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 |
|---|---|---|---|---|
279,238 | <p>I want to create a regular polygon from the initial two points <span class="math-container">$A$</span>, <span class="math-container">$B$</span> and number of vertices <span class="math-container">$n$</span>,<br />
<a href="https://i.stack.imgur.com/dKr9A.png" rel="noreferrer"><img src="https://i.stack.imgur.com/dKr9A.png" alt="enter image description here" /></a></p>
<p><code>regularPolygon[{0, 0}, {1, 0}, 3]</code> gives <code>{{0, 0}, {1, 0}, {1/2, Sqrt[3]/2}}</code></p>
<p><code>regularPolygon[{x1, y1}, {x2, y2}, 4]</code> gives <code>{{x1, y1}, {x2, y2}, {x2 + y1 - y2, -x1 + x2 + y2}, {x1 + y1 - y2, -x1 + x2 + y1}}</code></p>
<p>I found a related function <a href="http://reference.wolfram.com/language/ref/CirclePoints.html" rel="noreferrer">CirclePoints</a>, it seems not suitable. Is there a simple way to implement such a function? Maybe you can use iteration.</p>
| yode | 21,532 | <p>I will recommend <code>ResourceFunction["PolygonFromBase"]</code> here:</p>
<pre><code>base = {{0, 1}, {2, 3}};
Graphics[{Line[base], Table[{RandomColor[],
Line[ResourceFunction["PolygonFromBase"][base, i]]}, {i, 3, 10}]}]
</code></pre>
<p><a href="https://i.stack.imgur.com/3wSl2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3wSl2.png" alt="enter image description here" /></a></p>
|
4,112,011 | <p>Let <span class="math-container">$f:X\rightarrow Y$</span> be a map and let <span class="math-container">$T'$</span> be a topology on <span class="math-container">$Y$</span>.</p>
<p>Show that <span class="math-container">$$T=\{U\subseteq X \mid \exists V\in T' \text{ with }U=f^{-1}(V)\} $$</span> is a topology on <span class="math-container">$X$</span>.</p>
<p><span class="math-container">$T'$</span> is the coarsest topology on <span class="math-container">$X$</span> such that the map <span class="math-container">$f:(X, T) \rightarrow (Y, T') $</span> is continuous.</p>
<p><span class="math-container">$$$$</span></p>
<p>We have to show:</p>
<p>1.The set <span class="math-container">$X$</span> and the empty set are elements of <span class="math-container">$T$</span> .</p>
<ol start="2">
<li><p>Any union of elements of <span class="math-container">$T$</span> belongs to <span class="math-container">$T$</span>.</p>
</li>
<li><p>Any finite intersection of elements of <span class="math-container">$T$</span> belongs to <span class="math-container">$T$</span>.</p>
</li>
</ol>
<p><span class="math-container">$$$$</span></p>
<p>Toshow these axioms, Ihave done the following:</p>
<ol>
<li><p>It holds that <span class="math-container">$X\subseteq X$</span> and <span class="math-container">$f(X)\in T'$</span> where <span class="math-container">$T'$</span> is a topology, and so we get <span class="math-container">$X\in T$</span>.</p>
<p>We also have that <span class="math-container">$\emptyset\subseteq X$</span> and <span class="math-container">$f(\emptyset)\in T'$</span> where <span class="math-container">$T'$</span> is a topology, and so we get <span class="math-container">$\emptyset\in T$</span>.</p>
</li>
<li><p>We consider the union <span class="math-container">$O = \bigcup\limits_{\alpha \in I} U_{\alpha}$</span>.</p>
</li>
</ol>
<p>This set is in <span class="math-container">$T$</span> iff theimage under <span class="math-container">$f$</span> is in <span class="math-container">$T'$</span>.
So we consider <span class="math-container">$f \left( O \right)$</span>.</p>
<p>The union is well defined under images so we get <span class="math-container">\begin{equation*}f \left( O \right) = f \left( \bigcup\limits_{\alpha \in I} U_{\alpha} \right) = \bigcup\limits_{\alpha \in I} f \left( U_{\alpha} \right)\end{equation*}</span>
Since each <span class="math-container">$U_{\alpha}$</span> is in <span class="math-container">$T$</span>, the images <span class="math-container">$f\left( U_{\alpha} \right)$</span> are in <span class="math-container">$T'$</span>.</p>
<p>Since <span class="math-container">$T'$</span> is a topology, is the union again in <span class="math-container">$T'$</span> and so we get that <span class="math-container">$O$</span> in <span class="math-container">$T$</span>.</p>
<ol start="3">
<li>The respective argument of 2.for intersection.</li>
</ol>
<p>Is everything correct?</p>
| Henno Brandsma | 4,280 | <p><span class="math-container">$\emptyset = f^{-1}[\emptyset]$</span> and as <span class="math-container">$\emptyset \in T'$</span> we have by definition <span class="math-container">$\emptyset \in T$</span>. Same for <span class="math-container">$X=f^{-1}[Y]$</span> and <span class="math-container">$Y \in T'$</span>.</p>
<p>If <span class="math-container">$U,V \in T$</span> so <span class="math-container">$U = f^{-1}[U']$</span> for some <span class="math-container">$U' \in T'$</span> and <span class="math-container">$V = f^{-1}[V']$</span> for some <span class="math-container">$V' \in T'$</span>, we have <span class="math-container">$U \cap V = f^{-1}[U'] \cap f^{-1}[V']= f^{-1}[U' \cap V']$</span> and <span class="math-container">$U' \cap V' \in T'$</span>, as <span class="math-container">$T'$</span> is closed under finite intersections, so <span class="math-container">$U \cap V \in T$</span> as well.</p>
<p>If for all <span class="math-container">$i \in I, U_i \in T$</span> so that we have <span class="math-container">$U'_i \in T'$</span> such that <span class="math-container">$f^{-1}[U'_i]=U_i$</span>, then <span class="math-container">$\bigcup_i U'_i \in T'$</span> too and as <span class="math-container">$\bigcup_i U_i = \bigcup_i f^{-1}[U'_i] = f^{-1}[\bigcup_i U'_i]$</span> we get that <span class="math-container">$\bigcup_i U_i \in T$</span>, as required.</p>
<p>So you have to write everything as preimages from <span class="math-container">$T'$</span>, <strong>not</strong> check the image is in <span class="math-container">$T'$</span>, because that is how <span class="math-container">$T$</span> is defined.</p>
<p>If <span class="math-container">$U$</span> is any topology on <span class="math-container">$X$</span> that makes <span class="math-container">$f: (X,U) \to (Y,T')$</span> continuous then for any <span class="math-container">$O \in T'$</span> we must have <span class="math-container">$f^{-1}[O] \in U$</span>, but by definition, <span class="math-container">$f^{-1}[O]$</span> is the form of any set in <span class="math-container">$T$</span> so <span class="math-container">$T \subseteq U$</span>. This shows that <span class="math-container">$T$</span> is the smallest (coarsest) topology that makes <span class="math-container">$f$</span> continuous.</p>
|
3,486,480 | <p>Find all positive integers <span class="math-container">$n$</span> which are not perfect squares and such that the decimal representation of <span class="math-container">$\sqrt{n}$</span> has the following property: among the first six digits there are at least five equal.</p>
<p>For example <span class="math-container">$\sqrt{2020} = 44.944410108488\ldots$</span> has five <span class="math-container">$4$</span>s and one <span class="math-container">$9$</span> among its first six digits.</p>
| Oscar Lanzi | 248,217 | <p>Numbers consisting of all <span class="math-container">$1$</span> digits in base <span class="math-container">$10$</span> and even number of these digits will have square roots beginning with as many <span class="math-container">$3$</span>'s, thus <span class="math-container">$\sqrt{11}=3.the..., \sqrt{1111}=33.33..., \sqrt{111111}=333.333...$</span>.</p>
|
3,486,480 | <p>Find all positive integers <span class="math-container">$n$</span> which are not perfect squares and such that the decimal representation of <span class="math-container">$\sqrt{n}$</span> has the following property: among the first six digits there are at least five equal.</p>
<p>For example <span class="math-container">$\sqrt{2020} = 44.944410108488\ldots$</span> has five <span class="math-container">$4$</span>s and one <span class="math-container">$9$</span> among its first six digits.</p>
| Vepir | 318,073 | <p>Up to <span class="math-container">$n\le 10^6$</span>, there are <span class="math-container">$538$</span> non-squares whose square root has <span class="math-container">$5$</span> out of <span class="math-container">$6$</span> (decimal) digits equal:</p>
<p><a href="https://i.stack.imgur.com/K2z9R.png" rel="noreferrer"><img src="https://i.stack.imgur.com/K2z9R.png" alt="enter image description here"></a></p>
<p>There is a "staircase" pattern.</p>
<p>Here are all of the <span class="math-container">$538$</span> terms from above plot:</p>
<pre><code>2020, 6543, 7903, 8080, 9779, 9980, 9982, 9984, 9986, 9988, 9990, 9992, 9994, 9996, 9998, 9999, 12346, 12347, 12348, 12359, 12368, 12569, 13715, 14668, 25957, 29279, 31605, 35679, 40894, 44568, 48498, 49294, 49374, 49382, 49383, 49384, 49385, 49386, 49396, 49405, 49414, 49561, 49650, 50725, 51630, 52543, 59753, 63616, 71111, 74105, 85394, 92011, 98178, 109120, 109782, 110911, 110978, 111091, 111098, 111109, 111110, 111111, 111112, 111113, 111114, 111115, 111118, 111131, 111138, 111151, 111178, 111311, 111378, 111445, 111779, 113120, 113794, 115147, 117878, 118642, 126420, 132011, 134444, 139378, 142716, 151234, 154711, 163575, 169013, 171764, 178272, 180153, 187778, 188742, 193991, 194873, 195757, 196643, 197264, 197353, 197442, 197495, 197504, 197513, 197522, 197527, 197528, 197529, 197530, 197531, 197532, 197533, 197534, 197535, 197540, 197549, 197558, 197575, 197620, 197709, 197798, 197887, 197976, 198421, 199313, 200207, 202000, 206520, 207531, 215709, 217778, 225098, 228271, 234686, 239012, 244475, 250001, 255586, 261235, 265797, 272716, 276209, 284445, 286820, 296420, 297631, 303111, 304213, 304214, 305318, 306424, 307532, 308087, 308198, 308309, 308420, 308531, 308586, 308597, 308598, 308609, 308620, 308631, 308636, 308637, 308638, 308639, 308640, 308641, 308642, 308643, 308644, 308645, 308646, 308653, 308664, 308675, 308686, 308753, 308864, 308975, 309086, 309087, 309754, 310868, 311984, 313102, 319853, 321111, 331264, 333827, 342875, 346790, 354686, 359999, 360001, 368044, 373457, 380277, 380278, 387161, 392711, 401111, 405344, 415309, 418177, 418178, 429753, 431211, 436480, 437802, 437803, 439127, 440453, 441781, 441782, 443112, 443644, 443645, 443778, 443911, 444044, 444177, 444178, 444311, 444364, 444377, 444378, 444391, 444404, 444417, 444418, 444431, 444436, 444437, 444438, 444439, 444440, 444441, 444442, 444443, 444444, 444445, 444446, 444447, 444448, 444457, 444458, 444471, 444484, 444577, 444578, 444711, 444844, 445778, 445779, 447115, 448453, 457877, 457878, 459382, 459383, 471511, 474567, 474568, 485344, 489999, 490001, 500949, 505679, 505680, 515204, 515205, 521605, 521606, 529660, 537778, 544315, 544316, 554197, 554198, 559171, 570864, 574226, 574227, 587777, 587778, 589482, 589483, 594098, 595640, 595641, 597185, 598731, 598732, 600280, 601830, 601831, 603383, 603384, 603849, 603850, 604005, 604160, 604161, 604316, 604471, 604472, 604626, 604627, 604782, 604783, 604829, 604844, 604845, 604860, 604875, 604876, 604891, 604906, 604907, 604922, 604923, 604927, 604928, 604929, 604930, 604931, 604932, 604933, 604934, 604935, 604936, 604937, 604938, 604939, 604940, 604941, 604953, 604954, 604969, 605093, 605094, 605249, 606494, 606495, 608053, 620593, 620594, 622345, 636449, 639999, 640001, 654300, 654301, 657902, 670578, 670579, 676050, 687056, 694444, 694445, 703734, 713086, 713087, 720611, 720612, 731975, 731976, 737689, 737690, 751110, 751111, 754967, 754968, 770493, 770494, 772445, 775964, 775965, 777727, 777728, 779492, 781258, 781259, 783027, 783028, 784798, 784799, 786571, 786572, 788346, 788701, 788702, 788878, 788879, 789056, 789057, 789234, 789235, 789411, 789412, 789589, 789590, 789767, 789768, 789945, 789980, 789981, 789998, 789999, 790016, 790033, 790034, 790051, 790052, 790069, 790070, 790087, 790088, 790105, 790108, 790109, 790110, 790111, 790112, 790113, 790114, 790115, 790116, 790117, 790118, 790119, 790120, 790121, 790122, 790123, 790124, 790125, 790140, 790141, 790300, 790301, 791901, 791902, 808000, 808001, 809999, 810001, 828099, 830124, 830125, 846399, 850494, 850495, 864899, 871111, 871112, 883599, 891975, 891976, 902499, 913086, 913087, 921599, 934444, 934445, 940899, 956048, 956049, 960399, 977900, 977901, 980099, 982080, 984063, 986048, 988035, 990024, 992015, 994008, 996003, 998000, 998199, 998200, 998399, 998400, 998599, 998600, 998799, 998800, 998999, 999000, 999199, 999200, 999399, 999400, 999599, 999600, 999799, 999800, 999819, 999820, 999839, 999840, 999859, 999860, 999879, 999880, 999899, 999900, 999919, 999920, 999939, 999940, 999959, 999960, 999979, 999980, 999981, 999982, 999983, 999984, 999985, 999986, 999987, 999988, 999989, 999990, 999991, 999992, 999993, 999994, 999995, 999996, 999997, 999998, 999999
</code></pre>
<p>I'm not sure about a closed form for <em>all of them</em>. </p>
<p>There are <em>trivial</em> patterns such as <a href="https://math.stackexchange.com/a/3486526/318073">Oscar Lanzi</a> answered, but there are also <em>nontrivial</em> terms.</p>
|
301,038 | <p>If you want to show that a sequence $(a_{n})$ in $\mathbb{R}$ is convergent, when is it sufficient to show that there is a number $b\in\mathbb{R}$ such that
$$ \liminf a_{n} \geq b \geq \limsup a_{n}$$</p>
<p>In particular, I have a situation where my sequence is bounded and I wanted to use this approach, but I'm not sure I really understand what is going on and why or if this works.</p>
<p>Thanks for any illumination!</p>
| Ittay Weiss | 30,953 | <p>It is always sufficient to know such a $b$ exists. Since $\liminf a_n \le \limsup a_n$ always holds, if a $b$ as in the question exists then in particular holds that $\limsup a_n \le \liminf a_n$. But then the equality $\limsup a_n = \liminf a_n $ holds. Now, a sequence converges if, and only if, its limsup is equal to its liminf. So, the sequence converges. </p>
|
2,504,737 | <p>During one of my daily exercises, I was looking for properties of the elements of Cartan calculus. I stumbled on Wikipedia's page about interior products <a href="https://en.wikipedia.org/wiki/Interior_product" rel="nofollow noreferrer">(here)</a>, and I've noticed a property that sounds very useful:
$$
\iota_{[X,Y]}\omega=[\mathcal L_X,\iota_Y]\omega.
$$
In Wikipedia's notations, $\iota$ is the interior product, $\mathcal L_X$ is the Lie derivative with respect to the vector field $X$ (and $X$ and $Y$ are vector fields). $\omega$ is a differential form on a manifold $M$. As there is no source given, I'm trying to prove this equality, and failing due to a sign. Maybe I'm doing some stupid error somewhere.</p>
<p>My attempt so far follows.</p>
<p>First: I note that the operator on the right has the property
$$
[\mathcal L_X,\iota _Y](\omega\wedge\eta)=[\mathcal L_X,\iota_Y]\omega\wedge\eta+(-1)^k\omega\wedge[\mathcal L_X,\iota_Y]\eta,
$$
where $\omega$ is assumed to be a $k-$form, and $\eta$ is an arbitrary form. This is the same interaction with wedge product as the left hand side. It follows that I can decide the value of the right hand side locally, where I can expand any form as tensor product of the basis forms. Hence, if I prove that the equality holds for $0-$ and $1-$forms, I'm done.</p>
<p>I start with $0-$forms: I take a function $f$ from $M$ to the field, and compute left and right hand side. Well, "compute": the left hand side is the application of an inner product to a function, that is zero by definition, while on the right hand side I either have a contraction first, annulling $f$, or I have a Lie derivative acting first. The Lie derivative of a function is a function, so it follows that $\iota_Y\mathcal L f=0$, and I'm done for this case.</p>
<p>For $1-$forms: let $\alpha$ be such an $1-$form. The left hand side is
$$
\iota_{[X,Y]}\alpha=\alpha([X,Y]).
$$
I split the calculation of the right hand side in two, and use Cartan's formula $\mathcal L_X=d\iota_X+\iota_Xd$ whenever necessary. Lie derivatives are ugly, all hail Cartan.
$$
\mathcal L_X\iota_Y\alpha=\mathcal L_X(\alpha (Y))=X(\alpha(Y)),\\
\iota_Y\mathcal L_X\alpha=\iota_Yd\iota_X\alpha+\iota_Y\iota_Xd\alpha=Y(\alpha(X))+\underline{d\alpha(Y,X)}.
$$
Underlined for your convenience is the step in which it is most likely I've done an error, but I fail to see why. I state that $\iota_Y\iota_Xd\alpha=d\alpha(Y,X)$ as I am applying $X$ first, and $Y$ second, and $\iota$ places vectors at the beginning of a string. Is it correct? Was I stupid here?</p>
<p>Continuing: I use
$$
d\alpha(Y,X)=Y(\alpha(X))-X(\alpha(Y))-\alpha([Y,X]).
$$
Here comes the failure. I'd expect stuff to cancel out, but that's not happening. The signs of $Y(\alpha(X))$ agree, so they do not cancel.</p>
<p>Where am I doing wrong? I strongly suspect that it is something in the underlined passage, but I need clarification about what I did wrong.</p>
<p>Thanks all in advance for your time.</p>
<p>(p.s.: in some other places of this site there is a proof of that relying on a property of $\mathcal L_X$ acting on differential forms. As I said, I despise $\mathcal L_X$. I'd prefer to see the error in this proof, as it is short and nice. Lie has done many wonderful things, and an orrible derivative)</p>
| Anthony Carapetis | 28,513 | <p>Be careful - the interior product places vectors in the first slot, but repeated applications should be read right-to-left, as function composition always is. That is, by definition we have $(i_X d \alpha)(Y) = d \alpha(X,Y);$ so applying $i_Y$ to the 1-form $i_X d \alpha$ yields this same result.</p>
|
4,435,497 | <p>I have to prove the following:</p>
<p>Let <span class="math-container">$p$</span> be a prime and denote <span class="math-container">$\mathbb{Z}_p=\mathbb{Z}/p\mathbb{Z}$</span>. Show that <span class="math-container">$\mathbb{Z}_p[x]/(x^2+x+1)$</span> is not a field if and only if <span class="math-container">$p=3$</span> or <span class="math-container">$p \equiv 1 \pmod 3$</span>.</p>
<p>First I consider showing that <span class="math-container">$x^2+x+1$</span> is reducible, We may write <span class="math-container">$x^2+x+1 \equiv 0$</span> which implies that <span class="math-container">$x^2 \equiv -x -1$</span> which implies <span class="math-container">$x^2 \equiv -x-1 \equiv 2x+2 \pmod 3$</span>. Thus, finally showing that <span class="math-container">$x^2+x+1 \equiv x^2-2x-2$</span>. Now, <span class="math-container">$x^2-2x-2$</span> <strong>is</strong> reducible. Thus our original polynomial can be written as a reducible polynomial, thus we do not have a field?</p>
<p>Any help would be greatly appreciated.</p>
| David Lui | 445,002 | <p>By the quadratic formula, <span class="math-container">$x^2 + x + 1$</span> has roots <span class="math-container">$\frac{-1 \pm \sqrt{-3}}{2}$</span>.</p>
<p>Therefore, we consider when <span class="math-container">$-3$</span> has a square root mod <span class="math-container">$p$</span>. If it does, then <span class="math-container">$x^2+x+1$</span> is not prime and <span class="math-container">$\mathbb{F}_p [x] / (x^2 + x + 1)$</span> is not a field. Otherwise, <span class="math-container">$x^2+x+1$</span> is irreducible and <span class="math-container">$\mathbb{F}_p [x] / (x^2 + x + 1)$</span> is a field.</p>
<p>If <span class="math-container">$ p =3$</span> then <span class="math-container">$-3=0$</span> which has a square root <span class="math-container">$0$</span>. For <span class="math-container">$p = 2$</span>, we can check that neither <span class="math-container">$0$</span> nor <span class="math-container">$1$</span> is a root, so <span class="math-container">$x^2+x+1$</span> is irreducible.</p>
<p>Otherwise, use the law of quadratic reciprocity. Let <span class="math-container">$p \geq 5$</span> be a prime number. We have <span class="math-container">$\left( \frac{3}{p} \right) = \left( \frac{-1}{p} \right) \left( \frac{-3}{p} \right)$</span> and <span class="math-container">$\left( \frac{p}{3} \right) \left( \frac{3}{p} \right) = (-1)^{\frac{p-1}{2}}$</span>. Rearranging, we get <span class="math-container">$\left( \frac{-3}{p} \right) = \left( \frac{p}{3} \right)\left( \frac{-1}{p} \right)(-1)^{\frac{p-1}{2}}$</span>.</p>
<p><span class="math-container">$\left( \frac{-3}{p} \right)$</span> is what we are trying to find. By Euler's criterion, <span class="math-container">$\left( \frac{-1}{p} \right)$</span> is <span class="math-container">$1$</span> if <span class="math-container">$p = 1 \mod 4$</span> and <span class="math-container">$-1$</span> if <span class="math-container">$p = 3 \mod 4$</span>.</p>
<p><span class="math-container">$(-1)^{\frac{p-1}{2}}$</span> is <span class="math-container">$1$</span> if <span class="math-container">$p = 1 \mod 4$</span> and <span class="math-container">$-1$</span> if <span class="math-container">$p = 3 \mod 4$</span>. Therefore <span class="math-container">$\left( \frac{-1}{p} \right)$</span> and <span class="math-container">$(-1)^{\frac{p-1}{2}}$</span> cancel out, so <span class="math-container">$\left( \frac{p}{3} \right) = \left( \frac{-3}{p} \right)$</span></p>
<p>By periodicity of the Legendre symbol, <span class="math-container">$\left( \frac{p}{3} \right)$</span> is <span class="math-container">$1$</span> if <span class="math-container">$p = 1 \mod 3$</span>, and <span class="math-container">$-1$</span> if <span class="math-container">$p = 2 \mod 3$</span>.</p>
|
281,386 | <p>I have asked this question on stackexchange and have not received any answers or comments after 2 days of it being there.</p>
<p>I read somewhere that the following statement is correct. A proof or any hint as to how to prove it would be helpful.</p>
<p>Let $G$ be a connected reductive group defined over $k$(maynot be of characteristic 0). Let $H$ be a connected normal subgroup of $G\times \text{spec}\overline{k}$(apriori $H$ is only defined over $\overline{k}$). Then $H$ is defined over a finite 'separable' extension of $k$.</p>
<p>Also there must be counterexamples where this is not true.</p>
<p>Thanks.</p>
| anon | 114,662 | <p>After passing to a finite separable extension of the base field, we may suppose that the reductive group is split. After passing to a finite covering, we may suppose that it is the product of a simply connected semisimple group $G$ and a torus. The torus presents no problem. After passing to a finite separable extension, we may suppose that G is a product of absolutely almost-simple normal subgroups, so it all comes down to looking at the subgroups of the centre of $G$, but this presents no problem.</p>
<p>Added: All the statements used can be found, for example, in this
<a href="https://www.cambridge.org/core/books/algebraic-groups/48D122BAEC63C2B70304E20F34482C8F" rel="nofollow noreferrer">book</a></p>
|
362,944 | <blockquote>
<p>Consider a Brownian bridge <span class="math-container">$B: [0,1]\to \mathbb{R}$</span> with <span class="math-container">$B(0)=B(1)=0$</span>. Let <span class="math-container">$M[0, 1/2]=\max_{x\in[0,1/2]}B(x)$</span>. How to prove that
<span class="math-container">$$\mathbb{P}(M[0, 1/2]\geq s)\leq 2\mathbb{P}(B(1/2)\geq s/2)?$$</span></p>
</blockquote>
<p>Actually, here is a "no big max" argument that could be used in the proof of construction Airy line ensemble. The "no big max" means the top curve between <span class="math-container">$(a, b)$</span> cannot get too high. </p>
<p>(Definition of Brownian bridge) If <span class="math-container">$\{B(t): t\geq 0\}$</span> is standard Brownian motion, then <span class="math-container">$\{Z(t): 0\leq t\leq 1\}$</span> is a Brownian bridge process when <span class="math-container">$$Z(t)=B(t)-tB(1).$$</span></p>
| Iosif Pinelis | 36,721 | <p>Let <span class="math-container">$B_t:=B(t)$</span>. For <span class="math-container">$t\in[0,1]$</span>, we can write
<span class="math-container">$$B_t=W_t-tW_1,$$</span>
where <span class="math-container">$W$</span> is a standard Wiener process. We have to show that
<span class="math-container">$$P(M_{1/2}\ge s)\le2P(B_{1/2}\ge s/2)$$</span>
for <span class="math-container">$s\ge0$</span>, where <span class="math-container">$M_{1/2}:=\max_{0\le t\le1/2}B_t$</span>. The Brownian bridge <span class="math-container">$B$</span> is independent of <span class="math-container">$W_1$</span>. Therefore and by the reflection principle,
<span class="math-container">$$\tfrac12\,P(M_{1/2}\ge s)=P(M_{1/2}\ge s,W_1\ge0)
=P(\max_{0\le t\le1/2}(W_t-tW_1)\ge s,W_1\ge0)
\le P(\max_{0\le t\le1/2}W_t\ge s)
=2P(W_{1/2}\ge s)=2P(W_1\ge s\sqrt2)=2P(B_{1/2}\ge s/\sqrt2);$$</span>
here we also used the fact that <span class="math-container">$B_{1/2}$</span> equals <span class="math-container">$\tfrac12\,W_1$</span> in distribution.
So,
<span class="math-container">$$P(M_{1/2}\ge s)\le4P(B_{1/2}\ge s/\sqrt2).$$</span>
The upper bound <span class="math-container">$4P(B_{1/2}\ge s/\sqrt2)$</span> on <span class="math-container">$P(M_{1/2}\ge s)$</span> is better than your desired upper bound <span class="math-container">$2P(B_{1/2}\ge s/2)$</span> for all <span class="math-container">$s\ge0.992$</span>. </p>
|
1,037,632 | <p>How to find the last 2 digits of $2014^{2001}$? What about the last 2 digits of $9^{(9^{16})}$?</p>
| Krishnamurari | 195,422 | <p>By Euler's Theorem</p>
<p>$2014^{\phi(25)}=2014^{20}\equiv 1\pmod{25}\implies2014^{2000}\equiv 1\pmod{25}\implies2014^{2001}\equiv 2014\pmod{25}\implies2014^{2001}\equiv 14\pmod{25}$</p>
<p>Obviously $2014^{2001}\equiv 0\pmod{4}$</p>
<p>Let $2014^{2001}=25n+14\implies25n+14\equiv 0\pmod{4}\implies n\equiv 2\pmod{4}$</p>
<p>$2014^{2001}=25n+14=25(4k+2)+14=64+100k$</p>
|
1,761,273 | <p>Good morning. I have a problem with this:</p>
<p>Find the maximum and minimum distances from the origin to the curve* <span class="math-container">$$g\left(x,y\right)=5x^{2}+6xy+5y^{2}$$</span></p>
<p>I have done this:</p>
<p>Function to optimize:<span class="math-container">$f\left(x,y\right)=x^{2}+y^{2}$</span></p>
<p>Restriction: <span class="math-container">$g\left(x,y\right)=5x^{2}+6xy+5y^{2}=8
$</span></p>
<p>Applying Lagrange multipliers:
<span class="math-container">$\nabla f\left(x,y\right)=\lambda\nabla g\left(x,y\right)$</span></p>
<p>Then,
<span class="math-container">$\nabla f\left(x,y\right)=2x\hat{i}+2y\hat{j}$</span> and <span class="math-container">$\lambda\nabla g(x,y)=\lambda(2x+\frac{6}{5}y)\hat{i}+\lambda\left(2y+\frac{6}{5}x\right)\hat{j}
$</span></p>
<p>Making the ecuation system:</p>
<p><span class="math-container">$\begin{cases}
2x=(2x+\frac{6}{5}y)\lambda\\
2y=(2y+\frac{6}{5}x)\lambda\\
x^{2}+\frac{6}{5}xy+y^{2}=8
\end{cases}$</span></p>
<p>But I have serious problem solving the system. Any suggestions?</p>
| Rob | 350,976 | <p>Have you tried a software like Mathematica? There are four solutions to your system: $x=y=Sqrt[5/2]$ and the negative of that root (both with $λ=5/8$), and $x=-y=Sqrt[10]$ and the negative of that root (both with $λ=5/2$).</p>
|
1,831,250 | <p>I have come to conclusion that the most efficient and thorough way to prove whether or not a limit exists in three dimensions is to use polar coordinates. </p>
<p>$lim_{x,y \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$</p>
<p>Substituting for polar coordinates:</p>
<p>$lim_{r \to 0^+}$ $\frac{r^3(cos^3 \Theta + sin^3 \Theta)}{r^2(sin^2 \Theta + cos^2 \Theta)}$ </p>
<p>$lim_{r \to 0^+} r (cos^3 \Theta + sin^3 \theta)$</p>
<p>As we can see r is independent of Theta and therefore the limit exists and is in fact 0. However showing that it is independent using notation is a little bit of a grey area. </p>
<p>$\vert sin^3 \Theta + cos^3 \Theta \vert$ $\le$ $1$ (Since sin and cos are bound by 1). Then $\vert r sin^3 \Theta + cos^3 \Theta \vert$ $\le \vert r \vert$ $r \to 0$. Is this logical? To be honest I am not even sure what this is saying, if cos and sin are bound by one, how does this have anything to do with r? Also if I had $2x^3$ in the numerator how would that change this? I am not even sure how polar coordinates would be substituted with the 2 I would assume the constant would be pulled outside.</p>
| Mathemagician1234 | 7,012 | <p>Well,first of all, you really mean in 2 dimensions since the limit is defined by the domain,which lies in $\mathbb R^2$. The surface of the graph of the function lies in $\mathbb R^3$,but that's not relevant for the limit</p>
<p>Second, while r in this case <strong>is</strong> independent of $\theta$, I'm really not sure how that matters in the limit.$\vert sin^3 \Theta + cos^3 \Theta \vert$ $\le$ $1$ is correct regardless of whether or not the equations are separable and <strong>that's</strong> what matters in this particular limit. </p>
<p>What makes polar coordinates so useful in this particular example is that the equation is <strong>separable</strong> in polar coordinates and there's no clear way to do this in Cartesian coordinates. (Interestingly, if you did have $2x^3$ in the numerator, the 2 would in fact come out once you converted the coordinates. That's not in general true.) </p>
<p>An interesting historical note: The physicist Edwin Shroedinger, when he was devising his famous partial differential equation in $\mathbb R^3$ to describe quantum mechanical systems, was struggling with rearranging the equation. The domain and experimental data of the equation seemed to strongly suggest the equation was separable in it's variables, but he couldn't get it to separate no matter what he tried. In desperation,he turned to his friend, mathematician Hermann Weyl,and asked him to take a look at it. </p>
<p>He soon got a letter back from Weyl telling him the equation <strong>was</strong> separable, but only in polar coordinates! </p>
|
4,271,166 | <p>I have been studying probability and there are many results regarding the sum of random variables.</p>
<p>For example the sum of iid bernouli random variables is a binomial distributed
The sum of Geometric random variables is Negative binomial distributed
Sum of exponential random variables is Gamma distributed. Also what are the other important results like this?</p>
<p>Where can I find a list of such properties that would help me deal with the subject better?.
I am having to constantly search through my book for such references. If someone can provide me a a source for this I will be grateful.</p>
<p>This is not a question but rather a request for a good reading material. So please try and understand and not downvote. Thanks</p>
| G Cab | 317,234 | <p>A list of interrelation among distributions based on sum (like the examples you cited) would be much interesting. However I never came across a paper reporting it.<br />
Nor I saw any list reporting together pdf and characteristic function. That could be of some help, since the CF of the sum of random variables is the product of the single CF's.</p>
|
152,115 | <p>I would like to open the file "myfile.cdf" or "myfile.nb". One or the other will be in the directory but not both. Because I don't know which, I'm trying to use <code>StringExpression</code> to be general about opening the file. Here is the code below. It does't seem to work. Ideas?</p>
<pre><code>NotebookOpen[NotebookDirectory[]<>"myfile." ~~ __]
</code></pre>
| Chris Degnen | 363 | <pre><code>NotebookOpen[First@FileNames[NotebookDirectory[] <> "myfile.*"]]
</code></pre>
|
152,115 | <p>I would like to open the file "myfile.cdf" or "myfile.nb". One or the other will be in the directory but not both. Because I don't know which, I'm trying to use <code>StringExpression</code> to be general about opening the file. Here is the code below. It does't seem to work. Ideas?</p>
<pre><code>NotebookOpen[NotebookDirectory[]<>"myfile." ~~ __]
</code></pre>
| Itai Seggev | 4,848 | <p>StringExpression, like all patterns, are template which can be matched against. They don't automagically expand to match eveything--you do some function to do that. I'd write your code like this</p>
<pre><code>NotebookOpen[First[FileNames[NotebookDirectory[] <> "test." <> # & /@ {"cdf", "pdf"}]]]
</code></pre>
<p>This will give you a list of the 0, 1 or 2, files which match the two names, and no others. Of course, it will error if there are no matches (which you'll need to handle), but this also allows you to control which of the two files to prefer if both are present, just be reordering the list of extensions.</p>
<hr>
<p>Expansion: Since the first thing I wrote had so many typos, here's a bonus answer based on the same idea, which is probably how I'd write this. The code to find the files is the same, but rather than naively taking <code>First</code>, I use <code>Replace</code>. If the result is a list of one or more file names, I open the first one (which will return a <code>NotebookObject</code> pointing to what was just opened). Otherwise, I return <code>$Failed</code>. The code following the assignment can check <code>result</code> and decide what to do...</p>
<pre><code>result = Replace[
FileNames[NotebookDirectory[] <> "myfile." <> # & /@ {"cdf", "nb"}],
{
{file_String, ___} :> NotebookOpen[file],
_ -> $Failed
}
]
</code></pre>
|
152,115 | <p>I would like to open the file "myfile.cdf" or "myfile.nb". One or the other will be in the directory but not both. Because I don't know which, I'm trying to use <code>StringExpression</code> to be general about opening the file. Here is the code below. It does't seem to work. Ideas?</p>
<pre><code>NotebookOpen[NotebookDirectory[]<>"myfile." ~~ __]
</code></pre>
| m_goldberg | 3,066 | <p>This seems to be reasonably robust. You can even use wildcard * in the name string, but in that case you will open the file which is 1st match in the list returned by <code>FileNames</code>.</p>
<pre><code>openNBorCDF[name_String] :=
Module[{pathNames, file},
pathNames =
FileNames[{name <> ".nb", name <> ".cdf"}, NotebookDirectory[]];
file = First[pathNames, Return[$Failed]];
NotebookOpen[file];
file]
</code></pre>
<p>On my system, evaluating</p>
<pre><code>openNBorCDF["test"]
</code></pre>
<p>in a notebook loaded from the Desktop directory opens <code>test.cdf</code> and returns</p>
<blockquote>
<p><code>"/Users/oldmg/Desktop/test.cdf"</code></p>
</blockquote>
<p>while evaluating</p>
<pre><code>openNBorCDF["foo"]
</code></pre>
<p>opens nothing and returns</p>
<blockquote>
<p><code>$Failed</code></p>
</blockquote>
<p>because neither <code>foo.nb</code> nor <code>foo.cdf</code> exists in my Desktop directory.</p>
<h3>Edit</h3>
<p>The above code requires V10.3 or later. Those with earlier versions of Mathematica should try this:</p>
<pre><code>openNBorCDF[name_String] :=
Module[{pathNames, file},
pathNames =
FileNames[{name <> ".nb", name <> ".cdf"}, NotebookDirectory[]];
If[pathNames === {}, Return[$Failed], file = pathNames[[1]]];
NotebookOpen[file];
file]
</code></pre>
|
1,540,635 | <p>Given $$\min\limits_x \|x\|_2^2$$ $$\text{s.t.} Ax = b$$ <strong>show $x^* = A^T(AA^T)^{-1}b$ where $A \in \mathbb{R}^{m \times n}, m < n$</strong></p>
<p>This is projection $x$ onto the hyperplane $Ax - b = 0$</p>
<p>Multiplying $A^T$ on both sides</p>
<p>$A^TAx - A^Tb = 0$</p>
<p>So we have $\langle A^TAx - A^Tb, x \rangle = 0$</p>
<p>$(A^TAx - A^Tb)^Tx = 0$</p>
<p>$(x^TA^TA - b^TA)x = 0$</p>
<p>$x^TA^TAx = b^TAx = x^TA^Tb$</p>
<p>So $A^TAx = A^Tb$ (circular) </p>
<p>$x = (A^TA)^{-1}A^Tb$ (Wrong)</p>
<p>How can you do this correctly?</p>
| Omran Kouba | 140,450 | <p>You need the hypothesis that $A$ has rank $m$ so that $AA^T$ is invertible.</p>
<p>Since $x^*=A^T(AA^T)^{-1}b$ we have $Ax^*=b$, so for every $x$ such that $Ax=b$ we have $x-x^*\in {\rm ker}A$ and clearly $x^*\in {\rm Im}A^T$, but it is well-known that ${\rm Im}A^T=({\rm ker}A)^\bot$. Thus, by Pythagoras theorem
$$\Vert x\Vert^2=\Vert x-x^*\Vert^2+\Vert x^*\Vert^2\ge \Vert x^*\Vert^2$$
with equality if and only if $x=x^*$.</p>
|
4,148,930 | <p>Let's consider the function <span class="math-container">$\log x$</span>; how can I prove that it is a transcendental function on the function field of rational functions, i.e. that a polynomial in two variables <span class="math-container">$p(x,y)$</span> such that <span class="math-container">$p(x,\log x)=0$</span> identically does not exist?</p>
<p>I have been trying different approaches: seeing the logarithm on the real numbers, like a formal series or like a holomorphic function on an open in the complex plane, but I was not able to do this. I have also tried looking for the differences with <span class="math-container">$\sqrt{(1+x)}$</span> which is algebraic on the rational function, and can be defined on a subset of the reals, with a formal series or on an open subset of the complex plane.</p>
<p>Could you please help me?</p>
| Conrad | 298,272 | <p>A direct way to do this is to assume there is such a <span class="math-container">$p(x,\log x)=0, x\in \mathbb R_+$</span>
Since <span class="math-container">$\log$</span> takes infinitely many values, <span class="math-container">$p$</span> cannot be a polynomial in <span class="math-container">$\log x$</span> only, so let <span class="math-container">$n \ge 1$</span> the highest power of <span class="math-container">$x$</span> with non zero coefficients and write the equation as:</p>
<p><span class="math-container">$a_n(\log x)x^n+a_{n-1}(\log x)x^{n-1}+..a_0(\log x)=0$</span>, where <span class="math-container">$a_0,..a_n$</span> are polynomials in one variable.</p>
<p>Dividing by <span class="math-container">$x^n$</span> and noting that <span class="math-container">$\frac{a_k(\log x)}{x^{n-k}} \to 0, x \to \infty$</span> for all <span class="math-container">$0 \le k \le n-1$</span>, we get that <span class="math-container">$a_n(\log x) \to 0, x \to \infty$</span>.</p>
<p>But now if <span class="math-container">$a_n$</span> is non constant, obviously <span class="math-container">$a_n(\log x) \to \pm \infty, x \to \infty$</span> depending on the sign of its leading coefficient and that is a contradiction. Hence <span class="math-container">$a_n(\log x)$</span> is constant and then it must be zero, contradicting the original assumption that the coefficient of <span class="math-container">$x^n$</span> in <span class="math-container">$p$</span> is non-zero, so we are done!</p>
|
2,111,841 | <p>A geodesic is a line representing the shortest route between two points on a sphere, for example on the Earth treated here as a perfect sphere.
Two points on Earth having the same latitude can be also connected with the line being a part of a circle for selected constant latitude.
Differences between these two lines can be visualized with the use of
<a href="https://academo.org/demos/geodesics/" rel="nofollow noreferrer"> <strong>this Academo program</strong></a> presenting the situation in the context of the map of Earth.</p>
<p><strong>Question:</strong></p>
<ul>
<li>How to calculate <strong>the area</strong> between these two lines?</li>
</ul>
<p><em>(Assume for example that the starting point is $(\alpha, \beta_1)=(45^\circ, -120^\circ)$ and the destination $(\alpha, \beta_2)=(45^\circ, 0^\circ))$ - the arc of constant latitude $45^\circ$ has length $120^\circ$</em>.</p>
<p><a href="https://en.wikipedia.org/wiki/Spherical_trigonometry" rel="nofollow noreferrer"><strong>In wikipedia</strong></a> <em>a formula for an area of a spherical polygon is presented, but the polygon is limited in this case with parts of geodesics. Is it possible somehow transform these formulas of spherical geometry into the case of finding the area between geodesic and arc of constant latitude?</em></p>
| Larry B. | 364,722 | <p>Assume the sphere is of radius $1$, and we have our constant lattitude $\phi$ for two coordinates $(\phi, -\lambda_0), (\phi, \lambda_0) \in (-\pi/2,\pi/2)\times [-\pi,\pi]$. Note that the longitudes are positive and negative of a constant value. We can do this because we choose to measure the longitude angle $\lambda$ from the coordinates' midpoint. This will be useful for our great circle parametrization.</p>
<p>The great circle latitude function, $\gamma$, is tougher to parametrize by $\lambda$. First we should prove a property about this great circle: that the longitudes of its intersections with the equator are $\pi/2$ away from the midpoint of the two coordinates' longitudes. We can use <a href="https://en.wikipedia.org/wiki/Congruence_(geometry)#Congruent_triangles_on_a_sphere" rel="nofollow noreferrer">Side-Angle-Side Congruence</a> of spherical triangles using the great circle, equator, and midpoint longitude lines. [[Something about latitude lines having equal angle from great circle intersection to the longitude midpoint... but we can't use similar triangle arguments...]]</p>
<hr>
<p>To aid us in our parametrization, we'll keep using that type of spherical triangle. Keep the equator and great circle, and have a great circle (longitude line) for each longitude. Let the arc of the longitude be $a$, the arc of the equator be $b$, and the arc of the great circle be $c$. Corresponding angles are $A$ between great circle and equator, $B$ between great circle and longitude line, and $C$ between equator and longitude line. Note, this means $C = \pi/2$.</p>
<p><a href="https://en.wikipedia.org/wiki/Spherical_trigonometry#Napier.27s_rules_for_right_spherical_triangles" rel="nofollow noreferrer">With a little help from Napier</a>, we get $\tan(a) = \tan(A) \cdot \sin(b)$. We can do some translations into the language of our problem: $A$ is the constant angle between the great circle and the equator, $a$ is the latitude from the equator, and $b$ is the longitude measured from the intersection point of the great circle and the equator. We want to find $a$ in terms of our parametrized $\lambda$.</p>
<p>We know that $\lambda = b - \pi/2$, because we measure $\lambda$ from the longitudinal midpoint. We can also derive the constant $\tan(A)$ using known quantities in the great circle, our coordinates: $$\tan(A) = \frac{\tan(\phi)}{\sin(\pi/2 - \lambda_0)} = \frac{\tan(\phi)}{\sin(\pi/2 + \lambda_0)} = \frac{\tan(\phi)}{\cos(\lambda_0)}$$</p>
<p>Therefore, the latitude function $\gamma(\lambda)$ of the great circle is: $$\gamma(\lambda) = \tan^{-1}\left[ \tan(A) \cdot \sin(\pi/2 + \lambda) \right] = \tan^{-1}\left[ \frac{\tan(\phi)}{\cos(\lambda_0)} \cdot \cos(\lambda) \right] $$</p>
<hr>
<p>The constant-latitude function can be parametrized by longitude $\lambda$ as: $$\kappa(\lambda) = \phi$$</p>
<p>The solution can gotten by integrating the difference between latitudes, that integral being across longitude. In general:</p>
<p>$$\int_{-\lambda_0}^{\lambda_0} \gamma - \kappa d\lambda= \int_{-\lambda_0}^{\lambda_0} \tan^{-1}\left[ \frac{\tan(\phi)}{\cos(\lambda_0)} \cdot \cos(\lambda) \right] d\lambda - 2\phi\lambda_0 = -\cot(A)\int \frac{\tan^{-1}(v)}{\sin\left(\cot(A)\cdot v\right)} dv - 2\phi\lambda_0$$</p>
<p>I've checked around, and I can't seem to find an analytical solution for the integral $\int \tan^{-1}(x)/\sin(k\cdot x) dx$. For your example, the numerical solution is <a href="http://www.wolframalpha.com/input/?i=integral+from+-pi%2F3+to+pi%2F3++arctan+(cos(x)++*+tan(pi%2F4)%2Fcos(pi%2F3))" rel="nofollow noreferrer">$2.12618$</a>) $-\pi/6 \approx 1.60258$</p>
|
3,173,125 | <p>Let <span class="math-container">$f(x) = x^3 + a x^2 + b x + c$</span> and <span class="math-container">$g(x) = x^3 + b x^2 + c x + a\,$</span> where <span class="math-container">$a, b, c$</span> are integers and <span class="math-container">$c\neq 0\,$</span>. Suppose that the following conditions hold:</p>
<ol>
<li><span class="math-container">$f(1)=0$</span> </li>
<li>The roots of <span class="math-container">$g(x)$</span> are squares of the roots of <span class="math-container">$f(x)$</span>.</li>
</ol>
<p>I'd like to find <span class="math-container">$a, b$</span> and <span class="math-container">$c$</span>.</p>
<p>I tried solving equations made using condition 1. and relation between the roots, but couldn't solve.
The equation which I got in <span class="math-container">$c$</span> is <span class="math-container">$c^4 + c^2 +3 c-1=0$</span> (edit: eqn is wrong).
Also I was able to express <span class="math-container">$a$</span> and <span class="math-container">$b$</span> in terms of <span class="math-container">$c$</span>. But the equation isn't solvable by hand. </p>
| Servaes | 30,382 | <p>Let <span class="math-container">$u$</span>, <span class="math-container">$v$</span> and <span class="math-container">$w$</span> be the roots of <span class="math-container">$f$</span>, so that <span class="math-container">$u^2$</span>, <span class="math-container">$v^2$</span> and <span class="math-container">$w^2$</span> are the roots of <span class="math-container">$g$</span>. Then comparing the coefficients of
<span class="math-container">$$(x-u)(x-v)(x-w)=f(x)=x^3+ax^2+bx+c,$$</span>
<span class="math-container">$$(x-u^2)(x-v^2)(x-w^2)=g(x)=x^3+bx^2+cx+a,$$</span>
yields the equations
<span class="math-container">\begin{eqnarray*}
a&=&-u-v-w&=&-u^2v^2w^2,\\
b&=&uv+uw+vw&=&-u^2-v^2-w^2,\\
c&=&-uvw&=&u^2v^2+u^2w^2+v^2w^2.
\end{eqnarray*}</span>
This immediately shows that <span class="math-container">$a=-c^2$</span>, and the identities
<span class="math-container">\begin{eqnarray*}
u^2+v^2+w^2&=&(u+v+w)^2-2(uv+uw+vw),\\
u^2v^2+u^2w^2+v^2w^2&=&uvw(u+v+w)-(uv+uw+vw)^2,
\end{eqnarray*}</span>
show that <span class="math-container">$-b=a^2-2b$</span> and <span class="math-container">$c=ac-b^2$</span>, respectively, hence <span class="math-container">$b=a^2=c^4$</span> and so
<span class="math-container">$$f(x)=x^3-c^2x^2+c^4x+c,$$</span>
for some <span class="math-container">$c$</span>. Then <span class="math-container">$f(1)=1$</span> implies that
<span class="math-container">$$c^4-c^2+c+1=0,$$</span>
which has the clear root <span class="math-container">$c=-1$</span>. Then <span class="math-container">$a=-1$</span> and <span class="math-container">$b=1$</span>.</p>
|
1,614,111 | <p>Is there a sequence of positive real numbers $x_1,\ldots,x_n$ for which
$$
\sum_{1\leq i,j\leq n}\left[\frac{\sqrt{ij}}{2}-\min(i,j)\right]x_ix_j> 0?
$$</p>
| River Li | 584,414 | <p>Alternative solution:</p>
<p>There exists such <span class="math-container">$n$</span> and a sequence of positive real numbers <span class="math-container">$x_1, x_2, \cdots, x_n$</span>.</p>
<p>Let <span class="math-container">$A$</span> denote the matrix whose <span class="math-container">$(i,j)$</span>-th entry is <span class="math-container">$\frac{\sqrt{ij}}{2}$</span>.
Then, <span class="math-container">$A = uu^\mathsf{T}$</span> where
<span class="math-container">$u = \frac{1}{\sqrt{2}}[\sqrt{1}, \sqrt{2}, \cdots, \sqrt{n}]^\mathsf{T}$</span>.</p>
<p>Let <span class="math-container">$B$</span> denote the matrix whose <span class="math-container">$(i,j)$</span>-th entry is <span class="math-container">$\min(i, j)$</span>.
Note that <span class="math-container">$B^{-1}$</span> is a symmetric tridiagonal matrix of the form
<span class="math-container">\begin{align}
B^{-1} = \begin{pmatrix} 2 & -1 & 0 & \ldots & 0 & 0 \\ -1 & 2 & -1 & \ldots & 0 & 0 \\ 0 & -1 & 2 & \ldots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \ldots & 2 & -1 \\ 0 & 0 & 0 & \ldots & -1 & 1 \end{pmatrix}.
\end{align}</span>
(Note: The diagonal entries are <span class="math-container">$2, 2, \cdots, 2, 1$</span>. The subdiagonal and superdiagonal entries are all <span class="math-container">$-1$</span>. )</p>
<p>Let <span class="math-container">$v = B^{-1}u$</span>.</p>
<p><em>Fact 1</em>: <span class="math-container">$v_i > 0$</span> for <span class="math-container">$i=1, 2, \cdots, n$</span>.</p>
<p><em>Fact 2</em>: <span class="math-container">$u^\mathsf{T}B^{-1}u = \frac{n^2}{2} - \sum_{k=1}^{n-1} \sqrt{k(k+1)}$</span>.</p>
<p><em>Fact 3</em>: <span class="math-container">$v^\mathsf{T}(A - B)v = (u^\mathsf{T}B^{-1}u)^2 - u^\mathsf{T}B^{-1}u$</span>.</p>
<p>The proofs are easy and thus omitted.</p>
<p>Now, from Facts 1, 2, 3, if <span class="math-container">$u^\mathsf{T}B^{-1}u > 1$</span> or equivalently <span class="math-container">$\frac{n^2}{2} - \sum_{k=1}^{n-1} \sqrt{k(k+1)} > 1$</span>, then <span class="math-container">$v^\mathsf{T}(A - B)v > 0$</span>.
It is easy to prove that there exists <span class="math-container">$n$</span> such that <span class="math-container">$\frac{n^2}{2} - \sum_{k=1}^{n-1} \sqrt{k(k+1)} > 1$</span>.
Indeed, by using
<span class="math-container">$\sqrt{k(k+1)} \le k + \frac{1}{2} - \frac{1}{16k}$</span> for <span class="math-container">$k \ge 1$</span>, we have
<span class="math-container">$$\sum_{k=1}^{n-1} \sqrt{k(k+1)} \le \frac{n^2}{2} - \frac{1}{2} - \frac{1}{16}\sum_{k=1}^{n-1} \frac{1}{k}$$</span>
which results in
<span class="math-container">$$\frac{n^2}{2} - \sum_{k=1}^{n-1} \sqrt{k(k+1)} \ge \frac{1}{2} + \frac{1}{16}\sum_{k=1}^{n-1} \frac{1}{k}.$$</span>
The desired result follows.</p>
<p>For such <span class="math-container">$n$</span> and <span class="math-container">$v$</span>, we have
<span class="math-container">$$\sum_{1\le i, j\le n} \left[\frac{\sqrt{ij}}{2} - \min(i, j)\right] v_iv_j = v^\mathsf{T}(A - B)v > 0.$$</span></p>
<p>Remark: Denote <span class="math-container">$F(n) = \frac{n^2}{2} - \sum_{k=1}^{n-1} \sqrt{k(k+1)}$</span>. By Maple, <span class="math-container">$F(55) = 0.99974443 < 1$</span> and <span class="math-container">$F(56) = 1.00199623$</span>.</p>
|
1,125,891 | <p>Let $U=<\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0\end{pmatrix}, \begin{pmatrix}0 & 0 & -1 \\ 0 & 1 & 0\end{pmatrix}, \begin{pmatrix}2 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}>$ and $V= <\begin{pmatrix}2& 0 & 0 \\ 0 & 2 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix}$>. </p>
<p>Let $f:U \to V$ such that </p>
<p>$$f \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0\end{pmatrix}= \begin{pmatrix} 2 & 1 & 1 \\ 0 & 3 & 1\end{pmatrix}$$</p>
<p>$$f \begin{pmatrix}0 & 0 & -1 \\ 0 & 1 & 0\end{pmatrix} = \begin{pmatrix} 4 & 2 & 2 \\ 0 & 6 & 2\end{pmatrix}$$</p>
<p>$$f \begin{pmatrix}2 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 1/2 & 1/2 \\ 0 & 3/2 & 1/2\end{pmatrix}.$$ </p>
<p>I have to determine $ker(f)$ and $Im(f)$. Normally, if we work with $R^n$ and the standard basis, I've no problems, but this setting with matrices and non-common vector spaces confuses me: please, could you show me how to work out this problem?</p>
<p>Thank you very much in advance.</p>
| obataku | 54,050 | <p>Since $u_i$ generate $U$ it follows every element of $U$ can be written in the form $\sum c_iu_i$. As Bernard noted, $f$ acts such that $u_1\mapsto v_1+v_2,u_2\mapsto2(v_1+v_2),u_3\mapsto\frac12(v_1+v_2)$. It follows that $\operatorname{im}(f)$ is generated by $v_1+v_2$ alone.</p>
<p>Now, since we know the action of $f$ on these generators of $U$ in terms of the generators $v_i$ of $V$ we can represent $f$ as a matrix: $$\begin{bmatrix}1&2&1/2\\1&2&1/2\end{bmatrix}$$</p>
<p>Just find two independent $(x,y,z)$ such that $x+2y+\frac12z=0$; the easiest trivial examples are $(0,1/2,-2)$ and $(-2,1,0)$, which therefore span $\ker(f)$.</p>
|
204,087 | <p>What would be an isomorphism between
$\mathcal{F}(S;V / U)$ and $\mathcal{F}(S;V)/ \mathcal{F}(S;U)$, where
$S$ is a set, $V$ a vector space and $U$ a subspace of $V$. $\mathcal{F}(A,B)$ denotes the set of functions from $A$ to $B$ and $V/U$ is the quotient space. </p>
| Jair Taylor | 28,545 | <p>There is a nice formula for the partial sum of a geometric series:</p>
<p>$\sum_{i=0}^{n-1}{x^i} = \frac{1 - x^n}{1-x}$</p>
<p>So your expression will simplify if you let $\lambda_i = c^i$ for some choice of $c$. Since you want the $\lambda_i$'s to be increasing, you could choose $\lambda_i = 2^i$. Then you get:</p>
<p>$$ a^n (\sum_{i=0}^{n-1} \lambda_i \cdot b^i ) +(\sum_{i=0}^{n-1} \lambda_i \cdot a^i)\cdot b^n + \lambda_n\cdot a^n \cdot b^n$$</p>
<p>$$= a^n (\sum_{i=0}^{n-1} 2^i \cdot b^i ) +(\sum_{i=0}^{n-1} 2^i \cdot a^i)\cdot b^n + 2^n\cdot a^n \cdot b^n$$
$$=a^n\frac{ 1-(2b)^n}{1-2b} +b^n\frac{ 1-(2a)^n}{1-2a} + (2ab)^n.$$</p>
<p>That seems relatively simple to me - at least, it's closed form.</p>
|
184,575 | <p>I have a large jagged list, that is each sub-list has a different length. I would like to <code>Flatten</code> this list for <code>Histogram</code> purposes, but it seems to be taking an inordinate amount of time and memory</p>
<pre><code>jaggedList=Table[RandomReal[1,RandomSample[Range[400000,800000],1]],{n,100}];
</code></pre>
<p>Just to illustrate, length of each of elements of the main list</p>
<pre><code>ListPlot[Length/@jaggedList]
</code></pre>
<p><a href="https://i.stack.imgur.com/Io5a0.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Io5a0.png" alt="list lengths"></a></p>
<p>Full Flatten takes a long time, my real data is several times larger, it gets painfully slow</p>
<pre><code>fullFlatten=Flatten@jaggedList;//AbsoluteTiming
{10.0055,Null}
</code></pre>
<p>I noticed flattening non-jagged sub-lists is not a problem</p>
<pre><code>partialFlatten=Flatten/@jaggedList;//AbsoluteTiming
{0.289219,Null}
</code></pre>
<p>Memory usage is huge on the final result of the full list, even though number of elements is the same:</p>
<pre><code>ByteCount/@{fullFlatten,partialFlatten,jaggedList}
{1460378864,486808224,486808224}
</code></pre>
<p>Would super appreciate any tips on what I can change to make this faster / more memory compact !</p>
| kglr | 125 | <p><code>Apply</code>ing <code>Join</code> is much faster than <code>Flatten</code>:</p>
<pre><code>SeedRandom[1]
jaggedList = Table[RandomReal[1, RandomSample[Range[400000, 800000], 1]], {n, 100}];
fullFlatten = Flatten@jaggedList; // AbsoluteTiming // First
</code></pre>
<blockquote>
<p>8.2375848</p>
</blockquote>
<pre><code>fullFlatten2 = Join @@ jaggedList; // AbsoluteTiming // First
</code></pre>
<blockquote>
<p>0.29729</p>
</blockquote>
<pre><code>fullFlatten2 == fullFlatten
</code></pre>
<blockquote>
<p>True</p>
</blockquote>
<pre><code>ByteCount /@ {fullFlatten, fullFlatten2, jaggedList}
</code></pre>
<blockquote>
<p>{1462957016, 487652456, 487667608}</p>
</blockquote>
|
2,337,357 | <p><a href="https://i.stack.imgur.com/iWuCv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iWuCv.png" alt="enter image description here"></a></p>
<p>The derivative of $\frac1x$ is $\frac{-1}{x^2}$.</p>
<p>How do I find the $c$ if there is no zero in the derivative of the function?</p>
<p>I started with $-1/x^2= -0.0625$ but I'm confused from here on.</p>
| User8128 | 307,205 | <p>You're very close. Once the first one is rolled, there is a $(1-1/20)$ chance it is not replicated by the second die. Once the second has also been rolled, assuming it did not replicate the first, there is a $(1-2/20)$ that the third die replicates neither of the original two. Thus the probability is $$1 - (1-1/20)(1-2/20).$$ This gives an exact result of $29/200 = 0.145$.</p>
|
2,337,357 | <p><a href="https://i.stack.imgur.com/iWuCv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iWuCv.png" alt="enter image description here"></a></p>
<p>The derivative of $\frac1x$ is $\frac{-1}{x^2}$.</p>
<p>How do I find the $c$ if there is no zero in the derivative of the function?</p>
<p>I started with $-1/x^2= -0.0625$ but I'm confused from here on.</p>
| Sri-Amirthan Theivendran | 302,692 | <p>Consider the complement namely all three dice are distinct. There are $20\times19\times18$ possible such outcomes out of $20^3$ possible outcomes. Hence the probability is
$$
1-\frac{20\times19\times18}{20^3}=1-\frac{20}{20}\times\frac{19}{20}\times\frac{18}{20}.
$$</p>
|
2,607,333 | <p>Let $L$ be the splitting field of $x^3+2x+1$. I want to know how 59 splits in $L$. I calculated the discriminant of $\mathbb{Z}[\alpha]$ to be $-59$, (where $\alpha$ is a root of the polynomial), which is squarefree therefore $\mathcal{O}_K = \mathbb{Z}$ (because $d(\mathbb{Z}[\alpha]) = (\mathcal{O}_K:\mathbb{Z}[\alpha])^2d_K$). Since 59 divides the discriminant, it must be ramified. But I don't know how to get anything more than that, like what's the ramification index and how many primes does it split into?</p>
| Eric Wofsey | 86,856 | <p>I assume you want to require $C_i\neq \emptyset$ for all $i$ since otherwise this is trivial. It suffices to show that if $A$ is any topological space with more than $n$ components then it can be written as a union of $n+1$ disjoint nonempty clopen subsets (since then those subsets will be separated in $X$). The statement is trivial if $n=0$ so let us suppose $n>0$ and $A$ has at least $n+1$ components. In particular, $A$ is disconnected, so we can write $A=B\cup C$ where $B$ and $C$ are nonempty and clopen.</p>
<p>Each component of $A$ must be contained in either $B$ or $C$. Say $B$ has $b$ components and $C$ has $c$ components; then $b+c\geq n+1$. Since $B$ and $C$ are nonempty, $b>0$ and $c>0$. We can choose $b_0$ and $c_0$ such that $0<b_0\leq b$, $0<c_0\leq c$, and $b_0+c_0=n+1$. In particular, then $b_0<n+1$ and $c_0<n+1$, $B$ has more than $b_0-1$ components, and $C$ has more than $c_0-1$ components. By induction on $n$, then, we can write $B$ as a union of $b_0$ disjoint nonempty clopen sets and we can write $C$ as a union of $c_0$ disjoint nonempty clopen sets. Taking these sets together, we now have written $A$ as a union of $b_0+c_0=n+1$ disjoint nonempty clopen sets, as desired.</p>
|
299,140 | <p>Is there a closed form sum of </p>
<p>$\sum_{k=0}^{\infty} \frac{x^k}{(k!)^2}$</p>
<p>It is trivial to show that it is less than $e^x$ but is there a tighter bound?</p>
<p>Thanks</p>
| Iosif Pinelis | 36,721 | <p>$\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}
\newcommand{\ep}{\epsilon}
\newcommand{\ga}{\gamma}
\newcommand{\Ga}{\Gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\renewcommand{\th}{\theta}
\newcommand{\R}{\mathbb{R}}
\newcommand{\F}{\mathcal{F}}
\newcommand{\E}{\operatorname{\mathsf E}}
\newcommand{\PP}{\operatorname{\mathsf P}}
\newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}$</p>
<p>This answer is based on ideas quite different from those used in my previous answer to this question, and the result is much better. As in that answer, let
\begin{equation*}
S(x):=\sum_{k=0}^{\infty} \frac{x^k}{(k!)^2}.
\end{equation*}
Term-wise differentiation shows that
\begin{equation*}
(xS'(x))'=S(x).
\end{equation*}
This differential equation can be rewritten as
\begin{equation*}
16x^2a''(x)+8x(1+4\sqrt x\,\tanh(2\sqrt x))a'(x)+a(x)=0, \tag{1}
\end{equation*}
where
\begin{equation*}
a(x):=S(x)/S_*(x),
\end{equation*}
\begin{equation*}
S_*(x):=\cosh\sqrt{4x}\big/\sqrt{\pi\sqrt x}\sim e^{\sqrt{4x}}\big/\sqrt{4\pi\sqrt x}\sim S(x)
\end{equation*}
for large $x$, as was noted by Carlo Beenaker. </p>
<p>Note that $a(x)>0$ for $x>0$. So, by (1), if $a'(x)=0$ for some $x>0$, then $a''(x)<0$. So, the only local extrema of $a$ are local maxima; therefore and because between any two local maxima there is a local minimum, we see that there is at most one local maximum of $a$ on $(0,\infty)$. Since $a(0+)=0$, $a(1)>1$, and $a(\infty-)=1$, we conclude that $a$ has precisely one local (and hence global) maximum on $(0,\infty)$. </p>
<p>In fact, this maximum occurs at $x=x_*=0.7277\dots$, and $a(x_*)=1.0769\ldots<1.08$. Moreover, $a>1$ on $[0.2,\infty)$. Thus,
\begin{equation*}
S_*<S<1.08\, S_*\quad\text{on }[0.2,\infty).
\end{equation*}
These facts are illustrated by this graph of the function $a$: </p>
<p><a href="https://i.stack.imgur.com/VHUz7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VHUz7.png" alt="enter image description here"></a></p>
<hr>
<p>To completely complete this answer, let us verify the mentioned asymptotic equivalence
\begin{equation*}
S(x)\sim\cosh\sqrt{4x}\big/\sqrt{\pi\sqrt x}; \tag{2}
\end{equation*}
here and in what follows, $x\to\infty$.
Let $k_1,k_2,q_1,q_2$ be natural numbers such that
\begin{align*}
&k_1=x^{4/8}-\th x^{3/8},\quad q_1=x^{4/8}-\th x^{3/8}/2,\\
&k_2=x^{4/8}+\th x^{3/8},\quad q_2=x^{4/8}+\th x^{3/8}/2;
\end{align*}
here and elsewhere $\th$ denotes various expressions depending on $x$ (possibly different even in the same formula) such that $\th\to1$.<br>
Write
\begin{equation*}
S(x)=S_1+S_2+S_3,\quad \cosh\sqrt{4x}=T_1+T_2+T_3,
\end{equation*}
where
\begin{equation*}
S_1:=\sum_{k=0}^{k_1-1} \frac{x^k}{(k!)^2},\quad S_2:=\sum_{k=k_1}^{k_2-1} \frac{x^k}{(k!)^2},\quad
S_3:=\sum_{k=k_2}^\infty \frac{x^k}{(k!)^2},
\end{equation*}
\begin{equation*}
T_1:=\sum_{k=0}^{k_1-1} \frac{(4x)^k}{(2k)!},\quad T_2:=\sum_{k=k_1}^{k_2-1} \frac{(4x)^k}{(2k)!},\quad
T_3:=\sum_{k=k_2}^\infty \frac{(4x)^k}{(2k)!},
\end{equation*}</p>
<p>Asymptotic equivalence (2) is an immediate consequence of the following two lemmas. </p>
<blockquote>
<p><strong>Lemma 1.</strong> $S_1+S_3<<S_2$ and $T_1+T_3<<T_2$; here and elsewhere $A<<B$ means $A/B\to0$. </p>
<p><strong>Lemma 2.</strong> $S_2\sim T_2\big/\sqrt{\pi\sqrt x}$. </p>
</blockquote>
<p>It remains to prove the lemmas. </p>
<p><em>Proof of Lemma 1.</em> Let
\begin{equation*}
M:=M(x):=\max_{k\ge0}b_k,\quad b_k:=\frac{x^k}{(k!)^2},\quad r_k:=\frac{b_k}{b_{k-1}}=\frac{x}{k^2}.
\end{equation*}
Then $r_k$ is decreasing in $k$, $r_k\ge1$ for $k\le x^{4/8}$ and hence $b_k$ is nondecreasing in $k\le x^{4/8}$, $r_k\le1$ for $k\ge x^{4/8}$ and hence $b_k$ is nonincreasing in $k\ge x^{4/8}$. So, $M=b_{k_*}$ for some natural $k_*=k_*(x)=x^{4/8}+O(1)\in[k_1,k_2-1]$, and so,
\begin{equation*}
M\le \sum_{k=k_1}^{k_2-1}b_k=S_2.
\end{equation*}
Next, for $k\ge k_2$,
\begin{equation*}
b_k=b_{q_2}\prod_{j=q_2+1}^k r_j\le M r_{q_2}^{k-q_2}.
\end{equation*}
Also,
\begin{equation*}
r_{q_2}=\frac x{(x^{4/8}+\th x^{3/8}/2)^2}=(1+\th x^{-1/8}/2)^{-2}=1-\th x^{-1/8}
\end{equation*}
and $k_2-q_2=\th x^{3/8}/2$.
So,
\begin{multline*}
S_3=\sum_{k=k_2}^\infty b_k\le M r_{q_2}^{k_2-q_2}\,\frac1{1-r_{q_2}}
\sim M (1-\th x^{-1/8})^{\th x^{3/8}/2}\,x^{1/8} \\
=M \exp\{-\th x^{2/8}/2\}\,x^{1/8}<<M\le S_2. \tag{3}
\end{multline*}
Further,
\begin{equation*}
r_{k_1}=\frac x{(x^{4/8}-\th x^{3/8})^2}=(1-\th x^{-1/8})^{-2}=1+2\th x^{-1/8}
\end{equation*}
and $q_1-k_1=\th x^{3/8}/2$, whence
\begin{multline*}
b_{k_1}=b_{q_1}\Big/\prod_{j=k_1+1}^{q_1}r_j\le M/r_{k_1}^{q_1-k_1}
=M/ (1+2\th x^{-1/8})^{\th x^{3/8}/2}
=M /\exp\{\th x^{2/8}\}
\end{multline*}
and
\begin{equation*}
S_1=\sum_{k=0}^{k_1-1}b_k\le k_1 b_{k_1}\le x^{4/8}\,M /\exp\{\th x^{2/8}\}<<M\le S_2. \tag{4}
\end{equation*}
By (3) and (4), $S_1+S_3<<S_2$. That $T_1+T_3<<T_2$ is verified quite similarly; here instead of $r_k=\frac{x}{k^2}$, one will have to use $\frac{4x}{2k(2k-1)}=\frac{x}{k(k-1/2)}$. This completes the proof of Lemma 1. \qed </p>
<p><em>Proof of Lemma 2.</em> For $k\in[k_1,k_2-1]$, we obviously have $k\sim\sqrt x$. So, using Stirling's formula, it is easy to see that for natural $k\in[k_1,k_2-1]$<br>
\begin{equation*}
\frac{x^k}{(k!)^2}\Big/\frac{(4x)^k}{(2k)!}=\frac{(2k)!}{(k!)^2}\frac1{4^k}\sim\frac1{\sqrt{\pi k}}
\sim\frac1{\sqrt{\pi\sqrt x}}.
\end{equation*}
Now Lemma 2 immediately follows. \qed</p>
|
229,748 | <p>I have an integral given by <code>intS[zs,c]</code> and I want to plot the expression <code>functionS</code> (which contains <code>intS[zs,c]</code>) with respect to another integral function <code>inta[zs,c]</code> and <code>zs</code>. However, the values of <code>inta[zs,c]</code> are extremely small compared to <code>functionS</code> so in the <code>ParametricPlot3D</code> I got a badly looking plot that looks smashed.</p>
<p>My code is as follows,</p>
<pre><code>d = 4;(*dimensions*)
inta[zs_?NumericQ, c_?NumericQ] := NIntegrate[((c zs^(d + 1))/(2 d)) (y^((1 - d)/(2 d)))/(1 - c^2 zs^(2 d) y)^(1/2), {y, 0, 1}]
intS[zs_?NumericQ, c_?NumericQ] := NIntegrate[((c^2 zs^(2 d))/(d - 1)) (y^d)/(1 - c^2 zs^(2 d) y^(2 d))^(1/2), {y, 0, 1}]
functionS = ((-(1 - c^2 zs^(2 d))^(1/2))/(d - 1) - intS[zs, c] + 1/zs^(d - 1));
function = Log[10, functionS];
ParametricPlot3D[{zs, inta[zs, c], function}, {zs, 0, 10}, {c, 0, 10}, PlotStyle -> {LightBlue}, PlotRange -> Full, LabelStyle -> Directive[Bold, Medium], ImageSize -> Large] // Quiet
</code></pre>
<p><a href="https://i.stack.imgur.com/hUP27.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hUP27.jpg" alt="Image" /></a></p>
| cvgmt | 72,111 | <p>We can speed up the code by use <code>Newton-Leibniz Formula</code> instead of <code>NIntegrate</code> since all the functions have symbole expression say <code>Hypergeometric2F1</code></p>
<pre><code>Clear["`*"];
d = 4;(*dimensions*)
inta[zs_, c_] =
Subtract @@ (Integrate[((c zs^(d + 1))/(2 d)) (y^((1 -
d)/(2 d)))/(1 - c^2 zs^(2 d) y)^(1/2), y] /. y -> {1, 0});
intS[zs_, c_] =
Subtract @@ (Integrate[((c^2 zs^(2 d))/(d - 1)) (y^
d)/(1 - c^2 zs^(2 d) y^(2 d))^(1/2), y] /. y -> {1, 0});
functionS = ((-(1 - c^2 zs^(2 d))^(1/2))/(d - 1) - intS[zs, c] +
1/zs^(d - 1));
function = Log[10, functionS];
ParametricPlot3D[{zs, inta[zs, c], function}, {zs, 0, 10}, {c, 0, 10},
PlotStyle -> {LightBlue}, PlotRange -> Full,
LabelStyle -> Directive[Bold, Medium], ImageSize -> Large,
PlotPoints -> 100, BoxRatios -> 1]
</code></pre>
|
660,034 | <p>I wondered if all decimal expansions of $\frac{1}{n}$ could be thought of in such a way, but clearly for $n=6$,</p>
<p>$$.12+.0024+.000048+.00000096+.0000000192+...\neq.1\bar{6}$$</p>
<p>Why does it work for 7 but not 6? Is there only one such number per base, <em>i.e.</em> 7 in base 10? If so what is the general formula?</p>
| Henry | 6,460 | <p>The expansion in your title formula is effectively $\displaystyle \sum_1^\infty 7 \times \left(\frac{2}{100}\right)^n = 7 \times \frac {1}{49} = \frac {1}{7}$</p>
<p>Simpler similar ones might be $\displaystyle \sum_1^\infty 2 \times \left(\frac{2}{10}\right)^n = 2 \times \frac {1}{4} = \frac {1}{2}$, i.e. $0.4 + 0.08 +0.016 + \cdots = 0.5$ or $\displaystyle \sum_1^\infty 3 \times \left(\frac{1}{10}\right)^n = 3 \times \frac {1}{9} = \frac {1}{3}$, i.e. $0.3 + 0.03 +0.003 + \cdots = 0.333\ldots$ or $\displaystyle \sum_1^\infty 1 \times \left(\frac{5}{10}\right)^n = 1 \times \frac {1}{1} = 1$, i.e $0.5 + 0.25 +0.125 + \cdots = 1$.</p>
<p>To get $\frac {1}{6}$, a possibility, though not so pretty, is $\displaystyle \sum_1^\infty 4 \times \left(\frac{4}{100}\right)^n$ i.e. $0.16 + 0.0064 + 0.000256 + \cdots = 0.1666\ldots$.</p>
<p>One approach to getting $\frac {1}{k}$ is to look for a multiple of $k$ (say $mk$) which is one less than a number whose prime factors are $2$ or $5$. Then $mk+1$ will divide some power of $10$ (say $10^h =g (mk+1)$). You will then be able to write $\displaystyle \sum_1^\infty m \times \left(\dfrac{g}{10^h}\right)^n = m \times \frac {1}{mk} = \frac {1}{k}$. In your original example $k=7, m=7, g=2, h=2$, but apart from having $k=m$ as in my next three examples, there is nothing particularly special about it. </p>
|
660,034 | <p>I wondered if all decimal expansions of $\frac{1}{n}$ could be thought of in such a way, but clearly for $n=6$,</p>
<p>$$.12+.0024+.000048+.00000096+.0000000192+...\neq.1\bar{6}$$</p>
<p>Why does it work for 7 but not 6? Is there only one such number per base, <em>i.e.</em> 7 in base 10? If so what is the general formula?</p>
| lab bhattacharjee | 33,337 | <p>Please rectify me as I could not find following expression explicitly</p>
<p>$$\frac17=\frac{14}{98}=\frac{14}{100\left(1-\frac2{100}\right)}=\frac{14}{100}\left(1-0.02\right)^{-1}$$</p>
<p>$$=0.14[1+0.02+(0.02)^2+(0.02)^3+\cdots]$$</p>
|
199,842 | <p>I understand the reasoning behind $\pi r^2$ for a circle area however I'd like to know what is wrong with the reasoning below:</p>
<p>The area of a square is like a line, the height (one dimension, length) placed several times next to each other up to the square all the way until the square length thus we have height x length for the area.</p>
<p>The area of a circle could be thought of a line (The radius) placed next to each other several times enough to make up a circle. Given that circumference of a circle is $2 \pi r$ we would, by the same reasoning as above, have $2 \pi r^2$. Where is the problem with this reasoning?</p>
<p>Lines placed next to each other would only go straight like a rectangle so you'd have to spread them apart in one of the ends to be able to make up a circle so I believe the problem is there somewhere. Could anybody explain the issue in the reasoning above?</p>
| KeithS | 23,565 | <p>The problem with that reasoning is that when showing a rectangle's area by dividing the rectangle's length and height into units, the length and height units form squares. Rotating the radius around the circle by a set number of degrees to divide the circle does not produce squares; these slices are more like triangles. Recall that the area of a triangle is $\dfrac{bh}{2}$; the math's a bit more complex but you can draw a parallel here to the area of a circle versus a rectangle. All other things being equal, the rectangle is double the triangle's area, and so the area of a shape divided into a number of congruent triangles will be half the area of a shape divided into the same number of rectangles of the same length and height as the triangles.</p>
<p>The actual math to prove the area of a circle is very closely related to this, but incorporates an additional concept from calculus:</p>
<p>For any arbitrary $n$, draw an $n$-gon (hexagon, octagon, hectogon, etc) around a circle of radius $r$. Each side of this shape will have length $s$, and $s*n > 2\pi r$; the perimeter of the n-gon will be greater than the circumference of the circle (recall that $\pi = \dfrac{c}{d}, d=2r \therefore c=2\pi r$). However, as $n$ increases, $s$ decreases, and the perimeter of the n-gon will approach the circumference of the circle. It never quite gets there for any finite <em>n</em>, but it gets close enough, allowing us to define what's known in calculus as a limit: $\lim_{n\to \infty}ns = 2\pi r$.</p>
<p>Now, for each side of the n-gon, we can define an isoceles triangle between the vertices of the side and the center of the circle. The symmetrical sides of this triangle have length $l$ which is $>r$ (because the line connects to the vertex of the n-gon, outside the circle) but, similar to the way $ns$ approaches $2\pi r$, $l$ approaches $r$ as $n\to \infty$. This isoceles triangle of base $s$ and height $r$ can be split into two right triangles with base $s/2$ and height $r$. The area of a right triangle is $\dfrac{bh}{2}$ as previously stated, and so the area of the isoceles triangle is two of these, or $2*\dfrac{\dfrac{s}{2}r}{2}= \dfrac{sr}{2}$. There are $n$ of these triangles, one per side, so the area of the n-gon is $\dfrac{nsr}{2}$. Finally recall our limit; as $n\to\infty$, $ns\to 2\pi r$. The limit allows us to equate these two in the general case we are considering, where for our purposes $n=\infty$; $ns=2\pi r$. Plug that into the area of the n-gon, and behold: $\dfrac{nsr}{2} = \dfrac{(2\pi r)r}{2} = \dfrac{2\pi r^2}{2} = \pi r^2$.</p>
<p>QED.</p>
|
199,842 | <p>I understand the reasoning behind $\pi r^2$ for a circle area however I'd like to know what is wrong with the reasoning below:</p>
<p>The area of a square is like a line, the height (one dimension, length) placed several times next to each other up to the square all the way until the square length thus we have height x length for the area.</p>
<p>The area of a circle could be thought of a line (The radius) placed next to each other several times enough to make up a circle. Given that circumference of a circle is $2 \pi r$ we would, by the same reasoning as above, have $2 \pi r^2$. Where is the problem with this reasoning?</p>
<p>Lines placed next to each other would only go straight like a rectangle so you'd have to spread them apart in one of the ends to be able to make up a circle so I believe the problem is there somewhere. Could anybody explain the issue in the reasoning above?</p>
| cloudfeet | 45,136 | <p>OK - so the heart of your intuition is linking the area with the length of line required to fill in that area. It's like shading in a picture using a ballpoint pen.</p>
<p>Take your example with your square: at the end, when you've drawn all your lines, you've shaded the entire area. So if you know how much ink you've used from your pen, you know the area - you've shaded the area more or less evenly, so the amount of ink you've used is proportional to the area.</p>
<h3>Uneven shading:</h3>
<p>However when you draw the lines for your circle, you go over the bits in the middle quite a few times. In fact the very centre of the circle gets done a ridiculous amount, because every line you draw starts or ends there. However, at the edges of the circle, your lines are much more spaced out.</p>
<p>The intuition you're using relies on the idea that you're shading in the area evenly. If you're repeatedly scribbling over one area (the centre), then using the length of your lines does not give you an accurate estimate of the shape's area.</p>
<h3>How to fix it:</h3>
<p>So to get a similar sort of intuition for the circle, you need to sketch across the area evenly. Doing that with straight lines is more complicated, geometrically. Instead, try thinking of drawing concentric circles, from the inside out.</p>
<p>Now - imagine that every time you draw a circle, you draw a straight line on another piece of paper that's the same length. By the time you've drawn all your circles, your other piece of paper should have a triangle shaded on it - and you can even use the formula for the area of a triangle to get <span class="math-container">$\pi r^2$</span>.</p>
|
134,815 | <p>Assume I have a shuffled deck of cards (52 cards, all normal, no jokers) I'd like to record the order in my computer in such a way that the <em>ordering</em> requires the least bits (I'm not counting look up tables ect as part of the deal, just the ordering itself. </p>
<p>For example, I could record a set of strings in memory:</p>
<p>"eight of clubs", "nine of dimonds" </p>
<p>but that's obviously silly, more sensibly I could give each card an (unsigned) integer and just record that... </p>
<p>17, 9, 51, 33... </p>
<p>which is much better (and I think would be around 6 bits per number times 53 numbers so around 318 bits), but probably still not ideal.. for a start I wouldn't have to record the last card, taking me to 312 bits, and if I know that the penultimate card is one of two choices then I could drop to 306 bits plus one bit that was true if the last card was the highest value of the two remaining cards and false otherwise.... </p>
<p>I could do some other flips and tricks, but I also suspect that this is a branch of maths were there is an elegant answer... </p>
| carlop | 29,324 | <p>There are $52!$ possible ordering of the 52 cards. Each bit can store 2 value, so you need to find the smallest $n$ for which $2^n\geq52!$, $n\geq log_2(52!)$, that is $226$.</p>
<p>The algorithm for rebuild the ordering from the number is the one suggested by Ross Millikan.</p>
|
662,637 | <p>I have a quick question</p>
<p>If S is an orthonormal set of vectors in Rn, then S is a basis for the subspace it spans. I am not sure if this is true or not based off the definition of orthonormal basis which is a subspace w of Rn is a basis for w that is also an orthogonal set.</p>
| Berci | 41,488 | <blockquote>
<p>If $S$ is an orthonormal set of vectors in $\Bbb R^n$, then, <strong>yes</strong>, $S$ is a basis for the subspace it spans.</p>
</blockquote>
<p>Being a basis in the spanned subspace is equivalent to being <em>linearly independent</em>. So this claim means exactly that $S$ is then linearly independent. <br>
(Suppose $\sum_i\lambda_ie_i=0\ $ with some $\,e_i\in S$, then apply scalar multiplication with each $e_j$ to conclude $\lambda_j=0$.)</p>
<blockquote>
<p><strong>Def.</strong> $S$ is an orthonormal basis for a (sub-)space $W$, if $S$ spans the whole $W$ and length of each element of $S$ is $1$, and any $e,f\in S,\ e\ne f$ satisfies $e\perp f$.</p>
</blockquote>
|
1,105,787 | <p>I was given an excercise in my calculus class that i don't really understand, the problem says : Find the area limited by the curves $$ y = \frac{x+4}{x^2+1} ,\space x = -2 ,\space x = 3,\space y = 0 $$</p>
<p>I don't really know what approach to follow here, my guess would be to solve it using riemann sums or maybe definite integrals and using $ x = -2 $ and $ x = 3 $ as the interval but i'm totally lost.</p>
| Zubin Mukerjee | 111,946 | <p>Let $u = x^2+1$. The derivative of $u$ with respect to $x$ is $2x$. If $A$ is the area we want, then</p>
<p>\begin{align}
A= \int_{-2}^{3} \frac{x+4}{x^2+1}\,\mathrm{d}x&=\frac{1}{2}\left(\int_{-2}^{3} \frac{2x}{x^2+1} \,\mathrm{d}x\right)+4\left(\int_{-2}^{3} \frac{1}{x^2+1}\,\mathrm{d}x\right) \\\\
&=\frac{1}{2}\left(\int_{5}^{10} \frac{\mathrm{d}u}{u}\right)+4\left(\int_{-2}^{3} \frac{1}{x^2+1}\,\mathrm{d}x\right)
\end{align}</p>
<p>Remember, as Tim Raczkowski tells us, that $$\frac{\mathrm{d}}{\mathrm{d}x}\left(\arctan x\right) = \frac{1}{x^2+1}$$</p>
<p>Also:</p>
<p>$$\frac{\mathrm{d}}{\mathrm{d}x} \left(\ln \left|x\right|\right) = \frac{1}{x}$$</p>
<p>This <a href="http://www.wolframalpha.com/input/?i=%28%28log+10+-+log+5%29%2F2%29%2B4%28arctan%283%29-arctan%28-2%29%29" rel="nofollow">gives</a></p>
<p>\begin{align}
A&=\frac{\ln 10 - \ln 5}{2} +4 \left(\arctan 3 - \arctan \left(-2\right)\right)\\\\
&\approx\boxed{9.7714}
\end{align}</p>
|
4,318,842 | <p>Thanks for your time.</p>
<p>I am studying Euclidean Geometry through a text book and they have a corollary that afirms that given <span class="math-container">$r \parallel s$</span>, if t intersects r then it must intersect s, although the proof they give is not much convinceable. It says: "If one line can intersect only one of the two parallel lines, there is one parallel line to two non-parallel lines."</p>
<p>A proof that seems more reasonable to me: Given <span class="math-container">$r \parallel s$</span>. If t intersects r, but does not intersects s, then <span class="math-container">$t \parallel s$</span>. If <span class="math-container">$r \parallel s$</span> and <span class="math-container">$t \parallel s$</span> than <span class="math-container">$r \parallel t$</span>, which is a contradiction, therefore t must intersect both r and s.</p>
<p>Is my proof correct?</p>
| MarioPrix | 998,117 | <p>Yes, your proof is correct.</p>
<p>This is another alternative:</p>
<p>Let <span class="math-container">$m_l$</span> indicates the gradient of the line <span class="math-container">$l$</span>. Then <span class="math-container">$m_r=m_s$</span>.</p>
<p>Additionally, since <span class="math-container">$t$</span> intersects <span class="math-container">$r$</span>, <span class="math-container">$m_r \neq m_t$</span>.</p>
<p>It follows that <span class="math-container">$m_s \neq m_t$</span>, thus <span class="math-container">$t$</span> intersects <span class="math-container">$s$</span> at some point.</p>
|
3,836,662 | <p>I understand basic group theory. I would say that I've seen most of the standard stuff up to, say, the quotient group.</p>
<p>I feel like I've seen in more than one place the suggestion that group theory is the study of symmetries, or actions that leave something (approximately) unchanged. Unfortunately I can only find a couple sources. At 0:49 in this <a href="https://www.youtube.com/watch?v=mH0oCDa74tE" rel="nofollow noreferrer">3 Blue 1 Brown video</a>, the narrator says "[Group theory] is all about codifying the idea of symmetry." The whole video seems to be infused with the idea that every group represents the symmetry of something.</p>
<p>In <a href="https://www.youtube.com/watch?v=ihMyW7Z5SAs" rel="nofollow noreferrer">this video</a> about the Langlands Program, the presenter discusses symmetry as a lead-in to groups beginning around 33:00. I don't know if he actually describes group theory as being about the study of symmetry, but the general attitude seems pretty similar to that of the previous video.</p>
<p>This doesn't jive with my intuition very well. I can see perfectly well that <em>part</em> of group theory has to do with symmetries: one only has to consider rotating and flipping a square to see this. But is <em>all</em> of group theory about symmetry? I feel like there must be plenty of groups that have nothing to do with symmetry. Am I wrong?</p>
| markvs | 454,915 | <p>The best and standard way to prove it is to consider the map from <span class="math-container">$H\times K$</span> to <span class="math-container">$HK$</span> which sends <span class="math-container">$(h,k)$</span> to <span class="math-container">$hk$</span> and look at the equivalence classes of pairs mapped to the same element of <span class="math-container">$HK$</span>. That is for every <span class="math-container">$(h,k)$</span> count pairs <span class="math-container">$(h',k')$</span> such that <span class="math-container">$hk=h'k'$</span>.</p>
<p><strong>Edit</strong> To make it complete, <span class="math-container">$hk=h'k'$</span> is equivalent to <span class="math-container">$h^{-1}h'=k'k^{-1}$</span>. The LHS is in <span class="math-container">$H$</span>, the RHS is in <span class="math-container">$K$</span>, so both are in <span class="math-container">$K\cap H$</span>. So the number of pairs <span class="math-container">$(h',k')$</span> such that <span class="math-container">$hk=h'k'$</span> is the same as the number of elements in <span class="math-container">$h(H\cap K)$</span> which is <span class="math-container">$|H\cap K|$</span>.</p>
|
3,220,273 | <p>It seems the definition of a parallelogram is locked to quadrilaterals for some reason. Is there a reason for this? Why couldn't a parallelogram (given the way the word seems rather than as a mathematical/geometric construct) contain greater than two pairs of parallel sides? In a hexagon for example, all six sides are parallel to their opposing side. Is there a term for this kind of object?</p>
<p>It seems to me there must be some value in describing a polygon with even numbers of sides in which the opposing sides are parallel to each other. While a hexagon, octagon, decagon, etc. all match this rule, you could have polygons with unequal sides as well.</p>
<p><a href="https://i.stack.imgur.com/Ln3Yg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ln3Yg.png" alt="enter image description here"></a></p>
<p><strong>Edit 1:</strong> Object described by Mark Fischler</p>
<p><a href="https://i.stack.imgur.com/uEWA9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uEWA9.png" alt="Object described by Mark Fischler"></a></p>
<p><strong>Zonogon:</strong></p>
<p><a href="https://i.stack.imgur.com/KfV1w.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KfV1w.png" alt="enter image description here"></a></p>
| Mark Fischler | 150,362 | <p>I'm going to propose, out of the blue, terms like "hexaparallelogram", "octaparallelogram", and so forth.</p>
<p>I'm wondering whether, for more than <span class="math-container">$4$</span> sides, you would like your definition of hexaparallelogram to be restricted to having 3 pairs of parallel and pairwise equal sides (as in your picture - evidently these have a name, zonogon), or would you include a hexagon with vertices at <span class="math-container">$\{(0,0), (12,0), (16,6), (4,12), (0,12), (-6,3)\}$</span> which has three pairs of parallel sides but no two sides of equal length?</p>
<p>Euclid, in proposition 34, introduces the term (παραλληλόγραμμα χωρία) which we can translate to "parallelogrammic area." So much for the etymology sites that trace the word only to Middle French. Euclid himself restricted the word to just four-sided figures. Proclus credits Euclid with having introduced the term "parallelogram," as opposed to bringing down that term from earlier works. So that tells us who to blame.</p>
|
2,940,306 | <p>I understand that it would be n! permutations for the given amount of elements, but I am not sure calculate it with these parameters.</p>
| Paul Childs | 584,354 | <p>3 for the first</p>
<p>3 for the last</p>
<p>(n-2)! for the rest.</p>
<p>Multiply them together.</p>
|
4,041,595 | <p>I've been having lectures in group theory with Hungerford's book. We were presented with the following theorem:</p>
<blockquote>
<p><a href="https://i.stack.imgur.com/pkL0f.png" rel="noreferrer"><img src="https://i.stack.imgur.com/pkL0f.png" alt="enter image description here" /></a></p>
</blockquote>
<p>And then with:</p>
<blockquote>
<p><a href="https://i.stack.imgur.com/ZP9eR.png" rel="noreferrer"><img src="https://i.stack.imgur.com/ZP9eR.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/jpiHf.png" rel="noreferrer"><img src="https://i.stack.imgur.com/jpiHf.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/jYsqM.png" rel="noreferrer"><img src="https://i.stack.imgur.com/jYsqM.png" alt="enter image description here" /></a></p>
</blockquote>
<p>Previous to the lectures about them, I was understanding most of the stuff, that is: I kinda could figure out the motivation for things. But when I came into these theorems, things got pretty weird for me. I don't understand the motivation for them. I have the following guesses:</p>
<ul>
<li><p>Theorem 5.6 is used to prove Corollary 5.7 but I think it has a utility on its own: If given an homomorphism <span class="math-container">$f: G \to H$</span> with <span class="math-container">$N\lhd G$</span>, we have a unique homomorphism <span class="math-container">$\overline{f} : G/N\to H$</span>, then we could use this to know what homomorphisms could exist betwen <span class="math-container">$G$</span> and <span class="math-container">$H$</span> and this depends only on the normal subgroups of <span class="math-container">$G$</span>. Does that make sense?</p>
</li>
<li><p>For the isomorphism theorems, I noticed they are corollaries so they must be very important. One of the reasons I conjectured for them is that we can construct very interesting isomorphisms. But why are those isomorphisms interesting? It's not clear to me what interesting stuff one could do with them.</p>
</li>
</ul>
<p>I know my guesses are maybe too obvious or wrong, but I can't figure out on my own the answer to that question. Can you help me?</p>
| Arturo Magidin | 742 | <p>Let’s say you have a group you are interested in. One way to try to understand the group is to stare at it intently until you see interesting things about it.</p>
<p>But the big insight of abstract algebra in general (and of Category Theory) is that you can also learn a lot about the group by considering <em>morphisms</em>, both into and out of the group. In fact, the Yoneda embedding tells you that understanding all the ways that all groups can map into <span class="math-container">$G$</span> will completely determine the isomorphism type of <span class="math-container">$G$</span>, and that understanding all the ways in which you can map <em>from</em> <span class="math-container">$G$</span> to all other groups will completely determine the isomorphism type of <span class="math-container">$G$</span> as well.</p>
<p>In addition, one can view these maps as casting “shadows” of the object <span class="math-container">$G$</span>; with some luck, the different shadows will each be simpler than the group, and perhaps we can piece together enough “shadows” to get a full picture of what the group looks like without having to look at the full group (in turns out that is not always possible, as witnessed by simple groups, but it works for many other types of groups).</p>
<p>The First Isomorphism Theorem tell you that if you want to understand the way that <span class="math-container">$G$</span> maps into other groups, you don’t need to go scour the universe to find those groups: you can “realize” any map <span class="math-container">$f\colon G\to H$</span> by finding a suitable normal subgroup <span class="math-container">$K$</span> of <span class="math-container">$G$</span>, and then the image of <span class="math-container">$G$</span> under <span class="math-container">$f$</span> will behave exactly the same way as the image of <span class="math-container">$G$</span> under the canonical (and usually well-understood) quotient map <span class="math-container">$\pi\colon G\to G/K$</span> behaves. So in a sense you can understand all those images of <span class="math-container">$G$</span> by staring at suitable shadows of <span class="math-container">$G$</span>: just the quotient ones.</p>
<p>The Third Isomorphism Theorem tells you that you that iterating the process won’t get you any new information: if you take a quotient of <span class="math-container">$G/K$</span>, you are still just taking a quotient of <span class="math-container">$G$</span>.</p>
<p>The Second Isomorphism Theorem tells you how the subgroups of <span class="math-container">$G$</span> behave under the map <span class="math-container">$G\to G/N$</span>: namely the image of <span class="math-container">$K$</span> in the map will be <span class="math-container">$K/(N\cap K)$</span>, but this will be isomorphic to <span class="math-container">$KN/N$</span>. When you connect it with the Fourth (or Lattice) isomorphism theorem, which you did not quote, it provides complete information about what happens to the subgroups of <span class="math-container">$G$</span> under the map <span class="math-container">$G\to G/N$</span>: the subgroups that contain <span class="math-container">$N$</span> are mapped to the subgroups of <span class="math-container">$G/N$</span> in a one-to-one, inclusion preserving, normality preserving manner, while the subgroups of <span class="math-container">$G$</span> that do <em>not</em> contain <span class="math-container">$N$</span> correspond, under the Second Isomorphism Theorem, to certain subgroups that <em>do</em> contain <span class="math-container">$N$</span>.</p>
<p>These theorems are valid in a much more general setting: they apply, suitably modified to account for some technical issues, to any kind of algebraic structure: semigroups, monoids, rings, lattices, and more. They form a backbone of investigations, tools used over and over again, because one is always looking at homomorphisms to better understand the structures, and because one is always looking at quotients to simplify matters and try to get a better sense of the object one is studying.</p>
|
3,443,672 | <p>The equation is
<span class="math-container">$$\tan\frac{5\pi}{6} \cos x=1-\sin x$$</span>
<span class="math-container">$$\sin\frac{5\pi}{6} \cos x=\cos\frac{5\pi}{6}-\cos\frac{5\pi}{6} \sin x$$</span>
<span class="math-container">$$\sin\left(\frac{5\pi}{6}+x\right)=\cos \frac{5\pi}{6}$$</span> which looks weird to me. What am I doing wrong?</p>
| Quanto | 686,284 | <p>Note <span class="math-container">$\tan\frac{5\pi}{12}=2+\sqrt3$</span> to rewrite <span class="math-container">$(2+\sqrt 3)\cos x=1-\sin x$</span> as</p>
<p><span class="math-container">$$\sin\frac{5\pi}{12}\cos x+\cos\frac{5\pi}{12}\sin x=\cos\frac{5\pi}{12}$$</span></p>
<p>or</p>
<p><span class="math-container">$$\sin\left(\frac{5\pi}{12}+x\right)=\sin\frac{\pi}{12}\implies2\cos\left(\frac x2+\frac\pi4\right)\sin\left(\frac x2+\frac\pi6\right)=0$$</span></p>
<p>which yields <span class="math-container">$x=2k\pi-\frac\pi3$</span> and <span class="math-container">$x=2k\pi +\frac\pi2$</span>.</p>
|
1,827,612 | <p>I have to simplify $\sum_{i,j}\left[{n}\atop{i+j}\right]\binom{i+j}{i}$. I looks like we have $n$ children and we have to answer how many times we can arrange them into circles and color some of circles red.</p>
| RRL | 148,510 | <p>For the first lemma, since $\inf f(x) + \inf g(x) \leqslant f(x) + g(x)$ we have the inequality</p>
<p>$$\inf_{x \in [x_{i-1},x_i]} f(x) + \inf_{x \in [x_{i-1},x_i]} g(x) \leq \inf_{x \in [x_{i-1},x_i]} [f(x) + g(x)],$$</p>
<p>where $[x_{i-1},x_i]$ is a subinterval of some partition $P$.</p>
<p>This implies, for lower Darboux sums, that</p>
<p>$$\begin{equation*}L(P,f) + L(P,g)\leq L(P,f+g) \tag{1}\end{equation*}$$</p>
<p>Proving the desired inequality is a bit tricky, since taking suprema over lower sums directly leads to a dead end.</p>
<p>For example, we see immediately that </p>
<p>$$\sup_{P}[L(P,f) + L(P,g)]\leq \sup_{P}L(P,f+g)=\underline{\int}_a^b[f(x)+g(x)]\, dx \tag{2}$$</p>
<p>and</p>
<p>$$ L(P,f) + L(P,g) \leq \sup_{P}L(P,f) + \sup_{P}L(P,g) = \underline{\int}_a^bf(x)dx + \underline{\int}_a^bg(x)dx\\ \implies \sup_{P}(L(P,f) + L(P,g))\leq \underline{\int}_a^bf(x)dx + \underline{\int}_a^bg(x)dx \tag{3}$$</p>
<p>Unfortunately (2) and (3) do not lead to an ordering of the lower Darboux integrals on the RHS of each inequality. </p>
<p>Instead, we can arrive at the result indirectly. Assume on the contrary that</p>
<p>$$\underline{\int}_a^b [f(x)+g(x)] \, dx < \underline{\int}_a^b f(x) \, dx + \underline{\int}_a^b g(x) \,dx .$$</p>
<p>Rearrranging, we get</p>
<p>$$\underline{\int}_a^b [f(x)+g(x)] \, dx - \underline{\int}_a^b g(x) \,dx < \underline{\int}_a^b f(x) \, dx,$$</p>
<p>Since the lower integral on the RHS is a supremum over partitions, there exists a partition $P$ such that</p>
<p>$$\underline{\int}_a^b [f(x)+g(x)] \, dx - \underline{\int}_a^b g(x) \,dx < L(P,f) \leqslant \underline{\int}_a^b f(x) \, dx.$$</p>
<p>Rearranging again,</p>
<p>$$\underline{\int}_a^b [f(x)+g(x)] \, dx - L(P,f) < \underline{\int}_a^b g(x) \, dx.$$</p>
<p>Reasoning as before, there exists a partition $P’$ such that </p>
<p>$$\underline{\int}_a^b [f(x)+g(x)] \, dx - L(P,f) < L(P’,g) \leqslant \underline{\int}_a^b g(x) \, dx,$$
and
$$\underline{\int}_a^b [f(x)+g(x)] \, dx < L(P,f) + L(P’,g) .$$</p>
<p>Take a common refinement of the partitions $Q = P \cup P'$. Lower sums increase as partitions are refined and we have $L(Q,f) \geqslant L(P,f)$ and $L(Q,g) \geqslant L(P’,g).$</p>
<p>It follows that </p>
<p>$$L(Q,f+g) \leqslant \underline{\int}_a^b [f(x)+g(x)] \, dx < L(P,f) + L(P’,g) \leqslant L(Q,f) + L(Q,g).$$</p>
<p>This contradicts inequality (1) for lower sums, and, therefore</p>
<p>$$\underline{\int}_a^b f(x) \, dx + \underline{\int}_a^b g(x) \, dx \leqslant \underline{\int}_a^b [f(x) + g(x)] \, dx. $$</p>
<p><strong>Second Lemma</strong></p>
<p>Let $P$ be an arbitrary partition of $[a,b].$ Now $P$ may or may not include the point $c$. If not, then consider a refined partition $P'$ that includes $c$. Let $P' = P_1 \cup P_2$ where $P_1$ and $P_2$ are partitions of $[a,c]$ and $[c,b]$, respectively.</p>
<p>Then
$$\overline{\int}_a^c f(x) \, dx + \overline{\int}_c^b f(x) \, dx \leqslant U(P_1,f) + U(P_2,f) = U(P',f) \leqslant U(P,f).$$</p>
<p>Now take the infumum over all partitions $P$ to obtain</p>
<p>$$\overline{\int}_a^c f(x) \, dx + \overline{\int}_c^b f(x) \, dx \leqslant \inf_{P} \,\,U(P,f) = \overline{\int}_a^b f(x) \, dx .$$</p>
|
532,785 | <p>Let $f:[0, 1] \rightarrow \mathbb{R}$ a continuous function. If $a>0$, show that:</p>
<p>$$ f(0)\ln(\frac{b}{a})=\lim_{\epsilon\rightarrow 0}\int_{\epsilon a}^{\epsilon b} \frac{f(x)}{x}dx$$</p>
<p>Tried using Riemann sum, but did not succeed.</p>
| DonAntonio | 31,254 | <p>Another approach using , as you did, the bisectrix of the right angle: if we call $\;M\;$ the vertex where the right angle is, and the upper one in $\;N\;$ and the rightmost one is $\;Q\;$ (next time do <strong>call names</strong> to all the vertices and intersection points! Whithout that geometry exercises become cumbersome), we have that</p>
<p>$$MP=MB=r\;\;\text{(why?)}$$</p>
<p>Since both tangents to a circle from the same exterior point have the same length, we have that:</p>
<p>$$BN=NA=6-r\implies AQ=4+r=QP$$</p>
<p>But we also know that $\;QP=8-r\;$ , so we we find that</p>
<p>$$8-r=4+r\implies r=2$$</p>
|
4,644,847 | <p>Is there a formal fallacy to describe lack of an observation is not proof that it does not exist, or lack of an occurrence is not proof that it can never happen?</p>
| Bram28 | 256,001 | <p>This is the <a href="https://fallacyinlogic.com/appeal-to-ignorance/" rel="nofollow noreferrer">Appeal to Ignorance fallacy</a>: absence of evidence is not evidence of absence. It is not considered a formal fallacy though.</p>
|
2,013,650 | <p>I am trying to prove that
$$n^{n/2}<n!,\text{ for } n\ge2.$$
I can't really figure it out. </p>
| Najib Idrissi | 10,014 | <p>Your "functors" aren't actually functors. I'll stick to standard notations because "$\operatorname{Hom}_k(V,V)$" is too easy to mistake for the space of $k$-linears maps $V \to V$ for some field $k$... I'll also let the base field $\mathbb{R}$ be implied throughout.</p>
<p>Let's do the simplest case, $k=1$. Then you have a <em>bi</em>functor
$$\operatorname{Hom} : \mathsf{Vec}^{\mathrm{op}} \times \mathsf{Vec} \to \mathsf{Vec}, \; (V,W) \mapsto \operatorname{Hom}(V,W)$$
It's not possible to compose that with the functor $\mathsf{Vec} \to \mathsf{Vec} \times \mathsf{Vec}$, simply because the (co)domains don't match... So the mapping $V \mapsto \operatorname{Hom}(V,V)$ doesn't yield a functor a priori. I can't see any reasonable way to make that into a functor. Your question is doomed from the start...</p>
<hr>
<p>However you do have the bifunctor $\operatorname{Hom}$ as above. You also have another bifunctor, say $$\Phi : \mathsf{Vec}^{\mathrm{op}} \times \mathsf{Vec} \to \mathsf{Vec}, \; (V,W) \mapsto \operatorname{Hom}(V \otimes W^*, \mathbb{R}).$$</p>
<p>This is indeed a bifunctor; given $f : V' \to V$ and $g : W \to W'$, you get $\Phi(f,g) : \Phi(V,W) \to \Phi(V',W')$ given by $t \mapsto t \circ (f \otimes g^*)$ (where $g^* : \operatorname{Hom}(W', \mathbb{R}) \to \operatorname{Hom}(W,\mathbb{R})$ is precomposition by $g$).</p>
<p>These two functors are naturally isomorphic, when you restrict to finite-dimensional spaces. Indeed define a natural transformation $\eta : \operatorname{Hom} \to \Phi$ by:
\begin{align}
\eta_{(V,W)} : \operatorname{Hom}(V,W) & \to \operatorname{Hom}(V \otimes W^*, \mathbb{R}) \\
t & \mapsto (\eta(t) : v \otimes \psi \mapsto \psi(t(v)))
\end{align}</p>
<p>It's not hard (but it's a bit tedious) to check that this is a natural transformation. Both spaces have the same dimension, namely $(\dim V) (\dim W)$. So to check that this is an isomorphism for all $(V,W) \in \mathsf{Vec}^\mathrm{op} \times \mathsf{Vec}$, it suffices to check that this is surjective.</p>
<p>For a fixed couple $(V,W)$, let $(w_1, \dots, w_n)$ be a basis of $W$ (recall that it's finite dimensional), and let $(w_1^*, \dots, w_n^*)$ be the dual basis of $W^*$. Suppose given some $\beta \in \operatorname{Hom}(V \otimes W^*, \mathbb{R})$. Then you can let $\alpha \in \operatorname{Hom}(V, W)$ be defined by
$$\alpha(v) = \sum_{i=1}^n \beta(v \otimes w_i^*) \cdot w_i.$$</p>
<p>It is now a very tedious (but completely mechanical) check that $\eta_{(V,W)}(\alpha) = \beta$. The mapping is surjective between spaces of the same finite dimension, hence it's an isomorphism.</p>
<hr>
<p>Finally, to get the general case, consider the functor $\bigotimes : \mathsf{Vec}^{\times k} \to \mathsf{Vec}$ given by $\bigotimes(V_1, \dots, V_k) = V_1 \otimes \dots \otimes V_k$, and compose it with the natural isomorphism $\eta$ to get a natural isomorphism
$$\operatorname{Hom}(V_1 \otimes \dots \otimes V_k, W) \xrightarrow{\cong} \operatorname{Hom}(V_1 \otimes \dots \otimes V_k \otimes W^*, \mathbb{R}).$$</p>
<p>But again this is contravariant in the $V_i$ and covariant in $W$, so you can compose with $V \mapsto (V, \dots, V)$ because this is covariant in everything...</p>
|
3,648,315 | <p><a href="https://i.stack.imgur.com/IKnlD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IKnlD.png" alt="enter image description here"></a></p>
<p>I have an item, let's say sword. </p>
<p>As described above, <br/>
If my sword is at state 1, I can try upgrading it. There are 3 possibilities. <br/>
It can be upgraded with prob = 0.3, remain still with prob = 0.68, can be destroyed with prob = 0.02.</p>
<p>If my sword is at state 2, I still can try to upgrade it. <br/>
It can be upgraded with prob = 0.3, can be downgraded to state 1 with prob = 0.68, can be destroyed with prob = 0.02.</p>
<p>Once my sword destroyed, there is no turning back. <br/>
Once my sword reached at state 3, no need to do something else. I'm done.</p>
<p>I know it's a Markov chain problem. <br/>
I can express this situation with matrix, and if I multiply it over and over, it can reach equilibrium state.</p>
<pre><code>p2 = matrix(c(1, rep(0, 3),
0.02, 0.68, 0.3, 0,
0.02, 0.68, 0, 0.3,
rep(0, 3), 1), 4, byrow = T)
p2
## [,1] [,2] [,3] [,4]
## [1,] 1.00 0.00 0.0 0.0
## [2,] 0.02 0.68 0.3 0.0
## [3,] 0.02 0.68 0.0 0.3
## [4,] 0.00 0.00 0.0 1.0
matrix.power <- function(A, n) { # For matrix multiplication
e <- eigen(A)
M <- e<span class="math-container">$vectors
d <- e$</span>values
return(M %*% diag(d^n) %*% solve(M))
}
round(matrix.power(p2, 1000), 3)
## [,1] [,2] [,3] [,4]
## [1,] 1.000 0 0 0.000
## [2,] 0.224 0 0 0.776
## [3,] 0.172 0 0 0.828
## [4,] 0.000 0 0 1.000
</code></pre>
<p>But how can I get the <code>Pr(Reach state 3 without destroyed | currently at state 2)</code> using Markov chain?</p>
<p>I could get <code>Pr(Reach state 2 without destroyed | currently at state 1)</code> by using sum of geometric series.</p>
<p>Thank you.</p>
| Will Jagy | 10,400 | <p><span class="math-container">$$ (2x+1)(2y+1) = 167 $$</span>
which is prime, so we get
<span class="math-container">$$ (0,83),(83,0),(-84, -1),(-1,-84), $$</span></p>
|
2,698,555 | <p><strong>Question</strong></p>
<p>Find all the rational values of $x$ at which $y=\sqrt{x^2+x+3}$</p>
<p><strong>My attempt</strong></p>
<p>Since we only have to find the rational values of $x$ and $y$, we can assume that
$$ x \in Q$$
$$ y \in Q$$
$$ y-x \in Q $$
Let$$ d = y-x$$
$$d=\sqrt{x^2+x+3}-x$$
$$d+x=\sqrt{x^2+x+3}$$
$$(d+x)^2=(\sqrt{x^2+x+3})^2$$
$$d^2 + x^2 + 2dx =x^2+x+3$$
$$d^2 +2dx = x +3$$
$$x = \frac{3-d^2}{2d-1}$$</p>
<p>$$d \neq \frac{1}{2}$$</p>
<p>So $x$ will be rational as long as $d \neq \frac{1}{2}$.</p>
<p>Now
$$ y = \sqrt{x^2+x+3}$$
$$ y = \sqrt{(\frac{3-d^2}{2d-1})^2 + \frac{3-d^2}{2d-1} + 3}$$
$$ y = \sqrt{\frac{(3-d^2)^2}{(2d-1)^2} + \frac{(3-d^2)(2d-1)}{(2d-1)^2} + 3\frac{(2d-1)^2}{(2d-1)^2}}$$
$$ y = \sqrt{\frac{(3-d^2)^2 + (3-d^2)(2d-1) + 3(2d-1)^2}{(2d-1)^2}} $$
$$ y = \frac{\sqrt{(3-d^2)^2 + (3-d^2)(2d-1) + 3(2d-1)^2}}{(2d-1)}$$
$$ y = \frac{\sqrt{d^4-2d^3+7d^2-6d+9}}{(2d-1)}$$</p>
<p>I know that again $d \neq \frac{1}{2}$ but I don't know what to do with the numerator. Help</p>
| Michael Hoppe | 93,935 | <p>Apply the carpet theorem, see <a href="https://www.cut-the-knot.org/Curriculum/Geometry/CarpetsInSquare.shtml" rel="noreferrer">https://www.cut-the-knot.org/Curriculum/Geometry/CarpetsInSquare.shtml</a>. Consider the upper eighth of the big circle and the left halve circle. Since their areas are equal, the areas which are not covered by both must be equal as well, that is the upper half of the red area equals the lower half of the blue one.</p>
<p><a href="https://i.stack.imgur.com/Isu6d.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Isu6d.png" alt="enter image description here"></a></p>
|
2,983,231 | <p>I'm working out of Salamon's <em>Measure and Integration</em> in preparation for studying probability, and I came across the following exercise.</p>
<blockquote>
<p>Let <span class="math-container">$X$</span> be uncountable and let <span class="math-container">$\mathcal{A} \subset 2^X$</span> be the set of all subsets <span class="math-container">$A \subset X$</span> such that either <span class="math-container">$A$</span> or <span class="math-container">$A^C$</span> is countable. Define:
<span class="math-container">$$ \mu(A) := \begin{cases} 0 &\mbox{if } A \ \text{is countable} \\ 1 &\mbox{if } A^C \ \text{is countable} \end{cases}$$</span>
where <span class="math-container">$A \in \mathcal{A}$</span>. Show that <span class="math-container">$(X, \mathcal{A}. \mu)$</span> is a measure space. Describe the measurable functions and their integrals.</p>
</blockquote>
<p>I was able to attempt to show that <span class="math-container">$\mathcal{A}$</span> was a <span class="math-container">$\sigma$</span>-algebra with the following argument:</p>
<p>(<span class="math-container">$\mathcal{A}$</span> is a <span class="math-container">$\sigma$</span>-algebra.) From <span class="math-container">$X$</span>, we may construct a countable set <span class="math-container">$S = \bigcup_\limits{j \geq 1} S_j$</span> where each <span class="math-container">$S_j$</span> contains <span class="math-container">$j$</span> elements from <span class="math-container">$X$</span>. Thus if <span class="math-container">$X^C = S$</span>, then we may say <span class="math-container">$X \in \mathcal{A},$</span> and <span class="math-container">$\mathcal{A}$</span> is nontrivially nonempty. Let <span class="math-container">$T \in \mathcal{A},$</span> and suppose <span class="math-container">$T$</span> is countable. Then <span class="math-container">$T^C \in \mathcal{A},$</span> as <span class="math-container">$(T^C)^C = T$</span> is countable. Finally, to show closure under countable union, we note that if <span class="math-container">$Y_j$</span> is a countable collection of sets in <span class="math-container">$\mathcal{A}$</span>, then their union must be at most countable, and hence in <span class="math-container">$\mathcal{A}$</span>. Thus, <span class="math-container">$\mathcal{A}$</span> is a <span class="math-container">$\sigma$</span>-algebra.</p>
<p>(The triple is a measure space.) It is seen that all sets have finite measure in this space. It remains to show that this measure is countably additive with respect to disjoint sets. Let <span class="math-container">$A_j$</span> be a countable collection of disjoint sets in <span class="math-container">$\mathcal{A}.$</span> The measure counts the number of sets whose complements are countable.</p>
<p>I was then stuck there. How can I show that <span class="math-container">$\mu$</span> is countable additive? How can I also describe what the measurable functions are?</p>
| tomasz | 30,222 | <p><strong>Hint</strong>: If <span class="math-container">$(A_n)_n$</span> is a family of pairwise disjoint, countable or cocountable sets, then at most one of them is cocountable.</p>
|
569,484 | <p>A and B are events in a sample space with $p(A) > 0$ and $p(B) > 0$. Write $p(A|B)$ for the conditional probability of $A$ given that $B$ has occurred.</p>
<p>1) If $p(A|B) < p(A)$, show that $p(B|A) < p(B)$ </p>
<p>2) Show that $p(A|B) ≥ \frac{p(A)+p(B)-1}{p(B)}$</p>
<p>For the first part:
I subbed P(B|A) < P(B) into P(B|A)= P(A and B)/P(A) to get P(A and B)/P(A) < P(B) so basically P(A and B) < P(B). I think that is the correct way for part 1 but need confirmation if possible and i'm not quite sure how to start part 2.
Any help or tips would be much appreciated.</p>
| Community | -1 | <ol>
<li><p>If $P(A/B)<P(A)$ then $\frac{P(A\cap B)}{P(B)}<P(A)$ then $P(A\cap B)<P(B)P(A)$ and dividing by $P(A)$ to get $\frac{P(A\cap B)}{P(A)}<P(B)$. Thus, $P(B/A)<P(B)$.</p></li>
<li><p>Since $P(A\cup B)\leq 1$ then $P(A)+P(B)-(A\cap B)\leq 1$, this implies $P(A)+P(B)-1\leq (A\cap B)$. Dividing by P(B) to get $\frac{P(A)+P(B)-1}{P(B)}\leq \frac{(A\cap B)}{P(B)}$, which gives the result.</p></li>
</ol>
|
35,375 | <p>Good morning,
today I have read that "number theory is nothing but the study of $\mathrm{Gal}(\mathbb{\bar{Q}}/\mathbb{Q})$", here <a href="http://www.math.uconn.edu/~alozano/elliptic/finding%20points.pdf" rel="nofollow">http://www.math.uconn.edu/~alozano/elliptic/finding%20points.pdf</a>
can anyone give a very naive layman definition of what it actually means?</p>
<p>Furthermore, I got this doubt that $\bar {\mathbb{Q}}$ is the algebraic closure of $\mathbb{Q}$, and the thing that confuses me is
the field of rational numbers $\mathbb{Q}$ is not a algebraically closed as there exists a polynomial with $a_{1},a_{2},\dotsc,a_{n}\in \mathbb{Q}$ and $(x-a_{1})(x-a_{2})\cdots(x-a_{n})+1$ has no zero in $\mathbb{Q}$.</p>
<p>Then why are we considering the field extension of $\bar {\mathbb{Q}}/\mathbb{Q}$ when $\mathbb{Q}$ is not algebraically closed, won't it contradict the definition of algebraic closure?</p>
<p>But I am not getting an answer i was looking for ,i want what are the things going on behind the $\mathrm{Gal}(\mathbb{\bar{Q}}/\mathbb{Q})$ like what is the thing we get if we take the $\mathbb{\bar{Q}/\mathbb{Q}}$ and what does taking the $\mathrm{Gal}(\mathbb{\bar{Q}}/\mathbb{Q})$ give someone,</p>
<p>Thank you</p>
| G.. | 14,043 | <p>The object $\mathrm{Gal}(\mathbb{\bar{Q}}/\mathbb{Q})$ is not just an abstract group; it has a natural and intricate topology on it as well. Then one would consider various naturally occurring continuous actions of this topological group on various modules. Two examples are: The absolute Galois group of a number field acts naturally on the roots of unity contained in that field, or, given an elliptic curve over the field, the Galois group acts on the torsion points on the elliptic curve with co-ordinates contained in the field.</p>
<p>Also one considers some natural representations of this Galois group on some vector spaces. Examples can be constructed from the above description of Galois actions. A lot of number theory is encapsulated in the study of such <em>Galois actions</em> and <em>Galois representations</em>. This is perhaps what is meant by the statement that you quote.</p>
|
128,533 | <p>For the description, I have a simplified problem like the following:</p>
<pre><code>MapAt[f[1, #1], {a, b, c, d}, #2] & @@@ {{1, 2}, {3, 4}}
</code></pre>
<p>will give</p>
<blockquote>
<pre><code>{{a, f[1, 1][b], c, d}, {a, b, c, f[1, 3][d]}}
</code></pre>
</blockquote>
<p>But actually <code>{a, f[1, 1][b], c, f[1, 3][d]}</code> is what I expected. What happened? How to adjust the code?</p>
<hr>
<p><strong>Update:</strong></p>
<p>My real case is</p>
<pre><code>bigList = Range @ 10;
veryBigList = {{1, 3}, {1, 4}, {2, 7}, {2, 9}, {4, 10}};
Function[{binLevel, place},
MapAt[BitSet[#, binLevel] &, bigList, place]] @@@ veryBigList
</code></pre>
<blockquote>
<p>{{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, {1, 2, 3, 6, 5, 6, 7, 8, 9, 10}, {1,
2, 3, 4, 5, 6, 7, 8, 9, 10}, {1, 2, 3, 4, 5, 6, 7, 8, 13, 10}, {1,<br>
2, 3, 4, 5, 6, 7, 8, 9, 26}}</p>
</blockquote>
<p>In my case, the list size is very huge. If I use <code>Fold</code>, it will act on a very big result list every calculation, for the RAM so I want to avoid using <code>Fold</code>.</p>
| kglr | 125 | <pre><code>ClearAll[f1]
f1[func_, lst_] := Module[{funcs = func[1, #] & @@@ #, pos = #2 & @@@ #},
ReplacePart[lst, Thread[pos -> MapThread[# @ #2 &, {funcs, lst[[pos]]}]]]] &;
f1[f, {a, b, c, d}]@{{1, 2}, {3, 4}}
</code></pre>
<blockquote>
<p><code>{a, f[1, 1][b], c, f[1, 3][d]}</code></p>
</blockquote>
<p>or</p>
<pre><code>ClearAll[f2]
f2[func_, lst_] := Module[{funcs = func[1, #] & @@@ #, pos = #2 & @@@ #, l2 = lst},
l2[[pos]] = MapThread[#@#2 &, {funcs, lst[[pos]]}]; l2] &;
f2[f, {a, b, c, d}]@{{1, 2}, {3, 4}}
</code></pre>
<blockquote>
<p><code>{a, f[1, 1][b], c, f[1, 3][d]}</code></p>
</blockquote>
|
1,955,505 | <blockquote>
<p>$\sum_{n=0}^\infty \frac{n^2+3n+2}{4^n} = \frac{128}{27}$ Given hint: $(n^2+3n+2) = (n+2)(n+1)$</p>
</blockquote>
<p><strong>I've tried</strong> converting the series to a geometric one but failed with that approach and don't know other methods for normal series that help determine the actual convergence value. Help and hints are both appreciated</p>
| user246336 | 246,336 | <p>Here is another approach. Let us define the function </p>
<p>$$\phi(x):=\sum_{n=0}^\infty (n+1)(n+2) x^n.$$</p>
<p>If we differentiate we obtain</p>
<p>$$\phi'(x):=\sum_{n=0}^\infty n(n+1)(n+2) x^{n-1}=\sum_{n=0}^\infty(n+1)(n+2)(n+3)x^n.$$</p>
<p>Now, </p>
<p>$$\phi'(x)-3\phi(x)=\sum_{n=0}^\infty n(n+1)(n+2)x^{n}=\sum_{n=0}^\infty(n+1)(n+2)(n+3)x^{n+1}=x\phi'(x)$$</p>
<p>so</p>
<p>$$\phi'(x)(x-1)+3\phi(x)=0.$$</p>
<p>An elementary integration (with the initial condition $\phi(0)=2$) yields</p>
<p>$$\phi(x)=\frac2{(1-x)^3}.$$</p>
<p>Thus, the value you are seeking is $\phi(1/4)=128/27$.</p>
|
499,496 | <p>We have the diff. eq:</p>
<p>$$ y' = \dfrac{2y}{t}$$</p>
<p>I tried to do the following:</p>
<p>$$ y' = \dfrac{1}{t} \cdot 3y$$</p>
<p>$$3y = \dfrac{1}{t} \cdot y'$$</p>
<p>$$ \dfrac{3}{2} y^2 = \ln(t)$$</p>
<p>Here I stopped because I noticed the answer had to be $c \cdot t^3$. I don't understand how to get that answer.</p>
| Did | 6,179 | <p>$$\left(\frac{y}{t^\color{red}{\alpha}}\right)'=\frac{y'}{t^\color{red}{\alpha}}-\frac{\color{red}{\alpha}}{t^{\color{red}{\alpha}+1}}y=\frac1{t^\color{red}{\alpha}}\left(y'-\frac{\color{red}{\alpha}y}t\right)$$</p>
|
1,495,885 | <p>I'm trying to integrate this but I think the function can't be integrated? Just wanted to check, and see if anyone is able to find the answer (I used integration by parts but it doesn't work). Thanks in advance; the function I need to integrate is $$\int\frac{x}{x^5+2}dx$$</p>
| Michael Hardy | 11,667 | <p>Probably you'd want to use the calculus of residues to do this.</p>
<p>But below I do it using first-year calculus methods.</p>
<p>The cumbersome part may be the algebra, and that's what I concentrate on here.
\begin{align}
x^5 + 2 & = \left( x+\sqrt[5]{2} \right) \underbrace{\left( x - \sqrt[5]{2} e^{i\pi/5}\right)\left( x - \sqrt[5]{2} e^{-i\pi/5}\right)}_\text{conjugates}\ \underbrace{\left( x - \sqrt[5]{2} e^{i3\pi/5}\right)\left( x - \sqrt[5]{2} e^{-i3\pi/5}\right)}_\text{conjugates} \\[10pt]
& = \left( x+\sqrt[5]{2} \right) \left( x^2 - 2 \sqrt[5]{2} \cos\frac{\pi} 5 + \sqrt[5]{2}^2 \right)\left( x^2 - 2\sqrt[5]{2}\cos\frac{3\pi}5 + \sqrt[5]{2}^2 \right)
\end{align}
Then use partial fractions.</p>
<p>That the polynomials $x^2 - 2\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2$ and $x^2 - 2\sqrt[5]{2}\cos\frac{3\pi} 5 + \sqrt[5]{2}^2$ cannot be factored using real numbers can be seen from the way we factored them above. So you'll have
$$
\cdots + \frac{Bx+C}{x^2 - 2\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2} + \cdots
$$
etc. and you'll need to find $B$ and $C$.</p>
<p>Let $u=x^2 - 2x\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2$ then $du = (2x + \sqrt[5]{2}\cos\frac{\pi}5)\,dx$ and so
$$
\frac{Bx + C}{x^2 - 2x\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2}\,dx = \underbrace{\frac{\frac B 2 \left( 2x+\sqrt[5]{2}\cos\frac{\pi}5 \right)}{x^2 - 2x\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2}\,dx}_\text{Use the substitution for this part.} + \underbrace{\frac{\left( 1 - \frac B2 \right) \sqrt[5]{2}\cos\frac{\pi} 5}{x^2 - 2x\sqrt[5]{2}\cos\frac{\pi} 5 + 1}\,dx}_\text{See below for this part.}
$$</p>
<p>To integrate $\frac{\left( 1 - \frac B2 \right) \sqrt[5]{2}\cos\frac{\pi} 5}{x^2 - 2x\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2}\,dx$, complete the square:</p>
<p>\begin{align}
& x^2 - 2x\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2 = \left( x^2 - 2x\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2\cos^2\frac{\pi}5 \right) + \sqrt[5]{2}^2 \sin^2\frac{\pi} 5 \\[10pt]
= {} & \left( x - \sqrt[5]{2}\cos\frac{\pi} 5 \right)^2 + \sqrt[5]{2}^2\sin^2 \frac{\pi} 5 \\[10pt]
= {} & \left(\sqrt[5]{2}^2 \sin^2 \frac{\pi} 5 \right) \left( \left( \frac{x-\sqrt[5]{2}\cos\frac{\pi} 5}{\sqrt[5]{2}\sin\frac{\pi} 5} \right)^2 + 1 \right) = (\text{constant})\cdot(w^2 + 1) \\[10pt]
& \text{and } dw = \frac{dx}{\sqrt[5]{2}\sin\frac{\pi}5}.
\end{align}
So you get an arctangent from this term.</p>
|
3,080,566 | <p>please tell me how I can solve the following equation. </p>
<p><span class="math-container">$$z^3+\frac{(\sqrt2+\sqrt2i)^7}{i^{11}(-6+2\sqrt3i)^{13}}=0$$</span></p>
<p>What formula should I use? If possible, tell me how to solve this equation or write where I can find a formula for solving such an equation. I searched for it on the Internet, but could not find anything useful.</p>
| Bernard | 202,857 | <p><strong>Hint</strong></p>
<p>You have to use the exponential form of complex numbers: set <span class="math-container">$Z=r\mathrm e^{i\theta}$</span> and observe that, for instance,
<span class="math-container">$$\sqrt 2+\sqrt 2i=\sqrt 2(1+i)=\sqrt 2\cdot \sqrt2\mathrm e^{i\tfrac\pi 4}=2\,\mathrm e^{i\tfrac\pi 4}, \;\text{&c.}$$</span>
so that <span class="math-container">$\;(\sqrt 2+\sqrt 2i)^7=2^7 \mathrm e^{i\tfrac{7\pi}4}$</span>. </p>
<p>Can you proceed similarly for the denominator?</p>
|
4,371,244 | <blockquote>
<p>Given an independent set of vectors <span class="math-container">$\left\{v_1, v_2, v_3,v_4\right\}$</span> , then is <span class="math-container">$\left\{v_1+v_2+v_3, v_1+2v_2, 2v_3-v_2\right\}$</span> dependent or independent?</p>
</blockquote>
<p>I've tried this:
<span class="math-container">$$\left\{v_1+v_2+v_3, v_1+2v_2, 2v_3-v_2\right\}$$</span></p>
<p><span class="math-container">$$v_1+v_2+v_3+v_1+2v_2+2v_3-v_2=0$$</span></p>
<p><span class="math-container">$$2v_1+2v_2+3v_3=0$$</span>
The book says it's independent, but how can that be? If <span class="math-container">$3v_3=-2v_2-2v_1$</span>.
Please any help is welcome.</p>
| amitava | 566,278 | <p>Let <span class="math-container">$x(v_1+v_2+v_3)+y(v_1+2v_2)+z(2v_3-v_2) = 0$</span></p>
<p>we get
<span class="math-container">$(x+y)v_1+(x+2y-z)v_2+(x+2z)v_3=0$</span>, so</p>
<p><span class="math-container">$x+y = 0,x+2y-z=0,x+2z=0$</span></p>
<p>When we solve this we get <span class="math-container">$x=0,y=0,z=0$</span>
So they are independent.</p>
|
4,371,244 | <blockquote>
<p>Given an independent set of vectors <span class="math-container">$\left\{v_1, v_2, v_3,v_4\right\}$</span> , then is <span class="math-container">$\left\{v_1+v_2+v_3, v_1+2v_2, 2v_3-v_2\right\}$</span> dependent or independent?</p>
</blockquote>
<p>I've tried this:
<span class="math-container">$$\left\{v_1+v_2+v_3, v_1+2v_2, 2v_3-v_2\right\}$$</span></p>
<p><span class="math-container">$$v_1+v_2+v_3+v_1+2v_2+2v_3-v_2=0$$</span></p>
<p><span class="math-container">$$2v_1+2v_2+3v_3=0$$</span>
The book says it's independent, but how can that be? If <span class="math-container">$3v_3=-2v_2-2v_1$</span>.
Please any help is welcome.</p>
| Leox | 97,339 | <p>The rank of the matrix
<span class="math-container">$$
\left[ \begin {array}{cccc} 1&1&1&0\\ 1&2&0&0
\\ 0&-1&2&0\end {array} \right]
$$</span>
is <span class="math-container">$3$</span>, so the vectors are independent.</p>
|
3,113,850 | <blockquote>
<p><span class="math-container">$$f(x)=\sum_{n=1}^{\infty}{\frac{x^{n-1}}{n}},$$</span></p>
<p>Prove that <span class="math-container">$f(x)+f(1-x)+\log(x)\log(1-x)=\frac{{\pi}^2}{6}$</span></p>
</blockquote>
<p>In my mind though,I think that this is related to Basel problem<span class="math-container">$\left(\sum\limits_{n=1}^{\infty}{\frac{1}{n^2}}\right)$</span>,but I don't know how to solve this.</p>
<p>Any help would be greatly appreciated :-)</p>
<p>Edit:</p>
<p>My attempt:</p>
<p>I cannot use latex expertly,so I post image.The circled part<a href="https://i.stack.imgur.com/gFDdf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gFDdf.jpg" alt="enter image description here " /></a></p>
| Arthur | 15,500 | <p>If <span class="math-container">$A$</span> is an <span class="math-container">$n\times n$</span> matrix with <span class="math-container">$n>2$</span>, and all entries are equal, then the adjoint (adjugate) matrix will be zero.</p>
<p>A matrix with a <span class="math-container">$0$</span> adjugate matrix cannot have an inverse, as <span class="math-container">$A\times\operatorname{adj}(A) = \det(A)\cdot I$</span>. And if this product turns out to be <span class="math-container">$0$</span>, then that means that the determinant of <span class="math-container">$A$</span> is zero, which implies that <span class="math-container">$A$</span> is not invertible.</p>
|
3,161,874 | <p>Find the vector equation of the plane which contains points <span class="math-container">$A(0,1,1)$</span>, <span class="math-container">$B(-1,2,1)$</span> and <span class="math-container">$C(2,0,2)$</span>. </p>
<p>I did this by first finding <span class="math-container">$AB$</span> and <span class="math-container">$AC$</span> where I got <span class="math-container">$AB=(-1,1,0)$</span> and <span class="math-container">$AC=(2,-1,1)$</span>. Then I did cross product with this and I got <span class="math-container">$i+j-k$</span>. When I then used this to find equation I got <span class="math-container">$x+y-z=0$</span> but I am unsure why this is wrong. Can someone please help? </p>
| Arthur | 15,500 | <p>It has made infinitely many trips. Every trip will be shorter than the last, but the fly will always reach the wall before the car, so it will always have room for one more trip. And one more. And one more.</p>
<p>The total length the fly flies is 50km, as the car crashes into the wall exactly 30 minutes after the whole experiment started.</p>
|
4,627,133 | <p>We have <span class="math-container">$A$</span> <span class="math-container">$(3×3)$</span> matrix with real entries. We know that A is orthogonal and <span class="math-container">$\operatorname{trace}(A)>1$</span>. Show that matrix <span class="math-container">$A+I_{3}$</span> is invertible.</p>
<p>We can see that <span class="math-container">$\det(A)=1$</span> or <span class="math-container">$\det(A)=-1$</span>. We can easily find <span class="math-container">$\operatorname{trace}(A^{*})=\det(A)\operatorname{trace}(A)$</span>. Suppose <span class="math-container">$\det(A+I_{3})=0$</span>. If we take the characteristic polynomial of A
<span class="math-container">$$
P(x)=-\det(A-xI_{3})=-x^{3}+\operatorname{trace}(A)x^{2}-\operatorname{trace}(A^*)x+\det(A)
$$</span>
we can find that <span class="math-container">$P(-1)=0$</span> so <span class="math-container">$1+\operatorname{trace}(A)+\det(A) \operatorname{trace}(A)+\det(A)=0$</span>. If <span class="math-container">$\det(A)=1$</span> we get easily a contradiction, but in the case where <span class="math-container">$\det(A)=-1$</span> we get something right. I tried using eigen values to get in a contradiction with the fact that <span class="math-container">$ \operatorname{trace}(A)>1$</span>, but nothing.</p>
| Ben Grossmann | 81,360 | <p>One quick approach using eigenvalues: suppose that <span class="math-container">$A + I$</span> is not invertible. It follows that <span class="math-container">$A$</span> has <span class="math-container">$-1$</span> as an eigenvalue. On the other hand, because <span class="math-container">$A$</span> is orthogonal, all eigenvalues of <span class="math-container">$A$</span> have absolute value <span class="math-container">$1$</span>. If <span class="math-container">$A$</span> has eigenvalues <span class="math-container">$\lambda_1,\lambda_2,\lambda_3 = -1$</span>, then
<span class="math-container">\begin{align}
\operatorname{trace}(A) &= \lambda_1 + \lambda_2 - 1
\leq |\lambda_1 + \lambda_2| - 1
\\ & \leq |\lambda_1 | + |\lambda_2| - 1 = 1 + 1 - 1 = 1,
\end{align}</span>
contradicting the premise that <span class="math-container">$\operatorname{trace}(A) > 1$</span>.</p>
|
87,319 | <p>How might I show that there's no metric on the space of measurable functions on $([0,1],\mathrm{Lebesgue})$ such that a sequence of functions converges a.e. iff the sequence converges in the metric?</p>
| Davide Giraudo | 9,849 | <p>In fact, the almost converge doesn't correspond to any topology. We use the following fact: </p>
<blockquote>
<p>Let $(X,\mathcal T)$ a topological space. A sequence $\{x_n\}$ converges to $x$ on $(X,\mathcal T)$ if and only if for all subsequence $\{x_{n_k}\}$ we can extract a converging subsequence to $x$. </p>
</blockquote>
<p>Consider a sequence $\{X_n\}$ of random variables which converges in probability but not almost surely to $X$. For each subsequence of $\{X_n\}$, we can extract an almost everywhere converging subsequence, which yield a contradiction.</p>
<p>But there is a metric for the convergence in probability, namely
$$\delta(X,Y):= \int_{ \Omega}\dfrac{|X(\omega)-Y(\omega)|}{1+|X(\omega)-Y(\omega)|}d\,\mathbb P(\omega).$$</p>
|
301,106 | <p>In the game connect four with a $7 \times 6$ grid like in the image below, how many game situations can occur?</p>
<p><strong>Rules</strong>:</p>
<blockquote>
<p>Connect Four [...] is a two-player game in which the players first
choose a color and then take turns dropping colored discs from the top
into a seven-column, six-row vertically-suspended grid. The pieces
fall straight down, occupying the next available space within the
column. The object of the game is to connect four of one's own discs
of the same color next to each other vertically, horizontally, or
diagonally before your opponent.</p>
</blockquote>
<p><sup>Source: <a href="http://en.wikipedia.org/wiki/Connect_four" rel="noreferrer">Wikipedia</a></sup></p>
<p><img src="https://i.stack.imgur.com/pMjBU.gif" alt="Source: Wikipedia commons, File:Connect_Four.gif"></p>
<p><sup>Image source: <a href="http://commons.wikimedia.org/wiki/File:Connect_Four.gif" rel="noreferrer">http://commons.wikimedia.org/wiki/File:Connect_Four.gif</a></sup></p>
<p><strong>Lower bound</strong>:</p>
<p>$7 \cdot 6 = 42$, as it is possible to make the grid full without winning</p>
<p><strong>Upper bound</strong>:</p>
<p>Every field of the grid can have three states: Empty, red or yellow disc. Hence, we can have $3^{7 \cdot 6} = 3^{42} = 109418989131512359209 < 1.1 \cdot 10^{20}$ game situations at maximum. </p>
<p>There are not that much less than that, because you can't have four yellows in a row at the bottom, which makes $3^{7 \cdot 6 - 4} = 1350851717672992089$ situations impossible. This means a better upper bound is $108068137413839367120$</p>
<p>How many situations are there?</p>
<p>I think it might be possible to calculate this with the approach to subtract all impossible combinations. So I could try to find all possible combinations to place four in a row / column / vertically. But I guess there would be many combinations more than once.</p>
| Ivan Loh | 61,044 | <p>The number of possible Connect-Four game situations after $n$ plies ($n$ turns) is tabulated at <a href="https://oeis.org/A212693">OEISA212693</a>. The total is 4531985219092. More in-depth explanation can be found at the links provided by the OEIS site. (E.g. <a href="http://homepages.cwi.nl/~tromp/c4/c4.html">John's Connect Four Playground</a>)</p>
|
301,106 | <p>In the game connect four with a $7 \times 6$ grid like in the image below, how many game situations can occur?</p>
<p><strong>Rules</strong>:</p>
<blockquote>
<p>Connect Four [...] is a two-player game in which the players first
choose a color and then take turns dropping colored discs from the top
into a seven-column, six-row vertically-suspended grid. The pieces
fall straight down, occupying the next available space within the
column. The object of the game is to connect four of one's own discs
of the same color next to each other vertically, horizontally, or
diagonally before your opponent.</p>
</blockquote>
<p><sup>Source: <a href="http://en.wikipedia.org/wiki/Connect_four" rel="noreferrer">Wikipedia</a></sup></p>
<p><img src="https://i.stack.imgur.com/pMjBU.gif" alt="Source: Wikipedia commons, File:Connect_Four.gif"></p>
<p><sup>Image source: <a href="http://commons.wikimedia.org/wiki/File:Connect_Four.gif" rel="noreferrer">http://commons.wikimedia.org/wiki/File:Connect_Four.gif</a></sup></p>
<p><strong>Lower bound</strong>:</p>
<p>$7 \cdot 6 = 42$, as it is possible to make the grid full without winning</p>
<p><strong>Upper bound</strong>:</p>
<p>Every field of the grid can have three states: Empty, red or yellow disc. Hence, we can have $3^{7 \cdot 6} = 3^{42} = 109418989131512359209 < 1.1 \cdot 10^{20}$ game situations at maximum. </p>
<p>There are not that much less than that, because you can't have four yellows in a row at the bottom, which makes $3^{7 \cdot 6 - 4} = 1350851717672992089$ situations impossible. This means a better upper bound is $108068137413839367120$</p>
<p>How many situations are there?</p>
<p>I think it might be possible to calculate this with the approach to subtract all impossible combinations. So I could try to find all possible combinations to place four in a row / column / vertically. But I guess there would be many combinations more than once.</p>
| Kieleth | 379,373 | <p>I think that:</p>
<p>"Every field of the grid can have three states: Empty, red or yellow disc. Hence, we can have (...) game situations at maximum." </p>
<p>is not correct.</p>
<p>The reason is that by just capturing that each cell can have 3 states, we´re allowing "floating" discs in the board, and the game rules (and physics :) restrict the discs to be stacked.</p>
<p>That is, the "empty" state must be always "filling" any number of discs in the column, and every disk must reside upon another disk, besides the zero row.</p>
<p>So, the "Upper Bound" as defined is lower by a good chunk (<a href="https://oeis.org/A212693/b212693.txt" rel="nofollow">https://oeis.org/A212693/b212693.txt</a> seems to be like a good answer)...</p>
|
203,212 | <p>Q1: When a question asking me to show a set consists of countably many point, do it mean infinite countably many or both infinite or finite case. </p>
<p>Q2:Also, to show that a set consist of countably many point, is it enough to show that the set is countable or i have to show somethings more?</p>
| André Nicolas | 6,312 | <p>One of the official definitions of countable set is a set $A$ such that there is a bijection between $A$ and a <em>subset</em> of the natural numbers. That definition includes the finite sets.</p>
<p>If a set is countable but not finite, it should be called <em>countably infinite</em>.</p>
<p>But people often sloppily write "countable" when in principle they should write "countably infinite."
So if the phrase is in a problem on an assignment, you may have to ask the instructor. However, the answer may be clear from the context. And if, for example, you have shown that a set is in one to one correspondence with the natural numbers, you will have shown that it is countable. For you will have proved the stronger result that the set is countably infinite.</p>
<p>As to the second question, "countable" and having "countably many points" mean the same thing.</p>
|
11,882 | <p>Prove that $\lim\limits_{x \to 2} \frac{x^{2}-2x+9}{x+1}$ using an epsilon delta proof.</p>
<p>So I have most of the work done. I choose $\delta = min{\frac{1}{2}, y}$,<br>
$f(x)$ factors out to $\frac{|x-3||x-2|}{|x+1|}$
But $|x-3| \lt \frac{3}{2}$ for $\delta = \frac{1}{2}$ and also $|x+1| > 5/2$ (I'll spare you the details). </p>
<p>I'm not sure how to choose my y here. If I take $\lim\limits_{x \to 2} \frac{x^{2}-2x+9}{x+1}$ < $(3/5) \delta$ How do I choose my epsilon here (replace y with this) to satisfy this properly?</p>
<p>Thanks</p>
| Arturo Magidin | 742 | <p><strong>First:</strong> You <strong>don't</strong> choose $\epsilon$! </p>
<p>In an $\epsilon$-$\delta$ proof, you begin by taking an <em>arbitrary</em> $\epsilon$; you then need to show that it is possible to choose a $\delta$ that "works" for <em>that</em> $\epsilon$. So you are going at things <em>exactly</em> backwards, by first trying to work with $\delta$.</p>
<p><strong>Second:</strong> You cannot "prove $\lim\limits_{x\to 2}\frac{x^2-2x+9}{x+1}$" using $\epsilon$-$\delta$ proofs. What you <em>can</em> try to prove is that the limit is equal to <em>something</em> (in this case, to $3$). </p>
<p><strong>Third:</strong> $f(x)$ does not factor as $\frac{|x-2||x-3|}{|x+1|}$; what factors that way is $|f(x)-3|$. </p>
<p>So. Let's start from scratch, shall we?</p>
<p>An $\epsilon$-$\delta$ proof that $\lim\limits_{x\to 2}\frac{x^2-2x+9}{x+1}=3$ would work as follows: given some arbitrary $\epsilon\gt 0$, we must find a $\delta\gt 0$ (which may depend on $\epsilon$), with the property that if $0\lt |x-2|\lt \delta$, then we will have $\left|\frac{x^2-2x+9}{x+1} - 3\right|\lt \epsilon$. </p>
<p>So, let $\epsilon\gt 0$ be given. You want to make sure that
$$\left|\frac{x^2-2x+9}{x+1} - 3\right| = \left|\frac{x^2-2x+9-3x-3}{x+1}\right| = \left|\frac{x^2-5x+6}{x+1}\right| = \left|\frac{(x-3)(x-2)}{x+1}\right|$$
is small, by making sure that $|x-2|$ is small (that is, that $0\lt |x-2|\lt\delta$ for some chosen $\delta$). </p>
<p>Clearly, you also want to make sure that $\delta\lt \frac{|x+1|}{|x-3|}\epsilon$, because then we will have:
$$\left|\frac{x^2-x+0}{x+1}-3\right| = \frac{|x-3||x-2|}{|x+1|} \lt \delta\frac{|x-3|}{|x+1|} \lt \frac{|x+1|\epsilon}{|x-3|}\frac{|x-3|}{|x+1|} = \epsilon.$$
</p>
<p>You can make sure that $\frac{1}{|x+1|}$ is no larger than some constant by making sure that $x$ is close enough to $2$. You can do the same thing for $|x-3|$. So <strong>then</strong> you can pick a $\delta$ that is simultaneously small enough to ensure that $\frac{1}{|x+1|}$ is smaller than some $C$, that $|x-3|$ is smaller than some $D$, <em>and</em> that $|x-2|$ is smaller than $CD\epsilon$. Try that.</p>
|
1,105,056 | <p>There's something I've never understood about polynomials.</p>
<p>Suppose $p(x) \in \mathbb{R}[x]$ is a real polynomial. Then obviously,</p>
<p>$$(x-a) \mid p(x)\, \longrightarrow\, p(a) = 0.$$</p>
<p>The converse of this statement was used throughout high school, but I never really understood why it was true. I think <em>maybe</em> a proof was given in 3rd year university algebra, but obviously it went over my head at the time. So anyway:</p>
<blockquote>
<p><strong>Question.</strong> Why does $p(a)=0$ imply $(x-a) \mid p(x)$?</p>
</blockquote>
<p>I'd especially appreciate an answer from a commutative algebra perspective.</p>
| fiverules | 102,465 | <p>if $(x-a)$ does not divide $p(x)$, then every linear factor of $p(x)$ is not equal to $x-a$, that is, $p(a)$ is product of nonzero (complex) numbers, therefore $p(a) \neq 0$</p>
|
1,105,056 | <p>There's something I've never understood about polynomials.</p>
<p>Suppose $p(x) \in \mathbb{R}[x]$ is a real polynomial. Then obviously,</p>
<p>$$(x-a) \mid p(x)\, \longrightarrow\, p(a) = 0.$$</p>
<p>The converse of this statement was used throughout high school, but I never really understood why it was true. I think <em>maybe</em> a proof was given in 3rd year university algebra, but obviously it went over my head at the time. So anyway:</p>
<blockquote>
<p><strong>Question.</strong> Why does $p(a)=0$ imply $(x-a) \mid p(x)$?</p>
</blockquote>
<p>I'd especially appreciate an answer from a commutative algebra perspective.</p>
| Mark Bennet | 2,906 | <p>The proof using the division algorithm is a good one. Here is a more direct approach.</p>
<p>Note that $x^r-a^r=(x-a)(x^{r-1}+ax^{r-2}+a^2x^{r-3} \dots +a^{r-1})=(x-a)p_r(x)$ where $(x-a)$ is a factor.</p>
<p>Let $p(x)=c_rx^r+c_{r-1}x^{r-1}+\dots +c_0$. We are given $p(a)=0$ so $$p(x)=p(x)-p(a)=c_r(x^r-a^r)+c_{r-1}(x^{r-1}-a^{r-1})+\dots+c_0-c_0=(x-a)\left(c_rp_r(x)+c_{r-1}p_{r-1}(x)+\dots+c_1\right)$$</p>
|
2,755,785 | <p>I recently came across a problem in which I had to show that a discrete random variable $X$ has two modes $m_1$ and $m_2$. The information given was that $$\frac{P(X=n)}{P(X=n-1)} = \frac{0.9(n-1)}{n-3}$$</p>
<p>I need to show that this distribution has two modes, $m_1$ and $m_2$. My initial thoughts were to try and find the probability distribution, but could not do so from the data above. I don't really know how to approach this problem so any inputs would be appreciated.</p>
| The Integrator | 538,397 | <p>The given polynomial is $n^3+pn^2+qn+p=0$ $\qquad$</p>
<p>NOTE: im using $n$ as the variable to avoid confusion with it being the root</p>
<p><a href="https://en.wikipedia.org/wiki/Vieta's_formulas" rel="nofollow noreferrer">Vieta's Formula </a> gives us that $x+y+z= -p$</p>
<p>$ xy+yz+zx = q$</p>
<p>$xyz = -p$</p>
<p>$\tan^{-1}(x)+\tan^{-1}(y)+\tan^{-1}(z) = \tan^{-1}\bigg( \frac{x+y+z -xyz}{1-(xy+yz+zx)}\bigg)$</p>
<p>$\tan^{-1}(x)+\tan^{-1}(y)+\tan^{-1}(z) = \tan^{-1}\bigg( \frac{-p+p}{1-q}\bigg)$</p>
<p>$\tan^{-1}(x)+\tan^{-1}(y)+\tan^{-1}(z) = \tan^{-1}(0)$</p>
<p>$\implies\tan^{-1}(x)+\tan^{-1}(y)+\tan^{-1}(z) = n\pi$ $\qquad ,n\in Z$</p>
<p>Notice that the above is not defined when $1-q = 0\implies q =1$</p>
|
1,613,857 | <p>I'm reading <a href="https://en.wikipedia.org/wiki/The_Music_of_the_Primes" rel="nofollow">The Music of the Primes</a> by du Sautoy and I've come across a section that I'm having difficulty understanding:</p>
<blockquote>
<p>Euler fed imaginary numbers into the function $2^x$. To his surprise, out came waves which corresponded to a particular musical note. Euler showed that the character of each note depended on the coordinates of the corresponding imaginary number. The farther north one is, the higher the pitch. The farther east, the louder the volume.</p>
</blockquote>
<p>My understanding here is that the results are dependent on the sine function and that the real part of the exponent affects the amplitude and the imaginary part of the exponent affects the frequency.</p>
<p>I'd like to understand this more intuitively, which I tend to get through visualization. So I went to Wolfram Alpha and started with graphing $2^{x+iy}$. That wasn't very helpful.</p>
<p><a href="http://www.wolframalpha.com/input/?i=2%5E%28x%2Bi*y%29%20where%20x%20%3D%2010" rel="nofollow">So I tried graphing it with fixed $x$ values</a>, and indeed, I could see the amplitude of the (now 2D) graph <a href="http://www.wolframalpha.com/input/?i=2%5E%28x%2Bi*y%29%20where%20x%20%3D%20100" rel="nofollow">changing</a>. </p>
<p>I also see that $2^{x+iy}$ is also expressed as $2^x \cos(y \log(2))+i 2^x \sin(y \log(2))$ and I think I can see that changing the value of $x$ would affect the amplitude.</p>
<p>I'm unable to demonstrate the frequency changing by setting y to specific values. </p>
<p>What am I missing? (...Other than a semester in a Complex Analysis class!)</p>
<p>edit:</p>
<p>So while reading more online, I came across <a href="http://blog.echen.me/2011/03/14/prime-numbers-and-the-riemann-zeta-function/" rel="nofollow">this blog</a> that makes a similar claim. I suspect the book of oversimplifying, but wonder if this explains what was simplified?</p>
<blockquote>
<p>[...] But $x^{z-1} + x^{\bar{z} - 1}$ is just a wave whose amplitude depends on the real part of $z$ and whose frequency depends on the imaginary part (i.e., if $z=a+biz=a+bi$, then $x^{z-1} + x^{\bar{z}-1} = 2x^{a-1} cos (b \log x)$) [...]</p>
</blockquote>
<p>(I copied this from the blog, but removed some odd \'s ...)</p>
<p>Is it the inclusion of the conjugates that causes this amplitude/frequency?</p>
| SolUmbrae | 303,202 | <p>I don't have much to say but I had a complex analysis class and as far as I can tell that wouldn't be of any help.</p>
<p>To understand $2^x \cdot (\cos(y\ln 2) + i \sin(y\ln 2)$ as a pure-note-wave, which is of the form $A\cdot(\cos(\omega \cdot t + \phi_0)$ with so-called amplitude $A$, frequency $\omega$, phase-offset $\phi_0$ (these are just parameters) and time $t$ (which is your independent variable), $y$ would have to be seen as the "time", which is what you are doing when you fix values of $x$ (fixing $A$) and look at the 2D-graph with $y$ on the horizontal axis (if you look at the real part you get consider $\phi_0=0$, for example).</p>
<p>So if $y$ is the "time", then the frequency is always $\omega = \ln 2$, so the pitch is always the same (determined by the exponential base).</p>
<p>I think the book should say "If you go from south to north, you observe (in the real part) a single note wave, whose pitch is always the same and whose amplitude is higher, the further east you chose your path."</p>
|
199,235 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/9505/xy-yx-for-integers-x-and-y">$x^y = y^x$ for integers $x$ and $y$</a> </p>
</blockquote>
<p>Determine the number of solutions of the equation
$n^m = m^n$
where both m and n are integers.</p>
| Lubin | 17,760 | <p>This is a nice problem for a calculus class, to describe all pairs $(x,\xi)$ of positive real numbers with $x\ne\xi$ and $x^\xi=\xi^x$. From $\xi\log x=x\log\xi$ you get $(\log x)/x=(\log\xi)/\xi$, in other words, you’re looking for horizontal lines that intersect the graph of $f(x)=(\log x)/x$ twice. Since the function is defined and differentiable on $\langle0,\infty\rangle$ with a single maximum, all you need to do is spot where that maximum happens, and by differentiating, you see that it’s at $x=e$. Using the fact that $f(1)=0$, you see that for any $x$ in $\langle1,e\rangle$, there’s a unique $\xi>e$ for which $x^\xi=\xi^x$. And of course you see that there’s only one integer in the open interval $\langle1,e\rangle$.</p>
|
4,482,600 | <p>Prove
<span class="math-container">$$\int_{0}^{\frac{\pi}{2}} \sqrt[n]{\tan x} \,dx = \frac{\pi}{2} \sec \left(\frac{\pi}{2n}\right)$$</span></p>
<p>for all natural numbers <span class="math-container">$n \ge 2$</span>.</p>
<p>There are several answers (<a href="https://math.stackexchange.com/questions/4385116/how-to-evaluate-the-integral-int-0-frac-pi2-sqrtn-tan-theta-d-th?rq=1">A1</a> <a href="https://math.stackexchange.com/questions/1913325/real-analysis-methods-to-evaluate-int-0-infty-fracxa1x2-dx-a1/1913572#1913572">A2</a>) to this integral but they all involve the gamma function or the beta function or contour integration etc. Can one solve this using only 'real' 'elementary' techniques? For <span class="math-container">$n = 2$</span> and <span class="math-container">$n = 3$</span> it can be solved using only elementary substitutions and partial fractions.</p>
| Lai | 732,917 | <p><span class="math-container">\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \sqrt[n]{\tan x} d x&=\int_{0}^{\frac{\pi}{2}} \sin ^{\frac{1}{n}} x \cos ^{-\frac{1}{n}} x d x\\
&= \int_{0}^{\frac{\pi}{2}} \sin ^{2\left(\frac{1}{2 n}+\frac{1}{2}\right)-1} x \cos^ {2\left(-\frac{1}{2n}+\frac{1}{2}\right)-1 }x d x\\
&=\frac{1}{2} B\left(\frac{1}{2 n}+\frac{1}{2},-\frac{1}{2 n}+\frac{1}{2}\right)\\
&=\frac{1}{2} \pi \csc \left[\pi\left(\frac{1}{2 n}+\frac{1}{2}\right)\right]\\
&=\frac{\pi}{2} \sec \frac{\pi}{2 n}
\end{aligned}</span>
where the second last line using the property of Beta function:</p>
<p><span class="math-container">$$B(x, 1-x)=\pi \csc (\pi x) \quad x \notin \mathbb{Z}.$$</span></p>
|
77,485 | <p>Find the smallest integer $n$ such that $$\left(1-\frac{n}{365}\right)^n < \frac{1}{2}.$$</p>
<p>I cannot use a calculator, and I do not know where to begin.</p>
| Did | 6,179 | <p>To find an integer $n$ such that this holds, rewrite the inequality as $1-(n/365)<\exp(-(\ln2)/n)$ and use the fact that $\exp(-x)>1-x$ for every $x$. Then $n$ will do as soon as
$1-(n/365)<1-(\ln2/n)$, that is, $n^2>365\cdot\ln2$. Since $\ln2<.7$, one knows that $365\cdot\ln2<365\cdot.7=255.5<256=2^8$, hence every $n\geqslant2^4=16$ will do.</p>
<p>The numerical values the reasoning above requires to know to be performed without a calculator are the fact that $\ln2$ is (just) below $.7$ and the first powers of $2$.</p>
<p>It happens that the inequality does not hold for $n=15$ hence $n=16$ is the correct answer but at the moment I do not know how to prove this part without a calculator, except using the (alternating) expansion of the exponential at the second order to lower bound it. This is cumbersome, but here we go.</p>
<p>Since $\exp(-x)<1-x+\frac12x^2$ for every nonnegative $x$, it is enough to check that for $n=15$, $n^2<365\cdot\ln2\cdot(1-\ln2/n)$, which is true if $\ln2>\frac{n}2\left(1-\sqrt{1-\frac{4n}{365}}\right)$. Using $\sqrt{1+x}<1+\frac12x$ for $x=\frac{4n}{365-4n}$ yields $\sqrt{1-\frac{4n}{365}}>\frac{365-4n}{365-2n}$. Hence $\ln2>\frac{n^2}{365-2n}$ is enough, that is, for $n=15$, $\ln2>\frac{45}{67}$. Since $\frac{45}{67}\approx.672$, this proves the thing if one knows that $\ln2$ is greater than $.68$.</p>
<p>The numerical value the reasoning above requires to know to be performed without a calculator is the fact that $\ln2$ is about $.69$.</p>
|
2,053,255 | <p>Let $f_{X,Y}(x,y)=\frac{1}{8}$ for $-2<x<2$, $0<y<2$. Find $f(z)$ where $Z=X+Y$. </p>
<p>I am having difficulty solving the above problem. My attempt at a solution relies on a convolution formula, which says that the $pdf$ of $Z=X+Y$, given the joint $pdf$ $f(x,y)$ is given by $$f_Z(z)= \int_{s+t=z} f(s,t)ds=\int_{-\infty}^{\infty} f(s,z-s)ds$$. So first note that $f(s,z-s)=\frac{1}{8}$ if $-2<s<2$ and $0<z-s<2$. Then note that $0<z-s<2$ is the same as $z-2<s<z$. But these regions depend on the value of $z$. Hence,
$$\begin{equation}
f_Z(z)=
\begin{cases}
\int_{-2}^{z} \frac{1}{8}ds, & \text{if}\ 0\leq z \leq2\\
\int_{z-2}^{2} \frac{1}{8}ds, & \text{if}\ 2<z\leq4
\end{cases}
\end{equation}$$. After evaluating the integrals, I get </p>
<p>$$\begin{equation}
f_Z(z)=
\begin{cases}
\frac{z-2}{8}, & \text{if}\ 0\leq z \leq2\\
\frac{4-z}{8}, & \text{if}\ 2<z\leq4
\end{cases}
\end{equation}$$.</p>
<p>I have solved similar problems like this one but all of them have been defined over the intervals $0<x<1$ and $0<y<1$. In this problem, however, we have the intervals $(-2,2)$ for $x$ and $(0,2)$ for $y$.I have graphed these regions but I have no intuition on how to divide it. Any help would be appreciated. Thanks. </p>
| BruceET | 221,800 | <p>The joint distribution of $(X,Y)$ is uniform on a rectangle with
corners at $(-2,0)$ and $(2,2),$ and with area 8. So just by geometry, you should
be able to find the CDF of $S = X + Y.$ For example, draw the rectangle
and the line $x + y = 1.5.$ Then what is the value of $F_S(s) = P(S \le s = 1.5)?$
(Consider three cases: $s < 0,\, 0 < s < 2,\, s > 2.$)</p>
<p>Here is a simulation in R statistical software of the distribution of $S$
based on a million realizations. The histogram suggests the shape of the PDF
and the ECDF (empirical CDF) suggests the shape of the CDF. In one dimension, itegrate the CDF
over three separate regions to get the (piecewise) PDF. (The CDF is linear
between 0 and 2.)</p>
<pre><code>m = 10^6
x = runif(m, 0, 2); y = runif(m, -2, 2)
s = x + y
mean(s)
## 0.9980456 # aprx E(S) = 1
# plots
par(mfrow=c(1,2)) # 2 panels per figure
hist(s, prob=T, col="wheat", main="Simulated Distribution of S = X + Y")
plot(ecdf(s)); abline(v=c(0,2), col="green")
par(mfrow=c(1,1)) # returns to default single-panel plotting
</code></pre>
<p><a href="https://i.stack.imgur.com/exW37.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/exW37.png" alt="enter image description here"></a></p>
|
3,672,948 | <p>Consider the series
<span class="math-container">$$
S = \sum_{n=1}^\infty e^{-n^2x}
$$</span>
then I have to argue for that <span class="math-container">$S$</span> is convergent if and only if <span class="math-container">$x>0$</span>.</p>
<p>As this is if and only if I think I have to assume first that S is convergent and show that this implies that <span class="math-container">$x>0$</span> but I am not sure how to. It is easy for me see that if <span class="math-container">$x=0$</span> the series is divergent but if I were to assume that S is convergent and that for a contradiction that <span class="math-container">$x\leq 0$</span> how do I proceed? And how the other way around? </p>
<p>Do you mind helping me? </p>
| Gary | 83,800 | <p>If <span class="math-container">$x>0$</span>, then (since <span class="math-container">$0<e^{-x}<1$</span>)
<span class="math-container">$$
\sum\limits_{n = 1}^\infty {e^{ - n^2 x} } \le \sum\limits_{n = 1}^\infty {e^{ - nx} } = \frac{{e^{ - x} }}{{1 - e^{ - x} }} = \frac{1}{{e^x - 1}} < + \infty .
$$</span>
If <span class="math-container">$x\leq 0$</span>, the terms do not tend to zero, whence the series cannot converge.</p>
|
997,116 | <p>If one root of the equation $x^2 + ax + b = 0$ and $x^2 + bx + a = 0$ is common and $a \ne b$ then:</p>
<p>The options are as follows:
$$\begin{array}{ll}
(A)\quad& a + b = 0\\
(B)& a + b = -1\\
(C)& a - b = 1\\
(D)& a + b = 1
\end{array}$$</p>
<p>Idk how to solve this, please help me.</p>
| ajotatxe | 132,456 | <p>Let $c$ be the common root. Then
$$c^2+ac+b=0$$
$$c^2+bc+a=0$$
so $ac+b=bc+a$ or $(a-b)c=a-b$. Since $a\neq b$, $c$ must be $1$.</p>
<p>Now we have $a+b=-1$.</p>
|
1,618,952 | <p>How to prove this exercise?</p>
<p>Let $A$ be a $n \times n$ diagonal matrix with characterist polynominal</p>
<p>$$(x-c_1)^{d_1}...(x-c_k)^{d_k}$$</p>
<p>where $c_i$ are distinct. Let $V$ be the space of $n \times n$ matrixes $B$ such that $AB=BA$. Prove that dimension of $V$ is $d_{1}^{2}+...+d_{k}^{2}$.</p>
| Brian M. Scott | 12,042 | <p>HINT: Starting by writing out what the hypotheses mean is good. A good next step is to do the same kind of expansion of what you’re trying to prove: $\lim_{n\to\infty}\sum_{k=1}^nf(k)=a$ means that </p>
<blockquote>
<p>for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $$\left|a-\sum_{k=1}^nf(k)\right|<\epsilon\quad\text{whenever}\quad n\ge m_\epsilon\;.\tag{1}$$</p>
</blockquote>
<p>Thus, you want to start with an arbitrary $\epsilon>0$ and use the hypothesis that $\sum_{s\in S}f(s)=a$ to find a suitable $m_\epsilon$. That hypothesis tells you that there is a finite $T_\epsilon\subseteq S=\Bbb N$ such that </p>
<p>$$\left|a-\sum_{s\in T'}f(s)\right|<\epsilon\quad\text{whenever}\quad T'\text{ is finite and }T_\epsilon\subseteq T'\subseteq\Bbb N\;.\tag{2}$$</p>
<p>Somehow we should use $T_\epsilon$ to get $m_\epsilon$. </p>
<ul>
<li>Prove that we can replace $T_\epsilon$ with any finite subset of $\Bbb N$ that contains $T_\epsilon$, and $(2)$ will still be true. This is the key step in the argument.</li>
</ul>
<p>In particular, we can ‘fill in the holes’ in $T_\epsilon$ by replacing it with </p>
<p>$$\{k\in\Bbb N:k\le\max T_\epsilon\}=\{1,2,\ldots,\max T_\epsilon\}\;.$$</p>
<p>Can you see how to use this to find an $m_\epsilon$ making $(1)$ true?</p>
|
4,320,437 | <p>I was thinking about linear tramsformations and i came up with this example:
<span class="math-container">$$f:\mathbb{R}^n \to \mathbb{C}^n\\
f(x)=ix$$</span>
for this example, domain and co-domain are not defined over the same field and all linear transformations that i encountered by now had domain and co-domain defined over the same field. I was wondering that if this is a valid linear transformation or not? and if not, why did we put such a constraint?</p>
<p>also, if it is possible, keep the explanation simple because i'm pretty new in pure math. thank you in advance.</p>
| Daniel Wohlrath | 842,628 | <p><span class="math-container">$\mathbb{R}^n$</span> is by default just a set of points. The notation is often abbreviated when talking about vector spaces, so <span class="math-container">$\mathbb{R}^n$</span> can also be considered as a vector space (most often over the field of reals). Actually, ever field is a vector space over itself (with dimension <span class="math-container">$1$</span>).</p>
<p><span class="math-container">$\mathbb{C}^n$</span> is then also just a set of points. If the underlying field is taken to be real numbers, then we get a vector space of dimension <span class="math-container">$2n$</span>.</p>
<p>Actually, a linear transformation MUST ALWAYS BE mapping from a vector space <span class="math-container">$X$</span> to <span class="math-container">$Y$</span> which are over the same field. This is necessary because of the definition of a linear transformation. You should probably read it again in your book. Btw. welcome to "pure maths"</p>
|
743,465 | <p>Suppose $l,t\in[0,1]$ and $l+t\leq1$ I want to prove $1+l+t>6lt$. When $t=0$ or $l=0$, it is trivial, so I started with $l,t\neq0$ but I couldn't reach anywhere. I don't have time to write in detail what I have already tried, but I tried to manipulate $(l-t)^2$ mostly. Anyway, if anyone help me with the proof that would be great. Many thanks!</p>
| David | 119,775 | <p>If $l+t=a$ then $lt\le\frac{1}{4}a^2$ and we have
$$6lt-(l+t)\le{\textstyle\frac{3}{2}}a^2-a={\textstyle\frac{3}{2}}a(a-{\textstyle\frac{2}{3}})\ .$$
By sketching a graph it is easy to see that for $0\le a\le1$ the right hand side is at most $\frac{1}{2}$.</p>
|
743,465 | <p>Suppose $l,t\in[0,1]$ and $l+t\leq1$ I want to prove $1+l+t>6lt$. When $t=0$ or $l=0$, it is trivial, so I started with $l,t\neq0$ but I couldn't reach anywhere. I don't have time to write in detail what I have already tried, but I tried to manipulate $(l-t)^2$ mostly. Anyway, if anyone help me with the proof that would be great. Many thanks!</p>
| Macavity | 58,320 | <p>By AM-GM, $4lt \le (l+t)^2\le 1$. </p>
<p>Similarly, $2 lt \le 2\sqrt{lt} \le l+t$. Add these... and note equality is not possible for all the inequalities used simultaneously.</p>
|
1,973,724 | <p>Now this was the explanation on how to solve a prove in my book : </p>
<p>We will prove that the two sets complement (A∩B) and complement(A) ∪ complement(B) are equal by showing that each is a subset of the other. </p>
<p>First,we will show that complement (A∩B) ⊆ complement(A) ∪ complement(B). We do this by showing that if x is in complement (A∩B) ,then it must also be in complement(A) ∪ complement(B). Now suppose that x ∈ complement (A∩B).By the definition of complement,x ∈ complement (A∩B). Using the definition of intersection,we see that the proposition¬((x ∈ A)∧(x ∈ B))is true. By applying De Morgan’s law for propositions, we see that¬(x ∈ A) or¬(x ∈ B). Using the definition of negation of propositions, we have x ∈ A or x ∈ B. Using the definition of the complement of a set, we see that this implies that x ∈ A or x ∈ B. Consequently, by the definition of union, we see that x ∈ complement(A) ∪ complement(B) . We have now shown that complement (A∩B) ⊆ complement(A) ∪ complement(B). Next, we will show that complement(A) ∪ complement(B) ⊆ complement (A∩B).We do this by showing that if x is in complement(A) ∪ complement(B), then it must also be in A∩B.Now suppose that x ∈ A∪B. By the definition of union, we know that x ∈ A or x ∈ B.Using the definition of complement,we see that x ∈ A or x ∈ B.Consequently, the proposition¬(x ∈ A) ∨¬(x ∈ B)is true. By De Morgan’s law for propositions, we conclude that ¬((x ∈ A) ∧(x ∈ B)) is true. By the definition of intersection, it follows that ¬(x ∈ A∩B). We now use the definition of complement to conclude that x ∈ complement (A∩B). This shows that complement(A) ∪ complement(B) ⊆ complement (A∩B). Because we have shown that each set is a subset of the other,the two sets are equal,and the identity is proved. </p>
<p>Why do I have to negate the definition of an intersection to prove that complement (A∩B) = complement(A) ∪ complement(B)?
What does the definition of the intersection got to do with this?
Because this is my first time solving a proof. Am I just suppose to manipulate the equation complement (A∩B) and simplify it by definitions? </p>
| fleablood | 280,126 | <p>Because $(A\cap B)^c =\{a| a \not \in A\cap B\} $. Now to further understand and describe what it means to <em>not</em> be in the intersection it helps to know the definition of intersection. </p>
<p>$(A\cap B)^c =\{a| a \not \in A\cap B\}=\{a\not \in both A and B\}=\{a|\lnot(a\in A \land a \in B)\}=\{a|\lnot (a\in A)\lor\lnot (a\in B)\}=\{a|a\in A^c \lor a\in B^c\}=\{a|a\in A^c \cup B^c\}= A^c \cup B^c=$.</p>
|
2,006,927 | <p>I want to prove that $\cap_{i=1}^{\infty}\left(A_i \cap \left(\cup_{j=1}^{\infty}B_j\right)\right)=\cup_{j=1}^{\infty}\left(\left(\cap_{i=1}^{\infty}A_i\right) \cap B_j\right)$ for sets $A_i,B_j$ and natural numbers $i,j$. If an element $x$ belongs to the left hand side, then $x\in A_1$ and $x\in$ some of $B_j$ and $x\in A_2$ and $x\in$ some of $B_j$ and so forth. Then $x\in A_1$, $x\in A_2$, $x\in A_3$ etc so $x \in \cap_{i=1}^{\infty}A_i$ but I don't see how I can proceed with the B:s and get to $x \in B_1$ and $x\in$ all $A_i$ or $x \in B_2$ and $x\in$ all $A_i$ or $x \in B_3$ and $x\in$ all $A_i$ and so forth.</p>
| MarnixKlooster ReinstateMonica | 11,994 | <p>Here is a way to systematically calculate which elements $\;x\;$ are in both sides of this equality, just by expanding the definitions and simplifying.$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\Ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\when}{\Leftarrow}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$</p>
<p>For the left hand side, we calculate as follows:
$$\calc
x \in \bigcap_{i=1}^{\infty}\left(A_i \cap \left(\bigcup_{j=1}^{\infty}B_j\right)\right)
\op\equiv\hint{definition of $\;\bigcap\;$}
\langle \forall i : i \ge 1 : x \in A_i \cap \left(\bigcup_{j=1}^{\infty}B_j\right) \rangle
\op\equiv\hint{definition of $\;\cap\;$}
\langle \forall i : i \ge 1 : x \in A_i \;\land\; x \in \bigcup_{j=1}^{\infty}B_j \rangle
\op\equiv\hint{definition of $\;\bigcup\;$}
\langle \forall i : i \ge 1 : x \in A_i \;\land\; \langle \exists j : j \ge 1 : x \in B_j \rangle \rangle
\op\equiv\hints{logic: $\;{}\land\phi\;$ distributes over $\;\forall i\;$, for $\;\phi\;$ not containing $\;i\;$}\hint{-- to separate unrelated parts}
\langle \forall i : i \ge 1 : x \in A_i \rangle \;\land\; \langle \exists j : j \ge 1 : x \in B_j \rangle
\tag{*}
\endcalc$$</p>
<p>Now do the same with the right hand side, and observe that the result is also $\Ref{*}$.</p>
<p>In other words, both sides contain exactly the same $\;x\;$, and therefore by set extensionality they are equal.</p>
|
293,371 | <p>This is part of a homework assignment for a real analysis course taught out of "Baby Rudin." Just looking for a push in the right direction, not a full-blown solution. We are to suppose that $f(x)f(y)=f(x+y)$ for all real x and y, and that f is continuous and not zero. The first part of this question let me assume differentiability as well, and I was able to compose it with the natural log and take the derivative to prove that $f(x)=e^{cx}$ where c is a real constant. I'm having a little more trouble only assuming continuity; I'm currently trying to prove that f is differentiable at zero, and hence all real numbers. Is this an approach worth taking?</p>
| Community | -1 | <p>First note that $f(x) > 0$, for all $x \in \mathbb{R}$. This can be seen from the fact that
$$f(x) = f\left(\dfrac{x}2 + \dfrac{x}2\right) = f \left(\dfrac{x}2\right)^2$$ Further, you can eliminate the case $f(x) = 0$, since this would mean $f \equiv 0$.</p>
<p>One way to go about is as follows.</p>
<p>$1$. Prove that $f(m) = f(1)^m$ for $m \in \mathbb{Z}^+$.</p>
<p>$2$. Now prove that $f(m) = f(1)^m$ for $m \in \mathbb{Z}$.</p>
<p>$3$. Now prove that $f(p/q) = f(1)^{p/q}$ for $p \in \mathbb{Z}$ and $q \in \mathbb{Z} \backslash \{0\}$.</p>
<p>$4$. Now make use of the fact that rationals are dense in $\mathbb{R}$ and hence you can find a sequence of rationals $r_n \in \mathbb{R}$ such that $r_n \to r \in \mathbb{R} \backslash \mathbb{Q}$. Now use continuity to conclude that $f(x) = f(1)^x$ for all $x \in \mathbb{R}$. You will see that you need only continuity at one point to conclude that $f(x) = f(1)^x$.</p>
|
98,317 | <p>This comes from Artin Second Edition, page 219. Artin defined $G = \langle x,y\mid x^3, y^3, yxyxy\rangle$, and uses the Todd-Coxeter Algorithm to show that the subgroup $H = \langle y\rangle$ has index 1, and therefore $G = H$ is the cyclic group of order 3.</p>
<p>That being the case, $x$ cannot be either $y$ or $y^2$, for then the third relation would not be satisfied. So the relation $x=1$ must follow from the given relations. Is there another way of seeing this besides from the Todd-Coxeter algorithm?</p>
| djechlin | 79,767 | <p>I computed the powers of $z = xy$:</p>
<p>$xy$, $y^2$, $x$, $y$, $xy^2$, $x^2$, $xy$, $y^2$, $1$.</p>
<p>This in particular gives $z^7 = z$ so $z^6 = x^2 = 1$, and $x^3 = x^2 = 1$ gives $x = 1$.</p>
<p>All you need from this though is</p>
<p>$z^2 = y^2$<br>
$z^3 = x$</p>
<p>So $1 = y^6 = (z^2)^3 = (z^3)^2 = x^2$, then the last step of $x^3 = x^2 = 1 $ implies $x = 1$.</p>
<p>I posted this solution since the idea is just simplying that "playing around" with powers of $xy$ is algebraically easy and lets you do stuff. So I think computing up to its "period" of 9 is more insightful than just listing the final result.</p>
|
573,484 | <p>Prove that if $A \setminus B = \emptyset$, then $A \subseteq B$.</p>
<p>The Venn Diagram helped me to visualize what I'm trying to show (thanks @GA316), but the book asks for a written proof (step by step) by contradiction. Sorry if I wasn't more specific at first, is just that I've had many troubles in the past with proofs, somehow I have many ideas but I can't seem to connect them to get to the final proof. </p>
<p>This is what I have so far:
$P \rightarrow Q$ is equivalent to $\neg Q \rightarrow \neg P$ contraposition (thanks @The Chaz 2.0)</p>
<p>With P: $ A \setminus B = \emptyset$ and Q: $ A \subseteq B$</p>
<p>so $ \neg Q \equiv A \not\subseteq B\ , \exists x \in A : x \notin B $</p>
<p>be $ t: t \in A \wedge t \notin B $ ...is this right?</p>
<p>as this is the definition for $A \setminus B \ne \emptyset$ ...is this right?</p>
<p>$\therefore \neg Q \rightarrow \neg P \equiv A \not\subseteq B\ \rightarrow A \setminus B \ne\emptyset$</p>
<p>I have many concerns regarding if I'm using the correct notation. I am trying to learn this by myself and have nobody else to ask.</p>
<p>Also, sorry if it took me too long to update, I just started learning about this LaTEX notation.</p>
<p>Thank you very much in advance, you guys are so nice and helpful. You made me feel very welcomed and sure I need to read more about the rules and instructions for using this site. </p>
| The Chaz 2.0 | 7,850 | <p>Just for grins, you could prove the (logically equivalent) <strong>contrapositive</strong>, viz. </p>
<blockquote>
<p>If $A$ is <em>not</em> a subset of $B$, then [$A$ "toss" $B$] is nonempty.</p>
</blockquote>
<p>If $A$ is not a subset of $B$, then there is some element $x \in A$ that is <em>not</em> in $B$. Then when you "take away" all the things in $A$ that are/were in $B$, you have at least that element $x$ leftover.<br>
So [$A$ "toss" $ B$] is not the emply set. </p>
|
1,756,448 | <p>Suppose the function, $f$, is differentiable at $x = 1$. $$\lim_{h\rightarrow\ 0}\frac{f(1+h)}{h} = 5$$</p>
<p>Find a) $f(1)$ and, b) $f'(1)$. </p>
<p>I know b) (well at least I think it can) can be found by the definition of the derivative, i.e. </p>
<p>$$\lim_{h\rightarrow\ 0}\frac{f(1+h)-f(1)}{h} $$</p>
<p>Therefore, </p>
<p>$$f'(1) =5-\lim_{h\rightarrow\ 0}\frac{f(1)}{h} $$</p>
<p>However, I'm stuck for a). </p>
| drhab | 75,923 | <p>Hint:</p>
<p>if $f$ is differentiable at $1$ then it is continuous at $1$. For $h\neq 0$ we have:</p>
<p>$$f(1+h)=h\times\frac{f(1+h)}{h}$$</p>
<p>What equality will arise if $h\to0$?</p>
|
1,646,460 | <p>I was thinking about the following integral if I could solve it without using trigonometric formulas. If there is no other way to solve it, could you please explain me why do we replace $x$ with $2\sqrt 2 \sin(t)$? I'm really confused about these types of integrals.</p>
<p>$$\int \sqrt{8 - x^2} dx$$</p>
| Harish Chandra Rajpoot | 210,295 | <p>Let<br>
$$I=\int\sqrt{8-x^2}\ dx\tag 1$$
using integration by parts,<br>
$$I=\sqrt{8-x^2}\int 1\ dx-\int \left(\frac{-2x}{2\sqrt{8-x^2}}\right)\cdot x\ dx$$
$$I=\sqrt{8-x^2}(x)-\int \frac{(8-x^2)-8}{\sqrt{8-x^2}} \ dx$$
$$I=x\sqrt{8-x^2}-\int \left(\sqrt{8-x^2}-\frac{8}{\sqrt{8-x^2}} \right)\ dx$$
$$I=x\sqrt{8-x^2}-\int\sqrt{8-x^2}\ dx+8\int \frac{1}{\sqrt{8-x^2}}\ dx$$
setting the value from (1),
$$I=x\sqrt{8-x^2}-I+8\int \frac{1}{\sqrt{(2\sqrt 2)^2-x^2}}\ dx$$
$$2I=x\sqrt{8-x^2}+8\sin^{-1}\left(\frac{x}{2\sqrt 2}\right)+c$$
$$I=\color{red}{\frac{1}{2}\left(x\sqrt{8-x^2}+8\sin^{-1}\left(\frac{x}{2\sqrt 2}\right)\right)+C}$$</p>
|
176,613 | <p>Working on Harmonic numbers, I found this very interesting recurrence relation :
$$
H_n = \frac{n+1}{n-1} \sum_{k=1}^{n-1}\left(\frac{2}{k+1}-\frac{1}{1+n-k}\right)H_k
,\quad \forall\ n\in\mathbb{N},n>1$$
My proof of this is quite long and complicated, so I was wondering if someone knows an elegant or concise one. Any idea would be appreciated.</p>
<p>Alternatively, if someone knows a reference that talks about this kind of relation, it would be of great interest for me.</p>
<p>Thanks.</p>
| joriki | 6,622 | <p>It helps to visualize the terms in a square of products $1/(ij)$ with $i$ and $j$ running for $1$ to $n$. The sum over the first term contains all products with $i\ne j$ exactly once, whereas the sum over the second term roughly corresponds to the upper left half of the square, but with the left-most column, which adds up to $H_n$, excluded. Thus we have</p>
<p>$$
\def\sub#1{{\scriptstyle{i\ne j}\atop{\scriptstyle i,j\le #1}}}
\sum_{k=1}^{n-1}\frac{2}{k+1}H_k=\sum_{\sub n}\frac1{ij}$$</p>
<p>and</p>
<p>$$-\sum_{k=1}^{n-1}\frac{1}{1+n-k}=H_n-\sum_{i+j\le n+1}\frac1{ij}\;.$$</p>
<p>Substituting this into your equation, multiplying through by $n-1$ and simplifying leads to</p>
<p>$$\sum_{i+j\le n+1}\frac1{ij}-\sum_{\sub n}\frac1{ij}=\frac2{n+1}H_n\;,$$</p>
<p>$$\sum_{i+j\le n+1}\frac1{ij}-\sum_{\sub n}\frac1{ij}=2\sum_i\frac1{n+1}\frac1i\;,$$</p>
<p>$$\sum_{i+j\le n+1}\frac1{ij}=\sum_{\sub{n+1}}\frac1{ij}\;.$$</p>
<p>This we can prove by induction: The equation is satisfied for $n=0$, and going from $n$ to $n+1$ adds</p>
<p>$$\sum_{i+j=n+1}\frac1{ij}=\sum_{i=1}^n\frac1i\frac1{n+1-i}$$</p>
<p>to the left-hand side and also</p>
<p>$$2\frac1{n+1}\sum_{i=1}^n\frac1i=\frac1{n+1}\sum_{i=1}^n\left(\frac1i+\frac1{n+1-i}\right)=\sum_{i=1}^n\frac1i\frac1{n+1-i}$$</p>
<p>to the right-hand side.</p>
|
176,613 | <p>Working on Harmonic numbers, I found this very interesting recurrence relation :
$$
H_n = \frac{n+1}{n-1} \sum_{k=1}^{n-1}\left(\frac{2}{k+1}-\frac{1}{1+n-k}\right)H_k
,\quad \forall\ n\in\mathbb{N},n>1$$
My proof of this is quite long and complicated, so I was wondering if someone knows an elegant or concise one. Any idea would be appreciated.</p>
<p>Alternatively, if someone knows a reference that talks about this kind of relation, it would be of great interest for me.</p>
<p>Thanks.</p>
| dbot | 90,699 | <p>This page helped me prove the following related result, so I thought I'd share.</p>
<p>To begin,</p>
<p>$$
\sum_{i=1}^{n}\left ( \frac{1}{x_i} + \frac{1}{x_{n+1-i}}\right ) = 2 \sum_{i=1}^{n}\left ( \frac{1}{x_i} \right )
$$</p>
<p>$$
=\sum_{i=1}^{n}\left ( \frac{x_i+x_{n+1-i}}{x_i x_{n+1-i}} \right )
$$</p>
<p>Now, let $X$ be $n$ linear spaced numbers between $a$ and $b$ (and including them).</p>
<p>Then, since in that case $x_i+x_{n+1-i}=a+b$, we have</p>
<p>$$
\left ( a+b \right )\sum_{i=1}^{n}\left ( \frac{1}{x_i x_{n+1-i}} \right ) = 2 \sum_{i=1}^{n}\left ( \frac{1}{x_i} \right )
$$</p>
<p>In other words,</p>
<p>$$
\frac{\sum_{i=1}^{n}\left ( \frac{1}{x_i} \right )}{\sum_{i=1}^{n}\left ( \frac{1}{x_i x_{n+1-i}} \right )} = \frac{a+b}{2} = mean(X)
$$</p>
|
318,299 | <blockquote>
<p>Let <span class="math-container">$U$</span> be an open set in <span class="math-container">$\mathbb R$</span>. Then <span class="math-container">$U$</span> is a countable union of disjoint intervals. </p>
</blockquote>
<p>This question has probably been asked. However, I am not interested in just getting the answer to it. Rather, I am interested in collecting as many different proofs of it which are as diverse as possible. A professor told me that there are many. So, I invite everyone who has seen proofs of this fact to share them with the community. I think it is a result worth knowing how to prove in many different ways and having a post that combines as many of them as possible will, no doubt, be quite useful. After two days, I will place a bounty on this question to attract as many people as possible. Of course, any comments, corrections, suggestions, links to papers/notes etc. are more than welcome.</p>
| Trevor Wilson | 39,378 | <p>Let $U$ be an open subset of $\mathbb{R}$. Let $P$ be the poset consisting of collections $\mathcal{A}$ of disjoint open intervals where we say $\mathcal{A} \le \mathcal{A}'$ if each of the sets in $\mathcal{A}$ is a subset of some open interval in $\mathcal{A}'$. Every chain $C$ in this poset has an upper bound, namely $$\mathcal{B} = \left\{ \bigcup\left\{J \in \bigcup\bigcup C : I \subseteq J \right\}: I \in \bigcup\bigcup C\right\}.$$
Therefore by Zorn's lemma the poset $P$ has a maximal element $\mathcal{M}$. We claim that the union of the intervals in $\mathcal{M}$ is all of $U$. Suppose toward a contradiction that there is a real $x \in U$ that is not contained in any of the intervals in $\mathcal{M}$. Because $U$ is open we can take an open interval $I$ with $x \in I \subseteq U$.
Then the set
$$\mathcal{M}' = \{J \in \mathcal{M} : J \cap I = \emptyset\} \cup \left\{I \cup \bigcup \{J \in \mathcal{M} : J \cap I \ne \emptyset\}\right\}$$
is a collection of disjoint open intervals and
is above $\mathcal{M}$ in the poset $P$, contradicting the maximality of $\mathcal{M}$. It remains to observe that $\mathcal{M}$ is countable, which follows from the fact that its elements contain distinct rational numbers.</p>
<p>Note that the only way in which anything about order (or connectedness) is used is to see that $I \cup \bigcup \{J \in \mathcal{M} : J \cap I \ne \emptyset\}$ is an interval.</p>
|
318,299 | <blockquote>
<p>Let <span class="math-container">$U$</span> be an open set in <span class="math-container">$\mathbb R$</span>. Then <span class="math-container">$U$</span> is a countable union of disjoint intervals. </p>
</blockquote>
<p>This question has probably been asked. However, I am not interested in just getting the answer to it. Rather, I am interested in collecting as many different proofs of it which are as diverse as possible. A professor told me that there are many. So, I invite everyone who has seen proofs of this fact to share them with the community. I think it is a result worth knowing how to prove in many different ways and having a post that combines as many of them as possible will, no doubt, be quite useful. After two days, I will place a bounty on this question to attract as many people as possible. Of course, any comments, corrections, suggestions, links to papers/notes etc. are more than welcome.</p>
| sachin srivastava | 215,804 | <p>Let $G$ be a nonempty open set in $\mathbb{R}$. Write $a\sim b$ if the closed interval $[a, b]$ or $[b, a]$ if $b<a$, lies in $G$.This is an equivalence relation, in particular $a\sim a$ since $\{a\}$ is itself a closed interval. $G$ is therefore the union of disjoint equivalence classes. </p>
<p>Let $C(a)$ be the equivalence class containing $a$. Then $C(a)$ is clearly an interval. Also $C(a)$ is open, for if $k\in C(a)$, then $(k-\epsilon, k+\epsilon)\subseteq G$ for sfficiently small $\epsilon$.</p>
<p>But then $(k-\epsilon, k+\epsilon)\subseteq C(a)$; so $G$ is the union of disjoint intervals. These are at most countable in number by Lindel$\ddot{\rm o}$f's theorem. This completes the proof.</p>
<p>Reference:G. De Barra, Measure theory and Integration, Horwood Publishing.</p>
<p>(<a href="http://infoedu.ir/wp-content/uploads/2014/03/MeasureTheoryBook.pdf" rel="noreferrer">http://infoedu.ir/wp-content/uploads/2014/03/MeasureTheoryBook.pdf</a>)</p>
|
318,299 | <blockquote>
<p>Let <span class="math-container">$U$</span> be an open set in <span class="math-container">$\mathbb R$</span>. Then <span class="math-container">$U$</span> is a countable union of disjoint intervals. </p>
</blockquote>
<p>This question has probably been asked. However, I am not interested in just getting the answer to it. Rather, I am interested in collecting as many different proofs of it which are as diverse as possible. A professor told me that there are many. So, I invite everyone who has seen proofs of this fact to share them with the community. I think it is a result worth knowing how to prove in many different ways and having a post that combines as many of them as possible will, no doubt, be quite useful. After two days, I will place a bounty on this question to attract as many people as possible. Of course, any comments, corrections, suggestions, links to papers/notes etc. are more than welcome.</p>
| Alex | 560,714 | <p>Tell me if this has any potential... I understand the equivalence relation proof. However, since R is separable, we know that there exists a countable base. This means that there exists a family of open subsets of R such that every open set is equal to the union of a subcollection of this family.</p>
<p>But then by the disjointification lemma, we can generate disjoint sets whose union is the same as the union of the family (which need not be disjoint). But since this family is finite, we must be done, correct?</p>
|
2,056,209 | <p>We know that the union of countably many countable sets is countable.
What can se say about the union of infinitely many countable sets and its cardinality?
Thanks :)</p>
| Michael Hardy | 11,667 | <p>The set $\mathbb N \times \mathbb R$ is the set of all ordered pairs $(n,x)$ where $n\in \{1,2,3,\ldots\}$ and $x$ is a real number. The set of pairs $\{(1,x), (2,x), (3,x),\ldots\}$ where $x$ is <em>just one</em> real number is countably infinite, so the union $\displaystyle \bigcup_{x\,\in\,\mathbb R} \{(1,x), (2,x), (3,x),\ldots\}$ is that of infinitely many countable sets. That is has at <em>most</em> as many members as $\mathbb R$ is seen by considering that the mapping
$$
(n,x) \mapsto n + \frac{1}{1 + 2^x}.
$$
That that union has <em>at least</em> as many members as $\mathbb R$ is seen by the fact that $\{(n,x):x\in\mathbb R\}$ (with just one fixed value of $n$) has as many members are $\mathbb R$.</p>
<p>If it has at most as many members as $\mathbb R$ and also at least as many, does it then have exactly as many as $\mathbb R$? With finite sets, the answer to that is trivially "yes". With infinite sets, the affirmative answer is the <a href="https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem" rel="nofollow noreferrer">Schröder–Bernstein theorem</a>.</p>
|
768,925 | <p><img src="https://i.stack.imgur.com/SHRqU.png" alt="enter image description here"></p>
<p>This is quite a tricky question for me, but this is how far I got:<img src="https://i.stack.imgur.com/UV0ou.jpg" alt="enter image description here"></p>
<p>My drawing may not be precise, but I do know the points of tangency. I am a little stuck now, and I would appreciate it if someone can guide me through the answer. Thanks.</p>
| Jonas Granholm | 134,205 | <p>This construction occurs three times:
<img src="https://i.stack.imgur.com/gwcMM.png" alt="A square and a circle inscribed in a right triangle"></p>
|
1,824,638 | <p>The figure shows a piece of string tied to a circle with a radius of one unit. The string is just long enough to reach the opposite side of the circle. Find the area of the region, not including the circle itself that is traced out when the string is unwound counterclockwise and continues counterclockwise until it reaches the opposite side again.</p>
<p><a href="https://i.stack.imgur.com/KzJRQ.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KzJRQ.gif" alt="enter image description here"></a></p>
| Doug M | 317,162 | <p>There are 3 parts to the calculation.
The string unwinds until it points straight up.
A semi-circle.
The string winds back onto the wheel.
Part 1 and part 3 trace out equal area.</p>
<p>Part 1:
Let $(\cos\theta,\sin\theta)$ be the point where the string is starting to pull away from the wheel.</p>
<p>let (x,y) be the endpoint of the string.</p>
<p>the area traced inside of(x,y).</p>
<p>$-\int_0^{\pi} y(\theta) \frac {dx}{d\theta} d\theta$</p>
<p>It is negative, because x is moving from right to left.</p>
<p>Now we need to find functions for x and y in terms of theta.</p>
<p>The length of string on the free end = $(\pi - \theta)$
and the segment from $(\cos\theta, \sin\theta) \to (x,y)$ is tangent to the curve.</p>
<p>$(x,y) = (\cos\theta + (\pi-\theta) \sin \theta,\sin\theta - (\pi-\theta) \cos \theta)$</p>
<p>$\frac {dx}{d\theta} = -\sin\theta + (\pi-\theta) \cos\theta - \sin\theta$</p>
<p>$\int_0^\pi (\sin\theta - (\pi-\theta) \cos \theta)(2\sin\theta - (pi-\theta)\cos\theta) d\theta$</p>
<p>part 2: is a semi circle of radius $\pi$</p>
<p>$\frac 12\pi^3$</p>
<p>part 3 = part 1</p>
<p>By the way, this calculation has included the the are of the unit circle as being inside the area traced out by the end-point of the string.</p>
|
1,714,053 | <p>How can I prove that if $\sum\limits_{n\in\Bbb N} a_n^2 $ converges then also $\sum\limits_{n\in\Bbb N} \frac{\lvert a_n\rvert}{n}$ converges?</p>
<p>Thanks ! </p>
| DonAntonio | 31,254 | <p>By Cauchy-Schwarz$$\sum_{n=1}^\infty\frac{|a_n|}n\le\left(\sum_{n=1}^\infty a_n^2\right)^{1/2}\cdot\left(\sum_{n=1}^\infty\frac1{n^2}\right)^{1/2}$$</p>
|
29,926 | <p>Is there an algorithm which will allow me to find an isomorphism between two graphs if I have their adjacency lists?</p>
| Aaron | 9,863 | <p><B>Short answer:</B> using logs won't really help. Instead, you should multiply $p$ by something large so that matlab can deal with it, and then divide out in the end.</p>
<p><B>Long answer:</B>There is a $\log$ for matrices, but it doesn't behave quite the same as $\log$ for numbers, and so it's not quite suitable for computations in the way you think it is.</p>
<p>From Taylor series, we have that $e^x=\sum x^n/n!$ and $\log(1+x)=\sum (-1)^n x^{n+1}/(n+1)$. The fist formula holds for all $x$, the second for $x$ that are close to $0$, and they define inverse functions where they are defined.</p>
<p>Similarly, we can define a matrix $\log$ and matrix exponential by means of the exact same power series. $e^A$ will converge for all $A$, and $\ln(I+A)$ will converge for small matrices (take any linear norm on matrices such that $||AB||\leq ||A|| ||B||$, and the condition $||A||<1$ will work).</p>
<p>Unfortunately, these do not obey all the properties that you want. For example, $e^{A+B}=e^A e^B$ only when $A$ and $B$ commute with each other. When they don't commute, the <a href="http://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula" rel="nofollow">Campbell-Baker-Hausdorff formula</a> says what $\log(e^A e^B)$ equals. Because of this, we do not in general have that $\log AB=\log A + \log B$, even when everything is defined.</p>
<p>Even ignoring this problem, we would still need to define the $\log$ of a vector. I honestly don't know where one would begin, other than to take $\log$ of the individual entries, which wouldn't have any immediately useful properties as far as matrix actions are concerned.</p>
<p>If the problem is that Matlab has rounding errors, then you can exploit the fact that matrix multiplication is a linear operator and just multiply $p$ by some large constant $c$, do the calculation, and then divide by $c$ again. Additionally, if you have problems with $D$ or $K$ being too small/large you can do the same to them.</p>
|
64,905 | <p>Let's see if we could use MO to put some pressure on certain publishers...</p>
<p>Although it is wonderful that it has been put
<a href="http://www.jmilne.org/math/Books/DMOS.pdf" rel="nofollow">online</a>, I think it would make an even greater read if "Hodge Cycles, Motives and Shimura Varieties" by Deligne, Milne, Ogus and Shih would be (re)written in the latex typesetting (well, if I could understand its content..).</p>
<p>But enough about my opinion, what do you think? Which book(s) would you like to see "texified"? As customary in a CW question, one book per answer please.</p>
| Zev Chonoles | 1,916 | <p>Atiyah + Macdonald, <em>Introduction to Commutative Algebra</em>.</p>
|
64,905 | <p>Let's see if we could use MO to put some pressure on certain publishers...</p>
<p>Although it is wonderful that it has been put
<a href="http://www.jmilne.org/math/Books/DMOS.pdf" rel="nofollow">online</a>, I think it would make an even greater read if "Hodge Cycles, Motives and Shimura Varieties" by Deligne, Milne, Ogus and Shih would be (re)written in the latex typesetting (well, if I could understand its content..).</p>
<p>But enough about my opinion, what do you think? Which book(s) would you like to see "texified"? As customary in a CW question, one book per answer please.</p>
| Jim Conant | 9,417 | <p>"Rational Homotopy Theory and Differential Forms." by Griffiths and Morgan. </p>
|
2,747,074 | <p>I've seen multiple ways on how to solve it online (and most likely the majority are wrong), but I don't know how to solve this fully so that I could get the full grade during my exam.</p>
<p>The question is as follows:</p>
<blockquote>
<p>Prove that if $\sum a_n$ is convergent with $a_n > 0$ for all $n$,
then $\sum a^2_n$ is also convergent.</p>
</blockquote>
<p>I understand that the starting point to prove this is that $\lim a_n =0$, but after that I really don't know what I'm supposed to say.</p>
| user284331 | 284,331 | <p>Then for large $n$, $a_{n}<1$, then $a_{n}^{2}\leq a_{n}$ for all such $n$ and hence $\displaystyle\sum_{n\geq N}a_{n}^{2}\leq\sum_{n\geq N}a_{n}<\infty$.</p>
|
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