,
or use it to get an idea for a counterexample.</p>
<p><hr><hr></p>
<p><strong>Proof for the reflexive case:</strong>
Choose a sequence <span class="math-container">$r_n \to r_0$</span> such that for each <span class="math-container">$n$</span> there are <span class="math-container">$x_1^n,\dots,x_N^n \subset X$</span>
satisfying <span class="math-container">$A \subset \bigcup_{i=1}^N B_{r_n}(x_i^n)$</span>.
If <span class="math-container">$B_{r_n} (x_i^n) \cap A = \emptyset$</span> for some <span class="math-container">$i,n$</span>, replace <span class="math-container">$x_i^n$</span> by zero.
Note that this retains the property <span class="math-container">$A \subset \bigcup_{i=1}^N B_{r_n}(x_i^n)$</span>.</p>
<p>Since <span class="math-container">$A$</span> and the sequence <span class="math-container">$(r_n)_{n}$</span> are bounded, there is <span class="math-container">$R > 0$</span> such that <span class="math-container">$\| x \| \leq R$</span>
and <span class="math-container">$r_n \leq R$</span> for all <span class="math-container">$x \in A$</span> and <span class="math-container">$n \in \Bbb{N}$</span>:
There are now two cases for each <span class="math-container">$i,n$</span>:
1) There is some <span class="math-container">$x \in A \cap B_{r_n} (x_i^n)$</span>,
and hence <span class="math-container">$\| x_i^n \| \leq \| x_i^n - x \| + \| x \| \leq r_n + R \leq 2R$</span>.
2) There is no <span class="math-container">$x \in A \cap B_{r_n} (x_i^n)$</span>, and hence <span class="math-container">$x_i^n$</span>, whence <span class="math-container">$\| x_i^n \| \leq 2R$</span>.</p>
<p>Therefore, each of the sequences <span class="math-container">$(x_i^n)_{n \in \Bbb{N}} \subset X$</span> is bounded.
Since <span class="math-container">$X$</span> is assumed to be reflexive, we can choose a common subsequence
(which I will ignore in the notation below) such that <span class="math-container">$x_i^n \to x_i$</span> weakly for all <span class="math-container">$i = 1,\dots,N$</span>.</p>
<p>Now, let <span class="math-container">$x \in A$</span> be arbitrary.
For each <span class="math-container">$n \in \Bbb{N}$</span>, there is <span class="math-container">$i_n \in \{1,\dots,N\}$</span> satisfying <span class="math-container">$\| x - x_i^n \| \leq r_n$</span>.
Next, there is <span class="math-container">$\ell \in \{1,\dots,N\}$</span> such that <span class="math-container">$i_n = \ell$</span> for infinitely many <span class="math-container">$n \in \Bbb{N}$</span>,
say for <span class="math-container">$n = n_m$</span> with <span class="math-container">$n_m \to \infty$</span>.
Since <span class="math-container">$x - x_i^{n_m} \to x - x_i$</span> weakly and since the norm is lower semicontinuous with respect
to weak convergence, we see that
<span class="math-container">$\| x - x_\ell \| \leq \liminf_{m \to \infty} \| x - x_i^{n_m} \| \leq \liminf_{m \to \infty} r_{n_m} = r_0$</span>.
Since this holds for any <span class="math-container">$x \in A$</span>, we get <span class="math-container">$A \subset \bigcup_{\ell=1}^N B_{r_0} (x_\ell)$</span>,
as desired.</p>
| Selrach Dunbar | 278,932 | <p>Unfortunately, the infimum above is not actually a minimum. To see this let <span class="math-container">$A$</span> be any two point set in <span class="math-container">$\mathbb{R}^2$</span>. Then <span class="math-container">$r_0=0$</span> whenever <span class="math-container">$N \geq 2$</span>.</p>
|
2,693,143 | <p>For example, to convert $0.25$ to binary. Using this algorithm it gives the correct result $0.01$:</p>
<blockquote>
<ol>
<li>Multiply by two</li>
<li>take decimal as the digit</li>
<li>take the fraction as the starting point for the next step</li>
<li>repeat until you either get to 0 or a periodic number</li>
<li>read the number starting from the top - the first result is the first digit after the comma</li>
</ol>
</blockquote>
<p>Trying to understand why all this with the multiplying by two actually works, I do this: $$0.25=0.01=\frac{1}{2}(0+\frac{1}{2}(1+\frac{1}{2}*0))$$ So, the multiplying by two comes from dividing by $\frac{1}{2}$. But when I do the multiplying by two I cannot get the numbers in the brackets. This is what I get: $$0.25=0.5*\frac{1}{2}+0$$$$0.5=1*\frac{1}{2}+0$$$$1=2*\frac{1}{2}+0$$Can someone show me how to do that or point me if I make a mistake somewhere?</p>
| fleablood | 280,126 | <p>Cosider that $0 < x < 1$ and in base $10$, $x = 0.abcde....$ and in base $2$ $x = 0.uvwz......$ and let $2x = h.ijklmn.....$</p>
<p>Now $x \ge \frac 12 \iff a \ge 5\iff 2x \ge 1 \iff u = 1 \iff h = 1$.</p>
<p>And $x < \frac 12 \iff a < 5\iff 2x < 1 \iff u= 0 \iff h=0$.</p>
<p>So $u = h$ and that is why you can do this for the first decimal.</p>
<p>Now we get rid of the first digit $h$ (you weren't do doing that) and we let $y = 0.ijklm...$</p>
<p>Now if $x \ge \frac 12$ then $y = 0.ijklmn....2*(x -\frac 12)$. And if $x < \frac 12$ then $y = 0.ijklmn = 2*x$ and $0 \le y < 1$. Let $x' = x - \frac 12$ if $x \ge \frac 12$ or let $x' = x$ if $x < \frac 12$. Either way, $y = 2x'$.</p>
<p>And now let $2y = \alpha.\beta\gamma....$</p>
<p>Now $x' \ge \frac 14 \iff y \ge \frac 12 \iff 2y \ge 1 \iff i \ge 5 \iff v = 1\iff \alpha = 1$. And this is only if $\frac 14 \le x < \frac 12$ or $\frac 34 \le x < 1$ </p>
<p>And $x < \frac 14 \iff y < \frac 12 \iff 2y < 1 \iff i < 5 \iff v=0 \iff \alpha =0$. And this is only if $0 < x < \frac 14$ or $\frac 12 \le x < \frac 34$.</p>
<p>Either way $\alpha = v$ and that is why we can get the second digit.</p>
<p>Inductively we can carry that on forever (or until we hit $0$).</p>
<p>==== old=====</p>
<p>There are two things you are supposed to keep track of the binary decimal, which starts at $0.$ and the base $10$ decimal.</p>
<p>Watch:</p>
<p>A: $0. = 0$ B: $0.25$. </p>
<p>Multiply the B by 2:</p>
<p>A: $0. = 0$ B: $0.5 $ </p>
<p>Take the digit from B and put it on A: The digit is $0$.</p>
<p>A: $0.0 = 0$ B: $0.5$ (that doesn't look like much but the A has gotten one digit longer)</p>
<p>Take the fraction left in A as your new starting point of A. The fraction part is $0.5$.</p>
<p>A: $0.0 = 0$ B: $0.5$.</p>
<p>Multiple by 2:</p>
<p>A: $0.0 = 0$ B: $1.0$</p>
<p>Take the digit from B and put it on A: The digit is $1$.</p>
<p>A: $0.01 = \frac 14$ B: $0.0$ (But this time the $1$ goes from the $B$ to the $A$. And notice the B is reduced).</p>
<p>Take the fraction left in A as your new starting point of A. The fraction part is $.0$.</p>
<p>A: $0.01 = \frac 14$ B:$0.0$.</p>
<p>3)repeat until you either get to 0 or a periodic number.</p>
<p>We've reached $0$. We are done.</p>
<p>Let's do a more illuminating example:</p>
<p>A: $0.$ B: $0.375$</p>
<p>A: $0.$ B: $0.75$ (Double the B)</p>
<p>A: $0.0$ B: $0.75$ (Move the $0$ to the A)</p>
<p>A: $0.0$ B: $1.50$(Double the B).</p>
<p>A: $0.01$ B: $.50$(Move the $0$ to the A)</p>
<p>A: $0.01$ B: $1.0$(Double the B)</p>
<p>A: $0.011$ B: $.0$.(Move the $1$ to the A)</p>
<p>We've reached zero so we stop.</p>
<p>Basically with have $0.375 = 0.375$</p>
<p>$0.75 = 0.375*2 = 0.375*2 - 0$</p>
<p>$1.5 = 0.375*4 - 0*2$</p>
<p>$.5 = 0.375*4 - 0*2 - 1$</p>
<p>$1.0 = 0.375*8 - 0*4 - 1*2$</p>
<p>$.0 = 0.375*8-0*4 - 1*2 - 1$</p>
<p>So $0.375*8 = 0*4 + 1*2 + 1$</p>
<p>$0.375 = \frac {0*4 + 1*2+1}{8} = \frac 02 + \frac 14 + \frac 18$.</p>
<p>This works because way place a $1$ on the decimal if and only if what we have left is greater than $\frac 12$. But the important part is if it <em>is</em> we then remove that half and have double what is <em>left</em>.</p>
<p>"Trying to understand why all this with the multiplying by two actually works, I do this: "</p>
<p>$0.25=0.01=\frac{1}{2}(0+\frac{1}{2}(1+\frac{1}{2}*0))$</p>
<p>Nothing wrong with that.</p>
<p>$0.375 = 0.011 = \frac 12(0 +\frac 12(1 + \frac 12(1)))$</p>
|
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