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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3,510,156 | <p>This is a duplicate question of <a href="https://math.stackexchange.com/questions/2088815/find-integers-solutions-of-x27-y5#">Find integers solutions of $x^2+7=y^5$</a>, however there was no full answer. The solutions <span class="math-container">$(\pm5, 2)$</span> and <span class="math-container">$(\pm 181, 8)$</span> have been found. </p>
<p>The usual strategy for such a question is to work inside the ring of integers of <span class="math-container">$\mathbb{Q}(\sqrt{-7})$</span>, which is <span class="math-container">$\mathcal{O} = \mathbb{Z}[ \frac{1+\sqrt{-7}}{2}]$</span>. It turns out that this is a unique factorisation domain (which one can figure out by calculating its Class group). So it is natural to factor the equation as <span class="math-container">$(x - \sqrt{-7})(x+\sqrt{-7}) = y^5$</span>. If we assume that <span class="math-container">$x-\sqrt{-7}$</span> and <span class="math-container">$x+\sqrt{-7}$</span> are coprime, we find that <span class="math-container">$x+\sqrt{-7} = \beta^5$</span> for a certain <span class="math-container">$\beta = a + b\frac{1+\sqrt{-7}}{2}\in \mathbb{Z}[\frac{1+\sqrt{-7}}{2}]$</span>. Writing <span class="math-container">$c= 2a+b$</span> and expanding the fifth power, this gives the system of equations
<span class="math-container">$$ c^5 -70 c^3 b^2 + 245 c^4 b = 32 x, $$</span>
<span class="math-container">$$ 5 c^4 b -70 c^2 b^3 + 49 b^5 = 32. $$</span>
Now with enough patience, one can show that this system has no solutions with <span class="math-container">$b \equiv c \pmod{2}$</span>. </p>
<p>However this contradicts the solutions that we have found. And indeed there's no reason for <span class="math-container">$x \pm \sqrt{-7}$</span> to be coprime when <span class="math-container">$x$</span> is odd. </p>
<p>What is the approach to solve the remaining case of this diophantine equation?</p>
<p>One approach that I have tried is that the coprime condition holds inside the ring <span class="math-container">$\cal{O}[\frac{1}{2}]$</span>. This gives the equation <span class="math-container">$x + \sqrt{-7} = (a+b\sqrt{-7})^5$</span> with <span class="math-container">$a,b \in \mathbb{Z}[\frac{1}{2}]$</span>, which I am unable to solve.</p>
| Yong Hao Ng | 31,788 | <p>A possible factorization argument leading up till @Kevin's final form, for reference.</p>
<hr>
<p>Since
<span class="math-container">$$
(x+\sqrt{-7}) - (x-\sqrt{-7}) = 2\sqrt{-7}
$$</span>
The possible common factors are <span class="math-container">$\sqrt{-7}$</span> and the prime elements of norm <span class="math-container">$2$</span> (which works out to be <span class="math-container">$(1\pm \sqrt{-7})/2$</span>). The former will cause <span class="math-container">$x$</span> to be divisible by <span class="math-container">$7$</span> and hence fails in the original equation, so we exclude it since it can't happen. </p>
<p>If <span class="math-container">$x$</span> is even, then <span class="math-container">$x\pm \sqrt{-7}$</span> has odd norm so they can't have that common factor of norm <span class="math-container">$2$</span>. This is the part you wrote down. </p>
<hr>
<p>However if <span class="math-container">$x=2r+1$</span> is odd, then since
<span class="math-container">$$
\frac{x + \sqrt{-7}}{2} = r + \frac{1+\sqrt{7}}{2}\in\mathcal O, \frac{x - \sqrt{-7}}{2} = r+1 - \frac{1+\sqrt{7}}{2} \in \mathcal O,
$$</span>
the common factor between <span class="math-container">$x+\sqrt{-7}$</span> and <span class="math-container">$x-\sqrt{-7}$</span> is exactly <span class="math-container">$2$</span>. Since we also know that <span class="math-container">$y=2s$</span> must be even, this means
<span class="math-container">$$
\begin{align*}
2^2\left(r+\frac{1+\sqrt{-7}}{2}\right)\left(r+1-\frac{1+\sqrt{-7}}{2}\right) &= 2^5s^5\\
\left(r+\frac{1+\sqrt{-7}}{2}\right)\left(r+1-\frac{1+\sqrt{-7}}{2}\right) &= 2^3s^5 = \left(\frac{1+\sqrt{-7}}{2}\right)^3\left(\frac{1-\sqrt{-7}}{2}\right)^3s^5
\end{align*}
$$</span>
So by the coprime-ness (and absorbing any units into <span class="math-container">$\beta^5$</span>), you now have 4 possibilities
<span class="math-container">$$
r+\frac{1+\sqrt{-7}}{2} \in \left\{\beta^5,\left(\frac{1+\sqrt{-7}}{2}\right)^3\beta^5,\left(\frac{1-\sqrt{-7}}{2}\right)^3\beta^5,2^3\beta^5\right\}
$$</span>
With <span class="math-container">$\beta = a+b(1+\sqrt{-7})/2$</span> this works out to be 4 different (Thue) equations when comparing the real and imaginary parts:
<span class="math-container">$$
\begin{align*}
E1: 2r &= -1 + 2 a^5 + 5 a^4 b - 30 a^3 b^2 - 50 a^2 b^3 + 5 a b^4 + 11 b^5\\
1 &= 5 a^4 b + 10 a^3 b^2 - 10 a^2 b^3 - 15 a b^4 - b^5\\
E2: 2r &= -1 - 5 a^5 + 5 a^4 b + 110 a^3 b^2 + 90 a^2 b^3 - 65 a b^4 - 31 b^5\\
1 &= - a^5 - 15 a^4 b - 10 a^3 b^2 + 50 a^2 b^3 + 35 a b^4 - 3 b^5\\
E3: 2r &= -1 - 5 a^5 - 30 a^4 b + 40 a^3 b^2 + 160 a^2 b^3 + 40 a b^4 - 24 b^5\\
1 &= a^5 - 10 a^4 b - 40 a^3 b^2 + 40 a b^4 + 8 b^5\\
E4: 2r &= -1 + 16 a^5 + 40 a^4 b - 240 a^3 b^2 - 400 a^2 b^3 + 40 a b^4 + 88 b^5\\
1 &= 40 a^4 b + 80 a^3 b^2 - 80 a^2 b^3 - 120 a b^4 - 8 b^5
\end{align*}
$$</span>
Equation 4 clearly has no solutions modulo 2. </p>
<p>Equation 1 must have <span class="math-container">$b=\pm 1$</span>, then solving for <span class="math-container">$a$</span> (factoring over <span class="math-container">$\mathbb Z$</span>) gives only two integer solutions: <span class="math-container">$(a,b)= (0,-1),(1,-1)$</span>. Then <span class="math-container">$r=-6,5$</span> which corresponds to <span class="math-container">$x=-11,11$</span>, which both fails. </p>
<p>For equation 2, doing a substitution of
<span class="math-container">$$
(a,b,r) = (-u - v, v, -w-1)
$$</span>
will reveal that it's exactly the same form as equation 3. (Upon which <span class="math-container">$(u,v,w) = (a,b,r)$</span> in equation 3.) </p>
<hr>
<p>Equation 2 is given by @Kevin's solution. Alternatively Using Pari/GP to solve the Thue equation
<span class="math-container">$$
1 = - a^5 - 15 a^4 b - 10 a^3 b^2 + 50 a^2 b^3 + 35 a b^4 - 3 b^5
$$</span>
returns
<span class="math-container">$$
(a,b) = (-1, 0), (2, -1)
$$</span>
Then <span class="math-container">$r=90,2$</span>, so <span class="math-container">$x=181,5$</span>. Then for equation 3, using the earlier relationship of <span class="math-container">$(u,v,w) = (-a-b,b,-r-1)$</span> gives <span class="math-container">$w =-91,-3$</span>. Hence <span class="math-container">$x=-181,-5$</span>.</p>
|
1,557,688 | <p>I want to show that the two metrics are equivalent. </p>
<p>Suppose we have a metric space $X \times Y$. Two metrics are defined as:</p>
<p>$d_{X \times Y}((x, y), (x', y')) := \max\{d_X(x, x'), d_Y(y, y')\}$</p>
<p>$d'_{X \times Y}((x, y), (x', y')) := d_X(x, x')+d_Y(y, y')$</p>
<p>Here is my attempt at proof:</p>
<p>Define $\tau$ to be the collection of open sets with respect to $d_{X \times Y}((x, y), (x', y'))$ . Also define $\tau*$ to be the collection of open sets with respect to $d'_{X \times Y}((x, y), (x', y'))$.</p>
<p>First, assume an open set $U \in \tau$. We want to show that $U \in \tau*$.</p>
<p>Since $U \in \tau$, $\forall (x, y) \in U$, $\exists p > 0$ such that $B_{p}^{d} (x, y) \subset U$.</p>
<p>Then $\forall (x, y) \in U$, $B_{p/2}^{d} (x, y) \subset U.$
Since $d_X(x,x') + d_Y(y, y') \leqslant 2 \cdot \max\{d_X(x, x'), d_Y(y, y')\}$,</p>
<p>If we set $r := 2 \cdot \max\{d_X(x, x'), d_Y(y, y')\}$ then
$B_{r}^{d'}(x,y) \subset B_{p/2}^{d} \subset U$.</p>
<p>As you can see, this reasoning is confusing, in fact I got confused writing this. I can kind of see the direction I should head to but this is not it. Can someone enlighten me?</p>
<p>Thanks.</p>
<p>If I get this part correct I think I can handle the converse.</p>
| Em. | 290,196 | <p>Generally, I think, this is taught as or thought of as a box full of "good" items and "bad" items. I will use a slightly different notation. Suppose we have a box full of $N$ balls, and there are $G$ good balls, with $G\leq N$. Let $X$ represent the number of "good" balls I draw when I grab $n$ balls from the box. Then if I want to calculate the probability of getting $k$ good balls in $n$ draws, then in notation it is
$$P(X = k) = \frac{\binom{G}{k}\binom{N-k}{n-k}}{\binom{N}{n}}.$$
The denominator counts all the ways to choose $n$ balls from the $N$ balls in the box. The numerator counts all the ways to choose $n$ balls where there $k$ good balls and $n-k$ bad balls.</p>
<hr>
<p>So from your perspective, if you choose one of the $K$ items, or an event from $K$ is realized, that is considered a success. You can consider it a "good" ball.</p>
|
2,837,281 | <p>I know the splitting field is generated by $2^{1/4}$ and $i$, I could show $\mathbb{Q} [ 2^{1/4}, i] = \mathbb{Q}[i+2^{1/4}]$ using some algebra. </p>
<p>For the non trivial direction $\mathbb{Q} [ 2^{1/4}, i] \subset \mathbb{Q}[i+2^{1/4}]$. Let us call $\alpha = i+2^{1/4}$, then we know
$$(\alpha-i)^4 - 2 = 0$$
expand the 4th power
$$\alpha^4 - 4\alpha^3 i - 6 \alpha^2 + 4\alpha i + 1 - 2 = 0$$
then we can solve for $i$ in terms of $\alpha$, so $i\in \mathbb{Q}[\alpha]=:\mathbb{Q}[i+2^{1/4}]$. Then $2^{1/4} \in \mathbb{Q}[i+2^{1/4}]$ as well. </p>
<p><strong>However</strong> the hint is to show the orbit of $i+2^{1/4}$ has more than 5 elements under the action of the galois group $Gal(\mathbb Q[2^{1/4}, i]/\mathbb{Q})$. I calculated the Galois group which is $D_8$, and the orbit has more than 5 elements, but how would conclude from this?</p>
| The way of life | 373,966 | <p>To elaborate on @Eric Wofsey 's answer, or perhaps show a more direct view of the injection, define a $G=Gal(K/\mathbb{Q})$-action on $K=\mathbb{Q}[2^{1/4},i]$ by $g.x=g(x)$.</p>
<p>Let us look at the orbit $o(x)$ of $x=2^{1/4}+i$ and the stabilizer $stab_G(x)$.</p>
<p>By orbit-stabilizer theorem $|o(x)|=\frac{|G|}{|stab_G(x)|}$.</p>
<p>Say you prove that $|o(x)|\geq5$ and $|G|=8$. Then necessarily $|stab_G(x)|=1$ and so $H=Aut(K/\mathbb{Q}[x])=\{{e}\}$.</p>
<p>We know by the fundamental theorem of Galois theory that $\mathbb{Q}[x]=Fix(H)=K$, the latter being $K$ since $H$ is trivial.</p>
|
912,176 | <p>If $y,z$ are elements of an archimedean field $F$ and if $y<z$, then there is a rational element $r$ of $F$ such that $y<r<z$</p>
<p>The proof begins with saying that it is no loss of generality that we assume that $0<y<z$</p>
<p>I don't understand well why this the case. Please guide me .</p>
<p>I think there will be loss of generality because suppose : $z>0$ but still it's possible that $y<0$ i.e. $z \in P ~;~ y \notin P$ where $P$ is a positive class in $F$</p>
<p>Thank you for your help.</p>
| vociferous_rutabaga | 164,345 | <p>It seems to me that by "without loss of generality," the author really means "since the other case is much easier:" if $z>0$, $y<0$, it's clear that $0$ is the desired rational element. If both are negative, multiply by $-1$.</p>
|
3,756,436 | <p>Recently I was doing a physics problem and I ended up with this quadratic in the middle of the steps:</p>
<p><span class="math-container">$$ 0= X \tan \theta - \frac{g}{2} \frac{ X^2 \sec^2 \theta }{ (110)^2 } - 105$$</span></p>
<p>I want to find <span class="math-container">$0 < \theta < \frac{\pi}2$</span> for which I can later take the largest <span class="math-container">$X$</span> value that solves this equation, i.e. optimize the implicit curve to maximize <span class="math-container">$X$</span>.</p>
<p>I tried solving this by implicit differentiation (assuming <span class="math-container">$X$</span> can be written as a function of <span class="math-container">$\theta$</span>) with respect to <span class="math-container">$\theta$</span> and then by setting <span class="math-container">$\frac{dX}{d\theta} = 0$</span>:</p>
<p><span class="math-container">\begin{align}
0 &= X \sec^2 \theta + \frac{ d X}{ d \theta} \tan \theta - \frac{g}{2} \frac{ 2 \left( X \sec \theta \right) \left[ \frac{dX}{d \theta} \sec \theta + X \sec \theta \tan \theta \right]}{ (110)^2 } - 105 \\
0 &= X \sec^2 \theta - \frac{g}{2} \frac{ 2( X \sec \theta) \left[ X \sec \theta \tan \theta\right] }{ (110)^2 } \\
0 &= 1 - \frac{ Xg \tan \theta}{(110)^2} \\
\frac{ (110)^2}{ g \tan \theta} &= X
\end{align}</span></p>
<p>This is still not an easy equation to solve. However, one of my friends told we could just take the discriminant of the quadratic in terms of <span class="math-container">$X$</span>, and solve for <span class="math-container">$\theta$</span> such that <span class="math-container">$D=0$</span>.</p>
<p>Taking discriminant and equating to 0, I get</p>
<p><span class="math-container">$$ \sin \theta = \frac{ \sqrt{2 \cdot 10 \cdot 105} }{110}$$</span></p>
<p>and, the angle from it is, 24.45 degrees</p>
<p>I tried the discriminant method, but it gave me a different answer from the implicit differentiation method. I ended up with two solutions with the same maximum value of <span class="math-container">$X$</span> but different angles: <span class="math-container">$\theta =24.45^\text{o}$</span> and <span class="math-container">$X=1123.54$</span> (from discriminant method), and
<span class="math-container">$\theta = 47^\text{o}$</span> and <span class="math-container">$X=1123.54$</span> (from implicit differentiation).</p>
<p>I later realized the original quadratic can only have solutions if <span class="math-container">$D(\theta) > 0$</span>, where <span class="math-container">$D$</span> is the discriminant. Using the discriminant, I can find a lower bound on the angle. Once I have the lower bound, if I can prove that <span class="math-container">$X$</span> decreases monotonically as a function of <span class="math-container">$\theta$</span>, then I can use the lower bound for further calculations of <span class="math-container">$\theta$</span>.</p>
<p>So then I used the implicit function theorem and got</p>
<p><span class="math-container">$$ \frac{dX}{ d \theta} =- \frac{X \sec^2 \theta -\frac{g X^2}{2 (110)^2} 2 \sec \theta ( \sec \theta \tan \theta) } {\tan \theta - \frac{g \sec^2 \theta}{2 (110)^2} 2X }$$</span></p>
<p>Now the problem here is that I can't prove this function is in monotonic in terms of <span class="math-container">$\theta$</span> as the implicit derivative is a function of both <span class="math-container">$\theta$</span> and <span class="math-container">$X$</span>.</p>
| Narasimham | 95,860 | <p>We can proceed directly.. straightforward way. We have the given relation <span class="math-container">$ (x=X)$</span>:</p>
<p><span class="math-container">$$105 = x\tan\theta-\frac{g}{2} \dfrac{x^2\sec^2\theta}{(110)^2} \tag1 $$</span></p>
<p>To facilitate differentiation let us use symbols for the time being</p>
<p><span class="math-container">$$\ h=105,\ V=110 , \tan\theta =t,\dfrac{2v^2}{g}=a \tag2 $$</span></p>
<p><span class="math-container">$$ xt-x^2(1+t^2)/a -h =0 \tag3$$</span></p>
<p>We need to do <em><strong>implicit differentiation</strong></em> of the <em><strong>fourth order polynomial</strong></em> because <span class="math-container">$x,t$</span> are involved together.</p>
<p>Priming with respect to <span class="math-container">$t$</span> as independent variable</p>
<p><span class="math-container">$$x^{'} t + x -\big[ 2x/a \cdot (1+t^2) \cdot x ^{'} +x^2/a \cdot 2t \big]=0 \tag4 $$</span></p>
<p>Set <span class="math-container">$x^{'}=0 $</span> to find maximum <span class="math-container">$x,$</span> and simplify cancelling <span class="math-container">$x$</span> on LHS getting</p>
<p><span class="math-container">$$ x= \dfrac{a}{2t}\tag5 $$</span></p>
<p>From (3) and (5) we can find <span class="math-container">$(t,\theta) $</span>. Plugging this into (3) and simplifying</p>
<p><span class="math-container">$$\dfrac{1+t^2}{t^2}=\dfrac{2a-4h}{a} \tag6 $$</span></p>
<p><span class="math-container">$$\dfrac{t^2}{1}=\dfrac{a}{a-4h}\rightarrow t = \tan \theta =\sqrt{\dfrac{a}{a-4h}} =\dfrac{1}{\sqrt{1-2gh/V^2}}\tag7 $$</span></p>
<p>Plugging this into (5)</p>
<p><span class="math-container">$$x_{max}=\dfrac{a}{2}\sqrt{\dfrac{a-4h}{a}} = \dfrac{V^2}{g} {\sqrt{1-2gh/V^2}} \tag8 $$</span></p>
<p>You can verify sign of second derivative to check it is maximum. To check for <em>the special case</em></p>
<p><span class="math-container">$$ h=0, \; x_{max}= \dfrac{a}{2}= \dfrac{V^2}{g}\tag9 $$</span></p>
<p>a well known result in projectile motion that the maximum range is double the height reached when projected vertically at angle <span class="math-container">$\theta =\pi/2 $</span> to horizontal.</p>
<p>For this particular case put <span class="math-container">$h=0$</span> in (7) to tally <span class="math-container">$ \theta = 45^{\circ} $</span> okay.</p>
<p>With given <em>numerical values</em> <span class="math-container">$\; V=110, \; g=9.81,\; h=105,$</span> we calculate</p>
<p><span class="math-container">$$ X_{optimum}=1123.54,\;\theta_{optimum}=46.67^{\circ} \tag {10}$$</span></p>
<p>Projectile parabolic trajectories are shown below to scale.</p>
<p>By increasing and decreasing <span class="math-container">$\theta$</span> by <span class="math-container">$\pm 30^{\circ} $</span> the blue, green parabolas should cut <span class="math-container">$ y=h=105\;$</span> at values lower than 1123.54 for an optimum <span class="math-container">$X_m$</span> red parabola to be valid.. a fact that is graphically also verified.</p>
<p><a href="https://i.stack.imgur.com/AKSWL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AKSWL.png" alt="enter image description here" /></a></p>
|
493,102 | <p>I have a concern with nested quantifiers.</p>
<p>I have: $$ \forall x \exists y \forall z(x^2-y+z=0) $$ such that $$ x,y,z \in \Bbb Z^+$$ </p>
<p>My first question, can it be read like this:</p>
<p>$$ \forall x \forall z \exists y(x^2-y+z=0) $$</p>
<p>The way I did it, is I started off with $x=1, z=1 $ </p>
<p>$$ 2-y = 0 $$ $$ y =2 $$</p>
<p>Is this a good approach?</p>
| André Nicolas | 6,312 | <p>The original sentence says that for any $x$, there is a $y$, such that <strong>whatever</strong> $z$ we pick, we have $x^2-y+z=0$.</p>
<p>So the $y$ has to work for all $z$. But that's impossible. If it works for $z=1000$, it fails for all other $z$. The sentence is (very) false. </p>
<p>We <strong>cannot</strong> interchange the quantifiers $\forall$ and $\exists$ without altering the meaning of the sentence. </p>
|
3,178,860 | <p>I am trying to solve the question 27 of Section 10.4 of Dummit and Foote but I am stuck in the first problem: let me state the question and then I will attach the picture of the page of the corresponding book as well:</p>
<p><em>I am stuck at part c and d</em></p>
<p><strong>Prove that the map <span class="math-container">$f: \Bbb C \times \Bbb C \to \Bbb C \times \Bbb C$</span> by <span class="math-container">$f(z_1,z_2)=(z_1z_2,z_1\bar{z_2})$</span> is an <span class="math-container">$\Bbb R $</span> bilinear map.</strong></p>
<p>Now I am calculating <span class="math-container">$f(z_1+z_2, z_3)=((z_1+z_2)z_3,(z_1+z_2)\bar{z_3})=(z_1z_3+z_2z_3,z_1\bar{z_3}+z_2\bar{z_3})$</span> how is it <span class="math-container">$f(z_1, z_3)+f(z_2, z_3)$</span>?</p>
<p>Moreover, <span class="math-container">$f(az_1,z_2)=(az_1z_2,az_1\bar{z_2})$</span> why is it <span class="math-container">$af(z_1,z_2)$</span>?</p>
<p>Now part d) is <strong>Let <span class="math-container">$F$</span> be the <span class="math-container">$\Bbb R$</span> module homomorphism from <span class="math-container">$\Bbb C \otimes \Bbb C$</span> to <span class="math-container">$\Bbb C \times \Bbb C$</span> obtained from <span class="math-container">$f$</span>. Show that <span class="math-container">$F$</span> is <span class="math-container">$\Bbb C$</span> linear and deduce <span class="math-container">$F$</span> to be surjective.</strong></p>
<p>I have omitted the questions that I have proved. I need help basically on the part that I have mentioned in the question. I am attaching the full question as this will help you.</p>
<p><a href="https://i.stack.imgur.com/2Dgl5.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2Dgl5.jpg" alt="Page 1"></a></p>
<p><a href="https://i.stack.imgur.com/fXD1d.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fXD1d.jpg" alt="Page 2"></a></p>
| fGDu94 | 658,818 | <p>it's <span class="math-container">$-\Delta u$</span> (as a functional...). Why, you may ask...</p>
<p>We require that <span class="math-container">$E(u+h)-E(u) = \langle \nabla E(u) , h \rangle$</span> for any <span class="math-container">$h \in H_0^1 (\Omega)$</span>.</p>
<p><span class="math-container">$E(u+h)-E(u) = \frac{1}{2}\int_\Omega 2 \nabla u \nabla h$</span>. Now use integration by parts on this expression to get the answer.</p>
<p>We get that <span class="math-container">$\langle \nabla E(u) , h \rangle = \langle -\Delta u , h\rangle\>$</span>. This tells us we can associate the functional <span class="math-container">$\nabla E(u)$</span> acting on a function <span class="math-container">$h$</span> with integrating <span class="math-container">$h$</span> times <span class="math-container">$-\Delta u$</span>.</p>
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3,178,860 | <p>I am trying to solve the question 27 of Section 10.4 of Dummit and Foote but I am stuck in the first problem: let me state the question and then I will attach the picture of the page of the corresponding book as well:</p>
<p><em>I am stuck at part c and d</em></p>
<p><strong>Prove that the map <span class="math-container">$f: \Bbb C \times \Bbb C \to \Bbb C \times \Bbb C$</span> by <span class="math-container">$f(z_1,z_2)=(z_1z_2,z_1\bar{z_2})$</span> is an <span class="math-container">$\Bbb R $</span> bilinear map.</strong></p>
<p>Now I am calculating <span class="math-container">$f(z_1+z_2, z_3)=((z_1+z_2)z_3,(z_1+z_2)\bar{z_3})=(z_1z_3+z_2z_3,z_1\bar{z_3}+z_2\bar{z_3})$</span> how is it <span class="math-container">$f(z_1, z_3)+f(z_2, z_3)$</span>?</p>
<p>Moreover, <span class="math-container">$f(az_1,z_2)=(az_1z_2,az_1\bar{z_2})$</span> why is it <span class="math-container">$af(z_1,z_2)$</span>?</p>
<p>Now part d) is <strong>Let <span class="math-container">$F$</span> be the <span class="math-container">$\Bbb R$</span> module homomorphism from <span class="math-container">$\Bbb C \otimes \Bbb C$</span> to <span class="math-container">$\Bbb C \times \Bbb C$</span> obtained from <span class="math-container">$f$</span>. Show that <span class="math-container">$F$</span> is <span class="math-container">$\Bbb C$</span> linear and deduce <span class="math-container">$F$</span> to be surjective.</strong></p>
<p>I have omitted the questions that I have proved. I need help basically on the part that I have mentioned in the question. I am attaching the full question as this will help you.</p>
<p><a href="https://i.stack.imgur.com/2Dgl5.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2Dgl5.jpg" alt="Page 1"></a></p>
<p><a href="https://i.stack.imgur.com/fXD1d.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fXD1d.jpg" alt="Page 2"></a></p>
| Matematleta | 138,929 | <p>The Frechet derivative <span class="math-container">$DE$</span>, if it exists, is unique and satisfies</p>
<p><span class="math-container">$$E(u+h)=E(u)+DE(h)+r(h),\ $$</span> where <span class="math-container">$r(h)$</span> is <span class="math-container">$o(h).$</span> So, if we can find a candidate that satisfies the equation, we are done. </p>
<p>Claim (admittedly with the foreknowledge that the claim is true): </p>
<p><span class="math-container">$$DE(h)=\int_{\Omega}\langle \nabla u,\nabla h\rangle$$</span></p>
<p>The proof is a calculation:</p>
<p><span class="math-container">$$E(u+h)-E(u)=\frac{1}{2}\left (\int_{\Omega} | \nabla (u+h)|^2-\int_{\Omega} | \nabla (u)|^2\right )=\frac{1}{2}\left (\int_{\Omega} \langle\nabla (u+h),\nabla (u+h)\rangle-\int_{\Omega} | \nabla (u)|^2\right )=\int_{\Omega}\langle \nabla u,\nabla h\rangle+\frac{1}{2}\int_{\Omega}\langle \nabla h,\nabla h\rangle,$$</span></p>
<p>from which we see that, setting <span class="math-container">$r(h)=\frac{1}{2}\int_{\Omega}\langle \nabla h,\nabla h\rangle$</span> and noting that is is <span class="math-container">$o(h)$</span>, we have </p>
<p><span class="math-container">$$DE(h)=\int_{\Omega}\langle \nabla u,\nabla h\rangle.$$</span></p>
|
978,384 | <p>The following picture is constructed by connecting each corner of a square with the midpoint of a side from the square that is not adjacent to the corner. These lines create the following red octagon:</p>
<p><img src="https://i.stack.imgur.com/PZyGa.jpg" alt="enter image description here"></p>
<p>The question is, what is the ratio between the area of the octagon and the area of the square. One is supposed to find the solution without a ruler. </p>
<p>By removing some lines, I find it easy to see that the ratio between the yellow area and the square is 1/4. But I am not sure if this helps.</p>
<p><img src="https://i.stack.imgur.com/lpIZF.jpg" alt="enter image description here"></p>
| flawr | 109,451 | <p>I think the following image will say more than any text. You can divide the image into smaller squares that will allow you immediately to calculate the ratio.</p>
<p>The ratio of the red area within the whole square is the same as the red area in the big green square to the area of the whole green square. And this is (counting in units of the 9 small squares): $(1+1/4+1/4) : 9 = 1.5 : 9 = 1:6$.</p>
<p><img src="https://i.stack.imgur.com/1Gre3.png" alt="drawing"></p>
|
978,384 | <p>The following picture is constructed by connecting each corner of a square with the midpoint of a side from the square that is not adjacent to the corner. These lines create the following red octagon:</p>
<p><img src="https://i.stack.imgur.com/PZyGa.jpg" alt="enter image description here"></p>
<p>The question is, what is the ratio between the area of the octagon and the area of the square. One is supposed to find the solution without a ruler. </p>
<p>By removing some lines, I find it easy to see that the ratio between the yellow area and the square is 1/4. But I am not sure if this helps.</p>
<p><img src="https://i.stack.imgur.com/lpIZF.jpg" alt="enter image description here"></p>
| user514455 | 514,455 | <p>A calculated answer. Maybe not the simplest.</p>
<p><img src="https://i.stack.imgur.com/12crH.gif" alt="enter image description here"></p>
|
3,773,133 | <p>I have been thinking about this problem for a couple of months, and eventually failed. Could someone help me?</p>
<blockquote>
<p>Let <span class="math-container">$M$</span> and <span class="math-container">$X$</span> be two symmetric matrices with <span class="math-container">$M\succeq 0$</span> and <span class="math-container">$X=X^T$</span>, and let <span class="math-container">$p$</span> be a nonzero real number <span class="math-container">$|p|\le 10$</span>. What is the derivative of <span class="math-container">$f$</span> with respect to <span class="math-container">$M$</span>, where</p>
<p><span class="math-container">$$f(M) = \operatorname{Trace}(M^pX),$$</span></p>
<p>and <span class="math-container">$\operatorname{Trace}(A)$</span> is the trace operator to calculate the sum of elements in diagonal line of <span class="math-container">$A$</span>.</p>
</blockquote>
| greg | 357,854 | <p>Use a colon to denote the trace/Frobenius product
<span class="math-container">$$\eqalign{
A:B = {\rm Tr}(A^TB) = {\rm Tr}(AB^T)
}$$</span>
From this definition and the cyclic property
one can deduce the rearrangement rules
<span class="math-container">$$\eqalign{
A:B &= B:A = B^T:A^T \\
A:BC &= B^TA:C = AC^T:B = I:A^TBC = I:BCA^T \\
}$$</span>
If the matrices are symmetric, one can omit the transposes.</p>
<p>Consider the following function defined for integer <span class="math-container">$n$</span> values,
for which one can calculate the differential and gradient.
<span class="math-container">$$\eqalign{
f &= X:A^n \\
df
&= X:\left(\sum_{k=0}^{n-1} A^{k}\,dA\,A^{n-k-1}\right) \\
&= \left(\sum_{k=0}^{n-1} A^{k}\,X\,A^{n-k-1}\right):dA \\
\frac{\partial f}{\partial A}
&= \left(\sum_{k=0}^{n-1} A^{k}\,X\,A^{n-k-1}\right)
\;\doteq\; G \qquad\{{\rm the\,gradient}\} \\\\
}$$</span>
To extend this result to the matrix <span class="math-container">$M$</span> with a rational exponent <span class="math-container">$p=\ell/m$</span>
<br>define the matrix <span class="math-container">$A$</span> such that
<span class="math-container">$$A^n = M^{\ell/m}\;\implies\; A^{mn}=M^{\ell}$$</span>
Then find an expression for <span class="math-container">$dA$</span> in terms of <span class="math-container">$dM$</span>, i.e.
<span class="math-container">$$\eqalign{
\left(\sum_{k=0}^{mn-1} A^{k}\,dA\,A^{mn-k-1}\right)
&= \left(\sum_{j=0}^{\ell-1} M^{j}\,dM\,M^{\ell-j-1}\right) \\
}$$</span>
To avoid tensors, vectorize the matrix expressions
<span class="math-container">$$\eqalign{
&a = {\rm vec}(A),\qquad m = {\rm vec}(M) \\
&\left(\sum_{k=0}^{mn-1} A^{k}\otimes A^{mn-k-1}\right)\,da
= \left(\sum_{j=0}^{\ell-1} M^{j}\otimes M^{\ell-j-1}\right)\,dm \\
}$$</span>
which can be abbreviated to <span class="math-container">$\,\big(B\,da = C\,dm\big).\,$</span>
We'll also need two more vectorizations
<span class="math-container">$$\eqalign{
x &= {\rm vec}(X),\qquad
g &= {\rm vec}(G)
&= \left(\sum_{k=0}^{n-1} A^{k}\otimes A^{n-k-1}\right)x
\;\doteq\; Ex \\
}$$</span>
Substitute all of this into the previous result to obtain
<span class="math-container">$$\eqalign{
df &= g:da \\&= Ex:B^{-1}C\,dm \\&= CB^{-1}Ex:dm \\
\frac{\partial f}{\partial m} &= CB^{-1}Ex
\quad\implies\quad
\frac{\partial f}{\partial M} = {\rm vec}^{-1}\big(CB^{-1}Ex\big) \\
}$$</span>
For non-rational exponents, it'll be even harder.
<br><br></p>
<h3>Update</h3>
The comments pointed out that the above formula only works for <span class="math-container">$n>0$</span>.
To handle negative exponents, you can do the following.
<br><br>
Write the function in terms of the inverse matrix <span class="math-container">$V=A^{-1},\;$</span>
calculate its differential in terms of <span class="math-container">$dV.\;$</span>
Finally, substitute <span class="math-container">$dV=-V\,dA\,V,\;$</span> i.e.
<span class="math-container">$$\eqalign{
f &= X:A^{-n} \\&= X:V^n \\
df
&= \left(\sum_{k=0}^{n-1} V^{k}\,X\,V^{n-k-1}\right):dV \\
&= -\left(\sum_{k=0}^{n-1} V^{k}\,X\,V^{n-k-1}\right):V\,dA\,V \\
&= -\left(\sum_{k=0}^{n-1} V^{k+1}\,X\,V^{n-k}\right):dA \\
\frac{\partial f}{\partial A}
&= -\left(\sum_{k=0}^{n-1} V^{k+1}\,X\,V^{n-k}\right) \\
}$$</span>
|
2,797,902 | <p>AFAIK, every mathematical theory (by which I mean e.g. the theory of groups, topologies, or vector spaces), started out (historically speaking) by formulating a set of axioms that generalize a specific structure, or a specific set of structures. </p>
<p>For example, when people think of a “field” they AFAIK usually think of $\mathbb R$, or $\mathbb C$. A topology started out as a concept defined on $\mathbb R^n$ if I’m not mistaken. </p>
<p>But I’ve also seen cases where a certain structure has a natural topological structure, such as certain sets of propositions in first order logic. As far as I know, the people who formulated the axioms of a topology had no idea of this application. <strong>And the topological structure of a set of FOL statements is certainly conceptually vastly different from one on $\mathbb R^n$, certainly not two structures I would have expected to have such a deep commonality.</strong></p>
<p><strong>I would like to make a list of examples of mathematical structures that</strong></p>
<ol>
<li><p>Are interesting and well-behaved structures (e.g. not mere pathological counter examples)</p></li>
<li><p>satisfy the axioms of some mathematical theory in an interesting and nontrivial way,</p></li>
<li><p>But whose emergence is (conceptually/historically) very different from the structure of which those axioms were originally intended as a generalization.</p></li>
</ol>
| Gödel | 467,121 | <p>The Peano's arithmetic was thought to axiomatize the structure of natural numbers. However, exists the structure of <a href="https://en.wikipedia.org/wiki/Non-standard_model_of_arithmetic" rel="nofollow noreferrer">non standar naturals</a> where exists a biggest natural number and satisfies that axioms. </p>
|
165,069 | <p>I have a list of the following kind:</p>
<pre><code>{{1,0.5},{2,0.6},{3,0.8},{-4,0.9},{-3,0.95}}
</code></pre>
<p>The important property is, that somewhere in the list, the first element of the sublists changes sign (above is from + to -, but could be from - to +). How can I most efficiently split this into two lists:</p>
<pre><code>{{1,0.5},{2,0.6},{3,0.8}}
</code></pre>
<p>and</p>
<pre><code>{{-4,0.9},{-3,0.95}}
</code></pre>
<p>?</p>
| m_goldberg | 3,066 | <p>This works on your example data, but it might not be general enough to satisfy you, but it is the best I can do with only one example.</p>
<pre><code>data = {{1, 0.5}, {2, 0.6}, {3, 0.8}, {-4, 0.9}, {-3, 0.95}};
Column[{d1, d2} = SplitBy[data, Sign @* First]]
</code></pre>
<p><a href="https://i.stack.imgur.com/VpdEa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VpdEa.png" alt="split"></a></p>
<p><code>Column</code>, of course, is only used form display.</p>
|
1,227,419 | <p>$$\int\limits_6^{16}\left(\frac{1}{\sqrt{x^3+7x^2+8x-16}}\right)\,\mathrm dx=\frac{\pi }{k}$$</p>
<p><strong>Note:</strong> $k$ is a constant.</p>
| Prasun Biswas | 215,900 | <p>$$x^3+7x^2+8x-16=(x-1)(x+4)^2\implies \sqrt{x^3+7x^2+8x-16}=\sqrt{x-1}(x+4)$$</p>
<p>Denote the indefinite integral as $I$.</p>
<p>$$I=\int\frac{\mathrm dx}{(x+4)\sqrt{x-1}}$$</p>
<p>Make the substitution $u^2=x-1$ and $2u\,\mathrm du=\mathrm dx$ to get,</p>
<p>$$I=2\int\frac{\mathrm du}{u^2+5}$$</p>
<p>This is an elementary integral which can be evaluated easily. Put the limits and get the value of $k$</p>
|
180,647 | <p>Two persons have 2 uniform sticks with equal length which can be cut at any point. Each person will cut the stick into $n$ parts ($n$ is an odd number). And each person's $n$ parts will be permuted randomly, and be compared with the other person's sticks one by one. When one's stick is longer than the other person's, he will get one point. The person with more points will win the game. How to maximize the probability of winning the game for one of the person. What is the best strategy to cut the stick.</p>
| Jonathan Gleason | 10,109 | <p>You can't actually add a scalar and a matrix. In general, you can't add two matrices unless they are of the same dimension. However, it is often the case that we denote a scalar matrix (a diagonal matrix all of whose entries are the same) by a scalar. For example, you might write $4$ to denote the matrix $\begin{bmatrix}4 & 0 \\ 0 & 4\end{bmatrix}$. The dimension of the matrix has to be understood from context. This notation allows one to write, for example,
$$
4-\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}=\begin{bmatrix}4 & 0 \\ 0 & 4\end{bmatrix}-\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}=\begin{bmatrix}3 & -2 \\ -3 & 0\end{bmatrix}
$$
In your case, however, I have never seen a scalar used to denote a non-square matrix.</p>
|
120,177 | <p>here's a question I had in an exam today:</p>
<blockquote>
<p>Four people are checking 230 exams. In how many ways can you split the
papers between the four of them if you want each one to check at least
15?</p>
</blockquote>
<p>So, after checking $4 \cdot 15$ papers we are left with 170 and so the result is
$$\binom{170+(4-1)}{4-1}=\binom{173}{3}$$
And here's the twist: Following the last question, in how many ways can you split the papers between the four of them, only now the papers are different and its important to count who checks how many. Meaning - The number of ways to divide 230 elements to 4 different groups.</p>
<p>My initial attemp was $$\sum_{x_1,x_2,x_3,x_4=230}\binom{230}{x_1,x_2,x_3,x_4}=4^{230}$$
But that's obviously a mistake since I wasn't counting the first 60 papers correctly.
I think it probably involves using inclusion-exclusion, but I'm not sure how.</p>
<p>Would love to hear some ideas. Thanks!</p>
| Alex Becker | 8,173 | <p>One definition of continuity at a point $p$ is that the value of a function at $p$ is equal to the limit of the function as it approaches $p$. Thus what you want to do is set $$L=f(0)=\lim\limits_{x\to 0}\frac{\sin(.19x)}{x}$$
and I will let you compute that limit yourself.</p>
|
120,177 | <p>here's a question I had in an exam today:</p>
<blockquote>
<p>Four people are checking 230 exams. In how many ways can you split the
papers between the four of them if you want each one to check at least
15?</p>
</blockquote>
<p>So, after checking $4 \cdot 15$ papers we are left with 170 and so the result is
$$\binom{170+(4-1)}{4-1}=\binom{173}{3}$$
And here's the twist: Following the last question, in how many ways can you split the papers between the four of them, only now the papers are different and its important to count who checks how many. Meaning - The number of ways to divide 230 elements to 4 different groups.</p>
<p>My initial attemp was $$\sum_{x_1,x_2,x_3,x_4=230}\binom{230}{x_1,x_2,x_3,x_4}=4^{230}$$
But that's obviously a mistake since I wasn't counting the first 60 papers correctly.
I think it probably involves using inclusion-exclusion, but I'm not sure how.</p>
<p>Would love to hear some ideas. Thanks!</p>
| Arturo Magidin | 742 | <p>By definition, a function $f(x)$ is continuous at $a$ if and only if:</p>
<ol>
<li>$f(x)$ is <em>defined</em> at $x=a$;</li>
<li>$\lim\limits_{x\to a}f(x)$ exists; and</li>
<li>$\lim\limits_{x\to a}f(x) = f(a)$.</li>
</ol>
<p>So for your $f(x)$ to be continuous at $0$ you need it to be defined at $0$ (which it is), and you need
$$L = f(0) = \lim\limits_{x\to 0}f(x) = \lim\limits_{x\to 0}\frac{\sin(0.19x)}{x}$$
to be true.</p>
<p>Since you are free to decide what you want $L$ to be, what you need to do is figure out how much that limit is.</p>
<p><em>HINT</em>: you've probably recently seen the fact that $\lim\limits_{u\to 0}\frac{\sin u}{u} = 1$. Try to use that.</p>
|
3,936,676 | <p>Well this format of a limit <span class="math-container">$0^0$</span> is an indeterminate form.</p>
<p>I claim that whatever this limit is (which depends on the exact question) should always be in between <span class="math-container">$[0,1]$</span>.</p>
<p>Is my claim correct?</p>
<p>I have no mathematical proof for it but just a basic idea, that any number base <span class="math-container">$0$</span> should try to pull towards <span class="math-container">$0$</span>, whereas any number power <span class="math-container">$0$</span> should pull towards <span class="math-container">$1$</span>.</p>
| Théophile | 26,091 | <p>Your claim is wrong. You can choose any nonnegative number as the limit:
<span class="math-container">$$\lim_{x\to0^+}\left(e^{-1/x}\right)^{ax} = e^{-a}.$$</span></p>
<p>This example is from the <a href="https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero" rel="noreferrer">Wikipedia article</a> on <span class="math-container">$0^0$</span>, as @PrasunBiswas suggested. There are other examples there showing a limit of infinity, etc.</p>
|
2,867,479 | <p>From <a href="https://math.stackexchange.com/questions/2867457">ETS Major Field Test in Mathematics</a></p>
<blockquote>
<p>A student is given an exam consisting of
8 essay questions divided into 4 groups of
2 questions each. The student is required to
select a set of 6 questions to answer,
including at least 1 question from each of
the 4 groups. How many sets of questions
satisfy this requirement? </p>
</blockquote>
<p>I'm thinking $$\binom{2}{1}^4 \binom{4}{2}$$</p>
<p>because we have to pick 1 from each of the 4 groups of 2 and then from the remaining 4 questions we pick 2.</p>
| Bram28 | 256,001 | <blockquote>
<p>I'm thinking <span class="math-container">$$\binom{2}{1}^4 \binom{4}{2}$$</span></p>
<p>because we have to pick 1 from each of the 4 groups of 2 and then from the remaining 4 questions we pick 2.</p>
</blockquote>
<p>You are overcounting here: there will be two groups from which you end up picking both questions, and there is of course only <em>one</em> way to pick two questions from a group of two: pick both!</p>
<p>However, by first picking <span class="math-container">$1$</span> out of two (which can be done in two ways), and then picking the other one, your method ends up counting <em>two</em> ways to pick both questions from a group ... but those two ways are really just two different orderings of the group of two, and ordering is not a consideration for this question. Therefore, you overcount by a factor of <span class="math-container">$2$</span> for each of the groups where you end up picking both questions.</p>
<p>Finally, since there are two such groups, you overcount by a factor of <span class="math-container">$4$</span>.</p>
|
3,074,035 | <p>I am trying to find a simplified form for this summation:</p>
<p><span class="math-container">$$B(k,j) \equiv \sum_{i=1}^k (-1)^{k+i} {i \choose j} (k-1)_{i-1} \quad \quad \quad \text{for } 1 \leqslant j \leqslant k,$$</span></p>
<p>where the terms <span class="math-container">$(k-1)_{i-1} = (k-1) \cdots (k-i+1)$</span> are <a href="https://en.wikipedia.org/wiki/Falling_and_rising_factorials" rel="nofollow noreferrer">falling factorials</a> (and <span class="math-container">$(k-1)_0 = 1$</span> by convention). I have done some work on this by generating a matrix of values of this quantity (see below) and I can recognise simplified forms for some of the parts of the matrix, but I have been unable to see a simplified form for the whole thing.</p>
<hr>
<p><strong>Computing the terms and looking for a pattern:</strong> In case it helps to try to recognise a pattern, I have computed a matrix of terms using the <code>R</code> code below. From the matrix of values it is clear that <span class="math-container">$B(k,k) = (k-1)!$</span> and <span class="math-container">$B(k,k-1) = (k-1) (k-1)!$</span>. I also recognise that <span class="math-container">$B(k,1)$</span> is <a href="https://oeis.org/A000255" rel="nofollow noreferrer">A000255</a>. I do not recognise the overall matrix of numbers as any simple form.</p>
<pre><code>#Generate function
B <- function(k,j) { T1 <- (-1)^(k+1:k);
T2 <- choose(1:k, j);
TT <- c(1,(k-1):1);
T3 <- cumprod(TT);
sum(T1*T2*T3); }
#Create matrix of terms
M <- 10;
BBB <- matrix(0, nrow = M, ncol = M);
for (k in 1:M) {
for (j in 1:k) { BBB[k,j] <- B(k,j); }}
BBB; #k is the row, j is the column
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 1 0 0 0 0 0 0 0 0 0
[2,] 1 1 0 0 0 0 0 0 0 0
[3,] 3 4 2 0 0 0 0 0 0 0
[4,] 11 21 18 6 0 0 0 0 0 0
[5,] 53 128 156 96 24 0 0 0 0 0
[6,] 309 905 1420 1260 600 120 0 0 0 0
[7,] 2119 7284 13950 16080 11160 4320 720 0 0 0
[8,] 16687 65821 148638 210210 190680 108360 35280 5040 0 0
[9,] 148329 660064 1715672 2870784 3207120 2392320 1149120 322560 40320 0
[10,] 1468457 7275537 21381624 41278104 54701136 50394960 31872960 13245120 3265920 362880
</code></pre>
| robjohn | 13,854 | <p><strong>Some Asymptotic Approximations</strong></p>
<p>The AM-GM inequality says
<span class="math-container">$$
\frac{k!}{(k-i)!}\le\left(k-\frac{i-1}{2}\right)^i\tag1
$$</span>
The fact that
<span class="math-container">$$
\frac{k(k-i+1)}{\left(k-\frac{i-1}2\right)^2}=1-\left(\frac{\frac{i-1}2}{k-\frac{i-1}2}\right)^2\tag2
$$</span>
allows us to derive
<span class="math-container">$$
\frac{k!}{(k-i)!}\ge\left[1-\frac{\frac13\left(\frac{i+1}2\right)^3}{\left(k-\frac{i-1}2\right)^2}\right]\left(k-\frac{i-1}{2}\right)^i\tag3
$$</span>
and therefore,
<span class="math-container">$$
\frac{k!}{(k-i)!}\sim\left(k-\frac{i-1}{2}\right)^i\tag4
$$</span>
from which, we get
<span class="math-container">$$
\begin{align}
\frac{\binom{k-j}{i}}{\binom{k}{i}}
&\sim\left(\frac{k-j-\frac{i-1}2}{k-\frac{i-1}2}\right)^i\\
&\sim\left(\frac{k-j}{k}\right)^i\left(1-\frac{i(i-1)j}{2k(k-j)}\right)\\[6pt]
&=\left(\frac{k-j}{k}\right)^i-i(i-1)\left(\frac{k-j}{k}\right)^{i-2}\frac{j(k-j)}{2k^3}\tag5
\end{align}
$$</span>
and therefore,
<span class="math-container">$$
\begin{align}
\sum_{i=0}^\infty\frac{(-1)^i}{i!}\frac{\binom{k-j}{i}}{\binom{k}{i}}
&\sim\sum_{i=0}^\infty\frac{(-1)^i}{i!}\left[\left(\frac{k-j}{k}\right)^i-i(i-1)\left(\frac{k-j}{k}\right)^{i-2}\frac{j(k-j)}{2k^3}\right]\\
&=e^{-\frac{k-j}k}\left(1-\frac{j(k-j)}{2k^3}\right)\\[3pt]
&\sim e^{-\frac{k-j}k-\frac{j(k-j)}{2k^3}}\tag6
\end{align}
$$</span></p>
<hr>
<p><strong>Approximation of the Sum</strong></p>
<p>We don't get an exact closed form, but here is a pretty good asymptotic approximation:
<span class="math-container">$$
\begin{align}
\sum_{i=1}^k(-1)^{k+i}\binom{i}{j}(k-1)_{i-1}
&=\frac1k\sum_{i=1}^k(-1)^{k-i}\binom{i}{j}(k)_i\\
&=\frac1k\sum_{i=0}^{k-1}(-1)^i\binom{k-i}{j}(k)_{k-i}\\
&=(k-1)!\binom{k}{j}\sum_{i=0}^{k-1}\frac{(-1)^i}{i!}\frac{\binom{k-j}{i}}{\binom{k}{i}}\\
&\sim\bbox[5px,border:2px solid #C0A000]{(k-1)!\binom{k}{j}e^{-\frac{k-j}k-\frac{j(k-j)}{2k^3}}}\tag7
\end{align}
$$</span>
This is good when <span class="math-container">$k$</span> is large, exact when <span class="math-container">$j=k$</span>, and extremely close when <span class="math-container">$j=0$</span>. In fact, it looks as if the ratio of the approximation to the actual sum is asymptotically <span class="math-container">$1+\frac{(4j+k)j(k-j)}{6k^5}$</span>. The maximum of this ratio is when <span class="math-container">$\frac jk=\frac{3+\sqrt{21}}{12}\approx0.63188$</span>, where <span class="math-container">$\frac{(4j+k)j(k-j)}{6k^5}=\frac{27+7\sqrt{21}}{432k^2}\approx\frac{0.13675}{k^2}$</span> .</p>
<p>For example, when <span class="math-container">$k=10$</span> and <span class="math-container">$j=5$</span>, <span class="math-container">$(7)$</span> gives <span class="math-container">$54775664.104$</span>, which is about <span class="math-container">$0.136\%$</span> above the value in the table in the question.</p>
|
458,088 | <p>I would like to find an approximation when $ n \rightarrow\infty$ of $ \frac{n!}{(n-2x)!}(n-1)^{-2x} $. Using Stirling formula, I obtain $$e^{\frac{-4x^2+x}{n}}. $$ The result doesn't seem right!</p>
<p>Below is how I derive my approximation. I use mainly Stirling Approximation and $e^x =(1+\frac{x}{n})^n $.</p>
<p>$$\frac{n!}{(n-1)^{2 x} (n-2 x)!}\approx \frac{\left(\sqrt{2 \pi } n^{n+\frac{1}{2}} e^{-n}\right) (n-1)^{-2 x}}{\sqrt{2 \pi } e^{2 x-n} (n-2 x)^{-2 f+n+\frac{1}{2}}}\approx \frac{n^{n+\frac{1}{2}} e^{-2 x} n^{-2 x} \left(1-\frac{1}{n}\right)^{-2 x}}{(n-2 x)^{n-2 x+\frac{1}{2}}}\approx e^{-2 x} \left(\frac{n}{n-2 x}\right)^{n-2 x+\frac{1}{2}} \left(\left(1-\frac{1}{n}\right)^n\right)^{-\frac{2 x}{n}}\approx e^{-2 x} \left(\frac{n}{n-2 x}\right)^{n-2 x+\frac{1}{2}} e^{\frac{2 x}{n}}\approx \left(\frac{n-2 x}{n}\right)^{-n+2 x-\frac{1}{2}} \left(\frac{n-2 x}{n}\right)^n e^{\frac{2 x}{n}}\approx \left(\frac{n-2 x}{n}\right)^{2 x-\frac{1}{2}} e^{\frac{2 x}{n}}\approx \left(\left(1-\frac{2 x}{n}\right)^n\right)^{\frac{2 x}{n}-\frac{1}{2 n}} e^{\frac{2 x}{n}}\approx \left(e^{-2 x}\right)^{\frac{2 x}{n}-\frac{1}{2 n}}=e^{\frac{x-4 x^2}{n}}$$</p>
<p>I would appreciate your input!</p>
| Emily | 31,475 | <p><em>This is a long comment; hence, community wiki status.</em></p>
<p>Gamification is not the embedding of educational content into a game. Gamification is the construction of elements typically found in games around a traditional paradigm, whether it be education, training, marketing, etc. This mistaken understanding is a common fallacy and a big reason why there are so many failed attempts at "making learning fun."</p>
<p>Believe it or not, games aren't fun because they're interactive, or have good graphics, or involve knocking things over or shooting things, or anything else like that.</p>
<p>Games are fun because they happen to satisfied a user's intrinsic psychological needs. In fact, most gamers will encounter exceptional frustration at their favorite games. This is certainly not the description of "fun" that most people have!</p>
<p>What games do is they create a framework in which self-actualization is possible. Games synthesize a progression structure in which a user is able to demonstrate competence. Good games also allow a user to dictate their own courses of action, giving them autonomy. And great games allow us to be social with them, creating a sense of relatedness.</p>
<p>Consider any great student: aren't they proud of themselves when they solve a hard problem, even if it frustrated them for hours? Aren't they the ones that like to read three chapters ahead and stay with you after class? Aren't they the ones that like to show their peers the "tricks" they've discovered?</p>
<p>This is what games do. Any mathematical area can be gamified if you do it right. Doing it right is exceptionally hard, which is why most math games are exceptionally boring.</p>
<p>Does adding a colorful graphic make a cosine more fun? No. Does spinning a graph with a mouse make learning about graph theory more fun? No.</p>
<p>None of these things are fun and interesting unless the user is already motivated to see where it goes. To most students, they don't care, so a video game is just a different flavor of worksheet.</p>
<hr>
<p>If you want to gamify math, don't make the game math. Make the game something else. Make it a story. Give them something to care about, and make mathematics the gateway. Gate rewards behind problems. Scale rewards based on difficulty. The game should have almost nothing to do with mathematics, but rather should motivate students to use the mathematics to uncover the things they're intrigued about.</p>
<p>The star pupil is motivated by learning, and that's great. But the average student doesn't have this motivation, so we need to give it to them somehow. And if you're really clever, you can then tie that story into math, but don't feel like it has to be math. Make it feel like the student can achieve their own goals. Make them <strong>want</strong> to solve difficult problems and overcome frustration.</p>
<p>If the student hates math, the solution isn't to make the reward more math.</p>
|
458,088 | <p>I would like to find an approximation when $ n \rightarrow\infty$ of $ \frac{n!}{(n-2x)!}(n-1)^{-2x} $. Using Stirling formula, I obtain $$e^{\frac{-4x^2+x}{n}}. $$ The result doesn't seem right!</p>
<p>Below is how I derive my approximation. I use mainly Stirling Approximation and $e^x =(1+\frac{x}{n})^n $.</p>
<p>$$\frac{n!}{(n-1)^{2 x} (n-2 x)!}\approx \frac{\left(\sqrt{2 \pi } n^{n+\frac{1}{2}} e^{-n}\right) (n-1)^{-2 x}}{\sqrt{2 \pi } e^{2 x-n} (n-2 x)^{-2 f+n+\frac{1}{2}}}\approx \frac{n^{n+\frac{1}{2}} e^{-2 x} n^{-2 x} \left(1-\frac{1}{n}\right)^{-2 x}}{(n-2 x)^{n-2 x+\frac{1}{2}}}\approx e^{-2 x} \left(\frac{n}{n-2 x}\right)^{n-2 x+\frac{1}{2}} \left(\left(1-\frac{1}{n}\right)^n\right)^{-\frac{2 x}{n}}\approx e^{-2 x} \left(\frac{n}{n-2 x}\right)^{n-2 x+\frac{1}{2}} e^{\frac{2 x}{n}}\approx \left(\frac{n-2 x}{n}\right)^{-n+2 x-\frac{1}{2}} \left(\frac{n-2 x}{n}\right)^n e^{\frac{2 x}{n}}\approx \left(\frac{n-2 x}{n}\right)^{2 x-\frac{1}{2}} e^{\frac{2 x}{n}}\approx \left(\left(1-\frac{2 x}{n}\right)^n\right)^{\frac{2 x}{n}-\frac{1}{2 n}} e^{\frac{2 x}{n}}\approx \left(e^{-2 x}\right)^{\frac{2 x}{n}-\frac{1}{2 n}}=e^{\frac{x-4 x^2}{n}}$$</p>
<p>I would appreciate your input!</p>
| Ilmari Karonen | 9,602 | <p>Well, let me list some ideas from math games (using both words fairly loosely) that <em>I've</em> found interesting. Hopefully, I'm not a complete outlier here, so others may find them fun too.</p>
<h3><a href="http://en.wikipedia.org/wiki/Fractal" rel="noreferrer">Fractals</a></h3>
<p>There are a number of constructions, such as <a href="http://en.wikipedia.org/wiki/L-system" rel="noreferrer">L-systems</a> or <a href="http://en.wikipedia.org/wiki/Iterated_function_system" rel="noreferrer">iterated function systems</a> that allow one to specify a set of rules and generate interesting fractal shapes based on them. With a modern computer, one can even allow the user to vary the parameters in real time and see how the shape changes.</p>
<h3><a href="http://en.wikipedia.org/wiki/Cellular_automaton" rel="noreferrer">Cellular automata</a></h3>
<p>Another fertile area for experimenting with emergent complexity. There are a number of well known cellular automaton rules (<a href="http://en.wikipedia.org/wiki/Conway%27s_Game_of_Life" rel="noreferrer">Conway's Game of Life</a> being only the tip of the iceberg) that can produce surprisingly complex (and pretty!) behavior from simple rules and starting conditions, but of course, the <em>real</em> fun is in coming up with your <em>own</em> rules and seeing what they do.</p>
<h3><a href="http://en.wikipedia.org/wiki/Programming_game" rel="noreferrer">Programming games</a></h3>
<p>The most "game-like" idea in this list: take any traditional game but, instead of the players playing it directly, have them write a set of playing instructions in advance and let the game play out non-interactively according to these instructions. Then let the players refine their instructions and repeat.</p>
<p>The trick is coming up with games that are sufficiently simple mechanically to make the programming task easy to get started with, yet complex enough strategically to remain interesting.</p>
<p>One possible starting point might be to have the players design strategies for <a href="http://en.wikipedia.org/wiki/Iterated_prisoner%27s_dilemma" rel="noreferrer">iterated prisoner's dilemma</a> (which is <em>really</em> simple mechanically: on each turn you choose between two alternatives and then find out what your opponent chose) and see if anyone will come up with the <a href="http://en.wikipedia.org/wiki/Tit_for_tat" rel="noreferrer">tit for tat</a> strategy. Then change the payoffs or introduce some variations (like random errors, or nearest-neighbor competition and replication on a lattice) and see what happens.</p>
<hr>
<p>Common themes to all the ideas above are <em>indirect control</em> and <em>emergent complexity</em>: instead of having the player control what happens in the game directly in real time, have them specify a fairly simple set of rules in advance and watch complicated behavior arise. I don't think that's a coincidence. In a sense, that is exactly what math is <em>about</em>: starting with simple rules and seeing something complex and beautiful emerge.</p>
|
1,329,398 | <p>So, I've posted a question regarding Wikipedia's quartic page. This was from the first question.</p>
<blockquote>
<p>I'm trying to implement the general quartic solution for use in a ray tracer, but I'm having some trouble. The solvers I've found do cause some strange false negatives leaving holes in the tori I'm testing with.</p>
<p>Most implementations use the depressed quartic solutions, I don't understand the math involved and can't figure out why I'm having false non-intersections (link to layman explanation would be great). So I'm trying to implement the general solution at <a href="https://en.wikipedia.org/wiki/Quartic_function#General_formula_for_roots" rel="nofollow noreferrer">this wikipedia page</a>. I got the stuff up until the special cases implemented, but at that point I have an issue.</p>
</blockquote>
<p>With lots of rays being traced most of the special cases become common. I've found a set of coefficients that <a href="http://www.wolframalpha.com/input/?i=0.999999999x%5E4-24.591812430x%5E3%2B226.372338960x%5E2-924.44347730x%2B1412.89015972" rel="nofollow noreferrer">Wolfram Alpha tells me has two real roots</a>, but my code was just returning NaN, further searching I found my S was coming up as <span class="math-container">$\sqrt{-4.9 \times 10^{-11}}$</span> Floating point precision error, means this should equate to 0, so I need the special case for S=0, it says we need to "change choice of cubic root in Q" but it does not explain how to do this. I did try changing the sign of Q when S=0, but that doesn't work. Does anyone know what this means and how I can do it?</p>
| Rolf Hoyer | 228,612 | <p>If $\alpha$ is a root of $x^3 = A$, then so too is $\gamma \alpha$, where $\gamma$ is a complex cube root of unity. This follows since $(\gamma\alpha)^3 = \gamma^3\alpha^3 = \alpha^3 = A$.</p>
<p>The nontrivial values of these are given by $\gamma = -\frac{1}{2} + \frac{i\sqrt{3}}{2}$ or $\gamma = -\frac{1}{2} - \frac{i\sqrt{3}}{2}$.</p>
<p>In your case, you need to replace $Q$ with $\gamma Q = \left(-\frac{1}{2} + \frac{i\sqrt{3}}{2}\right)Q$ in this case, if I'm reading Wikipedia's algorithm correctly.</p>
<p>(Implementation might be easier if you used the note that this case is always accompanied by the depressed quartic being biquadratic, in which case the solutions follow from applying the quadratic formula.)</p>
|
1,329,398 | <p>So, I've posted a question regarding Wikipedia's quartic page. This was from the first question.</p>
<blockquote>
<p>I'm trying to implement the general quartic solution for use in a ray tracer, but I'm having some trouble. The solvers I've found do cause some strange false negatives leaving holes in the tori I'm testing with.</p>
<p>Most implementations use the depressed quartic solutions, I don't understand the math involved and can't figure out why I'm having false non-intersections (link to layman explanation would be great). So I'm trying to implement the general solution at <a href="https://en.wikipedia.org/wiki/Quartic_function#General_formula_for_roots" rel="nofollow noreferrer">this wikipedia page</a>. I got the stuff up until the special cases implemented, but at that point I have an issue.</p>
</blockquote>
<p>With lots of rays being traced most of the special cases become common. I've found a set of coefficients that <a href="http://www.wolframalpha.com/input/?i=0.999999999x%5E4-24.591812430x%5E3%2B226.372338960x%5E2-924.44347730x%2B1412.89015972" rel="nofollow noreferrer">Wolfram Alpha tells me has two real roots</a>, but my code was just returning NaN, further searching I found my S was coming up as <span class="math-container">$\sqrt{-4.9 \times 10^{-11}}$</span> Floating point precision error, means this should equate to 0, so I need the special case for S=0, it says we need to "change choice of cubic root in Q" but it does not explain how to do this. I did try changing the sign of Q when S=0, but that doesn't work. Does anyone know what this means and how I can do it?</p>
| Tito Piezas III | 4,781 | <p>Try this version. Given,</p>
<p>$$x^4+ax^3+bx^2+cx+d=0$$</p>
<p>then,</p>
<p>$$x_{1,2} = -\tfrac{1}{4}a+\tfrac{1}{2}\sqrt{u}\pm\tfrac{1}{4}\sqrt{3a^2-8b-4u+\frac{-a^3+4ab-8c}{\sqrt{u}}}\tag1$$</p>
<p>$$x_{3,4} = -\tfrac{1}{4}a-\tfrac{1}{2}\sqrt{u}\pm\tfrac{1}{4}\sqrt{3a^2-8b-4u-\frac{-a^3+4ab-8c}{\sqrt{u}}}\tag2$$</p>
<p>where,</p>
<p>$$u = \frac{3a^2-8b}{12} +\frac{1}{3}\left(v_1^{1/3}+\frac{b^2 - 3 a c + 12 d}{v_1^{1/3}}\right)$$</p>
<p>and $v_1$ is any non-zero root of the quadratic,</p>
<p>$$v^2 + (-2 b^3 + 9 a b c - 27 c^2 - 27 a^2 d + 72 b d)v + (b^2 - 3 a c + 12 d)^3 = 0$$</p>
<p><strong>P.S.</strong> This is essentially the method used by <em>Mathematica</em>, though much simplified for aesthetics.</p>
|
2,359,372 | <blockquote>
<p>Given that
$$\log_a(3x-4a)+\log_a(3x)=\frac2{\log_2a}+\log_a(1-2a)$$
where $0<a<\frac12$, find $x$.</p>
</blockquote>
<p>My question is how do we find the value of $x$ but we don't know the exact value of $a$? </p>
| Dhruv Kohli | 97,188 | <p>$x$ might come out as a function of $a$.</p>
<p>$$(3x-4a)(3x) = 4(1-2a) \implies 9x^2 - 12ax - 4(1-2a) = 0$$</p>
<p>Solve this quadratic equation.</p>
|
2,359,372 | <blockquote>
<p>Given that
$$\log_a(3x-4a)+\log_a(3x)=\frac2{\log_2a}+\log_a(1-2a)$$
where $0<a<\frac12$, find $x$.</p>
</blockquote>
<p>My question is how do we find the value of $x$ but we don't know the exact value of $a$? </p>
| Parcly Taxel | 357,390 | <p>$$\log_a(3x-4a)+\log_a(3x)=\frac2{\log_2a}+\log_a(1-2a)$$
$$\log_a(3x(3x-4a))=\frac2{\log_aa/\log_a2}+\log_a(1-2a)$$
$$\log_a(3x(3x-4a))=2\log_a2+\log_a(1-2a)$$
$$\log_a(3x(3x-4a))=\log_a4(1-2a)$$
$$3x(3x-4a)=4(1-2a)$$
$$9x^2-12ax+8a-4=0$$
$$x=\frac{12a\pm\sqrt{144a^2-4\cdot9(8a-4)}}{18}$$
$$=\frac{12a\pm12(a-1)}{18}=\frac{2a\pm2(a-1)}3=\frac{4a-2}3\lor\frac23$$</p>
|
114,754 | <p>I have several questions concerning the proof. I don't think I quite understand the details and motivation of the proof. Here is the proof given by our professor.</p>
<p>The space of polynomials $F[x]$ is not finite-dimensional.</p>
<p><em>Proof</em>. Suppose
$$F[x] = \operatorname{span}\{f_1,f_2,\dots,f_n\}$$</p>
<p>Let us choose a positive integer $N$ such that $N > deg (f_i)$ for all $i = 1,\dots,n$. As $\{f_1,f_2,\dots,f_n\}$
spans $F[x]$ we can find scalars $a_1, a_2, \dots,a_n$ such that $x^N = a_1f_1 + a_2f_2 + \cdots + a_nf_n$.
Then the polynomial </p>
<p>$$G(x) = x^N - a_1f_1 - a_2f_2 - \cdots - a_nf_n \equiv 0$$</p>
<p>is a polynomial of degree $N$ which is identically zero. This is a contradiction since $G(x)$ cannot
have more than $N$ roots. </p>
<p><strong>Questions</strong></p>
<ul>
<li>Why is $G(x)$ <strong>identically</strong> zero and what does it mean for it to be <strong>identically</strong> zero? </li>
<li>After obtaining that $G(x) = x^N - a_1f_1 - a_2f_2 - \cdots - a_nf_n \equiv 0$, why must we have more than $N$ roots?</li>
<li><p>What is the <strong>motivation</strong> behind choosing $N$ such that $N > deg (f_i)$ for all $i = 1,\dots,n?$ </p></li>
<li><p>What are the <strong>implications</strong> if we chose $N$ less than or equal to $f_i$ such that $f_i$ has the greatest degree? How will the proof fail in this case? </p></li>
<li><p>If there are any other important <strong>details</strong> and key <strong>insights</strong> that are worthy to be pointed out please let me know so that I can better my understanding of the argument presented.</p></li>
</ul>
| Yuval Filmus | 1,277 | <p>The idea of the proof is this. A basis for the set of all polynomials is
$$1,x,x^2,x^3,\ldots,x^n,\ldots$$
If you have a finite basis, there will be some polynomials that you will be missing, in particular polynomials with high degree. For example, if you take the span of
$$x^3+105x^4+x^7, 1+x^8$$
then for sure you'll be missing any polynomial with degree larger than $8$. The rest of the proof is just taking this idea and making it formal. You choose $N$ to be larger than all the degrees since you know that in that case $x^N$ isn't in the span. You prove that by claiming that you can't have the equality
$$x^N = A(x^3+105x^4+x^7) + B(1+x^8)$$
in $F[x]$. If there were an equality, then the difference between the two sides will be the zero polynomial. That's what we mean by <em>identically zero</em>. We know that the difference isn't identically zero since, well, it's not the zero polynomial, but rather a degree $N$ polynomial.</p>
|
49,074 | <p>It might sound silly, but I am always curious whether Hölder's inequality $$\sum_{k=1}^n |x_k\,y_k| \le \biggl( \sum_{k=1}^n |x_k|^p \biggr)^{\!1/p\;} \biggl( \sum_{k=1}^n |y_k|^q \biggr)^{\!1/q}
\text{ for all }(x_1,\ldots,x_n),(y_1,\ldots,y_n)\in\mathbb{R}^n\text{ or }\mathbb{C}^n.$$
can be derived from the Cauchy-Schwarz inequality.</p>
<p>Here $\frac{1}{p}+\frac{1}{q}=1$, $p>1$.</p>
| Peter Patzt | 10,856 | <p>I suggest the following consideration. We will prove the above inequality for rational $p,q\in(1,\infty)$ with $\frac1p+\frac1q= 1$, and the irrational cases follow by continuity.</p>
<p>If $p$ and $q$ are rational, let $p=\frac ab$ and $q=\frac ac$ with $b+c=a$ and $2^m\ge a$. Now by induction
$$ \sum |x^{(1)}\dots x^{(2^m)}| \le \left( \sum |x^{(1)}\cdots x^{(2^{m-1})}|^2 \right)^{\frac12}\cdot\left( \sum |x^{(2^{m-1}+1)}\cdots x^{(2^{m})}|^2 \right)^{\frac12}$$
$$\le \left( \sum |x^{(1)}|^{2^m} \right)^{\frac1{2^m}}\cdots \left( \sum |x^{(2^m)}|^{2^m} \right)^{\frac1{2^m}}$$
where $x^{(i)}$ are sequences of length $n$, whose indices are omitted. By plugging in
$$x^{(1)}= \dots= x^{(b)}= x^{\frac a{b2^m}},\quad x^{(b+1)}= \dots= x^{(b+c)}= y^{\frac a{c2^m}},\quad x^{(b+c+1)}= \dots = x^{(2^m)}=(xy)^{\frac1{2^m}}$$
we get
$$\sum |xy|= \sum |x^{\frac a{2^m}}y^{\frac a{2^m}}(xy)^{\frac {2^m-a}{2^m}}| \le \left( \sum |x|^{p} \right)^{\frac{b}{2^m}}\cdot \left( \sum |y|^{q} \right)^{\frac c{2^m}}\cdot \left( \sum |xy| \right)^{\frac {2^m-a}{2^m}}$$
implying the inequality we aimed for by dividing the third term of the RHS, and putting the inequality to the $\frac{2^m}{a}$-th power.</p>
|
161,029 | <p>I have not seen a problem like this so I have no idea what to do.</p>
<p>Find an equation of the tangent to the curve at the given point by two methods, without elimiating parameter and with.</p>
<p>$$x = 1 + \ln t,\;\; y = t^2 + 2;\;\; (1, 3)$$</p>
<p>I know that $$\dfrac{dy}{dx} = \dfrac{\; 2t\; }{\dfrac{1}{t}}$$</p>
<p>But this give a very wrong answer. I am not sure what a parameter is or how to eliminate it.</p>
| talmid | 19,603 | <p>One way to do this is by considering the parametric form of the curve:
$(x,y)(t) = (1 + \log t, t^2 + 2)$, so $(x,y)'(t) = (\frac{1}{t}, 2t)$
We need to find the value of $t$ when $(x,y)(t) = (1 + \log t, t^2 + 2) = (1,3)$, from where we deduce $t=1$. The tangent line at $(1,3)$ has direction vector $(x,y)'(1) = (1,2)$, and since it passes by the point $(1,3)$ its parametric equation is given by: $s \mapsto (1,2)t + (1,3)$.</p>
<p>Another way (I suppose this is eliminating the parameter) would be to express $y$ in terms of $x$ (this can't be done for any curve, but in this case it is possible). We solve for $t$: $x = 1 + \log x \Rightarrow x = e^{x-1}$, so $y = t^2 + 2 = (e^{x-1})^2 + 2 = e^{2x-2} + 2$. The tangent line has slope $\frac{dy}{dx}=y_x$ evaluated at $1$: we have $y_x=2e^{2x-2}$ and $y_x(1)=2e^0 = 2$, so it the line has equation $y=2x +b$. Also, it passes by the point $(1,3)$, so we can solve for $b$: $3 = 2 \cdot 1 + b \Rightarrow b = 1$. Then, the equation of the tangent line is $y = 2x + 1$.</p>
<p>Note that $s \mapsto (1,2)t + (1,3)$ and $y = 2x + 1$ are the same line, expressed in different forms.</p>
|
3,554,393 | <p>Namely, I need to prove <span class="math-container">${\max\limits_i} |\lambda_i| \leq {\max\limits_i}{\sum\limits_j} |M_{ij}|\mid$</span>, where <span class="math-container">$M$</span> is the matrix and <span class="math-container">$\lambda_i$</span> are its eigenvalues.</p>
<p>I'm not sure if there is any helpful theorem or lemma. Is there any hint on how to solve this?</p>
| Math1000 | 38,584 | <p>In fact a stronger statement is true. Let <span class="math-container">$\rho(M) = \max_i|\lambda_i|$</span> be the spectral radius of <span class="math-container">$M$</span>. Then for each positive integer <span class="math-container">$k$</span>, we have <span class="math-container">$\rho(M)\leqslant \|M^k\|^{\frac1k}$</span>. For if <span class="math-container">$v$</span> is an eigenvector of <span class="math-container">$M$</span> with associated eigenvalue <span class="math-container">$\lambda$</span>, then
<span class="math-container">$$
|\lambda|^k\|v\| = \|\lambda^k v\| = \|A^k v\| \leqslant \|A^k\|\|v\|,
$$</span>
and dividing by <span class="math-container">$\|v\|$</span> (which is positive since <span class="math-container">$v\ne0$</span>) yields
<span class="math-container">$$
|\lambda|^k\leqslant\|A^k\|.
$$</span>
Raising both sides to <span class="math-container">$\frac1k$</span> and taking the maximum over <span class="math-container">$|\lambda_i|$</span> yields the statement.</p>
<p>In fact it is true that <span class="math-container">$\rho(M) = \lim_{k\to\infty}\|M^k\|^{\frac1k}$</span>, a result known as Gelfand's formula, but that is beyond the scope of this question.</p>
|
440,242 | <p>I'm pretty sure almost all mathematicians have been in a situation where they found an interesting problem; they thought of many different ideas to tackle the problem, but in all of these ideas, there was something missing- either the "middle" part of the argument or the "end" part of the argument. They were stuck and couldn't figure out what to do.</p>
<blockquote>
<ol>
<li>In such a situation what do you do?</li>
<li>Is the reason for the "missing part" the incompleteness in the theory of the topic that the problem is related to? What can be done to find the "missing part"?</li>
</ol>
</blockquote>
<p>For tenure-track/tenure professors, maybe this is not a big deal because they have "enough" time and can let the problem "stew" in the "back-burner" of their mind, but what about limited-time positions, e.g. PhD students, postdocs, etc., where the student/employee has to prove their capability to do "independent" research so that they can be hired for their next position? I think for these people it is quite a bit of a problem because they can't really afford to spend a "lot" of time thinking about the same problem.</p>
| Dirk | 9,652 | <p>Here is an answer which may be math-specific: If you are stuck in some proof of some claim that you believe is true:</p>
<h3>Add the missing piece as assumption and continue as planned.</h3>
|
1,097,134 | <p>this is something that came up when working with one of my students today and it has been bothering me since. It is more of a maths question than a pedagogical question so i figured i would ask here instead of MESE.</p>
<p>Why is $\sqrt{-1} = i$ and not $\sqrt{-1}=\pm i$?</p>
<p>With positive numbers the square root function always returns both a positive and negative number, is it different for negative numbers? </p>
| egreg | 62,967 | <p>The square root function <em>doesn't</em> return two values for positive numbers, or it wouldn't be a function.</p>
<p>It's a fact that, if $x$ is a positive real number, there are two real numbers whose square is $x$. The positive one is denoted by $\sqrt{x}$, so the negative one is $-\sqrt{x}$.</p>
<p>In this way the function has the pleasant property that, for $x,y>0$, $\sqrt{xy\mathstrut}=\sqrt{x\mathstrut}\sqrt{y\mathstrut}$.</p>
<p>In the complex numbers, for every nonzero $x\in\mathbb{C}$ there are two complex numbers whose square is $x$. However, it's not possible to define a square root function with the property above, that is, $\sqrt{xy\mathstrut}=\sqrt{x\mathstrut}\sqrt{y\mathstrut}$.</p>
<p>Maybe you'd want to stretch the notion of function, to allow multiple values; but then, how many values should you assign to the expression
$$
\sqrt{2}+\sqrt{3}+\sqrt{5}
$$
and similar ones? You couldn't make the more obvious simplifications: from
$$
x+\sqrt{2}=\sqrt{2}
$$
you'd get <em>three</em> values for $x$.</p>
<p>When I introduce the complex numbers, I <em>never</em> use $\sqrt{-1}$, but rather I say that $i^2=-1$, which is a quite different statement, exactly because it's impossible to define a square root function that has sensible algebraic properties.</p>
|
3,306,341 | <p>Let <span class="math-container">$f\in Hom(R,R')$</span> be a surjective map and let <span class="math-container">$I$</span> be an ideal of <span class="math-container">$R$</span></p>
<p>Assume that <span class="math-container">$Ker(f)\subseteq I$</span> , prove that <span class="math-container">$f^{-1}(f(I))=I$</span></p>
<p>My work , </p>
<p><span class="math-container">$Ker(f)\subseteq I \Rightarrow f^{-1}(\left \{ 0_{R'} \right \})\subseteq I \Rightarrow \left \{ 0_{R'} \right \} \subseteq f(I)\Rightarrow f^{-1}(\left \{ 0_{R'} \right \})\subseteq f^{-1}(f(I))\Rightarrow Ker(f)\subseteq f^{-1}(f(I))$</span></p>
<p>Here I'm stucked</p>
<p>(<span class="math-container">$R$</span> and <span class="math-container">$R'$</span> are commutative and unitaty rings and <span class="math-container">$f$</span> is unital)</p>
| egreg | 62,967 | <blockquote>
<p>Let <span class="math-container">$f\colon R\to R'$</span> be a surjective ring homomorphism and <span class="math-container">$I$</span> an ideal of <span class="math-container">$R$</span>. Then <span class="math-container">$f^{-1}(f(I))=I$</span> if and only if <span class="math-container">$\ker f\subseteq I$</span>.</p>
</blockquote>
<p>One direction is easy: if <span class="math-container">$f^{-1}(f(I))=I$</span>, then clearly <span class="math-container">$\ker f\subseteq I$</span>, because <span class="math-container">$0\in f(I)$</span>. (This is what you attempted, so not really relevant for your problem.)</p>
<p>For the converse direction, note that <span class="math-container">$I\subseteq f^{-1}(f(I))$</span> holds regardless of <span class="math-container">$I$</span> being an ideal. Thus you just need to prove the converse inclusion.</p>
<p>Suppose <span class="math-container">$x\in f^{-1}(f(I))$</span>. Then <span class="math-container">$f(x)\in f(I)$</span>, by definition, so there is <span class="math-container">$y\in I$</span> such that <span class="math-container">$f(x)=f(y)$</span>.</p>
<p>Can you go on? What can you say about <span class="math-container">$x-y$</span>?</p>
|
668,959 | <p>I don't know if this is an already existing conjecture, or has been proven: There is at least one prime number between <span class="math-container">$N$</span> and <span class="math-container">$N-\sqrt{N}$</span>.</p>
<p>Some examples:
<span class="math-container">$N=100$</span></p>
<p><span class="math-container">$\sqrt{N}=10$</span>
Between and 90 and 100, there is a prime: 97</p>
<p><span class="math-container">$N=36$</span></p>
<p><span class="math-container">$\sqrt{N}=6$</span>
Between 30 and 36, there is a prime: 31</p>
<p><span class="math-container">$N=64$</span></p>
<p><span class="math-container">$\sqrt{N}=8$</span>
Between 56 and 64, there are the primes: 59 and 61</p>
<p>N=12</p>
<p><span class="math-container">$\sqrt{N}=3.46..$</span>
Between 8.54 and 12, there is a prime: 11</p>
<p>If this hasn't been brought up before, I'm calling this the Dwyer Conjecture.</p>
| Kevin Arlin | 31,228 | <p>I'm afraid this is false whether we consider the strong form in which endpoints are allowed, or not. $\sqrt{126}<12$ and $113$ is the next prime below $126$. </p>
|
1,680,269 | <p>Here $\mathbb{Z}_{n}^{*}$ means $\mathbb{Z}_{n}-{[0]_{n}}$</p>
<p>My attempt:</p>
<p>$(\leftarrow )$</p>
<p>$p$ is a prime, then, for every $[x]_{n},[y]_{n},[z]_{n}$ $\in (\mathbb{Z}_{n}^{*},.)$ are verified the following:</p>
<p>1) $[x]_{n}.([y]_{n}.[z]_{n}) = ([x]_{n}.[y]_{n}).[z]_{n}$, since from the operation . we have $[a]_{n}.[b]_{n}=[a.b]_{n}$ and . is associative in $\mathbb{Z}$.</p>
<p>2) There is an element $e$ such that $[x]_{n}.e = e.[x]_{n} = [x]_{n}$, since $ [x]_{n}.[1]_{n} = [x.1]_{n} = [x]_{n} = [x.1]_{n} = [x]_{n}[1]_{n}$</p>
<p>But I don't know how to check the inverse property, neither how to do the $(\rightarrow)$ part. </p>
<p>Thanks!</p>
| T. Eskin | 22,446 | <p><strong>Hint:</strong> Assume that $d=0$ would be true, and take sequences $(x_{n})\subset K$ and $(y_{n})\subset L$ so that $d(x_n,y_n)\to 0$. Use compactness to conclude a contradiction.</p>
|
4,549,300 | <p>Matrix C of size n<span class="math-container">$\times$</span>n is symmetric . Zero is a simple eigenvalue of C. The associated eigenvector is q. For <span class="math-container">$\epsilon$</span>>0, the equation <span class="math-container">$Cx+\epsilon x=d$</span> in x, where x and d are n-dimensional Column vectors and d is known, has a solution that depends on <span class="math-container">$\epsilon$</span>. Call this solution <span class="math-container">$x(\epsilon)$</span>. Express <span class="math-container">$ \lim\limits_{\epsilon \to 0^+} \epsilon x(\epsilon)$</span> in terms of vectors q and d.</p>
<p>I have some inspirations:</p>
<ol>
<li>When <span class="math-container">$\epsilon \to 0$</span>, <span class="math-container">$\epsilon q$</span> also comes to <span class="math-container">$0$</span>. So we have: <span class="math-container">$C(x+q)+\epsilon (x+q)=d$</span></li>
<li>Since C is a symmetrix matrix, we use diagonalization like: <span class="math-container">$U\Lambda U^Hx+\epsilon x=d$</span></li>
</ol>
<p>But to be honest, I have no clue on this and it bothers me a lot. I can't figure out how to use these conditions we have.</p>
<p>Thank you for reading my question!</p>
<p>Could you give me some clues on this problem?</p>
| snowman | 1,057,706 | <p>Thanks @<a href="https://math.stackexchange.com/users/71348/ted-shifrin">Ted Shifrin</a>! Here is my answer:</p>
<p>First,
<span class="math-container">$(C+\epsilon I)x = d$</span>, so <span class="math-container">$x=(C+\epsilon I)^{-1}d$</span> and then <span class="math-container">$\epsilon x=\epsilon (C+\epsilon I)^{-1}d$</span></p>
<p>As said by Ted,
<span class="math-container">$$(C+\epsilon I)^{-1} = \frac1{\epsilon} qq^\top + \sum_{i=1}^{n-1} \frac1{\lambda_i + \epsilon} v_iv_i^\top.$$</span>
So we have:
<span class="math-container">$$\epsilon x=\epsilon(C+\epsilon I)^{-1}d = qq^\top d + \sum_{i=1}^{n-1} \frac{\epsilon}{\lambda_i + \epsilon} v_iv_i^\top d$$</span>
The final result would be:
<span class="math-container">$$\lim\limits_{\epsilon\to 0^+}\epsilon x = qq^\top d$$</span></p>
|
2,067,003 | <p>(Mathematics olympiad Netherlands) Let $A,B$ and $C$ denote chess players in a tournament. The winner of each match plays the next match against the oponent that did not play the current. At the end of the tournament $A$, $B$ and $C$ played $10$, $15$ and $17$ times respectively. Each match only ended up in a win. <em>Question</em>: Which player lost the second match?</p>
<p><em>UPDATE</em>: So I think I got the answer. Denote $n$ as the amount of matches between $A$ and $B$. Since $A$ plays the same amount of matches against $B$ as the otherway around, we have $15 - n = 17 - (10-n) \implies n = 4$. So $A$ plays a total of 10 matches, while there are a total of 21 matches. This is only possible if $A$ plays all the even matches <strong>and</strong> he loses that match (else contradiction to amount of matches played).</p>
| Yiorgos S. Smyrlis | 57,021 | <p>$$
\sum_{k=1}^{n}\frac{(-1)^{k+1}}{k(k+1)}=\sum_{k=1}^{n}(-1)^{k+1}\left(\frac1k-\frac1{k+1}\right)
=\sum_{k=1}^n\frac{(-1)^{k+1}}{k}+\sum_{k=1}^n\frac{(-1)^{k+2}}{k+1}
\\=2\sum_{k=1}^n\frac{(-1)^{k+1}}{k}-1-\frac{(-1)^{n+1}}{n+1}.
$$
Hence,
$$
\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}=2\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}-1=2\ln 2-1,
$$
since
$$
\sum_{n=1}^n\frac{(-1)^{n+1}}{n}=\ln 2.
$$</p>
<p><em>Proof.</em> If $\lvert x\rvert<1$, then
$$
\frac{1}{1+x}=\sum_{k=0}^n(-1)^{k}x^k+\frac{(-1)^{n+1}x^{n+1}}{1+x},
$$
and hence
$$
\log(1+x)=\int_0^x\frac{dt}{1+t}=\sum_{k=0}^n(-1)^{k}\int_0^tt^k\,dt+\int_0^x\frac{(-1)^{n+1}t^{n+1}\,dt}{1+t} \\=\sum_{k=0}^n\frac{(-1)^kx^{n+1}}{n+1}+R_n(x).
$$
Clearly, for $x\in [0,1]$
$$
\lvert R_n(x)\rvert = \int_0^x\frac{t^{n+1}\,dt}{1+t}\le\int_0^xt^{n+1}\,dt\le \frac{1}{n+2}.
$$
Hence
$$
\ln 2=\lim_{x\to 1^-}\ln(1+x)=\lim_{x\to 1^-}\sum_{k=0}^n\frac{(-1)^kx^{n+1}}{n+1}+\lim_{x\to 1^-}R_n(x)=\sum_{k=0}^n\frac{(-1)^k}{n+1}+R_n(1),
$$
and hence
$$
\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}=\lim_{n\to\infty}\sum_{k=0}^n\frac{(-1)^k}{n+1}=\lim_{n\to\infty}\big(\ln 2+R_n(1)\big)=\ln 2.
$$</p>
|
2,239,240 | <p>I'm looking to do some independent reading and I haven't been able to find rough prerequisites for Differential Topology at the level of Milnor or Guillemin and Pollack.</p>
<p>Is a semester of analysis (Pugh) and a semester of topology (Munkres) enough to make sense of most of it or should I take a second semester of analysis first?</p>
| mrnovice | 416,020 | <p>If I'm understanding your question correctly, then note:</p>
<p>$(2+3)^2 = (5)^2=25 $</p>
<p>$(2+3)^2 = 2^2+2\cdot 2\cdot 3+3^2=4+12+9 = 25$</p>
<p>You can expand the brackets, or simplify the contents of the brackets and then square it and obtain the same answer via both methods, if that's what you mean. </p>
|
2,239,240 | <p>I'm looking to do some independent reading and I haven't been able to find rough prerequisites for Differential Topology at the level of Milnor or Guillemin and Pollack.</p>
<p>Is a semester of analysis (Pugh) and a semester of topology (Munkres) enough to make sense of most of it or should I take a second semester of analysis first?</p>
| celtschk | 34,930 | <p>It seems you confuse the <em>precedence</em> of operations, which describe how to interpret an expression (and thus which transformations are actually valid), with the order in which you <em>apply valid transformations</em> to an expression.</p>
<p>For your specific formula, $(2+3)^2$, the precedence rules say that the parentheses override the implicit precedences that otherwise would have been in effect. That is, the precedence rules say that the expression means that the sum $2+3$ is taken to the second power (i.e. squared), as opposed to $2+3^2$ which would mean that $2$ and the square of $3$ would be added.</p>
<p>However the precedence rules do <em>not</em> say that you need to calculate the term in parentheses first (although in this special case, it's the most economic choice). You have two valid transformations that you can apply (well, actually there are many more, but the others would only make the problem more complicated):</p>
<ul>
<li><p>You can do the sum first: It's a subexpression $2+3$ which you can replace by $5$, because $2+3=5$. Doing so, you'll get $(2+3)^2 = 5^2$. Then you can go on by using the fact that $5^2=25$ to finally arrive at the result $(2+3)^2 = 25$.</p></li>
<li><p>You can do the square first: Since it is the square of a sum, you can use the binomial formula to obtain $(2+3)^2 = 2^2 + 2\cdot 2\cdot 3 + 3^2$. Again, you can continue to find $2^2+2\cdot 2\cdot 3 + 3^2 = 4 + 12 + 9 = 25$. The result is the same again, as it of course has to be.</p></li>
</ul>
<p>However what you <em>cannot</em> do is to first replace $3^2=9$ because the original expression does not contain the subexpression $3^2$. Of course after applying the binomial formula, a subexpression of that form appears, which you then can replace by $9$.</p>
<p>Similarly, in $2+4\cdot 5$, the precedence rules say that this is the same as $2+(4\cdot 5)$, and therefore there's a subexpression $4\cdot 5$ which you can replace with $20$, but there is no subexpression $2+4$ which could be replaced by $6$. However you could, in principle, first write $2=2\cdot 1$ and $4=2\cdot 2$, and then use the distributive law to get $2\cdot 1 + 2\cdot 2\cdot 5 = 2\cdot (1+2\cdot 5) = 2\cdot 11 = 22$, and again, you get, necessarily, the same result as when simplifying in precedence order, $2+4\cdot 5 = 2 + 20 = 22$. However if you got the precedence wrong and started by replacing the non-subexpression $2+4$ with $6$, you'd arrive at $6\cdot 5 = 30\ne 22$, which is the wrong result.</p>
<p>Note that while in the cases above, applying the operations in precedence order is the most efficient, this is not always the case. Indeed, you can even have cases where first going to a more complicated expression simplifies the complete calculation. As an example, consider the expression $999^2$. While you can directly calculate it as $999\cdot 999$ using the standard multiplication algorithm, it is much easier to first split it into a difference and apply the binomial formula:
$$999^2 = (1000-1)^2 = 1000^2 - 2\cdot 1000\cdot 1 + 1^2
= 1\,000\,000 - 2000 + 1 = 998\,001$$
This is possible because each single step is a valid operation; note how the replacement of $999$ by $1000-1$ requires the addition of parentheses due to the precedence rules (we replace the <em>expression</em> $999$ by the <em>expression</em> $1000-1$, but without the parentheses we'd get $1000-1^2$ which doesn't even contain the subexpression $1000-1$). Also note that the first step would be futile if we then proceeded by doing the addition in parentheses first, as we'd just arrive at the starting point again.</p>
|
1,613,185 | <p>There are five red balls and five green balls in a bag. Two balls are taken out at random. What is the probability that both the balls are of the same colour</p>
| barak manos | 131,263 | <p>The number of ways to choose $2$ balls is $\binom{5+5}{2}=45$.</p>
<p>The number of ways to choose $2$ red balls is $\binom{5}{2}=10$.</p>
<p>The number of ways to choose $2$ green balls is $\binom{5}{2}=10$.</p>
<p>Hence the probability of choosing $2$ balls of the same color is $\frac{10+10}{45}=\frac49$.</p>
|
3,172,485 | <p>Consider the familiar trigonometric identity: <span class="math-container">$\cos^3(x) = \frac{3}{4} \cos(x) + \frac{1}{4} \cos(3x)$</span></p>
<p>Show that the identity above can be interpreted as Fourier series expansion.</p>
<p>so we know that cos is periodic between <span class="math-container">$\pi$</span> and <span class="math-container">$-\pi$</span> and <span class="math-container">$\cos$</span> is an even function, therefore, <span class="math-container">$\cos^3$</span> is even. </p>
<p>so we need to compute <span class="math-container">$a_0$</span> ( the integral of <span class="math-container">$f(x)$</span> and it will equal <span class="math-container">$0$</span>) and <span class="math-container">$a_n$</span> ( the integral from <span class="math-container">$\pi$</span> to <span class="math-container">$-\pi$</span> of <span class="math-container">$\cos^3(x) \cos(nx)$</span> )</p>
<p>how to compute <span class="math-container">$a_0$</span></p>
<p>thanks </p>
| John Doe | 399,334 | <p>Using the result given in your question,<span class="math-container">$$\begin{align}\int_{-\pi}^\pi\cos^3 x &= \int_{-\pi}^\pi\left(\frac34 \cos x + \frac 14 \cos 3x\right)\,dx\\
&=\left[\frac34\sin x+\frac1{12}\sin 3x\right]_{-\pi}^\pi\\
&=0\end{align}$$</span> since <span class="math-container">$\sin$</span> of multiples of <span class="math-container">$\pi$</span> is equal to <span class="math-container">$0$</span>.</p>
|
1,776,850 | <p>Given a square $ABCD$ such that the vertex $A$ is on the $x$-axis and the vertex $B$ is on the $y$-axis. The coordinates of vertex $C$ are $(u,v)$. Find the area of square in terms of $u$ and $v$ only.</p>
<p><strong>What I have done</strong></p>
<p>Let the coordinate of $A$ be $(x,0)$ and $B$ be $(0,y)$. Also let the side of the square be a units.</p>
<p>$2a^2=AC^2=(x-u)^2+v^2$</p>
<p>$a^2$ is the required area so if we write $x$ in terms of $u$ and $v$ then the job will be done. </p>
<p>Now from here I thought of two ways either using trigonometry or using rotation of axes but here none of them will work because some angles will be involved and we require just $u$ and $v$ and nothing else in the expression of the area of square.</p>
<p>So how to do it? Please help.</p>
| Roman83 | 309,360 | <p>Let $A(x;0), B(0,y), C(u,v)$, and $AB=a$. Then $$S=a^2$$.
$$\begin{cases}
AB=a=\sqrt{x^2+y^2}
\\
BC=a=\sqrt{u^2+(v-y)^2}{}
\\
AC=\sqrt2a=\sqrt{(u-x)^2+v^2}
\end{cases}$$</p>
<p>$$\begin{cases}
a^2=x^2+y^2 (1)
\\
a^2=u^2+(v-y)^2 (2)
\\
2a^2=(u-x)^2+v^2 (3)
\end{cases}$$</p>
<p>(3)+(2) $3a^2=u^2-2ux+x^2+v^2+u^2+v^2-2vy+y^2=2u^2-2ux+2v^2-2vy+(x^2+y^2)=$
$=2u^2-2ux+2v^2-2vy+a^2 \Rightarrow a^2=u^2-ux+v^2-vy$</p>
<p>$$\begin{cases}
a^2=x^2+y^2 (1)
\\
a^2=u^2-ux+v^2-vy (4)
\end{cases}$$</p>
|
1,581,161 | <p>Let the triangle $ABC$ and the angle $\widehat{ BAC}<90^\circ$ </p>
<p>Let the perpendicular to $AB$ passing by the point $C$ and the perpendicular to $AC$ passing by $B$ intersect the circumscribed circle of $ABC$ on $D$ and $E$ respectively .
We suppose that $DE=BC$</p>
<p>What is the angle $\widehat{BAC}$ </p>
<p>I tried using law of sines in triangle
Also , let O be center of circle so OD=OE=r</p>
| Narasimham | 95,860 | <p>Let $AECFBD $ (labeled anti-clockwise) be a <em>regular hexagon</em> inscribed in a circle center $O$. All the given conditions are satisfied by this assumption ( perpendicularities) and $(DE=BC).$ Simply by symmetry, angle BAC = $60^0.$ </p>
|
1,530,148 | <p>In $(C[0,1],d_\infty)$, consider $U=\{f\in C[0,1]: f(x)\neq 0, \forall x \in [0,1]\}$. Prove that $U$ is open and find his connected components.
I know that for proof the first thing, i have to show the existence of an $\epsilon\gt0$ so for all $f\in U$, $B(f,\epsilon)\subseteq U$. First of all, is that correct? Second, if it is ...how should i find that $\epsilon\gt 0$? For the connected components...i have no clue. Any suggestions?</p>
| Simon S | 21,495 | <p><em>Hint:</em></p>
<p>There is no such $\epsilon$ for <strong>all</strong> $f \in U$.</p>
<p>You want to show that for any $f \in U$ there exists an $\epsilon$ ball $B(f,\epsilon) \subset U$. </p>
<p>Note that as $f$ is continuous, $f$ is either strictly positive or strictly negative (candidate connected components?). Further, as $[0,1]$ is compact the image $f([0,1])$ is compact and $f$ attains a minimum and maximum (aka the first year calculus Extreme Value Theorem). Can you see now how to find such an $\epsilon$?</p>
<p>If you're still stuck, draw a graph of some $f \in U$: that's the mental image I am using in my head to see what kind of $B(f,\epsilon)$ work.</p>
|
1,530,148 | <p>In $(C[0,1],d_\infty)$, consider $U=\{f\in C[0,1]: f(x)\neq 0, \forall x \in [0,1]\}$. Prove that $U$ is open and find his connected components.
I know that for proof the first thing, i have to show the existence of an $\epsilon\gt0$ so for all $f\in U$, $B(f,\epsilon)\subseteq U$. First of all, is that correct? Second, if it is ...how should i find that $\epsilon\gt 0$? For the connected components...i have no clue. Any suggestions?</p>
| Georges Elencwajg | 3,217 | <p>The connected components of $U$ are the open subsets $$U_+=\{f\in C[0,1]: f(x)\gt 0, \forall x \in [0,1]\}$$ and $$U_-=\{f\in C[0,1]: f(x)\lt 0, \forall x \in [0,1]\}$$ Indeed if $f,g\in U_+$ (resp $f,g\in U_-$), then $tf+(1-t)g\in U_+$ (resp. $tf+(1-t)g\in U_-$) for all values of $t\in [0,1]$.<br>
This proves that $U_+,U_-$ are both pathwise connected and thus connected.<br>
Since they form a disjoint open partition of $U$ (notice that any $f\in U$ has constant sign since it never vanishes, so must belong to either $U_+$ or $U_-$) , they are the two connected components of $U$.</p>
|
1,876,708 | <p>Let $m<n$. Why $\mathbb R^m$ is closed in $\mathbb R^n$ ? For example, let us take $\mathbb R^3$ and the subspace $\mathbb R^2$. It looks weird to me that $\mathbb R^2$ is closed in $\mathbb R^3$. To me it looks impossible. It may be open, but not closed. Any explanation is welcome.</p>
| Christian Blatter | 1,303 | <p>Let $V$ be any plane in ${\mathbb R}^3$, e.g., the $(x_1,x_2)$-plane $x_3=0$. If ${\bf p}\notin V$ then we can drop a normal $n$ from ${\bf p}$ to $V$, which then will hit $V$ at some point ${\bf p}'$. Note that all points ${\bf x}\in V$ satisfy $$|{\bf x}-{\bf p}|\geq|{\bf p}'-{\bf p}|=:d>0\ .$$ It follows that the ball with radius ${d\over2}$ and center ${\bf p}$ does not intersect $V$. Since ${\bf p}\notin V$ was arbitrary we thus have proven that the complement ${\mathbb R}^3\setminus V$ is open, hence $V$ is closed in ${\mathbb R}^3$.</p>
|
1,876,708 | <p>Let $m<n$. Why $\mathbb R^m$ is closed in $\mathbb R^n$ ? For example, let us take $\mathbb R^3$ and the subspace $\mathbb R^2$. It looks weird to me that $\mathbb R^2$ is closed in $\mathbb R^3$. To me it looks impossible. It may be open, but not closed. Any explanation is welcome.</p>
| Bernard | 202,857 | <p>$\mathbf R^m$ is a finite intersection of hyperplanes in $\mathbf R^n$. Each of these is closed, as it's the kernel of a (continuous) linear form. </p>
|
8,816 | <p>What is the result of multiplying several (or perhaps an infinite number) of binomial distributions together?</p>
<p>To clarify, an example.</p>
<p>Suppose that a bunch of people are playing a game with k (to start) weighted coins, such that heads appears with probability p < 1. When the players play a round, they flip all their coins. For each heads, they get a coin to flip in the next round. This is repeated every round until they have a round with no heads.</p>
<p>How would I calculate the probability distribution of the number or coins a player will have after n rounds? Especially if n is extremely high and p extremely close to 1?</p>
| David E Speyer | 297 | <p>It is not always true that the automorphism group of an algebraic variety has a natural algebraic group structure. For example, the automorphism group of $\mathbb{A}^2$ includes all the maps of the form $(x,y) \mapsto (x, y+f(x))$ where $f$ is any polynomial. I haven't thought through how to say this precisely in terms of functors, but this subgroup morally should be a connected infinite dimensional object, and is thus not a subobject of an algebraic group.</p>
<p>On the other hand, I believe that the automorphism group of a projective algebraic variety, $X$, can be given the structure of algebraic group in a fairly natural way. This is something I've thought about myself, but not written down a careful proof nor found a reference for: For any automorphism $f$ of $X$, consider the graph of $f$ as a subscheme of $X \times X$, and thus a point of the Hilbert scheme of $X\times X$. In this way, we get an embedding of point sets from $\mathrm{Aut}(X)$ into $\mathrm{Hilb}(X\times X)$. </p>
<p>I believe that it should be easy to show that (1) $\mathrm{Aut}(X)$ is open in $\mathrm{Hilb}(X\times X)$, and thus acquires a natural scheme structure and (2) composition of automorphisms is a map of schemes.</p>
|
3,632,576 | <p>Considering that input <span class="math-container">$x$</span> is a scalar, the data generation process works as follows:</p>
<ul>
<li>First, a target t is sampled from {0, 1} with equal probability.</li>
<li>If t = 0, x is sampled from a uniform distribution over the interval
[0, 1]. </li>
<li>If t = 1, x is sampled from a uniform distribution over the
interval [0, 2].</li>
</ul>
<p>I'm trying to find the formula for <span class="math-container">$P(t=1)$</span>, <span class="math-container">$P(t=0)$</span>, <span class="math-container">$P(x|t=1)$</span> and <span class="math-container">$P(x|t=0)$</span> and then find the posterior probability <span class="math-container">$P(t = 0|x)$</span> as a function of <span class="math-container">$x$</span>.</p>
<p>So far I have that <span class="math-container">$P(t=1)$</span> and <span class="math-container">$P(t=0)$</span> <span class="math-container">$=$</span> <span class="math-container">$\frac12$</span> but I wasn't sure how to find <span class="math-container">$P(x|t=1)$</span> and <span class="math-container">$P(x|t=0)$</span>.</p>
<p>I know from there we can just use <span class="math-container">$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$</span> to compute the probability <span class="math-container">$P(t = 0|x)$</span> as a function of <span class="math-container">$x$</span>. Is that correct?</p>
| Duchamp Gérard H. E. | 177,447 | <p>In fact, you must combine your idea (cut the integral) and that of Reveillark. Indeed, you begin, as it was suggested by writing
<span class="math-container">$$
\left | \frac{1}{x}\int_0^x f(t)\,dt-a\right|=\left | \frac{1}{x}\int_0^x (f(t)-a) \,dt\right|
$$</span>
now, let us exploit the fact that <span class="math-container">$\lim_{x\to\infty} f(x)=a$</span> as you did.
For all <span class="math-container">$\epsilon_1>0$</span> (to be adjusted later), it exists <span class="math-container">$N$</span> such that <span class="math-container">$\forall t\geq N$</span>,
<span class="math-container">$|f(t)-a|\leq \epsilon_1$</span>, then you make your cut and for all <span class="math-container">$x\geq N$</span>
<span class="math-container">$$
\frac{1}{x}\int_0^x (f(t)-a) \,dt=
\frac{1}{x}\Big(\int_0^N (f(t)-a) \,dt+\int_N^x (f(t)-a) \,dt\Big)
$$</span><br>
Now, as <span class="math-container">$x$</span> grows, the contribution of
<span class="math-container">$\frac{1}{x}\int_0^N (f(t)-a) \,dt$</span> is less and less important, in fact
<span class="math-container">$$
\lim_{x\to \infty}\frac{1}{x}\int_0^N (f(t)-a) \,dt=0
$$</span>
and
<span class="math-container">$$
|\frac{1}{x}\int_N^x (f(t)-a) \,dt|=
\frac{x-N}{x}|\frac{1}{x-N}\int_N^x (f(t)-a) \,dt|\leq \frac{x-N}{x}\cdot \epsilon_1\leq \epsilon_1
$$</span>
so, take <span class="math-container">$\epsilon_1=\epsilon/2$</span>.</p>
<p>Can you finish ? </p>
|
1,282,486 | <p>Given the function $f(x) = |8x^3 − 1|$ in the set $A = [0, 1].$
Prove that the function is not differentiable at $x = \frac12.$ </p>
<p>The answer in my book is as follows:</p>
<p>$$\lim_{x \to \frac12-} \dfrac{f(x)-f(1/2)}{x-1/2} = -6$$
$$\lim_{x \to \frac12+} \dfrac{f(x)-f(1/2)}{x-1/2} = 6$$ </p>
<p>Can anyone explain how the $6$'s were derived. I understand that as $x$ tends to $\frac12$ from the negative side, the bottom will be negative, so thats why the first one is a minus.</p>
<p>But how do you get to the $6$, what am I missing? Obviously $f(\frac12)=0$ but what do you make $f(x)=$ as $x$ tends to $\frac12$</p>
<p>Thanks</p>
| Rob Arthan | 23,171 | <p><strong>Hint</strong>: $f(x) = |g(x)|$ where $g(x) = 8x^3 - 1$. What is the derivative of $g(x)$ when $x = \frac{1}{2}$?</p>
|
2,431,375 | <p>A continuous function $f$ on $[a,b]$, differentiable in $(a,b)$, has only 1 point where its derivative vanishes. What is true about this function?</p>
<p>A. $f$ cannot have an even number of extrema.</p>
<p>B. $f$ cannot have a maximum at one endpoint and minimum at the other.</p>
<p>C. $f$ might be monotonically increasing.</p>
<p>I think $A$ is true, since it has three extrema, but the right answer should be $C$, why?</p>
| Mathemagical | 446,771 | <p>To see that A is false, consider the function $f(x) = (x-0.5)^2$ on [-1,1]. To see that B is false and also that C is true, consider the function $f(x) = x^3$ on the same domain.</p>
|
3,620,612 | <p>I had a question in the exercises of a complex analysis course I couldn't solve, It asked me to evaluate this integral <span class="math-container">$$\int_{-\pi}^{\pi}\frac{dx}{\cos^2(x) + 1}$$</span></p>
<p>I tried to evaluate it without using residues, the antiderivative of this function contains tan, which is not defined at <span class="math-container">$\pm \pi/2 $</span></p>
<p>all the examples I have worked with so far are rational functions (a polynomial over another) </p>
| Elsa | 278,945 | <p>Here's an answer using residues.</p>
<p>Using
<span class="math-container">$cos(x) = \frac{1}{2}\left(e^{ix} + e^{-ix}\right)$</span>
and defining <span class="math-container">$z=e^{ix}$</span> which implies <span class="math-container">$dz = izdx$</span>, we write the integral as</p>
<p><span class="math-container">$$\int_{-\pi}^{\pi} \frac{1}{(cos(x))^2+1} dx$$</span>
<span class="math-container">$$=\int_{-\pi}^{\pi} \frac{4}{z^2+6+1/z^2} \frac{dz}{iz}$$</span>
<span class="math-container">$$=\frac{1}{i}\int_{-\pi}^{\pi} \frac{4z}{z^4+6z^2+1} dz. $$</span>
We so obtained an integral along the unit circle!</p>
<p>Next, we need to find all residues inside the unit circle.</p>
<p>For this, we define <span class="math-container">$g(z) = z^4+6z^2+1$</span>. Factoring out gives</p>
<p><span class="math-container">$$g(z) = (z^2 + 3 + 2\sqrt{2})\ldots\\ \ldots\left(z-i\sqrt{3-2\sqrt{2}}\right) \left(z+i\sqrt{3-2\sqrt{2}}\right). $$</span></p>
<p>Hence, there are two roots inside the unit circle,
<span class="math-container">$$z_1 =i\sqrt{3-2\sqrt{2}}$$</span>
and
<span class="math-container">$$z_2=-i\sqrt{3-2\sqrt{2}}$$</span>
(and two outside coming from the first bracket but we don't have to care about those). And it is for these two roots that we have to find the residues.</p>
<p>Defining <span class="math-container">$f(z) = 4z$</span>, we see that our integrand is a rational function, <span class="math-container">$f/g$</span>. Moreover, <span class="math-container">$z_1$</span> and <span class="math-container">$z_2$</span> are poles of order 1 of <span class="math-container">$f/g$</span>. For this case, there is a well known result giving us the residues for <span class="math-container">$f/g$</span> at these poles, namely
<span class="math-container">$$Res\left(\frac{f}{g}, z_k\right)=\frac{f(z_k)}{g'(z_k)}$$</span>
where <span class="math-container">$k\in\{1, 2\}$</span>. </p>
<p>Since <span class="math-container">$g'(z) = 4z^3 + 12z$</span>, we obtain:
<span class="math-container">$$Res\left(\frac{f}{g}, z_k\right)=\ldots=\frac{1}{z_k^2 + 3}=$$</span>
<span class="math-container">$$\ldots=\frac{1}{2\sqrt{2}}$$</span>
(yep, the same for both <span class="math-container">$k$</span>). </p>
<p>We conclude by applying the residue theorem:
<span class="math-container">$$\frac{1}{i}\int_{-\pi}^{\pi} \frac{4z}{z^4+6z^2+1} dz$$</span>
<span class="math-container">$$=\frac{1}{i}\left(2\pi i \sum_{k=1}^{2}Res\left(\frac{f}{g}, z_k\right)\right) $$</span>
<span class="math-container">$$=\frac{1}{i}\left(2\pi i \frac{2}{2\sqrt{2}}\right)$$</span>
<span class="math-container">$$=\pi \sqrt{2}. $$</span></p>
|
666,297 | <p>Find the value of $x$, what is the value of $x$ in this equation, step by step solution will be great.
\begin{equation}
0.4x+15=x
\end{equation}</p>
| Warren Hill | 86,986 | <p>As your asking for help on such a simple question. I'll explain each step</p>
<p>Starting with the original equation </p>
<p>$$0.4 x + 15 = x$$</p>
<p>First we want to get the 15 on its own which we can do by subracting $0.4 x$ from the left had side (LHS) of the equation because we do that on the LHS we need to do it to right hand side too; otherwise the equation no-longer balances.</p>
<p>$$\begin{align}
0.4 x + 15 & = x \\
0.4 x + 15 - 0.4 x & = x - 0.4 x \\
15 &= x - 0.4 x
\end{align}$$ </p>
<p>Now we note that $ x = 1x$ so </p>
<p>$$\begin{align}
15 &= x - 0.4 x \\
&= 1x - 0.4x \\
&= x(1-0.4) \\
&= 0.6 x \\
\frac{15}{0.6} & = x \\
x &= \frac{15}{0.6} \\
x &= 25
\end{align}$$ </p>
<p>This is the same result as given by dato and his answer is correct. I've just added a few extra steps to make it (I hope) a little clearer.</p>
<p>The key point to remember is that what you do to the LHS of the equation you also have to do to the RHS. </p>
|
481,086 | <blockquote>
<p>Find a formula (provide your answer in terms of $f$ and its derivatives) for the curvature of a curve in $\mathbb{R}^3$ given by
$\{(x,y,z)\ | \ x=y, f(x)=z\}$.</p>
</blockquote>
<p>How will I be able to do this problem? </p>
<p>I know that a regular parametrization of a curve then the curvature at $\mu(t)$ is given
by: $$\kappa(t) = \left|\left|\left(\frac{\mu'(t)}{||\mu'(t)||}\right)'\frac{1}{||\mu'(t)||}\right|\right|.$$</p>
<p>But since $f$ is not a parametrization I don't know how to continue this problem. </p>
<p>Since $x=y$ and there exists a function $f$ such that $f(x)=z$ and $f(y)=z$ is it safe to assume that it is a one to one function? </p>
| nbubis | 28,743 | <p><a href="http://en.wikipedia.org/wiki/Curvature" rel="nofollow">Wikipedia</a> gives the following for a curve $(x(t),y(t),z(t))$:
$$\kappa=\frac{\sqrt{(z''y'-y''z')^2+(x''z'-z''x')^2+(y''x'-x''y')^2}}{(x'^2+y'^2+z'^2)^{3/2}}$$</p>
|
1,988,420 | <p>An Ant is on a vertex of a triangle. Each second, it moves randomly to an adjacent vertex. What is the expected number of seconds before it arrives back at the original vertex?</p>
<p>My solution: I dont know how to use markov chains yet, but Im guessing that could be a way to do this. I was wondering if there was an intuitive way to solve this problem. I would have guessed 3 seconds as an answer. </p>
<p>I'm assuming that if it is at Vertex A, there is a 1/2 chance of going to Vertex B or C. So minimum number of seconds is 2 seconds. Max number could be infinite if it keeps bouncing back between B and C without returning to A.</p>
<p>I'm still not sure how to do this puzzle.</p>
| Dennis | 381,631 | <p>Let the 3 vertices be A, B, and C. Without loss of generality, assume that the ant starts at A.</p>
<p>Let <span class="math-container">$E_{A}$</span>, <span class="math-container">$E_{B}$</span>, and <span class="math-container">$E_{C}$</span> be the expected number of seconds to get back to vertex A when it's at A, B and C respectively.</p>
<p>We can write the following relation:</p>
<p><span class="math-container">$$E_{A} = 1 + \frac{1}{2}E_{B} + \frac{1}{2}E_{C}$$</span></p>
<p><span class="math-container">$$E_{B} = 1 + \frac{1}{2}(0) + \frac{1}{2}E_{C}$$</span></p>
<p><span class="math-container">$$E_{C} = 1 + \frac{1}{2}(0) + \frac{1}{2}E_{B}$$</span></p>
<p>Solving the set of equations above, we have</p>
<p><span class="math-container">$$E_{B} = 2 $$</span>
<span class="math-container">$$E_{C} = 2 $$</span>
<span class="math-container">$$E_{A} = 3 $$</span></p>
<p><span class="math-container">$E_{A} = 3$</span> is the answer to the question</p>
|
512,037 | <p>This is a question from our reviewer for our exam for linear algebra. I just want to have some ideas how to tackle the problem.</p>
<p>If $A$ is an $n\times n$ matrix with integer coefficients, such that the sum of each row's elements is equal to $m$, show that $m$ divides the determinant.</p>
| Casteels | 92,730 | <p>Here's a guide: Why is the "all 1's" vector an eigenvector of $A$? What is its eigenvalue? How does $\det(A)$ relate to the eigenvalues of $A$?</p>
|
3,280,095 | <p>Given the function
<span class="math-container">$$\int \frac{\sqrt{x}}{\sqrt{x}-3}dx $$</span>
You would need to have <span class="math-container">$u=\sqrt{x}-3$</span> and <span class="math-container">$du=\frac{1}{2 \sqrt{x}}$</span>, when I use a online calculator it suggests to rewrite the numerator as
<span class="math-container">$$\int \frac{2(u^2+6u+9)}{u}du $$</span>
and I was wondering how do you arrive to this trinomial mathematically/analytical?</p>
| azif00 | 680,927 | <p>If <span class="math-container">$u=\sqrt{x}-3$</span>, then <span class="math-container">$x=(u+3)^2$</span> and <span class="math-container">$dx=2(u+3)du$</span>. Therefore
<span class="math-container">$$\int \frac{\sqrt x}{\sqrt{x}-3}dx=\int \frac{u+3}{u}\cdot 2(u+3)du=\int \frac{2(u+3)^2}{u}du$$</span>
My recommendation is always to write <span class="math-container">$x$</span> according to <span class="math-container">$u$</span>, after perform the substitution.</p>
|
1,652,165 | <p>On a empty shelf you have to arrange $3$ cans of soup, $4$ cans of beans, and $5$ cans of tomato sauce. What is the probability that none of the cans of soup are next to each other?</p>
<p>I tried working this out but get very stuck because I'm not sure that I'm including all the possible outcomes. </p>
| Marco Disce | 17,010 | <p>You have 9 objets different from soup, put them in a row and consider all the spaces adiacent and in between (the spaces where you could put the soups avoiding to have 2 adjacent soups). There are 10 such spaces therefore:</p>
<ul>
<li>For the first soup you have $10$ choices.</li>
<li>For the second soup you have $9$ choices.</li>
<li>For the third soup $8$ choices.</li>
</ul>
<p>You have $10\cdot 9 \cdot 8$ choices to put the soups in non adjacent places.</p>
<p>If on the other hand you allow the soups to be in any positions you just freely choose a position over the 12 possible positions of 12 objects:</p>
<ul>
<li>a choice over $12$ positions for the first, </li>
<li>over $11$ for the second and </li>
<li>over $10$ for the third.</li>
</ul>
<p>Total choices: $12\cdot 11 \cdot 10$</p>
<p>So the probability is $\frac{10\cdot 9 \cdot 8}{12\cdot 11 \cdot 10}=\frac 6 {11}$</p>
|
1,684,741 | <p>I'm able to show it isn't absolutely convergent as the sequence $\{1^n\}$ clearly doesn't converge to $0$ as it is just an infinite sequence of $1$'s. How do I prove the series isn't conditionally convergent to prove divergence!</p>
| N. F. Taussig | 173,070 | <p>First, factor the equivalence as a difference of squares.<br>
\begin{align*}
y^4 & \equiv 4 \pmod{7}\\
y^4 - 4 & \equiv 0 \pmod{7}\\
(y^2 + 2)(y^2 - 2) & \equiv 0 \pmod{7}
\end{align*}
Hence, $$y^2 + 2 \equiv 0 \pmod{7} \implies y^2 \equiv -2 \equiv 5 \pmod{7}$$ or $$y^2 - 2 \equiv 0 \pmod{7} \implies y^2 \equiv 2 \pmod{7}$$
Observe that
\begin{align*}
4 & \equiv -3 \pmod{7}\\
5 & \equiv -2 \pmod{7}\\
6 & \equiv -1 \pmod{7}
\end{align*}
and that
\begin{align*}
0^2 & \equiv 0 \pmod{7}\\
1^2 & \equiv (-1)^2 \equiv 1 \pmod{7}\\
2^2 & \equiv (-2)^2 \equiv 4 \pmod{7}\\
3^2 & \equiv (-3)^2 \equiv 9 \equiv 2 \pmod{7}
\end{align*}
Hence, $y \equiv 3 \pmod{7}$ or $y \equiv -3 \equiv 4 \pmod{7}$, which you can check by direct substitution.</p>
|
4,050,893 | <p>Given a linear transformation <span class="math-container">$T: V \rightarrow W$</span> where <span class="math-container">$V$</span> and <span class="math-container">$W$</span> are finite dimensional, then is it true that nullity(<span class="math-container">$T$</span>) = nullity(<span class="math-container">$[T]_\beta$</span>) for any basis <span class="math-container">$\beta$</span> of <span class="math-container">$V$</span>?</p>
<p>A little embarrassing that I'm 20 weeks into studying linear algebra and I've forgotten how to prove this (if I ever knew how). I think this is true, and if it is, I feel like the reasoning is extremely simple, but I can't seem to rigorously spell it out.</p>
<p>Is there a rigorous (and I'm expecting pretty simple) explanation for why this is true?</p>
<p>Thank you!</p>
| Petrus1904 | 808,320 | <p>I have made an algebraic attempt that might help you somewhere. Lets start with the second and third equation:
<span class="math-container">$$AB(I+D) = -BCB \rightarrow AB = -BCB(I+D)^{-1}$$</span>
<span class="math-container">$$(I+D)CA = -CBC \rightarrow CA = -(I+D)^{-1}CBC$$</span>
Now plug these results in the first equation:
<span class="math-container">$$A^2 + BC - B(I+D)^{-1}CBC - BCB(I+D)^{-1}C+A = I$$</span>
From this, two things can be observed:
<span class="math-container">$$A = -B(I+D)^{-1}C, ~~ \text{and} ~~ BCA = ABC$$</span>
Next, lets postmultiply the second equation with <span class="math-container">$C$</span> and the premultiply the third with <span class="math-container">$B$</span>:
<span class="math-container">$$ABC + BCBC + ABDC = 0, ~~ BCA + BDCA + BCBC = 0$$</span>
<span class="math-container">$$BCA + BDCA + BCBC = ABC + BCBC + ABDC \rightarrow BDCA = ABDC$$</span>
Now if we substitute the definition for <span class="math-container">$A$</span> in this result the following can also be deduced:
<span class="math-container">$$BDCA = ABDC \rightarrow -BDCB(I+D)^{-1}C = -B(I+D)^{-1}CBDC$$</span>
<span class="math-container">$$DCB(I+D)^{-1}=(I+D)^{-1}CBD$$</span>
All these commutations are starting me to believe that <span class="math-container">$A$</span> is something like <span class="math-container">$\alpha I$</span>, and therefore <span class="math-container">$-(I+D)^{-1} = \alpha (CB)^{-1}$</span>. Hope this does help you get to an answer.</p>
|
51,246 | <p>In undergraduate courses we compute the sum $S$ of some series
of the form $\frac{1}{P(n)}$ where $P(x)$ is some simple
polynomial of degree $2$ with integer coefficients, by the following procedure:</p>
<p>(sketch)</p>
<p>(a) Choose an appropriate periodic function $f(x)$ defined over a domain $D.$</p>
<p>(b) Compute the Fourier series $S(x)$ of $f(x).$</p>
<p>(c) Choose a suitable $x$ in $D$ so that we obtain a linear equation for $S.$</p>
<p>(d) Solve the equation to get $S.$</p>
<p>Example: </p>
<p>When $P(x)=x^2+1$ we can take:</p>
<p>$f(x)= \exp(x),$
$D= [-\pi,\pi[$,
and $x=\pi.$ </p>
<p>$S$ is the sum from $n=1$ to infinity
of $\frac{1}{n^2+1}.$</p>
<p>We get the equation: </p>
<p>$$
ch(\pi) = S(\pi) = 2\frac{sh(\pi)}{\pi}(\frac{1}{2}+S)
$$
that gives
$$
S=\frac{1}{2}(\frac{\pi}{th(\pi)}-1).
$$</p>
<p>($ch,sh,th$ denote the classic hyperbolic functions)</p>
<p>Question:</p>
<p>Why this fails (in general) for polynomials $P(x)$ of degree $3.$ ? </p>
<p>Why this fails for the polynomial $P(x)=x^3.$ ?</p>
| Michael Renardy | 12,120 | <p>There is a systematic method for evaluating series of this type by residue calculus. It is explained in many texts on complex analysis. Using this method, certain sums over all integers can be evaluated. This makes use of functions like cotangent or cosecant, which have poles at all integers. An even function summed over the positive integers is easily reduced to a sum over all integers. For odd functions, this does not work. The method of summing series by residues is still applicable, but instead of the trig functions, we need to use a function that has poles only at integers of one sign. Such a function exists, namely the Gamma function, but it is not "elementary."</p>
|
43,611 | <p>I posted this on Stack Exchange and got a lot of interest, but no answer.</p>
<p>A recent <a href="http://people.missouristate.edu/lesreid/POW12_0910.html" rel="nofollow">Missouri State problem</a> stated that it is easy to decompose the plane into half-open intervals and asked us to do so with intervals pointing in every direction. That got me trying to decompose the plane into closed or open intervals. The best I could do was to make a square with two sides missing (which you can do out of either type) and form a checkerboard with the white squares missing the top and bottom and the black squares missing the left and right. That gets the whole plane except the lattice points. This seems like it must be a standard problem, but I couldn't find it on the web. <strong>Question:</strong> So can the plane be decomposed into unit open intervals? closed intervals?</p>
| S. Carnahan | 121 | <p>Here is a solution to the half-open interval problem:</p>
<ol>
<li>Start with the interval from $(0,0)$ to $(1,0)$ that contains the endpoint $(0,0)$.</li>
<li>Fill in the closed unit disc minus the point $(1,0)$ using half-open intervals that point inward.</li>
<li>Add the interval from $(1,0)$ to $(1,1)$ that contains the endpoint $(1,0)$.</li>
<li>Fill in the closed disc of radius $\sqrt{2}$ minus the point $(1,1)$ using half-open intervals that point along inward tangents. Now we have segments in all directions.</li>
<li>Fill in the ray $\{ (a,a) \mid a \geq \sqrt{2} \}$ using outward pointing radial segments.</li>
<li>Fill in the rest of the plane using inward pointing radial segments arranged in concentric annuli (with slits at $\arg z = \frac{\pi}{4}$).</li>
</ol>
|
4,031,476 | <p>I recently completed a variation of a problem I found from a mathematical olympiad which is as follows:</p>
<p>Prove that, for all <span class="math-container">$n \in \mathbb{Z}^+$</span>, <span class="math-container">$n \geq 1$</span>, <span class="math-container">$$\sum_{k=1}^n \frac{k}{2^k} < 2 $$</span></p>
<p>I would like to hear comments on my method of proving the inequality above as well as ways of improving it or making it more rigorous. Thank you!</p>
<p>Proof: We begin by noticing that the series above is a Riemann sum (or approximation by areas of rectangles) for the integral <span class="math-container">$$ \int_1^n \frac{x}{2^x} dx$$</span> where the right endpoints are taken, hence <span class="math-container">$$\sum_{k=1}^n \frac{k}{2^k} < \int_1^n \frac{x}{2^x} dx$$</span> for all <span class="math-container">$n$</span>.</p>
<p>By evaluating the original integral, it can be shown that <span class="math-container">$$ \int_1^n \frac{x}{2^x} dx = \frac{n \ln 2 + 1}{2^n (\ln 2)^2} + \frac{\ln 2 + 1}{2 (\ln 2)^2}$$</span></p>
<p>As <span class="math-container">$n$</span> approaches infinity, the integral approaches the value <span class="math-container">$\frac{\ln 2 + 1}{2 (\ln 2)^2}$</span>, which is less than <span class="math-container">$2$</span>. Hence, it can be deduced that <span class="math-container">$$ \int_1^n \frac{x}{2^x} dx < 2$$</span> for a large enough <span class="math-container">$n$</span>.</p>
<p>By inspection and calculation, it is obvious that <span class="math-container">$$ \int_1^n \frac{x}{2^x} dx < 2$$</span> when <span class="math-container">$n > 5, n \in \mathbb{Z}^+$</span>. Hence, when <span class="math-container">$1 \leq n \leq 5$</span>, the integral has a value that is greater than <span class="math-container">$2$</span>.</p>
<p>However, it can be verified by calculation that for <span class="math-container">$1 \leq n \leq 5$</span>, <span class="math-container">$$\sum_{k=1}^n \frac{k}{2^k} < 2 $$</span> Since the statement holds true for <span class="math-container">$1 \leq n \leq 5$</span> as well as for <span class="math-container">$n > 5$</span>, where <span class="math-container">$n \in Z^+$</span>, the statement must be true for all <span class="math-container">$n \geq 1$</span>. Hence, the proof is complete.</p>
<p>Hope to hear your thoughts on this!</p>
| heropup | 118,193 | <p>There are some major issues. First, when <span class="math-container">$n = 1$</span>, the integral is trivially <span class="math-container">$0$</span>, so you cannot make the claim that <span class="math-container">$$S(n) = \sum_{k=1}^n \frac{k}{2^k} < \int_{x=1}^n \frac{x}{2^x} \, dx$$</span> for all <span class="math-container">$n$</span>. In fact, this inequality is <em>never</em> true. I suspect you intended
<span class="math-container">$$S(n) < \int_{x=0}^n \frac{x}{2^x} \, dx = I(n)$$</span> but once again this is not true unless <span class="math-container">$n \ge 5$</span>, and more worrisome is that as <span class="math-container">$n \to \infty$</span>, <span class="math-container">$I(\infty) > 2$</span>, so the bound is simply too loose to show that <span class="math-container">$S(n) < 2$</span>, even if you attempt to patch the reasoning for small cases of <span class="math-container">$n$</span>.</p>
<p>When we use an integral to bound a sum, the behavior of the summand that is converted to an integrand must be monotone, otherwise you cannot make the argument. If you study the graph of <span class="math-container">$y = 2^{-x} x$</span> on <span class="math-container">$[0, \infty)$</span>, we see that it is increasing from <span class="math-container">$x = 0$</span> to <span class="math-container">$x = (\log 2)^{-1}$</span>, then decreasing afterward. This makes your approach fundamentally problematic.</p>
<p>Instead, what I would recommend is either the direct evaluation of <span class="math-container">$S$</span>, for which there are numerous methods. This can be accomplished by differentiation of geometric series, e.g. consider <span class="math-container">$$f(z) = \sum_{k=0}^\infty z^k = \frac{1}{1-z}, \quad |z| < 1,$$</span> hence <span class="math-container">$$z f'(z) = \sum_{k=1}^\infty kz^k = \frac{z}{(1-z)^2}.$$</span> Then <span class="math-container">$$S(n) < S(\infty) = \left[z f'(z) \right]_{z=1/2} = \frac{1/2}{(1-1/2)^2} = 2$$</span> with the first inequality a direct consequence of the fact that all terms in <span class="math-container">$S$</span> are positive.</p>
|
4,031,476 | <p>I recently completed a variation of a problem I found from a mathematical olympiad which is as follows:</p>
<p>Prove that, for all <span class="math-container">$n \in \mathbb{Z}^+$</span>, <span class="math-container">$n \geq 1$</span>, <span class="math-container">$$\sum_{k=1}^n \frac{k}{2^k} < 2 $$</span></p>
<p>I would like to hear comments on my method of proving the inequality above as well as ways of improving it or making it more rigorous. Thank you!</p>
<p>Proof: We begin by noticing that the series above is a Riemann sum (or approximation by areas of rectangles) for the integral <span class="math-container">$$ \int_1^n \frac{x}{2^x} dx$$</span> where the right endpoints are taken, hence <span class="math-container">$$\sum_{k=1}^n \frac{k}{2^k} < \int_1^n \frac{x}{2^x} dx$$</span> for all <span class="math-container">$n$</span>.</p>
<p>By evaluating the original integral, it can be shown that <span class="math-container">$$ \int_1^n \frac{x}{2^x} dx = \frac{n \ln 2 + 1}{2^n (\ln 2)^2} + \frac{\ln 2 + 1}{2 (\ln 2)^2}$$</span></p>
<p>As <span class="math-container">$n$</span> approaches infinity, the integral approaches the value <span class="math-container">$\frac{\ln 2 + 1}{2 (\ln 2)^2}$</span>, which is less than <span class="math-container">$2$</span>. Hence, it can be deduced that <span class="math-container">$$ \int_1^n \frac{x}{2^x} dx < 2$$</span> for a large enough <span class="math-container">$n$</span>.</p>
<p>By inspection and calculation, it is obvious that <span class="math-container">$$ \int_1^n \frac{x}{2^x} dx < 2$$</span> when <span class="math-container">$n > 5, n \in \mathbb{Z}^+$</span>. Hence, when <span class="math-container">$1 \leq n \leq 5$</span>, the integral has a value that is greater than <span class="math-container">$2$</span>.</p>
<p>However, it can be verified by calculation that for <span class="math-container">$1 \leq n \leq 5$</span>, <span class="math-container">$$\sum_{k=1}^n \frac{k}{2^k} < 2 $$</span> Since the statement holds true for <span class="math-container">$1 \leq n \leq 5$</span> as well as for <span class="math-container">$n > 5$</span>, where <span class="math-container">$n \in Z^+$</span>, the statement must be true for all <span class="math-container">$n \geq 1$</span>. Hence, the proof is complete.</p>
<p>Hope to hear your thoughts on this!</p>
| Community | -1 | <p><span class="math-container">$$2S_n-S_n=\sum_{k=1}^nk2^{1-k}-\sum_{k=1}^nk2^{-k}=\sum_{k=0}^{n-1}(k+1)2^{-k}-\sum_{k=1}^nk2^{-k}
\\=1-n2^{-n}+\sum_{k=0}^{n-1}2^{-k}<1+1.$$</span></p>
|
4,008,420 | <p>Suppose we had a differentiable curve <span class="math-container">$C$</span> in <span class="math-container">$\mathbb{R}^2$</span> that serves as our "light container". Light is shining in from all directions, so the space of incoming light-beams is <span class="math-container">$\mathbb{R} \times S^1$</span> where the <span class="math-container">$S^1$</span> parametrises the angle that the light is coming in from, and the <span class="math-container">$\mathbb{R}$</span> parametrises the offset from the origin. When the light hits <span class="math-container">$C$</span> it bounces as you would expect it to off a mirror. Some of the incoming light beams will be trapped inside <span class="math-container">$C$</span>, and keep bouncing forever without escaping, however most will just bounce off.</p>
<ul>
<li>Is there such a finite curve <span class="math-container">$C$</span> such that it actually captures a non-zero amount of light? (take the natural measure on <span class="math-container">$\mathbb{R} \times S^1$</span>, is there a curve <span class="math-container">$C$</span> that captures a non-null subset of the light?) My intuition seems to be telling me that there isn't (because I can't seem to write down any convincing examples), but I can't really think of a reason why not.</li>
<li>If so, what's the curve <span class="math-container">$C$</span> that captures the most light? (for a given length <span class="math-container">$L$</span>. Or, I guess any collection of curves <span class="math-container">$C_i$</span> whose total length is <span class="math-container">$L$</span>).</li>
</ul>
<hr />
<p><a href="https://i.stack.imgur.com/Cbiow.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/Cbiow.jpg" alt="enter image description here" /></a>
Sorry about the confusion, I mean that the light is shining in from infinity. Here for example, the black curve C is our jar, and the blue lines represent incoming light. Most just reflect off, some may get trapped forever. We can draw such C that captures a discrete amount of light, but can we draw C that captures a non-trivial amount of light?</p>
| Intelligenti pauca | 255,730 | <p>A possible answer is given by S. Tabachnikov in his book "Geometry and billiards" and originally proposed by R. Peirone (Reflections can be trapped. Amer. Math. Monthly 101 (1994), 259–260).</p>
<p>First of all, one can construct a trap for light rays parallel to a given direction, see figure below (taken from Tabachnikov's book).</p>
<p><a href="https://i.stack.imgur.com/1u8SB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1u8SB.png" alt="enter image description here" /></a></p>
<p>Curve <span class="math-container">$\gamma$</span> is an ellipse with foci <span class="math-container">$F_1$</span> and <span class="math-container">$F_2$</span>, curve <span class="math-container">$\Gamma$</span> is a parabola with focus <span class="math-container">$F_2$</span>. These curves are joined in a smooth way to produce a trap: light entering with a direction parallel to the axis of the parabola is reflected to focus <span class="math-container">$F_2$</span>, and one can prove (exercise 4.3 in that book) that the path of reflected light through the foci of an ellipse converges to its major axis, so that a captured ray never escapes (provided the hole in the ellipse is not too large).</p>
<p>Unfortunately, such a trap only works if incident rays exactly have the given direction: it is shown in the same book (p. 115) that it is not possible to trap a set of divergent rays of light.</p>
<hr />
<p>Edit by OP: the proof in the book that you can't capture a non-null amount of light is by Poincare Recurrence theorem; since the evolution of light-beams inside the container is given by a measure-preserving map. So Poincare Recurrence tells you that almost all ingoing light beams will eventually escape.</p>
<hr />
<p>Edit. Just to clarify the above observation: some light CAN be trapped, but only if coming from parallel rays. Such light has vanishing volume IN PHASE SPACE (because all rays have the same momentum), hence that doesn't contradict Poincaré's theorem.</p>
|
3,860,330 | <p>I am interested in proving what family of functions have the property
<span class="math-container">$$f'(x)=f^{-1}(x)$$</span>
I've never dealt with a differential equation of this form, hence I could only go as far as to gather a little data:</p>
<p><span class="math-container">$$f'(x)=f^{-1}(x)\implies f(f'(x))=x$$</span>
<span class="math-container">$$\implies f''(x)f'(f'(x))=1$$</span>
<span class="math-container">$$\implies f'(f'(x))=\frac{1}{f''(x)}$$</span></p>
<p>Let <span class="math-container">$f'=g$</span></p>
<p><span class="math-container">$$\implies g(g(x))=\frac{1}{g'(x)}$$</span></p>
<blockquote>
<p>Is this generally solvable?</p>
</blockquote>
<p>Any and all information would be much useful.</p>
| Yiorgos S. Smyrlis | 57,021 | <p>This is a partial answer.</p>
<p>In what follows we seek for a solution of the form <span class="math-container">$f(x)=ax^b$</span>.</p>
<p>Then <span class="math-container">$f'(x)=abx^{b-1}$</span> while <span class="math-container">$f^{-1}(x)=(x/a)^{1/b}=a^{-1/b}x^{1/b}$</span>.</p>
<p>So, setting, <span class="math-container">$f'(x)=f^{-1}(x)$</span>, we obtain
<span class="math-container">$$
abx^{b-1}=a^{-1/b}x^{1/b},
$$</span>
in which case <span class="math-container">$b-1=1/b$</span> equivalently <span class="math-container">$b^2-b-1=0$</span> or <span class="math-container">$b=\frac{1}{2}(1 \pm \sqrt{5})$</span>,
while <span class="math-container">$ab=a^{-1/b}$</span> or <span class="math-container">$a^{b}b=1$</span> or choosing the positive value for <span class="math-container">$b$</span>,
<span class="math-container">$$
a=b^{-1/b}=\left(\frac{1+\sqrt{5}}{2}\right)^{\frac{1-\sqrt{5}}{2}}.
$$</span>
Altogether,
<span class="math-container">$$
f(x)=\left(\frac{1+\sqrt{5}}{2}\right)^{\frac{1-\sqrt{5}}{2}}
x^{\frac{1 + \sqrt{5}}{2}}
$$</span>
possesses the required property: <span class="math-container">$f'=f^{-1}$</span>.</p>
<p>Note that this function defines a global <span class="math-container">$C^1-$</span>solution, if we define <span class="math-container">$x^b=|x|^{b-1}x$</span>.</p>
|
10,977 | <p>When I taught calculus, I posted my notes after the lecture. Then I had the students fill out a mid-quarter evaluation, and a lot of them wanted me to post my notes before class.</p>
<p>What I started doing was printing and handing out the notes to them, leaving the examples blank so they can fill those in. Many of them expressed to me that they liked it, so that they can concentrate on the problem instead of trying to write everything down. </p>
<p>I just read my course evaluations and I got mixed reviews about the notes. Some students said they liked them and found them helpful, and an equal amount said that they didn't like it and preferred to take their own notes, and learned better that way. I did tell them that they don't have to use me notes if they don't want them, and to let me know if they don't want to use them so I can save paper.</p>
<p><strong>In your experience, what has worked well, to provide notes, let the students make their own notes, or give them the option by posting it before class to they can print out notes if they want to use it?</strong> I usually print out the notes, a copy for each student, and pass it out to all of them individually, but I am wondering if I should change this.</p>
| Kenneth Laeremans | 6,516 | <p>Well, what I've got during my classes, like GeraldEdgar already said, is that you don't have one solution to solve it all. What you could do is posting your notes the night (or maybe two nights) before the lecture. That way the students who want to use your notes, can print it and take it with them. They're will always be some other students who wish to make their own notes, because they know by themselves that they will learn better from that. I had experienced that some students scored a little worse then usual after giving everyone the educator's notes. Furthermore I think it's not a bad idea to let sometimes all students make their own notes. If there's a more complex proof, you could give the student's one way to proof it by giving them your notes, but I'm sure they would understand the proof better if you could build up the proof on a different way during your lecture.</p>
|
1,579,528 | <p>You decide to play a holiday drinking game. You start with 100 containers of eggnog in a row. The 1st container contains 1 liter of eggnog, the 2nd contains 2 liters, all the way until the 100th, which contains 100 liters. You select a container uniformly at random and take a one liter sip from it. If the container is empty after taking this sip, you remove it from the row and select only from the remaining bottles. You continue this process until there is only 1 bottle remaining. What is the expected number of liters of eggnog in this last bottle? What is this as this as a function of n, the number of starting bottles?</p>
<p>I came up with this problem myself recently, and I'm not really sure how to approach it. I can find the conditional expectation of a bottle given that it is the last one remaining using linearity of expectations, but it's not clear to me how to use this to get the overall expectation. </p>
| fredq | 297,080 | <p><strong>EDIT</strong> This answer is valid only in the assumption that the probability of taking a sip from a bottle is proportional to the number of sips remaining. </p>
<p>Let $\alpha$ be the number of ways we can arrange <em>all</em> the sips (even counting the one in the end that are not taken)</p>
<p>$$\alpha = \frac{(\sum\limits_{i=1}^{100} i)!}{\prod\limits_{i=1}^{100} (i!)} $$</p>
<p>(there are $\sum\limits_{i=1}^{100} i$ sips, and for the bottle $i$, $i!$ sips are "identical")</p>
<p>Let $N(m)$ be the number of ways to arrange all sips in a way that exactly $m$ sips of one bottle remain in the end is :</p>
<p>$$N(m) = \sum\limits_{l=m}^{100} \frac{(C_l^m) m! (\sum\limits_{i=1}^{100}i - l) (\sum\limits_{i=1}^{100}i -m -1)!}{\prod\limits_{i=1}^{100} (i!)}$$</p>
<p>(To understand the formula, you have to work backward. Consider the case where the bottle $l$ is the last bottle. Then there are $(C_l^m)$ ways to select the $m$ sips from that bottle that are taken last, and there are $m!$ ways to order them. The sip before the last one can not be from bottle $l$ since there are exactly $m$ sips left, hence the factor $(\sum\limits_{i=1}^{100} - l)$. After that, there are $(\sum\limits_{i=1}^{100}i -m -1)$ sips that can be arranged in any way.)</p>
<p>Let $P(m)$ be the probability of having exactly $m$ sips in the last bottle.</p>
<p>$$P(m) = \frac{N(m)}{\alpha} = \sum\limits_{l=m}^{100} \frac{(C_l^m)m!(\sum\limits_{i=1}^{100}i - l) (\sum\limits_{i=1}^{100}i -m -1)!}{(\sum\limits_{i=1}^{100} i)!}$$</p>
<p>Now, the expected number of remaining sips in the last bottle is only :</p>
<p>$E = \sum\limits_{m = 1}^{100} m P(m)$</p>
|
438,263 | <p>Is there a concrete example of a <span class="math-container">$4$</span> tensor <span class="math-container">$R_{ijkl}$</span> with the same symmetries as the Riemannian curvature tensor, i.e.
<span class="math-container">\begin{gather*}
R_{ijkl} = - R_{ijlk},\quad R_{ijkl} = R_{jikl},\quad R_{ijkl} = R_{klij}, \\
R_{ijkl} + R_{iklj} + R_{iljk} = 0.
\end{gather*}</span>
for which there is no metric for which it is the Riemannian curvature tensor?</p>
<p>The existence of such a curvature was already shown by <a href="https://mathoverflow.net/questions/202211/equations-satisfied-by-the-riemann-curvature-tensor">Robert Bryant</a>, however, I'm looking for a concrete example.</p>
| Thomas Kojar | 99,863 | <p>The website <a href="https://www.scilag.net/" rel="nofollow noreferrer">https://www.scilag.net/</a> is also meant as a database.</p>
|
438,263 | <p>Is there a concrete example of a <span class="math-container">$4$</span> tensor <span class="math-container">$R_{ijkl}$</span> with the same symmetries as the Riemannian curvature tensor, i.e.
<span class="math-container">\begin{gather*}
R_{ijkl} = - R_{ijlk},\quad R_{ijkl} = R_{jikl},\quad R_{ijkl} = R_{klij}, \\
R_{ijkl} + R_{iklj} + R_{iljk} = 0.
\end{gather*}</span>
for which there is no metric for which it is the Riemannian curvature tensor?</p>
<p>The existence of such a curvature was already shown by <a href="https://mathoverflow.net/questions/202211/equations-satisfied-by-the-riemann-curvature-tensor">Robert Bryant</a>, however, I'm looking for a concrete example.</p>
| Alex W | 72,323 | <p>For open problems in group theory: <a href="https://arxiv.org/abs/1401.0300" rel="nofollow noreferrer">https://arxiv.org/abs/1401.0300</a></p>
|
754,583 | <p>Write <span class="math-container">$$\phi_n\stackrel{(1)}{=}n+\cfrac{n}{n+\cfrac{n}{\ddots}}$$</span> so that <span class="math-container">$\phi_n=n+\frac{n}{\phi_n},$</span> which gives <span class="math-container">$\phi_n=\frac{n\pm\sqrt{n^2+4n}}{2}.$</span> We know <span class="math-container">$\phi_1=\phi$</span>, the <a href="http://en.wikipedia.org/wiki/Golden_ratio" rel="nofollow noreferrer">Golden Ratio</a>, so let's take <span class="math-container">$\phi_n\stackrel{(2)}{=}\frac{n+\sqrt{n(n+4)}}{2}$</span>. (Is that justified?)</p>
<p><a href="http://m.wolframalpha.com/input/?i=%28n%2B%E2%88%9A%28n%28n%2B4%29%29%29%2F2&x=0&y=0" rel="nofollow noreferrer">Wolfram Alpha</a> states that, with <span class="math-container">$(2)$</span>, <span class="math-container">$$\lim\limits_{n\to -\infty}\phi_n=-1.$$</span> Why? Can I infer that this is true for <span class="math-container">$(1)$</span> and, if so, <em>why</em>?</p>
<p><strong>I wonder what happens in <span class="math-container">$(1)$</span> for <span class="math-container">$n\in\mathbb{C}\backslash\mathbb{Z}$</span> too</strong>. I got something horrendous looking in <span class="math-container">$(2)$</span> for <span class="math-container">$n=i$</span>.</p>
<hr>
<p><em>Clarification:</em> I'm trying to <strong>find <span class="math-container">$\phi_n$</span> in terms of <span class="math-container">$n$</span></strong>. See the comments below.</p>
| user88595 | 88,595 | <p>Found why it's $-1$.<br>
Rewrite the equation as :$$n\bigg(\frac{1+\frac1n\sqrt{n^2 + 4n}}{2}\bigg)$$
The square root can be written as $\sqrt{n^2(1+4/n)} = |n|\sqrt{1+4/n} = -n\sqrt{1+4/n}$ because $n<0$.
Then you obtain :
\begin{eqnarray*}
\phi(n) &=& \frac{n}{2}\big(1-\sqrt{1+4/n}\big)\\
&=& \frac n2(1-(1+\frac2n) + O(n^{-2}))\\
&\to&-1
\end{eqnarray*}</p>
|
680,364 | <p>I want some verification for my proof to a homework problem. (Is it correct? Is there a simpler way to do this?)</p>
<p>Let $G$ be a finite group of odd order and suppose there is an element $g$ that is conjugate to its own inverse. In other words, there is $h \neq e$ such that $h^{-1}gh = g^{-1}$. We will show $g=e$ by supposing it's not and finding a contradiction.</p>
<p>We see that $gh = hg^{-1}$ and $hg = g^{-1}h$. This means given any word composed of $g$, $h$, $g^{-1}$, and $h^{-1}$, we can always "push" the $g^r$'s to one side and the $h^s$'s to the other, giving us a canonical spelling $g^rh^s$. That is to say, $\langle g, h\rangle \cong \langle g \rangle \oplus \langle h \rangle$. </p>
<p>By Lagrange, finding any non-trivial even subgroup will prove that $|G|$ was even after all, giving us the desired contradiction. Since both $\langle g \rangle$ and $\langle h \rangle$ are nontrivial, neither may be even. But if that is the case, then they are both odd, and $\langle g \rangle \oplus \langle h \rangle \cong \langle g, h \rangle$ is now non-trivial even.</p>
<p><strong>EDIT</strong> Critical error at the end. The order of the direct sum is the product, not the sum.</p>
| Ted Shifrin | 71,348 | <p><strong>HINT:</strong> Note that $gh=hg^{-1}$. What is $(gh)^2$?</p>
|
150,809 | <p>An Iwasawa manifold is a compact quotient of a 3-dimensional complex Heisenberg group by a cocompact, discrete subgroup.
We can also refer to Griffiths and Harris's Principles of Algebraic Geometry p. 444 for simpler description.</p>
<p>I want to compute the automorphism of Iwasawa manifold,i.e.the group of biholomorphisms of X onto itself.
But I don't know how to set out to do this.Is there any already known result about this queation?If not,how should I start to study this problem?
Thanks in advance!</p>
| Robert Bryant | 13,972 | <p>I'm sure this can be found in the literature, though I don't know exactly where to look. On the other hand, it is easy to calculate the automorphism group directly from the following observations: As in Griffiths--Harris, let $M = G/\Gamma$ where $G$ is the $3$-dimensional complex Heisenberg group and $\Gamma\subset G$ is a discrete, cocompact subgroup. As Griffiths and Harris point out, there is a triple $\omega_1$, $\omega_2$, $\omega_3$ of holomorphic $1$-forms defined on $M$ such that $\omega_1\wedge\omega_2\wedge\omega_3$ is nowhere vanishing, $d\omega_1 = d\omega_2 = 0$, and $d\omega_3 = \omega_1\wedge\omega_2$. </p>
<p>Now, clearly, any holomorphic $1$-form on $M$ must be a constant linear combination of these three $1$-forms, so if $f:M\to M$ is an automorphism, then $f^*(\omega_i)$ must be a constant linear combination of these forms for each $i$. Moreover, because $\omega_3\wedge d\omega_3$ is a nonvanishing holomorphic $3$-form $M$, any automorphism $f$ must preserve this form up to multiplication by a unit complex number. In particular, $f^*(\omega_1\wedge\omega_2) = \lambda\ \omega_1\wedge\omega_2$ for some $\lambda\in\mathbb{C}$ with $|\lambda| = 1$. Pursuing this line, one sees that $f:M\to M$ must be covered by a biholomorphism $F:G\to G$ that preserves both the right-invariant holomorphic $1$-forms (up to constant linear combinations) and the lattice $\Gamma$ (up to an appropriate translation). </p>
<p>This is now a purely algebraic problem (whose solution might depend on which $\Gamma$ you use). In particular, if you let $z_1$, $z_2$, $z_3$ denote the coordinates on $G$ such that the forms $\omega_i$ pull back to $dz_1$, $dz_2$, and $dz_3 + z_1\ dz_2$, respectively, on $G$, then $F$ has to be of the form
$$
F(z_1,z_2,z_3) = (a_{11} z_1 {+} a_{12}z_2 {+} a_{10}, a_{21} z_1 {+} a_{22}z_2 {+} a_{20},
a_{33} (z_3{-}z_1z_2) {+} a_{31} z_1 {+} a_{32} z_2 {+} a_{30})
$$
for some constants $a_{ij}$ such that
$$
a_{33} = a_{11}a_{22}-a_{21}a_{12} = \lambda
$$
where $|\lambda| = 1$. It must also preserve $\Gamma$ up to an appropriate translation, which puts other restrictions on the $a_{ij}$. This will give you the automorphism group.</p>
|
2,292,713 | <blockquote>
<p><strong>Definition.</strong> Let <span class="math-container">$E$</span> be a nonempty subset of <span class="math-container">$X$</span>, and let <span class="math-container">$S$</span> be the set of all real numbers of the form <span class="math-container">$d(p, q)$</span>, with <span class="math-container">$p,q\in E$</span>. The sup of <span class="math-container">$S$</span> is called the diameter of <span class="math-container">$E$</span>.</p>
<p><strong>Theorem 3.10.</strong> If <span class="math-container">$\overline{E}$</span> is the closure of a set <span class="math-container">$E$</span> in a metric space <span class="math-container">$X$</span>, then <span class="math-container">$$\text{diam }\overline{E} = \text{diam }E.$$</span></p>
</blockquote>
<p>Proof: Fix <span class="math-container">$\varepsilon>0$</span>, and choose <span class="math-container">$p, q \in \overline{E}$</span>. By the definition of <span class="math-container">$\overline{E}$</span>, there are points <span class="math-container">$p',q' \in E$</span> such that <span class="math-container">$d(p,p') < \varepsilon$</span> and <span class="math-container">$d(q,q') < \varepsilon$</span>. Hence <span class="math-container">$$d(p, q) \le d(p,p') + d(p', q') + d(q', q) < 2\varepsilon + d(p', q') \le 2\varepsilon + \text{diam }E.$$</span></p>
<p>Ok until here. But then they use the inequality above to come up with <span class="math-container">$$\text{diam }\overline{E} \le 2\varepsilon + \text{diam }E$$</span></p>
<p>Where I can only see the strict inequality because of the strict inequality relation made in the inequalities above.</p>
| Ted Shifrin | 71,348 | <p>You're missing an important point. If $x<c$ for all $x\in S$, then $\sup S\le c$. (For example, take $S = (0,1)$. For every $x\in S$, we have $x<1$, but $\sup S = 1$.)</p>
|
418,724 | <p>This question arises in STEP 2011 Paper III, question 2. The paper can be found <a href="http://www.admissionstestingservice.org/our-services/subject-specific/step/preparing-for-step/" rel="nofollow">here</a>. </p>
<p>The first part of the question requires us to prove the result that if the polynomial
$$x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_{0}$$
where each of the $a_{n}$ are integers, has a rational roots if and only if that root is an integer. It does not give a name for this result.<br>
EDIT: user69810 has pointed out that this is in fact the rational root theorem. </p>
<p>We are then to use this result to prove that the polynomial $$x^{n}-5x+7=0$$
has no rational solutions for $n\ge 2$. My argument was the following:<br>
If there exists a rational root, then it must be an integer.<br>
If there exists an integer root, then $$x^{n}=5x-7$$
for some integers $x$ and $n\ge 2$. Then the LHS is divisible by $x$, meaning that $7$ must be divisible by $x$. Therefore $x \in \{-7,-1,1,7\}$. Checking each of these shows that there is no rational root. </p>
<p>Is this solution correct? It isn't the one in the solutions, but if it's right then I think it's more elegant than theirs.</p>
| Ross Millikan | 1,827 | <p>You have identified the possible rational roots correctly and shown that none of them work. It is a fine application of the rational root theorem.</p>
|
418,724 | <p>This question arises in STEP 2011 Paper III, question 2. The paper can be found <a href="http://www.admissionstestingservice.org/our-services/subject-specific/step/preparing-for-step/" rel="nofollow">here</a>. </p>
<p>The first part of the question requires us to prove the result that if the polynomial
$$x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_{0}$$
where each of the $a_{n}$ are integers, has a rational roots if and only if that root is an integer. It does not give a name for this result.<br>
EDIT: user69810 has pointed out that this is in fact the rational root theorem. </p>
<p>We are then to use this result to prove that the polynomial $$x^{n}-5x+7=0$$
has no rational solutions for $n\ge 2$. My argument was the following:<br>
If there exists a rational root, then it must be an integer.<br>
If there exists an integer root, then $$x^{n}=5x-7$$
for some integers $x$ and $n\ge 2$. Then the LHS is divisible by $x$, meaning that $7$ must be divisible by $x$. Therefore $x \in \{-7,-1,1,7\}$. Checking each of these shows that there is no rational root. </p>
<p>Is this solution correct? It isn't the one in the solutions, but if it's right then I think it's more elegant than theirs.</p>
| Key Ideas | 78,535 | <p>Yes, it's correct. Simpler: Rational Root Test $\,\Rightarrow\,x\in\Bbb Z\,\Rightarrow\, 7 = 5x\!-\!x^n\,$ is even, contradiction.</p>
|
445,816 | <p>I have to show that</p>
<blockquote>
<blockquote>
<p>$\mathbb{C}=\overline{\mathbb{C}\setminus\left\{0\right\}}$,</p>
</blockquote>
</blockquote>
<p>what is very probably an easy task; nevertheless I have some problems.</p>
<p>In words this means: $\mathbb{C}$ is the smallest closed superset of $\mathbb{C}\setminus\left\{0\right\}$.</p>
| Jay | 9,814 | <p>The elements of the sequence of complex numbers $1$, $\frac{1}{2}$, $\frac{1}{3}$,$\cdots$ are all contained in $\overline{\mathbb{C} \setminus \{ 0 \} }$ but the limit of the sequence is $0$; hence $0$ is in the closure.</p>
|
2,423,569 | <p>I am asked to show that if $T(z) = \dfrac{az+b}{cz+d}$ is a mobius transformation such that $T(\mathbb{R})=\mathbb{R}$ and that $ad-bc=1$ then $a,b,c,d$ are all real numbers or they all are purely imaginary numbers. </p>
<p>So far I've tried multiplying by the conjugate of $cz+d$ numerator and denominator and see if I get some information about $a,b,c,d$ considering that $T(z) \in \mathbb{R}$ whenever $z \in \mathbb{R}$ but this doesn't really work. Also I've considered $SL(2,\mathbb{C}) / \{ \pm I\}$ which is isomorphic to the group of Mobius transformations but this doesn't really help either.</p>
| Henno Brandsma | 4,280 | <p>Suppose $T$ is as asked for. </p>
<ul>
<li>Suppose $c\neq 0$. Then if $\frac{-d}{c} \in \mathbb{R}$ and we know that $T(\frac{-d}{c}) = \infty \notin \mathbb{R}$ unless the numerator is $0$ as well, in which $a(\frac{-d}{c}) = b = 0$ which means $\frac{-ad}{c} = -b$ or $-ad = -bc$ or $ad=bc$ and we have a contradiction with $ad-bc = 1$. </li>
</ul>
<p>So $c=0$ and so $ad=1$. What more can you do now?</p>
|
161,616 | <p>There is a well known result that every one dimensional topological manifold without boundary is homeomorphic either to the circle or to the whole real line. However there is one detail hidden: manifold is understood to be second countable (or paracompact). If we drop this assumption it becomes possible to construct different example, so called <em>open long line</em> or <em>Aleksandroff line</em>. It is defined as $\omega_1 \times [0,1) \setminus \{(0,0)\}$ with suitable order topology. What might be surprising, is that replacing $\omega_1$ by bigger ordinal does not produce manifold anymore (this would produce points with uncountable neighbourhood system). There is also a variant of long line "in both directions". So the natural question is: if we drop the assumption for (one dimensional) manifolds to be second countable, is it possible to characterise all of them?$
Edit: what about two dimensional case? </p>
| Mirko | 48,481 | <p>The long ray and the long line are the only non-metrizable 1-manifolds, see e.g. a paper by Peter Nyikos (who also discusses larger dimensions)
<a href="http://www.math.sc.edu/~nyikos/Manifolds.pdf"><code>here</code></a>
(p.2, just after Main Theorem). No proof is given in the above paper (just saying it is easy). Here is a sketch, say the manifold $M$ has an endpoint, and call it "the leftmost point", $a_0$, and then pick a sequence $a_n$ of points "going to the right". If $M=\cup_{n}[a_0,a_n)$ then we are done. Else keep adding points $a_\omega$, $a_{\omega+1}$, etc.
Then either we are done at some countable ordinal $\gamma$ (and then we are done, use that for every countable ordinal $\gamma$ there is a subset $T$ of the reals that is order-isomorphic to $\gamma$), or else we define $a_\beta$ for all $\beta<\omega_1$, so we get the long ray. It mist be the case that $M=\cup_{\beta<\omega_1}[a_0,a_\beta)$ since otherwise we may define $t=\sup_{\beta<\omega_1} a_\beta \in M$ and our manifold would not be first-countable at $t$, a contradiction. To see that each $\omega<\gamma<\omega_1$ could be thought a subset of $\mathbb R$ fix a bijection $f:\gamma\to \omega$ and any sequence $c_n>0$ with $\sum_n c_n<\infty$, and for each $\beta<\gamma$ define $r_\beta=\sum_{\delta\le\beta}c_{f(\delta)}\in\mathbb R$, then the set $\{r_\beta:\beta<\gamma\}\subset\mathbb R$ is order-isomorphic to $\gamma$. </p>
|
234,851 | <p>Find the length of the curve $x=0.5y\sqrt{y^2-1}-0.5\ln(y+\sqrt{y^2-1})$ from y=1 to y=2.</p>
<p>My attempt involves finding $\frac {dy}{dx}$ of that function first, which leaves me with a massive equation.</p>
<p>Next, I used this formula, </p>
<p>$$\int_1^2\sqrt{1+(\frac{dy}{dx})^2}$$</p>
<p>this attempt leaves me with such a messy long equation that eventually took up 2 pages, and still left me unsolved. I am convinced there must be an easier way.</p>
<p>Any hints please? thanks in advance.</p>
| Golob | 48,784 | <p>$$\frac{\mathrm{d}}{\mathrm{d}y} \Big[\frac{1}{2}y \sqrt{y^2-1}- \frac{1}{2} \ln(\sqrt{y^2-1}+y) \Big]=$$</p>
<p>$$=\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}y} \Big[y \sqrt{y^2-1} \Big] - \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}y} \Big[\ln(\sqrt{y^2-1}+y) \Big]=$$</p>
<p>$$=\frac{1}{2}\Big[\frac{2y^2-1}{\sqrt{y^2-1}} \Big] -\frac{1}{2} \Big[\frac{1}{\sqrt{y^2-1}} \Big]
=
\frac{y^2 - 1}{\sqrt{y^2-1}}
=
\sqrt{y^2-1}$$ for $y\not=1$</p>
<p>therefore
$$\int_1^2\sqrt{1+\left(\frac{\mathrm{d}x}{\mathrm{d}y}\right)^2}\mathrm{d}y
=
\lim_{x\to 1_+}\int_x^2\sqrt{1+\left(\sqrt{y^2-1}\right)^2}\mathrm{d}y
=
\lim_{x\to 1_+}\int_x^2 y \mathrm{d}y
=
1.5$$</p>
|
165,489 | <p>I have problem solving this equation, smallest n such that $1355297$ divides $10^{6n+5}-54n-46$. I tried everything using my scientific calculator, but I never got the correct results(!).and finally I gave up!. Could you help me find the first 2 solutions for this equation ? (thanks.)</p>
| Henrik Schumacher | 38,178 | <p>$n_1 = 2331259$, $n_2 = 3776127$, </p>
<p>Obtained from this Mathematica code:</p>
<pre><code>cf = Compile[{{m, _Integer}},
Block[{n, a, b, p, counter = 0,result},
result = ConstantArray[0, m];
p = 1355297;
n = 0;
a = Mod[10^(5), p];
b = 0;
While[counter < m,
n++;
a = Mod[a 1000000, p];
b = Mod[b - 54 , p];
While[Mod[a + b - 46, 1355297] != 0,
n++;
a = Mod[a 1000000, p];
b = Mod[b - 54 , p];
];
counter++;
result[[counter]] = n;
];
result
],
CompilationTarget -> "C"
];
cf[2]
</code></pre>
<blockquote>
<p>{2331259, 3776127}</p>
</blockquote>
|
1,476,847 | <p>I am a little puzzled by some notations in optimization community. Is there anyone who can explain why $f_1:\mathbb{R}^n\rightarrow\mathbb{R}$ is a finite valueed but $f_2:\mathbb{R}^n\rightarrow\mathbb{R}\cup\{\infty\}$ is not?? I have never have this kind of notations. For function $f_1$ I always calculated limit when $x\rightarrow \infty$ and nobody said you can't as infinity is not a member of real values. </p>
<p>Can anyone clarify this for me?
Thanks</p>
| PhoemueX | 151,552 | <p>This is false even for $2\times2$ diagonal matrices. For these, what
you want reduces to
\begin{eqnarray*}
\left(a_{1}+b_{1}\right)\left(a_{2}+b_{2}\right) & = & \det\left(\left(\begin{array}{cc}
a_{1}\\
& a_{2}
\end{array}\right)+\left(\begin{array}{cc}
b_{1}\\
& b_{2}
\end{array}\right)\right)\\
& \overset{!}{\leq} & \alpha\cdot\det\left(\begin{array}{cc}
a_{1}\\
& a_{2}
\end{array}\right)+\beta\cdot\det\left(\begin{array}{cc}
b_{1}\\
& b_{2}
\end{array}\right)\\
& = & \alpha a_{1}a_{2}+\beta b_{1}b_{2}
\end{eqnarray*}
with suitable $\alpha,\beta \in \Bbb{R}$ and all $a_1,a_2,b_1,b_2 >0$.</p>
<p>Now consider $a_{1}=b_{2}=n$ and $a_{2}=b_{1}=\frac{1}{n}$. Then
the desired inequality becomes
$$
n^{2}\leq\left(n+\frac{1}{n}\right)^{2}\leq\alpha+\beta
$$
for all $n\in\mathbb{N}$, which is absurd.</p>
|
1,558,256 | <p>The standard Normal distribution probability density function is $$p(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2},\int_{-\infty}^{\infty}p(t)\,dt = 1$$ i.e., mean 0 and variance 1. The cumulative distribution function is given by the improper integral $$P(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}e^{-t^2/2}\,dt$$ Describe a numerical method for approximating $P(x)$ given a value of $x$ to a prescribed absolute error $\tau$. Your solution should be as efficient as possible. Justify your answer.</p>
<p>I believe we need to consider the case where $x > 0$ and $x < 0$ and then use one a numerical approximating method that is the most accurate to use. That is the crux of the problem, there are many methods, does one need to try each and everyone to see which one is best. Is there way of determining which one is the best given this particular problem before actually performing any computations. Any suggestions is greatly appreciated.</p>
| m_gnacik | 182,603 | <p>First note that cumulative distribution of Normal distribution
$$ P(x) = \frac{1}{2}+\frac{1}{2}\mathrm{erf}\left(\frac{x}{\sqrt{2}}\right)$$
where $\mathrm{erf}$ is the error function. </p>
<p>Abramowitz and Stegun gave several approximations of the error function,
<a href="https://en.wikipedia.org/wiki/Error_function#Approximation_with_elementary_functions" rel="nofollow">check this</a> and <a href="http://people.math.sfu.ca/~cbm/aands/page_299.htm" rel="nofollow">that</a>. I hope that it was helpful. </p>
|
1,558,256 | <p>The standard Normal distribution probability density function is $$p(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2},\int_{-\infty}^{\infty}p(t)\,dt = 1$$ i.e., mean 0 and variance 1. The cumulative distribution function is given by the improper integral $$P(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}e^{-t^2/2}\,dt$$ Describe a numerical method for approximating $P(x)$ given a value of $x$ to a prescribed absolute error $\tau$. Your solution should be as efficient as possible. Justify your answer.</p>
<p>I believe we need to consider the case where $x > 0$ and $x < 0$ and then use one a numerical approximating method that is the most accurate to use. That is the crux of the problem, there are many methods, does one need to try each and everyone to see which one is best. Is there way of determining which one is the best given this particular problem before actually performing any computations. Any suggestions is greatly appreciated.</p>
| Math-fun | 195,344 | <p>Say we look at $x<0$ only. Then consider a change of variable of the form $u=\frac{t}{1-t}$ to have$$P(x)=\int_{-1}^{\frac{x}{1-x}}\frac{e^{-\frac{u^2}{2 (u+1)^2}}}{\sqrt{2 \pi } (u+1)^2}du$$ which is manageable stuff since $$\lim_{u\to -1}\frac{e^{-\frac{u^2}{2 (u+1)^2}}}{\sqrt{2 \pi } (u+1)^2}=0.$$</p>
|
3,886,523 | <p>Each of <span class="math-container">$3$</span> urns contains twenty balls. First urn contains ten white balls, second urn contains six white balls and third urn contains two white balls. All other balls are black.
One ball is drawn from the random urn with return in the same urn. The ball's color is white.
What is the probability that the second ball drawn from the same urn is white?</p>
<p>I think, this is <span class="math-container">$\frac{1}{9} \cdot \frac{2}{20} + \frac{3}{9} \cdot \frac{6}{20} + \frac{5}{9} \cdot \frac{10}{20}$</span> by Bayes'theorem and Law of total probability, but can't be sure.</p>
<p>Thanks for any help.</p>
| Community | -1 | <p>For the most general case in which the base doesn't even need to be a polygon, it is an application of <a href="https://en.wikipedia.org/wiki/Cavalieri%27s_principle" rel="nofollow noreferrer">Cavalieri's Principle</a>. Imagine a pyramid with a blob-shaped base and a square pyramid next to it with the same height and whose base has the same area as that of the first pyramid. Since any level cross-section of the two pyramids would have to have the same area, the volumes must also be equal.</p>
<p>There is a marvelous image showing that the area of a square pyramid whose height is equal to its base size and whose apex is above one corner of the base is 1/3 the volume of a unit cube. Long story short, you can construct three of those pyramids and see that they can be fit together to form a unit cube. From there, the only remaining step is to convince yourself that the volume of a pyramid whose base is a unit square is directly proportional to the height of the pyramid. At that point, you've proved it from a unit square pyramid with unit height to a unit square pyramid of arbitrary height to an arbitrary base and arbitrary height.</p>
|
1,318,462 | <p>I am struggling with the following problem.Any help will be appreciated.</p>
<p>If the following statement true then please give a proof otherwise give a counterexample.</p>
<ol>
<li><p>If $a^{27} \equiv 1 \pmod{37}$, then $a^9 \equiv 1 \pmod{37}$ </p></li>
<li><p>$a^{9} \equiv 1 \pmod{37}$, then $a^3 \equiv 1 \pmod{37}$ </p></li>
<li><p>$a^{5} \equiv 1 \pmod{37}$, then $a^3 \equiv 1 \pmod{37}$ </p></li>
</ol>
<p>Thank you.</p>
| tong_nor | 213,691 | <p>The first one is true: if $a^{27}\equiv 1\pmod{37}$, then $a\bmod{37}$ is one of numbers $1,7,9,10,12,16,26,33,34$. For each of them you have $a^{9}\equiv 1\pmod{37}$.</p>
<p>The second one is false: $7^{9}\equiv 1\pmod{37}$, but $7^3\equiv 10\pmod{37}$.</p>
<p>The third one is also false, try to find a counterexample.</p>
<p>Edit:
I thought there was $a^6$. With $a^5$ the statement is true.
If $a^5\equiv 1\pmod{37}$, then $a\equiv 1\pmod{37}$ (the only case) and then of course $a^3\equiv 1\pmod{37}$.</p>
|
1,119,634 | <p>Find the point on the curve $y=x^2+2$ where the tangent is parallel to the line $2x+y-1=0$</p>
<p>I understand the answer is $(-1,3)$ but I can't find a way to get there... Thanks </p>
| WW1 | 88,679 | <p>The line $2x+y-1=0$ has a slope of $-2$</p>
<p>the derivative of $x^2+2$ is $2x$</p>
<p>for the $x$-coordinate solve $2x=-2$</p>
<p>for the $y$-coordinate plug the value of $x$ into $y=x^2+2$</p>
|
11,973 | <p>I have a list of strings called <code>mylist</code>:</p>
<pre><code>mylist = {"[a]", "a", "a", "[b]", "b", "b", "[ c ]", "c", "c"};
</code></pre>
<p>I would like to split <code>mylist</code> by "section headers." Strings that begin with the character <code>[</code> are section headers in my application. Thus, I would like to split <code>mylist</code> in such a way as to obtain this output:</p>
<pre><code>{{"[a]", "a", "a"}, {"[b]", "b", "b"}, {"[ c ]", "c", "c"}}
</code></pre>
<p>(The <code>a</code>s, <code>b</code>s, and <code>c</code>s represent <em>any</em> characters; the string inside the section header does <em>not</em> necessarily match the strings that follow in that section. Also, the number of strings in each section can vary.</p>
<p>I have tried:</p>
<pre><code>SplitBy[mylist, StringMatchQ[#, "[" ~~ ___] &]
</code></pre>
<p>But this is not correct; I obtain:</p>
<pre><code>{{"[a]"}, {"a", "a"}, {"[b]"}, {"b", "b"}, {"[ c ]"}, {"c", "c"}}
</code></pre>
<p>Likewise, using <code>Split</code> (since it applies the test function only to adjacent elements) does not work. The command:</p>
<pre><code>Split[mylist, StringMatchQ[#, "[" ~~ ___] &]
</code></pre>
<p>yields:</p>
<pre><code>{{"[a]", "a"}, {"a"}, {"[b]", "b"}, {"b"}, {"[ c ]", "c"}, {"c"}}
</code></pre>
<p>Do you have any advice? Thanks.</p>
| rm -rf | 5 | <p>Here's one approach using <code>FixedPoint</code> and <code>Replace</code>:</p>
<pre><code>sectionQ := ! StringFreeQ[#, "["] &;
FixedPoint[
Replace[#, {h___, sec_?sectionQ, Longest[x___?(! sectionQ@# &)], t___} :> {h, t, {sec, x}}] &,
mylist]
(* {{"[a]", "a", "a"}, {"[b]", "b", "b"}, {"[ c ]", "c", "c"}} *)
</code></pre>
|
11,973 | <p>I have a list of strings called <code>mylist</code>:</p>
<pre><code>mylist = {"[a]", "a", "a", "[b]", "b", "b", "[ c ]", "c", "c"};
</code></pre>
<p>I would like to split <code>mylist</code> by "section headers." Strings that begin with the character <code>[</code> are section headers in my application. Thus, I would like to split <code>mylist</code> in such a way as to obtain this output:</p>
<pre><code>{{"[a]", "a", "a"}, {"[b]", "b", "b"}, {"[ c ]", "c", "c"}}
</code></pre>
<p>(The <code>a</code>s, <code>b</code>s, and <code>c</code>s represent <em>any</em> characters; the string inside the section header does <em>not</em> necessarily match the strings that follow in that section. Also, the number of strings in each section can vary.</p>
<p>I have tried:</p>
<pre><code>SplitBy[mylist, StringMatchQ[#, "[" ~~ ___] &]
</code></pre>
<p>But this is not correct; I obtain:</p>
<pre><code>{{"[a]"}, {"a", "a"}, {"[b]"}, {"b", "b"}, {"[ c ]"}, {"c", "c"}}
</code></pre>
<p>Likewise, using <code>Split</code> (since it applies the test function only to adjacent elements) does not work. The command:</p>
<pre><code>Split[mylist, StringMatchQ[#, "[" ~~ ___] &]
</code></pre>
<p>yields:</p>
<pre><code>{{"[a]", "a"}, {"a"}, {"[b]", "b"}, {"b"}, {"[ c ]", "c"}, {"c"}}
</code></pre>
<p>Do you have any advice? Thanks.</p>
| kglr | 125 | <pre><code>Split[mylist, StringFreeQ["["] @ #2 &]
</code></pre>
<blockquote>
<p>{{"[a]", "a", "a"}, {"[b]", "b", "b"}, {"[ c ]", "c", "c"}}</p>
</blockquote>
<pre><code>SequenceCases[mylist, a:{_?(!StringFreeQ[ "["]@#&),__?(StringFreeQ[ "["])}:> {a}]
</code></pre>
<blockquote>
<p>{{"[a]", "a", "a"}, {"[b]", "b", "b"}, {"[ c ]", "c", "c"}}</p>
</blockquote>
|
233,238 | <p>I am just practicing making some new designs with Mathematica and I thought of this recently. I want to make a tear drop shape (doesn't matter the orientation) constructed of mini cubes. I am familiar with the preliminary material, I am just having some difficulty getting it to work.</p>
| AsukaMinato | 68,689 | <pre><code>list // Map[Flatten]
</code></pre>
<blockquote>
<p>{{Position, Code}, {1, 0, 1}, {2, 100, 11}, {3, 110, 111}, {4, 1000,
1001}, {5, 1100, 1011}, {6, 1110, 1111}}</p>
</blockquote>
|
233,238 | <p>I am just practicing making some new designs with Mathematica and I thought of this recently. I want to make a tear drop shape (doesn't matter the orientation) constructed of mini cubes. I am familiar with the preliminary material, I am just having some difficulty getting it to work.</p>
| m_goldberg | 3,066 | <p><code>ReplacePart</code> can do it.</p>
<pre><code>data =
{{Position, {Code}},
{1, {0000, 0001}}, {2, {0100, 0011}}, {3, {0110, 0111}},
{4, {1000, 1001}}, {5, {1100, 1011}}, {6, {1110, 1111}}};
ReplacePart[data, {i_, 2} :> Sequence @@ data[[i, 2]]]
</code></pre>
<blockquote>
<pre><code>{{Position, Code},
{1, 0, 1}, {2, 100, 11}, {3, 110, 111},
{4, 1000, 1001}, {5, 1100, 1011}, {6, 1110, 1111}}
</code></pre>
</blockquote>
|
14,847 | <p>I thought a simple Mathematica kerning machine (for adjusting the space between characters) would be interesting, but I'm having trouble with the locators. (There are a number of other questions related to this, and I've read the answers, but as yet without finding a solution, or understanding them that well.)</p>
<pre><code> Manipulate[
LocatorPane[Dynamic[points],
text = "Wolfram";
fonts = FE`Evaluate[FEPrivate`GetPopupList["MenuListFonts"]];
Column[{
Button["Export", Export["/tmp/t.png", g]],
g = Graphics[{
MapIndexed[
Text[Style[#1, FontSize -> fontsize, FontFamily -> font],
points[[First[#2]]]] &, Characters[text]]}, ImageSize -> 500]
}]],
{points, {Table[{ x, 0}, {x, 1, Length[Characters[text]]}]},
ControlType -> None},
{fontsize , Table[x, {x, 96, 256, 12}]} ,
{font, fonts}
]
</code></pre>
<p>(The font menu gets populated once you use it.)</p>
<p><img src="https://i.stack.imgur.com/T1aNl.png" alt="kerning machine"></p>
<p>I want to be able to slide the letters right or left (but not up or down), but at the moment they don't move like they're supposed to.</p>
<h2>The solution</h2>
<p>Thanks to the fine answers, I've got something useful working. There are some kludges and hacks too, and some problems still to be ironed out (string length needs to be dynamic, 'canvas' needs resizing, and so on), but this is excellent for now. </p>
<pre><code>With[{fonts = FE`Evaluate[FEPrivate`GetPopupList["MenuListFonts"]]},
DynamicModule[{kernedText, points},
points = Table[{x, 0}, {x, 1, Length[Characters[text]] + 10}];
Column[{
(* input *)
InputField[Dynamic[text], String],
(* main panel *)
Panel@LocatorPane[
Dynamic[
points, (points = ReplacePart[#, {{_, 2} -> 0}]) &],
Dynamic[kernedText =
Graphics[{MapIndexed[
Text[Style[#1, FontSize -> fontsize, FontFamily -> font],
points[[First[#2]]]] &, Characters[text]]},
PlotRangePadding -> 0,
PlotRange -> {{0, 1 + Length[Characters[text]]}, Automatic},
Background -> None, ImageSize -> 800]], Appearance -> None],
(* controls *)
Row[
{
PopupMenu[Dynamic[fontsize], Table[x, {x, 72, 416, 12}]],
PopupMenu[Dynamic[font], fonts],
Button["Export",
Export[FileNameJoin[{$HomeDirectory, "Desktop",
"KernedText.png"}], kernedText],
ImageSize -> {Full, Automatic}]
}]
}, Left]]]
</code></pre>
<p><img src="https://i.stack.imgur.com/5DKID.png" alt="final"></p>
<p>I realised that the locator dots didn't need to be showing - just click on the characters. The result can be compared to the unkerned version:</p>
<pre><code>Text[Style[text, 192, FontFamily -> "Palatino"]]
</code></pre>
<p><img src="https://i.stack.imgur.com/lKUuh.png" alt="unkerned"></p>
<p>For me, this area of Mathematica (Manipulate/Dynamic) is gradually becoming less confusing (but only gradually).</p>
| m_goldberg | 3,066 | <p>This is just an addendum to jVincent's answer. In order to constrain the letters/locators to a horizontal line, you need to use the second argument of <code>Dynamic</code>. The following modified version of jVincent's answer adds the needed constraint: </p>
<pre><code>With[{text = "Wolfram", fontsize = 96, font = "Georgia"},
DynamicModule[{chars, kernedText, points},
chars = Length[Characters[text]];
points = Table[{x, 0}, {x, 1, chars}];
Column[{Panel@
LocatorPane[
Dynamic[points, (points = ReplacePart[#, {{_, 2} -> 0}]) &],
Dynamic[kernedText =
Graphics[
{
MapIndexed[
Text[Style[#1, FontSize -> fontsize,
FontFamily -> font], points[[First[#2]]]]&,
Characters[text]]
},
PlotRange -> {{0, 1 + chars}, Automatic},
ImageSize -> 450]]],
Button["Export",
Export[FileNameJoin[
{
$HomeDirectory, "Desktop", "KernedText.png"
}],
kernedText],
ImageSize -> {Full, Automatic}]},
Right]]]
</code></pre>
<h3>Edit</h3>
<p>I've cleaned up my code and incorporated kguler's fix. The resulting kerned letters look pretty and export nicely. </p>
<p><img src="https://i.stack.imgur.com/MTIh7.png" alt="kerned text"></p>
|
674,448 | <p>Prove $F: \mathbb{R}\to\mathbb{R}$ where $F(x) = \int_a^x f(t)\, dt$ ($a<x$) is surjective. </p>
<p>$f$ is continuous and bounded below by $m>0$. Also $a$ belongs to $\mathbb{R}$ (reals).</p>
| John Hughes | 114,036 | <p>OK. The $a < x$ condition is wrong, and makes the theorem false, as others have pointed out. So let's get rid of it. </p>
<p>Let $u \in \mathbb R$ be nonnegative. Let $x = a + u/m$. Now estimate $F(x)$:
\begin{align}
F(x) &= \int_a^x f(t) ~dt \\
&\ge \int_a^x m ~dt \\
&= mx - ma \\
&= m(a + u/m) - ma \\
&= ma + u - ma \\
&= u
\end{align}</p>
<p>On the other hand, $F(a) = 0$. So $F$, on the interval $[a, x]$ goes from less than $u$ to more than $u$; since $F$ is continuous (why? it's differentiable, by the Fundamental Theorem, hence continuous.) the intermediate value theorem applies, and there's a value $c \in [a, x]$ with $F(c) = u$. Since $u$ was an arbitrary nonnegative number, $F$ is surjective onto the nonnegative reals. </p>
<p>A corresponding argument, with $x$ again equaling $a + u/m$, where $u/m$ is now negative, applies to negative values of $u$. Thus $F$ is surjective onto the reals. </p>
<p>It also happens to be injective, because it's a strictly increasing function. </p>
|
424,209 | <p>I am a Computer Science student. While going through some random maths topics I came across Chaos Theory. I wanted to know if there are any applications of it in CS.
I tried searching on the internet about this but ended up only with <a href="https://security.stackexchange.com/questions/31000/does-chaos-theory-have-any-practical-application-in-computer-security">this</a> and <a href="http://www.slideshare.net/konakid/computer-science-student-uses-chaos-theory-to-create-new-routes" rel="nofollow noreferrer">this</a>.
But are there any other domains of computer science where this theory could be applied ?</p>
<p>Please help!</p>
| Angela Pretorius | 15,624 | <ol>
<li><p>A small world network introduced with time delay generates chaos. Databases that are linked in a small world network are faster to extract information from. Most neural networks are small world networks.</p></li>
<li><p>Computer graphics? Generating realistic-looking animations of flames, flowing water etc. involves mimicing the chaotic processes that form them.</p></li>
</ol>
|
13,166 | <p>I taught IT in an engineering school during three years in <a href="https://en.wikipedia.org/wiki/Problem-based_learning" rel="nofollow noreferrer">problem based learning</a> (PBL) only. Now I teach maths to pupils between 10 and 15 years old who have a lot of educational difficulties.</p>
<p>I'm thinking to use PBL to increase their motivation and, by the way, increase their learning.</p>
<p>But, I have a question: is PBL adapted to pupils who have already know a lot of failures (in school and elsewhere)?</p>
| James S. | 1,798 | <p>PBL, especially if you have a lot of students that struggle with mathematics, is definitely something that you will have to design carefully; ordinary project based approaches can sometimes exacerbate already existing difficulties. At the same time, it is important not to lower the rigor or expectations for those who struggle, and so the standard recommendation that folks might offer of differentiated instruction, where you give instruction at different levels and with different emphases to different students, is probably ill-advised, particularly if you are hoping to have all students work together on projects.</p>
<p>Having reviewed the literature in this area, I think there is one study that really stands out as an example of what you would ideally be striving for. Bottge et. al (2007) demonstrated the effectiveness of a method called anchored instruction with students with learning disabilities. In anchored instruction, students are presented with a video case (or "anchor") that serves as a source of data for a project-based lesson. More basic mathematical skills are presented to the students in the context of the problem, on an as-needed basis, rather than being taught prior to engaging in the project.</p>
<p>You do mention that you worry about your students being on different levels and trying to keep them together working on the same material. You might consider using an instructional system known as Complex Instruction (Cohen and Lotan, 2014). In Complex Instruction, teachers carefully design both intellectually rigorous and accessible tasks for students to work on in groups and then carefully observe (and sometimes intervene) to make sure that all students are actively participating.</p>
<p>Bottge, B. A., Rueda, E., LaRoque, P. T., Serlin, R. C., & Kwon, J. (2007). Integrating Reform-Oriented Math Instruction in Special Education Settings. Learning Disabilities Research & Practice, 22(2), 96–109. <a href="https://doi.org/10.1111/j.1540-5826.2007.00234.x" rel="nofollow noreferrer">https://doi.org/10.1111/j.1540-5826.2007.00234.x</a></p>
<p>Cohen, E. G., & Lotan, R. A. (2014). Designing Groupwork: Strategies for the Heterogeneous Classroom Third Edition. Teachers College Press.</p>
|
3,525,814 | <p>One reasonably well-known property of the Thue-Morse sequence is that it can be used to provide solutions to the <a href="https://en.wikipedia.org/wiki/Prouhet%E2%80%93Tarry%E2%80%93Escott_problem" rel="nofollow noreferrer">Prouhet–Tarry–Escott problem</a> - for example, splitting the first eight nonnegative integers into evil and odious numbers, we get</p>
<p><span class="math-container">$0^0+3^0+5^0+6^0=1^0+2^0+4^0+7^0$</span></p>
<p><span class="math-container">$0^1+3^1+5^1+6^1=1^1+2^1+4^1+7^1$</span></p>
<p><span class="math-container">$0^2+3^2+5^2+6^2=1^2+2^2+4^2+7^2$</span></p>
<p>If we increase <span class="math-container">$8$</span> in this example to <span class="math-container">$2^m$</span>, we can get two sets of numbers whose <span class="math-container">$0$</span>th, <span class="math-container">$1st$</span>, ... <span class="math-container">$m$</span>th powers are all equal to one another. </p>
<p>A natural generalization is to ask whether we can obtain this same kind of pattern with more than two sets of integers at once. More formally, we ask for which <span class="math-container">$k,m>0$</span> it is possible to have <span class="math-container">$k$</span> distinct multisets of nonnegative integers whose sum of <span class="math-container">$n$</span>th powers are equal for each <span class="math-container">$n=0,1,\ldots,m$</span>. As the Thue-Morse sequence shows, this can be done for all <span class="math-container">$m$</span> with <span class="math-container">$k=2$</span>. </p>
<p>With <span class="math-container">$m=1$</span>, it's trivial to see that all <span class="math-container">$k$</span> will work, but the case <span class="math-container">$(k,m)=(3,2)$</span> is already nonobvious; the smallest solution of (13, 11, 0), (15, 8, 1), and (16, 5, 3) takes some work to locate. </p>
<p>With <span class="math-container">$m=2$</span>, the maximal <span class="math-container">$k$</span>-value I have found is <span class="math-container">$16$</span>, with the 3-tuples (412, 389, 0), (430, 369, 2), (440, 357, 4), (444, 352, 5), (464, 325, 12), (474, 310, 17), (485, 292, 24), (497, 270, 34), (500, 264, 37), (510, 242, 49), (517, 224, 60), (522, 209, 70), (529, 182, 90), (530, 177, 94), (532, 165, 104), and (534, 145, 122), all of which sum to <span class="math-container">$801$</span> and whose squares sum to <span class="math-container">$321065$</span>.</p>
<p>I also have a solution to <span class="math-container">$k=3, m=3$</span>: (22, 21, 4, 3), (24, 18, 7, 1), (25, 15, 10, 0).</p>
<p>As the only barrier to locating these solutions has been computational power thus far, I expect that solutions exist for all <span class="math-container">$k, m$</span>, though this seems like a very difficult proposition to show; does anyone have pointers to existing work on this question? I was unable to locate any such problem in, e.g., <a href="https://en.wikipedia.org/wiki/Sums_of_powers" rel="nofollow noreferrer">Wikipedia's list of problems relating to sums of powers</a>, or listed generalizations of the Prouhet–Tarry–Escott problem.</p>
<p>I would also be curious to see larger <span class="math-container">$(k,m)$</span> tuples that solutions are located for, as I'm not sure my current algorithms are particularly efficient at finding solutions. </p>
| Sam | 746,289 | <p>Mathematician, "Chen Shuwen" has givn solution </p>
<p>for, <span class="math-container">$m=4$</span> & is shown below.</p>
<p><span class="math-container">$m=1,2,3,4$</span></p>
<p><span class="math-container">$(401,521,641,881,911)^m=(431,461,701,821,941)^m$</span></p>
<p>The link to his web page's is given below:</p>
<pre><code> http://eslpower.org
</code></pre>
|
3,858,517 | <p>Is it possible to count exactly the number of binary strings of length <span class="math-container">$n$</span> that contain no two adjacent blocks of 1s of the same length? More precisely, if we represent the string as <span class="math-container">$0^{x_1}1^{y_1}0^{x_2}1^{y_2}\cdots 0^{x_{k-1}}1^{y_{k-1}}0^{x_k}$</span> where all <span class="math-container">$x_i,y_i \geq 1$</span> (except perhaps <span class="math-container">$x_1$</span> and <span class="math-container">$x_k$</span> which might be zero if the string starts or ends with a block of 1's), we should count a string as valid if <span class="math-container">$y_i\neq y_{i+1}$</span> for every <span class="math-container">$1\leq i \leq k-2$</span>.</p>
<p>Positive examples : 1101011 (block sizes are 2-1-2), 00011001011 (block sizes are 2-1-2), 1001100011101 (block sizes are 1-2-3-1)</p>
<p>Negative examples : 1100011 (block sizes are <strong>2-2</strong>), 0001010011 (block sizes are <strong>1-1</strong>-2), 1101011011 (block sizes are 2-1-<strong>2-2</strong>)</p>
<p>The sequence for the first <span class="math-container">$16$</span> integers <span class="math-container">$n$</span> is: 2, 4, 7, 13, 24, 45, 83, 154, 285, 528, 979, 1815, 3364, 6235, 11555, 21414. For <span class="math-container">$n=3$</span>, only the string 101 is invalid, whereas for <span class="math-container">$n=4$</span>, the invalid strings are 1010, 0101 and 1001.</p>
| RobPratt | 683,666 | <p>I confirm your results for <span class="math-container">$n \le 16$</span>. It might be useful to compute the values by conditioning on <span class="math-container">$k\in\{1,\dots,\lfloor(n+3)/2\rfloor\}$</span>:
<span class="math-container">\begin{matrix}
n\backslash k & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
0 & 1 \\
1 & 1 & 1 \\
2 & 1 & 3 \\
3 & 1 & 6 & 0 \\
4 & 1 & 10 & 2 \\
5 & 1 & 15 & 8 & 0 \\
6 & 1 & 21 & 22 & 1 \\
7 & 1 & 28 & 48 & 6 & 0 \\
8 & 1 & 36 & 92 & 25 & 0 \\
9 & 1 & 45 & 160 & 77 & 2 & 0 \\
10 & 1 & 55 & 260 & 196 & 16 & 0 \\
11 & 1 & 66 & 400 & 437 & 74 & 1 & 0 \\
12 & 1 & 78 & 590 & 883 & 254 & 9 & 0 \\
13 & 1 & 91 & 840 & 1652 & 726 & 54 & 0 & 0 \\
14 & 1 & 105 & 1162 & 2908 & 1818 & 239 & 2 & 0 \\
15 & 1 & 120 & 1568 & 4869 & 4116 & 857 & 24 & 0 & 0 \\
16 & 1 & 136 & 2072 & 7819 & 8602 & 2627 & 156 & 1 & 0 \\
\end{matrix}</span></p>
<p>Maybe try inclusion-exclusion together with stars-and-bars? For fixed <span class="math-container">$k$</span>, the first term of inclusion-exclusion is the number of nonnegative integer solutions to
<span class="math-container">$$\sum_{j=1}^k x_j + \sum_{j=1}^{k-1} y_j = n - (k-2) - (k-1) = n-2k+3,$$</span>
which is
<span class="math-container">$$\binom{(n-2k+3) + (2k-1)-1}{(2k-1)-1} = \binom{n+1}{2k-2}.$$</span>
For <span class="math-container">$k\in\{1,2\}$</span>, this formula is correct. For <span class="math-container">$k \ge 3$</span>, it is only an upper bound.</p>
<hr />
<p>An alternative approach is to condition on the tail <span class="math-container">$(y_{k-1},x_k)$</span>. Explicitly,
let state space
<span class="math-container">$$S_n = \left\{k \in \{1,\dots,\lfloor(n+3)/2\rfloor\}, y \in \{[k\not=1],\dots,n\}, x \in \{0,\dots,n-y-2k+5\}\right\}.$$</span>
For <span class="math-container">$(k,y,x) \in S_n$</span>, let
<span class="math-container">$f_n(k,y,x)$</span> be the number of such binary strings that end in <span class="math-container">$1^y 0^x$</span>. Then <span class="math-container">$f$</span> satisfies the recursion
<span class="math-container">$$f_n(k,y,x) = \begin{cases}
1 &\text{if $n = 0$} \\
[y = 0 \land x = n] &\text{if $k = 1$} \\
\sum\limits_{\substack(k-1,y_{k-2},x_{k-1}) \in S_{n-y-x}:\\ y_{k-2} \not= y \land ((y_{k-2} \ge 1 \land x_{k-1} \ge 1) \lor k=2)} f_{n-y-x}(k-1,y_{k-2},x_{k-1}) &\text{otherwise}
\end{cases}$$</span></p>
<p>The desired values are then <span class="math-container">$\sum\limits_{(k,y,x) \in S_n} f_n(k,y,x)$</span>.</p>
|
2,416,671 | <p>Here is the problem: Let $K$ be a compact subset of $ \mathbb{R}^{m} $ ($m>1$) with empty interior and such that $\mathbb{R}^{m}\setminus K $ has no bounded component. For $n=1,2,...$, we define
$$K_{n}=\lbrace x\in \mathbb{R^{m}}: distance (x,K)=1/n\rbrace.$$ Prove that for all $x\in K$, there is a sequence $(y_{n})$ in $K_{n}$ converging to $x$, as $n\rightarrow\infty$.</p>
<p>Here is what I have done:
Fix $x\in K$. If for all $r>0$ the ball of center $x$ and radius $r$, $B(x,r)$, encounters some $K_{n}$, then we are done. If not $B(x,r_{0} )\cap K_{n}=\emptyset$, for all positive integer and some $r_{0} >0$. This implies $B(x,r_{0} )$ is included to the complement of $K_{n}$ for all $n$. Intuitively this is impossible since the complements of the $K_{n}$'s encounter each other. How can finish the argument?</p>
| Daniel Fischer | 83,702 | <p>Since $K$ has empty interior, for every $r > 0$ there is an $y \in B(x,r/2) \setminus K$. Let $\delta = \operatorname{dist}(y,K)$, and choose $z \in K$ with $\lVert y-z\rVert = \delta$. For $t \in (0,1]$, let $p(t) = ty + (1-t)z$. Then $$\operatorname{dist}(p(t),K) = \lVert p(t) - z\rVert = t\delta$$ since $B(p(t),t\delta)\subset B(y,\delta)$ by the triangle inequality. Hence $B(x,r) \cap K_n \neq \varnothing$ for all $n > 1/\delta$.</p>
<p>Now choose a sequence $(r_k)$ of radii tending to $0$, and corresponding $\delta_k$ such that $r_{k+1} < \delta_k$ for all $k$. For $n \leqslant 1/\delta_1$, choose $y_n \in K_n$ arbitrary. For $1/\delta_k < n \leqslant 1/\delta_{k+1}$ choose $y_n \in K_n \cap B(x,r_k)$. Then $(y_n)$ is a sequence with $y_n \in K_n$ for every $n$, and $y_n \to x$.</p>
|
2,637,812 | <p>Here is dice game question about probability.</p>
<p>Play a game with $2$ die. What is the probability of getting a sum greater than $7$?</p>
<p>I know how the probability for this one is easy, $\cfrac{1+2+3+4+5}{36}=\cfrac 5{12}$.</p>
<p>I don't know how to solve the follow-up question:</p>
<p>Play a game with $200$ die. What is the probability of getting a sum greater than $700$?</p>
| user | 505,767 | <p>Note that by Taylor's series</p>
<ul>
<li>$\log(1+x^2)=x^2-\frac{x^4}{2}+o(x^4)$</li>
<li>$\sin x^2=x^2+o(x^2)$</li>
</ul>
<p>thus
$$\frac{x^2-\log(1+x^2)}{x^2\sin^2x}=\frac{x^2-x^2+\frac{x^4}{2}+o(x^4)}{x^4+o(x^4)}=\frac{\frac12+o(1)}{1+o(1)}\to \frac12$$</p>
|
207,243 | <p>I am using FindFit function in order to fit my data and get two parameters: c and m.</p>
<p>The function that I am using has the following form:</p>
<pre><code>function = (m/x*(x/c)^m)*Exp[-1*(x/c)^m];
</code></pre>
<p>The answer should be c = 64.68 and m = 2.47, but I am constantly getting the error message Overflow.</p>
<p>My data and core are here:</p>
<pre><code>data = {{172.345, 0.000710716}, {136.899, 0.00238143}, {108.742,
0.00822192}, {86.3772, 0.0117385}, {68.6119, 0.0123856}, {54.5003,
0.0147863}, {43.2912, 0.0148628}, {34.3874, 0.0131214}, {27.3149,
0.00879578}, {21.697, 0.00634049}, {17.2345, 0.00310614}, {13.6899,
0.00231347}, {10.8742, 0.000858576}, {8.63772, 0.000286827}};
function = (m/x*(x/c)^m)*Exp[-1*(x/c)^m];
params = {c, m};
FindFit[data, function, params, x, MaxIterations -> 10000]
fitedfunc = function /. %
plot1 = ListPlot[data, PlotStyle -> Black, PlotMarkers -> "*"];
plot2 = Plot[fitedfunc, {x, 0, 200}, PlotStyle -> {Red, Thick}];
Show[plot1, plot2, Frame -> True]
</code></pre>
| kglr | 125 | <p>Using <code>pts</code> from mikado's answer:</p>
<pre><code>DelaunayMesh[pts]["Faces"]
</code></pre>
<blockquote>
<p>{{3, 4, 1}, {1, 4, 5}, {1, 2, 3}, {1, 5, 6}} </p>
</blockquote>
<pre><code>Grid @ %
</code></pre>
<p><a href="https://i.stack.imgur.com/7Nk5E.png" rel="noreferrer"><img src="https://i.stack.imgur.com/7Nk5E.png" alt="enter image description here"></a></p>
|
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