qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
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9,302 | <p>Say I have a symmetric matrix. I have the concept of 2-norm as defined on wikipedia. Now I want to prove (disprove?) that the norm of a symmetric matrix is maximum absolute value of its eigenvalue. I would really appreciate if this can be done only using simple concepts of linear algebra.</p>
<p>I am quite new to mathematics. </p>
| Jonas Meyer | 1,424 | <p>Using the <a href="http://en.wikipedia.org/wiki/Spectral_theorem#Hermitian_matrices" rel="nofollow">spectral theorem</a> to obtain an orthogonal basis of eigenvectors for the matrix is probably the best approach, similar to what Ross said, given your criterion of "simple concepts of linear algebra". Here is another approach I thought worth mentioning.</p>
<p>The largest of the absolute values of the eigenvalues of a matrix $A$ is called the spectral radius of $A$, which I'll denote $r(A)$. In general one allows complex eigenvalues (in part because there are real matrices with no real eigenvalues, like
$\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$), but fortunately it turns out that real symmetric matrices can have only real eigenvalues, so it makes no difference here. The inequality $r(A)\leq \|A\|$ is easy, because if $\lambda$ is an eigenvalue for $A$ and $v$ is a corresponding (nonzero) eigenvector, then $\|Av\|=\|\lambda v\| = |\lambda|\|v\|$, which shows that $\|A\|\geq |\lambda|$. </p>
<p>Remarkably, there is a <a href="http://en.wikipedia.org/wiki/Spectral_radius#Theorem_.28Gelfand.27s_formula.2C_1941.29" rel="nofollow">formula</a> for the spectral radius in terms of the norm, $r(A)=\lim_{k\to\infty}\|A^k\|^{1/k}$. I'll note before going on that everything I've said so far is independent of the norm on the space on which matrices act (that is, independent of which <a href="http://en.wikipedia.org/wiki/Operator_norm" rel="nofollow">operator norm</a> you choose for the matrices). But now, assuming the Euclidean norm (which is implicit in your question), the matrix norm has the property that $\|A^2\|=\|A\|^2$ for a real symmetric matrix $A$. Inductively you get $\|A^{2^k}\|=\|A\|^{2^k}$, and applying the spectral radius formula yields $r(A)=\lim_{k\to\infty}\|A^{2^k}\|^{1/2^k}=\lim_{k\to\infty}\|A\|=\|A\|$.</p>
|
4,560,624 | <p>Is there any elementary way to prove that <span class="math-container">$(\sin x + \cos x)(6 - \sin x)<9$</span>?</p>
<p>I've noticed that <span class="math-container">$(\sin x + \cos x)$</span> has to be positive so <span class="math-container">$x \in\left(-\dfrac{\pi}{4}, \dfrac{3\pi}{4}\right)$</span> and then <span class="math-container">$(6 - \sin x)\in\left(6-\dfrac{\sqrt2}{2},6+ \dfrac{\sqrt2}{2}\right)$</span> but since <span class="math-container">$(\sin x + \cos x) \in\left(0, \sqrt2\right]$</span> the maximum possible value of <span class="math-container">$(\sin x + \cos x)(6 - \sin x)$</span> can be <span class="math-container">$(6+ \dfrac{\sqrt2}{2}) * \sqrt2 = 1+6\sqrt2$</span> which is greater than <span class="math-container">$9$</span>.</p>
<p>My second attempt was to find the derivative, but I couldn't find its roots.</p>
| Zhanxiong | 192,408 | <p>Using the product-to-sum identity <span class="math-container">$\sin\theta\sin\phi = \frac{1}{2}(\cos(\theta - \phi) - \cos(\theta + \phi))$</span> to simplify:
<span class="math-container">\begin{align*}
& (\sin x + \cos x)(6 - \sin x) \\
= & \sqrt{2}\sin(x + \pi/4)(6 - \sin x) \\
= & 6\sqrt{2}\sin(x + \pi/4) - \sqrt{2}\sin x\sin(x + \pi/4) \\
= & 6\sqrt{2}\sin(x + \pi/4) - \sqrt2\times \frac{\cos(\pi/4) - \cos(2x+\pi/4)}{2} \\
= & 6\sqrt{2}\sin(x + \pi/4) + \frac{\sqrt{2}}{2}\cos(2x + \pi/4) - \frac{1}{2} \\
\leq & 6\sqrt{2} + \frac{\sqrt{2}}{2} - \frac{1}{2} \\
= & 8.69\ldots < 9.
\end{align*}</span></p>
|
4,560,624 | <p>Is there any elementary way to prove that <span class="math-container">$(\sin x + \cos x)(6 - \sin x)<9$</span>?</p>
<p>I've noticed that <span class="math-container">$(\sin x + \cos x)$</span> has to be positive so <span class="math-container">$x \in\left(-\dfrac{\pi}{4}, \dfrac{3\pi}{4}\right)$</span> and then <span class="math-container">$(6 - \sin x)\in\left(6-\dfrac{\sqrt2}{2},6+ \dfrac{\sqrt2}{2}\right)$</span> but since <span class="math-container">$(\sin x + \cos x) \in\left(0, \sqrt2\right]$</span> the maximum possible value of <span class="math-container">$(\sin x + \cos x)(6 - \sin x)$</span> can be <span class="math-container">$(6+ \dfrac{\sqrt2}{2}) * \sqrt2 = 1+6\sqrt2$</span> which is greater than <span class="math-container">$9$</span>.</p>
<p>My second attempt was to find the derivative, but I couldn't find its roots.</p>
| River Li | 584,414 | <p><strong>Proof 1</strong>: Note that <span class="math-container">$\sin x + \cos x = \sqrt 2 \, \sin (x + \pi/4) \le \sqrt 2$</span>
and <span class="math-container">$6 - \sin x > 0$</span>.</p>
<p>If <span class="math-container">$\sin x \ge 0$</span>, we have
<span class="math-container">$(\sin x + \cos x)(6 - \sin x) \le \sqrt 2 \cdot (6 - \sin x)
\le \sqrt 2 \cdot 6 < 9$</span>.</p>
<p>If <span class="math-container">$\sin x < 0$</span>, we have
<span class="math-container">$\sin x + \cos x \le \cos x \le 1$</span> and thus <span class="math-container">$(\sin x + \cos x)(6 - \sin x)
\le 6 - \sin x \le 7 < 9$</span>.</p>
<p>We are done.</p>
<p><span class="math-container">$\phantom{2}$</span></p>
<p><strong>Proof 2</strong>: <span class="math-container">$(a - b)^2 \ge 0$</span> yields <span class="math-container">$ab \le \frac{(a + b)^2}{4}$</span> for all real numbers <span class="math-container">$a, b$</span>.</p>
<p>We have
<span class="math-container">\begin{align*}
(\sin x + \cos x)(6 - \sin x) &= \frac14 \cdot 4(\sin x + \cos x)\cdot (6 - \sin x)\\
&\le \frac14 \cdot \frac{(4\sin x + 4\cos x + 6 - \sin x)^2}{4}\\
&=\frac{(3\sin x + 4\cos x + 6)^2}{16} \\
&\le \frac{(5 + 6)^2}{16}\\
& = \frac{121}{16} = 7.5625
\end{align*}</span>
where we have used <span class="math-container">$(3\sin x + 4\cos x)^2
\le (3^2 + 4^2)(\sin^2 x + \cos^2 x) = 25$</span> (C-S inequality)
to get <span class="math-container">$-5 \le 3 \sin x + 4\cos x \le 5$</span>.</p>
|
917,365 | <p>I hope to get the exact value of the following double series:</p>
<p>$$ \sum_{k \geq 0} \sum_{n \geq 0} \binom{2k}{ k} \frac{(-1)^n}{n! (2k+2n+1) 2^{4k+2n+1}}. $$</p>
<p>I am not sure it is possible or not.
I need your comments.</p>
| hkju | 28,848 | <p>This problem came from the question asked by one of my friends. What is the exact
value of the following definite integral </p>
<p>$$ \int_0^{1/2} \frac{exp(-x^2)}{\sqrt{1-x^2}} dx ? $$ </p>
<p>So, I got the corresponding Taylor series, multiply both of them.
I need closed form for the value, that is, which doesn't have the summation
notation.</p>
|
2,351,629 | <p>First of all I don't understand why we need Banach's theorem, as a result I can't make it intuitive for me to understand how it works but I tried to solve an example.</p>
<p><a href="https://en.wikipedia.org/wiki/Banach_fixed-point_theorem#Statement" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Banach_fixed-point_theorem#Statement</a></p>
<p>let $f(x) = x^3$ </p>
<p>a) for which fixed points will it converge ?</p>
<p>b) choose an interval $I$ such that all conditions for Banach's theorem are fulfilled (for function $f$)</p>
<p>My try :</p>
<p>a) we have 3 roots $( 0, 1, -1)$ with $0$ being a convergent point. (I think I solved a) but I don't understand what kind of interval are they asking about, and how to compute it)</p>
<p>PS: If some of the text doesn't make sense, please let me know, I translated it from another language.</p>
| Rodrigo de Azevedo | 339,790 | <p>Let</p>
<p><span class="math-container">$$\mathrm A := 1_n 1_n^\top - \mathrm I_n + \mathrm x \mathrm y^\top = - \mathrm I_n + \begin{bmatrix} | & |\\ \mathrm x & 1_n\\ | & |\end{bmatrix} \begin{bmatrix} | & |\\ \mathrm y & 1_n\\ | & |\end{bmatrix}^\top = - \left( \mathrm I_n - \begin{bmatrix} | & |\\ \mathrm x & 1_n\\ | & |\end{bmatrix} \begin{bmatrix} | & |\\ \mathrm y & 1_n\\ | & |\end{bmatrix}^\top \right)$$</span></p>
<p>Using the <a href="https://en.wikipedia.org/wiki/Weinstein%E2%80%93Aronszajn_identity" rel="nofollow noreferrer">Weinstein-Aronszajn determinant identity</a>,</p>
<p><span class="math-container">$$\begin{array}{rl} \det (\mathrm A) &= (-1)^n \cdot \det \left( \mathrm I_n - \begin{bmatrix} | & |\\ \mathrm x & 1_n\\ | & |\end{bmatrix} \begin{bmatrix} | & |\\ \mathrm y & 1_n\\ | & |\end{bmatrix}^\top \right)\\ &= (-1)^n \cdot \det \left( \mathrm I_2 - \begin{bmatrix} | & |\\ \mathrm y & 1_n\\ | & |\end{bmatrix}^\top \begin{bmatrix} | & |\\ \mathrm x & 1_n\\ | & |\end{bmatrix}\right)\\ &= (-1)^n \cdot \det \begin{bmatrix} (1 - \mathrm y^\top \mathrm x) & - \mathrm y^\top 1_n\\ -1_n^\top \mathrm x & (1-n)\end{bmatrix}\\ &= (-1)^n \cdot \left( (1 - \mathrm y^\top \mathrm x) (1-n) - \mathrm y^\top 1_n 1_n^\top \mathrm x \right)\\ &= (-1)^n \cdot \left( \mathrm x^\top \left( (n-1) \, \mathrm I_n - 1_n 1_n^\top \right) \mathrm y + 1-n \right)\end{array}$$</span></p>
|
2,706,504 | <p>In Lang's Algebra, I had hard times with this question:</p>
<blockquote>
<p>Let $K$ be field with characteristic $p$ (a prime). Let $L|K$ be a finite extension of $K$, and suppose $gcd([L:K],p)=1$. Show $L$ is separable over $K$.</p>
</blockquote>
<p>In order to have $L$ separable over $K$, one needs to have $K(\alpha)|K$ separable for each $\alpha \in L$. </p>
<p>We can say $L=K(\alpha_1,...,\alpha_m)$. It seems like I need to use the tower relation of these extensions and get some information about their degrees that will be relevant to the gcd thing. But I couldn't do these. A hint is welcomed.</p>
| faradawn | 1,152,902 | <p>For every <span class="math-container">$\alpha \in L$</span>, the minimum polynomial <span class="math-container">$f_\alpha(x)$</span> has a degree dividing the degree of the extension. As <span class="math-container">$[L : K]$</span> is relatively prime to <span class="math-container">$p$</span>, we get that the degree of <span class="math-container">$f_\alpha(x)$</span> must also be relatively prime to <span class="math-container">$p$</span>.</p>
<p>Then, as <span class="math-container">$f_\alpha(x)$</span> is irreducible, we can write
<span class="math-container">$$ f_\alpha (x) = a_q x^q + \cdots + a_0 $$</span>.</p>
<p>Then, its derivative is
<span class="math-container">$$ f'_\alpha (x) = q \cdot a_q x^{q-1} + \cdots + a_1 $$</span>.</p>
<p>As <span class="math-container">$F$</span> is an integral domain, <span class="math-container">$q \cdot a_q \neq 0$</span> for non-zero <span class="math-container">$q$</span> and <span class="math-container">$a_q$</span>. This shows that <span class="math-container">$f'_\alpha (x) \neq 0$</span>. Hence, <span class="math-container">$f_\alpha(x)$</span> is separable.</p>
<p>Reference: <a href="https://math.stackexchange.com/questions/1898477/separable-extension-given-prime-characteristic?newreg=7a4f0760887645c9a54f9d69c68ebd02">Separable Extension given prime characteristic</a>. The only thing left is that, in the extension <span class="math-container">$F/K$</span>, why the degree of any minimum polynomial divides the degree of the extension. Would appreciate any suggestions!</p>
<p>Hope it can help!</p>
|
1,249,730 | <p>let $ x \in G$ such that $(a^{-1})*(x^2)*(a) = x^3$ for some self inverse $a.$ Prove that $x$ has order $5.$</p>
<p>I don't know how to start this proof. Seems really difficult. </p>
| Alvaro Fuentes | 110,677 | <p>$a^{-1}x^2a=x^3 \implies a^{-1}x^4a=x^6 \implies a^{-1}x^6a=x^9$ but $x^6=x^3x^3=(a^{-1}x^2a)(a^{-1}x^2a)=a^{-1}x^4a \implies a^{-1}(a^{-1}x^4a)a=x^9 \implies a^{-1}a^{-1}x^4aa=x^9$ now using $a^2=a^{-2}=e$ we have $x^4=x^9$ so $x^5=e$ and because $5$ is prime $x=e$ or $x$ has order $5$.</p>
|
2,001,441 | <p>I am having the hardest time solving the following trigonometric equation. Can anyone help please? Thank you.</p>
<p>Solve for x. [Hint: Let $\\u = \tan^{-1}(x)$ and $\\v = tan^{-1}(2x)$. Solve the equation $\\u+v = \frac{π}{4}$ by taking the tangent of each side.]</p>
<p>$\tan^{-1}(x) +\tan^{-1}(2x)= \frac{π}{4}$</p>
| Claude Leibovici | 82,404 | <p>Among the useful trigonometric identities, try to memorize $$\tan^{-1}(a)\pm\tan^{-1}(b)=\tan^{-1}\frac{a\pm b}{1\mp ab}$$ So, using $a=x$ and $b=2x$ $$\tan^{-1}(x)+\tan^{-1}(2x)=\tan^{-1}\left(\frac{3x}{1-x^2}\right)$$ Then, the equation reduces to $$\frac{3x}{1-x^2}=1$$</p>
|
335,611 | <p>find this limit:
$$\displaystyle\lim_{n\to+\infty}\left[\sum_{k=1}^{n}\left(\dfrac{1}{\sqrt{k}}- \int_{0}^{\large {1/\sqrt k}}\dfrac{t^2}{1+t^2}dt\right)-2\sqrt{n}\right]$$</p>
| Community | -1 | <p>We have
$$\int_{0}^{\large {1/\sqrt k}}\dfrac{t^2}{1+t^2}dt=\frac{1}{\sqrt k}-\arctan\frac{1}{\sqrt k}$$
Now if we denote by
$$u_n=\sum_{k=1}^n\arctan\frac{1}{\sqrt k}-2\sqrt{n}$$
we have
$$u_n-u_{n-1}=\arctan\frac{1}{\sqrt n}-2\sqrt{n}+2\sqrt{n-1}\sim\frac{-7}{12n\sqrt{n}},$$
and since the series $\sum\frac{1}{n\sqrt{n}}$ is convergent then the sequence $(u_n)$ is also convergent to say $\ell$ and we we have
$$\sum_{k=n+1}^\infty u_k-u_{k-1}=\ell-u_n\sim\frac{-7}{12}\sum_{k=n+1}^\infty\frac{1}{k\sqrt{k}}\sim\frac{-7}{12}\int_{n+1}^\infty\frac{dx}{x\sqrt{x}}=\frac{-7}{6\sqrt{n}}$$
so we find the asymptotic equality
$$u_n=\ell+\frac{7}{6\sqrt{n}}+o(\frac{1}{\sqrt{n}})$$</p>
|
3,992,972 | <p>I want to draw the set <span class="math-container">$M_{4}$</span>={(x, y) ; |xy|< 1/4}</p>
<p>My attempt is that I evaluate the expression of |xy|<1/4 depending on the values of x an y, i.e depending on the quadrant.</p>
<p>For the second quadrant I get the following:</p>
<p>x < 0 <span class="math-container">$\implies$</span> |x| = -x</p>
<p>y <span class="math-container">$\ge$</span>0 <span class="math-container">$\implies$</span> |y| = y</p>
<p><span class="math-container">$\implies$</span> |xy| =-xy < <span class="math-container">$\frac{1}{4}$</span> <span class="math-container">$\implies$</span> y > - <span class="math-container">$\frac{1}{4x}$</span></p>
<p>Which does not make any sense as the set is enclosed by the lines y=-<span class="math-container">$\frac{1}{4}$</span> y = <span class="math-container">$\frac{1}{4x}$</span>, and thus for the second quadrant I should get y < - <span class="math-container">$\frac{1}{4x}$</span>. I don't understand where I go wrong, any help would be much appreciated!</p>
| DonAntonio | 31,254 | <p>An idea: draw first the two hyperbolas <span class="math-container">$\;y=\pm\cfrac1{4x}\;$</span> , and then "bound" the points with both coordinates different from zero belonging to <span class="math-container">$\;M_4\;$</span> by them. Of course, the above involves assuming <span class="math-container">$\;x\neq0\;$</span>, but together with <span class="math-container">$\;y=0\;$</span> (both axis are contained in <span class="math-container">$\;M_4\;$</span>) you shall get what you need.</p>
|
4,448,525 | <p>To do a proof by induction, it is necessary to prove</p>
<p>(1) the proposition <span class="math-container">$P(n)$</span> is true for <span class="math-container">$n=1$</span></p>
<p>(2) if the proposition <span class="math-container">$P(n)$</span> is true for <span class="math-container">$n=k$</span>, then <span class="math-container">$P(n)$</span> is true for <span class="math-container">$n=k+1$</span></p>
<p>My questions are</p>
<p>(1) What is the significance of showing the proposition <span class="math-container">$P(n)$</span> is true for <span class="math-container">$n=1$</span>? Why is it necessary to do this step?</p>
<p>(2) But based on which properties or axioms of number can we be sure that <span class="math-container">$P(n)$</span> is true for <span class="math-container">$n=k+1$</span> if <span class="math-container">$P(n)$</span> is true for <span class="math-container">$n=k$</span>?</p>
<p>(3) Can we use proof by induction when the proposition only involves natural numbers, i.e. no non-integer? For example, we cannot use proof by induction to show that <span class="math-container">$cos^2(x)+sin^2(x)=1$</span></p>
<p>(4) Also under which condition proof by contradiction cannot be used? Any example?</p>
<p>Thanks.</p>
| Shambhala | 673,139 | <p>(1) The base case is not necessarily <span class="math-container">$n=1$</span>. One might want to prove a statement is true for all numbers <span class="math-container">$n$</span> greater than or equal to a certain number <span class="math-container">$m$</span>.</p>
<p>To understand the importance of the base case you can suppose the following:</p>
<p><strong>The first domino falls right.</strong></p>
<p><strong>If a (fixed but arbitrary) domino falls right, then its next neighbor also falls right.</strong></p>
<p>(2) You might want to study the Peano axioms and the well-ordering principle, though induction is stronger than the latter.</p>
<p>(3) Mathematical induction is used to prove that a statement <span class="math-container">$P(n)$</span> holds for a certain subset of <span class="math-container">$\mathbb{N}$</span>. <span class="math-container">$\cos^2(x)+\sin^2(x)=1$</span> is a property that holds for all <span class="math-container">$x\in\mathbb{R}$</span>. Thus, cannot be proved using induction.</p>
<p>(4) There is no absolute algorithm for deciding which proof types can or cannot be used.</p>
|
4,448,525 | <p>To do a proof by induction, it is necessary to prove</p>
<p>(1) the proposition <span class="math-container">$P(n)$</span> is true for <span class="math-container">$n=1$</span></p>
<p>(2) if the proposition <span class="math-container">$P(n)$</span> is true for <span class="math-container">$n=k$</span>, then <span class="math-container">$P(n)$</span> is true for <span class="math-container">$n=k+1$</span></p>
<p>My questions are</p>
<p>(1) What is the significance of showing the proposition <span class="math-container">$P(n)$</span> is true for <span class="math-container">$n=1$</span>? Why is it necessary to do this step?</p>
<p>(2) But based on which properties or axioms of number can we be sure that <span class="math-container">$P(n)$</span> is true for <span class="math-container">$n=k+1$</span> if <span class="math-container">$P(n)$</span> is true for <span class="math-container">$n=k$</span>?</p>
<p>(3) Can we use proof by induction when the proposition only involves natural numbers, i.e. no non-integer? For example, we cannot use proof by induction to show that <span class="math-container">$cos^2(x)+sin^2(x)=1$</span></p>
<p>(4) Also under which condition proof by contradiction cannot be used? Any example?</p>
<p>Thanks.</p>
| Curious Goblin | 1,056,948 | <p>I'm not sure I can answer ##3 and 4 authoritatively. But I've never heard of proof by induction being used for n other than integers. And I've never heard of a circumstance where proof by contradiction is not a valid method of argument.</p>
<p>Regarding #1, you need to show that that <span class="math-container">$P(n)$</span> is true for at least some specific integer, preferably 1, in order to make the next proposition (i..e., that truth of <span class="math-container">$P(n)$</span> implies truth of <span class="math-container">$P(n+1)$</span>) relevant. Imagine, for instance, that <span class="math-container">$P(n)$</span> is false for all n. In that case, it would not matter much that the truth of <span class="math-container">$P(n)$</span> would imply the truth of <span class="math-container">$P(n+1)$</span>.</p>
<p>Regarding #2, There is no overarching property of integers that allows you to know that the truth of <span class="math-container">$P(n)$</span> implies truth of <span class="math-container">$P(n+1)$</span>. The reasoning is different for every proof and that it is why proof by induction isn't some automatic algorithm ... it requires thinking.</p>
|
4,299,417 | <p>Let <span class="math-container">$f: \mathbb{R} \rightarrow [0, \infty)$</span> be a function that:
<span class="math-container">$$f^2(x+y)+f^2(x-y)=2f^2(x)+2f^2(y) \forall x,y \in \mathbb{R}$$</span>
Prove that <span class="math-container">$f(x+y) \leq f(x)+f(y) \forall x,y \in \mathbb{R}$</span></p>
<p>This problem is from <a href="https://artofproblemsolving.com/community/c6h2304460p18259574" rel="nofollow noreferrer">this AOPS link</a>.</p>
<p>Let <span class="math-container">$x=y=0$</span>,then <span class="math-container">$f(0)=0$</span>. then let <span class="math-container">$x=y$</span>,we have
<span class="math-container">$$f^2(2x)=4f^2(x)\Longrightarrow f(2x)=2f(x)$$</span>
let <span class="math-container">$x=2y$</span>,then we have
<span class="math-container">$$f^2(3y)+f^2(y)=2f^2(2y)+2f^2(y)=10f^2(y)\Longrightarrow f(3x)=3f(x)$$</span>
let <span class="math-container">$x=3y$</span> then we have <span class="math-container">$f(4x)=4f(x)$</span>,and use induction we have
<span class="math-container">$$f(kx)=kf(x),\forall k\in N^{+}$$</span> then I can't it</p>
| DreamAR | 983,565 | <p>For an inner product space, we can define <span class="math-container">$\|x\|=\sqrt{(x,x)}$</span> to be a norm, which fulfill <em>triangular inequallity</em>: <span class="math-container">$\|x+y\|\le \|x\|+\|y\|$</span> by Cauchy-Schwartz inequallity. We have <em>parallelogram identity</em>:
<span class="math-container">$$
\|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2
$$</span>
And <em>polarization identity</em> (in real space):
<span class="math-container">$$
(x,y)=\frac{1}{4}(\|x+y\|^2-\|x-y\|^2)
$$</span>
A proposition in functional analysis is that, norm may not be induced by an inner product, but once your norm satisfies <em>parallelogram identity</em>, you can define an inner product by <em>polarization identity</em>. And this inner product will exactly induce your original norm.</p>
<p>So try to define an inner product by <span class="math-container">$f$</span>, verify it is an inner product, and then use property of inner product to prove the <em>triangular inequallity</em>.</p>
<hr />
<p>It takes me a lot of time to consider the following question: If the space we're working is <span class="math-container">$\mathbb{R},$</span> then if <span class="math-container">$f$</span> is a (semi-)norm, won't we directly have <span class="math-container">$$f(x+y)=f(x)+f(y)?$$</span></p>
<p>So we shall regard <span class="math-container">$\mathbb{R}$</span> as a <span class="math-container">$\mathbb{Q}$</span>-vector space, and <span class="math-container">$f$</span> will be a <strong>semi</strong>-norm following <span class="math-container">$\mathbb{Q}$</span>-linearity. You will have the <span class="math-container">$\mathbb{Q}$</span>-linearity of your <span class="math-container">$(-,-)$</span> by <em>parallelogram identity</em>. Prove the (<strong>semi</strong>) inner product satisfies Cauchy-Schwartz inequality, using <span class="math-container">$(x+\lambda y,x+\lambda y)\ge 0,$</span> <span class="math-container">$\lambda\in\mathbb{Q}.$</span> You will need to take limit of <span class="math-container">$\lambda\rightarrow \eta\in \mathbb{R}.$</span> However it doesn't need the continuity of <span class="math-container">$f.$</span> It only uses the topology in <span class="math-container">$\mathbb{R}$</span>.</p>
<hr />
<p>Above are all my thinking and backgrounds. The following is the answer to this question, which doesn't need many techniques.</p>
<p>You already have: <span class="math-container">$$f(0)=0, \qquad f(kx)=kf(x),\quad\forall k\in \mathbb{Q}$$</span></p>
<p>Let <span class="math-container">$(x,y)=\frac{1}{4}(f(x+y)^2-f(x-y)^2).$</span> Then it has the following properties:</p>
<ol>
<li><p><span class="math-container">$(x,x)=f(x)^2\ge 0,$</span> and equal to zero when <span class="math-container">$x=0$</span>;</p>
</li>
<li><p><span class="math-container">$(x,y)=(y,x),$</span> as <span class="math-container">$f(x-y)^2=f(y-x)^2$</span>;</p>
</li>
<li><p><span class="math-container">$(x+y,z)=(x,z)+(y,z).$</span> This is because:
<span class="math-container">$$
\begin{aligned}
(x,z)+(y,z)&=\frac{1}{4}(f(x+z)^2-f(x-z)^2+f(y+z)^2-f(y-z)^2)\\
&=\frac{1}{2}(f(\frac{x+y}{2}+z)^2+f(\frac{x-y}{2})^2-f(\frac{x+y}{2}-z)^2-f(\frac{x-y}{2})^2)\\
&=2(\frac{x+y}{2},z)\\
(0,z)&=\frac{1}{4}(f(z)^2-f(-z)^2)=0\\
(x+y,z)+(0,z)&=2(\frac{x+y}{2},z)
\end{aligned}
$$</span></p>
</li>
<li><p><span class="math-container">$\forall k\in \mathbb{Q},$</span> <span class="math-container">$(kx,y)=k(x,y).$</span> Denote <span class="math-container">$F(t)=(tx,y),$</span> then by property 3, <span class="math-container">$F(t_1+t_2)=F(t_1)+F(t_2).$</span> This shows that <span class="math-container">$F(t)=tF(1),$</span> <span class="math-container">$\forall t\in \mathbb{Q}.$</span></p>
</li>
</ol>
<p>Then we verify that <span class="math-container">$(-,-)$</span> satisfies Cauchy-Schwartz inequality:
<span class="math-container">$$(x,y)^2\le (x,x)(y,y).$$</span></p>
<p>Use <span class="math-container">$(x+\lambda y,x+\lambda y)\ge 0,$</span> <span class="math-container">$\lambda \in \mathbb{Q},$</span>
<span class="math-container">$$
(x,x)+2\lambda(x,y)+\lambda^2(y,y)\ge 0.
$$</span>
Let <span class="math-container">$\lambda\rightarrow -\frac{(x,y)}{(y,y)},$</span> the inequality keeps, so
<span class="math-container">$$
(x,x)-\frac{(x,y)^2}{(y,y)}\ge 0\Rightarrow (x,y)^2\le (x,x)(y,y).
$$</span></p>
<p>Finally, we have
<span class="math-container">$$
\begin{aligned}
f(x+y)^2&=(x+y,x+y)\\
&=(x,x)+2(x,y)+(y,y)\\
&\le(x,x)+2\sqrt{(x,x)(y,y)}+(y,y)\\
&=(\sqrt{(x,x)}+\sqrt{(y,y)})^2=(f(x)+f(y))^2
\end{aligned}
$$</span></p>
<p>Triangular inequality is proved.</p>
<p>So the main step is to prove <span class="math-container">$\mathbb{Q}$</span>-linearity of the inner product and Cauchy-Schwartz inequality. Retain this two step, you can find equivalent proof by only using <span class="math-container">$f,$</span> without defining <span class="math-container">$(-,-).$</span> However, I think it's pretty concise to prove it with the help of <span class="math-container">$(-,-).$</span></p>
|
3,755,638 | <p>Given a point <span class="math-container">$A$</span>, a circle <span class="math-container">$O$</span> and conic section <span class="math-container">$e$</span>, if <span class="math-container">$BC$</span> is a moving chord of the circle <span class="math-container">$O$</span> tangent to <span class="math-container">$e$</span>, then prove that<br />
<strong>the locus of △<span class="math-container">$ABC$</span>'s circumcenters <span class="math-container">$T$</span> is a conic section.</strong><br />
The question was posted in <a href="https://tieba.baidu.com/f?kw=%E7%BA%AF%E5%87%A0%E4%BD%95" rel="noreferrer">纯几何吧</a> by TelvCohl and remained unsolved for many years but regrettably I cannot provide the link because the post was deleted by Baidu accidentally.<br />
It seems that the locus related to circumcenter is often a conic section.Another example:<br />
The directions of two sides of a triangle is fixed and the third side passes through a fixed point, then the locus of the circumcenter is a conic section.(<em>The elementary geometry of conics</em>.1883)</p>
<p><a href="https://i.stack.imgur.com/bHkXD.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/bHkXD.jpg" alt="enter image description here" /></a></p>
| Blue | 409 | <p>Here's something of a brute-force vector proof.</p>
<p>In the figure, <span class="math-container">$P$</span> is a point on our given conic, <span class="math-container">$\bigcirc K$</span> of radius <span class="math-container">$r$</span> is our given circle, and <span class="math-container">$A$</span> is our given point. The tangent line at <span class="math-container">$P$</span> meets <span class="math-container">$\bigcirc K$</span> at <span class="math-container">$R$</span> and <span class="math-container">$R'$</span>, and the circumcenter of <span class="math-container">$\triangle ARR'$</span> is <span class="math-container">$Q$</span>. Points <span class="math-container">$A'$</span> and <span class="math-container">$K'$</span> are the respective projections of <span class="math-container">$A$</span> and <span class="math-container">$K$</span> onto the tangent line. Finally, <span class="math-container">$a:=|PA|$</span>, <span class="math-container">$k := |PK|$</span>, and <span class="math-container">$\alpha$</span> and <span class="math-container">$\kappa$</span> are the (signed) angles made by the tangent line and the respective vectors <span class="math-container">$\overrightarrow{PA}$</span> and <span class="math-container">$\overrightarrow{PK}$</span>.</p>
<p><a href="https://i.stack.imgur.com/6jLnb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6jLnb.png" alt="enter image description here" /></a></p>
<p>First, a bit of geometry. Ignoring the conic, and concentrating how <span class="math-container">$\bigcirc K$</span> meets the tangent line and determines circumcenter <span class="math-container">$Q$</span>, one can show that <span class="math-container">$|K'Q|$</span> is given by
<span class="math-container">$$\begin{align}
|K'Q| &= \frac{|AK'|^2 -|RK'|^2}{2|AA'|} \tag{1}\\[4pt]
&= \frac{\left(a^2+|PK'|^2-2a|PK'|\cos\alpha\right)-\left(r^2-|KK'|^2\right)}{2a\sin\alpha} \tag{2}\\[4pt]
&= \frac{a^2+k^2\cos^2\kappa-2ak\cos\alpha\cos\kappa-r^2+k^2\sin^2\kappa}{2a\sin\alpha} \tag{3}\\[4pt]
&= \frac{a^2+k^2-2ak\cos\alpha\cos\kappa-r^2}{2a\sin\alpha} \tag{4} \\[4pt]
&= \frac{a^2+k^2-2ak\cos(\alpha+\kappa)-2ak\sin\alpha\sin\kappa-r^2}{2a\sin\alpha} \tag{5}\\[4pt]
&= \frac{|AK|^2-r^2-2ak\sin\alpha\sin\kappa}{2a\sin\alpha} \tag{6}\\[4pt]
&= \frac{|AK|^2-r^2}{2a\sin\alpha}-k\sin\kappa \tag{7}\\[4pt]
&= \frac{|AK|^2-r^2}{2a\sin\alpha}-|KK'| \tag{8} \\[4pt]
\end{align}$$</span></p>
<p>Therefore, we can write, defining <span class="math-container">$s := |AK|$</span>,
<span class="math-container">$$Q = K + (P')^\perp (|KK'|+|K'Q|) = K + \frac{s^2-r^2}{2a\sin\alpha}\,(P')^\perp \tag{9}$$</span>
where <span class="math-container">$(P')^\perp := (P'_y,-P'_x)$</span> is perpendicular to <span class="math-container">$P'$</span>;that is, it's a unit normal to the conic. (We could take either normal. The calculation of <span class="math-container">$a\sin\alpha$</span> in <span class="math-container">$(12)$</span> ensures that <span class="math-container">$Q$</span> is properly offset from <span class="math-container">$K$</span>.)</p>
<p>Now, to bring the conic back into consideration ... Let us suppose that <span class="math-container">$P$</span> lies on an origin-focused conic, with corresponding vertex on the positive <span class="math-container">$x$</span>-axis, latus rectum <span class="math-container">$p$</span>, and eccentricity <span class="math-container">$e$</span>. Then <span class="math-container">$P$</span> is parameterized by
<span class="math-container">$$P = \frac{p}{1+e\cos\theta}\;(\cos\theta,\sin\theta) \tag{10}$$</span>
The unit tangent vector is then
<span class="math-container">$$P' = \frac1{\sqrt{1+e^2+2e\cos\theta}} (\sin\theta, -(e + \cos\theta)) \tag{11}$$</span>
We also have
<span class="math-container">$$
a \sin\alpha = (P')^\perp\cdot(A-P) =
\frac{p - A_x(e+\cos\theta) - A_y \sin\theta}{\sqrt{1 + e^2 + 2 e \cos\theta}} \tag{12}
$$</span>
Substituting into <span class="math-container">$(9)$</span> we find the locus point <span class="math-container">$(x,y)=Q$</span> parameterized as</p>
<p><span class="math-container">$$(x,y) = K + \frac{s^2-r^2}{2(p - A_x(e+\cos\theta) - A_y \sin\theta)}(e+\cos\theta,\sin\theta) \tag{13}$$</span>
Separating the components and clearing denominators gives equations that happen to constitute a linear system in <span class="math-container">$\cos\theta$</span> and <span class="math-container">$\sin\theta$</span>; solving and substituting into <span class="math-container">$\cos^2\theta+\sin^2\theta=1$</span> gives a second-degree polynomial equation in <span class="math-container">$x$</span> and <span class="math-container">$y$</span> which necessarily represents a conic.</p>
<p><a href="https://i.stack.imgur.com/wWTXS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wWTXS.png" alt="enter image description here" /></a></p>
<p>For completeness, the equation is as follows:</p>
<p><span class="math-container">$$\begin{align}
0 &= 4 x^2 (p + A_x (1-e)) (p - A_x(1+e))\\
&+ 4 y^2 ( p^2 - A_y^2(1-e^2)) \\
&-8 x y\, A_y (ep+A_x(1-e^2)) \\
&+4 x(A_x (s^2-r^2+2A_x K_x+2A_y K_y) (1-e^2)
+ ep( s^2 - r^2 + 4 A_x K_x + 2 A_y K_y ) - 2 K_x p^2 ) \\
&+4y ( A_y ( s^2 - r^2 + 2 A_x K_x + 2 A_y K_y ) (1-e^2) + 2 ep A_y K_x - 2 p^2 K_y ) \\
&-(s^2 - r^2 + 2 A_x K_x + 2 A_y K_y)^2 (1 - e^2)
- 4 e p K_x (s^2 - r^2 + 2 A_x K_x + 2 A_y K_y)
+ 4 p^2 (K_x^2 + K_y^2)
\end{align} \tag{$\star$}$$</span></p>
<p>Defining <span class="math-container">$m^2=s^2-r^2+2(A_xK_x+A_yK_y)=|OA|^2+|OK|^2-r^2$</span>, we can write this as</p>
<p><span class="math-container">$$\begin{align}
0 &= 4 x^2 (p + A_x (1-e)) (p - A_x(1+e)) \\
&+ 4 y^2 ( p^2 - A_y^2(1-e^2)) \\
&-8 x y\, A_y (ep+A_x(1-e^2)) \\
&+4 x(A_x m^2 (1-e^2)
+ ep( m^2 +2A_xK_x) - 2 K_x p^2 ) \\
&+4y ( A_ym^2 (1-e^2) + 2 ep A_y K_x - 2 p^2 K_y ) \\
&-m^4 (1 - e^2)- 4 e p K_x m^2+ 4 p^2 |OK|^2
\end{align} \tag{$\star'$}$$</span></p>
<p>which isn't much of an improvement, but I don't think the equation is the important thing here.</p>
<p>The discriminant (ignoring a factor of <span class="math-container">$64p^2$</span>) reduces to <span class="math-container">$|OA|^2 - (p - e A_x)^2$</span>, which indicates that the nature of the resulting conic (ellipse, parabola, hyperbola) depends only upon <span class="math-container">$A$</span>'s position relative to the focus.</p>
|
3,543,060 | <p>The ODE is defined on <span class="math-container">$[0,b]$</span> with Neumann boundary conditons.
<span class="math-container">$$y''(x)=\frac{c_1y(x)}{c_2+y(x)}$$</span>
<span class="math-container">$$y'(0)=0; y'(b)=0$$</span></p>
<p>How to solve the above ODE? Any help is appreciated!</p>
| Jan | 529,121 | <p>Reduce the order of the equation by substituting <span class="math-container">$w = w(y)$</span>. Then it yields </p>
<p><span class="math-container">$$ww' = \frac{c_1y}{c_2 + y},$$</span></p>
<p>which is a separable ODE. This equation can be integrated in order to get</p>
<p><span class="math-container">$$\frac{w^2}{2} = c_1 (y - c_2 \ln \lvert y + c_2 \rvert ) + C.$$</span></p>
<p>Solve this for <span class="math-container">$w$</span>, then use</p>
<p><span class="math-container">$$\int \frac{\mathrm dy}{w(y)} = \int dx$$</span></p>
<p>for calculating <span class="math-container">$y = y(x)$</span> and apply the boundary conditions.</p>
|
2,461,773 | <p>I would like some clarification about the Cantor Set:</p>
<ul>
<li>What are the elements in the Cantor Set?</li>
<li>How do I write the Cantor Set in mathematical terms (i.e in a summation)? I have seen online a formula but I do not understand how they got it so would be grateful if you could explain why it is this formula too.</li>
</ul>
<p>EDIT: The formula I have seen online is:
$$\sum_{i=0}^\infty (a(i)/3^i) $$ where a(i) is either 0 or 2.</p>
| Austin Weaver | 480,825 | <p>The Cantor Set $\mathcal C$ can be defined by the following recursive sequence of sets:</p>
<p>$$\mathcal C_0 = [0, 1]$$
$$\mathcal C_{n+1}= \frac{\mathcal C_n}3\cup\left(\frac23+\frac{C_n}3\right)$$
$$\mathcal C := \bigcap^\infty_{n=0}\mathcal C_n$$</p>
<p>This definition should reflect the intuitive notion of what the Cantor Set is and how it can be constructed.</p>
<hr>
<p>To explain this further, I will explain the math in laymen's terms.</p>
<p>Start with the closed unit interval, $[0, 1]$.</p>
<p>The next iteration is the previous one divided by three and the previous one divided by three with its origin shifted to be at $\frac23$. This yields $[0, \frac13]\cup[\frac23, 1]$, then $[0, \frac19]\cup[\frac29, \frac13]\cup[\frac23, \frac79]\cup[\frac89, 1]$, etc.</p>
<p>The set itself is the set of points that are present in every iteration of this process.</p>
<hr>
<p>In case you were wondering how the definition you gave ties into mine (which I find much more intuitive), Here is an explanation:</p>
<p>Observe that as $\mathcal C_n$ is iterated, the edges of each interval ($0, \frac13, \frac23, \frac19, $ etc) remain in the set.</p>
<p>All of these numbers can be expressed by the sum $$\sum^\infty_{n=1}\frac{a_n}{3^n}$$ where the sequence $a$ is all $0$s and $2$s.</p>
<p>For example, $\frac13$ is the sum when $a_n = \langle 0, 2, 2, 2, 2, \cdots\rangle$, or $\frac29+\frac2{27}+\frac2{81}+\cdots$</p>
<p>I'll leave it to you to find the reason for this connection, but a hint is that it has to do with the $\frac23$ being added.</p>
|
29,601 | <p>I know the order of the group is the number of elements in the set. For example the group of $U_{10}$ (units of congruence class of 20) has order 4. </p>
<p>Major Edit, kinda changed the question.
Lets say my element $a$ has a finite order $n$. Then what is the order of $a, a^2, a^3...a^{11}$?</p>
| MichalisN | 8,432 | <p><strong>Answer to the edited question:</strong>
That depends on n.
Generally the order of $a^i$ is $\frac{n}{\text{gcd}(n,i)}$. If you want to prove this, you have to check two things: </p>
<p>Firstly that $(a^i)^{\frac{n}{\text{gcd}(n,i)}}=a^{\frac{in}{\text{gcd}(n,i)}}=1$. This holds because we have a multiple of n in the exponent and $a^{kn}=(a^n)^k=1^k=1$.</p>
<p>Secondly that this is in fact the smallest exponent k with $(a^i)^k=a^{ik}=1$. Since n is the smallest number with $a^n=1$, we are looking for the smallest number k so that $ik$ is a multiple of n. This is $k=\frac{n}{\text{gcd}(n,i)}$.</p>
|
804,414 | <p>A Fair Dice is Thrown Repeatedly. Let $X$ be number of Throws required to get a '$6$' and $Y$ be number of throws required to get a '$5$'. Find $$E(X|Y=5)$$</p>
| James Dow Allen | 152,519 | <p>Since Y=5, we know the first four shakes are not 5, and the fifth is not 6.
The expectation is thus
(1/5) * 1 + (4/25) * 2 + (16/125) * 3 + (64/625) * 4 + (256/625) * (E(X) + 5)
with E(X) = 6.
This is 3637/625 = 5.8192</p>
|
223,087 | <p>Given a list of numbers in decimal form, what is the most efficient way to determine if there are any consecutive 1s in the binary forms of those numbers? My solution so far:</p>
<pre><code>dim = 3;
declist = Range[0, 2^dim - 1];
consecutiveOnes[binary_] := AnyTrue[Total /@ Split[binary], # > 1 &];
consecutiveOnes[#] & /@ IntegerDigits[declist, 2]
</code></pre>
<p>which gives <code>{False, False, False, True, False, False, True, True}</code>, in accordance with the binary representations <code>{{0}, {1}, {1, 0}, {1, 1}, {1, 0, 0}, {1, 0, 1}, {1, 1, 0}, {1, 1, 1}}</code>.</p>
<p>For <code>dim=15</code> this takes ~600ms on my machine, which seems a little high, and I just want to see if there is a cleaner way to do it. I've tried using BlockMap with Times but it was much slower.</p>
<p>Two "extras":</p>
<ol>
<li><p>I guess as a comment, it is also acceptable if your method simply returns all decimal numbers up to some max number for which the binary representations have no consecutive 1s. In other words, I'm just going to run <code>Pick</code> on the <code>declist</code> with the negated results of this function, so if your solution just cuts out the middle man, that is great/acceptable.</p></li>
<li><p>I also care about the possibility of "wrapping around", i.e. if the first and last binary digits are both 1s. Obviously I could just append the first digit to the end of each list, but perhaps this is not the most efficient way to proceed.</p></li>
</ol>
<p><strong>Addendum</strong>: Some great solutions! I took the liberty of implementing and speed testing them, with some minor modifications - hopefully I have not distorted your codes too badly:</p>
<pre><code>dim = 15;
declist = Range[0, 2^dim - 1];
m1[range_] :=
FromDigits[#, 2] & /@
DeleteCases[IntegerDigits[range, 2], {___, 1, 1, ___}];
m2helper[num_] := NoneTrue[Total /@ Split[num], # > 1 &];
m2[range_] := Pick[declist, m2helper[#] & /@ IntegerDigits[range, 2]];
m3helper[num_] :=
NestWhile[Quotient[#, 2] &, num, # > 0 && BitAnd[#, 3] != 3 &] > 0
m3[range_] := Pick[declist, Not[m3helper[#]] & /@ range];
m41 = (4^(Ceiling[dim/2]) - 1)/3;
m42 = 2 m41;
m4helper = Function[{n},
Evaluate[
Nor[BitAnd[BitAnd[n, m42], BitShiftLeft[BitAnd[n, m41], 1]] > 0,
BitAnd[BitAnd[n, m42], BitShiftRight[BitAnd[n, m41], 1]] >
0]], {Listable}];
m4[range_] := Pick[declist, m4helper[range]];
Clear[m5];
m5[0] = {0};
m5[1] = {0, 1};
m5[n_?(IntegerQ[#] && # > 1 &)] :=
m5[n] = Join[m5[n - 1], 2^(n - 1) + m5[n - 2]]
m6[range_] :=
Pick[range, Thread[BitAnd[range, BitShiftRight[range, 1]] == 0]];
aa = m1[declist] // RepeatedTiming;
bb = m2[declist] // RepeatedTiming;
cc = m3[declist] // RepeatedTiming;
dd = m4[declist] // RepeatedTiming;
ee = m5[dim] // AbsoluteTiming;
ff = m6[declist] // RepeatedTiming;
Column[{aa[[1]], bb[[1]], cc[[1]], dd[[1]],ee[[1]],ff[[1]]}]
aa[[2]] == bb[[2]] == cc[[2]] == dd[[2]] == ee[[2]]==ff[[2]]
</code></pre>
<p>yields</p>
<pre><code>0.0464
0.619
0.322
0.0974
0.00024
0.0086
True
</code></pre>
<p>So the direct construction method seems clearly the fastest - still this does "skip" the actual pruning step, which is not required for me, but maybe is in other use cases. If the actual pruning list is desired, it seems like the direct <code>BitAnd</code>+<code>BitShiftRight</code> method is fastest, followed by the <code>SelectCases</code>/<code>DeleteCases</code>. But if other people have other methods, certainly share them!</p>
| eyorble | 52,935 | <p>For the sake of getting answers for higher values of <code>dim</code>, I present to you some bit manipulation hacking for <code>dim=20</code>:</p>
<pre><code>dim = 20;
</code></pre>
<p>Find the binary numbers which encompass the range of interest in <code>dim</code> that are alternating 1s and 0s, one of which ends in 1 and one of which ends in 0.</p>
<pre><code>x1 = (4^(Ceiling[dim/2]-1)/3;
x2 = 2 x1;
</code></pre>
<p>Carefully define a function which uses <code>x1</code> and <code>x2</code> to filter binary digits out of an input <code>n</code>, and then determine if right-shifting or left-shifting the result from one of these by one place causes any digits to overlap with the other:</p>
<pre><code>f = Function[{n}, Evaluate[
Or[BitAnd[BitAnd[n, x2], BitShiftLeft[BitAnd[n, x1], 1]] > 0,
BitAnd[BitAnd[n, x2], BitShiftRight[BitAnd[n, x1], 1]] > 0]],
{Listable}]
</code></pre>
<p>Then run this <code>f</code> on the range in question:</p>
<pre><code>AbsoluteTiming[res = f[Range[0, 2^dim - 1]];]
</code></pre>
<p>On my machine this takes 2.5 seconds for <code>dim = 20</code>. It doesn't take too much longer before you're likely to run into RAM issues constructing the entirety of these lists, and if you're trying to apply this to very large numbers then <code>Compile</code> will restrict you to 128 bits or less (probably). I suspect this is fairly close to optimal time-wise, as a result.</p>
<p>This does not directly handle the 2nd case you provide, but you could construct the upper most bit of your <code>dim</code> of interest, add 1 to that, and use that to determine if both the highest and lowest bits are set:</p>
<pre><code>x3 = 2^(dim-1)+1;
f2 = Function[{n}, BitAnd[n, x3] >= x3, {Listable}];
</code></pre>
|
275,151 | <p><a href="https://i.stack.imgur.com/XxGxa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XxGxa.png" alt="enter image description here" /></a>I'm a beginner at Mathematica, and I'm trying to figure out how to fill between two lines horizontally. Consider the toy example</p>
<pre><code>Plot[{ConditionalExpression[2*x - 2, 2 < x < 6], ConditionalExpression[2*x + 2, x < 4]}, {x, 0, 10}, PlotRange -> {{0, 11}, {0, 11}}, Filling -> {1 -> {2}}]
</code></pre>
<p>Is there some easy way to fill horizontally between the two lines (so the whole area between the lines from y=2 to y=10 in the example is filled)? This whole thing is to be embedded in a bigger plot, so I cant for example add points outside of this plot to extend the top line and thus the coloring.</p>
| Bob Hanlon | 9,362 | <pre><code>$Version
(* "13.1.0 for Mac OS X x86 (64-bit) (June 16, 2022)" *)
Clear["Global`*"]
</code></pre>
<p>Split each of the lines into two segments and cycle the <code>PlotStyle</code></p>
<pre><code>Plot[{
ConditionalExpression[2*x - 2, 2 < x < 4],
ConditionalExpression[2*x + 2, 0 < x < 2],
ConditionalExpression[2*x - 2, 4 < x < 6],
ConditionalExpression[2*x + 2, 2 < x < 4]},
{x, 0, 10},
PlotRange -> {{0, 11}, {0, 11}},
Filling -> {1 -> {4}, 2 -> 2, 3 -> 10},
FillingStyle -> LightBlue,
PlotStyle -> (ColorData[97] /@ {1, 2}),
PlotLegends -> Placed[
{HoldForm[2*x - 2], HoldForm[2*x + 2], None, None},
{.7, .5}]]
</code></pre>
<p><a href="https://i.stack.imgur.com/KLyLy.png" rel="noreferrer"><img src="https://i.stack.imgur.com/KLyLy.png" alt="enter image description here" /></a></p>
|
4,036,117 | <p>In Joe Silverman's "The Arithmetic of Ecliptic Curves" he talks a lot about the integral domain <span class="math-container">$\overline{K}(C)$</span>. On page <span class="math-container">$27$</span> he suddenly decides to chose an element <span class="math-container">$f$</span> from some set <span class="math-container">$\overline{K}(C)^*$</span>. What does this notation mean? After page <span class="math-container">$27$</span> he begins using <span class="math-container">$\overline{K}(C)^*$</span> a lot, and I haven't been able to pick up what it is from context clues.</p>
| user10354138 | 592,552 | <p><span class="math-container">$\bar{K}(C)$</span> is the function field of <span class="math-container">$C$</span> over the algebraically closure of <span class="math-container">$K$</span> (as recalled in the beginning of chapter).</p>
<p><span class="math-container">$\bar{K}(C)^*$</span> is the multiplicative group of units of <span class="math-container">$\bar{K}(C)$</span>.</p>
<p>The notation <span class="math-container">$R^*$</span> for the multiplicative group of units of a ring <span class="math-container">$R$</span> is standard. If <span class="math-container">$R$</span> is a field then <span class="math-container">$R^*=R-\{0\}$</span>.</p>
|
4,036,117 | <p>In Joe Silverman's "The Arithmetic of Ecliptic Curves" he talks a lot about the integral domain <span class="math-container">$\overline{K}(C)$</span>. On page <span class="math-container">$27$</span> he suddenly decides to chose an element <span class="math-container">$f$</span> from some set <span class="math-container">$\overline{K}(C)^*$</span>. What does this notation mean? After page <span class="math-container">$27$</span> he begins using <span class="math-container">$\overline{K}(C)^*$</span> a lot, and I haven't been able to pick up what it is from context clues.</p>
| Álvaro Lozano-Robledo | 14,699 | <p>There is already a correct answer, but I will just add here that the notation is defined in Chapter I, in a Definition right after Example 1.3.3. where Silverman defines the affine coordinate ring and the function field of a variety.</p>
|
96,198 | <p>I'm trying to prove that the variance of a RV whose values are discrete 1's or 0's is greater than the variance of a RV who's values are 0's or continuous on the domain (0,1], where any "1" in the Bernoulli RV corresponds to a value on (0,1] in the other RV. Intuitively, I think this is the case, but I'm trying to demonstrate it.</p>
<p>For instance, I know that $Var[X] > Var[\alpha X]$ where $0 \le \alpha < 1$, but what if $\alpha$ is a RV on the interval $(0,1]$?</p>
<p>The reason I'm asking is that I'm sampling a RV with values on the interval (0,1] and I'm using a binomial confidence interval as an upper bound (you can read more of the <a href="https://stats.stackexchange.com/questions/15567/putting-a-confidence-interval-on-the-mean-of-a-very-rare-event">context from an old discussion here</a>), and I need to prove that that's a reasonable upper bound on the confidence interval, since the values could all be 1's or 0's, but have the possibility of being between those values.</p>
<p><strong>Edit</strong> Sasha provided a counterexample to my question as originally stated. The distribution in question is such that it's Bernoulli-like, however instead of having only 1's and 0's, the "true" values can take on values on the interval (0,1]. So, for Sasha's case of $p_{true} = 0.01$, the distribution in question would have a delta function at 0 of value $1-p_{true}$ and the rest of the distribution would be on (0,1].</p>
| Robert Israel | 8,508 | <p>I presume you're assuming $X$, $Y$, $Z$ independent.
Note that $(X+Y+Z)/\sqrt{3}$, $(Y - Z)/\sqrt{2}$ and $(2X-Y-Z)/\sqrt{6}$ are uncorrelated (and therefore independent) standard normal random variables.
Since $X = (X+Y+Z)/3 + (2X-Y-Z)/3$, $\text{Var}(X|X+Y+Z=1) = (1/9) \text{Var}(2X-Y-Z) = 2/3$.</p>
|
25,556 | <p>Why is the zero set in <span class="math-container">$\mathbb{C}\times\mathbb{C}$</span> of a polynomial <span class="math-container">$f(x,y)$</span> in two complex variables always non-discrete (no zero of <span class="math-container">$f$</span> is isolated)?</p>
| Mariano Suárez-Álvarez | 274 | <p>Write $f(x,y)$ as a polynomial in $y$ with coefficients in $\mathbb C[x]$, so that $f(x,y)=\sum_{i=0}^nf_i(x)y^i$. If $x\in\mathbb C$ is such that $f_n(x)\neq0$, then there exist $n$ values of $y$ such that $f(x,y)=0$. It follows that if $X\subseteq \mathbb C\times\mathbb C$ is the zero locus of $f$, then the first projection $\pi_x:X\to\mathbb C$ has an image with finite complement. This implies, in particular, that $X$ is uncountable, and no uncountable subset of $\mathbb C\times\mathbb C$ is discrete.</p>
|
1,134,323 | <p>Say you pick a number $x$, like $\frac 43$. Its inverse is of course $\frac 34$. $x$ is a distance of $\frac 13$ away from 1, and its inverse is a distance of $\frac 14$ away from 1. Is there any number $x$ that is a distance $d$ away from 1, whose inverse $\frac 1x$ is also a distance $d$ away from 1? I came up with the following equation and found the solution $x = 1$, but I was hoping there was another solution.</p>
<p>$x - 1 = 1 - \frac 1x$</p>
| malo_x | 213,118 | <p>Yes. Your equation only has the solution $x=1$ but that doesn't mean that the only solutions to the problem are solutions to said equation. You can easily check that -1 also satisfies your problem. Maybe consider the following equation, since you're measuring distance:
\begin{equation*}
|x-1| = \left|1-\frac{1}{x}\right|.
\end{equation*}</p>
<p>Edit in response to OPs comment: You can consider all possible cases. First rewrite the equation to
\begin{equation*}
|x-1| = \left|1-\frac{1}{x}\right| \iff
|x-1| = \left|\frac{1}{x}-1\right|.
\end{equation*}
Case 1: $x>1.$ Then
\begin{equation*}
|x-1| = x-1 = 1-\frac{1}{x} = \left|\frac{1}{x}-1\right| \iff x+\frac{1}{x}=2.
\end{equation*}
Therefore, there is no real solution for $x>1$.</p>
<p>We've already considered the case $x=1$. So let's consider $0<x<1$. Then
\begin{equation*}
|x-1| = 1-x = \frac{1}{x}-1 = \left|\frac{1}{x}-1\right| \iff x+\frac{1}{x}=2,
\end{equation*}
but we're only considering $0<x<1$, so the equation has no solution for $0<x<1$.</p>
<p>We're not going to consider the case $x=0$. So let's consider $-1<x<0$. Then
\begin{equation*}
|x-1| = 1-x = 1-\frac{1}{x} = \left|\frac{1}{x}-1\right| \iff x-\frac{1}{x} = 0,
\end{equation*}
which has no solution for $0<x<1$.</p>
<p>We've alread covered the case $x=-1$, so let's consider $x<-1$ as our last case. Then
\begin{equation*}
|x-1| = 1-x = 1-\frac{1}{x} = \left|\frac{1}{x}-1\right| \iff x-\frac{1}{x} = 0,
\end{equation*}
which, again, has no solution. So, yes, $\pm 1$ are the only real solutions to your problem.</p>
|
4,243,005 | <p>I was having trouble with this integral</p>
<blockquote>
<p>Prove
<span class="math-container">$$\int_{-\infty}^{\infty} e^{2x}x^2 e^{-e^{x}}dx=\gamma^2 -2\gamma+\zeta(2)$$</span></p>
</blockquote>
<p>Where <span class="math-container">$\gamma$</span> is the euler mascheroni constant. Let <span class="math-container">$u=e^x\rightarrow \ln(u)= x$</span> thus <span class="math-container">$du=e^x dx$</span></p>
<p><span class="math-container">$$\int_{-\infty}^{\infty} e^{2x}x^2 e^{-e^{x}}dx= \int_{0}^{\infty} u\ln^2(u) e^{-u}du$$</span>
Now
<span class="math-container">$$\Gamma(n)=\int_0^\infty x^{n-1}e^xdx$$</span>
<span class="math-container">$$\Gamma'(n)=\int_0^\infty x^{n-1}e^x\ln(x)dx$$</span>
<span class="math-container">$$\Gamma''(n)=\int_0^\infty x^{n-1}e^x\ln^2(x)dx\Rightarrow \Gamma''(2)=\int_0^\infty x e^x\ln^2(x)dx$$</span></p>
<p>How do I proceed from here? Do I have to use <span class="math-container">$\Gamma'(n)=\psi(n)\Gamma(n)$</span> ? How do I solve this integral ?</p>
<p>Thank you for your time</p>
| Diger | 427,553 | <p><span class="math-container">$$\Gamma''(2)=\Gamma''(s+2)\Big|_{s=0}=\left((s+1)\Gamma(s+1)\right)''\Big|_{s=0}=\left(\Gamma(s+1)+(s+1)\Gamma'(s+1)\right)'\Big|_{s=0} \\
=2\Gamma'(s+1)+(s+1)\Gamma''(s+1)\Big|_{s=0} = -2\gamma + \Gamma''(1) \, .$$</span>
Now <span class="math-container">$$\Psi(s+1)+\gamma = \sum_{n=1}^\infty \frac{1}{n}-\frac{1}{n+s} \\
\Rightarrow \quad \Psi'(s+1)=\frac{{\rm d}}{{\rm d}s} \frac{\Gamma'(s+1)}{\Gamma(s+1)}=\frac{\Gamma''(s+1)}{\Gamma(s+1)}-\frac{\Gamma'(s+1)^2}{\Gamma(s+1)^2} = \sum_{n=1}^\infty \frac{1}{(n+s)^2} \\
\stackrel{s=0}{\Rightarrow} \quad \Psi'(1)=\Gamma''(1)-\Gamma'(1)^2=\Gamma''(1)-\gamma^2=\zeta(2)$$</span>
and so
<span class="math-container">$$\Gamma''(2)=-2\gamma+\gamma^2+\zeta(2) \, .$$</span></p>
|
3,434,656 | <p>We are given a <span class="math-container">$3 \times 3$</span> real matrix <span class="math-container">$A$</span>, and we know it has three eigenvalues. One eigenvalue is <span class="math-container">$\lambda_1=-1$</span> with corresponding eigenvector <span class="math-container">$v_1=\left[\begin{matrix}
0 \\
1 \\
0 \\
\end{matrix}\right]$</span> and another eigenvalue <span class="math-container">$\lambda_2=1+i$</span> and corresponding eigenvector <span class="math-container">$v_2=\left[\begin{matrix}
1 \\
2 \\
i \\
\end{matrix}\right]$</span>. Given this, how can we find the third eigenvalue/eigenvector pair <span class="math-container">$(\lambda_3, v_3)$</span>? The point is ultimately to be able to find the general solution to the linear DE system <span class="math-container">$x'=Ax$</span>. </p>
<p>The context is that this problem came up in a qualifying exam. My linear algebra is incredibly rusty so I imagine there's just some eigenvalue/eigenvector related trick I'm not seeing. Now, considering the characteristic polynomial, it should be clear that the third eigenvalue is <span class="math-container">$\lambda_3 = 1-i$</span>. What's not clear to me is determining the corresponding eigenvector. Clearly, it must be linearly independent from the other two, but how can we use the given eigenvectors to deduce the third one?</p>
| Bernard | 202,857 | <p><strong>Hint</strong>:</p>
<p>The characteristic polynomial <span class="math-container">$\chi_A$</span> has real coefficients, hence its nonreal roots are pairwise conjugate, so <span class="math-container">$1-i$</span> is the third eigenvalue. Furthermore, if <span class="math-container">$v_2$</span> is an eigenvector associated to the eigenvalue <span class="math-container">$1+i$</span>, <span class="math-container">$\bar v_2$</span> is an eigenvector associated to the conjugate eigenvalue <span class="math-container">$1-i$</span>.</p>
|
95,341 | <p>I know that the complement of the zero set of a polynomial $P: \mathbb{C}^n \rightarrow \mathbb{C}$ is connected in $\mathbb{C}^n$ (by the way, can anybody suggest a reference?).</p>
<p>Is it possible to extend the proof also to polynomials
$P: SL(N,\mathbb{C})^n \rightarrow \mathbb{C}$ ?</p>
<p>Thanks!</p>
| ACL | 10,696 | <p>An answer to the more general question could be in three steps.</p>
<p>1) If $X$ is a complex algebraic variety, connected for the Zariski topology, then it is connected for the usual topology, an important and nontrivial fact that can be found in Mumford's Red Book, or in Shafarevich's one. </p>
<p>2) The complement to a closed Zariski subset in an irreducible variety is connected for the Zariski topology.</p>
<p>3) The varieties $SL(n,{\bf C})$ and $SL(N,{\bf C})^N$ are irreducible </p>
|
415,150 | <blockquote>
<p>Let $x$ and $y$ be positive numbers. Let $a_0=y$, and let $$a_n=\frac{(x/a_{n-1})+a_{n-1}}{2}$$Prove that the sequence $\{a_n\}$ has limit $\sqrt{x}$. Generalize to $n$th roots.</p>
</blockquote>
<p>I already solved the first part <a href="https://math.stackexchange.com/questions/415125/limit-of-recursive-sequence-a-n-fracx-a-n-1a-n-12">here</a>, but I have no idea how to generalize the $n$th roots. The limit should be $\sqrt[n]{x}$, but what is the recurrence supposed to look like?</p>
| Evariste | 29,168 | <p>The integral you've done is absolutely right, although you might always want to include the upper and lower limits before you evaluate the integral(i.e. never leave it in the indefinite form like this: $$-\frac{4}{25}\frac{x^{2+1}}{2+1}+x$$
).</p>
<p>You were probably wondering if you did everything correctly. I think it was correct. Next time you might want to put your stuff into Wolfram Alpha(www.wolframalpha.com). It's a useful website that will check your integrals.</p>
<p>Also in terms of your strategy, since you have already split your integrals into the sum of two integrals(which is easier to evaluate and less nasty), why did you put them back together when you actually got the the evaluation step? I suggest that you practice more on solving integrals and set up a systematical way of doing it. This will save you tons of time.</p>
<p>The final answer should be $51122500/81 m^2$ which is approximately $631141.98m^2$</p>
<p>Hope this helps.</p>
|
4,482,285 | <p>I'm currently making a game and have run into a problem I'm not quite sure how to solve, I'll try to lay it out as a maths question. None of the values are fixed, so I'm looking for an equation that solves the below question:</p>
<p>A plane lies on the position vector <code>p0</code> <code><x0, y0, z0></code> and has a normal unit vector <code>n</code> <code><a, b, c></code>.</p>
<p>Given a sphere with a radius of <code>R</code> forms a tangent with the plane at an unknown position vector <code>t</code> <code><tx, ty, tz></code>:</p>
<p>Find the highest height of the center of a sphere, when the sphere is located at the position vector <code>s</code> <code><sx, sy, sz></code>, and <code>sx</code> and <code>sz</code> are known values, but <code>sy</code> is not.</p>
<p>The known vectors are <code>p0</code>, <code>n</code>, and the other known values are <code>sx</code>, <code>sz</code>, and <code>R</code>. The position vector <code>t</code> is unknown, and the height of the sphere <code>sy</code> is also unknown. Should it be necessary, another point on the plane <code>p1</code> <code><x1, y1, z1></code> can be provided.</p>
<p>In my case, the Y axis is the up axis, the Z axis is the forward axis, and the X axis is the right axis.</p>
<p>I wanted to essentially do something similar to what is shown in this video, but in my case I know the length of the vector but not the positions: <a href="https://www.youtube.com/watch?v=zWMTTRJ0l4w" rel="nofollow noreferrer">https://www.youtube.com/watch?v=zWMTTRJ0l4w</a></p>
<p>Here is 2d visualisation of the problem:
<a href="https://i.stack.imgur.com/T0WbS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/T0WbS.png" alt="Sphere and plane" /></a></p>
<p>I was only able to find the positions due to trial and error, but these were approximately correct.</p>
<p>Here is what I do know:
The vector <code>t</code> to <code>s</code> will have the same direction as the normal vector, but with the magnitude being equal to R. It could then be defined as <code>Rn</code>, given that <code>n</code> is a unit vector.
Given that, the vector could be defined as a translation of the vector <code>Rn</code> along the plane.</p>
<p>Beyond this point, I am a little confused, so any help would be appreciated.
I would prefer to avoid cartesian form, and values can be stored in variables at later dates if necessary.
I have v1:Dot(v2) and v1:Cross(v2) methods available to me, and I expect this probably will use the dot product at some point.</p>
<p>Thank you.</p>
| Glorious Nathalie | 948,761 | <p>The center of the sphere lies on the line</p>
<p><span class="math-container">$ C(t) = (s_x, 0, s_z) + t (0, 1, 0) = (s_x, t, s_z) $</span></p>
<p>We want the distance of a point on this line to be equal to the known given value of <span class="math-container">$R$</span> (the radius of the sphere). The formula for the signed distance is</p>
<p><span class="math-container">$ d = \hat{n} \cdot ( C(t) - p_0) $</span></p>
<p>where <span class="math-container">$\hat{n}$</span> is the normalized normal vector of the plane.</p>
<p>Equating <span class="math-container">$d$</span> to <span class="math-container">$R$</span> results in the center of the sphere that on the side of the plane pointed to by <span class="math-container">$\hat{n}$</span>, while setting <span class="math-container">$d$</span> to <span class="math-container">$(-R)$</span> gives the center of the sphere that is on the opposite side as the one pointed to by <span class="math-container">$\hat{n}$</span>.</p>
<p>Now the tangency point is given by</p>
<p><span class="math-container">$ T = C - d \hat{n} $</span></p>
<p>where <span class="math-container">$d$</span> can be <span class="math-container">$+R$</span> or <span class="math-container">$-R$</span>, but <span class="math-container">$C$</span> is different between these two cases.</p>
<p>Now the height of the sphere is</p>
<p><span class="math-container">$ h = C_z + R $</span></p>
|
1,838,596 | <p>I need to compute the surface area of the torus $$T^2=\{(x,y,z)\subseteq\mathbb R^3 \left(\sqrt {x^2+y^2}- R\right)^2+z^2=r^2\}$$
where $0<r<R$.</p>
<p>I know I need to compute the metric tensor and the Gramian determinant etc, but in order to so, I need a regular global parametrization of $T^2$, I guess? How do obtain the latter for such a set?</p>
<p>Some help is appreciated.</p>
| Mark Viola | 218,419 | <p>We can describe the surface of the torus using parameters $(\phi,\alpha)$ by the position vector $\vec r(\phi,\alpha)$</p>
<p>$$\begin{align}
\vec r(\phi,\alpha)&=\hat \rho(\phi) R+(\hat \rho(\phi) r\cos(\alpha)+\hat z r\sin(\alpha))\\\\
&=\hat \rho(\phi)\,(R+r\cos(\alpha))+\hat z r\sin(\alpha)
\end{align}$$ </p>
<p>where $\hat \rho(\phi)=\hat x \cos(\phi)+\hat y \sin(\phi)$ is the radial unit vector in cylindrical coordinates, $0\le \phi<2\pi$, and $0\le \alpha<2\pi$.</p>
<p>Note that the angle $\alpha$ can be interpreted as the polar angle in a local cylindrical coordinate system centered at points on the axis of the torus.</p>
|
1,838,596 | <p>I need to compute the surface area of the torus $$T^2=\{(x,y,z)\subseteq\mathbb R^3 \left(\sqrt {x^2+y^2}- R\right)^2+z^2=r^2\}$$
where $0<r<R$.</p>
<p>I know I need to compute the metric tensor and the Gramian determinant etc, but in order to so, I need a regular global parametrization of $T^2$, I guess? How do obtain the latter for such a set?</p>
<p>Some help is appreciated.</p>
| David Jaramillo | 349,591 | <p>Note you have a sum of squares equal a constant. Hence you should use trigonometric functions.
$$
\left(\sqrt{x^2+y^2}-R\right)^2+z^2=r^2
$$
So we set $z=r\cos(\phi),\sqrt{x^2+y^2}-R=r\sin(\phi)$. Squaring the second relation we get
$$
x^2+y^2=(R+r\sin(\phi))^2
$$
So we may introduce another angle an set $x=(R+r\sin(\phi))\cos(\theta),y=(R+r\sin(\phi))\sin(\theta)$. And this gives you the natural parametrization of the torus (You can see for ur self that those angles actually have a geometric meaning)
$$
x=(R+r\cos(\phi))\cos(\theta)\\
y=(R+r\sin(\phi))\sin(\theta)\\
z=r\cos(\phi)
$$</p>
|
2,545,813 | <p>The question is:</p>
<p>If $R$ has an identity and $A$ is an $R$-module, then there are submodules $B$ and $C$ of $A$ such thatt $B$ is unitary $RC = 0$ and $A = B \oplus C.$
And if $A_{1}$is another $R$-module, with $A_{1} = B_{1} \oplus C_{1}$, $B_{1}$ unitary ,$RC_{1} = 0.$</p>
<p>Why if $f:A\rightarrow A_{1}$ is an $R$-module homomorphism then $f(B) \subset B_{1} $ and $f(C) \subset C_{1}.$? </p>
| 1ENİGMA1 | 255,913 | <p>Hint:Assume that $f(B) \not\subset B_{1}$. We have some $0\neq b\in B$ such that that $f(b)\in C_{1}$ (i.e., $f(b)\notin B_1$).<br/> So, $f(b)=f(1_Rb)=1_Rf(b)\in C_1$.But, we know that $RC_1=0$, so $1_Rf(b)=f(b)=0$.<br/> This is a contradiction of $b\neq 0$. ($f(b)=0=f(0)$).<br/>
You can use above idea for $f(C) \subset C_{1}$.</p>
|
3,946,998 | <p>I was doing a question. Suddenly I got stuck at this last part of the problem. It was to prove
<span class="math-container">$ 2^r +2 = a^2 +b^2$</span> where <span class="math-container">$r \neq 2$</span>, r is a prime and <span class="math-container">$ a \neq b$</span>. Also <span class="math-container">$r^2 -1$</span> is a mersenne prime. I tried to use fermat's little theorum, but to no avail. Thank you.</p>
<p>PS note: The problem I was solving was BMO2021 Q6.</p>
| NN2 | 195,378 | <p>The question is wrong. If <span class="math-container">$r=11$</span>, <span class="math-container">$2^{11}-7 = 2041$</span> is not a square number.</p>
|
4,645,950 | <p>The question is: Peter and Mary take turns rolling a fair die. If Peter rolls 1 or 2 he wins and the game stops. If Mary rolls 3,4,5, or 6, she wins and the game stops. They keep rolling until one of them wins. Suppose Peter rolls first.</p>
<p>a) What is the probability that Peter wins and rolls at most 4 times?</p>
<p>Here is the solution: We want to find <span class="math-container">$A=\\\{ \text{Peter wins and rolls at most 4 times} \\\}$</span>. We decompose the event into <span class="math-container">$A_k$</span>= {Peter wins on his kth roll}. Then <span class="math-container">$A=\cup^4_{k=1}$</span> and since the events <span class="math-container">$A_k$</span> are mutually exclusive, <span class="math-container">$P(A)=\sum^4_{k=1} P(A_k)$</span>. My book says that the ratio of favorable alternatives over the total number of alternatives yields <span class="math-container">$$P(A_k)=\frac{(4 \cdot 2)^{k-1}\cdot 2}{(6 \cdot 6)^{k-1} \cdot 6}=\left(\frac{8}{36} \right)^{k-1}\frac{2}{6}=\left( \frac{2}{9}\right)^{k-1}\frac{1}{3}$$</span> Then the solution if sound using the geometric sum.</p>
<p>I am having problems understanding this answer. Why is the 2/6 not to the k power? and why are we multiplying the probability of failure of Peter and Mary inside the k-1 power? I'd really appreciate it if someone could break down how we obtain <span class="math-container">$A_k$</span>.</p>
| user115350 | 334,306 | <p>For a roll, the probability that Peter fail is 4/6 and Mary fail 2/6.</p>
<p>If Peter wins on his kth roll, Peter should fail on his first k-1 rolls and Mary fail on her first k-1 rolls. The probability is <span class="math-container">$( \frac{2}{6} )^{k-1}(\frac{4}{6})^{k-1}$</span>.</p>
<p>And, on k-th roll, the probability that Peter wins is 2/6.</p>
<p>Thus the probability is, <span class="math-container">$P(A_k)=(\frac{2}{6})^{k-1}(\frac{4}{6})^{k-1}\frac{2}{6}$</span></p>
|
196,155 | <p>I have recently read some passage about nested radicals, I'm deeply impressed by them. Simple nested radicals $\sqrt{2+\sqrt{2}}$,$\sqrt{3-2\sqrt{2}}$ which the later can be denested into $1-\sqrt{2}$. This may be able to see through easily, but how can we denest such a complicated one $\sqrt{61-24\sqrt{5}}(=4-3\sqrt{5})$? And Is there any ways to judge if a radical in $\sqrt{a+b\sqrt{c}}$ form can be denested?</p>
<p>Mr. Srinivasa Ramanujan even suggested some CRAZY nested radicals such as:
$$\sqrt[3]{\sqrt{2}-1},\sqrt{\sqrt[3]{28}-\sqrt[3]{27}},\sqrt{\sqrt[3]{5}-\sqrt[3]{4}},
\sqrt[3]{\cos{\frac{2\pi}{7}}}+\sqrt[3]{\cos{\frac{4\pi}{7}}}+\sqrt[3]{\cos{\frac{8\pi}{7}}},\sqrt[6]{7\sqrt[3]{20}-19},...$$
Amazing, these all can be denested. I believe there must be some strategies to denest them, but I don't know how.</p>
<p>I'm a just a beginner, can anyone give me some ideas? Thank you.</p>
| Mr Pie | 477,343 | <p>There is also another radical identity, which generalises <span class="math-container">$\sqrt{a+\sqrt b}$</span> even further: <span class="math-container">\begin{align*}\sqrt{a\pm \sqrt{b+\sqrt{c}}}&= \frac{1}{\sqrt{2}}\left\{\sqrt{a^2+\sqrt{\frac{a^2-b+\sqrt{(a^2-b)^2-c}}{2}}-\sqrt{\frac{a^2-b-\sqrt{(a^2-b)^2-c}}{2}}}\right. \nonumber\\&\quad\left.\pm\,\sqrt{a^2-\sqrt{\frac{a^2-b+\sqrt{(a^2-b)^2-c}}{2}}+\sqrt{\frac{a^2-b-\sqrt{(a^2-b)^2-c}}{2}}}\right\}\end{align*}</span> such that the <span class="math-container">$\pm$</span> signs are not independent of each other.</p>
<p>We also have:</p>
<p>If <span class="math-container">$a^2+b^2=c^2$</span>, then <span class="math-container">$$(a+b-c)^2=2(c-a)(c-b).$$</span> Therefore, by letting <span class="math-container">$a=\sqrt x$</span> and <span class="math-container">$b=\sqrt{c^2-x}$</span>, it follows for all <span class="math-container">$c$</span> and <span class="math-container">$x$</span> such that <span class="math-container">$c\in \mathbb{R}\setminus \big({-\sqrt x}, \sqrt x\big)$</span>, <span class="math-container">$$\sqrt{2\big(c-\sqrt x\big)\big(c-\sqrt{c^2-x}\big)}=\sqrt x + \sqrt{c^2-x} -c.$$</span> Interestingly, given that <span class="math-container">$$(a+x)^2+(b+y)^2-(a+y)^2-(b+x)^2=2(a-b)(x-y)$$</span> then if <span class="math-container">$a^2+b^2=c^2$</span>, we also have <span class="math-container">$$(a+b-c)^2=(2c)^2+(a+b)^2-(b+c)^2-(a+c)^2.$$</span> Through one of Ramanujan's classic identities, <span class="math-container">$$\big\{\sqrt [3]{(a+b)^2}-\sqrt [3]{a^2-ab+b^2}\big\}^3=3\big(\sqrt [3]{a^3+b^3}-a\big)\big(\sqrt [3]{a^3+b^3}-b\big)$$</span> it can be found that if <span class="math-container">$a^3+b^3=c^3$</span>, then <span class="math-container">$$(a+b-c)^3=3(a+b)(c-a)(c-b)$$</span>
Therefore, again, for all <span class="math-container">$c$</span> and <span class="math-container">$x$</span> such that <span class="math-container">$c\in\mathbb{R}\setminus ({-\sqrt [3] x}, \sqrt [3] x)$</span>, it follows
<span class="math-container">$$\sqrt [3]{3\big(\sqrt [3] x+\sqrt [3]{c^3-x}\big)\big(c-\sqrt [3] x\big)\big(c-\sqrt [3]{c^3-x}\big)}=\sqrt [3] x + \sqrt [3]{c^3-x}-c.$$</span> Unfortunately, due to <em>Fermat's Last Theorem</em>, there are no positive integer solutions for unique <span class="math-container">$a$</span>, <span class="math-container">$b$</span> and <span class="math-container">$c$</span>.</p>
<hr>
<p>Try such substitutions yourself on the following one! :)</p>
<p>If <span class="math-container">$a^4+b^4=c^4$</span>, then <span class="math-container">$$(a+b-c)^4=6(ab-ac-bc)^2-4\big\{a^3(c-b)+b^3(c-a)+c^3(a+b)\big\}$$</span></p>
|
513,239 | <p>I've recently heard a riddle, which looks quite simple, but I can't solve it.</p>
<blockquote>
<p>A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "<em>Yes</em>", "<em>No</em>", or "<em>I don't know</em>," and after the girl answers it, he knows what the number is. What is the question?</p>
</blockquote>
<p>Note that the girl is professional in maths and knows EVERYTHING about these three numbers.</p>
<hr>
<p><strong>EDIT:</strong> The person who told me this just said the correct answer is:</p>
<blockquote class="spoiler">
<p> "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"</p>
</blockquote>
| Ben Millwood | 29,966 | <p>"I am thinking of a number which is either 0 or 1. Is the sum of our numbers greater than 2?"</p>
|
513,239 | <p>I've recently heard a riddle, which looks quite simple, but I can't solve it.</p>
<blockquote>
<p>A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "<em>Yes</em>", "<em>No</em>", or "<em>I don't know</em>," and after the girl answers it, he knows what the number is. What is the question?</p>
</blockquote>
<p>Note that the girl is professional in maths and knows EVERYTHING about these three numbers.</p>
<hr>
<p><strong>EDIT:</strong> The person who told me this just said the correct answer is:</p>
<blockquote class="spoiler">
<p> "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"</p>
</blockquote>
| Oleg567 | 47,993 | <p><strong>Question 1</strong>:</p>
<p>I'm thinking of a huge integer number with last digit $7$.</p>
<p>Is this ratio $~~\dfrac{\mbox{my number}}{\mbox{your number}}~~$ integer?</p>
<ul>
<li>$1$ (yes)</li>
<li>$2$ (no)</li>
<li>$3$ (I don't know)</li>
</ul>
<hr>
<p><strong>Question 2</strong>:</p>
<p>Consider series $\displaystyle\sum\limits_{n=1}^\infty a_n, \quad (a_n>0, ~~~ n \in \mathbb{N})$,
such that there exists limit (see <a href="http://en.wikipedia.org/wiki/Ratio_test">Ratio test</a>)
$$
L = \lim_{n\to \infty} \frac{a_{n+1}}{a_n}.
$$
If $L+1$ is equal to your number, is this series convergent?</p>
<ul>
<li>$1$ ($L=0$, yes)</li>
<li>$2$ ($L=1$, I don't know)</li>
<li>$3$ ($L=2$, no)</li>
</ul>
|
513,239 | <p>I've recently heard a riddle, which looks quite simple, but I can't solve it.</p>
<blockquote>
<p>A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "<em>Yes</em>", "<em>No</em>", or "<em>I don't know</em>," and after the girl answers it, he knows what the number is. What is the question?</p>
</blockquote>
<p>Note that the girl is professional in maths and knows EVERYTHING about these three numbers.</p>
<hr>
<p><strong>EDIT:</strong> The person who told me this just said the correct answer is:</p>
<blockquote class="spoiler">
<p> "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"</p>
</blockquote>
| Neal | 20,569 | <p>This was suggested by a friend:</p>
<blockquote>
<p>If $k$ is your number, does $\mathbb{S}^{3k-2}$ have an exotic smooth structure?</p>
</blockquote>
|
513,239 | <p>I've recently heard a riddle, which looks quite simple, but I can't solve it.</p>
<blockquote>
<p>A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "<em>Yes</em>", "<em>No</em>", or "<em>I don't know</em>," and after the girl answers it, he knows what the number is. What is the question?</p>
</blockquote>
<p>Note that the girl is professional in maths and knows EVERYTHING about these three numbers.</p>
<hr>
<p><strong>EDIT:</strong> The person who told me this just said the correct answer is:</p>
<blockquote class="spoiler">
<p> "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"</p>
</blockquote>
| Blue | 409 | <p>I get the sense that there's <a href="http://www.youtube.com/watch?v=kTcRRaXV-fg">a classic comedy routine</a> in here somewhere.</p>
<blockquote>
<p>"You know, they give baseball players these days very peculiar names. [...] Well, now, let's see ... On our team we have: <strong>NO</strong>'s on first, <strong>YES</strong>'s on second, and <strong>I-DON'T-KNOW</strong>'s on third ..." </p>
</blockquote>
|
513,239 | <p>I've recently heard a riddle, which looks quite simple, but I can't solve it.</p>
<blockquote>
<p>A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "<em>Yes</em>", "<em>No</em>", or "<em>I don't know</em>," and after the girl answers it, he knows what the number is. What is the question?</p>
</blockquote>
<p>Note that the girl is professional in maths and knows EVERYTHING about these three numbers.</p>
<hr>
<p><strong>EDIT:</strong> The person who told me this just said the correct answer is:</p>
<blockquote class="spoiler">
<p> "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"</p>
</blockquote>
| Bennett Gardiner | 78,722 | <p>Along the lines of "the open problem" method - </p>
<p>Define $$ f(n) = \pi^{n-1}\mathrm{e}^{\pi(n-1)}.$$</p>
<p>Where $n$ is the number chosen by the girl. The question is, is $f(n)$ irrational? </p>
<p>If $n=1$, $f(1) = 1$, so the answer is <strong>"No"</strong>. </p>
<p>If $n=2$, $f(2) = \pi\mathrm{e}^{\pi}$, the answer is <strong>"Yes"</strong>.</p>
<p>If $n=3$, $f(3) = \pi^2\mathrm{e}^{2\pi}$, so the answer is <strong>"I don't know"</strong>.</p>
|
4,036,704 | <blockquote>
<p><span class="math-container">$$(A ∧ (A → B)) ⇒ B$$</span></p>
</blockquote>
<p>I know that this is a tautology, but apart from setting up a truth-table I dont know how I would go about a formal proof in discrete mathematics. thanks</p>
<p><a href="https://i.stack.imgur.com/j6cHN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/j6cHN.png" alt="enter image description here" /></a></p>
| Gio | 872,828 | <p><span class="math-container">$(A \land (A\rightarrow B)) \rightarrow B $</span> is true when <span class="math-container">$A \land (A\rightarrow B)$</span> is false or <span class="math-container">$B$</span> is true. Therefore if it was false then <span class="math-container">$A \land (A\rightarrow B)$</span> would be true and <span class="math-container">$B$</span> would be false.</p>
<p>If <span class="math-container">$A \land (A\rightarrow B)$</span> is false then A is false and <span class="math-container">$A\rightarrow B$</span> is false.
But if <span class="math-container">$A\rightarrow B$</span> is false and <span class="math-container">$B$</span> is false then <span class="math-container">$A$</span> must be true, which is a contradiction.</p>
|
910,414 | <p>My question is about visualizing projective space, in particular the real projective plane $\mathbb{P}^2(\mathbb{R})$. I know there are different ways to define this space, but in each we can say that "two parallel lines intersect." If you type projective space into Wikipedia, or look it up in a textbook, a lot of the time you will see an image of two train tracks. The tracks are parallel, but as they go off into the distance (towards infinity) they appear to be approaching one another. </p>
<p>I believe I understand why we can say that two lines which are parallel in $\mathbb{R}^3$ intersect in $\mathbb{P}^2(\mathbb{R})$ (at least with the equivalence class construnction) but I do not see how this is related to that visual image of the two tracks appearing to approach a common point. </p>
<p>Maybe it is naive, but my question is: why do parallel lines, when looked at as in the image of the train tracks, appear to approach a common point? What is the mathematical reason for this?</p>
| John Hughes | 114,036 | <p>Suppose that your eye is at the origin, and the "canvas" on which you draw is in front of your eye, and is the $z = 1$ plane. Then the point $(2, 3, 5)$ in space will project, in your "drawing" or "seeing" of the world, to the point $(2/5, 3/5, 1)$ in the $z = 1$ plane. In general, any point $(x, y, z)$ will project to $(x/z, y/z, 1)$. </p>
<p>Now consider two parallel lines; the first one consisting of all points of the form $(1, t, 4)$ and the second consisting of points $(3, t, 2)$. The projections of these to the drawing plane will consist of points of the form $(1/4, t/4, 1)$ and $(3/2, t/2, 1)$, respectively, i.e., they'll still be parallel.</p>
<p>Now look at the lines $(-1, 0, t)$ and $(1, 0, t)$. These project to
$(-1/t, 0, 1)$ and $(1/t, 0, 1)$, which "meet" at the point $(0, 0, 1)$ when $t$ goes to $\infty$. </p>
<p>Does that help at all? </p>
|
2,173,586 | <p>Let $R$ be an unique factorization domain. Being an integral domain, it has a field of quotients $F$. We can consider $R[x]$ to be a subring of $F[x]$. </p>
<blockquote>
<p>Given any polynomial $f(x)\in F[x]$, then $f(x)=(f_0(x)/a)$, where $f_0(x)\in R[x]$ and where $a\in R$.</p>
</blockquote>
<p>I don't understand why $a\in R$, in the proof of the theorem that Every integral domain can be imbedded in a field, we used $a/b$, where $a,b$ where in the integral domain. So why in this example above we have $a/b$ where $a\in R[x]$ and $b\in R$? </p>
| Patrick Stevens | 259,262 | <p>The countable intersection and union are also limits. $$\bigcup_{n \in \mathbb{N}} A_n = \lim_{n \to \infty} \bigcup_{i=1}^n A_n$$</p>
<p>Set operations can be defined without reference to the size of the domain of "summation": $$\bigcup A := \{ x: x \in X \in A \}$$
$$\bigcap A := \{x : (\forall X \in A)(x \in x) \}$$
(no countability needed).</p>
<p>Note that infinite sums are only really defined when the domain of summation is countable (see <a href="https://math.stackexchange.com/questions/20661/the-sum-of-an-uncountable-number-of-positive-numbers">this question</a>). So infinite sums are really very restricted objects compared to sets.</p>
|
2,173,586 | <p>Let $R$ be an unique factorization domain. Being an integral domain, it has a field of quotients $F$. We can consider $R[x]$ to be a subring of $F[x]$. </p>
<blockquote>
<p>Given any polynomial $f(x)\in F[x]$, then $f(x)=(f_0(x)/a)$, where $f_0(x)\in R[x]$ and where $a\in R$.</p>
</blockquote>
<p>I don't understand why $a\in R$, in the proof of the theorem that Every integral domain can be imbedded in a field, we used $a/b$, where $a,b$ where in the integral domain. So why in this example above we have $a/b$ where $a\in R[x]$ and $b\in R$? </p>
| Asaf Karagila | 622 | <p>Sets are not real numbers. They are part of a much larger universe. The real numbers are bound by $\Bbb R$, they all have to be members of $\Bbb R$, and the summation has to obey the rules of summations which extend addition (under the usual definition of a sum, that is).</p>
<p>On the other hand, an infinite family of sets is itself just a set of sets. One of the axioms of set theory tells us that given a set $A$ of sets—finite, countable, uncountable, <em>any</em> set—we can take the union of all the sets in $A$.</p>
|
308,014 | <p>First of all, on this Matlab exercise sheet that I am currently working through what does the term 'the lower $2 \times 2$ block' mean in the question below?</p>
<p>$A =
\left[\begin{array}\
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9 \\
\end{array}\right]$,</p>
<p>Define a $2 \times 2$ matrix that is the lower $2 \times 2$ block in $A$.</p>
| Dennis Jaheruddin | 45,737 | <p>If the size of <code>A</code> can vary, you can always get the lower 2x2 block as such:</p>
<pre><code>A(end-1:end,end-1:end)
</code></pre>
<p>Note that this always works as long as the width and height of <code>A</code> are both at least 2.</p>
|
15,013 | <p>I am doing a plot where I have multiple shaded regions, and I want the line that separates the two regions to be dashed with dashes being alternating colors (so the demarcation stands out from both regions).</p>
<p>For example, say I am plotting the two regions shown here</p>
<pre><code>Plot[{1, Abs[BesselJ[1, x]]}, {x, 0, 20}, Filling -> Axis]
</code></pre>
<p><img src="https://i.stack.imgur.com/z26zI.jpg" alt="enter image description here"></p>
<p>The only way I could think to add the dashing was using <code>ColorFunction</code>, but it doesn't give what I'm looking for: </p>
<pre><code>bgplot = Plot[{1, Abs[BesselJ[1, x]]}, {x, 0, 20},
Filling -> Axis, PlotStyle -> {Automatic, None}];
dashplot = Plot[Abs[BesselJ[1, x]], {x, 0, 20},
PlotStyle -> Thickness[.01],
ColorFunction -> (If[EvenQ[Floor[#]], Black, White]&),
ColorFunctionScaling -> False, PlotPoints -> 500];
Show[bgplot, dashplot]
</code></pre>
<p><img src="https://i.stack.imgur.com/QgojW.jpg" alt="enter image description here"></p>
<p>This is almost what I want, but the dashes are all the same length in the x-coordinate, whereas I'd prefer they be the same total length. Also, I have to have <code>PlotPoints</code> set to an unreasonably high value to avoid any gray regions. </p>
<p>Any ideas how to do this better?</p>
| Michael E2 | 4,999 | <p>Here's a way to use the <code>"ArcLength"</code> setting to get equal-length segments. One has to scale the coordinates so the arc length in the coordinate geometry is proportional to the arc length in the screen geometry. This can be done with <a href="http://reference.wolfram.com/language/ref/ScalingFunctions.html" rel="noreferrer"><code>ScalingFunctions</code></a>, if the <code>PlotRange</code> is known; otherwise, the plot range will need to be computed and the plot redrawn.</p>
<pre><code>Plot[Abs[BesselJ[1, x]],
{x, 0, 20},
Filling -> {1 -> Axis, 1 -> Top},
FillingStyle -> {Opacity[1/5, ColorData[1, 1]], Opacity[1/5, ColorData[1, 2]]},
PlotRange -> {0, 1}, PlotStyle -> Thickness[.01],
Mesh -> 18, MeshStyle -> None, MeshShading -> {Black, White},
MeshFunctions -> {"ArcLength"},
ScalingFunctions ->
{{GoldenRatio #/20 &, 20 #/GoldenRatio &}, (* scale by plot range & aspect ratios *)
None}
]
</code></pre>
<p><img src="https://i.stack.imgur.com/wRG1u.png" alt="Mathematica graphics"></p>
<p>Here's a two-pass function that computes the plot range and aspect ratio and inserts the appropriate scaling functions:</p>
<pre><code>ClearAll[arclengthmesh];
SetAttributes[arclengthmesh, HoldAll];
arclengthmesh[plot_] := Module[{p, pr, ar},
p = plot;
pr = Ratios[Differences /@ PlotRange@p][[1, 1]];
ar = AspectRatio /. Options[p, AspectRatio] /. Automatic -> 1/GoldenRatio;
ReleaseHold[
Insert[
Hold[plot],
{MeshFunctions -> {"ArcLength"},
ScalingFunctions -> {{pr*#/ar &, ar*#/pr &}, None}},
{1, 3}
]]
]
arclengthmesh@
Plot[Abs[BesselJ[1, x]],
{x, 0, 20},
Filling -> {1 -> Axis, 1 -> Top},
FillingStyle -> {Opacity[1/5, ColorData[1, 1]],
Opacity[1/5, ColorData[1, 2]]},
PlotRange -> {0, 1}, PlotStyle -> Thickness[.01],
Mesh -> 18, MeshStyle -> None, MeshShading -> {Black, White},
AspectRatio -> 1/3]
</code></pre>
<p><img src="https://i.stack.imgur.com/42DLZ.png" alt="Mathematica graphics"></p>
|
867,938 | <p>I was reading the book Linear Algebra Done Right by Axler. In the chapter on inner product space (Ch.6), he defines the norm of x on $R^n$ space as:</p>
<p>$||x|| = \sqrt{x_1^2 + ... + x_n^2}$</p>
<p>and says:</p>
<p>"The norm is not linear on $R^n$. To inject linearity into the discussion, we introduce the dot product."</p>
<p>I don't see why the norm is not linear.</p>
<p>If I check a multiplicity for $R^2$, using a scalar of 3 for example</p>
<p>$3||x|| =? ||3x||$ </p>
<p>$3 \sqrt{x_1^2 + x_2^2} =? \sqrt{(3 x_1)^2 + (3 x_2)^2} $</p>
<p>$3 \sqrt{x_1^2 + x_2^2} =? \sqrt{9(x_1^2 + x_2)^2} $</p>
<p>$3 \sqrt{x_1^2 + x_2^2} = 3 \sqrt{x_1^2 + x_2^2} $</p>
<p>Why is the norm not linear? </p>
<p>Axler then says:</p>
<p>"Also, if $y \in R^n$ is fixed, then clearly the map from $R^n$ to $R$ that sends $x \in R^n$ to $x \cdot y$ is linear."</p>
<p>Why is the dot product linear if the norm isn't?</p>
<p>Regards,
Madeleine.</p>
| mdp | 25,159 | <p>Try checking linearity of the norm with $-3$ as the scalar instead, or by checking the additive property - you should see that it fails.</p>
<p>The dot product is linear in the sense that the map $x\mapsto x\cdot y$ for some fixed $y$ (independent of $x$) is linear; the map $x\mapsto x\cdot x$ is <em>not</em> linear.</p>
|
1,287,692 | <p>To show whether or not the 3 planes
$$x+y-2z=5\tag 1$$
$$x-y+3z=6 \tag2$$
$$x+5y-12z=12 \tag 3$$ all have a common line of intersection.</p>
<p>Can I do $(3)-(2)$ to get the line $6y-15z=6$ and $(1)-(2)$ to get the line $2y-5z=-1$ which is $6y-15z=-3$ , and say that as these aren't the same line, they don't have a common line of intersection?</p>
<p>Another thing that is confusing me is that if instead of eliminating $x$, I chose to eliminate $z$, I would get different lines in terms of $x$ and $y$. But how can I get the equations of two different lines by eliminating from the same pair of plane equations? There's only one line of intersection between any pair of planes, so surely I should only be able to get one unique line if I eliminate a variable from a pair of planes?
Any help would be appreciated</p>
| Vlad | 229,317 | <p><strong>HINT</strong>: Find normal vectors of the planes and check if three of them are linearly independent. If there is a common line for all the planes, then their normal vectors will lie within the same plane, therefore three of them will <strong>not</strong> be linearly independent.</p>
<hr>
<p>In your specific case,
$$
x+y-2z=5\implies \vec{n}_1 = \begin{bmatrix} 1\\ 1\\ -2\end{bmatrix}\\
x-y+3z=6\implies \vec{n}_2 = \begin{bmatrix} 1\\ -1\\ 3\end{bmatrix}\\
x+5y-12z=5\implies \vec{n}_3 = \begin{bmatrix} 1\\ 5\\ -12\end{bmatrix}
$$
First, we need to check if the system of vector $\left\{\vec{n}_1, \vec{n}_2, \vec{n}_3 \right\}$ is clearly independent or not. The simplest way to do that is to compute rank of the matrix $\left[\vec{n}_1 \ \vec{n}_2\ \vec{n}_3 \right]$:
$$
\operatorname{rank}\Big(\left[\vec{n}_1 \ \vec{n}_2\ \vec{n}_3 \right]\Big) = \operatorname{rank}
\begin{pmatrix}
1 & 1& 1 \\
1 & -1 & 5 \\
-2& 3 & -12
\end{pmatrix}
$$</p>
<p>If $\ \operatorname{rank}\!\left(\vec{n}_1 \ \vec{n}_2\ \vec{n}_3 \right)=2$, then the normal vectors are linearly dependent, yet still span a plane. </p>
<p>Second, we need to find out if there is a point common for all three planes.</p>
<p>If you can find a common point <strong>and</strong> the rank of system of normal vectors is 3, then there is a line shared by all three planes.</p>
|
1,287,692 | <p>To show whether or not the 3 planes
$$x+y-2z=5\tag 1$$
$$x-y+3z=6 \tag2$$
$$x+5y-12z=12 \tag 3$$ all have a common line of intersection.</p>
<p>Can I do $(3)-(2)$ to get the line $6y-15z=6$ and $(1)-(2)$ to get the line $2y-5z=-1$ which is $6y-15z=-3$ , and say that as these aren't the same line, they don't have a common line of intersection?</p>
<p>Another thing that is confusing me is that if instead of eliminating $x$, I chose to eliminate $z$, I would get different lines in terms of $x$ and $y$. But how can I get the equations of two different lines by eliminating from the same pair of plane equations? There's only one line of intersection between any pair of planes, so surely I should only be able to get one unique line if I eliminate a variable from a pair of planes?
Any help would be appreciated</p>
| zoli | 203,663 | <p><strong>An elementary solution and notes to the OP</strong></p>
<p>If $3$ planes have a unique common point then they don't have a common straight line. In order to see if there is a common line we have to see if we can solve the following system of equations:</p>
<p>$$
\begin{matrix}
x+y&-2z&=&5\\
x-y&+3z&=&6\\
\ x+5y&-12z&=&12.
\end{matrix}
$$
Adding the first equation to the second one we get $$2x+z=11.$$
Multiplying the second equation by $5$ and then adding it to the third equation we get $$3x+z=21.$$
These two equations have a unique solution:
$$x=10\text { and } z=-9.$$
Substituting these numbers back to any of the original equations we get $y=-23$.</p>
<p>So, the three planes have a unique common point; no common line exists.</p>
<p><strong>Notes to the OP</strong></p>
<p>If you take, say, $(1)$ and $(2)$ and eliminate one of the variables, say $x$ then you get an equation of a straight line in the plane $zy$.</p>
<p>This line is a perpendicular projection of the common line of $(1)$ and $(2)$ to $yz$. <strong>If</strong> the planes $(1)$, $(2)$, and $(3)$ have a unique point then all of the possible eliminations will result in a triplet of straight lines in the different coordinate planes. </p>
<p>By erecting a perpendiculars from the common points of the said line triplets you will get back to the common point of the three planes. </p>
<p>I hope that this brief explanation helped you to understand better your own efforts.</p>
|
4,584,609 | <p>Which is bigger</p>
<p><span class="math-container">$$ \int_0^{\frac{\pi}{2}}\frac{\sin x}{1+x^2}dx$$</span> or <span class="math-container">$$ \int_0^{\frac{\pi}{2}}\frac{\cos x}{1+x^2}dx~?$$</span></p>
<p>I let <span class="math-container">$x=\frac{\pi}{2}-t$</span> in the second integral, and I obtain this
<span class="math-container">$$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1+(\frac{\pi}{2}-x)^2}dx$$</span>
But it is still to decide which is the bigger.</p>
| Claude Leibovici | 82,404 | <p>Suppose that you use the two <span class="math-container">$1,400^+$</span> years old approximations
<span class="math-container">$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$</span>
<span class="math-container">$$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad (-\frac \pi 2 \leq x\leq\frac \pi 2)$$</span>
the antiderivatives are simple and you obtain
<span class="math-container">$$\int_0^{\frac{\pi}{2}}\frac{\sin (x)}{1+x^2}dx\sim 0.527077$$</span>
<span class="math-container">$$\int_0^{\frac{\pi}{2}}\frac{\cos (x)}{1+x^2}dx\sim 0.748683$$</span> wile the exact values are respectively <span class="math-container">$0.526979$</span> and <span class="math-container">$0.749042$</span>.</p>
|
4,584,609 | <p>Which is bigger</p>
<p><span class="math-container">$$ \int_0^{\frac{\pi}{2}}\frac{\sin x}{1+x^2}dx$$</span> or <span class="math-container">$$ \int_0^{\frac{\pi}{2}}\frac{\cos x}{1+x^2}dx~?$$</span></p>
<p>I let <span class="math-container">$x=\frac{\pi}{2}-t$</span> in the second integral, and I obtain this
<span class="math-container">$$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1+(\frac{\pi}{2}-x)^2}dx$$</span>
But it is still to decide which is the bigger.</p>
| GEdgar | 442 | <p>Hardy-Littlewood-Polya ...<br />
Consider two nonnegative functions <span class="math-container">$f(x),g(x)$</span> on <span class="math-container">$[a,b]$</span>. Among all rearrangements
<span class="math-container">$f_1$</span> of <span class="math-container">$f$</span> and all rearrangements <span class="math-container">$g_1$</span> of <span class="math-container">$g$</span>, the largest value
of <span class="math-container">$\int_a^b f_1(x)g_1(x)\;dx$</span> occurs when both <span class="math-container">$f_1$</span> and <span class="math-container">$g_1$</span> are increasing (or both decreasing). The smallest value of <span class="math-container">$\int_a^b f_1(x)g_1(x)\;dx$</span> occurs when one of <span class="math-container">$f_1,g_1$</span> is increasing and the other decreasing, since <span class="math-container">$\sin x$</span> is increasing and <span class="math-container">$\cos x$</span> is decreasing.</p>
<p>In this case, since <span class="math-container">$1/(1+x^2)$</span> is decreasing on <span class="math-container">$[0,\pi/2]$</span> and <span class="math-container">$\sin(x), \cos(x)$</span> are rearrangements of each other, we see that
<span class="math-container">$$
\int_0^{\frac{\pi}{2}}\frac{\sin x}{1+x^2}dx
\quad\text{ is smallest, and }\quad
\int_0^{\frac{\pi}{2}}\frac{\cos x}{1+x^2}dx \quad\text{ is largest}
$$</span>
among all
<span class="math-container">$$
\int_0^{\pi/2}\frac{f_1(x)}{1+x^2}\;dx
$$</span>
as <span class="math-container">$f_1$</span> ranges over all rearrangements of <span class="math-container">$\sin x$</span> on <span class="math-container">$[0,\pi/2]$</span>.</p>
<hr />
<p>For the technical definition of "rearrangement" see the text</p>
<p><em>Hardy, G. H.; Littlewood, J. E.; Pólya, G.</em>, Inequalities., Cambridge: University Press. xii, 314 p. (1934). <a href="https://zbmath.org/?q=an:60.0169.01" rel="nofollow noreferrer">ZBL60.0169.01</a>.</p>
|
455,642 | <p>I'm coming from the programming world , and I need to create unique number for each element in a matrix. Say I have a $4\times4$ matrix $A$. I want to find a simple formula that will give each of the $16$ elements a unique number id. Can you suggest me where to start ? </p>
| pathfinder | 23,431 | <p>How about $$f(r,c) = Kr+c,$$ where $K$ is the number of columns in the matrix and $r$ and $c$ correspond to the row and column you want an ID for?</p>
<p>For your 4 by 4 matrix, we would have $$f(r,c) = 4r+c,$$ where we have $r,c\in\{0,1,2,3\}$. </p>
|
455,642 | <p>I'm coming from the programming world , and I need to create unique number for each element in a matrix. Say I have a $4\times4$ matrix $A$. I want to find a simple formula that will give each of the $16$ elements a unique number id. Can you suggest me where to start ? </p>
| Emily | 31,475 | <p>If you're coming from a programming background, you should know about column-major and row-major orderings.</p>
<p>In row-major order, the matrix $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ is laid out in memory as <code>[1, 2, 3, 4]</code>. In column-major order, it is <code>[1 3 2 4]</code>.</p>
<p>It should be easy to see how one can reverse the vector <code>[1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16]</code> into a matrix using row-major ordering (or column-major ordering--whichever!).</p>
|
264,130 | <p>I have been trying to find an analytical expression for the following:</p>
<p>$\frac{\partial {X^{+}}}{\partial {X}}$</p>
<p>In my case, $X$ has a constant rank.
I've found the formula for differentiating a pseudoinverse in <a href="http://i.stanford.edu/pub/cstr/reports/cs/tr/72/261/CS-TR-72-261.pdf" rel="nofollow noreferrer">Golub's paper</a> (equation 4.12): </p>
<p>$$
\frac{\mathrm d}{\mathrm d x} A^+(x) =
-A^+ \left( \frac{\mathrm d}{\mathrm d x} A \right) A^+
+A^+ A{^+}^T \left( \frac{\mathrm d}{\mathrm d x} A^T \right) (1-A A^+)
+ (1-A^+ A) \left( \frac{\mathrm d}{\mathrm d x} A^T \right) A{^+}^T A^+
$$</p>
<p>but I can't see how to input the original matrix.</p>
| rych | 45,979 | <blockquote>
<p>but I can't see how to input the original matrix.</p>
</blockquote>
<p>I think when people write $\tfrac {\partial f} { \partial X}$ or $\tfrac {df}{dX}$ they really mean to find the Fréchet derivative, denoted by $df$, or $Df$. Your formula becomes (appears also in <a href="https://math.stackexchange.com/questions/1689434/derivative-of-the-frobenius-norm-of-a-pseudoinverse-matrix">this answer</a>)
$$\mathrm d A^+ = -A^+ \left( \mathrm d A \right) A^+ +A^+ A{^+}^T \left( \mathrm d A^T \right) (1-A A^+) + (1-A^+ A) \left( \mathrm d A^T \right) A{^+}^T A^+$$</p>
<p>Since your differentiation is with respect to $A$ itself as a variable, then no chain rule needed, $dA=id$ and the derivative (at the given element $A$) is a linear map $H\mapsto [dA^+](H)$:</p>
<p>$$[\mathrm d A^+](H) = -A^+ \left( H \right) A^+ +A^+ A{^+}^T \left( H^T \right) (1-A A^+) + (1-A^+ A) \left( H^T \right) A{^+}^T A^+$$</p>
|
1,832,887 | <p>Consider the conjunction introduction and implication elimination rules of natural deduction:</p>
<p>$$\frac{\Gamma\vdash\alpha \quad \Gamma\vdash\beta}{ \Gamma\vdash \alpha \land \beta} (\land I)
\qquad
\frac{ \Gamma \vdash \alpha \to \beta \quad \Gamma \vdash \alpha} {\Gamma,\vdash\beta} (\to E) \qquad \text{(single)}$$</p>
<p>and note that the context $\Gamma$ of both premises of $(\to E)$ and $(\land I)$ must be the same.</p>
<p>Because this <em>need not be the case in general</em>, why not to write those rules like this instead:</p>
<p>$$\frac{\Gamma\vdash\alpha \quad \Delta\vdash\beta}{ \Gamma,\Delta\vdash \alpha \land \beta} (\land I') \qquad
\frac{ \Gamma \vdash \alpha \to \beta \quad \Delta \vdash \alpha} {\Gamma, \Delta\vdash\beta} (\to E') \qquad \text{(multiple)} $$</p>
<p>i.e. with the rules stated like this one might allow premises with distinct contexts.</p>
<p><strong>Questions</strong>:</p>
<ol>
<li><p>Should multiple premises of a natural deduction inference rule always have the same context?</p></li>
<li><p>In spite of their generality, why most (if not all) textbook or canonical presentations of the inference rules of the natural deduction refrain from using $\text{(multiple)}$-like rules? Because they are less didactical?</p></li>
<li><p>Aren't $\text{(multiple)}$-like rules valid as well in the natural deduction? </p></li>
</ol>
<p>Thanks!</p>
| lab bhattacharjee | 33,337 | <p>HINT:</p>
<p>Let $$\dfrac{p(x)}{(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)}=\sum_{i=1}^6\dfrac{A_i}{x-i}$$</p>
<p>Multiply both sides by $(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)$ and put $x=1,2,3,4,5,6$ one by one in the resultant identity.</p>
|
1,832,887 | <p>Consider the conjunction introduction and implication elimination rules of natural deduction:</p>
<p>$$\frac{\Gamma\vdash\alpha \quad \Gamma\vdash\beta}{ \Gamma\vdash \alpha \land \beta} (\land I)
\qquad
\frac{ \Gamma \vdash \alpha \to \beta \quad \Gamma \vdash \alpha} {\Gamma,\vdash\beta} (\to E) \qquad \text{(single)}$$</p>
<p>and note that the context $\Gamma$ of both premises of $(\to E)$ and $(\land I)$ must be the same.</p>
<p>Because this <em>need not be the case in general</em>, why not to write those rules like this instead:</p>
<p>$$\frac{\Gamma\vdash\alpha \quad \Delta\vdash\beta}{ \Gamma,\Delta\vdash \alpha \land \beta} (\land I') \qquad
\frac{ \Gamma \vdash \alpha \to \beta \quad \Delta \vdash \alpha} {\Gamma, \Delta\vdash\beta} (\to E') \qquad \text{(multiple)} $$</p>
<p>i.e. with the rules stated like this one might allow premises with distinct contexts.</p>
<p><strong>Questions</strong>:</p>
<ol>
<li><p>Should multiple premises of a natural deduction inference rule always have the same context?</p></li>
<li><p>In spite of their generality, why most (if not all) textbook or canonical presentations of the inference rules of the natural deduction refrain from using $\text{(multiple)}$-like rules? Because they are less didactical?</p></li>
<li><p>Aren't $\text{(multiple)}$-like rules valid as well in the natural deduction? </p></li>
</ol>
<p>Thanks!</p>
| Ghartal | 83,884 | <p>Let's do it in the most elementary way. Let $$Q(x)=P(x+1)-P(x)-x+2 \tag{1}$$Observe that $Q$ is of degree $4$ and $Q(3)=Q(4)=Q(5)=0$. Therefore we can write$$Q(x)=a(x-3)(x-4)(x-5)(x-b) \tag{2}$$You have also from $(1)$ that $Q(1)=Q(2)=1$, which after substitution in $(2)$ you get $a=-1/8$ and $b=2/3$. So $$Q(6)=-\frac{1}{8}(6-3)(6-4)(6-5)\left(6-\frac{2}{3} \right)=-4$$And finally$$P(7)=Q(6)+P(6)+6-2=-4+8+6-2=8$$</p>
|
881,013 | <p>I am still an undergraduate student, and so perhaps I just haven't seen enough of the mathematical world. </p>
<p><strong>Question:</strong> What are some examples of mathematical logic solving open problem outside of mathematical logic?</p>
<p>Note that the <a href="//en.wikipedia.org/wiki/Ax%E2%80%93Grothendieck_theorem" rel="nofollow noreferrer">Ax-Grothendieck Theorem</a> <em>would have been</em> a perfect example of this (namely, If $P$ is a polynomial function from $\mathbb{C}^n$ to $\mathbb{C}^n$ and $P$ is injective then $P$ is bijective). However, even though there is a beautiful logical proof of this result, it was first proven not specifically using mathematical logic. I'm curious as to whether there are any results where "the logicians got there first".</p>
<p><strong>Edit 1:Bonus</strong>: I am quite curious if one can post an example from Reverse Mathematics. </p>
<p><strong>Edit 2:</strong><a href="https://math.stackexchange.com/questions/886848/why-exactly-is-whiteheads-problem-undecidable">This post</a> reminded me that the solution to <a href="http://en.wikipedia.org/wiki/Whitehead_problem" rel="nofollow noreferrer">Whitehead's Problem</a> came from logic (a problem in group theory). According to the wikipedia article, the proof by Shelah was 'completely unexpected'. It utilizes the fact that <strong>ZFC+(V=L)</strong> implies every Whitehead group is free while <strong>ZFC+$\neg$CH+MA</strong> implies there exists a Whitehead group which is not free. Since these two separate axiom systems are equiconsistent, hence Whitehead's problem is undecidable. </p>
<p><strong>Edit 3:</strong> A year later, we have some more examples: </p>
<p>1) Hrushovski's Proof of the Mordell-Lang Conjecture for functional fields in all characteristics. </p>
<p>2) The Andre-Óort conjecture by Pila and Tsimerman.</p>
<p>3) Various results in O-minimality including work by Wilkie and van den Dries (as well as others). </p>
<p>4) Zilber's Pseudo-Exponential Field as work towards Schanuel's conjecture. </p>
| bof | 111,012 | <p>The <a href="https://en.wikipedia.org/wiki/James_Earl_Baumgartner" rel="nofollow noreferrer">Baumgartner</a>–<a href="https://en.wikipedia.org/wiki/Andr%C3%A1s_Hajnal" rel="nofollow noreferrer">Hajnal</a> theorem says that, for any countable ordinal $\alpha,$ and for any coloring of the edges of the complete graph on the vertex set $\omega_1$ (the first uncountable ordinal) with finitely many colors, there is a monochromatic complete subgraph of order type $\alpha.$ This theorem of classical set theory was proved using the method of <a href="https://en.wikipedia.org/wiki/Forcing_(mathematics)" rel="nofollow noreferrer">forcing</a> (in the form of <a href="https://en.wikipedia.org/wiki/Martin%27s_axiom" rel="nofollow noreferrer">Martin's axiom</a>) and <a href="https://en.wikipedia.org/wiki/Absoluteness" rel="nofollow noreferrer">absoluteness</a>. It seems reasonable to consider forcing and absoluteness as logical, and classical set theory as non-logical.</p>
|
672,265 | <p>Find the normal form of the matrix $A$: $$\begin{bmatrix}2 & 0 & 0 & 0\\1 & 2 & 0 & 0\\0 & 0 & 1 & 1\\0 & 0 & 0 & 1\end{bmatrix}$$</p>
<p>It looks like A's jordan decomposition should be: </p>
<p>$$\begin{bmatrix}2 & 1 & 0 & 0\\0 & 2 & 0 & 0\\0 & 0 & 1 & 1\\0 & 0 & 0 & 1\end{bmatrix}$$</p>
<p>And from here it's easy to find the change of basis matrix that Jordanizes' $A$.</p>
<p>My question is this: What do I add to make this proof formal and acceptable?</p>
<p>Thanks for your time.</p>
| EuYu | 9,246 | <p>A quick argument would be something along these lines.</p>
<p>You have a block matrix
$$C=\begin{pmatrix}A & 0 \\ 0 & B\end{pmatrix}$$
So the Jordan form of $C$ is just the direct sum of the Jordan forms of $A$ and $B$. Since $B$ is already in Jordan form, we only have to worry about $A$.</p>
<p>Now $A$ is the transpose of a Jordan block, and a matrix is similar to its transpose. Therefore you can conclude that the Jordan form of $A$ is the same as the Jordan form of $A^\mathrm{T}$, which is just $A^\mathrm{T}$ itself.</p>
|
4,242,877 | <p>I was reading the book of Real Analysis by H.L.Royden and they have given the definition of Riemann sums and Riemann integral as follows:<a href="https://i.stack.imgur.com/cFfRc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cFfRc.jpg" alt="enter image description here" /></a></p>
<p>After this definition they have given the following footnote<a href="https://i.stack.imgur.com/qhafB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qhafB.jpg" alt="enter image description here" /></a></p>
<p>Now I am not getting what the footnote 1 is saying. I think the statement of that note 1 is not always true. For example if we take the Dirichlet function (which is 0 on rationals and 1 on irrationals ) then the footnote 1 is not true. Am I correct or am I missing something?</p>
<p>(Edit:) My thoughts on why the foot note does not work for Dirichlet function:</p>
<p>We define <span class="math-container">$f:[0,1]\rightarrow \mathbb R$</span> by <span class="math-container">$f(x)=0$</span> if <span class="math-container">$x\in\mathbb Q$</span> and <span class="math-container">$f(x)=1$</span> otherwise.
Now consider any partition <span class="math-container">$P$</span> of <span class="math-container">$[0,1]$</span>. Then for any subinterval <span class="math-container">$[x_{i-1},x_i]$</span> we get <span class="math-container">$m_i=0$</span> because each subinterval will definitely have infinitely may rationals. Also <span class="math-container">$M_i=1$</span> because each subinterval will definitely have infinitely many irrationals. Then Lower sum will be 0 and upper sum will be 1.</p>
| Danny | 184,313 | <p>If <span class="math-container">$f$</span> is the Dirichlet function, then we get the same values for the upper and lower integrals if we use the first definitions of <span class="math-container">$m_i$</span> and <span class="math-container">$M_i$</span>. If <span class="math-container">$x_{i-1} < x_i$</span>, then any interval from <span class="math-container">$x_{i-1}$</span> to <span class="math-container">$x_i$</span> (whether it is (half) open or (half) closed) will contain both a rational and an irrational number; in fact infinitely many of both. If you are not convinced that this is the case for open intervals, then an argument would be: If <span class="math-container">$[x_{i-1}, x_i]$</span> contains infinitely many (ir)rationals, then <span class="math-container">$(x_{i-1}, x_i)$</span> does as well, since we have only removed finitely many points from the interval, namely the endpoints.</p>
<hr />
<p>Let us call the first definition using strict inequalities "definition 1" and the other definition "definition 2", and let us write <span class="math-container">$m_i$</span> and <span class="math-container">$m_i'$</span> respectively in the two definitions. Similarly, write <span class="math-container">$L$</span> and <span class="math-container">$L'$</span> for lower sums and <span class="math-container">$\underline I$</span> and <span class="math-container">$\underline I'$</span> for lower integrals, where we use <span class="math-container">$m_i$</span> and <span class="math-container">$m_i'$</span> respectively. The footnote is saying that, while <span class="math-container">$m_i(f)$</span> and <span class="math-container">$m_i'(f)$</span> might not generally agree, we still have <span class="math-container">$\underline I(f) = \underline I'(f)$</span>.</p>
<p>We need a few facts about these integrals: Clearly we have <span class="math-container">$L'(f,P) \leq L(f,P)$</span>, since we take infimums over larger sets on the left-hand side. It follows that <span class="math-container">$\underline I'(f) \leq \underline I(f)$</span>.</p>
<p>Let <span class="math-container">$f \colon [a,b] \to \mathbb{R}$</span> be bounded. If <span class="math-container">$P$</span> is a partition of <span class="math-container">$[a,b]$</span>, let <span class="math-container">$g_P \colon [a,b] \to \mathbb{R}$</span> be the function that agrees with <span class="math-container">$f$</span> on the interiors of each subinterval of <span class="math-container">$P$</span>, and let <span class="math-container">$g_P(x_i) = \sup_{x \in [a,b]} f(x)$</span>. Then <span class="math-container">$m_i(f) = m_i'(g_P)$</span>, using obvious notation, and so <span class="math-container">$L(f,P) = L'(g_P,P)$</span>.</p>
<p>Now let <span class="math-container">$\epsilon > 0$</span>, and choose a partition <span class="math-container">$P$</span> such that <span class="math-container">$\underline I(f) - L(f,P) < \epsilon$</span>. Since <span class="math-container">$g_P$</span> arises from <span class="math-container">$f$</span> by modifying <span class="math-container">$f$</span> in finitely many points, we have <span class="math-container">$\underline I(f) = \underline I(g_P)$</span> (see below). Furthermore, <span class="math-container">$\underline I(f) = \underline I(g_P) \geq \underline I'(g_P)$</span>, so <span class="math-container">$0 \leq \underline I'(g_P) - L'(g_P,P) < \epsilon$</span>. It follows that</p>
<p><span class="math-container">$$ \begin{align}
0 \leq \underline I(f) - \underline I'(f)
& = \underline I(f) - \underline I'(g_P) \\
& = \bigl( \underline I(f) - L(f,P) \bigr) - \bigl( \underline I'(g_P) - L'(g_P,P) \bigr) \\
& < \epsilon.
\end{align} $$</span></p>
<p>And since <span class="math-container">$\epsilon$</span> was arbitrary, <span class="math-container">$\underline I(f) = \underline I'(f)$</span>. And clearly we can do the same for upper integrals, defining <span class="math-container">$g_P$</span> slightly differently.</p>
<hr />
<p>We just need to justify that modifying a function in finitely many points does not change the upper/lower integrals. It suffices to show that the integrals do not change if we modify the function at a single point. The case of finitely many points then follows by induction.</p>
<p>Let me sketch the proof for the upper integral. It is not difficult to show that</p>
<p><span class="math-container">$$ \overline {\int_a^b} (f + g) \leq \overline {\int_a^b} f + \overline {\int_a^b} g, $$</span></p>
<p>for bounded <span class="math-container">$f$</span> and <span class="math-container">$g$</span>. Let <span class="math-container">$g$</span> be identically zero except at a single point <span class="math-container">$c \in (x_{i-1}, x_i)$</span>, at which <span class="math-container">$g(c) > 0$</span>. Then by choosing partitions such that <span class="math-container">$x_i - x_{i-1}$</span> is very small we can make <span class="math-container">$M_i(g)$</span> as small as we want. Hence <span class="math-container">$\overline {\int_a^b} g = 0$</span>. It follows that</p>
<p><span class="math-container">$$ \overline {\int_a^b} f \leq \overline {\int_a^b} (f + g) \leq \overline {\int_a^b} f, $$</span></p>
<p>where the first inequality is due to the upper integral being increasing in the integrand.</p>
|
339,289 | <blockquote>
<p>Prove that if $f$ is defined for $x\ge 0$ by $f(x)=\sqrt x$, then $f$
is continuous at every point of its domain.</p>
</blockquote>
<p>Definition of a continuous function is:</p>
<p>Let $A\subseteq\mathbb{R}$ and let $f:A\to\mathbb{R}$. Denote $c\in A$.</p>
<p>Then $f(x)$ is continuous at $c$ iff for every $\varepsilon>0$, $\exists$ $\delta>0$ such that </p>
<p>$|x-c|<\delta\implies |f(x)-f(c)|<\varepsilon.$</p>
<p>My attempt:</p>
<p>We know that the function $f: x\to \mathbb{R}$, where $x\in [0,\infty)$ is defined to be $f(x)=\sqrt x$. So, for $0\le x<\infty$, then
$|f(x)-f(c)|=|\sqrt x - f(c)|$ and I can't continue since I don't necessarily know what $c$ is in this case.</p>
| user62089 | 62,089 | <p>Let us go from the definition you gave:</p>
<p><span class="math-container">$f(x)$</span> is continuous at <span class="math-container">$c$</span> iff for every <span class="math-container">$\varepsilon>0$</span>, <span class="math-container">$\exists$</span> <span class="math-container">$\delta>0$</span> such that </p>
<p><span class="math-container">$|x-c|<\delta\implies |f(x)-f(c)|<\varepsilon.$</span></p>
<p>Given <span class="math-container">$\varepsilon > 0$</span> we must show that <span class="math-container">$\vert \sqrt x - \sqrt c \vert < \varepsilon$</span> provided that <span class="math-container">$x$</span> and <span class="math-container">$c$</span> are close enough.</p>
<p><span class="math-container">$$\vert \sqrt x - \sqrt c \vert = \frac {\vert x - c \vert} {\vert \sqrt x + \sqrt c \vert} < \frac {\vert x - c \vert} {\sqrt c} $$</span>
So, <span class="math-container">$\delta = \varepsilon \sqrt{c}$</span> will do.</p>
<p>As pointed out by Ian, the above argument fails for <span class="math-container">$c = 0$</span>. For <span class="math-container">$c = 0$</span>, given <span class="math-container">$\varepsilon >0$</span>, <span class="math-container">$\delta = \varepsilon^2$</span> would suffice.</p>
|
3,118,282 | <p>I'm trying to do the following problem in my book, but I don't understand how the book got their answer.</p>
<p>The problem:
Determine whether the following relations are equivalence relations:<span class="math-container">$\newcommand{\relR}{\mathrel{R}}$</span></p>
<p>The relation <span class="math-container">$\relR$</span> on <span class="math-container">$\mathbb{R}$</span> given by <span class="math-container">$x\relR y$</span> if and only if <span class="math-container">$|x-y|\leq1$</span>. </p>
<p>The answer only says it isn't transitive and gives this example:
<span class="math-container">$(1\relR2)\wedge(2\relR3)$</span>, but <span class="math-container">$1\not\relR 3$</span>. Where did they get those numbers from?</p>
<p>As for the problem being reflexive and symmetric, please correct me if I'm wrong but here is what I assume it to be:</p>
<p>Reflexive: For any <span class="math-container">$x$</span> such that <span class="math-container">$x\relR x\Rightarrow x\leq1$</span></p>
<p>Symm: For any <span class="math-container">$x$</span>, <span class="math-container">$y$</span> such that <span class="math-container">$x\relR y\Rightarrow|x-y|\leq1\text{ and }1\geq|x-y|$</span></p>
| Nico | 101,332 | <p>Let <span class="math-container">$xRy$</span> if and only if <span class="math-container">$|x-y|\le 1$</span>.</p>
<p><strong>Reflexivity</strong>: Notice <span class="math-container">$|x-x|=0\le 1$</span>. Thus <span class="math-container">$xRx$</span>.</p>
<p><strong>Symmetry</strong>: Let <span class="math-container">$xRy$</span>. Then <span class="math-container">$|x-y|=|y-x|=1$</span>, which implies <span class="math-container">$yRx$</span>.</p>
<p>The numbers for the proof that <span class="math-container">$R$</span> is not transitive are cooked up (there are many such examples), but you can see why there is a problem:</p>
<p><span class="math-container">$|1-2|=1\le 1$</span>, which proves that <span class="math-container">$1R2$</span>. Similarly, <span class="math-container">$|2-3|\le 1$</span>, so <span class="math-container">$2R3$</span>. But then we can compute <span class="math-container">$|1-3|=2\not\le 1$</span>, so <span class="math-container">$1\not R 3$</span>.</p>
<p>This is contradicts the requirement of transitivity, which would imply (in particular) that
<span class="math-container">$$1R2\;\wedge\; 2R3\quad\Rightarrow\quad 1R3.$$</span></p>
|
907,055 | <p>I've begun a course in "Real Analysis" recently and I have this trivial exercise. Could someone check if my proof is correct?</p>
<p>Proposition: There exists Injective function $ f: A \rightarrow B \iff $ there exists function $ g: B \rightarrow A $ is surjective</p>
<p>Proof: Firstly, we prove injective function $f: A \rightarrow B \Longrightarrow g: B \rightarrow A$ is surjective Suppose $\exists f: A \rightarrow B, $ such that $ f$ is injective, i. e., $ \forall x_{1}, x_{@} \in A, x_{1} \neq x_{2} \rightarrow f(x_{1}) \neq f(x_{2})$. </p>
<p>By hypothesis, $\exists g: B \rightarrow A$ such that $g$ is not surjective. Then,there is at least one $ x \in A $ such that $ \forall y \in B, g(y) \neq x $. But, that is not possible, because if $f$ is injective, then all $x \in A$ correspond to some $y \in B$. Contradiction!</p>
<p>Now, we prove surjective function $g: B \rightarrow A \Longrightarrow f: A \rightarrow B$ is injective. Suppose $g: B \rightarrow A $ is surjective, i. e., $\forall y \in B, \exists x \in A$, such that $ g(y) = x$. By hypothesis, $\exists f: A \rightarrow B$ such that f is not injective. Then, there are $x_{1}, x_{2} \in A$ such that for $x_{1} \neq x_{2}$, there are $f(x_{1}) = f(x_{2})$. By the definition of function, that only could happen, if there is $ y \in B $ such that $ y \notin Dom(g) $. Contradiction!</p>
<p>So, There exists Injective function $ f: A \rightarrow B \iff $ there exists function $ g: B \rightarrow A $ is surjective. Q.E.D.</p>
| Steve Kass | 60,500 | <p>No, this proof is not correct. First of all, you misstate what you set out to prove in each half. “Firstly, we prove...” says you will prove that something implies $g$ is surjective. But at this point, $g$ is not something that has been defined. There is a big difference between “$g$ is surjective” and “There exists a surjective function $g$.” You need to prove the latter.</p>
<p>Next, in each part of your proof, you say “By hypothesis, there exists...” You have no hypothesis that allows you to say this. In the “Firstly” part, for example, your only hypothesis is that there exists an injective function $f$ from $A$ to $B$. This does allow you to say there exists a non-surjective function from $B$ to $A$.</p>
|
2,329,475 | <p>Google returns that response, not sure why is a complex number.</p>
| Evargalo | 443,536 | <p>If you remember the famous 'Euler' formula:</p>
<p>$e^{\pi i}+1=0$</p>
<p>And assuming $\ln$ verify $\ln(\exp(z)) = z$, you get the result $\ln(-1)=\pi i$.</p>
|
1,428,319 | <p>If we have a non-zero matrix $A\in\mathrm{Mat}_n(\mathbb{R})$ that is non-invertible. How do we prove that $A$ is both a left and right zero divisor?</p>
| ajotatxe | 132,456 | <p>If $a$ is not a multiple of $2$ or $3$, then $a=6m\pm 1$ and then
$$a^2+23=36m^2\pm12m+1+23=12(3m^2\pm m+2)$$
so it suffices to show that $3m^2\pm m$ is even. And indeed it is, because $3m^2\pm m=m(3m\pm 1)$, and one of these two factors is even.</p>
|
1,428,319 | <p>If we have a non-zero matrix $A\in\mathrm{Mat}_n(\mathbb{R})$ that is non-invertible. How do we prove that $A$ is both a left and right zero divisor?</p>
| Gabriel Romon | 66,096 | <p>$a^2+23=(a-1)(a+1)+24$</p>
<p>Since $a$ is odd; $a-1$ and $a+1$ are even. </p>
<p>Moreover, since $4$ does not divide $a$, 4 divides either $a-1$ or $a+1$.</p>
<p>Since $3$ does not divide $a$, $3$ divides either $a-1$ or $a+1$.</p>
<p>Hence $2\times 4\times 3$ divides $(a-1)(a+1)$.</p>
|
2,465,803 | <p>Given a random permutation <span class="math-container">$\sigma \in S_n$</span> from <span class="math-container">$[n] \to [n]$</span> in a uniform probability space, what is the probability that <span class="math-container">$\sigma $</span> has exactly <span class="math-container">$k$</span> fixed points for a given <span class="math-container">$k$</span> between <span class="math-container">$1$</span> and <span class="math-container">$n$</span>?</p>
<p>In other words: what is the probability that <span class="math-container">$\exists x_1 ,...,x_k \in [n] : \sigma (x_i) = x_i $</span> for <span class="math-container">$\ i\in \{1,...,k\}$</span> and for every <span class="math-container">$y \notin \{x_1 , ... , x_k\}$</span> we get <span class="math-container">$\sigma(y) \neq y$</span>.</p>
<p>I saw that <span class="math-container">$\lim_{n \to \infty } prob(A_0) = e^{-1}$</span> using Inclusion–exclusion principle and i belive that for a given k : <span class="math-container">$\lim_{n \to \infty} prob(A_k) = \frac{e^{-1}}{k!}$</span> but I am not sure how to show it.</p>
<p>*<span class="math-container">$A_k$</span> stands for the event "k".</p>
| Jaroslaw Matlak | 389,592 | <ol>
<li>Number of dearrangements of $k$-elements set is
$$!k = k!\sum\limits_{m=0}^{k}\frac{(-1)^m}{m!}$$</li>
<li>In n-elements set we can select $(n-k)$ elements to dearrange them (all remaining $k$ points are fixed) in $\binom{n}{k}$ ways</li>
</ol>
<p>Thus
$$A_k^n = \binom{n}{k}(n-k)!\sum\limits_{m=0}^{n-k}\frac{(-1)^m}{m!}$$
And probability is
$$P(A_k^n) = \frac{\binom{n}{k}(n-k)!\sum\limits_{m=0}^{n-k}\frac{(-1)^m}{m!}}{n!}=\frac{\frac{n!}{k!}\sum\limits_{m=0}^{n-k}\frac{(-1)^m}{m!}}{n!}=\frac{1}{k!}\sum\limits_{m=0}^{n-k}\frac{(-1)^m}{m!}$$</p>
|
3,132,404 | <p>For the purpose of this question, <span class="math-container">$A'$</span> is the derived set of set <span class="math-container">$A$</span>, <span class="math-container">$A^\circ$</span> is the interior of set <span class="math-container">$A$</span>, and <span class="math-container">$A^{c}$</span> is the complement of set <span class="math-container">$A$</span>.</p>
<p>As we know, <span class="math-container">$A$</span> is closed if and and only if <span class="math-container">$A^{c}$</span> is open. To paraphrase,</p>
<p><span class="math-container">$$\begin{align}
A' \subset A &\iff A^{c} \subset (A^{c})^\circ \\
A^{c} \subset (A')^{c} &\iff A^{c} \subset (A^{c})^\circ
\end{align}$$</span></p>
<p>As a result, we have</p>
<p><span class="math-container">$$\forall A, (A')^{c} = (A^{c})^\circ$$</span></p>
<p>This doesn't seem to be right however, because</p>
<p><span class="math-container">$$\begin{align}
A' &= \{x : \forall \delta > 0, U^\circ(x, \delta) \cap A \ne \varnothing \} \\
(A')^{c} &= \{x : \exists \delta > 0, U^\circ(x, \delta) \cap A = \varnothing \} \\
(A^{c})^\circ &= \{x : \exists \delta > 0, U(x, \delta) \cap A = \varnothing \} \\
\end{align}$$</span></p>
<p>Obviously, <span class="math-container">$U^\circ(x, \delta)^\ddagger$</span> is not equivalent to <span class="math-container">$U(x, \delta)$</span>, yet according to my reasoning <span class="math-container">$(A')^{c}$</span> and <span class="math-container">$(A^{c})^\circ$</span> are equal for all <span class="math-container">$A$</span>. Are these two sets really equal?</p>
<p><sub><span class="math-container">$\ddagger$</span>: <span class="math-container">$U^\circ$</span> stands for a <a href="https://en.wikipedia.org/wiki/Neighbourhood_(mathematics)#Deleted_neighbourhood" rel="nofollow noreferrer">deleted neighbourhood</a>.</sub></p>
| little o | 543,867 | <p><span class="math-container">$A' \subseteq \overline A.$</span> So <span class="math-container">$(\overline A)^c \subseteq (A')^{c}.$</span> Now we know that <span class="math-container">$(\overline A)^c = (A^c)^{\circ}.$</span> So we have <span class="math-container">$(A^c)^{\circ} \subseteq (A')^{c}.$</span></p>
|
2,915,786 | <p>I started writing a proof using the method of proof by contradiction and encountered a situation which was true. More specifically, the hypothesis that I set out to prove was:</p>
<p>If the first 10 positive integer is placed around a circle, in any order, there exists 3 integer in consecutive locations around the circle that have a sum greater than or equal to 17. (From Discrete Mathematics and its Applications - K. Rosen)</p>
<p>This is how I proceeded:</p>
<p>Let $a_i$ denote the $i^{th}$ integer on the boundary of the circle. To proceed with proof by contradiction, we assume that $\forall i$</p>
<p>$a_i + a_{i+1} + a_{i+2} < 17$</p>
<p>Then, </p>
<p>$a_1 + a_2 + a_3 < 17$</p>
<p>$a_2 + a_3 + a_4 < 17$</p>
<p>$\vdots$</p>
<p>$a_{10} + a_1 + a_2 < 17$</p>
<p>$\therefore\ 3 \cdot (a_1 + a_2 + \dots + a_{10}) < 17 \cdot10$</p>
<p>$\Rightarrow\ 3 \cdot 55 < 170$ </p>
<p>$\Rightarrow\ 165 < 170$ </p>
<p>which is true. What does this mean? </p>
<p>P.S. I am not looking for the solution to this problem. I am aware of how to prove the claim. I am just curious about what it means to arrive at a truth after assuming the negation of the hypothesis. </p>
| Siong Thye Goh | 306,553 | <p>Your goal is to show that $p$ is false. </p>
<p>If $p \implies q$ and $q$ is true. </p>
<p>We can't conclude if $p$ is true or false. Hence, we get an inconclusive situation. </p>
|
2,296,968 | <p>So far I have this: Let P(n) be the statement "$n2^n \lt n!$". $k_{0}=6$. $(6)2^6 = 384 <720=6!$. $P(k_{0})$ is true. Let $n \geq 6$ and assume P(n) to be true. By the induction hypothesis, $(n+1)2^{n+1}=(n+1)(2)2^n ...$ Somehow this gets to be $n!(n+1)(2)2^n$. I am clearly missing a few steps in my proof. </p>
| Community | -1 | <p>Assume this holds:</p>
<p>$$
n2^n < n!\qquad(1)
$$</p>
<p>Multiplying (1) by $n+1$:</p>
<p>$$n(n+1)2^n < (n+1)!\qquad(2)$$</p>
<p>But $n > 2$ so from (2):</p>
<p>$$(n+1)2*2^n < n(n+1)2^n < (n+1)!$$</p>
|
3,845,756 | <p>I am studying sequences in <span class="math-container">$\mathbb{R}$</span>. I would like to know if a sequence satisfies that
"For all <span class="math-container">$n \in \mathbb{N}, |a_{n+1} - a_{n}| \leq \frac{1}{3^{n}}$</span>" then sequence <span class="math-container">$\left\{a_{n}\right\}_{n \in \mathbb{N}}$</span> is convergent is true. I have looked for lots of examples and I have the same conditions for convergent sequences. Any counterexample?</p>
| J. W. Tanner | 615,567 | <p>It's not true.</p>
<p>A counterexample is <span class="math-container">$a_n=\sqrt n$</span>.</p>
|
1,415,214 | <p>I'm supposed to calculate the derivative of $\frac{d}{dx}\int_{x^{2}}^{x^{8}}\sqrt{8t}dt$ the answer I got is $8x^7\cdot \sqrt{8x^8}$ but when I put this into the grading computer it is marked wrong. I've tried solving in several different ways and always got the same answer so I used my graphing calculator to test.</p>
<p>I solved for $\frac{d}{dx}\int_{x^{2}}^{x^{8}}\sqrt{8t}dt$ with my graphing calculator's derivative and integral functions, with $x=3$. The result was $40083336.292$. The I tried solving with my formula, again plugging in $3$ for $x$. The result was $4008379.039$.</p>
<p>Very similar, but not the same. Is my formula wrong or is it a bug in the grading program?</p>
<p>Thanks for any help. If I solved this wrongly please explain where I went wrong. Thanks.</p>
| randomgirl | 209,647 | <p>$g(x)=\int_{x^2}^{x^8} f(t) dt=F(t)|_{x^2}^{x^8}=F(x^8)-F(x^2)$ and so differentiating we have $g'(x)=(x^8)'f(x^8)-(x^2)'f(x^2)$</p>
|
1,415,214 | <p>I'm supposed to calculate the derivative of $\frac{d}{dx}\int_{x^{2}}^{x^{8}}\sqrt{8t}dt$ the answer I got is $8x^7\cdot \sqrt{8x^8}$ but when I put this into the grading computer it is marked wrong. I've tried solving in several different ways and always got the same answer so I used my graphing calculator to test.</p>
<p>I solved for $\frac{d}{dx}\int_{x^{2}}^{x^{8}}\sqrt{8t}dt$ with my graphing calculator's derivative and integral functions, with $x=3$. The result was $40083336.292$. The I tried solving with my formula, again plugging in $3$ for $x$. The result was $4008379.039$.</p>
<p>Very similar, but not the same. Is my formula wrong or is it a bug in the grading program?</p>
<p>Thanks for any help. If I solved this wrongly please explain where I went wrong. Thanks.</p>
| Bernard | 202,857 | <p>The derivative is
$$8x^7\sqrt{8x^8}-2x\sqrt{8x^2}=4\sqrt 2\bigl(4x^{11}-x\lvert x\rvert\bigr).$$
You have to derive also w.r.t. the lower bound of integration.</p>
|
252,511 | <p>I am interested in the details of Elie Cartan's thesis, and, more specifically the explicit construction of the exceptional Lie groups as groups of symmetries of some specific homogeneous polynomials (according to what I have read in many places). I am interested in the details. For instance, what does one such a polynomial look like? What do all such polynomials (with groups of symmetry a fixed exceptional Lie group) look like? Is the set of all such polynomials dense in the relevant polynomial space? </p>
<p>Also, I am interested in different realizations of the exceptional Lie groups: for instance, $G_2$ is the group of symmetries of a skew-symmetric trilinear form on $\mathbb{C}^7$. And in this case, the space of such skew-symmetric trilinear forms is dense in the set of all skew-symmetric trilinear forms on $\mathbb{C}^7$.</p>
<p>So I guess in general, I am interested in concrete and explicit realizations of the exceptional Lie groups as groups of symmetries of a specific algebraic object. Also what is the set of all such objects with this property? And, is the set of all such objects dense in the relevant vector space? I hope my questions are sufficiently clear!</p>
| Alexandre Eremenko | 25,510 | <p>All details about Cartan's thesis can be found in the thesis itself:</p>
<p><a href="https://archive.org/details/surlastructured00bourgoog" rel="nofollow">https://archive.org/details/surlastructured00bourgoog</a></p>
<p>There is also a German translation for those who do not read French:</p>
<p>Ueber die einfachen Transformationsgruppen. (German) JFM 25.0638.01
Leipz. Ber. XLV. 395-420 (1893).</p>
|
3,623,724 | <p>The game works as follows: assume that there are two players, 1 and 2. 1 decides to lie or tell the truth. If 1 tells the truth, 2 needs to decide to take her medications or not. Later, in case 2 did not take her medications or 1 lied to her, 2 needs to decide to go to the hospital or not.
If 1 lied to her and she went to the hospital, then the utility of the 1 player is 0 and the second player's utility is 1. If 1 lied to her but she did not go to the hospital, then the first player's utility is 1 and the second player's utility is 0. If 1 told her the truth and she did not take her medications, then her going to hospital generates 1 utility to the first players and 0 utility to the second; not going to the hospital gives both of them 0. If 2 takes her medications after 1 told the truth, then they both get 2.
Okay, I used backward induction and this is what I got: if player 1 lies, 2 goes to the hospital. If player 1 tells the truth, player 2 is better off taking the meds. My question is: when stating my SPNE, should I mention that if 2 player decides not to take meds, then she is indifferent between going and not going to the hospital? Because it's clear that the decision to take the meds clearly dominates the decision not to take meds, should I even mention it when stating SPNE??
THANKS FOR YOUR HELP!!</p>
| Hagen von Eitzen | 39,174 | <p>The strategy of 2 consists of a few "bits": </p>
<ul>
<li>When told the truth, will she take her medicine? Or not take her medicine, but go to the hospital? Or neither? Symbolize these by <span class="math-container">$M, H_T, 0_T$</span>.</li>
<li>When lied to, will she go to the hospital or not? Symbolize these by <span class="math-container">$H_L, 0_L$</span>.</li>
</ul>
<p>In total, 2 has six strategies, and the payouts are as follows:
<span class="math-container">$$\begin{matrix}&0_T0_L&0_TH_L&M0_L&MH_L&H_T0_L&H_TH_L\\
T&(0,0)&(0,0)&(2,2)&(2,2)&(1,0)&(1,0)\\
L&(1,0)&(0,1)&(1,0)&(0,1)&(1,0)&(0,1)\end{matrix} $$</span></p>
<p>From 2's perspective, <span class="math-container">$MH_L$</span> is strictly better than <span class="math-container">$0_T0_L$</span> or <span class="math-container">$H_T0_L$</span>, and at least as good as the other three, hence rationally, 2 will follow strategy <span class="math-container">$MH_L$</span>.
With that in mind, 1 will prefer <span class="math-container">$T$</span> over <span class="math-container">$L$</span>. This will lead to utility <span class="math-container">$2$</span> for both players.</p>
|
10,992 | <p>Alright, I'm trying to figure out how to calculate a critical value using t-distribution in Microsoft Excel... ex. a one-tailed area of 0.05 with 39 degrees of freedom: t=1.685</p>
<p>I know the answer, but how do I get this? I've tried TDIST() TINV() and TTEST() but they all give me different answers. This web calculator: <a href="http://www.danielsoper.com/statcalc/calc10.aspx" rel="nofollow">http://www.danielsoper.com/statcalc/calc10.aspx</a> always gives me what I'm looking for but I cannot manage to get Excel to do the same.</p>
<p>Any help would be greatly appreciated!</p>
| Mike Spivey | 2,370 | <p>The right function is <code>TINV()</code>, but the problem is that <code>TINV()</code> assumes that you're inputting the two-tailed probability, not the one-tailed probability. In other words, <code>TINV(p,d)</code> outputs the value of $x$ such that $P(|X| \geq x) = p$, where $X$ has a $t$ distribution with $d$ degrees of freedom.</p>
<p>Thus if you want <code>TINV()</code> to give you the critical value associated with a one-tailed probability, you have to double the probability first to account for the other tail. Thus, for example, <code>TINV(0.1,39)</code> yields an output of <code>1.684875122</code>, which is the critical value you want.</p>
|
1,709,915 | <p>Given an algebraic category, Birkhoff's Variety Theorem gives a categorical characterization of the full subcategories whose object-class forms a variety (i.e. can be defined by equations in the sense of Model Theory).</p>
<p>The theorem is often stated as being of fundamental importance to Universal Algebra. As far as its importance for metamathematical questions is concerned, this does not surprise me, as it describes a connection between Model Theory and Universal Algebra. But what about its “internal” importance for Universal Algebra? Suppose we are studying a certain class of objects in some algebraic category. In how far could it be useful to know whether this class forms a variety?</p>
<p>My question only concerns the one implication of Birkhoff's theorem of course. It is clear that the result that varieties are closed under the taking of products, subalgebras and homomorphic images has a wide range of possible applications. But what about the converse?</p>
| Alex Kruckman | 7,062 | <p>This may not be the best answer, but it is an interesting sort of application of Birkhoff's theorem.</p>
<p>Let's say you have some class $V$ of algebras which is defined "semantically" (e.g. as all the algebras arising from some construction), and you want to give an equivalent "syntactic" description (i.e. an axiomatization $T$). Classical examples of this include showing that the group axioms pick out the class of subgroups of permutation groups, or that the Boolean algebra axioms pick out the class of subalgebras of powerset algebras [note that I'm not claiming that Birkhoff's theorem is useful for these classical examples].</p>
<p>If you can show directly that $V$ is closed under the HSP operations, then you can conclude that <em>some</em> equational axiomatization $T^*$ exists, and this knowledge can actually help you prove that a <em>particular</em> equational axiomatization $T$ works. </p>
<p>How? Well, for example, it's clear that an algebra $A$ satisfies an equation if and only if every finitely generated subalgebra of $A$ satisfies the equation. So if you want to check that $A$ satisfies $T$ if and only if it's in $V$ (if and only if $A$ satisfies the mysterious $T^*$), then you can assume without loss that $A$ is finitely generated. And finiteness assumptions like this can be very useful.</p>
|
1,709,915 | <p>Given an algebraic category, Birkhoff's Variety Theorem gives a categorical characterization of the full subcategories whose object-class forms a variety (i.e. can be defined by equations in the sense of Model Theory).</p>
<p>The theorem is often stated as being of fundamental importance to Universal Algebra. As far as its importance for metamathematical questions is concerned, this does not surprise me, as it describes a connection between Model Theory and Universal Algebra. But what about its “internal” importance for Universal Algebra? Suppose we are studying a certain class of objects in some algebraic category. In how far could it be useful to know whether this class forms a variety?</p>
<p>My question only concerns the one implication of Birkhoff's theorem of course. It is clear that the result that varieties are closed under the taking of products, subalgebras and homomorphic images has a wide range of possible applications. But what about the converse?</p>
| J.-E. Pin | 89,374 | <p>Birkhoff's Variety Theorem is indeed of fundamental importance to Universal Algebra, notably for the stream of research it has generated, inside and outside of Universal Algebra. Here are a few examples.</p>
<p>The book [4] is entirely devoted to varieties of groups. See also the paper <a href="http://projecteuclid.org/download/pdf_1/euclid.bams/1183529007" rel="nofollow">Varieties of groups</a> by B.H. Neumann. One important result in this field states that the variety of groups generated by a finite group is finitely based. This result does not hold for arbitrary algebras (groupoids and semigroups are counter-examples).
But there is a large literature on the following question:</p>
<blockquote>
<p>Is it decidable whether the variety generated by a given finite algebra is finitely based?"</p>
</blockquote>
<p>Birkhoff's variety theorem has also been extended in various directions. For instance, it has been extended to ordered algebras in [2] and to (pseudo)varieties of finite algebras in [1, 6] (the equations are now profinite equations) and to (pseudo)varieties of finite first-order structures in [5].
These results in turn have found important applications in the study of regular languages through Eilenberg's variety theorem [3, p. 194].</p>
<p>Finally, let me mention the little known but very nice book [7], which contains some interesting material on varieties and equational theories.</p>
<p>[1] B. Banaschewski, The Birkhoff theorem for varieties of finite algebras, <em>Algebra Universalis</em> <strong>17</strong> (1983), 360–368.</p>
<p>[2] S. Bloom, Varieties of ordered algebras, <em>J. Computer System Sciences</em> <strong>13</strong> (1976), 200-212.</p>
<p>[3] S. Eilenberg, Automata, Languages and Machines, Vol. B, Academic Press, coll. Pure and Applied Mathematics (no 59), 1976, xiii+387 p.</p>
<p>[4] Hanna Neumann, <em>Varieties of groups</em>, Ergebnisse der Mathematik und ihrer Grenzgebiete, Band 37, Springer-Verlag, Berlin, 1967.</p>
<p>[5] J.-E. Pin and P. Weil, A Reiterman theorem for pseudovarieties of finite first-order structures, <em>Algebra Universalis</em> <strong>35</strong> (1996), 577–595.</p>
<p>[6] J. Reiterman, The Birkhoff theorem for finite algebras, <em>Algebra Universalis</em> <strong>14</strong> (1982), 1–10.</p>
<p>[7] Wechler, <em>Universal Algebra for Computer Scientists</em>, EATCS Monographs on Theoretical Computer Science <strong>25</strong> (1992)</p>
|
982,259 | <p>Let $n\in\mathbb{N}$.</p>
<p>So far I have: If the sum of the digits of $n$ is $k$, then $n = 9m + k$, where $m$ element of an integer (not sure why). Now consider $5n-n$.</p>
<p>Help?</p>
| mookid | 131,738 | <p><strong>Hint:</strong> start to prove that if $S(n)$ is the sum of the digits of $n$, then
$
n - S(n)
$ is a multiple of 9.</p>
|
2,894,315 | <p>Let $a;b;c>0$ such that $a+b+c=6$. Find the maximum value of $A=a^2bc+a^2+2b^2+2c^2$</p>
<hr>
<p>WLOG $b\ge c$. I see maximum value of $A=36$ at $(a;b;c)=(2;1;3)$</p>
<p>So i need to prove $A\le 36$. Or I will prove </p>
<p>$(a+b+c)^4\ge 36a^2bc+(a^2+2b^2+2c^2)(a+b+c)^2$</p>
<p>Or $(2a-b-c)(b^3+c^3+a^2b+a^2c+2ab^2+2ac^2-12abc-b^2c-bc^2)\ge 0$</p>
<p>Then Im stuck here, help me solve it.</p>
| Cesareo | 397,348 | <p>With the help of the Lagrange multipliers and $f(a,b,c) = a^2 b c + a^2 + 2 b^2 + 2 c^2$</p>
<p>$$
L(a,b,c,\lambda) =f(a,b,c)+\lambda(a+b+c-6)
$$</p>
<p>the stationary points are the solutions for</p>
<p>$$
\nabla L = \left\{
\begin{array}{c}
2 b c a+2 a+\lambda =0 \\
c a^2+4 b+\lambda =0 \\
b a^2+4 c+\lambda =0 \\
a+b+c-6=0 \\
\end{array}
\right.
$$</p>
<p>giving</p>
<p>$$
\left[
\begin{array}{ccccc}
a & b & c & \lambda & f(a,b,c)\\
-2 & -1 & 9 & -32 & \color{red}{132} \\
-2 & 9 & -1 & -32 & \color{red}{132} \\
2 & 1 & 3 & -16 & \color{green}{36} \\
2 & 3 & 1 & -16 & \color{green}{36} \\
3 & \frac{3}{2} & \frac{3}{2} & -\frac{39}{2} & \color{green}{\frac{153}{4}} \\
3-\sqrt{5} & \frac{1}{2} \left(3+\sqrt{5}\right) & \frac{1}{2} \left(3+\sqrt{5}\right) & -12 & \color{green}{32} \\
3+\sqrt{5} & \frac{1}{2} \left(3-\sqrt{5}\right) & \frac{1}{2} \left(3-\sqrt{5}\right) & -12 & \color{green}{32} \\
\end{array}
\right]
$$</p>
<p>In red non feasible solutions and in green feasible solutions. The maximum corresponds to $\frac{153}{4}= 38.25$</p>
<p>NOTE</p>
<p>The same solution can be obtained under the tangency condition between $f(a,b,c)-\mu=0$ and $a+b+c-6=0$ by calculating the values for $\mu$</p>
|
1,700,457 | <p>Assume $H \leq G$ and $K \leq G$ and assume $aH=bK$ for some $a,b \in G$. Prove that $H=K$</p>
<p>would someone give me a hint. I know that definition of being a subgroup. I know that the element look like this $ah_1=bk_1$ for all $h_1 \in H$ and for all $k_1 \in K$. I do not have an idea from where should I get started. </p>
| Mark Fischler | 150,362 | <p>Assume $H \subseteq G$ and $K \subseteq G$ and assume $aH=bK$ for some $a,b \in G$. </p>
<p>Then $\forall h \in H \exists k \in k :ah = bk
$ and of course that $k$ may depend on $h$, so let us label one of the (possibly multiple) values of $k$ that satisfies $ah = bk$ as $k(h)$.</p>
<p>Consider the $e$ (the group identity) which must be an particular element $h$:
$$
a e = bk(e) \implies a = bk(e)
$$ </p>
<p>Then for any arbitrary $h\in H$,
$$ ah = bk(h) \implies (bk(e) )h = bk(h) \implies b^{-1}(bk(e) )h =b^{-1}bk(h)
\implies k(e) h =k(h)
$$
But $k(e) \in k$ so $[k(e)]^{-1}\in k$. </p>
<p>Left-multiply both sides by $[k(e)]^{-1}$:</p>
<p>$$h =[k(e)]^{-1}k(h)$$</p>
<p>Can you see now that $h\in k$?</p>
<p>So for all $h$, $h\in H \implies h \in K$. Similarly you show that for all $k$, $k\in K \implies k \in H$. You can take it from there... </p>
|
99,018 | <p>If $g$ is Lie algebra over field char(k)=0,
then the following facts are well-known:</p>
<p>1) S(g) and U(g) are isomorphic as $g$-modules. (Symmetrization map S(g)->U(g) gives isomorphism).</p>
<p>2) S(g)^g and ZU(g)=U(g)^g are isomorphic as commutative algebras. (The <a href="http://en.wikipedia.org/wiki/Duflo_isomorphism" rel="nofollow noreferrer">Duflo map</a>
defines is isomorphism which is combination of symmetrization map with some intricate corrections by terms of smaller degree). </p>
<p>Both facts are based on the symmetrization.
If we consider $g$ over field char(k)$\ne$0, there is NO symmetrization map.
So I wonder the following:</p>
<p><strong>Question</strong> are the facts above true for Lie algebras over char(k)$\ne$0 ?</p>
<hr>
<p>Notations</p>
<p>U(g) - <a href="http://en.wikipedia.org/wiki/Universal_enveloping_algebra" rel="nofollow noreferrer">universal enveloping algebra</a> (non-commutative associative algebra defined by relations $[x_i, x_j] = \sum_k c_{ij}^k x_k$, for any linear bases $x_k$ of $g$.</p>
<p>S(g) - symmetric algebra of $g$ (defined as $k[x_1...x_n]$ for any bases $x_k$ of $g$.</p>
<p>S(g)^g, U(g)^g means subspaces of g-invariants (g act by zero).</p>
<p>Symmetrization map is defined as $S(x_1..x_k) = 1/k! \sum_{\sigma} \prod_l {x_{\sigma(l)} $.</p>
<p>Duflo map is not so easy to write so let me just mention some MO questions:</p>
<p><a href="https://mathoverflow.net/questions/80025/is-the-duflo-map-for-lie-algs-unique">Is the Duflo map for Lie algs. unique ?</a></p>
<p><a href="https://mathoverflow.net/questions/92348/capelli-determinant-duflo-determinant-was-it-known">Capelli determinant = Duflo ( determinant) - was it known ? </a></p>
<hr>
<p>Remark</p>
<p>For the case of gl_n there are so-called <a href="http://en.wikipedia.org/wiki/Capelli's_identity" rel="nofollow noreferrer">Capelli generators</a> of the center of U(g),
which can be defined over any char. So I think that the center of U(gl) is the same as
for char(k)=0 the result is true. (It is NOT the same as in char=0, as I mistakenly wrote first, since there should be generators corresponding to a^p (which are S(g)^g for any a)).</p>
<p>I think the same is true for other classical semi-simples - there are analogous Capelli like formulas. </p>
| Alexander Premet | 24,386 | <p>Question 2 has a negative answer. Indeed, let $L =sl(2,k)$ where $k$ is an algebraically closed field of characteristic $p>3$ and let {$e,h,f$} be the standard basis of $L$. Then it is well-known (and was first proved by Rudakov-Shafarevich in the late 60s) that the centre $Z(L)$ of $U(L)$ is generated by $E=e^p$, $F=f^p$, $H=h^p-h$ and a Casimir element $C$ subject to the relation
$$4EF+H^2=\prod_{i\in{\mathbb F}_p}(C-i^2)=C^p-2C^{(p+1)/2}+C.$$ Furthermore, the maximal spectrum of $Z(L)$ is a hypersurface in ${\mathbb A}^4$ given by the above equation. On the other hand, it follows from a result of Brown-Goodearl that the Azumaya locus of $U(L)$ coincides with the smooth locus of ${\rm Specm\ }Z(L)$. So the modular representation theory of $sl(2,k)$ yields that ${\rm Specm\ }Z(L)$
has exactly $(p-1)/2$ singular points (one can also see this directly by differentiating the above equation).</p>
<p>On the other hand, it is also known (and not difficult to see directly) that $S(L)^L$ is generated by
$X=e^p$, $Y=f^p$, $Z=h^p$ and $\Omega=4ef+h^2$ subject to the relation $4XY+Z^2=\Omega^p$. Moreover, ${\rm Specm\ }S(L)^L$ is a hypersurface in ${\mathbb A}^4$ given by the above equation. This hypersurface has a unique singular point at the origin.
Since $(p-1)/2>1$ under our assumptions on $p$, we see that the $k$-algebras $Z(L)$ and $S(L)^L$ cannot be isomorphic.</p>
<p>As far as I know the answer to Question 1 is unknown, in general, although there do exist finite dimensional Lie algebras $L$ for which $L\subset U_1(L)$ is NOT a direct summand of $U(L)$ (if $U(L)$ is isomorphic to $S(L)$ as $ad(L)$-modules, then $U(L)$ must contain an isomorphic copy of $L$ as a direct summand, but that copy doesn't have to lie in the first component $U_1(L)$ of the canonical filtration of $U(L)$). </p>
<p>If we require the stronger condition that an isomrphism $S(L)\rightarrow U(L)$ sends $S^1(L)$ onto $L\subset U_1(L)$ (which is obviously true for the symmetrisation map), then the answer to Question 1 is NO. For that condition to hold the Lie algebra $L$ must admit at least one $[p]$-th power map, i.e. must be ${\it restrictable}$ (this observation is due to Michel Duflo). The majority of finite dimensional simple Lie algebras are non-restrictable, and the smallest example is the $3$-dimensional simple Lie algebra over an algebraically closed field of characteristic $2$.</p>
|
3,214,255 | <p>How to simplify <span class="math-container">$$\frac{\sqrt{6+4\sqrt{2}}}{4+2\sqrt{2}}?$$</span></p>
<p>Rationalise the denominator </p>
<p><span class="math-container">$$\frac{\sqrt{6+4\sqrt{2}}}{4}(2-\sqrt{2})$$</span></p>
<p>This is still not simplify.</p>
| Dr. Sonnhard Graubner | 175,066 | <p>We have <span class="math-container">$$6+4\sqrt{2}=4+2+4\sqrt{2}$$</span> so <span class="math-container">$$2\sqrt{6+4\sqrt{2}}=4+2\sqrt{2}$$</span> and we get</p>
<p><span class="math-container">$$\frac{\sqrt{6+4\sqrt{2}}}{4+2\sqrt{2}}=\frac{1}{2}$$</span></p>
|
2,386,864 | <p>Six persons P, Q, R, S, T and U play in a tournament called "High Rollers". Every game involved two players. Each of the participants played with every other participant exactly once. In the game both the players rolled an unbiased die each. The player who gets the larger number on the top surface of the die wins the game. Every game resulted in a win/loss. The player who wins, gets as many coins as the number on his die. The total number of games won by each person is distinct.
Further it is known that:</p>
<ol>
<li>Q won his game against P.</li>
<li>The number of coins won by R was more than any one else.</li>
<li>U lost only two games, one against S and the other against R.</li>
<li>T wins at least two games.</li>
</ol>
<p>R won 11 coins and he lost one game. If the total number of coins won by all the six players in the tournament was 47, then what was the number of coins won by T?</p>
<p>What I did till now:</p>
<p>Number of games won by P is 0<br>
Number of games wom by Q is 1<br>
Number of games won by R is 4<br>
Number of games won by S is 5<br>
Number of games won by T is 2<br>
Number of games won by U is 3 </p>
<p>How to find the number of coins? </p>
| Piquito | 219,998 | <p>COMMENT.-This is another way, maybe interesting for some people, of stating the same problem.</p>
<p>Given an odd natural number, $2n + 1$, there are $n$ different ways to express it as the sum of two natural $$2n+1=(2n-k)+(k+1);\space k=1,2,....,n$$ Then the problem can be stated as follows equivalently: $$\text{ For all natural 2n+1 greater than 1}\text{ at least one of the n numbers}\\\begin{cases}M_1=4n^2+1\\M_2=(2n-1)^2+2^2\\M_3=(2n-3)^2+3^3\\...........\\...........\\M_n=(n+1)^2+n^2\end{cases}\\ \text{ is a prime}$$</p>
<p><strong>NOTE</strong>.- <em>It is known that such a prime (if it exists) is necessarily of the form $p=4m+1$. Besides each $M_k$ has a factorization of the form $$M_k=\prod p_i^{\alpha_i}\prod q_j^{2\beta_j}$$ where $\alpha_i,\space \beta_j$ are non-negative integers,the primes $p_i$ and $q_j$ being of the form $4m+1$ and $4m-1$ respectively.</em></p>
<p><em>While larger 2n + 1 is, more likely to exist such a prime number. It would seem that the conjecture is true</em></p>
|
803,488 | <p>Imagine Rock Paper Scissors, but where winning with a different hand gives a different reward.</p>
<ul>
<li><p>If you win with Rock, you get \$9. Your opponent loses the \$9.</p>
</li>
<li><p>If you win with Paper, you get \$3. Your opponent loses the \$3.</p>
</li>
<li><p>If you win with Scissors, you get \$5. Your opponent loses the \$5.</p>
</li>
<li><p>If you tie, you get $0</p>
<p>My first intuition would be that you should play Rock with a probability of <code>9/(9+3+5)</code>, Paper with <code>3/(9+3+5)</code> and Scissors with <code>5/(9+3+5)</code> however this seems wrong, as it doesn't take into consideration the risk you expose yourself to (if you play <code>Paper</code>, you have an upside of \$3 but a downside of \$5).</p>
</li>
</ul>
<p>So I put the question to you, in such a game -- what is the ideal strategy.</p>
<p>Edit: By "ideal" strategy, I mean playing against an adversarial player who knows your strategy.</p>
| mjqxxxx | 5,546 | <p>Let $(x_1,x_2,x_3)$ be the first player's strategy (i.e., his probabilities of playing rock, paper, and scissors respectively), and let $(y_1,y_2,y_3)$ be the second player's strategy. The expected payoff to the first player is
$$
P(x,y)=9(x_1y_3-x_3y_1)+3(x_2y_1-x_1y_2)+5(x_3y_2-x_2y_3).
$$
To constrain the probability sums to be $1$, we take $x_3=1-x_1-x_2$ and $y_3=1-y_1-y_2$. So
$$
P(x,y)=9\left(x_1(1-y_1-y_2)-(1-x_1-x_2)y_1\right)+3(x_2y_1-x_1y_2)+5\left((1-x_1-x_2)y_2-x_2(1-y_1-y_2)\right) \\
=9(x_1-y_1)+ 17(x_2 y_1 -x_1y_2) + 5(y_2-x_2).
$$
The first derivatives are zero when
$$
\frac{\partial}{\partial x_1}P(x,y)= 9 -17y_2=0 \\
\frac{\partial}{\partial x_2}P(x,y)= 17y_1 -5=0 \\
\frac{\partial}{\partial y_1}P(x,y)=-9+17x_2 = 0\\
\frac{\partial}{\partial y_2}P(x,y)=-17x_1+5=0,
$$
or at $(x_1,x_2,x_3)=(y_1,y_2,y_3)=(5/17, 9/17, 3/17)$. The Nash equilibrium is to play to beat each move with probability proportional to that move's reward.</p>
<p>To check that this is indeed a Nash equilibrium, suppose $y_1=5/17$ and $y_2=9/17$. Then
$$
P(x)=9(x_1-5/17)+17\left(x_2 (5/17)-x_1 (9/17)\right)-5(x_2-9/17)=0;
$$
that is, the first player's expected payoff is zero with any strategy. So the first player cannot improve his payoff by changing his strategy unilaterally, and by symmetry, neither can the second player; this is the definition of a Nash equilibrium.</p>
|
758,135 | <p>A visiting speaker in Economics recently happened to mention that John Maynard Keynes' <a href="http://www.gutenberg.org/ebooks/32625">A Treatise on Probability</a> revolutionized probability theory. I have not heard any such claim before and it struck me as strange. The <a href="http://en.wikipedia.org/wiki/A_Treatise_on_Probability#cite_note-russell-3">Wikipedia</a> page contains some effusive praise from Russell but nothing specific. This leads me to ask:</p>
<p>1) Is this claim approximately true?</p>
<p>2) In what specific ways did it impact probability theory?</p>
<p>3) What are some specific citations which demonstrate this?</p>
| Michael Emmett Brady | 160,180 | <p>J M Keynes did not revolutionize probability theory.J M Keynes revolutionized decision theory by showing that the mathematical laws of the calculus of probabilities were a special case requiring that the weight of the evidence,w,where w is defined on the unit interval[ 0,1],must equal ,approximate ,tend,or approach 1 in order for the probability calculus to be operationalized where 0 is less than or equal to w ,which is less than or equal to 1.The weight of the evidence ,w,measures the completeness of the relevant evidence upon which the probability estimates are based.w=1 requires that the sample space of all possible outcomes be specified before any decision is made or,what amounts to the same thing,a specific ,unique probability distribution is known before any decision is made. A w <1 requires the use of interval estimates ,which contain an upper and lower bound.
The first economist to specify an interval estimate of probability was Adam Smith in the Wealth of Nations on pp.106-109 of the Modern Library(Cannan) edition See p.714 for Smith's clearcut statement that probabilities can't be estimated precisely .</p>
|
2,828,636 | <p>I wanted to show that both sets are equal. My Textbook says following:</p>
<p>$\in$ means "is element of", $\land$ is the and operator, $\lnot$ is the not operator, $\notin$ means "is not element of", $\lor$ is the or operator</p>
<p>$$\def\-{\setminus}\begin{split} x \in A\-A\-B &\iff\\
(x \in A) \land (\lnot(x \in (A\-B)) &\iff\\
(x \in A) \land (\lnot(x \in A \land (\lnot(x \in B))) &\iff\\
(x \in A) \land (\lnot(x \in A \land x \notin B)) &\iff\\
(x \in A) \land (x \notin A \lor x \in B)\end{split}$$</p>
<p>Let $(x \in A)$ be $C$ and $(x \in B)$ be $D$ </p>
<ul>
<li>Case 1 $C$ True $D$ True -> Contradiction $x \in A \land x \notin A$ cannot be true </li>
<li>Case 2 $C$ True $D$ True -> Contradiction $x \in A \land x \notin A$ cannot be true </li>
<li>Case 3 $C$ False $D$ True -> True</li>
</ul>
<p>Thus
$$(x \in A) \land (x \notin A \lor x \in B) \iff\\
(x \in A) \land (x \in B) \iff\\
x \in (A \cap B)$$</p>
<p>What is the subsetproof? Because I have not fully understood the negation of $\land$ </p>
| user | 505,767 | <p>In $x-y$ plane the equation of the line is given by</p>
<p>$$ax+by=d$$</p>
<p>beeing the normal $n=(a,b)$ such that $|n|=1$, indeed by dot product between $OP$ and $n$ for any point $P=(x,y)$ on the line we have</p>
<p>$$(x,y)\cdot (a,b)=ax+by=|OP|\cos \theta=d$$</p>
|
3,132,039 | <p>I am trying to create a set of 10-tuples, where each tuple consists of elements taken from a set {defective, nondefective). Each sample must have precisely 1 defective part and 9 nondefective parts. My attempt</p>
<p>Let the sample be a 10-tuple <span class="math-container">$$S:=(X,\Omega,F)$$</span> <span class="math-container">$$X=\{i \ \epsilon \ Z:1\leq i\leq10\}$$</span> </p>
<p><span class="math-container">$$\Omega=\{\text{defective, nondefective}\} $$</span></p>
<p><span class="math-container">$$F:X\to\Omega$$</span></p>
<p><span class="math-container">$$Q=\{x \ \epsilon \ S:(\exists i \ \epsilon \ X) F(i)=\text{defective} \ \land \ (\forall n \ \epsilon \ X)n\ne i \to \ F(n)=\text{nondefective}\}$$</span></p>
<p>Assuming my notation is correct, the issue I see is that my quantifier for i does not preclude the existence of multiple defective parts being part of the tuple. What quantifier can I use that specifies existence and uniqueness? </p>
| Exit path | 161,569 | <p>A sequence of sheaves
<span class="math-container">$$0 \to \mathcal{F} \to \mathcal{G} \to \mathcal{H} \to 0$$</span>
is exact if and only if the induced sequence on stalks is exact for every <span class="math-container">$x \in X$</span>. It's easy to show this fact is equivalent to the statement that taking stalks takes <em>any</em> exact sequence to an exact sequence. This means that a sequence
<span class="math-container">$$0 \to \mathcal{F} \to \mathcal{G} \to \mathcal{H}$$</span>
is exact if and only if it's exact on the level of stalks.</p>
|
3,477,152 | <p>I am trying to prove that</p>
<p><span class="math-container">$$\sum_{n=1}^\infty \frac{n}{\sqrt{n+1}}$$</span></p>
<p>diverges without checking the limit, bounds or doing any other lengthy steps, as it should be seen as divergent "immediately", but I have no clue about how I would quickly prove this.</p>
<p>So far I thought about using the P-series convergence test where it only converges for <span class="math-container">$p>1$</span> but it does not seem to make any sense for this one. I also thought about comparing it to other series but nothing comes to my mind.</p>
| Francesco_Trig | 731,440 | <p>You can notice that this series has the same behavior of <span class="math-container">$$\sum_{n=1}^\infty \frac{n}{\sqrt n} =\sum_{n=1}^\infty \sqrt n$$</span> that diverges.</p>
|
3,477,152 | <p>I am trying to prove that</p>
<p><span class="math-container">$$\sum_{n=1}^\infty \frac{n}{\sqrt{n+1}}$$</span></p>
<p>diverges without checking the limit, bounds or doing any other lengthy steps, as it should be seen as divergent "immediately", but I have no clue about how I would quickly prove this.</p>
<p>So far I thought about using the P-series convergence test where it only converges for <span class="math-container">$p>1$</span> but it does not seem to make any sense for this one. I also thought about comparing it to other series but nothing comes to my mind.</p>
| Deepesh Thakur | 639,689 | <p><span class="math-container">$$n+1 \le 2n\ \ \forall \ n \ge 1$$</span>
<span class="math-container">$$\Rightarrow \frac{n}{\sqrt{2n}} \le \frac{n}{\sqrt{n+1}}$$</span>
<span class="math-container">$$\Rightarrow \sum_{n=1}^{\infty}\frac{n}{\sqrt{2n}} \le \sum_{n=1}^{\infty}\frac{n}{\sqrt{n+1}}$$</span>
<span class="math-container">$$\Rightarrow \frac{1}{\sqrt{2}}\sum_{n=1}^{\infty}{\sqrt{n}} \le \sum_{n=1}^{\infty}\frac{n}{\sqrt{n+1}}$$</span>
Hence this won't converge!</p>
|
3,321,324 | <p>In doing some old exam questions I came across this. The results are all excruciatingly obvious intuitively, but I have no idea how to formalize a proof for any of them.</p>
<p>Let <span class="math-container">$(M,d)$</span> be a metric space. For <span class="math-container">$A\subseteq M, A \ne \emptyset$</span> and <span class="math-container">$x\in M$</span> we define the distance from x to A as
<span class="math-container">$$dist(x,A):=\inf\{d(x,a):a\in A\}$$</span>
Show the following:</p>
<p>a) If <span class="math-container">$x,y\in M$</span> and <span class="math-container">$r:=\frac{d(x,y)}{2}$</span>, then <span class="math-container">$dist(x,U_r(y))\ge r$</span></p>
<p>b) Let <span class="math-container">$A_1, ..., A_i$</span> be non-empty subsets of M and let <span class="math-container">$x\in M$</span> then
<span class="math-container">$$dist(x,\bigcup_{k=0}^i A_k) = \min \{dist(x,A_j): j\le i\}$$</span></p>
<p>c) If <span class="math-container">$K\subseteq M$</span> is compact and <span class="math-container">$x\in M\setminus K$</span>, then <span class="math-container">$dist(x,K)>0$</span></p>
<p>My best attempt at (a) was using that obviously for every <span class="math-container">$z\in M\setminus U_r(y)$</span> <span class="math-container">$$d(y,z)\ge r$$</span>
but then i still need to prove <span class="math-container">$x\in M\setminus U_r(y)$</span></p>
<p>I'm stumped on (b) and (c). I would be grateful for any hints on how to solve these kinds of problems in general as well. </p>
| Lázaro Albuquerque | 85,896 | <p>a) Take an arbitrary <span class="math-container">$z \in U_r(y)$</span> and play with the triangle inequality.</p>
<p>b) For bounded sets <span class="math-container">$A$</span> and <span class="math-container">$B$</span>, <span class="math-container">$\inf A \cup B = \min (\inf A, \inf B)$</span>. Apply induction.</p>
<p>c) For a closed subset <span class="math-container">$A \subset M$</span>, <span class="math-container">$d(x, A)=0$</span> iff <span class="math-container">$x \in A$</span>.</p>
|
3,321,324 | <p>In doing some old exam questions I came across this. The results are all excruciatingly obvious intuitively, but I have no idea how to formalize a proof for any of them.</p>
<p>Let <span class="math-container">$(M,d)$</span> be a metric space. For <span class="math-container">$A\subseteq M, A \ne \emptyset$</span> and <span class="math-container">$x\in M$</span> we define the distance from x to A as
<span class="math-container">$$dist(x,A):=\inf\{d(x,a):a\in A\}$$</span>
Show the following:</p>
<p>a) If <span class="math-container">$x,y\in M$</span> and <span class="math-container">$r:=\frac{d(x,y)}{2}$</span>, then <span class="math-container">$dist(x,U_r(y))\ge r$</span></p>
<p>b) Let <span class="math-container">$A_1, ..., A_i$</span> be non-empty subsets of M and let <span class="math-container">$x\in M$</span> then
<span class="math-container">$$dist(x,\bigcup_{k=0}^i A_k) = \min \{dist(x,A_j): j\le i\}$$</span></p>
<p>c) If <span class="math-container">$K\subseteq M$</span> is compact and <span class="math-container">$x\in M\setminus K$</span>, then <span class="math-container">$dist(x,K)>0$</span></p>
<p>My best attempt at (a) was using that obviously for every <span class="math-container">$z\in M\setminus U_r(y)$</span> <span class="math-container">$$d(y,z)\ge r$$</span>
but then i still need to prove <span class="math-container">$x\in M\setminus U_r(y)$</span></p>
<p>I'm stumped on (b) and (c). I would be grateful for any hints on how to solve these kinds of problems in general as well. </p>
| Eric Towers | 123,905 | <p>Hints: a) uses the triangle inequality, b) uses that the infimum of a finite set is the minimum of that set, and c) proceed by contradiction using an open cover.</p>
<p>For a), you want to use the idea that <span class="math-container">$d(x,a) + d(a,y) \leq d(x,y)$</span> is true for every <span class="math-container">$a \in U_r(a)$</span> (which I assume is the open ball of radius <span class="math-container">$r$</span> centered at <span class="math-container">$a$</span>).</p>
<blockquote class="spoiler">
<p> In particular, manipulate it to the form <span class="math-container">$d(x,b) \geq d(x,y) - d(b,y)$</span>.
<span class="math-container">\begin{align*} dist(x,U_r(a)) &= \inf \{d(x,b) : b \in U_r(a)\} \\ &\geq \inf \{d(x,y) - d(b,y): b \in U_r(a)\} \\ &= \inf \{2r - d(b,y): b \in U_r(a)\} \\ &= 2r - \sup \{d(b,y): b \in U_r(a)\} \\ &\geq r \text{.} \end{align*}</span>
(We can't finish with "<span class="math-container">${} = r$</span>" because we do not know that the point of the ball that would be closest to <span class="math-container">$x$</span> is actually in the metric space. All we know is that the closest point is a distance <span class="math-container">${} \leq r$</span> from <span class="math-container">$y$</span>.)</p>
</blockquote>
<p>For b) use that the infimum of the union of finitely many sets is the minimum of their infima: <span class="math-container">$\inf(A \cup B) = \min\{\inf A, \inf B\}$</span>. (Warning: need not be true for an infinite collection of sets. Consider <span class="math-container">$\cup A_k$</span>, <span class="math-container">$A_k = \{1/k\} \subseteq \mathbb{R}$</span>.)</p>
<blockquote class="spoiler">
<p> In detail,
<span class="math-container">\begin{align*} dist(x,\bigcup_{k=0}^i A_k) &= \inf\{d(x,b) : b \in \bigcup_{k=0}^i A_k\} \\ &= \min\{\inf\{d(x,b) : b \in A_k\} : k \leq i\} \\ &= \min\{dist(x, A_k) : k \leq i\} \text{,} \end{align*}</span>
as desired.</p>
</blockquote>
<p>For c), cover <span class="math-container">$K$</span> by open balls of radius <span class="math-container">$r$</span> (part a) might help us subsequently) and note that since <span class="math-container">$K$</span> is compact, finitely many balls suffice (part b) might help us subsequently). So if <span class="math-container">$dist(x,K) = 0$</span>, it is always in this cover.</p>
<blockquote class="spoiler">
<p> In particular, suppose <span class="math-container">$dist(x,K) = 0$</span>, let <span class="math-container">$r > 0$</span>, and let <span class="math-container">$A = \bigcup_{k=0}^i U_r(x_k)$</span> be a (finite sub-)cover of <span class="math-container">$K$</span> by open balls of radius <span class="math-container">$r$</span>. By parts a) and b), there is a ball in the union defining <span class="math-container">$A$</span> containing <span class="math-container">$x$</span>. Let <span class="math-container">$y_r$</span> be a point in that ball. Repeating, with a decreasing sequence of <span class="math-container">$r$</span>, <span class="math-container">$r_i$</span>, we get a sequence <span class="math-container">$y_{r_i}$</span> converging to <span class="math-container">$x$</span> with each <span class="math-container">$y_{r_i} \in K$</span>. Since in a metric space compact is equivalent to sequentially compact, and <span class="math-container">$x$</span> is a convergent of <span class="math-container">$y_{r_i}$</span>, <span class="math-container">$x \in K$</span>. With this contradiction, we conclude <span class="math-container">$dist(x,K) > 0$</span>.</p>
</blockquote>
|
1,829,086 | <p>So far, I've tried out to reformulate: $$\int{\frac{1}{\cos(x)}}dx$$
to: $$\int{\frac{\sin(x)}{\cos(x)\sin(x)}}dx$$</p>
<p>which is basically: $$\int{\frac{\tan(x)}{\sin(x)}}dx$$
But I'm not sure if this is the right way to go, or if I try something else.</p>
<p>Any tips or methods would be very helpful.</p>
| Sophie Agnesi | 346,183 | <p><strong>Hint:</strong></p>
<p>If $u=\tan\left(x\over2\right)$, then $\cos x={1-u^2\over1+u^2}$ and $dx={2\ du\over1+u^2}$. Hence</p>
<p>\begin{equation}
\int {1\over\cos x}\ dx=\int{2\over1-u^2}\ du=\int\left[\frac{1}{1+u}+\frac{1}{1-u}\right]\ du
\end{equation}</p>
|
1,829,086 | <p>So far, I've tried out to reformulate: $$\int{\frac{1}{\cos(x)}}dx$$
to: $$\int{\frac{\sin(x)}{\cos(x)\sin(x)}}dx$$</p>
<p>which is basically: $$\int{\frac{\tan(x)}{\sin(x)}}dx$$
But I'm not sure if this is the right way to go, or if I try something else.</p>
<p>Any tips or methods would be very helpful.</p>
| Strategy Thinker | 112,976 | <p>These are the things I would try.
I would find $\cos(x)$ in terms of $x/2$ because I would have to do that eventually anyway, here $\cos(x) = 2\cos^2(x/2) - 1$.</p>
<p>Then I would replace the $dx$ with an expression in $du$ as that also must be done eventually.
\begin{align*}
du = & 0.5 \sec^2(x/2)\, dx\\
dx = & 2 \cos^2(x/2)\, du
\end{align*}
Thus the integral becomes
$$\int \frac{2 \cos^2(x/2)}{1 - 2 \cos^2(x/2)} du$$
This reminds me of the rule $\sec^2(x/2) = 1 + \tan^2(x/2)$,
thus I divide the top and the bottom by $\cos^2(x/2)$, I rewrite it in terms of $\sec^2(x/2)$ and then in terms of $\tan^2(x/2)$ and then in terms of $u$ and I solve it.</p>
|
1,829,086 | <p>So far, I've tried out to reformulate: $$\int{\frac{1}{\cos(x)}}dx$$
to: $$\int{\frac{\sin(x)}{\cos(x)\sin(x)}}dx$$</p>
<p>which is basically: $$\int{\frac{\tan(x)}{\sin(x)}}dx$$
But I'm not sure if this is the right way to go, or if I try something else.</p>
<p>Any tips or methods would be very helpful.</p>
| JamesJ | 474,394 | <p>\begin{align}
I&=\int\frac{dx}{\cos x}=\int\frac{1-\sin x}{\cos x}\frac{dx}{1-\sin x}=\int\frac{1-\sin x}{\cos x}d\left(\frac{\cos x}{1-\sin x}\right)=\ln\left|\frac{\cos x}{1-\sin x}\right|+C
\end{align}</p>
|
969,601 | <p>For the function whose graph is a paraboloid given by</p>
<p>$z = x^2 + y^2/4$</p>
<p>I know that the level curve represents an ellipse. I also know that the parametrization of this curve in the form $x = x(t)$, $y = y(t)$ is </p>
<p>$x(t) = \cos(t)$
and<br>
$y(t) = 2\sin(t)$</p>
<p>In class, my professor said to compute the derivative of $z$ as a function of $t$, or $∂z/∂t.$
however, he stopped right there, saying the answer is too simple.</p>
<p>Is the derivative with respect to t 0? because deriving</p>
<p>$z= \cos^2(t) + 2\sin^2(t)/4$</p>
<p>with respect to $t$ yields $0$, because both of those sums are a constant number.</p>
<p>Is this the simplicity he was referring to?</p>
| Community | -1 | <p>My attempt at calculating the probability that there is a mine in $D$:</p>
<p>If there is one mine in $\lbrace A, B, C, D \rbrace$ and one mine in $\lbrace D, E \rbrace$, then there are two cases:</p>
<ol>
<li>There are two mines, placed in either E and A, or E and B, or E and C</li>
<li>There is one mine, placed in D</li>
</ol>
<p>Let $S$ denote the number of mines in $\lbrace A, B, C, D, E \rbrace$.</p>
<p>Define $p_1 = P(S=1 | S \in \lbrace 1,2 \rbrace)$. Let $P(D)=P(\text{mine in D})$.</p>
<p>Since $P(D|\text{one mine})=P(\text{one mine}|D)=1$, we have $P(D)=p_1$. Meaning, the probability of the mine being in $D$ is the same as the probability of the no. of mines being $1$. So we calculate the latter:</p>
<p>$$
p_1 = P(S=1 | S \in \lbrace 1,2 \rbrace) = \frac{P(S=1 \cap S \in \lbrace 1,2 \rbrace)}{P(S \in \lbrace 1,2 \rbrace)} = \frac{P(S=1)}{P(S \in \lbrace 1,2 \rbrace)} = \frac{P(S=1)}{P(S=1)+P(S=2)}
$$</p>
<p>The probabilities in the last fraction are 'a-priori' probabilities, which are calculated thus: Let $N$ denote the total number of mines, and $T$ the total number of squares.
$$
P(S=1)=\frac{ {5 \choose 1} {N\choose 1} {T-N \choose 4}}{T\choose 5} \hspace{1cm}\text{and} \hspace{1cm} P(S=2)=\frac{ {5 \choose 2} {N\choose 2} {T-N \choose 3}}{T\choose 5}
$$
Now caluclating $p_1$, after substituting the formulas and do some algebra, we get
$$
p_1 = \frac{T-N-3}{3N+T-7}
$$
So for example for a 'medium' sized minesweeper with $N=40$ mines and $T = 16 \times 16$ squares, one gets $P(D) = 0.577$</p>
|
729,373 | <p>I am to evaluate $\displaystyle\int_0^{\infty} \dfrac{\sin x}{x(x^2+1)}dx$ via contour integration.</p>
<p>Now I used an indented semicircular contour, and the parts lying on the real line and the big arc were no problem, but the small arc is being resistant, and I'm not sure what to do. Usually, on the small arc from $-\varepsilon$ to $\varepsilon$ I can take a laurent expansion of the integrand, and consider integrating its principle part over the arc, letting the rest go to zero in the limit $\varepsilon \to 0$ as the "holomorphic part". My issue is this particular integrand doesn't have a principle part...</p>
<p>The end result is $\dfrac{(e-1)\pi}{2e}$, and so far I have the integral over the whole contour as $\dfrac{-\pi i}{e}$ (I'm not sure why this came out imaginary..) so this part is going to have to contribute something. What should I do to get something out?</p>
| Ellya | 135,305 | <p>My guess is you should have two integrals, one for the residue at $i$ and the other one at $-i$</p>
<p>For the $i $ residue take the angle $3\pi/4$ for the other take $-3\pi/4$</p>
|
146,973 | <p>I want to find the expected value of $\text{max}\{X,Y\}$ where $X$ ist $\text{exp}(\lambda)$-distributed and $Y$ ist $\text{exp}(\eta)$-distributed. X and Y are independent.
I figured out how to do this for the minimum of $n$ variables, but i struggle with doing it for 2 with the maximum.</p>
<p>(The context in which this was given is waiting for the later of two trains, with their arrival times being exp-distributed).</p>
<p>Thanks!</p>
| Julius | 24,547 | <p>Let <span class="math-container">$V=\max\{X,Y\}$</span>. Then
<span class="math-container">$$\mathbb{P}(V\leq t)=\mathbb{P}(X\leq t,Y\leq t)=\mathbb{P}(X\leq t)\mathbb{P}(Y\leq t).$$</span>
Now find <span class="math-container">$f_V(t)$</span> and then <span class="math-container">$\int_{0}^{+\infty}tf_V(t)dt$</span>, which should be <span class="math-container">$\frac{1}{\lambda}+\frac{1}{\eta}-\frac{1}{\lambda+\eta}$</span>.</p>
|
3,938,933 | <p>I am interested in the general case, but let us start with a smaller example. Suppose that cars arrive to a street with intensity <span class="math-container">$\lambda$</span> per minute. We would like to know the probability that at least two cars have arrived on the street during any five minute interval in the next hour. How can we find this?</p>
<p>My initial thought was to use complementary event: 1 - the probability that at most one car is on the street in any five minute period during the next hour. Hence my reasoning was something like <span class="math-container">$1 - \int_0^{55}\mathbb{P}(N(s + 5) - N(s) \leq 1)ds - \int_{55}^{60}\mathbb{P}(N(60) - N(55 + s) \leq 1)ds$</span>.</p>
<p>But I quickly realized that finding this probability might not be that easy, since the intervals we are considering overlap, namely <span class="math-container">$N(5)$</span> and <span class="math-container">$N(5 + s)$</span> overlap except for the infinitesimal point <span class="math-container">$s$</span>. So then, is the correct way to integrate just <span class="math-container">$\mathbb{P}(N(s) \leq 1$</span> over the region? Moreover, I think that my line of reasoning is missing something critical, since I do not see a reason, why the summation would not end up being negative.</p>
| robjohn | 13,854 | <p><strong>Generating Function Approach</strong></p>
<p>Let's compute the probability that no <span class="math-container">$5$</span> minute span has more than <span class="math-container">$1$</span> car entry.</p>
<hr />
<p><strong>Poisson with <span class="math-container">$\bf{1}$</span> Minute Resolution</strong></p>
<p>Poisson says the probability that <span class="math-container">$0$</span> cars enter the street in a given minute is <span class="math-container">$e^{-\lambda}$</span>, and the probability that <span class="math-container">$1$</span> car enters the street in a given minute is <span class="math-container">$\lambda e^{-\lambda}$</span>.</p>
<p>Let a minute with one car be represented by '<span class="math-container">${+}$</span>', which has a probability of <span class="math-container">$\lambda e^{-\lambda}$</span>. Let a minute with no cars be represented by '<span class="math-container">${-}$</span>', which has a probability of <span class="math-container">$e^{-\lambda}$</span>.</p>
<p>The <span class="math-container">$60$</span> minute span can be uniquely built from atoms of '<span class="math-container">${-}$</span>' and '<span class="math-container">${+}{-}{-}{-}{-}$</span>' followed by an optional terminal atom of '<span class="math-container">$\text{+}$</span>', '<span class="math-container">$\text{+}{-}$</span>', '<span class="math-container">$\text{+}{-}{-}$</span>', '<span class="math-container">$\text{+}{-}{-}{-}$</span>'</p>
<p>This gives a generating function of
<span class="math-container">$$
f_\lambda(x)=\frac{1+\overbrace{\ \ \lambda e^{-\lambda}x\ \ }^{+}+\overbrace{\lambda e^{-2\lambda}x^2}^{+-}+\overbrace{\lambda e^{-3\lambda}x^3}^{+--}+\overbrace{\lambda e^{-4\lambda}x^4}^{+---}}{1-\underbrace{\ \ \ e^{-\lambda}x\ \ \ }_{-}-\underbrace{\lambda e^{-5\lambda}x^5}_{+----}}\tag1
$$</span>
The probability that in no <span class="math-container">$5$</span> minute span of the hour there is more than <span class="math-container">$1$</span> car is
<span class="math-container">$$
\begin{align}
\left[x^{60}\right]f_\lambda(x)
&=e^{-60\lambda}\left(1+60\lambda+1540\lambda^2+22100\lambda^3+194580\lambda^4\right.\\
&\left.+1086008\lambda^5+3838380\lambda^6+8347680\lambda^7+10518300\lambda^8\right.\\[2pt]
&\left.+6906900\lambda^9+1961256\lambda^{10}+167960\lambda^{11}+1820\lambda^{12}\right)
\end{align}\tag2
$$</span>
Computing the complementary probability, we get the probability that two or more cars will enter in a single <span class="math-container">$5$</span> minute period during an hour:</p>
<p><a href="https://i.stack.imgur.com/hNjNk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/y5Xvq.png" alt="enter image description here" /></a></p>
<hr />
<p><strong>The Generating Function</strong></p>
<p>In order to prevent any <span class="math-container">$5$</span> minute span from having more than <span class="math-container">$1$</span> car entry, above, we break down the <span class="math-container">$60$</span> minutes into <span class="math-container">$1$</span> minute atoms:</p>
<blockquote>
<p>The <span class="math-container">$60$</span> minute span can be uniquely built from atoms of '<span class="math-container">${-}$</span>' and '<span class="math-container">${+}{-}{-}{-}{-}$</span>' followed by an optional terminal atom of '<span class="math-container">$\text{+}$</span>', '<span class="math-container">$\text{+}{-}$</span>', '<span class="math-container">$\text{+}{-}{-}$</span>', '<span class="math-container">$\text{+}{-}{-}{-}$</span>'</p>
</blockquote>
<p>and we compute the probabilities for each of these atoms using the Poisson Distribution:</p>
<blockquote>
<p>Poisson says the probability that <span class="math-container">$0$</span> cars enter the street in a given minute is <span class="math-container">$e^{-\lambda}$</span>, and the probability that <span class="math-container">$1$</span> car enters the street in a given minute is <span class="math-container">$\lambda e^{-\lambda}$</span>.</p>
</blockquote>
<p>In the generating function, we represent an '<span class="math-container">${+}$</span>' by <span class="math-container">$\lambda e^{-\lambda}x$</span> and an '<span class="math-container">${-}$</span>' by <span class="math-container">$e^{-\lambda}x$</span>. Thus, the power of <span class="math-container">$x$</span> counts the minutes and the coefficients accumulate the probabilities. Thus, for each of these atoms, we include a factor of
<span class="math-container">$$
\overbrace{\phantom{i^5}e^{-\lambda}x\phantom{i^5}}^{-}+\overbrace{\lambda e^{-5\lambda}x^5}^{+----}\tag3
$$</span></p>
<p>To account for any number of the standard atoms ('<span class="math-container">${+}$</span>' and '<span class="math-container">${+}{-}{-}{-}{-}$</span>'), we total the product of all powers of the standard atom:
<span class="math-container">$$
\sum_{k=0}^\infty\left(e^{-\lambda}x+\lambda e^{-5\lambda}x^5\right)^k=\frac1{1-e^{-\lambda}x-\lambda e^{-5\lambda}x^5}\tag4
$$</span>
Since products of the standard atoms of length <span class="math-container">$4$</span> or more will end in at least <span class="math-container">$4$</span> '<span class="math-container">${-}$</span>'s, we add optional atoms of '<span class="math-container">${+}$</span>', '<span class="math-container">${+}{-}$</span>', '<span class="math-container">${+}{-}{-}$</span>', or '<span class="math-container">${+}{-}{-}{-}$</span>'. This is included in the generating function as a factor of
<span class="math-container">$$
\overbrace{\phantom{iw^2}1\phantom{iw^2}}^\text{none}+\overbrace{\phantom{i^2}\lambda e^{-\lambda}x\phantom{i^2}}^{+}+\overbrace{\lambda e^{-2\lambda}x^2}^{+-}+\overbrace{\lambda e^{-3\lambda}x^3}^{+--}+\overbrace{\lambda e^{-4\lambda}x^4}^{+---}\tag5
$$</span>
Thus, we arrive at the generating function, <span class="math-container">$f_\lambda(x)$</span>, given in <span class="math-container">$(1)$</span>. To get the probability for a span of <span class="math-container">$60$</span> minutes, we look at the coefficient of <span class="math-container">$x^{60}$</span>: <span class="math-container">$\left[x^{60}\right]f_\lambda(x)$</span>.</p>
<hr />
<p><strong>Poisson with <span class="math-container">$\boldsymbol{\frac1n}$</span> Minute Resolution</strong></p>
<p>We can handle finer resolutions using atoms of '<span class="math-container">${-}$</span>' and '<span class="math-container">${+}\overbrace{{-}{-}\cdots{-}}^{5n-1}$</span>'. To simplify notation, we will write <span class="math-container">$e^{-\lambda/n}x=u$</span>.
<span class="math-container">$$
\begin{align}
g_{\lambda,n}(x)
&=\frac{1+\frac\lambda{n}e^{-\lambda/n}x+\frac\lambda{n}e^{-2\lambda/n}x^2+\cdots+\frac\lambda{n}e^{-(5n-1)\lambda/n}x^{5n-1}}{1-e^{-\lambda/n}x-\frac\lambda{n}e^{-5\lambda}x^{5n}}\tag{6a}\\
&=\frac{1+\frac\lambda{n}u\,\frac{1-u^{5n-1}}{1-u}}{1-u-\frac\lambda{n}u^{5n}}\tag{6b}\\
&=\frac{\left(1-u-\frac\lambda{n}u^{5n}\right)+\frac\lambda{n}u}{\left(1-u-\frac\lambda{n}u^{5n}\right)\left(1-u\right)}\tag{6c}\\
&=\frac1{1-u}+\frac{\frac\lambda{n}u}{\left(1-u-\frac\lambda{n}u^{5n}\right)(1-u)}\tag{6d}\\
&=\frac1{1-u}+\left(\frac1{1-u-\frac\lambda{n}u^{5n}}-\frac1{1-u}\right)\frac1{u^{5n-1}}\tag{6e}
\end{align}
$$</span>
Since all coefficients of <span class="math-container">$\frac1{1-u}$</span> are <span class="math-container">$1$</span>, we get
<span class="math-container">$$
\begin{align}
\left[x^{60n}\right]g_{\lambda,n}(x)
&=e^{-60\lambda}\!\left[u^{60n}\right]g_{\lambda,n}(x)\tag{7a}\\[9pt]
&=e^{-60\lambda}\!\left[u^{65n-1}\right]\frac1{1-u-\frac\lambda{n}u^{5n}}\tag{7b}\\
&=e^{-60\lambda}\!\left[u^{65n-1}\right]\sum_{k=0}^\infty\left(u+\frac\lambda{n}u^{5n}\right)^k\tag{7c}\\
&=e^{-60\lambda}\!\left[u^{65n-1}\right]\sum_{k=0}^\infty\sum_{j=0}^k\binom{k}{j}u^{k-j}\left(\frac\lambda{n}\right)^ju^{5nj}\tag{7d}\\
&=e^{-60\lambda}\sum_{j=0}^{12}\binom{65n-1-(5n-1)j}{j}\left(\frac\lambda{n}\right)^j\tag{7e}
\end{align}
$$</span>
For <span class="math-container">$n=1$</span>, <span class="math-container">$\text{(7e)}$</span> agrees with <span class="math-container">$(2)$</span>.</p>
<hr />
<p><strong>Infinite Resolution</strong></p>
<p>The limit as <span class="math-container">$n\to\infty$</span> of <span class="math-container">$\text{(7e)}$</span> is
<span class="math-container">$$
\begin{align}
p(\lambda)
&=e^{-60\lambda}\sum_{j=0}^{12}\frac{(65-5j)^j}{j!}\lambda^j\tag{8a}\\
&=e^{-60\lambda}\left(\vphantom{\frac11}\right.
1+60\lambda+\frac{3025}2\lambda^2+\frac{62500}3\lambda^3+\frac{1366875}8\lambda^4\\
&+\frac{2560000}3\lambda^5+\frac{367653125}{144}\lambda^6+\frac{30375000}7\lambda^7+\frac{30517578125}{8064}\lambda^8\\
&+\frac{800000000}{567}\lambda^9+\frac{284765625}{1792}\lambda^{10}+\frac{15625000}{6237}\lambda^{11}+\frac{9765625}{19160064}\lambda^{12}
\left.\vphantom{\frac11}\right)\tag{8b}
\end{align}
$$</span>
<a href="https://i.stack.imgur.com/M3psL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nGvs8.png" alt="enter image description here" /></a></p>
<p>At <span class="math-container">$\frac14$</span> car per minute, the probability is <span class="math-container">$99.91\%$</span> that two or more cars will enter in a single <span class="math-container">$5$</span> minute period during an hour.</p>
|
3,938,933 | <p>I am interested in the general case, but let us start with a smaller example. Suppose that cars arrive to a street with intensity <span class="math-container">$\lambda$</span> per minute. We would like to know the probability that at least two cars have arrived on the street during any five minute interval in the next hour. How can we find this?</p>
<p>My initial thought was to use complementary event: 1 - the probability that at most one car is on the street in any five minute period during the next hour. Hence my reasoning was something like <span class="math-container">$1 - \int_0^{55}\mathbb{P}(N(s + 5) - N(s) \leq 1)ds - \int_{55}^{60}\mathbb{P}(N(60) - N(55 + s) \leq 1)ds$</span>.</p>
<p>But I quickly realized that finding this probability might not be that easy, since the intervals we are considering overlap, namely <span class="math-container">$N(5)$</span> and <span class="math-container">$N(5 + s)$</span> overlap except for the infinitesimal point <span class="math-container">$s$</span>. So then, is the correct way to integrate just <span class="math-container">$\mathbb{P}(N(s) \leq 1$</span> over the region? Moreover, I think that my line of reasoning is missing something critical, since I do not see a reason, why the summation would not end up being negative.</p>
| robjohn | 13,854 | <p><strong>Conditional Probability Approach</strong></p>
<p>Poisson says that in an hour, the probability that <span class="math-container">$n$</span> cars have entered is
<span class="math-container">$$
e^{-60\lambda}\frac{(60\lambda)^n}{n!}\tag1
$$</span>
Given that <span class="math-container">$n$</span> cars have entered in the hour, the probability that none is within <span class="math-container">$5$</span> minutes of another is the ratio of the volumes of the simplices
<span class="math-container">$$
\frac{\left|\left\{x_k\ge0:\sum\limits_{k=1}^nx_k\le60-5(n-1)\right\}\right|}{\left|\left\{x_k\ge0:\sum\limits_{k=1}^nx_k\le60\right\}\right|}=\left(\frac{13-n}{12}\right)^n\tag2
$$</span>
Each <span class="math-container">$x_k$</span> is the time from car <span class="math-container">$k-1$</span>, or the beginning of the hour, to car <span class="math-container">$k$</span>. With <span class="math-container">$n-1$</span> buffers of <span class="math-container">$5$</span> minutes removed, we get the numerator; without the buffers removed, we get the denominator.</p>
<p>Thus, <a href="https://en.wikipedia.org/wiki/Bayes%27_theorem" rel="nofollow noreferrer">Bayes' Theorem</a> gives
<span class="math-container">$$
\begin{align}
p(\lambda)
&=\sum_{n=0}^{12}e^{-60\lambda}\frac{(60\lambda)^n}{n!}\left(\frac{13-n}{12}\right)^n\tag{3a}\\
&=e^{-60\lambda}\sum_{n=0}^{12}\frac{(65-5n)^n}{n!}\lambda^n\tag{3b}\\
&=e^{-60\lambda}\left(\vphantom{\frac11}\right.
1+60\lambda+\frac{3025}2\lambda^2+\frac{62500}3\lambda^3+\frac{1366875}8\lambda^4\\[3pt]
&+\frac{2560000}3\lambda^5+\frac{367653125}{144}\lambda^6+\frac{30375000}7\lambda^7+\frac{30517578125}{8064}\lambda^8\\[3pt]
&+\frac{800000000}{567}\lambda^9+\frac{284765625}{1792}\lambda^{10}+\frac{15625000}{6237}\lambda^{11}+\frac{9765625}{19160064}\lambda^{12}
\left.\vphantom{\frac11}\right)\tag{3c}
\end{align}
$$</span>
This matches the result from <span class="math-container">$(8)$</span> of <a href="https://math.stackexchange.com/a/3939173">my previous answer</a>.</p>
<p><a href="https://i.stack.imgur.com/5ffko.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/susS7.png" alt="enter image description here" /></a></p>
|
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