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 |
|---|---|---|---|---|
173,107 | <p>are there any examples of functions $f:x\in\mathbb{R}_0^+\rightarrow\mathbb{R}_0^+$ and intervals $(a,b), 0\le a \lt b \le \infty$ , for which $$\Big(\int_a^b{|f(x)|^p dx}\Big)^\frac{1}{p} = f(p)$$ $$\forall p\in(a,b)$$<br>
This question came up when thinking about $L_p$ norms as functions of $p$ and thus as a mapping of a real function to another one. </p>
<p>In view of the comment of Mark Meckes, I would like to know, if there are also examples of non-trivial functions, i.e. that are not identical to $1$</p>
| Matthew Daws | 406 | <p><strong>Edit:</strong> This is correct, except for the actual application to the question at hand, which is wrong; see below for details.</p>
<p>I learnt about this technique from:</p>
<p>MR2526788 (2010m:46098)<br/>
Haagerup, Uffe(DK-SU-CS); Musat, Magdalena(1-MEMP)<br/>
Classification of hyperfinite factors up to completely bounded isomorphism of their preduals.<br/>
J. Reine Angew. Math. 630 (2009), 141–176. </p>
<p>See pages 149--150. Let $M,N$ be von Neumann algebras. Recall that any function $\phi\in M^*$ has a unique decomposition as $\phi=\phi_s+\phi_n$ where $\phi_n\in M_*$ and $\phi_s$ is "singular". For $T:M\rightarrow N$ a bounded linear map, we can similarly decompose $T=T_n+T_s$ where $T_n,T_s:M\rightarrow N$ are bounded linear maps with
$$ \phi\circ T_n = (\phi\circ T)_n, \quad \phi\circ T_s = (\phi\circ T)_s
\qquad (\phi\in N_*). $$
It's easy to see that if $T$ is (C)P then so are $T_s,T_n$. An original reference for this claim is:</p>
<p>MR0107825 (21 #6547)<br/>
Tomiyama, Jun<br/>
On the projection of norm one in W∗-algebras. III.<br/>
Tôhoku Math. J. (2) 11 1959 125–129.<br/>
<a href="http://projecteuclid.org/euclid.tmj/1178244633" rel="nofollow">http://projecteuclid.org/euclid.tmj/1178244633</a></p>
<p>So, for your question, let $\Phi_1:\mathcal{A}\rightarrow \mathcal L(H)$ be any UCP extension, and let $\Phi = (\Phi_1)_n$. So $\Phi$ is normal and UCP. As $\Phi_0$ is normal, $(\Phi_0)_n = \Phi_0$ (if there is a problem, it's at this point: what precisely is your definition of "normal" for a UCP map from an operator system?) and so $\Phi$ extends $\Phi_0$ as required.</p>
<p><strong>Edit:</strong> This is nonsense as Taka points out. I was claiming that $(\Phi_1)_n=\Phi_0$ on $\mathcal R$, or equivalently that
$$ \langle \omega, (\Phi_1)_n(a) \rangle = \langle \omega, \Phi_0(a) \rangle
\qquad (a\in \mathcal R,\omega\in\mathcal{B}(H)_*), $$
which is in turn equivalent to
$$ \langle (\omega\circ\Phi_1)_n, a \rangle =
\langle \omega\circ (\Phi_1)_n, a \rangle = \langle \omega\circ\Phi_0 , a \rangle
\qquad (a\in \mathcal R,\omega\in\mathcal{B}(H)_*). $$
That $\Phi_0$ is normal might be taken to mean that $\omega\circ\Phi_0 \in \mathcal{R}_*$ (if $\mathcal R$ is weak$^*$-closed, say). But all we know about $\Phi_1$ is that $\Phi_1(a)=\Phi(a)$ for $a\in\mathcal R$, which seems to tell us nothing about $(\omega\circ\Phi_1)_n\in\mathcal{A}_*$ restricted to $\mathcal R$.</p>
|
2,216,723 | <p>A player at draw poker holds the seven of spades and the eight, nine,
ten, and ace of diamonds. Aware that all the other players are drawing three cards, he figures that any hand he could win with a flush he could also win with a straight. For which should he draw? </p>
| Misha Lavrov | 383,078 | <p>The determinant is the product of the eigenvalues. We know that $\frac12$ is an eigenvalue with multiplicity at least $m-2$, because $M - \frac12I$ has rank $1$. In addition, the sum of the eigenvalues is the trace of $M$, which is $m-1$. So if $x$ is the remaining eigenvalue, then $x + \frac12(m-2) = m-1$, which means that $x = \frac m2$.</p>
<p>Taking the product of the eigenvalues, we get $\frac m2 \cdot \left(\frac12\right)^{m-2} = \frac{m}{2^{m-1}}$.</p>
<hr>
<p>Also, <a href="https://en.wikipedia.org/wiki/Sylvester%27s_determinant_identity" rel="nofollow noreferrer">Sylvester's determinant identity</a> applies to $2$ times this matrix. If $\mathbf u$ is the vector of all $1$'s, then $2M = I + \mathbf{u}\mathbf{u}^T$, so $\det(2M) = \det(I + \mathbf{u}\mathbf{u}^T) = 1 + \mathbf{u}^T\mathbf{u} = m$. Therefore $\det(M) = \frac1{2^{m-1}} \det(2M) = \frac{m}{2^{m-1}}$.</p>
|
2,680,744 | <p>If image of a simple loop is contained in an open connected set $D\subseteq R^2$ such that this loop is homotopic to a point in $D$, will interior of curve be contained in $D$.</p>
<p>What if we take closed contour instead of simple loop?</p>
<p>Definition: Simple loop is homeomorphic to $S^1$.</p>
<p>A similar question: <a href="https://math.stackexchange.com/questions/2680441/inside-of-boundary-of-an-open-region">Inside of boundary of an open region</a></p>
| Mauro ALLEGRANZA | 108,274 | <p>For the second task, you need <a href="http://integral-table.com/downloads/logic.pdf" rel="nofollow noreferrer">logical equivalence rules</a>, like e.g. <a href="https://en.wikipedia.org/wiki/De_Morgan%27s_laws#Informal_proof" rel="nofollow noreferrer">De Morgan's laws</a> and the equivalence between $(p \to q)$ and $(\lnot p \lor q)$ (called: <a href="https://en.wikipedia.org/wiki/Material_implication_(rule_of_inference)" rel="nofollow noreferrer">Material implication rule</a>).</p>
<hr>
<p>For the first task, you can show that the formula:</p>
<blockquote>
<p>$(\lnot Q) \to (R \to \lnot (P \land Q))$</p>
</blockquote>
<p>is a <em>tautology</em>, arguing by contradiction.</p>
<p>I.e. assume not: this means that there is a <em>valuation</em> $v$ such that:</p>
<blockquote>
<p>$v(\lnot Q)=$ TRUE and $v((R \to \lnot (P \land Q)))=$ FALSE.</p>
</blockquote>
|
2,680,744 | <p>If image of a simple loop is contained in an open connected set $D\subseteq R^2$ such that this loop is homotopic to a point in $D$, will interior of curve be contained in $D$.</p>
<p>What if we take closed contour instead of simple loop?</p>
<p>Definition: Simple loop is homeomorphic to $S^1$.</p>
<p>A similar question: <a href="https://math.stackexchange.com/questions/2680441/inside-of-boundary-of-an-open-region">Inside of boundary of an open region</a></p>
| Bram28 | 256,001 | <p>I am not sure what you mean by 'contradiction theory' ... but I am guessing the idea is to show the statement is a tautology by showing that the assumption that it is not leads to a contradiction. So, below I'll show you a semi-formal technique to do something like this; it's known as a 'short truth-table'.</p>
<p>So, let's assume the statement is not a tautology. This means that it should be possible for it to be false. Let's indicate that by putting a False (F) under its main connective:</p>
<p>\begin{array}{ccccccccc}
\neg & Q & \rightarrow & (R & \rightarrow & \neg & (P & \land & Q))\\
&&F
\end{array}</p>
<p>Now, the only way for a conditional to be false is if the antecedent is true and the consequent is false, so:</p>
<p>\begin{array}{ccccccccc}
\neg & Q & \rightarrow & (R & \rightarrow & \neg & (P & \land & Q))\\
T&&F&&F
\end{array}</p>
<p>And the same holds for the second conditional:</p>
<p>\begin{array}{ccccccccc}
\neg & Q & \rightarrow & (R & \rightarrow & \neg & (P & \land & Q))\\
T&&F&T&F&F
\end{array}</p>
<p>For negations we of course flip the truth-value, so this forces:</p>
<p>\begin{array}{ccccccccc}
\neg & Q & \rightarrow & (R & \rightarrow & \neg & (P & \land & Q))\\
T&F&F&T&F&F&&T
\end{array}</p>
<p>And finally, a true conjunction means both conjuncts are true:</p>
<p>\begin{array}{ccccccccc}
\neg & Q & \rightarrow & (R & \rightarrow & \neg & (P & \land & Q))\\
T&F&F&T&F&F&T&T&T
\end{array}</p>
<p>But now we see that $Q$ is forced to be both False and True ... which is a contradiction. Hence, out assumptions that the statement is not a tautology is incorrect: it <em>is</em> a tautology.</p>
<p>Now that we're done maybe you can understand why this technique is called a 'short truth-table': the line of truth-values we created looks like one of the lines you may find in a full truth-table ... except of course that this actual line of values would never appear on a truth-table since $Q$ is both true and False. However, if we would have been able to make the statement false <em>without</em> running into such a contradiction, then we would have shown that the statement is not a tautology, and the line we would heave created really would have been one of the lines in a full truth-table. But, we would have found this line much more quickly than if we had done a full truth-table.</p>
<p>Finally, now that you have seen how the method works, you can show the order in which you assign truth-values simply by indexing those truth-values:</p>
<p>\begin{array}{ccccccccc}
\neg & Q & \rightarrow & (R & \rightarrow & \neg & (P & \land & Q))\\
T_2&F_6&F_1&T_4&F_3&F_5&T_8&T_7&T_9
\end{array}</p>
|
1,703,100 | <p>A car starts from rest and accelerates in $a = \frac{2\cdot m}{3\cdot s^3}t$,</p>
<p>After $3$ seconds, The car will be $27$ metres from beginning.</p>
<p>Find distance as function of time.</p>
<p>I know i have to integral the acceleration,But i don't know how.</p>
<p>I found the Equation is $x_{t} = 24 + \frac{1}{3} t^2$.</p>
<p>What do you think ?</p>
| Cm7F7Bb | 23,249 | <p>The result of the first throw can be $1,2,3,4,5,6$ with equal probabilities. If the result of the first throw is $1$, then we throw the dice again and can obtain $1,2,3,4,5,6$ with equal probabilities. The random variable can be equal to $2,3,4,5,6$ with the probabilities $1/6$ after the first throw or it can be equal to $2,3,4,5,6,7$ with probability $(1/6)^2$ after the second throw. There are two ways to obtain $2,3,4,5,6$ and the probability is equal to $1/6+(1/6)^2=7/36$. Hence, the probability distribution table looks like this
$$
\left.\begin{array}{c|c|c|c|c|c|c|c}\text{Probability} & 7/36 & 7/36 & 7/36 & 7/36 & 7/36 & 1/36 \\\hline \text{Value} & 2 & 3 & 4 & 5 & 6 & 7 \end{array}\right..
$$</p>
|
42,053 | <p>I have this problem to solve:</p>
<blockquote>
<p>Hockey teams 1 and 2 score goals at times of Poisson process with rates 1 and 2. Suppose that $N_1(0)=3$ and $N_2(0)=1$.
What is the probability that $N_1(t)$ will reach 5 before $N_2(t)$ does?</p>
</blockquote>
<p>I've re-worded this to: What is the prob that in the next 5 goals at least 2 of them are scored by team 1? </p>
<p>The only problem I have is finding out the probability of team one scoring a goal. Can we use the rates to work this out?</p>
| Robert Israel | 8,508 | <p>You must assume that the two Poisson processes are independent. Goal-scoring (without regard to which team scores) is then a Poisson process $N(t) = N_1(t) + N_2(t)$ with rate $1+2=3$, and each individual goal that is scored has probability $1/3$ of coming from team 1.</p>
|
42,053 | <p>I have this problem to solve:</p>
<blockquote>
<p>Hockey teams 1 and 2 score goals at times of Poisson process with rates 1 and 2. Suppose that $N_1(0)=3$ and $N_2(0)=1$.
What is the probability that $N_1(t)$ will reach 5 before $N_2(t)$ does?</p>
</blockquote>
<p>I've re-worded this to: What is the prob that in the next 5 goals at least 2 of them are scored by team 1? </p>
<p>The only problem I have is finding out the probability of team one scoring a goal. Can we use the rates to work this out?</p>
| Did | 6,179 | <p>Yes, each scored goal has probability 1/3 of being scored by team 1 and probability 2/3 of being scored by team 2, and by the magics of Poisson proceses, each goal is <em>colored</em> independently from the others. </p>
<p>Hence you look for the probability of a random walk starting from (3,1) and with step (+1,0) with probability 1/3 and (0,+1) with probability 2/3 to reach the line (5,something) before reaching the line (something,5). </p>
<p>Writing u(x,y) for this probability starting from (x,y), you look for u(3,1) and you know that u(5,something)=1, u(something,0)=0 and 3u(x,y)=u(x+1,y)+2u(x,y+1). This shows that u(3,1) is part of the unique solution of a linear system of equations whose unknowns are the u(x,y) for (x,y) any of the points (3,1), (4,1), (3,2), (4,2), (3,3), (4,3), (3,4), (4,4).</p>
<p>Starting backwards, one gets u(4,4)=1/3, u(3,4)=u(4,4)/3=1/9, u(4,3)=1/3+2u(4,4)/3=5/9, u(3,3)=u(4,3)/3+2u(3,4)/3=7/27, u(4,2)=1/3+2u(4,3)/3=19/27, u(3,2)=u(4,2)/3+2u(3,3)/3=11/27, u(4,1)=1/3+2u(4,2)/3=65/81, and finally u(3,1)=u(4,1)/3+2u(3,2)/3=131/243.</p>
<p><strong>Edit</strong> An alternative solution is, as suggested by the OP, to consider the five first goals and to note that team 1 loses the game if and only if team 1 scores one or zero of these five goals. The number of goals scored by team 1 amongst the five first goals is binomial (5,1/3), hence the probability that team 1 wins is
1-P(Bin(5,1/3)=0)-P(Bin(5,1/3)=1). This is 1-(2/3)^5-5(2/3)^4(1/3)=1-32/243-80/243=131/243.</p>
|
237,496 | <p>I have a complex data set and I want to fit with a mathematical model using <code>NonlinearModelFit</code>, but an error appears. Following is the data I'm using and the code that I have tried.</p>
<pre><code> dataset = {{0.1, 78.2356 - 0.8683 I}, {0.1998, 78.4897 - 0.5794 I}, {0.2996,
78.1675 - 1.0737 I}, {0.3994, 78.2344 - 1.4989 I}, {0.4992,
78.1883 - 1.8812 I}, {0.599, 78.1934 - 2.2289 I}, {0.6988,
78.1711 - 2.5836 I}, {0.7986, 78.1692 - 3.0154 I}, {0.8984,
78.1315 - 3.3294 I}, {0.9982, 78.0316 - 3.6955 I}, {1.098,
78.0287 - 4.0985 I}, {1.1978, 77.9932 - 4.4796 I}, {1.2976,
77.9667 - 4.8723 I}, {1.3974, 77.8871 - 5.2557 I}, {1.4972,
77.7778 - 5.6587 I}, {1.597, 77.5958 - 5.9334 I}, {1.6968,
77.7181 - 6.2427 I}, {1.7966, 77.661 - 6.6707 I}, {1.8964,
77.6011 - 7.0251 I}, {1.9962, 77.4963 - 7.394 I}, {2.096,
77.4624 - 7.8034 I}, {2.1958, 77.3749 - 8.1873 I}, {2.2956,
77.2339 - 8.5655 I}, {2.3954, 77.1747 - 8.8844 I}, {2.4952,
76.9861 - 9.2161 I}, {2.595, 76.9674 - 9.4487 I}, {2.6948,
77.0184 - 9.9153 I}, {2.7946, 76.8488 - 10.3242 I}, {2.8944,
76.7229 - 10.6691 I}, {2.9942, 76.6241 - 11.0533 I}, {3.094,
76.4483 - 11.3778 I}, {3.1938, 76.3633 - 11.7194 I}, {3.2936,
76.2347 - 12.0944 I}, {3.3934, 76.117 - 12.3783 I}, {3.4932,
76.0038 - 12.7369 I}, {3.593, 75.897 - 13.1024 I}, {3.6928,
75.7642 - 13.4812 I}, {3.7926, 75.6128 - 13.8315 I}, {3.8924,
75.4548 - 14.1748 I}, {3.9922, 75.3121 - 14.5079 I}, {4.092,
75.112 - 14.846 I}, {4.1918, 74.9628 - 15.1255 I}, {4.2916,
74.7993 - 15.4578 I}, {4.3914, 74.6315 - 15.8132 I}, {4.4912,
74.4547 - 16.1097 I}, {4.591, 74.2938 - 16.4089 I}, {4.6908,
74.152 - 16.7319 I}, {4.7906, 73.9702 - 16.9752 I}, {4.8904,
73.8144 - 17.2465 I}, {4.9902, 73.6398 - 17.5256 I}, {5.09,
73.5109 - 17.8629 I}, {5.1898, 73.3688 - 18.1599 I}, {5.2896,
73.2115 - 18.4785 I}, {5.3894, 73.0502 - 18.7804 I}, {5.4892,
72.8459 - 19.0808 I}, {5.589, 72.6908 - 19.4046 I}, {5.6888,
72.5409 - 19.711 I}, {5.7886, 72.3442 - 20.0245 I}, {5.8884,
72.1652 - 20.2782 I}, {5.9882, 71.9966 - 20.6356 I}, {6.088,
71.7659 - 20.9083 I}, {6.1878, 71.5486 - 21.2 I}, {6.2876,
71.3843 - 21.4641 I}, {6.3874, 71.1986 - 21.7711 I}, {6.4872,
70.9554 - 22.0703 I}, {6.587, 70.724 - 22.3318 I}, {6.6868,
70.4961 - 22.6017 I}, {6.7866, 70.2746 - 22.8812 I}, {6.8864,
70.0688 - 23.1126 I}, {6.9862, 69.8475 - 23.4055 I}, {7.086,
69.6181 - 23.6201 I}, {7.1858, 69.4073 - 23.8674 I}, {7.2856,
69.1847 - 24.1005 I}, {7.3854, 68.9453 - 24.3616 I}, {7.4852,
68.7403 - 24.5436 I}, {7.585, 68.5239 - 24.8114 I}, {7.6848,
68.2914 - 25.0535 I}, {7.7846, 68.0972 - 25.2649 I}, {7.8844,
67.8553 - 25.5043 I}, {7.9842, 67.6882 - 25.7208 I}, {8.084,
67.4227 - 25.9499 I}, {8.1838, 67.1762 - 26.1625 I}, {8.2836,
66.9512 - 26.3742 I}, {8.3834, 66.7394 - 26.6041 I}, {8.4832,
66.5167 - 26.842 I}, {8.583, 66.3146 - 27.0334 I}, {8.6828,
66.0724 - 27.2423 I}, {8.7826, 65.8313 - 27.4527 I}, {8.8824,
65.6043 - 27.6363 I}, {8.9822, 65.3769 - 27.9247 I}, {9.082,
65.1264 - 28.1023 I}, {9.1818, 64.8719 - 28.2838 I}, {9.2816,
64.6268 - 28.4543 I}, {9.3814, 64.4335 - 28.6643 I}, {9.4812,
64.1462 - 28.892 I}, {9.581, 63.9215 - 29.021 I}, {9.6808,
63.6482 - 29.2224 I}, {9.7806, 63.4241 - 29.4026 I}, {9.8804,
63.2 - 29.5862 I}, {9.9802, 62.9566 - 29.749 I}, {10.08,
62.7357 - 29.9395 I}, {10.1798, 62.4715 - 30.1185 I}, {10.2796,
62.1795 - 30.2156 I}, {10.3794, 61.96 - 30.4348 I}, {10.4792,
61.7823 - 30.5786 I}, {10.579, 61.4458 - 30.7737 I}, {10.6788,
61.2525 - 30.9524 I}, {10.7786, 60.9821 - 31.029 I}, {10.8784,
60.7528 - 31.2795 I}, {10.9782, 60.497 - 31.3811 I}, {11.078,
60.2911 - 31.4382 I}, {11.1778, 60.0138 - 31.6786 I}, {11.2776,
59.7766 - 31.7127 I}, {11.3774, 59.5339 - 31.9257 I}, {11.4772,
59.3229 - 32.0314 I}, {11.577, 59.0414 - 32.1797 I}, {11.6768,
58.7069 - 32.3222 I}, {11.7766, 58.6335 - 32.4309 I}, {11.8764,
58.2855 - 32.6102 I}, {11.9762, 57.9934 - 32.5999 I}, {12.076,
57.8657 - 32.8212 I}, {12.1758, 57.5263 - 32.79 I}, {12.2756,
57.43 - 33.0486 I}, {12.3754, 56.9925 - 33.0071 I}, {12.4752,
56.9032 - 33.357 I}, {12.575, 56.6686 - 33.2642 I}, {12.6748,
56.4106 - 33.4076 I}, {12.7746, 56.1114 - 33.6666 I}, {12.8744,
55.9125 - 33.7541 I}, {12.9742, 55.7078 - 33.7603 I}, {13.074,
55.3646 - 33.919 I}, {13.1738, 55.1957 - 33.9628 I}, {13.2736,
54.8702 - 34.0163 I}, {13.3734, 54.6922 - 34.2249 I}, {13.4732,
54.4734 - 34.3232 I}, {13.573, 54.2027 - 34.3559 I}, {13.6728,
53.9333 - 34.4731 I}, {13.7726, 53.6691 - 34.5305 I}, {13.8724,
53.5039 - 34.4862 I}, {13.9722, 53.27 - 34.6338 I}, {14.072,
52.8795 - 34.7439 I}, {14.1718, 52.6861 - 34.7778 I}, {14.2716,
52.503 - 34.8645 I}, {14.3714, 52.4029 - 34.9377 I}, {14.4712,
52.2769 - 35.0561 I}, {14.571, 51.9974 - 35.1729 I}, {14.6708,
51.6274 - 35.2887 I}, {14.7706, 51.3093 - 35.3206 I}, {14.8704,
51.1775 - 35.2592 I}, {14.9702, 50.9441 - 35.2813 I}, {15.07,
50.6445 - 35.2355 I}, {15.1698, 50.387 - 35.4794 I}, {15.2696,
50.2707 - 35.462 I}, {15.3694, 49.9586 - 35.5747 I}, {15.4692,
49.6906 - 35.4405 I}, {15.569, 49.5962 - 35.6865 I}, {15.6688,
49.2882 - 35.7023 I}, {15.7686, 48.9925 - 35.7119 I}, {15.8684,
48.9255 - 35.7546 I}, {15.9682, 48.8472 - 35.9351 I}, {16.068,
48.4388 - 35.908 I}, {16.1678, 48.4093 - 35.9757 I}, {16.2676,
48.0541 - 35.8649 I}, {16.3674, 47.932 - 35.9882 I}, {16.4672,
47.6193 - 35.9926 I}, {16.567, 47.423 - 36.2149 I}, {16.6668,
47.1304 - 36.1237 I}, {16.7666, 46.8657 - 36.0961 I}, {16.8664,
46.8233 - 36.1513 I}, {16.9662, 46.4927 - 36.3335 I}, {17.066,
46.282 - 36.1248 I}, {17.1658, 46.0527 - 36.2736 I}, {17.2656,
45.7096 - 36.1602 I}, {17.3654, 45.6516 - 36.1769 I}, {17.4652,
45.5632 - 36.3304 I}, {17.565, 45.2815 - 36.3742 I}, {17.6648,
45.1287 - 36.4635 I}, {17.7646, 44.8639 - 36.384 I}, {17.8644,
44.5912 - 36.4433 I}, {17.9642, 44.3884 - 36.4687 I}, {18.064,
44.2389 - 36.4418 I}, {18.1638, 44.1411 - 36.5143 I}, {18.2636,
43.9103 - 36.559 I}, {18.3634, 43.6563 - 36.4571 I}, {18.4632,
43.4871 - 36.5651 I}, {18.563, 43.3273 - 36.5756 I}, {18.6628,
42.9873 - 36.6038 I}, {18.7626, 42.9083 - 36.5295 I}, {18.8624,
42.6972 - 36.3761 I}, {18.9622, 42.3707 - 36.5342 I}, {19.062,
42.2738 - 36.5439 I}, {19.1618, 42.0595 - 36.5423 I}, {19.2616,
41.9375 - 36.4932 I}, {19.3614, 41.7015 - 36.5501 I}, {19.4612,
41.5701 - 36.4344 I}, {19.561, 41.2793 - 36.4456 I}, {19.6608,
41.0975 - 36.544 I}, {19.7606, 40.9272 - 36.5619 I}, {19.8604,
40.8173 - 36.4263 I}, {19.9602, 40.5276 - 36.4277 I}, {20.06,
40.5298 - 36.3943 I}, {20.1598, 40.2316 - 36.3789 I}, {20.2596,
40.1209 - 36.4193 I}, {20.3594, 39.8604 - 36.5787 I}, {20.4592,
39.7385 - 36.4455 I}, {20.559, 39.5266 - 36.522 I}, {20.6588,
39.3144 - 36.4394 I}, {20.7586, 39.3321 - 36.4143 I}, {20.8584,
39.0912 - 36.4885 I}, {20.9582, 38.9314 - 36.2492 I}, {21.058,
38.5343 - 36.24 I}, {21.1578, 38.5108 - 36.3822 I}, {21.2576,
38.4282 - 36.3555 I}, {21.3574, 38.1504 - 36.4566 I}, {21.4572,
38.0167 - 36.2047 I}, {21.557, 37.7961 - 36.2409 I}, {21.6568,
37.5403 - 36.2281 I}, {21.7566, 37.3251 - 36.2135 I}, {21.8564,
37.258 - 36.1749 I}, {21.9562, 37.174 - 36.1892 I}, {22.056,
37.0593 - 36.2818 I}, {22.1558, 36.7329 - 36.2821 I}, {22.2556,
36.6355 - 36.2482 I}, {22.3554, 36.466 - 36.1053 I}, {22.4552,
36.281 - 36.0756 I}, {22.555, 36.2407 - 36.1418 I}, {22.6548,
36.0274 - 36.1273 I}, {22.7546, 35.9257 - 36.0706 I}, {22.8544,
35.7085 - 36.0424 I}, {22.9542, 35.487 - 35.9645 I}, {23.054,
35.3758 - 35.8514 I}, {23.1538, 35.24 - 35.909 I}, {23.2536,
35.1003 - 36.0131 I}, {23.3534, 35.0614 - 35.9159 I}, {23.4532,
34.9312 - 35.866 I}, {23.553, 34.429 - 35.9661 I}, {23.6528,
34.4772 - 35.6764 I}, {23.7526, 34.1887 - 35.7948 I}, {23.8524,
34.0702 - 35.7791 I}, {23.9522, 34.0382 - 35.7418 I}, {24.052,
33.7731 - 35.8373 I}, {24.1518, 33.7543 - 35.5773 I}, {24.2516,
33.6515 - 35.5328 I}, {24.3514, 33.3638 - 35.7687 I}, {24.4512,
33.2197 - 35.5497 I}, {24.551, 33.1856 - 35.429 I}, {24.6508,
32.8109 - 35.4585 I}, {24.7506, 32.8435 - 35.3567 I}, {24.8504,
32.6808 - 35.4156 I}, {24.9502, 32.5051 - 35.3273 I}, {25.05,
32.2659 - 35.1744 I}, {25.1498, 32.4265 - 35.3344 I}, {25.2496,
32.1252 - 35.2248 I}, {25.3494, 31.9565 - 35.2524 I}, {25.4492,
32.0133 - 35.1773 I}, {25.549, 31.8092 - 35.1778 I}, {25.6488,
32.0878 - 35.0699 I}, {25.7486, 31.5873 - 34.6076 I}, {25.8484,
31.2771 - 35.0794 I}, {25.9482, 31.5152 - 34.5532 I}, {26.048,
31.303 - 34.9144 I}, {26.1478, 30.9818 - 34.9513 I}, {26.2476,
30.8336 - 34.773 I}, {26.3474, 31.0272 - 34.896 I}, {26.4472,
30.6672 - 34.7709 I}, {26.547, 30.5985 - 34.8338 I}, {26.6468,
30.5358 - 34.8296 I}, {26.7466, 30.2887 - 34.7839 I}, {26.8464,
30.2015 - 34.5392 I}, {26.9462, 30.0512 - 34.6196 I}, {27.046,
29.9653 - 34.541 I}, {27.1458, 29.804 - 34.5801 I}, {27.2456,
29.6835 - 34.5962 I}, {27.3454, 29.6406 - 34.502 I}, {27.4452,
29.4564 - 34.4098 I}, {27.545, 29.3452 - 34.4996 I}, {27.6448,
29.1932 - 34.3772 I}, {27.7446, 29.1953 - 34.2989 I}, {27.8444,
29.0024 - 34.1343 I}, {27.9442, 28.8577 - 34.1517 I}, {28.044,
28.703 - 34.0196 I}, {28.1438, 28.6418 - 34.0946 I}, {28.2436,
28.58 - 34.0431 I}, {28.3434, 28.3131 - 33.9986 I}, {28.4432,
28.2246 - 33.9749 I}, {28.543, 28.1047 - 34.0503 I}, {28.6428,
28.0789 - 33.7741 I}, {28.7426, 27.8409 - 33.7681 I}, {28.8424,
27.8007 - 33.8519 I}, {28.9422, 27.5743 - 33.7117 I}, {29.042,
27.8188 - 33.6009 I}, {29.1418, 27.6161 - 33.5073 I}, {29.2416,
27.4297 - 33.6119 I}, {29.3414, 27.3634 - 33.5724 I}, {29.4412,
27.3038 - 33.5951 I}, {29.541, 27.1626 - 33.2034 I}, {29.6408,
27.0371 - 33.3796 I}, {29.7406, 26.8387 - 33.268 I}, {29.8404,
26.8122 - 33.2622 I}, {29.9402, 26.6344 - 33.2716 I}, {30.04,
26.6695 - 33.3534 I}, {30.1398, 26.5725 - 32.789 I}, {30.2396,
26.3733 - 33.226 I}, {30.3394, 26.2388 - 32.951 I}, {30.4392,
26.5557 - 33.177 I}, {30.539, 26.1338 - 33.0525 I}, {30.6388,
25.8534 - 32.933 I}, {30.7386, 26.0376 - 32.9274 I}, {30.8384,
26.0954 - 32.9956 I}, {30.9382, 25.7592 - 32.9014 I}, {31.038,
25.4958 - 32.8915 I}, {31.1378, 25.6263 - 32.6875 I}, {31.2376,
25.3777 - 32.7197 I}, {31.3374, 25.4505 - 32.852 I}, {31.4372,
25.2404 - 32.7977 I}, {31.537, 25.3987 - 32.4925 I}, {31.6368,
25.1363 - 32.3624 I}, {31.7366, 24.9608 - 32.3569 I}, {31.8364,
24.6876 - 32.5949 I}, {31.9362, 24.8773 - 32.361 I}, {32.036,
24.7508 - 32.3924 I}, {32.1358, 24.7399 - 32.16 I}, {32.2356,
24.5998 - 32.0557 I}, {32.3354, 24.5433 - 32.1212 I}, {32.4352,
24.3455 - 32.3712 I}, {32.535, 23.9694 - 32.0721 I}, {32.6348,
24.1629 - 31.8856 I}, {32.7346, 24.0315 - 32.056 I}, {32.8344,
23.7825 - 31.8144 I}, {32.9342, 24.0951 - 31.867 I}, {33.034,
23.6539 - 31.6875 I}, {33.1338, 23.7025 - 31.7136 I}, {33.2336,
23.5243 - 31.7632 I}, {33.3334, 23.6693 - 31.6976 I}, {33.4332,
23.2781 - 32.0396 I}, {33.533, 23.2774 - 31.6915 I}, {33.6328,
23.4158 - 31.588 I}, {33.7326, 23.2458 - 31.0499 I}, {33.8324,
23.385 - 31.3526 I}, {33.9322, 23.1822 - 31.9303 I}, {34.032,
23.0974 - 30.9541 I}, {34.1318, 22.9448 - 30.8666 I}, {34.2316,
22.9013 - 31.1484 I}, {34.3314, 23.2542 - 31.2115 I}, {34.4312,
22.397 - 31.0947 I}, {34.531, 22.9416 - 31.5076 I}, {34.6308,
21.9635 - 31.074 I}, {34.7306, 22.3843 - 31.0943 I}, {34.8304,
22.3009 - 31.1258 I}, {34.9302, 22.1042 - 31.1654 I}, {35.03,
22.5128 - 30.9664 I}, {35.1298, 22.339 - 31.1127 I}, {35.2296,
21.9851 - 30.8603 I}, {35.3294, 22.1375 - 30.9703 I}, {35.4292,
21.9631 - 30.7983 I}, {35.529, 22.286 - 30.9338 I}, {35.6288,
21.458 - 30.3835 I}, {35.7286, 21.3285 - 30.7709 I}, {35.8284,
21.8479 - 30.4151 I}, {35.9282, 21.6401 - 30.5023 I}, {36.028,
21.6042 - 30.496 I}, {36.1278, 21.461 - 30.0798 I}, {36.2276,
21.2528 - 29.9247 I}, {36.3274, 21.2585 - 30.5237 I}, {36.4272,
21.3279 - 30.3249 I}, {36.527, 21.4263 - 30.1211 I}, {36.6268,
21.4058 - 30.3662 I}, {36.7266, 21.1253 - 30.1514 I}, {36.8264,
21.2413 - 30.4202 I}, {36.9262, 21.1598 - 30.0798 I}, {37.026,
20.9465 - 30.0659 I}, {37.1258, 20.6829 - 30.0269 I}, {37.2256,
20.6885 - 29.8712 I}, {37.3254, 20.4891 - 30.1871 I}, {37.4252,
20.7532 - 30.1849 I}, {37.525, 20.7161 - 29.8535 I}, {37.6248,
20.735 - 29.5632 I}, {37.7246, 20.2155 - 29.6935 I}, {37.8244,
20.0769 - 29.622 I}, {37.9242, 20.4147 - 29.9926 I}, {38.024,
20.9966 - 29.8357 I}, {38.1238, 20.4021 - 29.9879 I}, {38.2236,
20.136 - 29.2182 I}, {38.3234, 20.0021 - 29.7264 I}, {38.4232,
20.063 - 29.5102 I}, {38.523, 19.7925 - 29.852 I}, {38.6228,
20.2011 - 29.744 I}, {38.7226, 19.6156 - 29.6467 I}, {38.8224,
19.9306 - 28.4294 I}, {38.9222, 19.3085 - 28.5398 I}, {39.022,
19.1026 - 29.2675 I}, {39.1218, 19.1628 - 29.3752 I}, {39.2216,
19.605 - 28.9693 I}, {39.3214, 19.6677 - 28.804 I}, {39.4212,
18.8673 - 28.5751 I}, {39.521, 18.5662 - 28.675 I}, {39.6208,
18.6279 - 28.8541 I}, {39.7206, 18.8227 - 28.9262 I}, {39.8204,
19.4427 - 28.9469 I}, {39.9202, 19.5005 - 28.3647 I}, {40.02,
18.4979 - 28.1689 I}, {40.1198, 18.2211 - 28.3022 I}, {40.2196,
18.569 - 28.2742 I}, {40.3194, 18.7859 - 28.6754 I}, {40.4192,
18.5712 - 28.4534 I}, {40.519, 19.3936 - 27.9846 I}, {40.6188,
18.4431 - 27.5675 I}, {40.7186, 18.1889 - 28.0742 I}, {40.8184,
18.5421 - 28.2055 I}, {40.9182, 18.6898 - 28.4374 I}, {41.018,
18.3297 - 28.0645 I}, {41.1178, 18.8042 - 27.8506 I}, {41.2176,
18.3059 - 27.344 I}, {41.3174, 18.5508 - 27.7339 I}, {41.4172,
17.9299 - 28.1919 I}, {41.517, 17.9836 - 27.7716 I}, {41.6168,
18.2556 - 27.7537 I}, {41.7166, 17.9775 - 28.0211 I}, {41.8164,
18.3221 - 27.786 I}, {41.9162, 17.7802 - 27.4491 I}, {42.016,
17.9248 - 27.6812 I}, {42.1158, 17.9283 - 27.5206 I}, {42.2156,
18.1385 - 27.7968 I}, {42.3154, 17.5093 - 27.5915 I}, {42.4152,
18.0288 - 27.3542 I}, {42.515, 17.8391 - 27.5258 I}, {42.6148,
18.2251 - 27.4179 I}, {42.7146, 17.4823 - 27.427 I}, {42.8144,
17.5969 - 27.0717 I}, {42.9142, 17.2438 - 27.2268 I}, {43.014,
17.5722 - 26.9162 I}, {43.1138, 17.0512 - 27.4752 I}, {43.2136,
17.731 - 27.7474 I}, {43.3134, 17.9637 - 27.0977 I}, {43.4132,
17.8519 - 26.5461 I}, {43.513, 16.4758 - 27.4136 I}, {43.6128,
17.5664 - 27.4024 I}, {43.7126, 17.3127 - 26.7502 I}, {43.8124,
17.0702 - 28.1142 I}, {43.9122, 16.944 - 25.6304 I}, {44.012,
17.6921 - 26.3435 I}, {44.1118, 16.7857 - 26.9177 I}, {44.2116,
17.8315 - 26.765 I}, {44.3114, 16.7895 - 26.8332 I}, {44.4112,
16.9196 - 26.2703 I}, {44.511, 16.7878 - 26.4833 I}, {44.6108,
16.5801 - 26.8623 I}, {44.7106, 16.9558 - 26.4815 I}, {44.8104,
16.7723 - 26.5304 I}, {44.9102, 16.6461 - 26.5571 I}, {45.01,
16.445 - 26.2377 I}, {45.1098, 16.3595 - 26.1725 I}, {45.2096,
16.6057 - 26.3542 I}, {45.3094, 16.8819 - 26.7169 I}, {45.4092,
16.7706 - 26.4544 I}, {45.509, 16.186 - 26.0565 I}, {45.6088,
16.5001 - 26.0794 I}, {45.7086, 16.2941 - 26.4279 I}, {45.8084,
16.0685 - 26.4002 I}, {45.9082, 16.5396 - 26.0268 I}, {46.008,
15.7409 - 26.6186 I}, {46.1078, 16.4295 - 26.1331 I}, {46.2076,
15.5538 - 26.457 I}, {46.3074, 16.4832 - 25.7581 I}, {46.4072,
15.8374 - 26.0055 I}, {46.507, 15.7819 - 26.8736 I}, {46.6068,
15.8253 - 25.7855 I}, {46.7066, 16.385 - 26.3928 I}, {46.8064,
16.2869 - 26.1208 I}, {46.9062, 15.5076 - 25.7764 I}, {47.006,
15.9992 - 26.1976 I}, {47.1058, 15.8232 - 25.8068 I}, {47.2056,
15.6416 - 25.5574 I}, {47.3054, 15.7279 - 25.4791 I}, {47.4052,
15.23 - 24.9536 I}, {47.505, 16.2428 - 24.9651 I}, {47.6048,
15.4889 - 26.3522 I}, {47.7046, 15.2959 - 24.9988 I}, {47.8044,
15.3342 - 25.6946 I}, {47.9042, 15.3437 - 25.544 I}, {48.004,
15.3685 - 25.4386 I}, {48.1038, 15.2888 - 25.6183 I}, {48.2036,
15.6667 - 25.3608 I}, {48.3034, 15.5084 - 24.8011 I}, {48.4032,
15.3141 - 25.2034 I}, {48.503, 15.0334 - 25.3849 I}, {48.6028,
15.3935 - 24.5825 I}, {48.7026, 15.4125 - 25.0509 I}, {48.8024,
14.9718 - 25.1981 I}, {48.9022, 14.7825 - 25.0575 I}, {49.002,
15.2234 - 25.1159 I}, {49.1018, 14.9325 - 25.1141 I}, {49.2016,
14.9938 - 25.3016 I}, {49.3014, 14.9617 - 24.3614 I}, {49.4012,
14.8526 - 25.004 I}, {49.501, 14.8749 - 24.9538 I}, {49.6008,
15.208 - 24.7995 I}, {49.7006, 14.6468 - 25.1923 I}, {49.8004,
14.7013 - 24.5138 I}, {49.9002, 15.0272 - 24.3538 I}, {50.,
14.9279 - 24.6538 I}};
</code></pre>
<p>Here is the the code:</p>
<pre><code>Subscript[ϵ, 0] = 8.854*10^-12
modelfit =
NonlinearModelFit[dataset,
Subscript[ϵ, ∞] + (Subscript[ϵ, 1] -
Subscript[ϵ, ∞])/(
1 + I*2*π*f*Subscript[τ, 1]) + σ/(
I*2*π*f*Subscript[ϵ, 0]), {{Subscript[ϵ, 1],
78.88}, {Subscript[ϵ, ∞],
5.2}, {Subscript[τ, 1], 0.008}, {σ, 0.2}}, f];
</code></pre>
<p>This error appears:</p>
<blockquote>
<pre><code>NonlinearModelFit::nrlnum: The function value {0.642538 -3.5951*10^10 I,0.382869
-1.79935*10^10 I,0.695794 -1.19997*10^10 I,<<45>>,0.867068 -7.35134*10^8
I,0.878805 -7.20431*10^8 I,<<451>>} is not a list of real numbers with dimensions
{501} at {Subscript[ϵ, 1],Subscript[ϵ, ∞],Subscript[τ, 1],σ} = {78.88,5.2,0.008,0.2}.
</code></pre>
</blockquote>
<p>Can anyone please help me to get rid of this? Thank you.</p>
| Anton Antonov | 34,008 | <p>Can you use model fitting that is done separately for the real and imaginary parts?</p>
<p>Example is given below.</p>
<h2>Separate fitting of real and imaginary parts</h2>
<p>Fit for the real parts:</p>
<pre><code>e0 = 8.854*10^-12;
modelfitRe = NonlinearModelFit[Re /@ dataset, Re[eInf + (e1 - eInf)/(1 + I*2*\[Pi]*f*t1) + \[Sigma]/(I*2*\[Pi]*f*e0)], {{e1, 78.88}, {eInf, 5.2}, {t1, 0.008}, {\[Sigma], 0.2}},f]
</code></pre>
<p><a href="https://i.stack.imgur.com/EIpV1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EIpV1.png" alt="enter image description here" /></a></p>
<p>Fit for the imaginary parts:</p>
<pre><code>modelfitIm = NonlinearModelFit[Map[{#[[1]], Im[#[[2]]]} &, dataset], Im[eInf + (e1 - eInf)/(1 + I*2*\[Pi]*f*t1) + \[Sigma]/(I*2*\[Pi]*f*e0)], {{e1, 78.88}, {eInf, 5.2}, {t1, 0.008}, {\[Sigma], 0.2}},f]
</code></pre>
<blockquote>
<p>NonlinearModelFit::sszero: The step size in the search has become less than the tolerance prescribed by the PrecisionGoal option, but the gradient is larger than the tolerance specified by the AccuracyGoal option. There is a possibility that the method has stalled at a point that is not a local minimum.</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/Yo8gD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Yo8gD.png" alt="enter image description here" /></a></p>
<p>Make the complex numbers list of fits:</p>
<pre><code>datasetFit = Map[{#, modelfitRe[#] + I*modelfitIm[#]} &, dataset[[All, 1]]];
</code></pre>
<p>See the errors for the real parts:</p>
<pre><code>{ResourceFunction["RecordsSummary"][Abs[dataset[[All, 2]] - datasetFit[[All, 2]]]], ResourceFunction["RecordsSummary"][Abs[dataset[[All, 2]] - datasetFit[[All, 2]]]/Abs[dataset[[All, 2]]]]}
</code></pre>
<p><a href="https://i.stack.imgur.com/oqMnc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oqMnc.png" alt="enter image description here" /></a></p>
<p>See the errors for the imaginary parts:</p>
<pre><code>{ResourceFunction["RecordsSummary"][Abs[Im[dataset[[All, 2]] - datasetFit[[All, 2]]]]], ResourceFunction["RecordsSummary"][Abs[Im[dataset[[All, 2]] - datasetFit[[All, 2]]]/Im[dataset[[All, 2]]]]]}
</code></pre>
<p><a href="https://i.stack.imgur.com/0zQxo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0zQxo.png" alt="enter image description here" /></a></p>
<p>Plot the original data and the fits:</p>
<pre><code>gr1 = MapThread[
ListPlot[#, PlotRange -> All, PlotTheme -> "Detailed", PlotLegends -> {"Original"}, PlotStyle -> {{PointSize[0.015], GrayLevel[0.7]}}, PlotLabel -> #2, ImageSize -> Large] &, {{Map[{#[[1]], Re[#[[2]]]} &, dataset], Map[{#[[1]], Im[#[[2]]]} &, dataset]}, {"Real parts", "Imaginary parts"}}];
gr2 = ListLinePlot[#, PlotRange -> All, PlotTheme -> "Detailed", PlotLegends -> {"Fit"}, PlotStyle -> {Red}, PlotLabel -> "Imaginary parts", ImageSize -> Large] & /@ {Map[{#[[1]], Re[#[[2]]]} &, datasetFit], Map[{#[[1]], Im[#[[2]]]} &, datasetFit]};
MapThread[Show, {gr1, gr2}]
</code></pre>
<p><a href="https://i.stack.imgur.com/952jV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/952jV.png" alt="enter image description here" /></a></p>
|
2,316,246 | <p>I was pondering on how to answer this question. It looks like this:</p>
<p>In a certain region of the country it is known from past experience that the probability of selecting an adult over 40 years of age with cancer is 0.05. If the probability of a doctor correctly diagnosing a person with cancer as having the disease is 0.78 and the probability of incorrectly diagnosing a person without cancer as having the disease is 0.06, what is the probability that a person is diagnosed as having cancer?</p>
<p>I thought it was very obvious that the answer above would be 0.05 because it talks about a probability that a person having a cancer is 0.05. The answer written in the book was 0.0960.</p>
<p>My head is messed up right now, pondering too hard on how to answer it. How do you answer it?</p>
<p>Does the problem above can be visualized using Venn diagrams?</p>
| Michael Hardy | 11,667 | <p>This is the sort of situation for which the normal approximation was first introduced (by Abraham de Moivre in the 18th century). I have up-voted the answer by Big Agness, but I want to give some detail here that was not in that answer.</p>
<p>The difficulty is that computing the exact answer arithmetically is prohibitively expensive in this kind of situation, and would gain nothing of practical import over the normal approximation.</p>
<p>We have
\begin{align}
\text{expected value} & = np = 5000 \times 0.5 = 250 \\
\text{variance} & = npq = 5000 \times 0.5 \times 0.95 = 237.5 \\
\text{so standard deviation} & = \sqrt{npq} \approx 15.411\ldots \\[12pt]
\Pr(199.5 < X < 200.5) & \approx \Pr\left( \frac{199.5-250}{15.411} < \frac{X-250}{15.411} < \frac{200.5 - 250}{15.411} \right) \\[10pt]
& \approx \Pr\left( -3.277 < Z < -3.212 \right) = \Phi(-3.212) - \Phi(-3.277).
\end{align}
You get those numbers from a table or from standard software packages (unless you want to work with numerical algorithms used in creating those tables or software).</p>
<p>Not surprisingly, you get a very small probability because $200$ is so far from the mean.</p>
<p>If the binomial distribution were not a better model than the Poisson distirbution in this case, you would use the fact that the expected value and variance of $\operatorname{Poisson}(\lambda)$ are both $\lambda,$ and the standard normal distribution would approximate the distribution of
$$
\frac{X-\lambda}{\sqrt{\lambda}}.
$$
The Poisson approximation to the binomial should be used when $\lambda$ is small and $n$ is big.</p>
|
886,337 | <p>I need to find $$\int x\sqrt{(a^2 - x^2)}dx$$</p>
<p>I tried putting $x=a cos(t)$ but I ended up getting a very complicated expression, so any tips?</p>
| Shine | 157,361 | <p>$-\frac{1}{3}(a^2-x^2)^\frac{3}{2}$ is one of the origional function because $\frac{d}{dx}\bigg(-\frac{1}{3}(a^2-x^2)^\frac{3}{2}\bigg)=x\sqrt{a^2-x^2}.$</p>
<p>Why did I come up with it because $\int x\sqrt{a^2-x^2}dx=\int\sqrt{a^2-x^2}d(x^2)=-\frac{1}{3}(a^2-x^2)^\frac{3}{2}$+C.</p>
|
4,315,926 | <p>This question is in the context of a robotics problem. The goal is to track a robot using both its onboard odometry system and a VR system (HTC Vive Pro) using a VR controller mounted to the robot.</p>
<p>What is <strong>known</strong> is the transformation between odometry origin and the robot (measurements <span class="math-container">$A_n$</span>) and between the VR origin and the controller (measurements <span class="math-container">$C_n$</span>). Ignoring both systems' inaccuracies for now, we can also assume the robot's pose according to both systems is identical (<span class="math-container">$I$</span>). Driving around will result in many pairs of measurements (<span class="math-container">$A_n$</span>, <span class="math-container">$C_n$</span>).</p>
<p>What is <strong>unknown</strong> is the fixed transformation <span class="math-container">$B$</span> between the two coordinate systems and the fixed transformation <span class="math-container">$D$</span> between the robot's origin and the VR controller. All transformations involved are proper rigid.</p>
<p>The chain of transformation looks as follows.</p>
<p><a href="https://i.stack.imgur.com/4Qc2s.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4Qc2s.png" alt="Transformation graph for a robot tracked by both a VR system and wheel odometry" /></a></p>
<p>This leaves us with an equation system of <span class="math-container">$4 \times 4$</span> homogeneous transformation matrices</p>
<p><span class="math-container">$$A_n \cdot B \cdot C_n \cdot D = I$$</span></p>
<p>with <span class="math-container">$I$</span> being the <span class="math-container">$4 \times 4$</span> identity matrix and <span class="math-container">$(A_n, C_n)$</span> being (many) pairs of measurements.</p>
<p>I am looking for the <strong>optimal</strong> (least-squares, I suppose) solution for <span class="math-container">$B$</span> and <span class="math-container">$D$</span>, so that the equation holds approximately true for all pairs of measurements.</p>
| Bertrand87 | 953,938 | <p>One way using complex analysis:</p>
<p>Since the integrand is even:
<span class="math-container">$$
I=\int_{0}^{\pi} \frac{d x}{(a-\cos x)^n} = \frac{1}{2} \int_{-\pi}^{\pi} \frac{d x}{(a-\cos x)^n}
$$</span></p>
<p>Since <span class="math-container">$a>1$</span>, we can expand the integrand with the generalized binomial theorem:</p>
<p><span class="math-container">$$ I = \frac{1}{2} \int_{-\pi}^{\pi} \frac{d x}{(a-\cos x)^n} = \frac{1}{2} \int_{-\pi}^{\pi} \sum_{j=0}^{\infty} \binom{-n}{j} a^{-n-j}(-1)^j\cos^j(x) dx $$</span></p>
<p>From Fubini-Tonelli we can interchange integral and series:</p>
<p><span class="math-container">$$ I = \frac{1}{2} \sum_{j=0}^{\infty} \binom{-n}{j}a^{-n-j}(-1)^j\int_{-\pi}^{\pi} \cos^j(x) dx $$</span></p>
<p>Do the following substitution</p>
<p><span class="math-container">$$ \cos x = \frac{z+z^{-1}}{2}$$</span></p>
<p><span class="math-container">$$ dx = \frac{dz}{zi}$$</span></p>
<p>The integral is now a contour integral round the unit complex circle:</p>
<p><span class="math-container">$$ I = \frac{1}{2} \sum_{j=0}^{\infty} \binom{-n}{j}\frac{a^{-n-j}(-1)^j}{2^j}\oint_{|z|=1} \frac{(z^2+1)^j}{z^{j+1} }dz$$</span></p>
<p>The integral has a pole at <span class="math-container">$z=0$</span>. To find the residue, expand the integrand:</p>
<p><span class="math-container">$$\frac{(z^2+1)^j}{z^{j+1} }= \sum_{k=0}^{j}\binom{j}{k} z^{2k-j-1}$$</span></p>
<p>The residue is the coefficient of <span class="math-container">$z^{2k-j-1}= z^{-1}$</span></p>
<p>Then</p>
<p><span class="math-container">$$ 2k-j-1 = -1 \Longrightarrow k = \frac{j}{2}$$</span></p>
<p>Therefore, the residue exists if <span class="math-container">$j$</span> is <span class="math-container">$\textbf{divisible by}$</span> <span class="math-container">$2$</span>:</p>
<p><span class="math-container">$$\oint_{|z|=1} \frac{(z^2+1)^j}{z^{j+1} }dx = 2\pi i \operatorname{Res}\left(\frac{(z^2+1)^j}{z^{j+1}},0\right) = 2\pi i \binom{j}{\frac{j}{2}} $$</span></p>
<p>Hence, we have</p>
<p><span class="math-container">$$ I = \pi\sum_{j=0}^{\infty} \binom{-n}{2j}\binom{2j}{j}\frac{a^{-n-2j}}{2^{2j}} $$</span></p>
<p>Note</p>
<p><span class="math-container">$$\binom{-n}{2j} = \frac{(-n-2j+1)_{2j}}{(2j!}$$</span></p>
<p>where <span class="math-container">$(x)_{n} = x(x+1)\cdots(x+n-1)$</span> is the rising factorial (Pochhammer polynomial)</p>
<p>and</p>
<p><span class="math-container">$$\binom{2j}{j} = \frac{(2j)!}{j!^2}$$</span></p>
<p>Hence</p>
<p><span class="math-container">$$\binom{-n}{2j}\binom{2j}{j} = \frac{(-n-2j+1)_{2j}}{j!^2}$$</span></p>
<p>From the reflection formula for the Pochhammer polynomial:</p>
<p><span class="math-container">$$(-x)_{m} = (-1)^m(x-m+1)_{m}$$</span></p>
<p>we have
<span class="math-container">$$(-n-2j+1)_{2j}= (n)_{2j}$$</span></p>
<p>and from the duplication formula for the degree of the Pochhammer polynomial:</p>
<p><span class="math-container">$$(x)_{2m} = 4^m\left(\frac{x}{2}\right)\left(\frac{1+x}{2}\right)$$</span></p>
<p>we have</p>
<p><span class="math-container">$$(n)_{2j} = 4^{j}\left(\frac{n}{2}\right)\left(\frac{n+1}{2}\right)$$</span></p>
<p>Hence</p>
<p><span class="math-container">$$ I = \pi\sum_{j=0}^{\infty} \binom{-n}{2j}\binom{2j}{j}\frac{a^{-n-2j}}{2^{2j}} = \frac{\pi}{a^n}\sum_{j=0}^{\infty}\frac{\left(\frac{n}{2}\right)\left(\frac{n+1}{2}\right)}{(1)_{j}} \frac{\left(\frac{1}{a^2}\right)^j}{j!}$$</span></p>
<p><span class="math-container">$$\Longrightarrow I = \frac{\pi}{a^n}{}_{2}F_{1}\left({\frac{n}{2},\frac{n+1}{2}\atop 1};\frac{1}{a^2}\right)$$</span></p>
<p>where <span class="math-container">${}_{2}F_{1}({a,b\atop c};z)$</span> is the Guassian function</p>
<p>Now using the <a href="http://dlmf.nist.gov/15.9.E17" rel="nofollow noreferrer">following forumula</a>:</p>
<p><span class="math-container">$${}_{2}F_{1}\left({a,a+\tfrac{1}{2}\atop c};z\right)=2^{c-1}z^{\frac{(1-c)}{2}}%
(1-z)^{-a+\left(\frac{c-1}{2}\right)}P^{1-c}_{2a-c}\left(\frac{1}{\sqrt{1-z}}\right) $$</span></p>
<p>where <span class="math-container">$P_{n}^{(\alpha,\beta)}$</span> is the Jacobi polynomial</p>
<p>we have</p>
<p><span class="math-container">$$ I =\frac{\pi}{a^n}{}_{2}F_{1}\left({\frac{n}{2},\frac{n+1}{2}\atop 1};\frac{1}{a^2}\right)= \frac{\pi}{a^n}\left(1-\frac{1}{a^2}\right)^{-\frac{n}{2}}P_{n-1}\left(\frac{a}{\sqrt{a^2-1}}\right)$$</span></p>
<p>where <span class="math-container">$P_{n}$</span> is the standard Legendre polynomial which satisfies the <em>Rodrigues' formula</em>:</p>
<p><span class="math-container">$$ P_{n}(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2-1)^n$$</span></p>
<p>Finally, we can conclude</p>
<p><span class="math-container">$$\boxed{\int_{0}^{\pi} \frac{d x}{(a-\cos x)^n} = \frac{\pi}{2^{n-1}(n-1)!a^n}\left(1-\frac{1}{a^2}\right)^{-\frac{n}{2}}\left( \frac{d^{n-1}}{dx^{n-1}} (x^2-1)^{n-1} \right)_{x = \frac{a}{\sqrt{a^2-1}}}}$$</span></p>
<p>This is similar to the formula you found. However, Legendre polynomials also have an explicit formula:</p>
<p><span class="math-container">$$P_{n} (x) = \sum_{j} (-1)^{\frac{j}{2}} \frac{(2n-j-1)!!}{j!!(n-j)!} x^{n-j} \quad j=0,2,4,...,n-\frac{1}{2}\pm \frac{1}{2}$$</span></p>
<p>(ends with <span class="math-container">$n$</span> in the case <span class="math-container">$n$</span> even and ends with <span class="math-container">$n-1$</span> in the case <span class="math-container">$n$</span> odd).</p>
<p>Hence</p>
<p><span class="math-container">$$ \boxed{I =\int_{0}^{\pi} \frac{d x}{(a-\cos x)^n} = \frac{\pi}{a^n}\left(1-\frac{1}{a^2}\right)^{-\frac{n}{2}}\sum_{j} (-1)^{\frac{j}{2}} \frac{(2n-j-3)!!}{j!!(n-1-j)!} \left(\frac{a}{\sqrt{a^2-1}}\right)^{n-1-j} \quad j=0,2,4,...,n-\frac{3}{2}\pm \frac{1}{2} \textrm{ ($n-1$ even or odd)}} $$</span></p>
<p>It turned out that the polynomial you are looking for is this Legendre polynomial.</p>
|
1,514,028 | <p>It's pretty obvious to me as I see the plot of the two functions but how can I prove it with some algebra?</p>
| Yes | 155,328 | <p>By mean-value theorem, if $-1 \leq x < 0$, then
$$
|\sin x - \sin 0| = |\sin x| \leq |x|\sup_{0 < t < x}|\cos t| \leq |x|.
$$</p>
|
4,149,316 | <p>Is <span class="math-container">$f(x,y) = x^2 - 2e^y$</span> concave or convex?</p>
<p>My thought process is as such:</p>
<p><span class="math-container">$f'(x,y)=2x - 2e^y$</span>;</p>
<p><span class="math-container">$f''(x,y)=2 - 2e^y$</span></p>
<p>Thus when <span class="math-container">$y<0$</span> (<span class="math-container">$2e^y<2$</span>), <span class="math-container">$f''(x,y)<0 \implies f(x,y)$</span> is convex.</p>
<p>Likewise, when <span class="math-container">$y>0$</span>, <span class="math-container">$f''(x,y)>0 \implies f(x,y)$</span> is concave.</p>
<p>But the answer says <span class="math-container">$f(x,y)$</span> is neither concave nor convex - why is that?</p>
| Siong Thye Goh | 306,553 | <p>The hessian is <span class="math-container">$\begin{bmatrix} 2 & 0 \\ 0 & -2e^{-y}\end{bmatrix}$</span>. The eigenvalues are <span class="math-container">$2$</span> (positive) and <span class="math-container">$-2e^{-y}$</span> (negative). Hence it is indefinite.</p>
<p>If you fix <span class="math-container">$y$</span> and move along <span class="math-container">$x$</span>, you trace out a convex path.</p>
<p>If you fix <span class="math-container">$x$</span> and move along <span class="math-container">$y$</span>, you trace out a concave path.</p>
<p>It is like a saddle.</p>
<p><a href="https://i.stack.imgur.com/EVAxB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EVAxB.jpg" alt="enter image description here" /></a></p>
|
2,079,246 | <p>Is (A vector +B vector) always perpendicular to (A vector - B vector)?</p>
| Vidyanshu Mishra | 363,566 | <p>Try to get resultant from the parallelogram law, and you will see that $(\vec A + \vec B)$ and $(\vec A - \vec B)$ are perpendicular only when their magnitude is equal.</p>
<p>Resultant of $(\vec A + \vec B)=\sqrt{a^2+b^2+2AB\cos\theta}$ and resultant of $(\vec A - \vec B)=\sqrt{a^2+b^2-2AB\cos\theta}$ where $\theta$ is the angle between $\vec A $ and $\vec B$.</p>
|
1,977,081 | <p>You play a game using a standard six-sided die. You start with 0 points. Before every roll, you decide whether you want to continue the game or end it and keep your points. After each roll, if you rolled 6, then you lose everything and the game ends. Otherwise, add the score from the die to your total points and continue/stop the game.</p>
<p>When should one stop playing this game? Obviously, one wants to maximize total score.</p>
<p>As I was asked to show my preliminary results on this one, here they are:</p>
<p>If we simplify the game to getting 0 on 6 and 3 otherwise, we get the following: </p>
<p>$$\begin{align}
EV &= \frac{5}{6}3+\frac{25}{36}6+\frac{125}{216}9+\ldots\\[5pt]
&= \sum_{n=1}^{\infty}\left(\frac{5}{6}\right)^n3n
\end{align}$$ </p>
<p>which is divergent, so it would make sense to play forever, which makes this similar to the St. Petersburg paradox. Yet I can sense that I'm wrong somewhere!</p>
| Timothy Shields | 49,879 | <p>The question is missing the concept of <a href="https://en.wikipedia.org/wiki/Utility" rel="noreferrer">utility</a>, a function that specifies how much you value each possible outcome. In the game, the utility of ending the game with a certain score would be the value you place on that outcome. Although you could certainly argue it is implied in the question that the utility of a score is simply the score, I would like to add an answer that takes a non-trivial utility into account. If each point translated to 1000 USD, for example, you might have a utility that looks more like $U(x) = \log(x)$ than $U(x) = x$.</p>
<p>Let's say that $U(x)$ is the utility of score $x$ for $x \ge 0$ and assume that $U$ is monotonically non-decreasing. Then we might say that the optimal strategy is that which maximizes $E[U(X)]$, where $X$ is a random variable representing the final score if one plays with the policy where you roll the die if and only if your current score is less than $t \in \mathbb{Z}_{\ge 0}$. (It is clear that the optimal policy must have this form because the utility is non-decreasing.)</p>
<p>Let $Z$ denote current score. Suppose we are at a point in the game where our current score is $z \ge 0$. Then</p>
<p>$$E[U(X)|Z = z] = \frac{1}{6} \left( U(0) + \sum_{i=1}^5 E[U(X)|Z=z+i] \right) \text{ if } z < t$$</p>
<p>$$E[U(X)|Z = z] = U(z) \text{ if } z \ge t$$</p>
<p>Note that for many choices of $U(x)$ the recurrence relation is very difficult to simplify, and that, in the case of choosing to roll the die, we must consider the expected change in utility from that roll <em>and all future rolls</em>. The figures below are examples of what the above recurrence relation gives for $U(x) = x$ and $U(x) = \log_2(x + 1)$. The expression $E[U(X)]$ means $E[U(X)|Z=0]$, because at the start we have $0$ points. The horizontal axis corresponds to different policies, and the vertical axis corresponds to expected utility under each policy.</p>
<p><a href="https://i.stack.imgur.com/t0SI8.png" rel="noreferrer"><img src="https://i.stack.imgur.com/t0SI8.png" alt="enter image description here"></a></p>
<p><a href="https://gist.github.com/timothy-shields/aa3e295df8827fda2f4363edeeecb0a4" rel="noreferrer">Gist with Python code</a></p>
|
1,977,081 | <p>You play a game using a standard six-sided die. You start with 0 points. Before every roll, you decide whether you want to continue the game or end it and keep your points. After each roll, if you rolled 6, then you lose everything and the game ends. Otherwise, add the score from the die to your total points and continue/stop the game.</p>
<p>When should one stop playing this game? Obviously, one wants to maximize total score.</p>
<p>As I was asked to show my preliminary results on this one, here they are:</p>
<p>If we simplify the game to getting 0 on 6 and 3 otherwise, we get the following: </p>
<p>$$\begin{align}
EV &= \frac{5}{6}3+\frac{25}{36}6+\frac{125}{216}9+\ldots\\[5pt]
&= \sum_{n=1}^{\infty}\left(\frac{5}{6}\right)^n3n
\end{align}$$ </p>
<p>which is divergent, so it would make sense to play forever, which makes this similar to the St. Petersburg paradox. Yet I can sense that I'm wrong somewhere!</p>
| user381133 | 381,133 | <p>I interpreted this such that you're in competition against another player.</p>
<p>If playing second your strategy is simple - play until you've won.</p>
<p>If playing first, you must accumulate enough points such that the next player is statistically more likely to lose than win - that means making them roll more than 6 times (since they have a 1/6 chance of losing every roll). To achieve this you must amass at least the value of the average scoring roll (3) multiplied by 6, which is 18. Any score above this is statistically more likely to win than lose.</p>
|
1,977,081 | <p>You play a game using a standard six-sided die. You start with 0 points. Before every roll, you decide whether you want to continue the game or end it and keep your points. After each roll, if you rolled 6, then you lose everything and the game ends. Otherwise, add the score from the die to your total points and continue/stop the game.</p>
<p>When should one stop playing this game? Obviously, one wants to maximize total score.</p>
<p>As I was asked to show my preliminary results on this one, here they are:</p>
<p>If we simplify the game to getting 0 on 6 and 3 otherwise, we get the following: </p>
<p>$$\begin{align}
EV &= \frac{5}{6}3+\frac{25}{36}6+\frac{125}{216}9+\ldots\\[5pt]
&= \sum_{n=1}^{\infty}\left(\frac{5}{6}\right)^n3n
\end{align}$$ </p>
<p>which is divergent, so it would make sense to play forever, which makes this similar to the St. Petersburg paradox. Yet I can sense that I'm wrong somewhere!</p>
| Mike Earnest | 177,399 | <p>It's not hard to see that the optimal strategy in this game is "stop on $k$ or higher," for some positive integer $k$. This is because what is optimal is only determined by your current score, not your history of rolls, and if rolling again on $k$ is unwise, then rolling again on $k+1$ is unwise as well (you stand to gain the same amount from future rolls, but have more to lose).</p>
<p>Let's compare the strategies "stop on $k$ or higher" with "stop on $k+1$ or higher". If your score is never $k$, then these strategies result in the same final score. If you do reach $k$, then the first strategy will result in $k$, while the second will result in $\frac56k+\frac16(1+2+3+4+5)$. Comparing these, we see that the stop on $k$ strategy is better precisely when $k>15$, is worse when $k<15$, and you should be indifferent between then two when $k=15$. </p>
|
1,977,081 | <p>You play a game using a standard six-sided die. You start with 0 points. Before every roll, you decide whether you want to continue the game or end it and keep your points. After each roll, if you rolled 6, then you lose everything and the game ends. Otherwise, add the score from the die to your total points and continue/stop the game.</p>
<p>When should one stop playing this game? Obviously, one wants to maximize total score.</p>
<p>As I was asked to show my preliminary results on this one, here they are:</p>
<p>If we simplify the game to getting 0 on 6 and 3 otherwise, we get the following: </p>
<p>$$\begin{align}
EV &= \frac{5}{6}3+\frac{25}{36}6+\frac{125}{216}9+\ldots\\[5pt]
&= \sum_{n=1}^{\infty}\left(\frac{5}{6}\right)^n3n
\end{align}$$ </p>
<p>which is divergent, so it would make sense to play forever, which makes this similar to the St. Petersburg paradox. Yet I can sense that I'm wrong somewhere!</p>
| Adrian | 98,081 | <p>Here's an answer using value function iteration:</p>
<p>Let $v(x)$ denote the value of state $x$. The Bellman equation for this problem is $$v(x) = \max\left\{u(x), \frac{1}{6} \left[u(0) + \sum^5_{k=1}v(x+k)\right]\right\},$$
where $u$ is the utility of ending the game with $x$.</p>
<p>Several of the answers above point out that the optimal strategy must be "keep gambling until $x \geq x^\star$, then stop", and solve for $x^\star$ analytically.</p>
<p>Here's some R code that solves for $v$ (under linear utility) using value function iteration, i.e. by starting with an incorrect guess for $v$ and then iterating on the Bellman equation until convergence:</p>
<pre><code>## State x is number of points
## Each time period, either stop and enjoy utility u(X), or roll a six-sided die
## If roll a 6, lose everything, game ends, get u(0)
## If roll k < 6, move to state x' = x + k
max_x <- 30 # Maximum state to keep things simple -- in "true" problem this is infinite
value <- rep(0, max_x)
utility <- function(x) {
return(x) # Linear utility function
}
n_iterations <- 100 # Value function iteration
for(iteration in seq_len(n_iterations)) {
value_next <- value
for(x in seq_along(value)) {
x_next_if_roll <- pmin(x + seq_len(5), max_x) # Next period's states if roll 1, 2, ... , 5
value_next[x] <- max(utility(x), (1/6) * (utility(0) + sum(value[x_next_if_roll])))
}
distance <- mean(abs(value - value_next))
value <- value_next
message("iteration ", iteration, " value function distance ", round(distance, 4))
if(distance < 10^-8) break
}
max(which(value > seq_along(value))) # Last index at which it is optimal to keep playing
plot(value, type="b", xlab="state", ylab="value")
abline(a=0, b=1, lty=2, col="red")
abline(v=max(which(value > seq_along(value))) + 1, lty=2, col="grey") # Stop when reach this state
sum(value[1:5]) / 6 # Value before starting game (i.e. starting from state x=0): around 6.1537
</code></pre>
<p>The value function $v(x)$ when $u(x) = x$:</p>
<p><a href="https://i.stack.imgur.com/X0ehF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/X0ehF.png" alt="value fn"></a></p>
<p>When $v(x) = u(x)$ (i.e. when you reach a score greater than or equal to 15, the vertical line in the plot above), the optimal action is to stop playing and "consume" your $x$. </p>
|
113,266 | <p>One of the problems of my Statistics assignment requires me to calculate the expected value of a minimum. Here's the situation:</p>
<p>The following density function is given: $f(x) = \frac{\theta}{x^2}$ where $x\ge\theta$ and $\theta>0$</p>
<p>I have to calculate: E[min{$X_i$}]</p>
<p>My initial guess was that the smallest possible value of $X$ is $\theta$ sense $x\ge\theta$ so the expected value of the minimum would be $\theta$, but then again, in an acquired sample, you can't be 100% certain that the smallest possible value of $X$ will be one of the observations. So what then?</p>
| Did | 6,179 | <blockquote>
<p>Minima and expectations of nonnegative random variables are both well suited to complementary CDF, two remarks which together make for a painless solution. </p>
</blockquote>
<p>In the present case, $\mathrm P(X_i\geqslant x)=\theta/x$ for every $i$ and every $x\geqslant\theta$ hence $M_n=\min\limits_{1\leqslant i\leqslant n}X_i$ is such that $\mathrm P(M_n\geqslant x)=\mathrm P(X_1\geqslant x)^n$ is $(\theta/x)^n$ if $x\geqslant\theta$ and $1$ if $x\leqslant\theta$. </p>
<p>Now, $\mathrm E(Y)=\int\limits_0^{+\infty}\mathrm P(Y\geqslant x)\mathrm dx$ for every nonnegative random variable $Y$, hence
$$
\mathrm E(M_n)=\int_0^\theta 1\cdot\mathrm dx+\int_\theta^{+\infty}(\theta/x)^n\mathrm dx=\theta+\theta\int_1^{+\infty}\mathrm dx/x^n=\theta+\theta/(n-1),
$$
that is,
$$
\mathrm E(M_n)=n\theta/(n-1).
$$
One sees that $M_1=X_1$ is not integrable, that $M_n$ is integrable for every $n\geqslant2$, and that $M_n\to\theta$ almost surely and in $L^1$ when $n\to\infty$.</p>
|
2,130,527 | <p>I am trying to plot the function $(-2(\alpha-1)/\alpha)^{(-2(\alpha+2)/\alpha)}$ in maple,
but I got the following warning</p>
<pre><code>plot( (-2*(alpha-1)/alpha)^(-2*(alpha+2)/alpha) ,alpha=5..10);
</code></pre>
<blockquote>
<p>Warning, unable to evaluate the function to numeric values in the
region; complex values were detected.</p>
</blockquote>
<p>Actually the function have real values as roots but maple does not plot the values.</p>
<p>How do I plot it? </p>
| Community | -1 | <p><strong>Hint:</strong> Show that $(a_n)_n$ is a Cauchy sequence. Therefore:</p>
<ol>
<li>show $|a_{n+2}-a_{n+1}|\leq c^{n+1}|a_1-a_0|$</li>
<li>estimate $|a_n-a_m|$ by using step 1 and dont forget about the geometric sum and series, where you have to use $0<c<1$.</li>
</ol>
|
1,011 | <p>None of these are critical, just polishing stuff.</p>
<hr>
<p>When breaking out the up/down votes on a post, the top number crashes into the separator line:</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/OApgh.png" alt="vote crash"></p>
</blockquote>
<hr>
<p>For some badges, e.g. taxonomy on the parent site, the list of users gets clipped at the right edge:</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/5cHyg.png" alt="taxonomy clipped"></p>
</blockquote>
<hr>
<p>The the leading for the heading style in a post is too tight:</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/qi7Zr.png" alt="heading-style leading"></p>
</blockquote>
<hr>
<p>In a user's profile, on the activity page, the checkmark for accepted is clipped:</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/pfZBr.png" alt="clipped check in user activity"></p>
</blockquote>
<hr>
<p><s>There's also some strangeness with some links being blue and others red on the 10k tools page on the parent site, but I don't have a good screenshot of that.</s> (by-design)</p>
| theTuxRacer | 2,843 | <p><a href="http://i56.tinypic.com/2mweedv.jpg" rel="nofollow">http://i56.tinypic.com/2mweedv.jpg</a></p>
<p>there are two glitches in the image, so i ddnt need to post two images. the other glitch in the upper right corner.</p>
|
1,011 | <p>None of these are critical, just polishing stuff.</p>
<hr>
<p>When breaking out the up/down votes on a post, the top number crashes into the separator line:</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/OApgh.png" alt="vote crash"></p>
</blockquote>
<hr>
<p>For some badges, e.g. taxonomy on the parent site, the list of users gets clipped at the right edge:</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/5cHyg.png" alt="taxonomy clipped"></p>
</blockquote>
<hr>
<p>The the leading for the heading style in a post is too tight:</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/qi7Zr.png" alt="heading-style leading"></p>
</blockquote>
<hr>
<p>In a user's profile, on the activity page, the checkmark for accepted is clipped:</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/pfZBr.png" alt="clipped check in user activity"></p>
</blockquote>
<hr>
<p><s>There's also some strangeness with some links being blue and others red on the 10k tools page on the parent site, but I don't have a good screenshot of that.</s> (by-design)</p>
| Jonas Meyer | 1,424 | <p>Apparently these have been fixed, so we probably don't need this on the unanswered list anymore. (I'm posting this because the question was just bumped by <a href="https://math.meta.stackexchange.com/users/-1/community">Community</a>. Anyone who agrees with me can upvote either of the answers until at least one of them has a positive vote count.)</p>
|
1,044,096 | <p>Let $\mathscr { S } $ the class of all closed intervals $[a,b]$, where $a$ and $b$ are rational , i.e. $a,b\in \mathbb{Q}$ and $a<b$ . Show that $\mathscr{S}\cup \{\{p\}:p\in\mathbb{ Q }\}$ is a basis of the topology $\tau$ of the real line $\mathbb{R}$ generated by $\mathscr {S}$ .</p>
<p>Having problems show that the intersection of two base elements can be expressed as a union of elements of the base.</p>
| egreg | 62,967 | <p>The class $\mathscr{S}'=\mathscr{S}\cup\{\{p\}:p\in\mathbb{Q}\}\cup\{\emptyset\}$ is closed under binary intersections, because the intersection of two closed proper intervals is either a closed proper interval or a singleton, unless it is empty.</p>
<p>Every element of $\mathscr{S}'$ is an intersection of elements of $\mathscr{S}$.</p>
<p>What is $\bigcup\mathscr{S}$?</p>
|
5,197 | <p>Is it possible to split a list into two lists at a specific position? The main list is for example: </p>
<pre><code>data={{xa,ya},{xb,yb},{xc,yc},...,{xz,yz}}.
</code></pre>
<p>I want to split this list into two new lists: </p>
<pre><code>data1={{xa,ya},{xb,yb},...,{xi,yi}}
</code></pre>
<p>and </p>
<pre><code>data2={{xj,yj},{xk,yk},...,{xz,yz}}
</code></pre>
<p>at a specific y-value at position <code>i</code>. I was not successful in using <code>Part</code>. Maybe there is another possibility? </p>
| Dr. belisarius | 193 | <p>This can be accomplished easily using <code>Part</code> (<code>[[ ]]</code>) and <a href="http://reference.wolfram.com/mathematica/ref/Span.html"><code>Span</code></a> (<code>;;</code>), as follows:</p>
<pre><code>data = {{x1, y1}, {x2, y2}, {x3, y3}, {x4, y4}, {xz, yz}};
data[[;; 3]]
data[[4 ;;]]
(* ->
{{x1, y1}, {x2, y2}, {x3, y3}}
{{x4, y4}, {xz, yz}}
*)
</code></pre>
|
751,293 | <p>I'm nearing the end of the semester of an introductory-level complex variables class. (Very introductory -- it's the version of the class that's only required for engineering and physics majors, as it doesn't require two semesters of undergrad analysis that are prerequisite to the complex variables class for math majors, at my school.)</p>
<p>One of the many fascinating things I've seen this semester has been, speaking in broad terms, the behavior of analytic function, and the way that a harmonic function and its conjugate 'synchronize' (for lack of a better word) to create analyticity.</p>
<p>Despite the examples I've seen of harmonic functions being steady-state solutions to heat problems and showing up in descriptions of electric fields and whatnot, I feel like I lack any sense of what the 'harmonicity' of harmonic functions is all about.</p>
<p>On a side note: I do recall one day, however, where I was working through an example having to do with the level curves of a harmonic function and its conjugate, where I believe the significance what they points of intersection where always orthogonal. This 'mesh' notion created, for me, a visual image of how the the two functions work together to give an analytic function its synchronized, predictable nature. (But, as with most things, I could be mistaken in my understanding of this; these weren't points being stressed in the book, and it was in a chapter later in the book than what we'll cover in the class.)</p>
<blockquote>
<p>So, my question is that of how one ought finish this statement:<br />"I was considering a problem, and I intuitively knew the solution would need to be a harmonic function because the problem had the property..."</p>
</blockquote>
<p>I stress the word 'intuitively,' by the way.</p>
<p>If you feel this misses the point of harmonic functions and how I should think of them, then by all means please answer however you feel is appropriate.</p>
| Stephen Montgomery-Smith | 22,016 | <p>If you take a circular piece of wire, and dip it into a soap solution to make a bubble surface across the circle, and then place the circular piece of wire in the $xy$-plane, if the circular wire has small perturbations of order $\epsilon$, then the first order solutions of the $z$-coordinate as functions of $x$ and $y$ are harmonic.</p>
<p>This also intuitively explains the maximum principle - that a harmonic function can not attain its maximum on the interior of the domain.</p>
|
999,227 | <p>I have convinced myself that this true, however I'm at a loss of where I should start with this proof. Looking at a similar proof with 3 instead of 9, I saw the use of the basis representation theorem, but I'm not 100% comfortable with that, so it was hard for me to follow. Is there a way to do this with mod?</p>
| Clayton | 43,239 | <p><strong>Hint:</strong> $10\equiv1\pmod{9} $.</p>
|
2,483,254 | <p>I have to compute the limit</p>
<p>$$
\lim\limits_{N \to +\infty} \sqrt{N+1} \log \left(1+\frac{x}{N+1}\right)
$$
where $x \ge 0$ is fixed. I tried to see the previous as</p>
<p>$$
\log \lim\limits_{N \to +\infty} \left(1+\frac{x}{N+1}\right)^{\sqrt{N+1}}
$$
and to change variable, but it doesn't work. Intuitively, this limit is 0, but I have no clue on how to solve it. Can you help me?</p>
| the_candyman | 51,370 | <p>Let's pose $z = \frac{1}{\sqrt{N+1}}$. Then, when $N$ goes to $+\infty$, then $z$ goes to $0$. You can rewrite your limit as follows:</p>
<p>$$\lim_{z \to 0} \frac{1}{z} \log\left(1 + xz^2\right).$$</p>
<p>It is well known that:</p>
<p>$$\lim_{a \to 0} \frac{\log\left(1 + a\right)}{a} = 1.$$</p>
<p>Starting from this, we can rewrite the original limit as follows:</p>
<p>$$\lim_{z \to 0} xz \left(\frac{\log\left(1 + xz^2\right)}{xz^2}\right) = 0 \cdot 1 = 0.$$</p>
<p>The idea here is that "$a = xz^2$", since both goes to $0$...</p>
<hr>
<p>If the limit was</p>
<p>$$\lim\limits_{N \to +\infty} (N+1) \log \left(1+\frac{x}{N+1}\right),$$</p>
<p>then, passing to $z$, we get:</p>
<p>$$\lim_{z \to 0} \frac{1}{z^2} \log\left(1 + xz^2\right),$$</p>
<p>or equivalently</p>
<p>$$\lim_{z \to 0} x \left(\frac{\log\left(1 + xz^2\right)}{xz^2}\right) = x \cdot 1 = x.$$</p>
|
879,051 | <p>How do topologists prove continuity of a function with the usual topology at the endpoints of a closed interval? For instance, how would a topologist prove continuity for $f(x)=x^2$ on the closed interval $[0,1]$ at the point $x = 1$ using open sets (not using calculus)?</p>
| Pedro | 23,350 | <p>It is simply a matter of translating the $\varepsilon$-$\delta$ definition to a metric language. Saying that given any $\varepsilon >0$ there exists a $\delta >0$ such that $|x-a|<\delta\implies |f(x)-f(a)|<\varepsilon$ is saying that, given any open ball $V=B(f(a),\varepsilon)$ there is an open ball $U=B(a,\delta)$ such that $f(U)\subseteq V$. One then uses that open balls are a basis for the open sets in any metric space. This matches the general definition, that a map $X\to Y$ is open iff the preimage of any open set in $Y$ is open in $X$. Indeed, suppose that $U$ is open in $Y$, and consider $f^{-1}U$. Pick a point $a\in f^{-1}U$. This means $f(a)\in U$. Since $U$ is open, there is a ball $B(f(a),\varepsilon)\subseteq U$. By definition, there exists a ball $B(a,\delta)$ such that $f(B(a,\delta))\subseteq B(f(a),\varepsilon)\subseteq U$, so $B(a,\delta)\subseteq f^{-1}U$, i.e. $f^{-1}U$ is open. I leave the other direction to you.</p>
|
1,055,229 | <p>Consider the following exercise from a book I'm reading:</p>
<p>If $n$ is odd show that
$$ f: O(n) \to SO(n) \times \{1,-1\}, A \mapsto (A \operatorname{det}{A}, \operatorname{det}{A})$$</p>
<p>is an isomomorphism.</p>
<p>But why does $n$ have to be odd? I can easily show that it's a group homo., injective and surjective without using any information about $n$. What am I missing here?</p>
| Saibal | 191,019 | <p>If $n$ is even, multiplying $A\in O(n)\backslash SO(n)$ by $\det(A)=-1$ should not give a matrix in $SO(n)$.</p>
|
518,794 | <p>$f:R^2$ \{y=0} $\Rightarrow R$ , $f:(x,y)\Rightarrow x/y$.<br>
Prove (formally) that $f$ is continuous. </p>
<p>I think what I should show is that any point that belongs to an open ball of radius $\epsilon$ of image, has a pre-image that belongs to an open ball around (x,y), and since image and pre-image are both open, then $f$ is continuous. But this doesn't seem correct to me.</p>
<p>Any help is appreciated. </p>
| rfauffar | 12,158 | <p>You can use limits to show it if you want or open balls. Using limits, the following facts is easy: The quotient of two continuous functions that go to $\mathbb{R}$ is continuous, wherever the functions are defined and the denominator is not 0.</p>
<p>After proving that, it is easy to see that the projections $\pi_1:(x,y)\mapsto x$ and $\pi_2:(x,y)\mapsto y$ are continuous, since if $B_r(x)$ is an open ball of radius $r$ around $x$, then $\pi_1^{-1}(B_r(x))=B_r(x)\times\mathbb{R}$, which is an open set in $\mathbb{R}^2$. We then have that $x/y$ is the quotient of two continous functions: if $\phi(x,y)=x/y$, then $\phi(x,y)=\pi_1(x,y)/\pi_2(x,y)$, and so is continuous.</p>
|
4,458,244 | <p>To factorize <span class="math-container">$xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)$</span> I used the fact that <span class="math-container">$x=-y$</span> and <span class="math-container">$y=z$</span> and <span class="math-container">$x=-z$</span> make the expression zero. Hence it factorize to <span class="math-container">$\lambda (x+y)(y-z)(x+z)$</span> and we can check that the number <span class="math-container">$\lambda$</span> is equal to <span class="math-container">$1$</span>.</p>
<p>I'm looking for other approaches/ideas to factorize the expression.</p>
| Ivan Kaznacheyeu | 955,514 | <p>I think OP approach is the simplest one because it can be done just in mind without paper writing. One of possible another ways</p>
<p><span class="math-container">$$A=xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)=x^2(y-z)+x(y^2-z^2)+yz(y-z)$$</span></p>
<p><span class="math-container">$(y-z)$</span> is common factor of <span class="math-container">$(y-z)$</span> and <span class="math-container">$(y^2-z^2)=(y-z)(y+z)$</span>, so we can factor it out:</p>
<p><span class="math-container">$$A=(y-z)(x^2+x(y+z)+yz)$$</span></p>
<p>Then we need only factorize <span class="math-container">$$B=x^2+x(y+z)+yz$$</span></p>
<p>If we don't see that</p>
<p><span class="math-container">$$B=x^2+xy+xz+yz=x(x+y)+z(x+y)=(x+z)(x+y)$$</span></p>
<p>we can use standard way of factorizing quadratic expression with full square separation</p>
<p><span class="math-container">$$B=x^2+x(y+z)+yz=x^2+2x\cdot\frac{y+z}2+\frac{(y+z)^2}4-\frac{(y+z)^2-4yz}4$$</span>
<span class="math-container">$$B=\left(x+\frac{y+z}2\right)^2-\frac{y^2-2yz+z^2}4=\left(x+\frac{y+z}2\right)^2-\left(\frac{y-z}2\right)^2$$</span>
<span class="math-container">$$B=\left(x+\frac{y+z}2+\frac{y-z}2\right)\left(x+\frac{y+z}2-\frac{y-z}2\right)=(x+y)(x+z)$$</span>
Then <span class="math-container">$$A=(y-z)B=(y-z)(x+y)(x+z)$$</span></p>
|
4,458,244 | <p>To factorize <span class="math-container">$xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)$</span> I used the fact that <span class="math-container">$x=-y$</span> and <span class="math-container">$y=z$</span> and <span class="math-container">$x=-z$</span> make the expression zero. Hence it factorize to <span class="math-container">$\lambda (x+y)(y-z)(x+z)$</span> and we can check that the number <span class="math-container">$\lambda$</span> is equal to <span class="math-container">$1$</span>.</p>
<p>I'm looking for other approaches/ideas to factorize the expression.</p>
| User | 1,044,391 | <p>This method can be faster.</p>
<ul>
<li>
<ol>
<li>Write the expression as a general quadatic respect to <span class="math-container">$y$</span> ( or <span class="math-container">$x$</span>, or <span class="math-container">$z$</span>)</li>
</ol>
</li>
</ul>
<p>You get,</p>
<p><span class="math-container">$$A=(x+z)y^2+y(\color {red}{x^2-z^2})-xz(x+z)=(x+z)(y^2-y(z-x)-zx)$$</span></p>
<ul>
<li>
<ol start="2">
<li>Apply the <a href="https://en.m.wikipedia.org/wiki/Vieta%27s_formulas" rel="nofollow noreferrer">Vieta's formulas</a></li>
</ol>
</li>
</ul>
<p><span class="math-container">$z$</span> and <span class="math-container">$-x$</span> are the root of the quadatic: <span class="math-container">$$y^2-y(z+(-x))+z\times (-x)$$</span></p>
<p>Hence we have</p>
<p><span class="math-container">$$A=(x+z)(y+x)(y-z).$$</span></p>
|
3,350,074 | <p>Say, I have a circle </p>
<p><span class="math-container">$$\left(x – \frac 12 \right)^2 + \left( y – \frac 12 \right)^2 = \frac 14$$</span></p>
<p>If I have a point <span class="math-container">$(0,0)$</span> and a point on the circumference of the circle <span class="math-container">$(x_1,y_1)$</span>, how would I find the second intersection point between the circle and the line that goes through these two points <span class="math-container">$(0,0)$</span> and <span class="math-container">$(x_1,y_1)$</span>? </p>
<p>I wrote the equation of the line </p>
<p><span class="math-container">$$y=\frac{y_1}{x_1}x$$</span></p>
<p>and tried substituting it into the equation of the circle. After expanding everything I got,</p>
<p><span class="math-container">$$\frac{y_1^2 x^2 - y_1 x_1 x + x_1^2 x^2 - x_1^2 x + 0.5 x_1^2}{x_1^2} = 0.25$$</span></p>
<p>But, I'm not sure how to proceed from here.</p>
| Con | 682,304 | <p>Take <span class="math-container">$B = \lbrace b \rbrace$</span> and <span class="math-container">$C = \lbrace c \rbrace$</span> for some integers <span class="math-container">$b \neq c$</span>. Now define a function <span class="math-container">$f \colon \mathbb{Z} \rightarrow \mathbb{Z}$</span>, such that <span class="math-container">$f(b) = f(c)$</span> (for example a constant function). Then we get what you want as <span class="math-container">$$f(B \cap C) = f(\emptyset) = \emptyset \subset \lbrace f(b) \rbrace = f(B) \cap f(C).$$</span></p>
<p>You can also pretty much do the same thing without relying on an empty intersection.</p>
|
299,156 | <p>We say that paths $\alpha, \beta: I \to X$ with common initial point $\alpha(0)=\beta(0)$ and common terminal point $\alpha(1)=\beta(1)$ are homotopic provided that there is a continuous function $H:I \times I \to X$ such that $H(t,0)=\alpha(t)$ and $H(t,1)=\beta(t)$ for $t \in I$ and $H(0,s)=\alpha(0)=\beta(0)$ and $H(1,s)=\alpha(1)=\beta(1)$ for $s \in I$.</p>
<p>Since a path is a continuous function, I would expect that each map between two homotopic paths, $\alpha$ and $\beta$, is a path, and therefore continuous.</p>
<p>My questions are: is the above true, is it guaranteed by continuity of $H$, and if so, how do we prove it?</p>
<p>On a related note, intuitively we pass from $\alpha$ to $\beta$ as t varies over $[0,1]$. In general, can we explicitly realize a map 'in between' $\alpha$ and $\beta$ by fixing a $t$?</p>
| Dylan Wilson | 423 | <p>Of course.</p>
<p>The inclusion map $I \times \{t\} \rightarrow I \times I$ is continuous, therefore the composite
$$
I \times \{t\} {\rightarrow} I \times I \stackrel{H}{\rightarrow} X
$$</p>
<p>is continuous. The assumption on endpoints shows this is a path from $\alpha(0)$ to $\alpha(1)$. </p>
|
2,678,895 | <p>I was wondering how to map the set $\mathbb{Z}^+$ to the sequence $1, 2, 3, 1, 2, 3, \ldots$. I thought it would be easy, but I was only able to obtain an answer through trial and error.</p>
<p>For a function $f \colon \mathbb{Z}^+ \rightarrow \mathbb{Z}$, we have that</p>
<p>$f(x) = x \bmod 3$ gives the numbers $1, 2, 0, 1, 2, 0, \ldots$</p>
<p>$f(x) = (x \bmod 3) + 1$ gives the numbers $2, 3, 1, 2, 3, 1, \ldots$</p>
<p>After a bit of experimenting, I finally found that</p>
<p>$f(x) = ((x + 2) \bmod 3) + 1$ gives the numbers $1, 2, 3, 1, 2, 3, \ldots$</p>
<p>More generally, if I want to map the set $\mathbb{Z}^+$ to the sequence $\{1, 2, \ldots, n, 1, 2, \ldots, n, \ldots\}$, I need to use the function</p>
<p>$$f(x) = ((x + n - 1) \bmod n) + 1$$</p>
<p>I was only able to come to this result by trial and error. I was not able to find a solution to this relatively simple question online (although perhaps my search terms were off).</p>
<p>How would one come to this result in a more systematic way?</p>
| chepner | 78,700 | <p>I'm surprised no one has mentioned composition. You've identified that you need to shift the range</p>
<p>$$f_1(x) = x - 1$$</p>
<p>then take the value <em>modulo</em> $n$</p>
<p>$$f_2(x) = x \mod n$$</p>
<p>and finally shift the result again</p>
<p>$$f_3(x) = x + 1$$</p>
<p>The function you want is just the composition $f = f_3 \circ f_2\circ f_1$.</p>
|
231,027 | <p>I have the following complicated expression:</p>
<pre><code>(3 (Sqrt[6] a^13 + a^12 e0 + Sqrt[6] a^11 e0^2 + a^10 e0^3 + Sqrt[6] a^9 e0^4 -
a^8 e0^5 - Sqrt[6] a^7 e0^6 - a^6 e0^7 + Sqrt[6] a^5 e0^8 - a^4 e0^9 -
Sqrt[6] a^3 e0^10 + a^2 e0^11 -Sqrt[6] a e0^12 + e0^13)) /
(256 e0 (a^6 + Sqrt[6] a^5 e0 + a^4 e0^2 + Sqrt[6] a^3 e0^3 - a^2 e0^4 +
Sqrt[6] a e0^5 - e0^6)^2)
</code></pre>
<p>The numbers are huge and not very important. In fact, I am only interested in the behavior of the expression in the limit <span class="math-container">$a/e_0 \to 0$</span>, so I would like to write it as a series of the form</p>
<pre><code>constant + A (a/e_0) + O((a/e_0)^2),
</code></pre>
<p>for some coefficient A.</p>
<p>How can I do that automatically with Mathematica? I tried to use the Series function, but it cannot to an expansion for the variable <span class="math-container">$a/e_0$</span>.</p>
| Daniel Huber | 46,318 | <p>We give your expression a name: <code>y== expression</code>, then we add a second equation <code>x== a/e0</code> and eliminate a. Next, we solve for y to get the original function back. Finally we have an expression in x that we can expand around zero.</p>
<pre><code>ex = (3 (Sqrt[6] a^13 + a^12 e0 + Sqrt[6] a^11 e0^2 + a^10 e0^3 +
Sqrt[6] a^9 e0^4 - a^8 e0^5 - Sqrt[6] a^7 e0^6 - a^6 e0^7 +
Sqrt[6] a^5 e0^8 - a^4 e0^9 - Sqrt[6] a^3 e0^10 + a^2 e0^11 -
Sqrt[6] a e0^12 +
e0^13))/(256 e0 (a^6 + Sqrt[6] a^5 e0 + a^4 e0^2 +
Sqrt[6] a^3 e0^3 - a^2 e0^4 + Sqrt[6] a e0^5 - e0^6)^2)
el = Eliminate[{y == ex, a/e0 == x}, a]
yx = y /. Solve[el, y]
Series[yx, {x, 0, 2}]
</code></pre>
<p><a href="https://i.stack.imgur.com/plhcq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/plhcq.png" alt="enter image description here" /></a></p>
|
1,349,720 | <p>Say we have two irreducible polynomials in $Q [x] $. We call them $f, g$. Say one of the roots of $f $ is $a$. Is it possible that $g$ satisfies a root of the form $a^n$ for some natural number $n $? </p>
<p>Thanks </p>
| Lucian | 93,448 | <p><strong>Hint:</strong> Take a root of the form $A\sqrt\alpha+B\sqrt\beta$, where both <em>a</em> and <em>b</em> are square-free. For instance, for $A=B=1$, $\alpha=2$, $\beta=3$, and $n=2$, we have $x^2-10x+1$ and $x^4-10x^2+1$.</p>
|
3,005,576 | <p>I want to show explicitly that <span class="math-container">$L^2[0,1]$</span> subspace of <span class="math-container">$L^1[0,1]$</span> has no internal points. </p>
<p>If S subspace of X, I have the following result:</p>
<p>S = X if and only if the interior of S <span class="math-container">$\neq \emptyset$</span></p>
<p>Then to prove that the interior is empty, I just need to prove S is different from X. </p>
<p>Can I just pick <span class="math-container">$f(x) = \frac{1}{\sqrt{x}}$</span> and say that it belongs to <span class="math-container">$L^1$</span> but not to <span class="math-container">$L^2$</span> and be done? </p>
<p>Is there something that I am missing?</p>
| uniquesolution | 265,735 | <p>No, you are not missing anything. If <span class="math-container">$X,Y$</span> are normed spaces, and <span class="math-container">$X$</span> is a proper subspace of <span class="math-container">$Y$</span>, then <span class="math-container">$X$</span> has empty interior relative to <span class="math-container">$Y$</span>. So if you show that <span class="math-container">$L^2[0,1]$</span> is a proper subspace of <span class="math-container">$L^1[0,1]$</span>, you are done. The function <span class="math-container">$f(x)=\frac{1}{\sqrt{x}}$</span> is indeed an element of <span class="math-container">$L^1[0,1]$</span> that does not belong to <span class="math-container">$L^2[0,1]$</span>.</p>
|
3,005,576 | <p>I want to show explicitly that <span class="math-container">$L^2[0,1]$</span> subspace of <span class="math-container">$L^1[0,1]$</span> has no internal points. </p>
<p>If S subspace of X, I have the following result:</p>
<p>S = X if and only if the interior of S <span class="math-container">$\neq \emptyset$</span></p>
<p>Then to prove that the interior is empty, I just need to prove S is different from X. </p>
<p>Can I just pick <span class="math-container">$f(x) = \frac{1}{\sqrt{x}}$</span> and say that it belongs to <span class="math-container">$L^1$</span> but not to <span class="math-container">$L^2$</span> and be done? </p>
<p>Is there something that I am missing?</p>
| Kavi Rama Murthy | 142,385 | <p>I think 'explicitly' here means you have to show that if <span class="math-container">$f \in L^{2}$</span> then <span class="math-container">$B(f,\epsilon)$</span> is not contained in <span class="math-container">$L^{1}$</span> for any <span class="math-container">$\epsilon >0$</span>. General theory is not supposed to be used here. For this just consider <span class="math-container">$f+\frac 1 {n\sqrt x}$</span> and show that this function lies in the ball <span class="math-container">$B(f,\epsilon)$</span> for <span class="math-container">$n$</span> sufficiently large but it does not belong to <span class="math-container">$L^{2}$</span> for any <span class="math-container">$n$</span>. </p>
|
1,064,534 | <p>$$x^3-3y^2x=-1$$
$$3yx^2 -y^3=1$$</p>
<p>This was the real part and imaginary part on a previous question I asked, instead of the system it was easier to just use polar coordinates to solve, but if this was just a system unrelated to the problem, how would one solve it? I seem to have trouble when the terms are like this. All I can see is that x and y are nonzero so we can divide by them.</p>
<p>How would you approach this system?</p>
<p>I have
$$x(x^2-3y^2)=-1$$
$$y(3x^2 -y^2)=1$$</p>
| Ishfaaq | 109,161 | <ul>
<li>When $x = y$ the system reduces to $ x = y= t $ and $ 2t^3 = 1 $ hence the solution is $x = y= \dfrac{1}{\sqrt[3]2}$</li>
<li>Suppose $x \neq y$ then the system reduces to (after adding both equations) $$ x^3 - y^3 = 3y^2x - 3yx^2 $$</li>
</ul>
<p>$$ \require{cancel} \cancel {(x - y)} (x^2 + xy + y^2) = - 3xy \cancel {(x - y)} $$</p>
<p>$$ x^2 + 4xy + y^2 = 0 $$</p>
<p>$$ [x + (2 - \sqrt 3) y] [x + (2 + \sqrt 3) y ] = 0 $$</p>
<p>Hence the set of solutions is; </p>
<p>$$ (x, y ) \in \{ (\frac{1}{\sqrt[3]2}, \frac{1}{\sqrt[3]2}) \} \bigcup \{ (x, y) \ | \ x \neq y \; \text{and} \; x = (\sqrt 3 - 2) y \} \bigcup \{ (x, y) \ | \ x \neq y \; \text{and} \; x = - (\sqrt 3 + 2) y \} $$</p>
|
1,254,275 | <p>We are given a number $$K(n) = (n+3) (n^2 + 6n + 8)$$ defined for integers n. The options suggest that the number K(n) should either always be divisible by 4, 5 or 6. </p>
<p>Factorizing the second bracket term, we get $$ K(n) = (n+3) (n + 2) (n+4)$$</p>
<p>i.e the number is always divisible by -4 -2 and -3. IMO that means that the number should always be divisible by 4 and 6. However the answer book states that the number is only divisible by 6. Why is not always divisible by 4?</p>
| AlexR | 86,940 | <p>$K(n)$ is the product of three consecutive integers. Therefor it's automatically divisible by $2$ and $3$, thus by $6$. $K(n)$ need not be divisible by $4$. Consider
$$K(3) = 5\cdot6\cdot7 = 210$$</p>
<p>In general, the product of $k$ consecutive integers is divisible by all natural numbers $n\le k$.</p>
|
1,254,275 | <p>We are given a number $$K(n) = (n+3) (n^2 + 6n + 8)$$ defined for integers n. The options suggest that the number K(n) should either always be divisible by 4, 5 or 6. </p>
<p>Factorizing the second bracket term, we get $$ K(n) = (n+3) (n + 2) (n+4)$$</p>
<p>i.e the number is always divisible by -4 -2 and -3. IMO that means that the number should always be divisible by 4 and 6. However the answer book states that the number is only divisible by 6. Why is not always divisible by 4?</p>
| ajotatxe | 132,456 | <p>$K(n)$ is the product of three consecutive numbers. If the first and the third are odd and the second is even (of course) but not a multiple of $4$, then $K(n)$ will not be a multiple of $4$ either.</p>
|
3,104,936 | <p>I was wondering if anyone could explain to me why this holds true?</p>
<p><span class="math-container">$$C\left(n, k\right) = \frac{C\left(n, k-1\right) \cdot C(n - (k - 1), 1)}k$$</span></p>
<p>or</p>
<p><span class="math-container">$$C\left(3, 2\right) = \frac{C\left(3, 1\right) \cdot C(2, 1)}2$$</span></p>
<p>I was playing around with combinatorics, and realized that this holds true for all values that I put into it.</p>
| Community | -1 | <p>Expanded with factorials,</p>
<p><span class="math-container">$$\frac{n!}{k!(n-k)!}=\frac1k\frac{n!}{(k-1)!(n-k+1)!}\frac{(n-k+1)!}{1!(n-k)!}.$$</span></p>
<p>The simplifications are obvious.</p>
|
4,529,873 | <p>I am currently dealing with the two terms "Gateaux-derivation" and "Frechet-derivation" and would like to know if there is a Frechet differentiable function <span class="math-container">$f$</span> and Gateaux differentiable function <span class="math-container">$g$</span>, so that <span class="math-container">$(g\circ f)$</span> is no longer Gateaux differentiable?</p>
<p>For the definitions, I'll link to the Wikipedia article:</p>
<p>Definition of Frechet derivative: <a href="https://en.wikipedia.org/wiki/Fr%C3%A9chet_derivative" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Fr%C3%A9chet_derivative</a></p>
<p>Definition of Gateaux derivative: <a href="https://en.wikipedia.org/wiki/Gateaux_derivative" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Gateaux_derivative</a> .</p>
| daw | 136,544 | <p>Take
<span class="math-container">$$
g(x,y) = \begin{cases}
1 & \text{ if } x\ne0 \text{ and } y=x^2\\
0 & \text{ else }
\end{cases}
$$</span>
and
<span class="math-container">$$
f(x) = (x,x^2).
$$</span>
Then <span class="math-container">$g:\mathbb R^2 \to \mathbb R$</span> is Gateaux, <span class="math-container">$f:\mathbb R \to \mathbb R^2$</span> is Frechet differentiable, but <span class="math-container">$g\circ f : \mathbb R \to \mathbb R$</span> is discontinuous, hence it cannot be (Gateaux) differentiable.</p>
|
3,118,410 | <p>I've been trying to use the change of base property but I'm not having much luck. Can anyone give me any ideas on how I should approach this problem? The answer is 49/5</p>
<p>Thanks.</p>
| Haris Gušić | 450,231 | <p><strong>Hint:</strong>
<span class="math-container">$$\log_x(5x)=\frac{2}{\log_7 x} = \frac{\log_7 49}{\log_7 x}$$</span>
I hope that you have tested the conditions for the logarithms to be defined.</p>
|
4,441,421 | <p>I am studying PDEs, in particular hyperbolic conservation laws.
In particular we are using the method of characteristic to solve some of the problems.</p>
<h1>Setting</h1>
<p>Given the quite general PDE
<span class="math-container">$$
U_t + \partial_x \big( a(x,t)U(x,t)\big) = 0\iff U_t + a\cdot U_x = - a_x\cdot U
$$</span>
where we assume <span class="math-container">$a:\mathbb{R}\times\mathbb{R}_{\geq 0}\to \mathbb{R}$</span> to be <span class="math-container">$C^1$</span>.
Say that we have initial condition <span class="math-container">$U(x,0) = F(x)$</span>.</p>
<hr />
<h1>Method</h1>
<p>We want to introduce a characteristic, we do so by adding a dependency to a new variable <span class="math-container">$s$</span> and computing:
<span class="math-container">$$
\frac{d}{ds}U\big(x(s),t(s)\big) = \frac{dt}{ds}\cdot U_t + \frac{dx}{ds}\cdot U_x
$$</span>
By imposing <span class="math-container">$\frac{dt}{ds}=1$</span> and <span class="math-container">$\frac{dx}{ds}=a$</span> we should be able to rewrite the above equation as
<span class="math-container">$$
\frac{d}{ds}U\big(x(s),t(s)\big) = -(\partial_x a)\ U\big(x(s),t(s)\big)
$$</span></p>
<p>The condition <span class="math-container">$\frac{dt}{ds}=1\implies t(s) = s + t_0 $</span> can be reduced to <span class="math-container">$t = s$</span> since we want <span class="math-container">$t(0) = 0$</span>.<br />
Cleaning up things we get two equations
<span class="math-container">\begin{gather}
\frac{d}{dt}U\big(x(t),t\big) = -(\partial_x a)\ U\big(x(t),t\big) \tag{1}\\
\frac{dx}{dt} = a \tag{2}
\end{gather}</span>
Knowing this, if we are given <span class="math-container">$(x,t)$</span> in the domain, we just have to get back along the characteristics to get <span class="math-container">$x_0$</span>, then use <span class="math-container">$(1)$</span> to get the value of our solution.</p>
<h1>Question(s)</h1>
<ul>
<li><p><strong>How do you interpret <span class="math-container">$(2)$</span>?</strong><br />
Should we interpret <span class="math-container">$a$</span> as <span class="math-container">$a(x,t)$</span> or as <span class="math-container">$a(x(t),t)$</span>?
I'm not sure if it makes a lot of difference.</p>
</li>
<li><p><strong>How do you get back along the characteristics to the initial value?</strong><br />
Assuming we interpret <span class="math-container">$a$</span> as <span class="math-container">$a(x(t),t)$</span> above, do we just work with <span class="math-container">$(2)$</span> as an ODE?<br />
Let the solution to <span class="math-container">$(2)$</span> be a family <span class="math-container">$\varphi(t,c)$</span> where <span class="math-container">$c\in\mathbb{R}$</span> should depend on initial condition of <span class="math-container">$(2)$</span> (Note that we don't have initial condition imposed).
Then does getting <span class="math-container">$x_0$</span> for a given pair <span class="math-container">$(x,t)$</span> mean we have to find <span class="math-container">$c$</span> such that
<span class="math-container">$\varphi(t,c) = x$</span>, meaning <span class="math-container">$x_0$</span> is the number such that
<span class="math-container">$\varphi(t,x_0) = x $</span>?</p>
</li>
</ul>
| Alan Cai | 966,080 | <p>I don't think the statement you are trying to prove is true for any right triangle. Here is why.</p>
<p>Let the two adjacent sides of the right triangle be represented by vectors, A = (0,a), B = (b,0), then the hypotenuse of the triangle would be B-A = (b,-a) by the parallelogram law. Let the vector V = (1,1)(in fact, any vector with angle 45 degrees). Note that V is an angle bisector of the right angle.</p>
<p>Now, if we take the dot product of V and B-A, and set it equal to 0(since we want them to be perpendicular). We get b-a = 0. Thus V is perpendicular to the other side if and only if b = a.</p>
|
750,468 | <p>Show that if $f: U \to \mathbb{C}$ is analytic on an open set $U$ in $\mathbb{C}$, then the Cauchy Riemann equations for $f$ hold in $U$. </p>
<p>My question is, how do the Cauchy-Riemann equations differ on an open set $U$ versus a function $f$ on $\mathbb{C}$? Not sure how to prove this.</p>
| Jair Taylor | 28,545 | <p>This is one of those cases where the proof strategy is just to write down your conclusion and your premise, and at every step just write down the first thing you can do with the tools you are given. </p>
<p>Always start by writing down the premise: </p>
<blockquote>
<p>Suppose that $g \circ f = id$ and $f$ is onto.</p>
</blockquote>
<p>We would like to show that $g$ is one-to-one, which means $g(a) = g(b) \implies a = b$. To show this, start with the premise:</p>
<blockquote>
<p>Suppose that $b_1,b_2 \in B$ and $g(b_1) = g(b_2)$. </p>
</blockquote>
<p>Now what? Well, there aren't very many possible options at this point as we only have a few symbols. We have some $b_1,b_2$ - what can we do with them? We have to use the premise somewhere, which is that $f$ is onto. Even if we don't see immediately how it's useful, there's only one thing we can say:</p>
<blockquote>
<p>Since $f$ is onto, there exists $a_1,a_2 \in A$ with $f(a_1) = b_1$, $f(a_2) = b_2$.</p>
</blockquote>
<p>What now? There's only one thing we know about $b_1,b_2$, and one thing we know about $a_1,a_2$, so put them together and rewrite what we have.</p>
<blockquote>
<p>Then $g(f(a_1)) = g(f(a_2))$.</p>
</blockquote>
<p>But then we're already finished.</p>
<blockquote>
<p>But $g\circ f = id$, so $a_1 = g(f(a_1)$ and $a_2 = g(f(a_2)$, and so $a_1 = a_2$.</p>
</blockquote>
<p>Tim Gowers has written some nice <a href="http://gowers.wordpress.com/2014/02/03/how-to-work-out-proofs-in-analysis-i/" rel="nofollow">blog</a> posts about this kind of proof. You don't have to be clever - just write down the premise and then keep writing down something you can do with the symbols you currently have according to what you know at that point. In my experience students mainly have trouble getting the logical form of the argument right rather than coming up with the individual steps.</p>
|
1,896,399 | <p>Im asked to solve find the derivative of: $$ \frac{\ln x}{e^x}$$
<strong>my attempt</strong>
$$D\frac{\ln x}{e^x} = \frac{\frac{1}{x}e^x + \ln (x) e^x}{e^x} = e^x \frac{\frac{1}{x}+\ln x}{e^{2x}} = \frac{\frac{1}{x}+\ln x}{e^x}$$</p>
<p>But this is apparently wrong and the correct answer is:
$$\frac{\frac{1}{x} - \ln x}{e^x}$$</p>
<p><strong>Where do I go wrong?</strong></p>
| Uri Goren | 203,575 | <p>Note that:
$$ \frac{\ln x}{e^x}=e^{-x}\ln {x}$$
And
$$(uv)'=u'v+uv'$$
Thus
$$ (\frac{\ln x}{e^x})'=(e^{-x}\ln {x})'=(-e^{-x}\ln {x})+(\frac{e^{-x}}{x})=\frac{\frac{1}{x} - \ln x}{e^x}$$</p>
|
4,062,300 | <p>Given 16 Trials equally distributed to 4 individuals ( 4 Trials per person).</p>
<p>How many combinations are there where these conditions are met:<br />
Condition A: 5 Total Successes<br />
Condition B: 2 or Fewer Successes per Individual.</p>
<p>I understand Condition A in exclusivity would be:<br />
<span class="math-container">$$\binom{16}{5} = 4368$$</span></p>
<p>I believe Condition B in exclusivity would be the following but please correct me if I am wrong:<br />
<span class="math-container">$$\left(\binom42 + \binom41 + \binom40\right)^4 = 14641$$</span></p>
<p>My problem is I do not know how to solve for the number of combinations when both conditions are applied.</p>
| Sid | 428,996 | <p>uhm the following way isn't that easy... you have to consider several properties of functions...</p>
<p>first off: <span class="math-container">$\sin(\dfrac{\pi x}{9}) = \dfrac{2x}{9} \iff \sin^{-1}(\dfrac{2x}{9}) = \dfrac{\pi x}{9}$</span></p>
<p>Let: <span class="math-container">$f(x) =\sin^{-1}(\dfrac{2x}{9}) - \dfrac{\pi x}{9} $</span></p>
<p>Our goal is to find where <span class="math-container">$f(x) = 0$</span></p>
<p>Notice <span class="math-container">$f(x) = -f(-x) \implies f(x)$</span> is an odd function. <strong>If an odd function is defined at zero, then its graph must pass through the origin.</strong> <a href="https://math.stackexchange.com/questions/892154/do-odd-functions-pass-through-the-origin/892176">Proof</a>. This means <span class="math-container">$x=0$</span> is a solution.</p>
<p>Now lets examine the domain of <span class="math-container">$f(x)$</span>. Note <span class="math-container">$\sin^{-1}(x)$</span> is defined only when <span class="math-container">$-1\leq x\leq1$</span></p>
<p>This means domain of <span class="math-container">$f(x) =[-\dfrac{9}{2},\dfrac{9}{2}]$</span></p>
<p>Now lets examine the derivative of f(x):</p>
<p><span class="math-container">$f'(x) = \dfrac{2}{9\sqrt{1-\dfrac{4x^2}{81}}} - \dfrac{\pi}{9}$</span></p>
<p>Now <span class="math-container">$f'(x) =0 \implies x = \pm \dfrac{9\sqrt{1-\dfrac{4}{\pi^2}}}{2} = \pm a, a>0$</span>.
After drawing a sign diagram one can see:</p>
<p>f(x) is increasing <span class="math-container">$-\dfrac{9}{2}\leq x\leq -a$</span></p>
<p>f(x) is decreasing <span class="math-container">$-a\leq x\leq a$</span></p>
<p>f(x) is increasing <span class="math-container">$a\leq x\leq \dfrac{9}{2}$</span></p>
<p>Since f(x) is strictly decreasing in the interval <span class="math-container">$[-a,a]$</span> where <span class="math-container">$a>0$</span> and <span class="math-container">$f(-a) >0$</span> and <span class="math-container">$f(a)<0$</span>, then there must be only one root from <span class="math-container">$[-a,a]$</span> and we already saw that <span class="math-container">$x = 0$</span> is the solution and now we can confirm that this is the only root in <span class="math-container">$[-a,a]$</span> Also notice the boundary points are <span class="math-container">$\pm \dfrac{9}{2}$</span> where the function is zero. With the same increasing/decreasing function logic we can confirm these two are also the only roots in their respective interval.</p>
|
3,807,293 | <p>I do apologise if this question has been asked before, but I am having trouble trying to get from <code>!A & B || A & !B</code> to <code>A XOR B</code>.</p>
<p>I've tried doing it in reverse by starting with <code>A XOR B</code> (<code>(A || B) & !(A & B)</code>), and I reached <code>(A || B) & (!A || !B)</code> and <code>!((!A & !B) || (A & B))</code> using 2 different paths, and I know that the results of the 2 paths are the same because of the de morgan's law, yet I just can't figure out how to get to <code>!A & B || A & !B</code>.</p>
<p>EDIT: Just thought I'd explain my use of symbols. I'm not familiar with MathJax, and while I could use the proper symbols of ¬, ∧, ∨, it's just easier for me to type !, &, ||.</p>
| Paco Adajar | 477,061 | <p>To answer the question as to why it doesn't work: draw the entire circle. Your cuts are not actually removing the entirety of two quarter circles, but only a section thereof.</p>
<p><a href="https://i.stack.imgur.com/i6ALW.png" rel="noreferrer"><img src="https://i.stack.imgur.com/i6ALW.png" alt="Whole circle with two quarters cut out." /></a></p>
|
682,057 | <p>I'm seeing a physics paper about percolation (<a href="http://arxiv.org/abs/cond-mat/0202259" rel="nofollow">http://arxiv.org/abs/cond-mat/0202259</a>). In the paper the following asymptotic relation is used without derivation.
$$
\sum_{k=0}^{\infty} k P(k) (1 - \epsilon)^{k-1} \sim \left<k\right> - \left<k(k - 1)\right> \epsilon + \cdots + c \Gamma(2 - \lambda) \epsilon^{\lambda - 2},
$$
where $P(k) = c k^{-\lambda}$ with $\lambda > 2$ is a power-law probability mass function of $k$ and the bracket means average. It's the equation number (12) of the paper. I have no idea how to get the relation. Especially, from where does the $\epsilon^{\lambda - 2}$ term come?</p>
| Kai Sikorski | 127,462 | <p>$$
\sum_{k=0}^{\infty} k P(k) (1 - \epsilon)^{k-1} = \sum_{k=0}^\infty k P(k) \left(\sum_{j=0}^{k-1} {k-1 \choose j}(-\epsilon)^j \right)\\
= \sum_{k=0}^\infty k P(k) { k-1 \choose 0} - \sum_{k=0}^\infty k P(k) { k-1 \choose 1} \epsilon + \ldots\\
= \langle k \rangle - \langle k(k-1) \rangle\epsilon + \ldots
$$</p>
|
328,021 | <p><span class="math-container">$f(x_1,x_2,\ldots x_n)$</span> is polynomial with integer coefficients.
<span class="math-container">$f$</span> is rather large to be computed explicitly, but an algorithm can
compute it efficiently at integers and complex number and "lazy" form using
parenthesis.
The degree of <span class="math-container">$f$</span> is <span class="math-container">$n$</span>.
Let <span class="math-container">$C$</span> be the coefficient of the monomial <span class="math-container">$\prod_{i=1}^n x_i$</span>.</p>
<blockquote>
<p>Q1: What is the complexity of deciding if <span class="math-container">$C$</span> is zero or not?</p>
<p>Q2: What is the complexity of computing <span class="math-container">$C$</span>?</p>
</blockquote>
<p>Can we do better than <span class="math-container">$\exp(o(n())$</span> (small oh)?</p>
<p>Possible approaches:</p>
<ul>
<li>Find some domain <span class="math-container">$K$</span> with many nilpotent elements
and work over <span class="math-container">$K[x_1,x_2,\ldots x_n]/(x_1^2,x_2^2 \ldots x_n^2)$</span></li>
<li>Using <a href="https://en.wikipedia.org/wiki/Automatic_differentiation" rel="nofollow noreferrer">Automatic differentiation</a> compute the partial derivative.</li>
</ul>
| Aaron Meyerowitz | 8,008 | <p>I think these comments may be helpful.</p>
<p>I will assume that we have some polynomial or multinomial (perhaps with monomials from some restricted set) and an oracle which instantly returns the value at any given point (perhaps from a restricted domain). We wish to find a specific coefficient and wonder how many queries we need to make and how hard the calculation would be. Each query gives a linear combination of the various coefficients so we wonder how small a system of equations (drawn from the restricted class available to us) suffices. I would expect that in general (but not always) <span class="math-container">$m$</span> queries would be required where <span class="math-container">$m$</span> is the potential number of monomials. So it (often) takes as many queries to get one particular coefficient as to get them all.</p>
<p>Consider first this problem: Given a polynomial <span class="math-container">$\sum_0^na_nx^n$</span> find <span class="math-container">$a_i.$</span> Then <span class="math-container">$a_0=f(0).$</span> But it would take <span class="math-container">$n+1$</span> queries to specify any other coefficient or even to find <span class="math-container">$a_0$</span> if we were not allowed to ask about <span class="math-container">$f(0).$</span> At that point we could find them all. If we had an upper bound <span class="math-container">$\max_i(|a_i|)\lt B$</span> and wanted <span class="math-container">$a_n$</span> then <span class="math-container">$|a_n-\frac{f(N)}{N^n}| \lt \varepsilon$</span> for <span class="math-container">$N \gt \frac{nB}{\varepsilon}.$</span> If the coefficients were know to be integers this might be effective.This isn't the given problem so I will not discuss finding <span class="math-container">$f(e^{2j\pi/n}),$</span> precomputing the inverse of a Vandermonde matrix or doing successive differences.</p>
<p>Next suppose that we know in advance that the vector of variables is <span class="math-container">$\mathbf{x}=(x_1,\cdots,x_n)$</span> and <span class="math-container">$f(\mathbf{x})=\sum_Sa_S\prod_{i \in S}x_i$</span> where <span class="math-container">$S$</span> varies over the subsets of <span class="math-container">$[n]=\{1,\cdots ,n\}.$</span> Then there are <span class="math-container">$2^n$</span> coefficients. If our queries need are restricted to <span class="math-container">$0-1$</span> vectors then to find <span class="math-container">$a_S$</span> would could do <span class="math-container">$2^{|S|}$</span> queries (setting all irrelevant variable to <span class="math-container">$0$</span>) and use inclusion exclusion. If done correctly we could have all <span class="math-container">$a_T$</span> for <span class="math-container">$T \subseteq S$</span> for the same work. This would seem best possible.</p>
<p><strong>Minimality</strong> Suppose I tell you in advance that <span class="math-container">$f=\prod_1^n z_i$</span> where either <span class="math-container">$z_i=x_i$</span> or <span class="math-container">$z_i=1-x_i.$</span> Then the coefficient of <span class="math-container">$x_1x_2\cdots x_n$</span> is <span class="math-container">$\pm 1$</span> and we wish to determine which it is. Then each of the <span class="math-container">$2^n$</span> possible queries will return <span class="math-container">$0$</span> except a particular one which will tell us everything. If we are lucky we find out in a few queries, but it could take <span class="math-container">$2^n$</span> queries. </p>
<p>In this rather restricted case one can technically do it with one evaluation: Set <span class="math-container">$$x_i=\frac{2^{2^i}}{1+2^{2^i}}.$$</span> </p>
<p>Then the result is a fraction with an odd denominator and numerator <span class="math-container">$2^{2^k}$</span> where the bits of <span class="math-container">$k$</span> tell us which <span class="math-container">$z_i=x_i.$</span></p>
<p>The question as asked concerns linear combinations of monomials in <span class="math-container">$n$</span> variables all of degree at most <span class="math-container">$n.$</span> There are <span class="math-container">$\binom{n+k-1}k$</span> such monomials of degree <span class="math-container">$k$</span> so potentially <span class="math-container">$\binom{2n}n \approx \frac{4^n}{\sqrt{\pi n}}$</span> summands. In this case, for each <span class="math-container">$T \subseteq [n]$</span>, <span class="math-container">$2^{|T|}$</span> queries suffice to determine the sum of all the coefficients for monomials using positive powers of exactly <span class="math-container">$\{x_t \mid t \in T\}.$</span> For <span class="math-container">$T=[n]$</span> this sum is, as remarked above, just what we want.</p>
<p>I suspect that if we can't use <span class="math-container">$0-1$</span> vectors , for example all the <span class="math-container">$x_i$</span> in a query need to be integers at least <span class="math-container">$2$</span>, then <span class="math-container">$2^n$</span> queries would not be enough. Maybe then it would take <span class="math-container">$\binom{2n}n$</span> queries.</p>
<hr>
<p>Here is an interesting case I wondered about. Suppose we explicitly see that <span class="math-container">$$f=\prod_{i=1}^n\left(\sum_{j=1}^na_{ij}x_j+c_i\right).$$</span> Then the finding the coefficient in question amounts to <a href="https://en.wikipedia.org/wiki/Computing_the_permanent" rel="nofollow noreferrer">computing the permanent</a> of the matrix <span class="math-container">$A={\huge(}a_{ij}{\huge)}.$</span> Naively that takes <span class="math-container">$n!n$</span> operations. This can perhaps be sped up to <span class="math-container">$2^nn^2$</span> or <span class="math-container">$2^nn.$</span> Much more than for the determinant. Of course each of our allowed queries instantly gives us the result of many arithmetic operations so it is not contradictory that <span class="math-container">$2^n$</span> suffice.</p>
<p>For that matter, we don’t need the information about where <span class="math-container">$f$</span> came from. I was just emphasizing that knowing that was no real help. The method here merely uses that <span class="math-container">$f$</span> is some linear combination of <span class="math-container">$\binom{2n}{n}$</span> monomials.</p>
|
3,618,340 | <p>I read a set of notes on projective curves, where it introduced discrete valuations on a field <span class="math-container">$K$</span> as a surjective homomorphism <span class="math-container">$v:K^\times \to \mathbb Z$</span> where <span class="math-container">$v(x+y) \geq \min(v(x), v(y))$</span> for all <span class="math-container">$x,y$</span>. Then proceeds to give <span class="math-container">$\mathcal O_P$</span>, the local ring at <span class="math-container">$P$</span>, the structure of a discrete valuation ring by counting the order of vanishing of rational functions at <span class="math-container">$P$</span>.</p>
<p>The way that I see it, I don't think the structure of the discrete valuation or the DVR had any significance in what's to follow (counting degrees of functions, constructing the divisor class group and Jacobian). One could have said to consider <span class="math-container">$v_P(f)$</span> as the order of vanishing of <span class="math-container">$f$</span> at <span class="math-container">$P$</span>, without mentioning the structure of discrete valuation, and proceed without any problems. </p>
<p>So I am wondering what am I missing here. Is there a deeper significance of discrete valuation that I'm not seeing? And are there examples in other settings of mathematics where DVRs are used (where the valuation is not order of vanishing)?</p>
| Hagen Knaf | 2,479 | <p>Historically one/the(?) reason for the prominent appearance of discrete valuations in the case of algebraic curves is that one gets a theory that possesses strong similarities to algebraic number theory. Here are some points of the story:</p>
<ul>
<li>In algebraic number theory the <span class="math-container">$p$</span>-adic valuations on <span class="math-container">$\mathbb{Q}$</span> and their extensions to extension fields <span class="math-container">$K|\mathbb{Q}$</span> of finite degree are a tool of central importance. All these valuations are discrete.</li>
<li>Along with every algebraic curve <span class="math-container">$C$</span> comes its field of rational functions <span class="math-container">$K(C)$</span>, which is an extension of finite degree of a rational function field <span class="math-container">$K(x)$</span> in one variable <span class="math-container">$x$</span>.</li>
<li>Vice versa: Given an extension field <span class="math-container">$F$</span> of finite degree over a rational function field <span class="math-container">$K(x)$</span> there exists a unique projective algebraic curve <span class="math-container">$C$</span> without singularities such that <span class="math-container">$F=K(C)$</span>. The points of this curve are in bijection with the discrete valuations on <span class="math-container">$F$</span> that are trivial on <span class="math-container">$K$</span>. (In fact more is true here.)</li>
<li>As a consequence of the vague points just stated one could say that doing geometry with projective algebraic curves without singularities is the same as doing arithmetic in function fields <span class="math-container">$F|K$</span> of one variable. </li>
<li>In the case of a finite field <span class="math-container">$K$</span> there are strong similarities doing arithmetic in finite extensions of <span class="math-container">$\mathbb{Q}$</span> or in finite extensions of <span class="math-container">$K(x)$</span>. </li>
</ul>
|
1,604,226 | <p>Let $a,b,c$ be positive real numbers. Prove that </p>
<p>$\frac{bc}{a^2+bc}+\frac{ca}{b^2+ca}+\frac{ab}{c^2+ab}\le{\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}}$.</p>
<p>I have no idea how to solve this question.</p>
| Michael Rozenberg | 190,319 | <p>We can use SOS here.
$\sum\limits_{cyc}\left(\frac{a}{b+c}-\frac{bc}{a^2+bc}\right)=\sum\limits_{cyc}\frac{a^3+abc-b^2c-bc^2}{(b+c)(a^2+bc)}=\sum\limits_{cyc}\frac{a^3-abc+2abc-b^2c-bc^2}{(b+c)(a^2+bc)}=$
$=\frac{1}{2}\sum\limits_{cyc}\frac{2a(a^2-bc)+2bc(2a-b-c)}{(b+c)(a^2+bc)}=\frac{1}{2}\sum\limits_{cyc}\frac{(a-b)(a^2+ac+2bc)-(c-a)(a^2+ab+2bc)}{(b+c)(a^2+bc)}=$
$=\frac{1}{2}\sum\limits_{cyc}(a-b)\left(\frac{a^2+ac+2bc}{(b+c)(a^2+bc)}-\frac{b^2+bc+2ac}{(a+c)(b^2+ac)}\right)=\frac{1}{2}\sum\limits_{cyc}\frac{(a-b)^2(a^2b^2+(a^3-a^2b-ab^2+b^3)c+abc^2+ac^3+bc^3)}{(a+c)(b+c)(a^2+bc)(b^2+ac)}\geq0$</p>
|
1,917,852 | <p>Suppose $(\Omega, \mathcal{F}, \mu)$ is a probability space and $f$ is integrable function. Then, is it true that $\mu \{ \omega: |f(w)|=\alpha\}=0$ for all but countably many $\alpha$ ? \</p>
| detnvvp | 85,818 | <p>Suppose that there exists an uncountable set $A$ such that, if $a\in A$, then $\mu\left(\{\omega\in\Omega\Big{|}|f(\omega)|=a \}\right)>0$. For every $n\in\mathbb N$, set $A_n$ to be the set of $a\in A$ such that $$\frac{1}{n}\leq\mu\left(\{\omega\in\Omega\Big{|}|f(\omega)|=a \}\right)<\frac{1}{n-1},$$ then $\bigcup_{n\in\mathbb N}A_n=A$. Since the $A_n$ are countably many and $A$ is uncountable, there exists $N\in\mathbb N$ such that $A_N$ is uncountable. Therefore, we can find $\varepsilon>0$ and a sequence $(a_m)$ in $A_N$, such that $a_m>\varepsilon$ for all $m\in\mathbb N$.</p>
<p>Set $$B_m=\{\omega\in\Omega\Big{|}|f(\omega)|=a_m \}.$$ We then compute $$\int_X|f|\,d\mu\geq\sum_{m\in\mathbb N}\int_{B_m}|f|\,d\mu=\sum_{m\in\mathbb N}\int_{B_m}a_m\,d\mu>\sum_{m\in\mathbb N}\varepsilon\mu(B_m)\geq\sum_{m\in\mathbb N}\frac{\varepsilon}{N}=\infty,$$ which is a contradiction.</p>
|
3,335,850 | <p>I know that if <span class="math-container">$x$</span> is an element of A by definition of subsets then it is an element of <span class="math-container">$B$</span>, but I don't know how to use the complement to prove the latter half.</p>
| Jean Lagace | 91,358 | <p>Pick an element <span class="math-container">$x \in C - B$</span>. By definition, it is in <span class="math-container">$C$</span>, and also it is not in <span class="math-container">$B$</span>. Since <span class="math-container">$A \subseteq B$</span>, <span class="math-container">$x$</span> is not in <span class="math-container">$A$</span> either. Since <span class="math-container">$x \in C$</span> and <span class="math-container">$x \not \in A$</span>, <span class="math-container">$x \in C - A$</span> and <span class="math-container">$C - B \subseteq C - A$</span>.</p>
|
824,255 | <p>Suppose I have a polynomial (of any order) and I'm not able to calculate the roots. Is there a way to get at least some information about the roots such as how many of them are complex, negative or positive? For example, I can safely identify the behavior (and therefore roots' character) of $f(x)=ax+b$ or even a quadratic expression just by inspection. </p>
<p>I'm aware of Descartes' sign rule <a href="http://en.wikipedia.org/wiki/Descartes%27_rule_of_signs" rel="nofollow">http://en.wikipedia.org/wiki/Descartes%27_rule_of_signs</a> but it apparently provides only an upper bound on the number of positive/negative roots. Is there something more general giving an exact number (of positive roots) preferably without using the methods of calculus?</p>
| Tom | 103,715 | <p>Let $g(x) = x$, $h(x) = \sin(x)$ and $k(x) = \frac{1}{x}$ defined for all $x \neq 0$. Then what you have is that $f(x) = g(x) \cdot h(k(x))$. In other words, $f$ is created by the product and composition of "nice" functions. So, if you know that $g, h,$ and $k$ are differentiable (when $x \neq 0$) and if you know the product rule and chain rule, then the differentiability of $f$ should follow.</p>
|
901,549 | <p>I am struggling with part c of this question. Could someone please tell me how to approach and solve this type of questions?</p>
<p><img src="https://i.stack.imgur.com/SIpJ8.png" alt="enter image description here"></p>
| callculus42 | 144,421 | <p>ad a)</p>
<p>$P(Y=y_j)=\sum_{i \in I} \ p(X=x_i,Y=y_j) $</p>
<p>$x_i \in \{-1,0,2 \}; \ y_j \in \{-2,1,2 \}$ </p>
<p>$P(Y=-2)=0.1+0.2+0.2=0.5$ </p>
<p>Sum of the first row. For the whole marginal probability function you have additional calculate $P(Y=1)$ and $P(Y=2)$</p>
<p>ad b)</p>
<p>$E(x\cdot y)=\sum_{i \in I} \sum_{j \in J} \ \ x_i\cdot y_i\cdot P(X=x_i,Y=y_i)=(-1) \cdot (-2) \cdot 0.1 + \ldots$</p>
<p>You have to finish the calculation.</p>
|
2,155,201 | <p>By a real ODE I mean an ordinary differential equation with only real coefficients and the resulting function is a function of a real argument. If such a solution exists, can you give an example?</p>
<p>Edit: To add to this, is it still possible if the initial conditions must also be real?</p>
| John Hughes | 114,036 | <p>$$
f(x)^2 + 1 = 0
$$
That's a differential equation where the "derivative" coefficient is zero; as it happens, the solution is one of the constant functions $f(x) = \pm i$. </p>
<p>If you want one with a derivative term, consider
$$
f(x)^2 + 1 = xf'(x)
$$
At $x = 0$, any solution must have $f(0) = \pm i$. </p>
|
3,009,919 | <p>Give <span class="math-container">$n$</span>-th derivatives, can we find a function ?<br>
For example, if <span class="math-container">$f^{(i)}(x)=i,i=1,2,\cdots,n.$</span> , can we construct a function such that it satisfies the condition.<br>
It just occurs in my mind. I didn't know if it has a solution.</p>
| ShapeOfMatter | 618,435 | <p>It sounds like you're asking for a function <span class="math-container">$f$</span>, such that
<span class="math-container">$$f'=1$$</span>
<span class="math-container">$$f''=2$$</span>
<span class="math-container">$$f'''=3$$</span>
<span class="math-container">$$...$$</span>
But if <span class="math-container">$f'$</span> is a constant, then <span class="math-container">$f''=0$</span> (and so are all higher derivatives), so no such function can exist.</p>
<p>I suppose we can't rule out functions with domains/codomains other than the Real numbers without a little more work, but we'd still need <span class="math-container">$f$</span> to be differentiable.</p>
<p>Am I answering the right question? </p>
|
623,873 | <p>I'm given with the following function:</p>
<p>$f(x) = x^n \arctan\left(\dfrac{1}{x}\right) , x\ne0 $ , and $f(0)=0$ . </p>
<p>I'm asked to:</p>
<ol>
<li>determine when is it continuous in $x=0$ </li>
<li>determine when is it differentiable in $x=0$</li>
<li>determine when is it continuously differentiable in $x=0$</li>
</ol>
<p>I have a problem with proving everything:</p>
<ol>
<li><p>When $n>0$ , we obviously have continuity because of the Squeeze Theorem. But how can I prove that when $n<0$ I don't have continuity ? </p></li>
<li><p>When $n>0$, I don't have differentiability at $n=1$ . But how can I prove that in general , for $0<n<1$, the function isn't differentiable? and for $n>1$ it is? </p></li>
<li><p>I guess that if I'll understand $1$ and $2$ I'll be able to understand $3 $. </p></li>
</ol>
<p>Hope someone will help me</p>
<p>Thanks ! </p>
| Charles | 1,778 | <p>$$
\sum_{p\le x}p = \frac{x^2}{2\log x} + \frac{x^2}{2\log^2x} + \frac{x^2}{4\log^3x} + \frac{3x^2}{8\log^4x} + O\left(\frac{x}{\log^5x}\right).
$$</p>
|
623,873 | <p>I'm given with the following function:</p>
<p>$f(x) = x^n \arctan\left(\dfrac{1}{x}\right) , x\ne0 $ , and $f(0)=0$ . </p>
<p>I'm asked to:</p>
<ol>
<li>determine when is it continuous in $x=0$ </li>
<li>determine when is it differentiable in $x=0$</li>
<li>determine when is it continuously differentiable in $x=0$</li>
</ol>
<p>I have a problem with proving everything:</p>
<ol>
<li><p>When $n>0$ , we obviously have continuity because of the Squeeze Theorem. But how can I prove that when $n<0$ I don't have continuity ? </p></li>
<li><p>When $n>0$, I don't have differentiability at $n=1$ . But how can I prove that in general , for $0<n<1$, the function isn't differentiable? and for $n>1$ it is? </p></li>
<li><p>I guess that if I'll understand $1$ and $2$ I'll be able to understand $3 $. </p></li>
</ol>
<p>Hope someone will help me</p>
<p>Thanks ! </p>
| Kusavil | 78,968 | <p>There are formulas for computing prime numbers, the problem lies in the costs and time needed for that and there is often (or we don't know) no closed formula for that, like <a href="http://www.matematyka.pl/159928.htm#p624705" rel="nofollow noreferrer">these ones:</a></p>
<p>1.
<span class="math-container">$$p_n= 2+ \sum_{j=2}^{2^n} \left(\Bigg[\frac{n-1}{ \sum_{m=2}^{j} \left[\frac{1} { \sum_{k=2}^{m} [1-\frac{m}{k}+[\frac{m}{k}]] } \right]}\Bigg]-\left| \frac{n-1}{ \sum_{m=2}^{j} [\frac{1} { \sum_{k=2}^{m} [1-\frac{m}{k}+[\frac{m}{k}]] } ]}-1\right|\,\right)$$</span></p>
<p>2.</p>
<p><span class="math-container">$$p_n=\left[ 1- \frac{1}{\log(2)} \log\left(-\frac{1}{2} + \sum_{d | P_{n-1}} \frac{\mu(d)}{2^d -1}\right)\right]$$</span>
Where <span class="math-container">$[x ] = floor(x)$</span> is the largest integer not greater than x and
<span class="math-container">\begin{cases}
\mu(1)=1 ,& \\
\mu(n)=(-1)^r, & \text{if $n$ is product of $r$ distinct prime numbers} \\
\mu(n)=0, & \text{if $n$ has one or more repeated prime factors} \\
\end{cases}</span></p>
<p>The second one is from "My Numbers, My Friends - Popular Lectures on Number Theory - Paulo Ribenboim", don't know from where is the first one.</p>
<p>I am sure there are more of formulas like that, even for exact n-th prime number, but why we don't use them? </p>
<p>Because we don't know any "effective" ones. Even if we compute some prime numbers with that, it would just take too long to find the big ones. So we use special algorithms for finding prime numbers, which are much faster, for example <a href="http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes" rel="nofollow noreferrer">Sieve of Eratosthenes</a>, and still seek for better and better ones :) </p>
<p>You can also find something here: </p>
<p><a href="http://mathworld.wolfram.com/PrimeFormulas.html" rel="nofollow noreferrer">http://mathworld.wolfram.com/PrimeFormulas.html</a></p>
<p><a href="https://oeis.org/A000040" rel="nofollow noreferrer">https://oeis.org/A000040</a>(look at formulas at the bottom of page)</p>
|
1,562,416 | <p>Suppose $32$ objects are placed along a circle at equal distances. In how many ways can $3$ objects be chosen from among them so no two of the three chosen objects are adjacent nor diametrically opposite.
This is again a problem in a math contest in India and this is how I tried it:</p>
<p><strong>Number of ways of choosing 3 objects from 32 objects</strong>$={32 \choose 3 }$</p>
<p><strong>Number of ways to choose 3 points such that they are adjacent</strong>$=32$</p>
<p><strong>Number of ways to choose $3$ points such that two of them are adjacent</strong>$=(32×2×30)$.......</p>
<p>for each point there are two ways to choose an adjacent point. For each choice there are $30$ options to choose the third point.</p>
<p><strong>Number of ways to choose $3$ points such that two of them are diametrically opposite</strong>$=(32×30)$</p>
<p><strong>Number of ways to choose $3$ points from $32$ equidistant points on a circle with the restrictions placed by the problem</strong>
$={32 \choose 3}-(32+(32×2×30)+(32×30)$
Am I correct in my approach?</p>
| Smayan das | 296,724 | <p>Suppose I denote the objects as 1-32.
Then suppose I select object 1 first. Then there would be 2 cases.</p>
<p>CASE 1 :-If I choose 3 or 31 (as I can't select 32 or 2 but I can choose those adjacent to 2 and 32) then there would be 25 possibilities for each .
Thus there being a total of 25×2 = 50 possibilities.</p>
<p>CASE 2 :-Then again leaving those numbers that are considered in first case (3 and 31) I am left with 26 possibilities for 2nd object and for third I will have 23 chances.(excluding the adjacent and the opposite ones that are 3 in number) and excluding the very same digit I took in the second place.</p>
<p>Hence in 2nd case total possibilities are 26 × 23 = 598</p>
<p>So,Total in both cases = 598 + 50 = 648 </p>
<p>But I completely ignored those numbers adjacent to 1 and diametrically opposite to 1 that are 2,32,17 .</p>
<p>So again each of them will have **648 ** total possibilities thus making final answer:-
=> 648 × 4 = 2592.
Please rectify me if I'm wrong.</p>
|
2,899,422 | <p>I need some hints for solving $yy''-(y')^2=xy^2$. </p>
<p>I noticed that the left hand side is close to $(yy')'$:</p>
<p>$yy''-(y')^2=xy^2\ \Leftrightarrow\ yy''+(y')^2-2(y')^2=xy^2\ \Leftrightarrow\ (yy')'-2(y')^2=xy^2$.</p>
<p>But I don't know how to continue expressing the terms as derivatives of some functions.</p>
<p>Thanks</p>
| Community | -1 | <p><strong>By indeterminate coefficients:</strong></p>
<p>Assume the solution to be of the form $P(x)e^{x^2}$ for some polynomial $P$. Then</p>
<p>$$(P(x)e^{x^2})'=(P'(x)+2xP(x))e^{x^2}.$$</p>
<p>For the left factor to equal $x^2$, $P$ must be of the first degree, let $ax+b$.</p>
<p>Now</p>
<p>$$P'(x)+2xP(x)=a+2x(ax+b)$$</p>
<p>and by identification</p>
<p>$$a=\frac12,b=0,a=0.$$</p>
<p>Hey, what is going on ? Identification is not possible, we are in a dead end. The fact is that there is no solution of the form $P(x)e^{x^2}$!</p>
<p>If we consider the partial solution $\dfrac x2e^{x^2}$, the derivative is</p>
<p>$$ x^2e^{x^2}+\frac12e^{x^2}.$$</p>
<p>Hence we remain with the extra term $e^{x^2}$, which is known to have no closed-form antiderivative and requires the so-called error function.</p>
|
2,200,672 | <p>Find the roots of $f(x)=x^2+x+1$ modulo $7$, and modulo $13$, and modulo $91$ </p>
<p>I think for mod $7$ and $13$, it can be done by trial and error. But how about mod $91$?</p>
| fleablood | 280,126 | <p>To find the solutions to $x^2 + x +1 \mod k$ we can factor $x^2 + x + 1$ and to factor that we need to find the $nm \equiv 1 \mod k$ so that $n + m \equiv 1 \mod k$.</p>
<p>For $k = 7$, 7 is prime so every term has a multiplicative inverse. $1^{-1} = 1$ and $1 +1 \equiv 2$ so that won't do. $2^{-1}=4$ but $2+4 \equiv 6\equiv -1$. So </p>
<p>$x^2 + x+1 \equiv (x-2)(x-4) \mod 9$</p>
<p>And the solutions to that are $x =2, 4$</p>
<p>Likewise in $modulo 13$ $2^{-1} = 7$ but $2+7 \equiv 9$. $3^{-1} = 9$ and $3+9 \equiv -1$ so </p>
<p>$x^2 + x + 1 \equiv (x-3)(x-9) \mod 9$</p>
<p>And the solutions to that are $x=3, x=9$.</p>
<p>So we just need to use the chinese remainder theorem to solve the following four equations modulo $91 = 7*13$</p>
<p>$x \equiv 2 \mod 7; x \equiv 3 \mod 13$</p>
<p>$x \equiv 2 \mod 7; x \equiv 9 \mod 13$</p>
<p>$x \equiv 4 \mod 7; x \equiv 3 \mod 13$</p>
<p>$x \equiv 4 \mod 7; x \equiv 9 \mod 13$</p>
<p>$1 = 2*7 - 13$</p>
<p>So $-13*2 + 3*14 = 16$ is solution $2\mod 7; 3\mod 13$</p>
<p>$-13*2 + 9*14 \equiv 9 \mod 91$ is solution to $2\mod 7; 9 \mod 13$</p>
<p>$-13*4 + 3*14 \equiv 81 \mod 91$ is solution to $4\mod 7; 3 \mod 13$</p>
<p>$-13*4 + 9 *14 \equiv 74 \mod 91$ is solution to $4 \mod7; 3\mod 13$</p>
<p>=======</p>
<p>$x^2 + x +1 \mod 7\equiv$</p>
<p>$x^2 -6x + 8\mod 7 \equiv$</p>
<p>$(x-4)(x-2)\mod 7 \equiv$</p>
<p>$7$ is prime so there are not zero divisors.</p>
<p>So $4,2$ are the solutions modulo 7.</p>
<p>$x^2 + x + 1 \mod 13 \equiv$</p>
<p>$x^2 -12x + 27 \mod 13 \equiv$</p>
<p>$(x-3)(x-9)\mod 13$</p>
<p>So $3$ and $9$ are the solutions mod 13.</p>
<p>$91 =13*7$ solutions are</p>
<p>$4 + 7k \equiv 3 + 13j \mod 91$</p>
<p>Or $4 + 7k \equiv 9 + 13j \mod 91$</p>
<p>or $2 + 7k \equiv 3 + 13j \mod 91$</p>
<p>or $2 + 7k \equiv 9 + 13j \mod 91$</p>
<p>$4 + 7k \equiv 3 + 13j \mod 91$ is solved by $81=4 + 7*11 = 3+ 6*13$ and, indeed, $81^2 + 81 + 1 \equiv (-10)^2 -10 + 1 =91 \equiv 0 \mod 91$.</p>
<p>$4+7k \equiv 9 + 13j \mod 91$ is solved by $74 = 4+7*10 = 9 + 5*13$ and, indeed, $74^2 + 74 + 1 \equiv (-17)^2 - 17 + 1\equiv 16*17 + 1 \equiv 8*17*2 + 1 \equiv 136*2 + 1 \equiv 45*2 + 1 \equiv 91 \equiv 0 \mod 91$.</p>
<p>$2+7k \equiv 3 + 13j \mod 91$ is solved by $16=2 + 14 = 3 + 13$ and, indeed $16^2 + 16 +1= 17*16 + 1 $ ... didn't we just do this?</p>
<p>So $2+7k\equiv 9 + 13j$ is solved by $9 = 7+2 = 9$ and $9^2 + 9 + 1 = 91 \equiv 0 \mod 91$</p>
|
3,230,922 | <p>I was working through some textbook problems for my Number Theory class and needed some help with the following question:</p>
<p>Describe the set of quadratic integers α in Q[sqrt−3] for which α ̄ and α are associates.</p>
<p>I'd really like some help with this problem. Thank you!</p>
| José Carlos Santos | 446,262 | <p>As a first step, you can search for an element <span class="math-container">$at+bu\in K$</span> such that<span class="math-container">$$(\forall ct+du\in K):(at+bu)(ct+du)=(ct+du)(at+bu)ct+du.$$</span>After a few computations (Linear Algebra stuff), you will get <span class="math-container">$a=b=\frac12$</span>. So, your ring isomorphism should map <span class="math-container">$1$</span> into <span class="math-container">$\frac12u+\frac12v$</span>.</p>
<p>The second step would consists in finding an element <span class="math-container">$ct+du\in K$</span> whose square is <span class="math-container">$-\left(\frac12t+\frac12u\right)$</span>. Afer a few more computations, you will be able to prove that <span class="math-container">$-\frac12t+\frac12u$</span> has that property.</p>
<p>So, define<span class="math-container">$$\begin{array}{rccc}\varphi\colon&\mathbb C&\longrightarrow&K\\&a+bi&\mapsto&a\left(\frac12t+\frac12u\right)+b\left(-\frac12t+\frac12u\right)\end{array}$$</span>and prove that it is indeed an isomorphism.</p>
|
2,458,259 | <p>Are there any sequences s.t. $\{a_{n}\}^{\infty}_{n=1}$ and $\{b_{n}\}^{\infty}_{n=1}, b_{n}\neq0, \forall n$ which diverge, but $\{\frac{a_{n}}{b_{n}}\}^{\infty}_{n=1}$ converges to $0$.</p>
<p>I have tried to plug in a bunch of examples, but failing to come up with one. </p>
| Randall | 464,495 | <p>$a_n = (-1)^n$ and $b_n =n$ should work. </p>
|
1,770,858 | <p>I am trying to use the integral test on the series $$\sum\limits_{n=2}^{\infty} \frac{1}{\ln(n)^2}.$$ I am not sure how to evaluate the integral. Any hints?</p>
| Vik78 | 304,290 | <p>The sum is clearly bounded below by the harmonic series starting from 2 (since $\ln(n)^2 < n$), so it diverges.</p>
|
1,770,858 | <p>I am trying to use the integral test on the series $$\sum\limits_{n=2}^{\infty} \frac{1}{\ln(n)^2}.$$ I am not sure how to evaluate the integral. Any hints?</p>
| Olivier Oloa | 118,798 | <p>One may onserve that
$$
0<\ln x<\sqrt{x} ,\qquad x>2,
$$ giving
$$
0<(\ln x)^2<x, \qquad x>2,
$$ then</p>
<blockquote>
<p>$$
\int_2^M \frac{dx}x < \int_2^M \frac{dx}{(\ln x)^2}, \qquad M>2,
$$ </p>
</blockquote>
<p>thus, by letting $M \to \infty$, the initial integral <strong>diverges</strong>.</p>
|
142,180 | <p>First take a question as an example:</p>
<p>Let $f:L\to M$ be an irreducible morphism in $\mathrm{mod}-A$ and $X$ be a right $A$-module.
Show that $\mathrm{Ext}_A(X,f):\mathrm{Ext}_A(X,L)\to\mathrm{Ext}_A(X,M)$ is a monomorphism,if $\mathrm{Hom}_A(M,X)=0$. </p>
<p>What is the special meaning of $\mathrm{Ext}$ functor in the quiver representation?</p>
| Aaron | 28,841 | <p>As Matt has said in the comment, representation of quivers, and modules over the corresponding path algebra are really the same thing. Since you said you understand what it means to take Ext in homological algebra, i.e. taking Ext of modules over an algebra. So what you want for the answer is just $Rep(kQ/I) \simeq A-mod$. Here $A\cong kQ/I$, or in general, can be taken as Morita equivalent to $kQ/I$ (which is a basic algebra, meaning simple modules are all 1 dimensional).</p>
<p>We prove simply $Rep(Q)\simeq kQ-mod$ here.</p>
<p>If you have $M\in kQ-mod$, to get a representation of $Q$, take $M_x = e_x M$ where $e_x$ is the trivial path (i.e. primitive idempotent) corresponding to $x\in Q_0$.</p>
<p>If you have $V=\bigoplus_{x\in Q_0} V_x\in Rep(Q)$, then take the vector space $M=V$, and define the action of path $x_1\to \cdots \to x_k$ on $V$ (this lies in the path algebra $kQ$) as composition $V \twoheadrightarrow V_{x_1}\cdots V_{x_k} \hookrightarrow V$. This then gives you a module over the algebra $kQ$.</p>
<p>Hope this helps. Being able to think of a module in terms of representation of quiver is exactly why quiver is so useful to study representation theory. One should never be afraid of representation of quiver. IMO, I find it much harder to study modules over arbitrary algebra.</p>
|
637,099 | <p>I'm having trouble verifying my proof, would appreciate some input on this one.</p>
<p>Let $(X,d)$ be a metric space with $E\subset X$.</p>
<p>Suppose $E$ is closed in $X$, which means that $E=\overline{E}$.
By defintion of the <em>boundary</em> of $E$, denoted $\partial E$ we have that</p>
<p>$\partial E=\overline{E}\cap\overline{E^{c}}\subseteq \overline{E}=E$ which easily establishes the first part. </p>
<p>Moving on to the converse, where I'm having my main doubts, we assume that</p>
<p>$\partial E \subseteq E$ or equivalently $\overline{E}\cap \overline{E^{c}}\subseteq E$</p>
<p>Now $\forall x\in E^{c}$ we have that $x \notin \partial E$.</p>
<p>Noticing that $\partial E$ is closed, since it is an intesection of two closed sets. </p>
<p>I can't get any further than this... would appreciate a detailed help of completing the last part. Thanks</p>
| gerw | 58,577 | <p>One possibility is to take
$$f_n(x) = a_n \, \chi_{[0,1/n]}(x)$$
with suitable chosen $a_n$.</p>
|
2,033,470 | <blockquote>
<p>Assume that an integer <span class="math-container">$a$</span> has a multiplicative inverse modulo an integer <span class="math-container">$n$</span>. Prove that this inverse is unique modulo <span class="math-container">$n$</span>.</p>
</blockquote>
<p>I was given a hint that proving this Lemma:
<span class="math-container">\begin{align}
n \mid ab \ \wedge \ \operatorname{gcd}\left(n,a\right) = 1
\qquad \Longrightarrow \qquad
n \mid b
\end{align}</span>
should help me in finding the answer.</p>
<p>Here are my steps in trying to solve the problem:
<span class="math-container">\begin{align}
\operatorname{gcd}\left(n,a\right) = 1
& \qquad \Longrightarrow \qquad
sn + ta = 1
\qquad \Longrightarrow \qquad
sn = 1 - ta \\
& \qquad \Longrightarrow \qquad
1 \equiv ta \mod n
\qquad \Longrightarrow \qquad
ta \equiv 1 \mod n .
\end{align}</span>
I know that having the GCD of m and a equal to 1 proves there is a multiplicative inverse mod n, but I'm not sure on how to prove <span class="math-container">$n \mid b$</span> with and how it helps prove the multiplicative inverse is unique.</p>
| Bill Dubuque | 242 | <p>If <span class="math-container">$\,b,\bar b$</span> are inverses of <span class="math-container">$\,a\,$</span> then <span class="math-container">$\ b\equiv b(a\bar b)\equiv (ba)\bar b\equiv \bar b\, $</span> by <a href="https://math.stackexchange.com/a/879262/242">Congruence Product Rule</a>. We used only commutativity & associativity of products so the proof works in any commutative ring or monoid.</p>
<p><strong>Note</strong> <span class="math-container">$ $</span> The proof is just a slick way of cancelling <span class="math-container">$\,\color{#c00}a\,$</span> from <span class="math-container">$\,\color{#c00}ab\equiv 1\equiv \color{#c00}a\bar b\,$</span> by scaling by <span class="math-container">$\,\color{#c00}a^{-1},\,$</span> yielding <span class="math-container">$\,b\equiv \bar b.\,$</span> Hence we see that the uniqueness of inverses is simply a special case of the general fact that <strong>invertible elements are cancellable</strong>, via scaling by the inverse.</p>
<p>Proofs like this can often be <em>discovered</em> by a general method of "overlapping" equations to find a term that both equations apply to (above the overalapped term is <span class="math-container">$\,ba\bar b\,$</span> which can be reduced by both rules <span class="math-container">$\,ab\to 1,\ a\bar b\to 1).\, $</span> Below is another example with further explanation.</p>
<hr />
<p>Below I will explain a general way to <em>discover</em> a simpler proof (vs. pulling it out of a hat like magic). The <em>key idea</em> is very simple: one can discover <em>consequences</em> of identities (axioms) by "overlapping" them: <span class="math-container">$ $</span>looking for a "unified" term that they both apply to. Let's try that to prove the uniqueness of additive identity elements. Suppose that <span class="math-container">$\,0\,$</span> and <span class="math-container">$\,0'$</span> are both additive identities. This means that</p>
<p><span class="math-container">$$\begin{eqnarray}
x = && \color{#0a0}0 + \color{#c00}x\\
&& \color{#0a0}y + \color{#c00}{0'} = y\\
\hline
\Rightarrow\ \ 0' = && 0 + 0' = 0\end{eqnarray}$$</span></p>
<p>where we chose the values of the specialization <span class="math-container">$\,\color{#0a0}{y=0},\ \color{#c00}{x = 0'}\,$</span> in order unify <span class="math-container">$\,0+x\,$</span> with <span class="math-container">$\,y+0', \,$</span> yielding the "unified" term <span class="math-container">$\,0+0'\,$</span> that both axioms apply to. Applying both axioms to the unified term we can rewrite it in two different ways, deducing the new consequence <span class="math-container">$\,0' = 0.\,$</span> Presto!</p>
<p>This is a very widely applicable method of deriving consequences of axioms, i.e. by "unifying" or "overlapping" terms of both so that both axioms apply, yielding a rewriting of the term in two different ways (e.g. <a href="https://math.stackexchange.com/a/616715/242">another example)</a>. In fact in some cases it can be used to algorithmically derive <em>all</em> of the consequences of the axioms, so yielding algorithms for deciding equality, e.g. see the <a href="http://en.wikipedia.org/wiki/Knuth%E2%80%93Bendix_completion_algorithm" rel="nofollow noreferrer">Knuth-Bendix equational completion algorithm</a> and the <a href="http://en.wikipedia.org/wiki/Gr%C3%B6bner_basis" rel="nofollow noreferrer">Grobner basis algorithm,</a> and see George Bergman's classic paper <a href="http://dx.doi.org/10.1016/0001-8708%2878%2990010-5" rel="nofollow noreferrer">The Diamond Lemma in Ring Theory.</a></p>
|
1,928,142 | <p>I need help with this geometric problem. </p>
<p>$ \overline {AB}$ is a diameter in circle $M$ , $C \in$ $\overrightarrow{AB}$ lying outside the circle , $\overline{CD}$ is drawn tangent to the circle at point $D$ , then $\overrightarrow{DH} \bot \overrightarrow{AB} $ to intersect it at $H$.</p>
<p><strong>Prove that :</strong> $(CD)^2 = CH \times CM = CB \times CA $</p>
<p>What I have proved to far that the $(CD)^2 = CH \times CM $,
but I couldn't prove the rest.</p>
| operatorerror | 210,391 | <p>Shortcut: In the first quadrant take $\arctan(y/x)$ </p>
|
257,368 | <p>I have the following system of ODEs:</p>
<p>dx/dt = a/(1 + z) - Q*z</p>
<p>dy/dt = Qx - qy</p>
<p>dz/dt = qy - cz/(K + z).</p>
<p>Assuming K = 1, Q = q < c, and a = c*(Sqrt[c/Q] - 1), is there a way to use Mathematica to test whether a Hopf bifurcation exists?</p>
<p>Any help would be greatly appreciated!</p>
<p>Thank you</p>
| E. Chan-López | 53,427 | <p>The following answer is just a guide to detect the Hopf bifurcation in your system:</p>
<p>Considering <span class="math-container">$a=\frac{3}{8}$</span>, <span class="math-container">$c=\frac{4}{3}$</span>, <span class="math-container">$k=\frac{3}{2}$</span>, <span class="math-container">$Q=\frac{1}{2}$</span> and <span class="math-container">$q$</span> as a bifurcation parameter, your system becomes as follows:</p>
<pre><code>F[{x_, y_, z_}][q_] := {-(z/2) + 3/(8 (1 + z)), 1/2 (x - 2 q y), q y - (8 z)/(9 + 6 z)}
X = {x, y, z};
</code></pre>
<p>The nontrivial equilibrium point:</p>
<pre><code>X0[q_] := First@SolveValues[F[X][q] == 0, X]
X0[q]
(*{2/3, 1/(3 q), 1/2}*)
</code></pre>
<p>Jacobian matrix:</p>
<pre><code>J[{x_, y_, z_}][q_] := Evaluate[D[F[X][q], {X}]]
J[X][q]
(*{{0, 0, -(1/2) - 3/(8 (1 + z)^2)}, {1/2, -q, 0}, {0, q, (48 z)/(9 + 6 z)^2 - 8/(9 + 6 z)}}*)
</code></pre>
<p>The linear approximations at <span class="math-container">$X_{0}(q)$</span>:</p>
<pre><code>J0[q_] := J[X0[q]][q]
J0[q]
(*{{0, 0, -(2/3)}, {1/2, -q, 0}, {0, q, -(1/2)}}*)
</code></pre>
<p>Local stability using <code>BialternateSum</code>:</p>
<pre><code>Reduce[Det[BialternateSum[J0[q]]] < 0 && q > 0, q]
(*q > 1/6*)
</code></pre>
<p>Looking for the Hopf bifurcation value using <code>BialternateSum</code>:</p>
<pre><code>q0 = First@SolveValues[Det[BialternateSum[J0[q]]] == 0 && q > 0, q]
(*1/6*)
</code></pre>
<p>The eigenvalues at <span class="math-container">$q=q_0$</span>:</p>
<pre><code>Eigenvalues[J0[q0]]
(*{-(2/3), I/(2 Sqrt[3]), -(I/(2 Sqrt[3]))}*)
</code></pre>
<p>The limit cycle appears for <span class="math-container">$q<q_0$</span>:</p>
<p><a href="https://i.stack.imgur.com/C63RK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C63RK.png" alt="enter image description here" /></a></p>
<p>There is a condition that makes the two non-trivial equilibria collapse into one, so it is possible to study a Bogdanov-Takens bifurcation in your system.</p>
<p>I hope you enjoy it!</p>
|
3,924,376 | <p>Given <span class="math-container">$i,a_i,b_i\in\{1...n\},\space a_i\neq a_j,b_i\neq b_j,\forall i\neq j$</span> prove that
<span class="math-container">$$\sum_{i=1}^n|a_i-b_i|\le\big\lfloor \frac{n^2}{2}\big\rfloor$$</span>
This problem was proposed by the new contributor @user3458994 and it was closed by five users. I find it somewhat challenging (it does not have an immediate answer), but it is sufficiently well posed and, indeed, it can be solved by answering correctly.</p>
<p>There are many possible sums <span class="math-container">$\sum_{i=1}^n|a_i-b_i|$</span>; in fact there are <span class="math-container">$n!$</span> possibilities (number of permutations of the set <span class="math-container">$\{1,2,\cdots,n\}$</span>). The minimum for these sums is <span class="math-container">$0$</span> corresponding to the identity permutation <span class="math-container">$a_i\rightarrow b_i=a_i$</span> in which case the inequality is trivially verified. We expose one of these sums having a maximum value <span class="math-container">$M$</span> exactly equal to <span class="math-container">$\big\lfloor \frac{n^2}{2}\big\rfloor$</span>. I believed no other sum has a value greater than <span class="math-container">$M$</span> in which case the problem would be false (am I wrong?).</p>
| Neat Math | 843,178 | <p>I have already given an answer in that post. I will post it again here. It's somewhat similar to the rearrangement inequality: When <span class="math-container">$\{a_i\}$</span> and <span class="math-container">$\{b_i=i\}$</span> have opposite order, the sum of the absolute difference reaches maximum (there could be other cases which also reach this maximum). The rest is just easy calculation.</p>
<hr />
<p><strong>Lemma:</strong> If <span class="math-container">$x>y,z>w$</span> then <span class="math-container">$|x-w|+|y-z|\geqslant |x-z|+|y-w|.$</span></p>
<p>WLOG we can assume <span class="math-container">$y\geqslant w$</span>. Then <span class="math-container">$x>w$</span>.</p>
<p><span class="math-container">$$|x-w|+|y-z|\geqslant |x-z|+|y-w| \iff x-w+|y-z| \geqslant |x-z|+y-w \\
\iff |x-y|+|y-z|\geqslant |x-z|$$</span></p>
<p>which follows from the triangle inequality.</p>
<p>WLOG assume <span class="math-container">$b_i=i$</span>. Then from the lemma the sum of absolute differences obtains its maximum value when <span class="math-container">$a_i$</span> is decreasing, i.e. <span class="math-container">$$\sum_{i=1}^n|a_i-i| \leqslant \sum_{i=1}^n |n+1-2i|.$$</span></p>
<p>If <span class="math-container">$n=2m$</span>, <span class="math-container">$$\sum_{i=1}^n |n+1-2i|=2(2m-1) + 2(2m-3)+\cdots + 2(1)=2m^2 = \lfloor \frac{n^2}{2} \rfloor.$$</span></p>
<p>If <span class="math-container">$n=2m+1$</span>, <span class="math-container">$$\sum_{i=1}^n |n+1-2i|=2(2m) + 2(2m-2)+\cdots + 2(0)=2m(m+1) = \lfloor \frac{n^2}{2} \rfloor.\blacksquare$$</span></p>
|
105,577 | <p>The analytic rank of the Mordell elliptic curve $y^2=x^3-86069^5$ indicates that it has rank 2. However, deriving a set of generators, and hence the regulator, is proving to be a little bit of an intractable problem.</p>
<p>I'm posting this question in the hope that someone may have investigated this curve already and, as such, may be able to help me out with some of the points on the curve.</p>
<p>Any help with this would be very much appreciated.</p>
<p>As further background, I have been involved in a small group searching for values of #Sha > 1 up through the ranks of Mordell curves, currently from ranks 0 to 9.</p>
<p>To date we have been able to progress to a rank 9 curve were #Sha would seem to be at least equal to 9. The details of which may be seen at the <a href="https://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind1208&L=NMBRTHRY&P=23667" rel="nofollow">NMBRTHRY Archives</a>. </p>
<p>Kevin.</p>
| Igor Rivin | 11,142 | <p>Take a look at <a href="https://mathoverflow.net/questions/42016/algorithms-for-finding-rational-points-on-an-elliptic-curve">this question and answers.</a> I am conjecturing that Cremona's mwrank (which, I believe, is available in sage) will find what you are looking for.</p>
<p><strong>EDIT</strong> Yes, <a href="http://www.sagenb.org/doc/live/reference/sage/schemes/elliptic_curves/ell_rational_field.html" rel="nofollow noreferrer">available in sage.</a></p>
|
2,244,321 | <p>Let $f, g$ absolutely continuous on $[a,b]$ with $f'(x)=g'(x)$ a.e. on $[a,b]$. Prove if $f=g$ at a single point in $[a,b]$ then $f=g$ everywhere</p>
<p>This is not homework, this is part of a list of problems that it would be advisable to solve before a general exam. Any suggestion is welcome, I have tackled it on my own but after hours of work I'm blind, I know I'm asking a lazy question but I need to move on, so any help is appreciated. </p>
| Rafael | 303,887 | <p>Let $M'$ be a finitely generated submodule of $M$ and $v:M'\rightarrow M$ the canonical inclusion. Then there are $y'_1,y'_2,...,y'_n\in M$ such that $M'=<y'_1,...,y'_n>$ and, since $M$ is the direct limit of $M_i$, if $u_i:M_i\rightarrow M$ is the canonical map, there are $l_1,...,l_n\in I$ such that $y'_i=u_{l_i}(y_{l_i})$ with $y_{l_i}\in M_{l_i}$. Since $I$ is a directed set, there is $k\in I$ such that $l_i\leq k$ and $y'_i=u_k(y_i)$ for all $i\in\{1,...,n\}$.</p>
<p>Take $z\in \ker(1_F\otimes v)$. There are $x_1,...,x_n\in F$ such that $\displaystyle z=\sum_{i=1}^{n} x_i\otimes u_k(y_i)$. Then $$\left(1_F\otimes u_k\right)\left(\sum x_i\otimes y_i\right)=0$$ in $F\otimes M$.</p>
<p>It is clear that $\left(F\otimes M_i,1_F\otimes u_{ij}\right)$ form a direct system of modules with direct limit has been $F\otimes M$ and $1_F\otimes u_i$ the canonical projections. From the last equality we have that there is $l\in I$ such that $k\leq l$ and $$\left(1_F\otimes u_{kl}\right)\left(\sum x_i\otimes y_i\right)=0$$ in $F\otimes M_l$.</p>
<p>Now, define, for each $r\in I$ such that $k\leq r$, $M'_r=<u_{rk}(y_1),...,u_{rk}(y_n)>$. Let $u'_{sr}$ the restriction of the map $u_{sr}$ to $M'_s$. It is clear that $(M'_r,u'_{sr})$ form a direct system of modules and so $(F\otimes M'_r,1_F\otimes u'_{rs})$ also. Let $\gamma$ be the canonical map $\displaystyle\gamma:\lim_{\longrightarrow}M'_r\rightarrow M$ and consider the canonical composition $$(1_F\otimes\gamma)\circ(1_F\otimes u'_l):F\otimes M'_l\longrightarrow F\otimes\left(\lim_{\longrightarrow}M'_r\right)\longrightarrow F\otimes M$$ Note that $$\left(1_F\otimes\gamma\right)\circ\left(1_F\otimes u'_l\right)\left(1\otimes u_{kl}\left(\sum x_i\otimes y_i\right)\right)=z$$ and, since $$\left(1\otimes u_{kl}\right)\left(\sum x_i\otimes y_i\right)=0$$ in $F\otimes M'_l$, we have $$z=\left(1_F\otimes\gamma\right)\circ\left(1_F\otimes u'_l\right)\left(1\otimes u_{kl}\left(\sum x_i\otimes y_i\right)\right)=0.$$ Therefore the map $1_F\otimes v$ is injective.</p>
|
1,822,314 | <p>I think of two numbers, $x$ and $y$.
When I add them together I get $5$ and when I find the difference I get $13$. What numbers did I think of?</p>
<p>I need to know how to write this down in simultaneous equation form</p>
<p>many thanks</p>
| Soham | 242,402 | <p>Let the two numbers be $a$ and $b$.From the question,we have </p>
<p>$$a+b=5$$ and $$a-b=13$$.</p>
<p>Adding them we get,$$(a+b)+(a-b)=18$$ </p>
<p>$$\implies2a=18$$</p>
<p>$$\implies a=9$$</p>
<p>Substituting $a=9$ in the first equation,we have $9+b=5$ or $b=-4$.</p>
<p>So,$a=9,b=-4$ is the solution.</p>
|
417,855 | <p>Let <span class="math-container">$a>1$</span> and define <span class="math-container">$G_a(x)=\sum\limits_{n=0}^{+\infty} \frac{\Gamma(\frac{2n+1}{a})}{\Gamma(2n+1)\Gamma(\frac{1}{a})}x^n$</span> where <span class="math-container">$\Gamma$</span> is the Gamma function. This series is convergent on <span class="math-container">$\mathbb{R}$</span> thanks to a ratio test and Stirling's formula.</p>
<p>For <span class="math-container">$a=2$</span>, using Legendre duplication formula <span class="math-container">$\Gamma(z)\Gamma(z+\frac{1}{2}) = 2^{1-2z} \Gamma(\frac{1}{2})\Gamma(2z)$</span> with <span class="math-container">$z=n+\frac{1}{2}$</span> shows that <span class="math-container">$G_2(x) = e^{\frac{x}{4}}$</span>.</p>
<p>Can we get a somewhat explicit formula for other integer values of <span class="math-container">$a$</span>, or even for arbitrary real values <span class="math-container">$a>1$</span>? Has this series been studied somewhere?</p>
<p>For context, I am computing the Laplace transform of generalised Gaussian distributions [1], i.e their log-densities are <span class="math-container">$\propto -\lvert x\rvert^{a}$</span> (hence the case <span class="math-container">$a=2$</span> corresponds to the usual Gaussian measures). The above power series is related to the moment-generating function of such distributions.</p>
<p>I tried my luck a bit with hypergeometric functions and the multiplication theorem for the Gamma function (generalisation of the duplication formula) for small integer values of <span class="math-container">$a$</span> but did not quite make it.</p>
<p>References:
[1] <a href="https://jsdajournal.springeropen.com/track/pdf/10.1186/s40488-018-0088-5.pdf" rel="nofollow noreferrer">https://jsdajournal.springeropen.com/track/pdf/10.1186/s40488-018-0088-5.pdf</a></p>
| Jorge Zuniga | 141,375 | <p>For arbitrary real <span class="math-container">$a > 0$</span> this is a special case of the generalized <span class="math-container">$_p\Psi_q(A;B;ζ)$</span> Fox-Wright function, where <span class="math-container">$A=[(a_1,\alpha_1),(a_2,\alpha_2),...,(a_p,\alpha_p)]$</span> and <span class="math-container">$B=[(b_1,\beta_1),(b_2,\beta_2),...,(b_q,\beta_q)]$</span> being <span class="math-container">$a_j, j=1,..,p$</span> and <span class="math-container">$b_k, k=1,..,q$</span> complex parameters and <span class="math-container">$\alpha_j, \beta_k$</span> are positive. <span class="math-container">$$_p\Psi_q(A;B;ζ)=\sum_{n=0}^\infty \frac{ζ^n}{n!}\frac{\prod_{j=1}^{p}\Gamma(a_j+\alpha_jn)}{\prod_{k=1}^{q}\Gamma(b_k+\beta_kn)}$$</span> None gamma function in the numerator is singular. This means <span class="math-container">$$a_{j}+α_{j}m≠-ℓ,$$</span> with <span class="math-container">$j=1,2,..,p ∧ ℓ,m∈ℕ₀$</span>. Series convergence depends on <span class="math-container">$\kappa, \rho, ϑ$</span> <span class="math-container">$$κ=∑_{j=1}^{q}β_{j}-∑_{j=1}^{p}α_{j}+1$$</span> <span class="math-container">$$ρ=∏_{j=1}^{p}α_{j}^{-α_{j}}⋅∏_{j=1}^{q}β_{j}^{β_{j}}$$</span> <span class="math-container">$$ϑ=½(q-p)+∑_{j=1}^{p}a_{j}-∑_{j=1}^{q}b_{j}$$</span> If <span class="math-container">$κ>0$</span> the series has an infinite radius of convergence and <span class="math-container">$_p\Psi_q(ζ)$</span> is an entire function. Series is uniformly and absolutely convergent for all finite <span class="math-container">$ζ$</span> . If <span class="math-container">$κ<0$</span> the sum is divergent for all nonzero values of <span class="math-container">$ζ$</span> whereas for <span class="math-container">$κ=0$</span> the function series has a finite radius of convergence <span class="math-container">$ρ$</span>. Convergence on the boundary <span class="math-container">$|ζ|=ρ$</span> depends on parameter <span class="math-container">$ϑ$</span> converging absolutely if <span class="math-container">$ℜ(ϑ)<-½$</span>.</p>
<p>For |arg<span class="math-container">$(-ζ)$$|<π-ε$</span>, the Mellin-Barnes Integral <span class="math-container">$$_{p}Ψ_{q}(ζ)=\frac{1}{2πi}\int_{L}\Gamma(s)⋅\frac{\prod_{j=1}^{p}\Gamma(a_{j}-α_{j}s)}{\prod_{k=1}^{q}\Gamma(b_{k}-\beta_{k}s)}(-ζ)^{-s}ds$$</span> defines a wider representation of Wright function. <span class="math-container">$L$</span> is a contour separating the poles of <span class="math-container">$\Gamma(s)$</span> to the left from those of <span class="math-container">$\Gamma(a_{j}-α_{j}s)$</span> to the right. For contour <span class="math-container">$L$</span> going from <span class="math-container">$-i\infty$</span> up to <span class="math-container">$+i\infty$</span> (possibly non-parallel to the vertical axis) this integral provides an analytical continuation of <span class="math-container">$_{p}Ψ_{q}(ζ)$</span> in <span class="math-container">$ζ∈ℂ\backslash [ρ,∞)$</span> when <span class="math-container">$ κ=0$</span>.</p>
<p>This function is a special case of FoxH function, (See Wiki's or Wolfram's sites)
<span class="math-container">$$_p\Psi_q(A;B;ζ)=H_{1+q,p}^{p,1}((1,1),B;A;-ζ^{-1})$$</span> For this particular case <span class="math-container">$A=[(1,1),(1/a,2/a)]$</span> and <span class="math-container">$B=[(1,2)]$</span>. Thus <span class="math-container">$G_a$</span> function is <span class="math-container">$$G_a(x)=\frac{_2\Psi_1([(1,1),(1/a,2/a)];[(1,2)];x)}{\Gamma(1/a)}$$</span> <span class="math-container">$$G_a(x)=\frac{H_{2,2}^{2,1}([[(1,1)],[(1,2)]];[[(1,1),(1/a,2/a)],[\cdot]];-x^{-1})}{\Gamma(1/a)}$$</span> Note that <span class="math-container">$κ=2(1-1/a)$</span> and series converges for <span class="math-container">$a>1$</span> to an entire function. You can set this expression using FoxH function in Wolfram's Mathematica v13.0 in symbolic mode to see if there are some explicit formulae for other values of parameter <span class="math-container">$a$</span>. I suggest try with <span class="math-container">$a\in \mathbb{Q}$</span> where <span class="math-container">$a>1$</span></p>
|
9,719 | <p>I have proved this using Venn diagram but when I am trying to prove this using the rule that <strong><em>"If $ A \subset B \text{ and } B \subset A $ then $ A = B $"</em></strong>, I am having some problems with my understanding of the same,here is how I did so far:</p>
<p>Let $x \in A \cap (B-C) \Rightarrow x \in A \text{ and } x \in (B-C) \Rightarrow x \in A \text{ and } (x \in B \text{ and } x \notin C) $</p>
<p>How to proceed next? Since if I am do something like this: <strong>$ x \in A \text{ and } x \in B \text{ and } x \in A \text{ and } x \notin C $</strong>, it's not giving the correct results, what exactly I am missing here?</p>
| Aryabhata | 1,102 | <p>Hint: $x \notin C \Rightarrow x \notin D \cap C$ for any set $D$.</p>
|
335 | <p>I looked at the “accept rate” of some of the site’s top users and found that it varies widely in the range 45%–100%. I was surprised with that, as it may be taken as an indicator that either:</p>
<ol>
<li>hard questions don’t tend to get very good answers (let's assume that top users ask difficult questions), or</li>
<li>we could better reward people who wrote good answers.</li>
</ol>
<p>So, what do you think of these two hypotheses, and do we want to do something about it?</p>
<p>Regarding the statistics, the accept rate of the users with ≥ 3k reputation and ≥ 10 questions are: 55%, 86%, 57%, 47%, 89%, 100%, 68%, 52%.</p>
<hr>
<p>PS: I know that there has been debate over whether the “accept rate” is generally a good thing and what its optimal value should be is, but I think a community consensus should emerge on this topic.</p>
| David | 103 | <p>When I don't get a satisfying answer to my question I don't accept. I don't see how that could possibly be negotiable.</p>
|
2,556,993 | <p>I've come across two limits which are reported to take on a range of values rather than a unique one.</p>
<p>They are:</p>
<p>$$\lim \limits_{x \to \infty} \space\frac{1+\cos x}{1-\sin x} = 0 \space to\space \infty$$</p>
<p>$$\lim \limits_{x \to \infty} \space \frac{2+2x+\sin(2x)}{(2x+\sin(2x))e^{\sin x}} = \frac{1}{e} \space to \space e$$</p>
<p>These answers seem to contradict what I know and understand as the definition of a limit. I'm self-studying maths so am clearly missing something here.</p>
<p>My questions:</p>
<ol>
<li>How are these limits solved?</li>
<li>Why do they exist? Why is a range of values allowed here?</li>
<li>What kind of books should I read to learn more about this? If you could provide references that would be perfect.</li>
</ol>
<p>ETA results from Wolfram:</p>
<p><a href="https://www.wolframalpha.com/input/?i=limit%20x-%3Einf%20((1%2Bcosx)%2F(1-sinx))" rel="nofollow noreferrer">The First</a>
<a href="https://www.wolframalpha.com/input/?i=limit%20x-%3Einf%20((2%2B2x%2Bsin2x)%2F(2x%2Bsin2x)e%5E(sinx))" rel="nofollow noreferrer">The Second</a></p>
| StackTD | 159,845 | <p>In the usual sense (and common definition), these limits <strong>do not</strong> exist (since, loosely speaking, the function values don't come arbitrarily close to a specific value for sufficiently large values of $x$).</p>
<p>For a simpler example, consider:
$$\lim_{x \to +\infty} \sin x$$
The fact that $\sin x$ keeps taking (all) values in the interval $[-1,1]$, even for arbitrary large values of $x$, can be used as an argument <em>against</em> the existence of this limit.</p>
<p>Of course, saying that $\sin x$ keeps taking values in $[-1,1]$ contains <em>more information</em> than simply saying "the limit does not exist", so it can be useful to consider (and determine) this range - although I would never say "the limit is [a range]".</p>
|
3,753,788 | <p>In his book Serge Lang provides the following proof that every bounded set that is closed is also compact.</p>
<p>" Let S be a closed and bounded set. This implies that there exists <span class="math-container">$B$</span> s.t. <span class="math-container">$|z|\leq B$</span> where <span class="math-container">$z\in s$</span>. For all <span class="math-container">$z$</span> write <span class="math-container">$z=x+iy$</span>, then <span class="math-container">$|x|\leq B$</span> and <span class="math-container">$|y| \leq B$</span>.</p>
<p>Let <span class="math-container">${z_n}$</span> be a sequence in <span class="math-container">$S$</span>, and write <span class="math-container">$z_n=x_n+iy_n$</span>.</p>
<p>There is a subsequence {<span class="math-container">$z_{n_1}$</span>} s.t. {<span class="math-container">${x_{n_1}}$</span>} converges to a real number <span class="math-container">$a$</span> and similarly there is a subsequence {<span class="math-container">${z_{n_2}}$</span>} s.t. {<span class="math-container">${y_{n_2}}$</span>} converges to a real number <span class="math-container">$b$</span>.</p>
<p>Then {<span class="math-container">$z_{n_2}= x_{n_2}+i{y_2}$</span>} converges to <span class="math-container">$a+ib$</span> which implies that <span class="math-container">$S$</span> is compact."</p>
<p>What I don't understand is two things.</p>
<ol>
<li>How does proving {<span class="math-container">$x_{n_1}$</span>} converges to <span class="math-container">$a$</span> prove that {<span class="math-container">${x_{n_2}}$</span>} converges to <span class="math-container">$a$</span> aswell?</li>
<li>How is it implied that there exist such subsequences such that {<span class="math-container">${x_{n_1}}$</span>} and {<span class="math-container">${y_{n_2}}$</span>} converge to these values? Afterall isn't this what we are suppose to prove?</li>
</ol>
<p>If anybody could shine some light, I will greatly appreciate it.</p>
| Mark | 470,733 | <p>This is indeed a mistake. You should take a convergent subsequence of <span class="math-container">$y_{n_1}$</span> which converges, not just a subsequence of <span class="math-container">$y_n$</span>. In that case <span class="math-container">$x_{n_2}$</span> is a subsequence of <span class="math-container">$x_{n_1}$</span> and hence it converges to <span class="math-container">$a$</span> as well.</p>
<p>As for your second question, I believe the author assumes you already know that a bounded sequence of real numbers has a convergent subsequence (the Bolzano-Weierstrass theorem), and now he uses it to prove this property holds for complex sequences as well.</p>
|
3,753,788 | <p>In his book Serge Lang provides the following proof that every bounded set that is closed is also compact.</p>
<p>" Let S be a closed and bounded set. This implies that there exists <span class="math-container">$B$</span> s.t. <span class="math-container">$|z|\leq B$</span> where <span class="math-container">$z\in s$</span>. For all <span class="math-container">$z$</span> write <span class="math-container">$z=x+iy$</span>, then <span class="math-container">$|x|\leq B$</span> and <span class="math-container">$|y| \leq B$</span>.</p>
<p>Let <span class="math-container">${z_n}$</span> be a sequence in <span class="math-container">$S$</span>, and write <span class="math-container">$z_n=x_n+iy_n$</span>.</p>
<p>There is a subsequence {<span class="math-container">$z_{n_1}$</span>} s.t. {<span class="math-container">${x_{n_1}}$</span>} converges to a real number <span class="math-container">$a$</span> and similarly there is a subsequence {<span class="math-container">${z_{n_2}}$</span>} s.t. {<span class="math-container">${y_{n_2}}$</span>} converges to a real number <span class="math-container">$b$</span>.</p>
<p>Then {<span class="math-container">$z_{n_2}= x_{n_2}+i{y_2}$</span>} converges to <span class="math-container">$a+ib$</span> which implies that <span class="math-container">$S$</span> is compact."</p>
<p>What I don't understand is two things.</p>
<ol>
<li>How does proving {<span class="math-container">$x_{n_1}$</span>} converges to <span class="math-container">$a$</span> prove that {<span class="math-container">${x_{n_2}}$</span>} converges to <span class="math-container">$a$</span> aswell?</li>
<li>How is it implied that there exist such subsequences such that {<span class="math-container">${x_{n_1}}$</span>} and {<span class="math-container">${y_{n_2}}$</span>} converge to these values? Afterall isn't this what we are suppose to prove?</li>
</ol>
<p>If anybody could shine some light, I will greatly appreciate it.</p>
| Moe Sarah | 787,944 | <p>As <span class="math-container">$\mathbb{R}^2=\mathbb{C}$</span>, we may proceed by noting that:</p>
<p><strong>Every closed and bounded set in <span class="math-container">$\mathbb{R}^2$</span> is compact.</strong></p>
<p>Here is a proof:</p>
<p>Let <span class="math-container">$F\subseteq \mathbb{R}^2$</span> be both, closed and bounded. Let <span class="math-container">$(x_n)_n\subseteq F$</span> be a sequence. As <span class="math-container">$(x_n)_n$</span> is bounded, it follows from the Bolzano Weirstrass theorem that <span class="math-container">$(x_n)_n$</span> has a convergent subsequence, which converges to some <span class="math-container">$x\in \mathbb{R}^2$</span>. As <span class="math-container">$F$</span> is closed, <span class="math-container">$x\in F$</span>. <span class="math-container">$\square$</span></p>
|
3,997,676 | <p>Let <span class="math-container">$1\leq n\in \mathbb{N}$</span>, <span class="math-container">$E=\left (e_1, \ldots , e_n\right )$</span> the standardbasis of <span class="math-container">$\mathbb{R}^n$</span>, <span class="math-container">$B=\left (b_1, \ldots , b_n\right )$</span> a basis of <span class="math-container">$\mathbb{R}^n$</span> and <span class="math-container">$a:=\left (b_1\mid b_2\mid \ldots \mid b_n\right )\in M_n(\mathbb{R}^n)$</span>.</p>
<p>Show the following:</p>
<ol>
<li><p><span class="math-container">$a\gamma_B(v)=v$</span> for all <span class="math-container">$v\in \mathbb{R}^n$</span>.</p>
</li>
<li><p><span class="math-container">$a$</span> is invertible and it holds that <span class="math-container">$a^{-1}=\left (\gamma_B(e_1)\mid \gamma_B(e_2)\mid \ldots \mid \gamma_B(e_n)\right )$</span>.</p>
</li>
</ol>
<p>I have done an example and I have seen that it holds that <span class="math-container">$\left (b_1\mid b_2\mid b_3\right )\left (\gamma_B(e_1)\mid \gamma_B(e_2)\mid \gamma_B(e_3)\right )=I_{3\times 3}$</span>.</p>
<p>But how can we show that it holds?</p>
<p>We have that <span class="math-container">$$\gamma_B(v)=v \Rightarrow c_1b_1+c_2b_2+\ldots c_nb_n=v$$</span> then multiplying with the matrix <span class="math-container">$a$</span> and knowing that <span class="math-container">$B$</span> is a basis we get the result, but how exactly can we show that?</p>
| Adam Rubinson | 29,156 | <p><span class="math-container">$0 \notin \mathbb{N}$</span>.</p>
<p>Let <span class="math-container">$f:\mathbb{N} \to \mathbb{Z}$</span> be a surjective and injective function. Define:</p>
<p><span class="math-container">$$g(n) = \cases {f(1),\quad n =1\\
f(1),\quad n =2\\
f(n-1),\quad n\geq3}$$</span></p>
<p>Then <span class="math-container">$g:\mathbb{N} \to \mathbb{Z}$</span> is surjective but not injective.</p>
|
2,934,472 | <p>Let <span class="math-container">$X_1,X_2,\dots$</span> denote a sequence of identically distributed, exchangeable, but not independent, Bernoulli<span class="math-container">$(p)$</span> random variables. If there exists <span class="math-container">$n>0$</span> such that</p>
<p><span class="math-container">$$
\sum_{i=1}^{n}X_i
$$</span></p>
<p>is not uniformly distributed on <span class="math-container">$\{0,1,\dots,n\}$</span>, how can we show that this implies</p>
<p><span class="math-container">$$
\sum_{i=1}^{n}X_i + X_{n+1}
$$</span></p>
<p>is not uniformly distributed on <span class="math-container">$\{0,1,\dots,n+1\}$</span>? If <span class="math-container">$p \ne 1/2$</span> then the claim follows by expectations, so we can consider <span class="math-container">$p=1/2$</span>.</p>
| Did | 6,179 | <p>First, some notations: for every <span class="math-container">$n$</span>, let <span class="math-container">$X_{1:n}=(X_1,X_2,\ldots,X_n)$</span>, and, for every word <span class="math-container">$w=(w_1,w_2,\ldots,w_n)$</span> in <span class="math-container">$\{0,1\}^n$</span>, let <span class="math-container">$|w|=w_1+w_2+\cdots+w_n$</span>. Now, to the proof, which uses crucially the exchangeability hypothesis.</p>
<p>Assume that, for some <span class="math-container">$n\geqslant1$</span>, <span class="math-container">$|X_{1:n+1}|$</span> is uniformly distributed on <span class="math-container">$\{0,1,\ldots,n+1\}$</span>. </p>
<p>Then, by exchangeability, for every word <span class="math-container">$w$</span> in <span class="math-container">$\{0,1\}^{n+1}$</span>, <span class="math-container">$P(X_{1:n+1}=w)$</span> depends only on <span class="math-container">$|w|$</span>. For each <span class="math-container">$k$</span> in <span class="math-container">$\{0,1,\ldots,n+1\}$</span>, there are <span class="math-container">${n+1\choose k}$</span> words <span class="math-container">$w$</span> in <span class="math-container">$\{0,1\}^{n+1}$</span> such that <span class="math-container">$|w|=k$</span>, hence, for every word <span class="math-container">$w$</span> in <span class="math-container">$\{0,1\}^{n+1}$</span> such that <span class="math-container">$|w|=k$</span>, <span class="math-container">$$P(X_{1:n+1}=w)={n+1\choose k}^{-1}P(|X_{1:n+1}|=k)=(n+2)^{-1}{n+1\choose k}^{-1}$$</span>
For every word <span class="math-container">$v$</span> in <span class="math-container">$\{0,1\}^n$</span>, the event <span class="math-container">$[X_{1:n}=v]$</span> is the disjoint union of the events <span class="math-container">$[X_{1:n}=v,X_{n+1}=0]$</span> and <span class="math-container">$[X_{1:n}=v,X_{n+1}=1]$</span>.
If <span class="math-container">$|v|=k$</span> for some <span class="math-container">$k$</span> in <span class="math-container">$\{0,1,\ldots,n\}$</span>, then <span class="math-container">$|v0|=k$</span> and <span class="math-container">$|v1|=k+1$</span>, hence <span class="math-container">$$P(X_{1:n}=v)=(n+2)^{-1}{n+1\choose k}^{-1}+(n+2)^{-1}{n+1\choose k+1}^{-1}$$</span>
Now, it happens that <span class="math-container">$$(n+2)^{-1}{n+1\choose k}^{-1}+(n+2)^{-1}{n+1\choose k+1}^{-1}=(n+1)^{-1}{n\choose k}^{-1}\tag{$\ast$}$$</span> hence <span class="math-container">$$P(X_{1:n}=v)=(n+1)^{-1}{n\choose k}^{-1}$$</span> Summing these over the <span class="math-container">${n\choose k}$</span> words <span class="math-container">$v$</span> in <span class="math-container">$\{0,1\}^n$</span> such that <span class="math-container">$|v|=k$</span>, one gets, for every <span class="math-container">$k$</span> in <span class="math-container">$\{0,1,\ldots,n\}$</span>, <span class="math-container">$$P(|X_{1:n}|=k)=(n+1)^{-1}$$</span>
Thus, if <span class="math-container">$|X_{1:n+1}|=X_1+X_2+\cdots+X_{n+1}$</span> is uniformly distributed on <span class="math-container">$\{0,1,\ldots,n+1\}$</span>, then <span class="math-container">$|X_{1:n}|=X_1+X_2+\cdots+X_n$</span> is uniformly distributed on <span class="math-container">$\{0,1,\ldots,n\}$</span>.</p>
<p>By contraposition, this proves the desired statement -- and also that, as soon as <span class="math-container">$|X_{1:n}|=X_1+X_2+\cdots+X_{n}$</span> is uniformly distributed on <span class="math-container">$\{0,1,\ldots,n\}$</span> for some <span class="math-container">$n\geqslant1$</span>, then <span class="math-container">$|X_{1:1}|=X_1$</span> is uniformly distributed on <span class="math-container">$\{0,1\}$</span>, and, again by exchangeability, every <span class="math-container">$X_n$</span> is uniformly distributed on <span class="math-container">$\{0,1\}$</span>, that is, necessarily, <span class="math-container">$p=\frac12$</span>.</p>
<p><em>Exercise:</em> Prove <span class="math-container">$(\ast)$</span>.</p>
|
2,910,643 | <p>This question was in my maths paper . I tried to prove that the series converges but failed to do so. Any hint are welcomed .</p>
| b00n heT | 119,285 | <p>Hint: $$(n+1)\cdot(n+2)\cdots(n+k)=\frac{(n+k)!}{n!}$$
from here you only need to get to $\displaystyle \sum \frac1{k!}=e$, which is easy by adding some missing terms.</p>
|
3,173,766 | <p>If you have the option to roll 2 separate 10-sided dice or a single 10-sided die that you add 2 to the result with a goal of at least one die getting a specific number or higher (say 6 for the purposes of this question) is one of these options inherently better?</p>
<p>It seems that the single die with a modifier is statistically better, but I'm not sure if that accounts for the 2 dice being more consistent.</p>
| Brian Tung | 224,454 | <p>Suppose in option (a), you have <span class="math-container">$k$</span> <span class="math-container">$n$</span>-sided dice, and in option (b), you have a single <span class="math-container">$n$</span>-sided die, and you add a modifier <span class="math-container">$m$</span>. Furthermore, suppose that you are aiming for a target score of at least <span class="math-container">$t$</span> (with <span class="math-container">$1 \leq t \leq n$</span>).</p>
<p>In case (a), you achieve your target unless each of the <span class="math-container">$k$</span> dice is below <span class="math-container">$t$</span>, which happens with probability</p>
<p><span class="math-container">$$
P(\text{all $k$ dice are below $t$}) = \left(\frac{t-1}{n}\right)^k
$$</span></p>
<p>so the probability of success here is</p>
<p><span class="math-container">$$
P(\text{success using (a)}) = 1-\left(\frac{t-1}{n}\right)^k
$$</span></p>
<p>In case (b), you achieve your target so long as you roll at least <span class="math-container">$t-m$</span>, which happens with probability</p>
<p><span class="math-container">$$
P(\text{success using (b)}) = \min\left\{1, \frac{n-t+m+1}{n}\right\}
$$</span></p>
<p>As Ross Millikan indicates in his answer, there is no simple formula that tells you which is higher, other than comparing the two quantities above. For example, with <span class="math-container">$k = 2$</span> dice, each with <span class="math-container">$n = 10$</span> sides, versus a single die with a bonus of <span class="math-container">$m = 2$</span>, we get</p>
<p><span class="math-container">$$
1-\left(\frac{t-1}{10}\right)^2 \qquad \text{vs} \qquad \min\left\{1, \frac{13-t}{10}\right\}
$$</span></p>
<p>and we find that option (a) is better for targets <span class="math-container">$t = 4, 5, 6, 7, 8$</span>, but option (b) is better for targets <span class="math-container">$t = 1, 2, 3, 9, 10$</span>.</p>
|
201,103 | <p>When comparing datasets using <code>DistributionChart</code>, is it possible to add a horizontal line to the chart that lies over all of the individual distributions within the chart?</p>
| MelaGo | 63,360 | <p>You can also use <code>Show</code>:</p>
<pre><code>dc = DistributionChart[
RandomVariate[NormalDistribution[2, .5], {2, 100}]];
Show[dc, Graphics[Line[{{.5, 2}, {2.5, 2}}]]];
</code></pre>
<p><a href="https://i.stack.imgur.com/i7cAi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/i7cAi.png" alt="enter image description here"></a></p>
|
3,517,376 | <p>A friend of mine presented a problem I found interesting:</p>
<blockquote>
<p>Compute the following: <span class="math-container">$$\sum_{n=0}^\infty\left(\prod_{k=1}^j(1+x^k)\right)[x^{mn}]$$</span> where <span class="math-container">$P(x)[x^n]$</span> denotes the <span class="math-container">$x^n$</span> coefficient of <span class="math-container">$P$</span>.</p>
</blockquote>
<p>This problem is meant to be solved with the assistance of a program for large <span class="math-container">$j$</span> (larger than 1000) and <span class="math-container">$m\ll j$</span>. I'm interested in general, but if the case of <span class="math-container">$m~|~j$</span> is special, then that suffices for me.</p>
<p>My first thought was to brute force compute all of the coefficients, but this takes <span class="math-container">$\mathcal O(j^3)$</span> time, which is too slow.</p>
<p>My second thought was that this is essentially searching for <a href="https://en.wikipedia.org/wiki/Partition_(number_theory)#Odd_parts_and_distinct_parts" rel="noreferrer">distinct partitions</a>. The issue is the finite upper bound, meaning that this becomes inaccurate after than <span class="math-container">$x^{j+1}$</span> coefficient. I'm not super familiar with partitions to know how to adjust for this without brute force like above.</p>
<p>My third thought was to tackle this more generally with:</p>
<p><span class="math-container">$$\sum_{n=0}^\infty P(x)[x^{mn}]=\frac1m\sum_{n=0}^{m-1}P(e^{2\pi in/m})$$</span></p>
<p>But this almost completely ignores the form of <span class="math-container">$P$</span>, so I don't feel like this is how the problem should be tackled.</p>
<p>How can I tackle this problem?</p>
<hr>
<p>Update:</p>
<p>I just realized that when <span class="math-container">$m~|~j$</span> we have</p>
<p><span class="math-container">$$\sum_{n=0}^\infty\left(\prod_{k=1}^j(1+x^k)\right)[x^{mn}]=\frac1m\sum_{n=0}^{m-1}\left[\prod_{k=0}^{m-1}(1+e^{2\pi ink/m})\right]^{j/m}$$</span></p>
<p>which is far fewer computations. <a href="https://tio.run/##VY2xDsIwDER3vuLEVEIJoQsCqSxMDEjsiAFoKlpIUkwjFSG@PbiBhclnvycf@dMzkL77ijSGZ3NsL8NQOod8soJJUeM1ADrkWG@ZSd01SVaJPi@Xu83UjBg/vGFBcSodwaKyUFJKwwegIVcwncWl59d//jMEKxijEyKx4jqKSNsizv7/OI@eJH28SXLeFmzW3/6vR7r1ZKM8hRm8Q8N1br9Ikam5OoQP" rel="noreferrer">Ruby code</a>.</p>
<p>Still interested in alternative solutions.</p>
| Simply Beautiful Art | 272,831 | <p>Since the only criterion for a term to contribute to the sum is for the power to be a multiple of <span class="math-container">$m$</span>, we can consider all exponents mod <span class="math-container">$m$</span> i.e. <span class="math-container">$\mathbb R[x]/(x^m-1)$</span>. Thus if <span class="math-container">$m~|~j$</span> then we have</p>
<p><span class="math-container">$$\sum_{n=0}^\infty\left(\prod_{k=1}^j(1+x^k)\right)[x^{mn}]=\sum_{n=0}^\infty\left(\prod_{k=0}^{m-1}(1+x^k)\right)^{j/m}[x^0]$$</span></p>
<p>This can then be interpreted as matrix exponentiation:</p>
<p><span class="math-container">$$\sum_{n=0}^\infty\left(\prod_{k=1}^j(1+x^k)\right)[x^{mn}]=(A^{j/m})_{0,0}$$</span></p>
<p>where</p>
<p><span class="math-container">$$A_{a,b}=\left(\prod_{k=0}^{m-1}(1+x^k)\right)[x^{(a-b)\%m}]$$</span></p>
<p>which is probably the intended solution.</p>
|
984,154 | <p>Let say my girlfriend makes $2000$ Euros per month and I make $3400$ Euros, let say all our living costs sum up to $1540$ Euros.
How can i calculate how much each of us must pay (off the $1540$ Euros) so as our participation is proportional to our income?</p>
<p>Thanks</p>
| 5xum | 112,884 | <p>Write down the equations that must be satisfied. Let $x$ be the amount you pay and $y$ the amount your girlfriend pays.</p>
<p>One equation is simple: $x+y=1540$, since you must pay the whole costs. The other equation comes from the proportions that must be satisfied:</p>
<p>$$x:y = 3400 : 2000$$
Which can also be written as $$\frac xy = \frac{3400}{2000}$$</p>
|
679,343 | <p>I have been researching this for the past hour or so, but I have not really found a sufficient answer. What I have found is that vaguely, math education PhDs set policies for math education. However, universities almost uniformly have math professors teach mathematics courses. I believe that usually the math professor runs his course the way he wants to. Do their policies affect the university level, and if so, why do math professors teach the courses instead of math education professors? </p>
| Fly by Night | 38,495 | <p>Having a PhD in anything doesn't suddenly make you into anything. Lots of people with Mathematics PhDs don't work in universities and don't do mathematical research. </p>
<p>People who have a PhD in mathematical education will have spent time researching mathematical pedagogy, more often than not, at pre-university level. In the UK, people who have done this are then often involved in teaching and training those who are training to be a school teacher. </p>
<p>Academic reform, e.g. introducing new curricular, is often done by panels of maths teachers, maths professors, politicians and educationalists. </p>
|
1,740,221 | <p>If I have $x^{\lfloor\frac{c}{a-b}\rfloor}$ is this equivalent to $(x^{\lfloor\frac{1}{a-b}\rfloor})^{\lfloor c \rfloor}$ if a,b,c are all integers and x is between 0 and 1?</p>
<p>I'm concerned with this expression in the context of a geometric series. Normally if we have $\sum_{i=0}^{\infty} r^{q}$ with |r| < 1 then this sum converges to $\frac{1}{1-q}$. So I'm wondering if I can do something like $\sum_{i=0}^{\infty} (x^{\lfloor\frac{1}{a-b}\rfloor})^{\lfloor i \rfloor}$</p>
<p>It seems that the answer is no, because we could have something like $\frac{2}{3}$ which is .75 but the floor is 0. Splitting it up we would end up with .75</p>
| Graham Kemp | 135,106 | <p>No. Witness:</p>
<p>$$\left\lfloor \tfrac{4}{5-2}\right\rfloor = 1 \\ \lfloor 4\rfloor\left\lfloor \tfrac 1{5-2} \right\rfloor=0$$</p>
|
1,869,843 | <p>We know that:
$$\left(\sum_{k=1}^n k\right)^2 = \sum_{k=1}^n k^3 $$ </p>
<p>My question is there other examples that satisfies:
$$\left(\sum_{k=1}^n k\right)^a = \sum_{k=1}^n k^b $$</p>
| Rosie F | 344,044 | <p>Put $n=2.$ Then the RHS is$$S=\sum k^b=2^b+1=\frac{2^{2b}-1}{2^b-1}.$$ By <a href="http://mathworld.wolfram.com/ZsigmondyTheorem.html" rel="nofollow noreferrer">Zsigmondy's Theorem</a>, if $b\ne 3$, there is a prime $p$ where the order of 2 mod $p$ is $2b$, and so $p\mid S$. (If $b=3$, the RHS is $9=3^2$ and there is no such $p$, for the order of 2 mod 3 is 2, not 6.)</p>
<p>Further, $p\mid (2^c+1)$ only if $b\mid c$, thus not for any $0<c<b$. But the only primes that divide $\left[\sum k\right]^a$ are those that divide $3^a$, viz 3.</p>
<p>Even if the LHS were generalised to $S=\left[\sum k^c\right]^a$, then, for a non-trivial solution, $a>1$ and $c<b$, so the above prime $p\nmid S$.</p>
|
271,846 | <p>In his <em>"Some topological properties..." (1955)</em>, Klee gave a construction (simple and beautiful) of an isotopy $h_t\colon\mathbb{R}^{2\cdot n}\to \mathbb{R}^{2\cdot n}$ which moves any compact set $K$ in the coordinate $\mathbb{R}^n$-subspace to any other homeomorphic compact $K'$ set in this subspace. </p>
<p>The idea is to extend the homeomorphisms $K\to K'$ and $K'\to K$ to continuous functions $f,g\colon\mathbb{R}^n\to\mathbb{R}^n$ and construct the needed isotopy as concatenation of $(x,y)\mapsto (x,y+t\cdot f(x))$ and $(x,y)\mapsto (x-t\cdot g(y),y)$ </p>
<p>Later in <em>"Plane separation" (1968)</em>, Doyle used this idea to give 5 line proof of the Jordan separation theorem.</p>
<blockquote>
<p>Do you know any other places where this idea shows up?</p>
</blockquote>
| Dan Ramras | 4,042 | <p>This answer is a correction to my comment above. In <a href="http://math.iupui.edu/%7Edramras/M-Cohen-Klee-trick-notes.pdf" rel="noreferrer">these notes</a> from Marshall Cohen's course (from Spring 2001), he uses the Klee trick to give a simple proof that if $f, g: X \to S^n$ are non-surjective embeddings with $X$ compact, then the integral homology of $S^n - f(X)$ is the same as that of $S^n - g(X)$. From this he gave simple deductions of Invariance of Domain (showing that $\mathbb{R^k}$ does not embed into $\mathbb{R^n}$ if $k>n$ and the Jordan-Brouwer separation theorem for embeddings of $S^{n-1}$ into $S^n$. </p>
|
3,837,815 | <blockquote>
<p>Prove <span class="math-container">$\log_{4}6$</span> is irrational.</p>
</blockquote>
<p>I've seen proofs of this which boil down to:
<span class="math-container">$$4^{m} = 6^{n}$$</span></p>
<p>But how does this prove that it is irrational?
For example it is possible to have <span class="math-container">$n$</span> be:
<span class="math-container">$$x=\frac{\log 4}{\log 6}$$</span></p>
| redroid | 411,951 | <p>It proves it's irrational because for it to be rational would require integers <span class="math-container">$n,m$</span> that make the equation you've written true. However, that equation decomposes into:</p>
<p><span class="math-container">$$ 2^{2m} = 3^n 2^n $$</span></p>
<p>or if you like:</p>
<p><span class="math-container">$$ 2^{2m-n} = 3^n $$</span></p>
<p>For this to be true, an integer power of <span class="math-container">$3$</span> would have to be equal to some other integer power of <span class="math-container">$2$</span>. Since prime decompositions are unique, and therefore any power of <span class="math-container">$3$</span> (or <span class="math-container">$2$</span>) cannot be decomposed into any other primes, this is impossible.</p>
|
12,239 | <p>I know this is a trivial problem but am stuck with this thing.</p>
<p>$A \in \mathbb{R}^{m \times m}$ is a symmetric positive definite matrix and $C \in \mathbb{R}^{m \times p}$
</p>
<p>How do I show that the matrix $$B = \begin{bmatrix} A & C \\ C^T & 0 \end{bmatrix}$$ is indefinite?</p>
<p>I can show that there exists a vector $v$ such that $v^TBv > 0$.
(This is trivial. For instance, $v = [x , 0]^T$ where $x \in \mathbb{R}^{1 \times m}$ and $0 \in \mathbb{R}^{1 \times p}$)</p>
<p>Now how do I find a vector such that $v^TBv < 0$.</p>
<p>I do not need the answer. A clue/hint is welcome.</p>
<p>You can assume that $p \leq m$ and the matrix $C$ is full rank that is to say that all the constraints are linearly independent.</p>
<p>Thanks</p>
| Hans Lundmark | 1,242 | <p>Hint for a less convoluted argument: Unless $C$ is zero (in which case the statement is false), there are vectors $x$ and $y$ such that $x^T C y$ is a nonzero $1\times 1$ matrix. Now consider $v=[x,ky]^T$ for large $k$ (positive or negative, depending on the sign of $x^T C y$).</p>
|
3,638,206 | <p>Is it true that if <span class="math-container">$m$</span> is a Lebesgue measure on <span class="math-container">$[0,1]$</span> and that <span class="math-container">$\lambda$</span> is the counting measure on <span class="math-container">$[0,1]$</span>, and that both defined on the Lebesgue <span class="math-container">$\sigma$</span>-algebra, then there exists <span class="math-container">$h\in L^1(\lambda)$</span> such that <span class="math-container">$m(E)=\int_Eh\,d\lambda$</span>?</p>
<p>I've been thinking about this for awhile now and have been going back and forth convincing myself it's true, then convincing myself that it's not!!! Would really appreciate if somebody settled this issue for me!! Thanks!</p>
| Physical Mathematics | 592,278 | <p>Fix <span class="math-container">$E \subseteq \mathbb{R}$</span> Lebesgue measurable. Let <span class="math-container">$h = 1_{\{0\}} m(E)$</span> (i.e. <span class="math-container">$h(x)=0$</span> for all <span class="math-container">$x \neq 0$</span> and <span class="math-container">$h(0) = m(E)$</span>). Then <span class="math-container">$\int h d\lambda = m(E)\int 1_{\{0\}} d\lambda = m(E) \lambda(\{0\}) = m(E)$</span>.</p>
|
221,867 | <p>For a Hermitian nonnegative-definite matrix $A$, if $Ax$ is always real for any real vector $x$, can we conclude that $A$ is also real?</p>
| copper.hat | 27,978 | <p>Since $e_i^T A e_j = [A]_{i,j}$, and $Ax$ is real for real $x$, then $A$ must be real.</p>
|
619,519 | <p>The equation $\log_3(\sqrt{x+1}+1)=(3^{x+1}-1)^2\,$
has two solutions, but I can't solve the equation.</p>
| Claude Leibovici | 82,404 | <p>After the nice remark from Constructor and the simpler formulation given by labuwx, let see how to solve the equation </p>
<p>3^t - Sqrt[t] - 1 = 0 </p>
<p>I think that Newton iterative scheme is very simple. Starting with a guess (say t_old), the iterates are given by </p>
<p>t_new = t_old - f(t_old) / f'(t_old) </p>
<p>In your case, f(t) = 3^t - Sqrt[t] - 1 and labuwx showed that there is a root close to 0.5. So, start the iterations with t_old = 0.5. According to Newton scheme, the successive iterates will then be 0.479139, 0.478604 which is the solution. </p>
<p>For sure, you could do the same with the original equation.</p>
|
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