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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4,245,475 | <p>Evaluate <span class="math-container">$ \, \displaystyle \int _{0}^{4} \frac{1}{(2x+8)\, \sqrt{x(x+8)}}\, dx. $</span>
<br /><br /><span class="math-container">$My\ work:-$</span><br />
by completing the square and substitution i.e. <span class="math-container">$\displaystyle \left(\begin{array}{rl}x+4 & = 4\sec (\theta )\\ dx & = 4\sec \theta \tan \theta \, d\theta \end{array}\right) \qquad$</span> <br /> <span class="math-container">$\Rightarrow \displaystyle \int \frac{4\sec \theta \tan \theta \, d\theta }{2(4\sec (\theta ))( |4\tan (\theta )|) }$</span> <br /> <br /> <span class="math-container">$\Rightarrow \displaystyle \frac{1}{8} \int \frac{\tan \theta \, d\theta }{|\tan (\theta )|}$</span> <br /> <br /> now because my limits are positive so <span class="math-container">$sec\ \theta \geq 0\ $</span> and <span class="math-container">$sec\ \theta\ $</span> is positive in <span class="math-container">$\ Ist\ $</span> and <span class="math-container">$\ IVth\ $</span> Quadrant. At this stage i have 2 options either i consider <span class="math-container">$\ Ist\ $</span> Quadrant and take postive <span class="math-container">$|tan\ \theta|\ =\ tan\ \theta\ $</span> or i consider <span class="math-container">$\ IVth\ $</span> Quadrant where <span class="math-container">$\ |tan\ \theta|\ = -tan\ \theta$</span> <br /> <br /> So when in 1st quadrant i.e. <span class="math-container">$\ |tan\ \theta|\ = tan\ \theta ,$</span> <span class="math-container">$\ 0\ \geq\ \theta\ \geq\ \pi/2\ $</span> i get</p>
<p><span class="math-container">$\Rightarrow \displaystyle \frac{1}{8} \theta +C$</span> <br /><br /> <span class="math-container">$\Rightarrow \displaystyle \frac{1}{8} \mathrm{arcsec}\left(\frac{x+4}{4}\right)+C$</span> <br /><br /> <span class="math-container">$\Rightarrow \displaystyle \left. \frac{1}{8} \mathrm{arcsec}\left(\frac{x+4}{4}\right)\, \right|_{x=0}^{x=4}$</span> <br /><br /> <span class="math-container">$\Rightarrow \displaystyle \frac{1}{8} \left(\mathrm{arcsec}(2)-\mathrm{arcsec}(1)\right)$</span></p>
<p><br /> Now if i consider 4th quadrant i.e. <span class="math-container">$\ |tan\ \theta|\ = -tan\ \theta ,$</span> <span class="math-container">$\ 3 \pi /2 \ \geq\ \theta\ \geq\ 2\pi\ $</span> i get<br /><br /> <span class="math-container">$\Rightarrow \displaystyle -\frac{1}{8} \theta +C$</span> <br /><br /> <span class="math-container">$\Rightarrow \displaystyle -\frac{1}{8} \mathrm{arcsec}\left(\frac{x+4}{4}\right)+C$</span> <br /><br /> <span class="math-container">$\Rightarrow \displaystyle \left. -\frac{1}{8} \mathrm{arcsec}\left(\frac{x+4}{4}\right)\, \right|_{x=0}^{x=4}$</span> <br /><br /> <span class="math-container">$\Rightarrow \displaystyle -\frac{1}{8} \left(\mathrm{arcsec}(2)-\mathrm{arcsec}(1)\right)$</span> <br /><br /> So i am getting positive value in 1st case whereas in 4th quadrant my answer is Negative. Why is so ? am i making some mistake when considering 4th quadrant ? or both are acceptable answer ? also in MIT lecture they said whatever acceptable quadrant you choose you'll get the same answer. so why i m getting Negative answer ?</p>
| egreg | 62,967 | <p>Let <span class="math-container">$x\in V$</span>, <span class="math-container">$x\ne0$</span>. Then <span class="math-container">$\{x\}$</span> is a linearly independent set, so there exists a basis <span class="math-container">$\mathscr{B}$</span> of <span class="math-container">$V$</span> such that <span class="math-container">$x\in\mathscr{B}$</span> (see final note). Then, by surjectivity, <span class="math-container">$T(\mathscr{B})$</span> spans <span class="math-container">$W$</span>.</p>
<p>If <span class="math-container">$T(x)=0$</span>, then also <span class="math-container">$T(\mathscr{B}\setminus\{x\})$</span> spans <span class="math-container">$V$</span>, so <span class="math-container">$\mathscr{B}\setminus\{x\}$</span> spans <span class="math-container">$V$</span>: contradiction.</p>
<p>Therefore <span class="math-container">$T(x)\ne0$</span>.</p>
<p><strong>Final note.</strong> I don't think you can avoid existence of bases, hence the axiom of choice, if you want the result for arbitrary vector spaces; no problem for finite dimensional spaces.</p>
|
1,817,542 | <p><strong>Problem:</strong> Let $(X, Y)$ be uniformly distributed on the unit disk $\{ (x,y) : x^2 + y^2 \le 1\}$. Let $R = \sqrt{X^2 + Y^2}$. Find the CDF and PDF of $R$.</p>
<p><strong>Attempted Solution:</strong> First note that $r \in R = \sqrt{X^2 + Y^2}$ represents a point on $\mathbb{R}^2$ with radius $r$ about the origin. Since only points with radius $1$ had probability greater than $0$ on $(X, Y)$ we have that</p>
<p>$$
F_R(r) = \begin{cases} 0 & r < 1 \\ 1 & r \ge 1 \end{cases}
$$</p>
<p>so that</p>
<p>$$
f_R(r) = F_R'(r) = \begin{cases} 0 & r < 1 \\ \text{undefined} & r = 0 \\ 0 & r > 1 \end{cases}
$$</p>
<p>since </p>
<ol>
<li>$F'_R$ is discontinuous at $r = 0$.</li>
<li>$F_R$ is constant everywhere else (so that the derivative of a constant is $0$, and hence $F_R'$ is $0$).</li>
</ol>
<p><strong>Question:</strong> Is my reasoning correct here?</p>
| Community | -1 | <p>As the random variable is uniformly distributed, the probability of $R$ not exceeding a given $r$ is proportional to the enclosed area.</p>
<p>$$P(R\le r)\propto r^2.$$</p>
<p>As the probability is exactly $1$ for the radius $r=1$, the constant of proportionality is $1$.</p>
<p>For $r\le1$,</p>
<p>$$F_R(r)=r^2,\\f_R(r)=2r.$$</p>
|
1,817,542 | <p><strong>Problem:</strong> Let $(X, Y)$ be uniformly distributed on the unit disk $\{ (x,y) : x^2 + y^2 \le 1\}$. Let $R = \sqrt{X^2 + Y^2}$. Find the CDF and PDF of $R$.</p>
<p><strong>Attempted Solution:</strong> First note that $r \in R = \sqrt{X^2 + Y^2}$ represents a point on $\mathbb{R}^2$ with radius $r$ about the origin. Since only points with radius $1$ had probability greater than $0$ on $(X, Y)$ we have that</p>
<p>$$
F_R(r) = \begin{cases} 0 & r < 1 \\ 1 & r \ge 1 \end{cases}
$$</p>
<p>so that</p>
<p>$$
f_R(r) = F_R'(r) = \begin{cases} 0 & r < 1 \\ \text{undefined} & r = 0 \\ 0 & r > 1 \end{cases}
$$</p>
<p>since </p>
<ol>
<li>$F'_R$ is discontinuous at $r = 0$.</li>
<li>$F_R$ is constant everywhere else (so that the derivative of a constant is $0$, and hence $F_R'$ is $0$).</li>
</ol>
<p><strong>Question:</strong> Is my reasoning correct here?</p>
| Em. | 290,196 | <p>One of the reasons I love probability is that there are usual a million ways to do a problem.</p>
<p>I provide a heuristic alternative.</p>
<p>We recognize that we are interested in the event $R\in dr$. In words, this means we want the radius to fall in an infinitesimal annulus, with infinitesimal width $dr$, and area $2\pi rdr$. Since we were told that the points are uniformly distributed, then density is flat over the region of interest, call this $h$. We know the entire volume must be $1$, hence
$$h\cdot \pi 1^2 = 1\implies h = \frac{1}{\pi}.$$
This tells us that
$$P(R\in dr) = \frac{1}{\pi}\cdot 2\pi r\,dr = 2r\,dr.$$
This implies that the density is $f_R(r) = 2r$ over the region of interest.</p>
<p>Thus, for $0\leq r\leq 1$,
$$F_R(r) = \int_0^r f(t)\,dt = r^2.$$</p>
|
1,302,990 | <p>I want to ask basic question. In our mathematics classes ,while teaching the Fourier series and transform topic,the professor says that when the signal is periodic ,we should use Fourier series and Fourier transform for aperiodic signals.</p>
<p>My question is can't we use Fourier transform formula in case of periodic signal ? Also, Is Fourier series used always for periodic signals and Fourier transform for aperiodic signals only ?</p>
| Kishan | 247,422 | <p>Fourier Series (FS) exists only for periodic signals.</p>
<p>Fourier Transform (FT) is derived from FS, i.e. FT is the envelope of the FS. Thus as the frequency domain became more finer, time domain enlarges making it a aperiodic signal.</p>
<p>So when FT is applied on a periodic signal, the result is just, T*X(k) where T is the time period of the signal and X(k) is the FS of the signal.</p>
<p>Fourier is just awesome!</p>
|
306,011 | <p>Does anyone have a proof for $$\int_0^{\infty}\frac{\sin(x^2)}{x^2}\,dx=\sqrt{\frac{\pi}{2}}.$$
I tried to get it from contour integrating $$\frac{e^{iz^2}-1}{z^2},$$ but failed.
Thanks.</p>
| Kaster | 49,333 | <p>I'll extend <strong>@sos440</strong>'s answer a bit.
$$
I = \int_o^\infty \frac {\sin x^2}{x^2}dx = -\left. \frac{\sin x^2}x \right|_0^\infty + 2\int_0^\infty \cos x^2 dx \\
\left .\frac{\sin x^2}x \right |_0^\infty = \lim_{x \rightarrow \infty} \frac {\sin x^2}x - \lim_{x \rightarrow 0} \frac {\sin x^2}x = 0-\lim \frac 1x \left ( x^2 + O(x^4)\right ) = -\lim_{x \rightarrow 0}\left( x + O(x^3)\right ) = 0 \\
$$
To take Fresnel's integral $C(x) = \int_0^x \cos x^2 dx$ at $\infty$ just take a contour like <a href="http://en.wikipedia.org/wiki/Fresnel_integral#Evaluation">here</a>, and you're done.</p>
<p>PS: $I = 2\lim_{x \rightarrow \infty} C(x) = 2 \sqrt{\frac \pi 8} = \sqrt{\frac \pi 2}$</p>
|
3,754,548 | <p>Suppose there is a strictly convex continuous function <span class="math-container">$f$</span>: <span class="math-container">$R^n$</span> <span class="math-container">$\rightarrow$</span> <span class="math-container">$R$</span>.</p>
<p>Is the supremum of <span class="math-container">$f$</span> always infinity? How can we prove it?</p>
<p>I am trying to come up with proof. If <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are two points in <span class="math-container">$R^n$</span>, strictly convex implies <span class="math-container">$f(\alpha x_1 + (1-\alpha) x_2) $</span> < <span class="math-container">$\alpha f(x_1) + (1-\alpha)f(x_2)$</span> . Suppose <span class="math-container">$f$</span> is bounded.</p>
<p>Case 1:
The bound is attained at a point, say <span class="math-container">$x_0$</span>. Then for some <span class="math-container">$\alpha$</span>, some <span class="math-container">$x_1$</span> and <span class="math-container">$x_2$</span> s.t. <span class="math-container">$ (\alpha x_1 + (1-\alpha) x_2) = x_0$</span>:<br />
<span class="math-container">$f(x_0)$</span>< <span class="math-container">$\alpha f(x_1) + (1-\alpha)f(x_2)$</span><br />
Therefore a contradiction.<br />
Case 2: The bound is not attained. Since the function is strictly convex, we know <span class="math-container">$f(x)$</span> approaches this bound as <span class="math-container">$x$</span> approaches <span class="math-container">$ \infty $</span><br />
I don't know how to proceed after this step. Where can I find a contradiction in this case?</p>
| Paul | 138,918 | <p>The tabular method will work, in just the same way as the traditional by parts method works. The last row in the table is the integral still to be done. You will find that the third row in your table (the integral still to do) is a multiple of the original integral. If you call the original integral <span class="math-container">$I$</span> then you obtain <span class="math-container">$$I= e^{ax}( -\frac{1}{b}cos(bx))+ ae^{ax}( -\frac{1}{b^2}sin(bx))-\frac{a^2}{b^2}I$$</span> and solve for <span class="math-container">$I$</span>.</p>
|
1,252,414 | <p>In rectangle ABCD below, points F and G lie on segment AB such that AF = FG = GB and E is the midpoint of segment DC. Also, segment AC intersects segment EF at H and segment EG at J. The area of rectangle ABCD is 70. Find the area of triangle AHF.</p>
<p>(Note: This question has been slightly changed from the original AMC 12 problem.)</p>
<p><img src="https://i.stack.imgur.com/neZJw.png" alt="Diagram"></p>
<p><strong>Work</strong></p>
<ul>
<li>Triangle AHF is similar to triangle CHE with ratio 2:3</li>
<li>Triangle AJG is similar to triangle CJE with ratio 4:3</li>
</ul>
| the_candyman | 51,370 | <p>Let's say that $AB=3x$, $CD=2y$ and $BC=h$. Then, $$3xh = 2yh = 70,$$and
$$x = \frac{2y}{3}.$$
The triangles $AFH$ and $HEC$ are similar and have area:</p>
<p>$$S_{AFH} = \frac{x h_1}{2}, S_{HEC} = \frac{y h_2}{2}$$</p>
<p>where $h_1$ and $h_2$ are the heights of the 2 triangles, with $h_1+h_2 = h$ and $h_1 = \frac{2h_2}{3}$.
Then:
$$h_1 = \frac{2h}{5}, h_2 = \frac{3h}{5},$$
and since $xh = \frac{70}{3}$, then:</p>
<p>$$S_{AFH} = \frac{x h}{5} = \frac{70}{15} = \frac{14}{3}$$</p>
|
1,698,039 | <p>Alright, so let's say I have $$\frac{x^{-6}}{-x^{-4}}$$ The answer is $\dfrac{1}{x^2}$, but why isn't it $\dfrac{1}{-x^2}$?</p>
| Tae Hyung Kim | 94,401 | <p>Let's think about what negative exponents really mean. Consider the following list
$$2^3 = 8$$
$$2^2 = 4$$
$$2^1 = 2$$
Notice that every time we decrease an exponent, we divide by 2. Continuing this list, we have that
$$2^3 = 8$$
$$2^2 = 4$$
$$2^1 = 2$$
$$2^0 = 1$$
$$2^{-1} = 1/2$$
$$2^{-2} = 1/4$$
Rather interesting, we see a pattern: $2^{-n} = \frac{1}{2^n}$. Therefore there is no reason for a negative sign on the quantity itself, since we're successively dividing by <strong>positive</strong> numbers.</p>
|
655,378 | <p>I'm new to discrete mathematics and was wondering whether the following functions are one to one:</p>
<p>$$f(x) = x - 1$$
$$f(x) = x^2 + 1$$</p>
<p>The reason I stand by this is because for the first equation:</p>
<p>$$x - 1 = y - 1\\x = y$$</p>
<p>and for the second one:</p>
<p>$$x^2 +1 = y^2 +1\\x = y$$</p>
| RE60K | 67,609 | <p>$$\frac{dy}{dx} = \frac{1}{x\cos(y) +\sin(2y)}=\frac{1}{x\cos(y) +2\sin(y)\cos y}\\
\frac{d(\sin y)}{dx}=\frac{1}{x+2\sin y}$$
Then $$\frac{dx}{d(\sin y)} -x=2\sin y\\xe^{\int -d(\sin y)}=\int 2\sin ye^{\int -d(\sin y)}d(\sin y)\\xe^{-\sin y}=2\int \sin ye^{-\sin y}d(\sin y)=-2e^{-\sin y}(\sin y+1)+c\\x+2\sin y+2=ce^{\sin y}$$</p>
|
2,814,703 | <p>I am reading <a href="https://en.wikipedia.org/wiki/Lower_limit_topology" rel="nofollow noreferrer">lower limit topology</a> on Wikipedia, which states that the lower limit topology </p>
<blockquote>
<p>[...] is the topology generated by the basis of all half-open intervals $[a,b)$, where a and b are real numbers. [...] The lower limit topology is finer (has more open sets) than the standard topology on the real numbers (which is generated by the open intervals). The reason is that every open interval can be written as a (countably infinite) union of half-open intervals.</p>
</blockquote>
<p>I cannot see how to write $(a,b)$ as a countably infinite union of half-open intervals.</p>
| qualcuno | 362,866 | <p>One can write </p>
<p>$$
(a,b) = \bigcup_{n\geq 1}\left[a+\frac{b-a}{2n},b\right)
$$</p>
<p>One inclusion is straighforward. For the non trivial one, pick $t \in (a,b)$, and $n$ large enough such that $t > a +\frac{b-a}{2n}$, so that $t \in \left[a+ \frac{b-a}{2n},b\right)$.</p>
|
3,710,018 | <p>Problem:
Find prime solutions to the equation
<strong><span class="math-container">$p^2+1=q^2+r^2$</span></strong></p>
<p>I welcome you to post your own solutions as well</p>
<p>I have found a <em>strange solution</em> which I can't understand why it works(or what's the math behind it.) Here it is through examples</p>
<p>Put <strong><span class="math-container">$r=17$</span></strong>(prime)
Now <span class="math-container">$17^2-1=16\times 18=288=2 \times 144$</span> (a <strong>particular factorization</strong>)</p>
<p><span class="math-container">$\frac {2+144}{2}=73$</span></p>
<p><span class="math-container">$\frac {144-2}{2}=71$</span></p>
<p>Solution pair <span class="math-container">$(p,q,r)=(73,71,17)$</span></p>
<p>Put <strong><span class="math-container">$r=23$</span></strong>,
<span class="math-container">$23^2-1=22\times 24=8\times 66$</span></p>
<p><span class="math-container">$\frac {8+66}{2}=37$</span></p>
<p><span class="math-container">$\frac {66-8}{2}=29$</span></p>
<p>Solution pair <span class="math-container">$(37,29,23)$</span></p>
<p>It works for each prime except for <strong><span class="math-container">$2,3,5$</span></strong> Which generate
<span class="math-container">$(2,2,1),(3,3,1),5,5,1)$</span></p>
<p>Please explain me how it's working </p>
| Edward Porcella | 403,946 | <p>This seems to hold much more broadly than just for primes.</p>
<p>Let<span class="math-container">$$r^2-1=ab$$</span>where <span class="math-container">$b>a$</span>, and let <span class="math-container">$p=\frac{b+a}{2}$</span> and <span class="math-container">$q=\frac{b-a}{2}$</span>.</p>
<p>Then the equation <span class="math-container">$p^2+1=q^2+r^2$</span> becomes<span class="math-container">$$\left(\frac{b+a}{2}\right)^2+1=\left(\frac{b-a}{2}\right)^2+ab+1$$</span>and we have<span class="math-container">$$\frac{b^2+2ab+a^2}{4}+1=\frac{b^2-2ab+a^2+4ab}{4}+1=\frac{b^2+2ab+a^2}{4}+1$$</span></p>
<p><strong>Addendum:</strong> If <span class="math-container">$r$</span> is an odd prime, then <span class="math-container">$p$</span> and <span class="math-container">$q$</span>, that is <span class="math-container">$\frac{b+a}{2}$</span> and <span class="math-container">$\frac{b-a}{2}$</span>, can be prime only if both are odd, i.e. only if <span class="math-container">$b\equiv 2\pmod 4$</span> and <span class="math-container">$a\equiv 0\pmod 4$</span>, or <em>vice-versa</em>.</p>
<p>This restricts the number of divisor pairs <span class="math-container">$a, b$</span> that need to be considered in determining whether <span class="math-container">$r^2-1=ab$</span> for some odd primes <span class="math-container">$p$</span> and <span class="math-container">$q$</span>.</p>
<p>To illustrate, let <span class="math-container">$r=47$</span>. Then<span class="math-container">$$r^2-1=ab=2208=2^5\cdot3\cdot23$$</span>Thus the possible <span class="math-container">$ab$</span> are <span class="math-container">$2\cdot 1104$</span>, <span class="math-container">$6\cdot 368$</span>, and <span class="math-container">$2^4\cdot 138$</span>.</p>
<p>For <span class="math-container">$b=1104$</span>, <span class="math-container">$a=2$</span>, <span class="math-container">$p=\frac{b+a}{2}=553=7\cdot79$</span>, not prime.</p>
<p>For <span class="math-container">$b=368$</span>, <span class="math-container">$a=6$</span>, <span class="math-container">$p=\frac{b+a}{2}=187=11\cdot17$</span>, not prime.</p>
<p>For <span class="math-container">$b=138$</span>, <span class="math-container">$a=2^4$</span>, <span class="math-container">$p=\frac{b+a}{2}=77=7\cdot11$</span>, not prime.</p>
<p>It seems <span class="math-container">$47$</span> is the least prime value of <span class="math-container">$r>3$</span> for which<span class="math-container">$$p^2+1=q^2+r^2$$</span>is true for no primes <span class="math-container">$p$</span>, <span class="math-container">$q$</span>.</p>
<p>As @Gerry Myerson shows, <span class="math-container">$r=193$</span> is another instance. I've yet to find other primes between these two, but have not looked much beyond <span class="math-container">$r=97$</span>.</p>
|
3,808,575 | <p>Assuming I have the statement ∀x(∀y¬Q(x,y)∨P(x)), can I pull the universal quantifier ∀y out of the parenthesis? Meaning, is this statement equivalent to ∀x∀y(¬Q(x,y)∨P(x)) ?</p>
<p>An approach I tried so far:</p>
<ol>
<li>∀x((∃y Q(x,y) ) => P(x)). (original eq.)</li>
<li>∀x((∀y¬Q(x,y))∨P(x)) (De Morgan's application)</li>
<li>∀x∀y(¬Q(x,y)∨P(x)). (Working off the assumption that taking out the ∀y is a valid operation).</li>
<li>∀x∀y (Q(x,y) => P(x)) (Going backwards from the ¬P v Q definition of implication)</li>
</ol>
<p>Statement 4 does not seem to be equivalent to statement 1, which suggests that pulling out the universal quantifier is not acceptable. I would greatly appreciate any confirmation of whether this is the case, and if so, what governs when quantifiers can be brought to the outside of the parenthesis.</p>
| Graham Kemp | 135,106 | <p>The original expression: <span class="math-container">$\forall x~((\exists y~Q(x,y))\to P(x))$</span> says "For any <span class="math-container">$x$</span> it holds that if some <span class="math-container">$y$</span> satisfies <span class="math-container">$Q(x,y)$</span>, then <span class="math-container">$P(x)$</span> is satisfied."</p>
<p>Now either consequent is true for all <span class="math-container">$x$</span> or, <em>whenever it is false</em>, the antecedent is also false (ie for <em>that</em> <span class="math-container">$x$</span> <em>no</em> <span class="math-container">$y$</span> can satisfy <span class="math-container">$Q(x,y)$</span>). Thus the expression equates to: <span class="math-container">$\forall x~(\neg P(x)\to\forall y~\neg Q(x,y))$</span></p>
<hr />
<p>The final expression: <span class="math-container">$\forall x~\forall y~(Q(x,y)\to P(x))$</span> says: "For any <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, it holds that if <span class="math-container">$Q(x,y)$</span> then <span class="math-container">$P(x)$</span>."</p>
<p>Now either consequent is true for all <span class="math-container">$x$</span> or, <em>whenever it is false</em>, the antecedent is also false; moreover false for all <span class="math-container">$y$</span> when <span class="math-container">$P(x)$</span> is false for some <span class="math-container">$x$</span>. Thus the expression equates to: <span class="math-container">$\forall x~\forall y~(\neg P(x)\to \neg Q(x,y))$</span></p>
<hr />
<p>Therefore the original and final expressions are equivalent.</p>
<hr />
|
3,682,987 | <blockquote>
<p>Let <span class="math-container">$a$</span>, <span class="math-container">$b$</span>, and <span class="math-container">$c$</span> be positive real numbers. What is the smallest possible value of <span class="math-container">$(a+b+c)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)$</span>?</p>
</blockquote>
<hr>
<p>I don't know how to approach this problem, though I think it might use the AM-GM inequality. Can someone please help? </p>
| Calvin Lin | 54,563 | <p><strong>Hint:</strong> Do you know why <span class="math-container">$ ( a + b+ c) ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} ) \geq 9 $</span>? </p>
<blockquote class="spoiler">
<p> Arithmetic Mean - Harmonic Mean.</p>
</blockquote>
<p>Hence, conclude that <span class="math-container">$ 2 ( a + b + c) ( \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} ) \geq 9 $</span></p>
<p>When does equality occur? </p>
|
3,684,917 | <p>Let <span class="math-container">$C_{1}$</span> and <span class="math-container">$C_{2}$</span> be polytopes in <span class="math-container">$\mathbb{R}^{n}$</span> such that
<span class="math-container">$C_{1}=conv\left( V\right) $</span> with <span class="math-container">$V$</span> being a set of vertices. If
<span class="math-container">$V\subseteq C_{2}$</span>, my question is <span class="math-container">$C_{1}\subseteq C_{2}$</span>?</p>
| Jotadiolyne Dicci | 780,918 | <p>With the help of the answers you sent, I was able to find a form that seems rather simple to me. Then I'm not 100% sure it works all the time.</p>
<p>Let <span class="math-container">$S$</span> be the number of integers congruent to <span class="math-container">${n}\pmod p$</span> in the interval <span class="math-container">$a$</span> inclusive and <span class="math-container">$b$</span> inclusive </p>
<p><span class="math-container">$T=n-a+p\lfloor\frac{b-n}{p}\rfloor$</span></p>
<p><span class="math-container">$S =\lfloor\frac{T}{p}\rfloor+1$</span></p>
|
1,419,897 | <blockquote>
<p><strong>Theorem:</strong> Let $A$ be a bounded infinite subset of $\mathbb{R}^l$. Then
it has a limit point.</p>
</blockquote>
<p>So this is the Euclidean version of the Bolzano-Weierstrass theorem, the thing is that I was trying to prove it by induction, but it doesn't help because in the case $l=1$ we constructed a sequence of intervals, such that each iteration has infinite number of points of the set, but what is the anlogy in $\mathbb{R}^l$?, Can someone help me to prove this please?</p>
<p>Thanks a lot in advance :)</p>
| Kitegi | 120,267 | <p>Proving it for $l=1$ is enough to get the rest.</p>
<p>Let $X_n = (x_{1,n},\dots,x_{l,n})$ be an injective infinite sequence of points in the set $A$.</p>
<p>$x_{1,n}$ is bounded</p>
<p>So there is a subsequence s.t $x_{1,\phi_1(n)}$ converges</p>
<p>Now consider $X_{\phi_1(n)} = (x_{1,\phi_1(n)},\dots,x_{l,\phi_1(n)})$</p>
<p>$x_{2,\phi_1(n)}$ is bounded</p>
<p>So there is a subsequence s.t $x_{2,\phi_1\circ\phi_2(n)}$ converges</p>
<p>(Note that $x_{1,\phi_1\circ\phi_2(n)}$ also converges)</p>
<p>We repeat that until we find a subsequence that makes all the coordinates converge.</p>
<p>$(x_{1,\phi_1\circ\phi_2\dots\circ\phi_l(n)},\dots,x_{l,\phi_1\circ\phi_2\dots\circ\phi_l(n)})$ all converge, so $X_{\phi_1\circ\phi_2\dots\circ\phi_l(n)}$ also converges and its limit is a limit point of $A$</p>
|
909,228 | <p>I'm trying to find a closed form for the following sum
$$\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n},$$
where $H_n=\displaystyle\sum_{k=1}^n\frac{1}{k}$ is a harmonic number.</p>
<p>Could you help me with it?</p>
| Ali Shadhar | 432,085 | <p><strong>Alternative approach:</strong></p>
<p>first we start with proving the following equality that appeared as Problem 11921 in The American Mathematical Monthly 2016 proposed by <strong>Cornel Ioan Valean</strong>:
<span class="math-container">\begin{equation*}
S=\ln^22\sum_{n=1}^{\infty}\frac{H_n}{(n+1) 2^{n+1}}+\ln2\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^2 2^n}+\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^3 2^n}=\frac14\ln^42+\frac14\zeta(4)
\end{equation*}</span>
<strong><em>Proof:</em></strong> lets start with the following integral <span class="math-container">$ I=\displaystyle \int_{1/2}^{1} \frac{\ln(1-x)\ln^2x}{1-x}\,dx $</span></p>
<p>By using </p>
<p><span class="math-container">$$\frac{\ln(1-x)}{1-x}=-\displaystyle \sum_{n=1}^{\infty}H_n x^n$$</span> </p>
<p>we can write</p>
<p><span class="math-container">$$I=-\sum_{n=1}^{\infty}H_n\int_{1/2}^{1}x^n \ln^2x\,dx$$</span></p>
<p><span class="math-container">$$=-\sum_{n=1}^{\infty}H_n\left( -\frac{\ln^22}{(n+1)2^{n+1}}-\frac{\ln2}{(n+1)2^{n+1}}-\frac{1}{(n+1)^32^n}+\frac{2}{(n+1)^3}\right)$$</span></p>
<p><span class="math-container">$$=S-2\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^3} \tag{1}$$</span></p>
<p>On the other hand, upon integrating by parts we obtain
<span class="math-container">\begin{equation*}
I=\frac12\ln^42+\int_{1/2}^1 \frac{\ln^2x\ln(1-x)}{x}\,dx\overset{x\mapsto 1-x}{=}\frac12\ln^42+\int_0^{1/2}\frac{\ln^2x\ln(1-x)}{1-x}\,dx
\end{equation*}</span>
Adding the integral <span class="math-container">$I=\int_{1/2}^{1}\frac{\ln^2x\ln(1-x)}{1-x}\,dx\ $</span> to both sides</p>
<p><span class="math-container">$$2I=\frac12\ln^42+\int_0^1 \frac{\ln^2x\ln(1-x)}{1-x}\,dx=\frac12\ln^42-\sum_{n=1}^{\infty}H_n\int_{0}^{1}x^n\ln^2x\,dx$$</span>
<span class="math-container">$$=\frac12\ln^42-2\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^3}\Longrightarrow I=\frac14\ln^42-\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^3} \tag{2}$$</span></p>
<p>combining <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span> yields</p>
<p><span class="math-container">$$S=\frac14\ln^42+\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^3}=\frac14\ln^42-\zeta(4)+\sum_{n=1}^{\infty}\frac{H_n}{n^3}$$</span></p>
<p>subbing <span class="math-container">$\sum_{n=1}^{\infty}\frac{H_n}{n^3}=\frac54\zeta(4)$</span> completes the proof.</p>
<hr>
<p>Using the proved equality:
<span class="math-container">\begin{align*}
\frac14\ln^42+\frac14\zeta(4)&=\ln^22\sum_{n=1}^{\infty}\frac{H_n}{(n+1) 2^{n+1}}+\ln2\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^2 2^n}+\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^3 2^n}\\
&=\ln^22\sum_{n=1}^{\infty}\frac{H_{n-1}}{n 2^n}+2\ln2\sum_{n=1}^{\infty}\frac {H_{n-1}}{n^2 2^n}+2\sum_{n=1}^{\infty}\frac{H_{n-1}}{n^3 2^n}\\
&=\ln^22\sum_{n=1}^{\infty}\frac{H_n}{n 2^n}+2\ln2\sum_{n=1}^{\infty}\frac {H_n}{n^2 2^n} +2\sum_{n=1}^{\infty}\frac{H_n}{n^3 2^n}-\ln^22\sum_{n=1}^{\infty}\frac{1}{2^n n^2}\\
&\quad -2\ln2\sum_{n=1}^{\infty}\frac{1}{ n^32^n}-2\sum_{n=1}^{\infty}\frac{1}{n^42^n}
\end{align*}</span>
rearrange the terms to get</p>
<blockquote>
<p><span class="math-container">$$\sum_{n=1}^{\infty}\frac{H_n}{n^3 2^n}=-\ln2\sum_{n=1}^{\infty}\frac{H_n}{n^2 2^n}-\frac12\ln^22\sum_{n=1}^{\infty}\frac{H_n}{n 2^n}+\operatorname{Li_4}\left( \frac12\right)+\ln2\operatorname{Li_3}\left( \frac12\right)\\+\frac12\ln^22\operatorname{Li_2}\left( \frac12\right)+\frac18\zeta(4)+\frac18\ln^42$$</span></p>
</blockquote>
<p>plugging the values of the first and second sum proved <a href="https://math.stackexchange.com/questions/604316/infinite-series-sum-n-1-infty-frach-nn22n/3196287#3196287">here</a> and <a href="https://math.stackexchange.com/questions/2333713/elementary-way-to-calculate-the-series-sum-limits-n-1-infty-frach-nn2/3196314#3196314">here</a> respectively, along with the values of <span class="math-container">$\displaystyle\operatorname{Li_3}\left(\frac12\right)$</span> and <span class="math-container">$\displaystyle\operatorname{Li_2}\left(\frac12\right)$</span> we obtain
<span class="math-container">\begin{align}
\sum_{n=1}^\infty \frac{H_n}{2^nn^3}=\color{blue}{\operatorname{Li}_4\left(\frac12\right)+\frac18\zeta(4)-\frac18\ln2\zeta(3)+\frac1{24}\ln^42},
\end{align}</span></p>
|
658,761 | <p>Sorry if this question is too basic.</p>
<p>We can suppose that we have a matrix that reduces to the identity matrix in reduced row-echelon form. My question is fairly simple: Can we multiply one or more columns by a constant and still be able to reduce the matrix to the identity?</p>
<p>I'd like a proof of this, but for some reason the proof eludes me.</p>
<p><strong>My ideas for a proof</strong>
I am thinking that we can start by examining the determinant of the matrix. Since we are multiplying a column by a constant, the determinant is simply multiplied by a constant. Then we can use Cramer's rule to finish off the proof.</p>
| heropup | 118,193 | <p>In <em>Mathematica</em> (version 6 or newer)</p>
<pre><code>Plot3D[x/(1 - y), {x, -1, 1}, {y, -1, 1},
RegionFunction -> Function[{x, y}, x^2 + y^2 < 1]]
</code></pre>
|
658,761 | <p>Sorry if this question is too basic.</p>
<p>We can suppose that we have a matrix that reduces to the identity matrix in reduced row-echelon form. My question is fairly simple: Can we multiply one or more columns by a constant and still be able to reduce the matrix to the identity?</p>
<p>I'd like a proof of this, but for some reason the proof eludes me.</p>
<p><strong>My ideas for a proof</strong>
I am thinking that we can start by examining the determinant of the matrix. Since we are multiplying a column by a constant, the determinant is simply multiplied by a constant. Then we can use Cramer's rule to finish off the proof.</p>
| Mikasa | 8,581 | <p>In Maple, you can use the following codes:</p>
<pre><code>[> with(plots):
plot3d(x/(1-y), x = -1 .. 1, y = -sqrt(1-x^2) .. sqrt(1-x^2), axes = boxed, filled = true,numpoints = 1000,color=green);
</code></pre>
<p><img src="https://i.stack.imgur.com/ysKzx.png" alt="enter image description here"></p>
|
2,030,116 | <p>How can i prove that $\sqrt[12]{2}$ is irrational number? </p>
<p>I'm trying: </p>
<p>$$\sqrt[12]{2} = \frac{p}{q}$$ where $p$, $q$ are integers</p>
<p>it follows that :</p>
<p>$$p^{12} = 2q^{12} $$</p>
<p>What is argument of irrationality in this case?
From what we know that the right-hand side has an even number of 2s and the right hand side has an odd number of 2s? </p>
<p>I will be grateful for your help Best regards</p>
| Fred | 380,717 | <p>If $\sqrt[12]{2} = \frac{p}{q}$, then $\sqrt[]{2}=(\sqrt[12]{2})^6=\frac{p^6}{q^6}$.</p>
<p>Your turn !</p>
|
2,509,308 | <p>My friend asked me a so called modified version of <a href="https://math.stackexchange.com/questions/96826/the-monty-hall-problem#">Monty Hall problem</a> in his opinion.
But I find the description a bit spooky and maybe someone here can enlighten
us with what is the problem with the description of the problem, or maybe it is me missing something.</p>
<p>Imagine exactly same setting as in monthy hall problem with one difference.
Imagine I randomly pick one card.
Now, as opposed to the original problem, the show host doesn't know where the car is, and of the remaining two cards, he
randomly picks a card: if it happens to be a goat the host shows us this card (as in original game),
if it happens to be a car however, my friend said you "quit" the game (or don't consider such case).
Now the question is same as in original problem, is it better for me to change my initial choice?</p>
<p>One problem with this description is mainly with the "quit" part. Does not this hinder the process of calculating whether we should switch or not? How do you model the case the host quit the game (in calculating whether you should switch or not)?</p>
<p>He claimed that in such case, switching my initial choice doesn't give me any advantage anymore and made a simulation program as follows. He made 10000 experiments, where he was skipping the part where
the host chose a car, so of course he was remained with 2/3 of the experiment test cases, and
he claimed that now, since the number of matches the user made (1/3rd of 1000) is half of the number of remaining
test cases(2/3rd of 1000- the ones where host didn't choose a car) - it is not advantageous anymore to change the card.</p>
<p>I am failing to find a flaw either in description in problem or in the implementation I mentioned above.
Can someone help figure what is wrong with either the problem description or implementation?</p>
<p>I would appreciate some help because I got confused overall with the whole thing now (whereas I understand original monthy hall problem well).</p>
<p>PS. Here is the Java implementation actually: <a href="http://codepad.org/rt7fOqei" rel="nofollow noreferrer">http://codepad.org/rt7fOqei</a>, where he deduces that since <code>match</code> is one half of <code>count</code>, then it makes no sense to change my initial choice.</p>
| Doug M | 317,162 | <p>If the host has no inside information, then with no new information there should be nothing to be gained by switching.</p>
<p>However you can model the outcomes explictly.</p>
<p>scenario 1. You choose the right card (P = 1/3). The host reveals a goat probability (always).</p>
<p>scenario 2. you choose the wrong card P = 2/3 the host reveals the car (P=1/2) game over (net P = 1/3)</p>
<p>scenario 3. you choose the wrong card (P = 2/3) the host reveals the goat (P=1/2) game continues P = 1/3.</p>
<p>If you have survived scenario 2. the conditional probability is that there is a 50% you have the right card and a 50% chance you have the wrong card.</p>
<p>Here is a picture
<a href="https://i.stack.imgur.com/YfgS1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YfgS1.png" alt="enter image description here"></a> </p>
<p>Suppose the car is on card A (you don't know that).</p>
<p>You choose, A,B or C. (the black lines). The host makes the red line choice.
You make the final keep or switch choice.</p>
<p>All of the keep/switch choices are equally likely. Half win, half lose.</p>
|
105,857 | <p>Let $\mathcal{O}$ be the ring of integers in an algebraic number field. Is $\text{SL}_2(\mathcal{O})$ generated by elementary matrices? If it isn't, is there any other natural generating set for it?</p>
<p>The usual argument shows that this is true for $\mathcal{O} = \mathbb{Z}$ (or, more generally, a Euclidean domain). However, I haven't been able to generalize this to other rings of integers.</p>
| Luc Guyot | 84,349 | <p>L. N. Vaserstein's theorem [2] asserts that if <span class="math-container">$R$</span> is a Dedekind ring of arithmetic type with infinitely many units then <span class="math-container">$SL_2(R) = E_2(R)$</span> holds. In other words <span class="math-container">$R$</span> is a <span class="math-container">$GE_2$</span>-ring in the sense of P. M. Cohn [1]. Here <span class="math-container">$E_2(R)$</span> denotes the subgroup of <span class="math-container">$SL_2(R)$</span> generated by the elementary matrices (also called transvections), i.e., the matrices of the form <span class="math-container">$\begin{pmatrix} 1 & r \\ 0 & 1\end{pmatrix}$</span> and <span class="math-container">$\begin{pmatrix} 1 & 0 \\ r & 1\end{pmatrix}$</span> with <span class="math-container">$r \in R$</span>.
Since the class of Dedekind rings of arithmetic type comprises the class of rings of integers of algebraic number fields, Vaserstein’s theorem answers OP’s question in the positive when the unit group is infinite.</p>
<p>But you may object, like B. Liehl [3], that there is a gap in Vaserstein’s proof. Fortunately A. Leutbecher [Section 2, 3] corrected this gap and B. Liehl [3] subsequently extended Vaserstein’s theorem to orders of arithmetic type with infinitely many units.</p>
<p>By Dirichlet’s unit theorem, a ring of integers <span class="math-container">$R$</span> of an algebraic number field <span class="math-container">$K$</span> has finitely many units if and only if <span class="math-container">$K$</span> is either the field of rationals <span class="math-container">$\mathbb{Q}$</span> or an imaginary quadratic number field. Under this assumption P. M. Cohn proved that <span class="math-container">$SL_2(R) = E_2(R)$</span> if and only if <span class="math-container">$R$</span> is Euclidean with respect to complex modulus. Thus <span class="math-container">$\mathbb{Z}[\frac{1 + \sqrt{-19}}{2}]$</span> is a PID which is not a <span class="math-container">$GE_2$</span>-ring. Remarkably, the subgroup <span class="math-container">$E_2(R)$</span> is non-normal in <span class="math-container">$SL_2(R)$</span> and has infinite index if <span class="math-container">$R$</span> is the ring of integers of an imaginary quadratic number field which is not Euclidean, see [Theorem 1.5, 6]. The proof of the latter exhibits infinitely many distinct coset representatives for the quotient <span class="math-container">$SL_2(R)/E_2(R)$</span>, namely the so-called <span class="math-container">$S_{x, y}$</span> matrices, but not a transversal, a priori. The ring of integers <span class="math-container">$R$</span> of <span class="math-container">$\mathbb{Q}(\sqrt{-D})$</span> for <span class="math-container">$D = 5, 10$</span> and <span class="math-container">$14$</span> is scrutinized in [5] where a group presentation of <span class="math-container">$SL_2(R)$</span> is derived. I leave here the second part of OP's question, only partially answered.</p>
<p>These facts were already mentioned under the form of comments to OP’s question and are thoroughly discussed in B. Nica’s paper [6], which I warmly recommend. I put them in this answer because the question seems to be considered unsettled in a <a href="https://mathoverflow.net/questions/241840/factorisation-in-special-linear-groups-over-rings-of-integers?lq=1">closely related MO post</a>.</p>
<p>The following may help clear any doubt: If <span class="math-container">$R$</span> is an order in an algebraic number field <span class="math-container">$K$</span> which is not imaginary quadratic then <span class="math-container">$E_2(R)$</span> is a normal subgroup of <span class="math-container">$SL_2(R)$</span> of finite index. This is a result of [4], extracted under this form in [Theorem 1.6, 6]. The fact that the index of <span class="math-container">$E_2(R)$</span> in <span class="math-container">$SL_2(R)$</span> is actually <span class="math-container">$1$</span> if <span class="math-container">$R$</span> is moreover a maximal order, i.e., <span class="math-container">$R$</span> is the (full) ring of integers of <span class="math-container">$K$</span>, may easily slip out of one’s mind.</p>
<hr />
<p>[1] «On the structure of the <span class="math-container">$GL_2$</span> of a ring», P. M. Cohn, 1966.<br />
[2] «On the group <span class="math-container">$SL_2$</span> over Dedekind rings of arithmetic type», L. N. Vaserstein, 1972.<br />
[3] «Euklidischer Algorithmus und die Gruppe <span class="math-container">$GL_2$</span>», A. Leutbecher, 1972.<br />
[4] «On the group <span class="math-container">$SL_2$</span> over orders of arithmetic type», B. Liehl, 1981.<br />
[5] "On the groups <span class="math-container">$SL_2(\mathbb{Z}[x])$</span> and <span class="math-container">$SL_2(k[x, y])$</span>", F. Grunewald et al., 1994.<br />
[6] «The Unreasonable Slightness of <span class="math-container">$E_2$</span> over Imaginary Quadratic Rings», B. Nica, 2013.</p>
|
1,123,050 | <p>This is the same problem asked here. - <a href="https://math.stackexchange.com/questions/1105927/next-step-to-take-to-reach-the-contradiction">Next step to take to reach the contradiction?</a>
Here is it again.</p>
<p><img src="https://i.stack.imgur.com/onqzq.png" alt="enter image description here"></p>
<p>I understand the solution - how you want to get to the fact 100 divides n^2 and then go through all the possibilities to show that the integer k for which n^2 * k = 100 is not n + 1.</p>
<p>Here is my instructor's solution
<img src="https://i.stack.imgur.com/wu6xx.png" alt="enter image description here"></p>
<p>The n^2 + n^3 > n^3 part of the proof makes sense to me. After all any positive integer squared will make an expression higher than itself. How did she get to 5^3 >= n^3 -> n < 5 though? I can't figure out the algebra step she took to get to that conclusion. I tried setting n^2 + n^3 equal to n^3 but that just got me zero. I am just to understand different ways of doing a proof</p>
| nelv | 136,824 | <p>Think of it physically: each measure assigns different <em>weights</em> to given sets: consider for example the particular case $d\mu=df(x)=f'(x)dx$ for a well behaved $f(x)$. Here you can really see the difference between the "ordinary" measure $dx$, which does not care about the location of the set, and $f'(x)dx$, which indeed does! In formulas:
\begin{equation}
\int_{[0,1]}dx =1= \int_{[1,2]}dx
\end{equation}
But in general
\begin{equation}
\int_{[0,1]}df(x) = f(1)-f(0) \neq f(2)-f(1) = \int_{[1,2]}df(x)
\end{equation}
This is just an example, but the idea applies with general measures $\mu$; it is in this sense that this perspective conveys the concept - in my view - of a weighted measure. It allows far more general and powerful structures than the old Riemann integration idea with shrinking rectangles.</p>
<p>One can then go further and (try to) relate different measures, for example via the Radon-Nikodym theory. There's a new world out there. :-)</p>
|
1,351,350 | <p>Assume that probability of $A$ is $0.6$ and probability of $B$ is at least $0.75$. Then how do I calculate the probability of both $A$ and $B$ happening together?</p>
| lulu | 252,071 | <p>You can't calculate that from the information given. </p>
<p>As an illustration: Imagine you choose a value N at random from 1, ..., 20. at one extreme, suppose A is the event {N ≤ 12} and B is the event {N ≤ 15}. Then A $\Rightarrow$ B and the probability of both is .6.
Now suppose A is the event {N ≤ 12} but B is now the event {N > 5}. In this case, the only over lap is N = 6, 7, 8, 9, 10, 11,12 so the probability of both occurring is $\frac{7}{20}$</p>
|
4,442,223 | <p>How does one show this?
<span class="math-container">$$
\exp(-x) \sum_{k=0}^\infty x^k \frac{(k+m)!}{(k!)^2} = L_m(-x) m!,
$$</span> where <span class="math-container">$m$</span> is a positive integer, and <span class="math-container">$L_{m}(x)$</span> is the <span class="math-container">$m$</span>th order Laguerre polynomial.</p>
| Paul Frost | 349,785 | <p>The definition you found in <a href="https://mathworld.wolfram.com/SimplicialComplex.html" rel="nofollow noreferrer">Wolfram MathWorld</a> describes simplicial complexes as a collection <span class="math-container">$K$</span> of simplices in some <span class="math-container">$\mathbb R^n$</span> such that</p>
<ol>
<li>Every face of a simplex of <span class="math-container">$K$</span> is in <span class="math-container">$K$</span>.</li>
<li>The intersection of any two simplices of <span class="math-container">$K$</span> is a face of each of them.</li>
</ol>
<p>This is a quotation from §2 Simplicial Complexes and Simplicial Maps, p.7, of</p>
<blockquote>
<p>Munkres, James R. Elements of algebraic topology. CRC press, 2018.</p>
</blockquote>
<p>Wolfram does not say anything about topology; and in fact, a simplicial complex is <strong>not</strong> a topological space. Yes, it is tempting to identify <span class="math-container">$K$</span> with the subset <span class="math-container">$\lvert K \rvert = \bigcup_{\sigma \in K} \sigma \subset \mathbb R^n$</span>, but the set <span class="math-container">$\lvert K \rvert$</span> does not allow to reconstruct the collection <span class="math-container">$K$</span>. In particular the set <span class="math-container">$K^0$</span> of all <span class="math-container">$0$</span>-simplices of <span class="math-container">$K$</span> does not have a topology, thus it does not make sense to say it is a <em>discrete set</em>. <span class="math-container">$K^0$</span> is a set of <em>single point subsets of <span class="math-container">$\mathbb R^n$</span></em>, and we must not confuse it with the set <span class="math-container">$\lvert K^0 \rvert = \bigcup_{\sigma \in K^0} \sigma \subset \mathbb R^n$</span>. As an example consider any subset <span class="math-container">$S \subset \mathbb R^n$</span> and define a simplicial complex
<span class="math-container">$$K_S = \{ \{x\} \mid x \in S\}$$</span>
which has only <span class="math-container">$0$</span>-simplices. We have <span class="math-container">$K_S^0 = K_S$</span> and <span class="math-container">$\lvert K_S^0 \rvert = S$</span>. This applies in particular to your set <span class="math-container">$\{0, 1/n\}_{n=1,...}$</span>.</p>
<p>Concerning topology let us read what Munkres writes:</p>
<p><a href="https://i.stack.imgur.com/cInPt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cInPt.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/Ox89S.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ox89S.jpg" alt="enter image description here" /></a></p>
<p>Using the above topology of <span class="math-container">$\lvert K \rvert$</span>, it is easy to see that the subspace <span class="math-container">$\lvert K^0 \rvert \subset \lvert K \rvert$</span> is always discrete. Our above example nicely illustrates that the "simplicial topology" on <span class="math-container">$\lvert K \rvert$</span> is in general different from the subspace topology induced from <span class="math-container">$\mathbb R^n$</span> on the subset <span class="math-container">$\lvert K \rvert \subset \mathbb R^n$</span>.</p>
|
1,749,730 | <p>What is the maximum number of faces of totally convex solid that one can "see" from a point? </p>
<p>...and, more importantly, how can I ask this question better? (I'm a college student with little experience in asking well formed questions, much less answering them.) </p>
<p>By "see" I mean something like this: you point a camera from a point at the solid, and look at the picture. How many of the faces of the solid look like faces and not just lines? Let's assume that the lens is just a point in space (no lenses wider than the solid itself) and that the camera is a finite distance from the solid. I know this is a crude definition... if you have any ideas for a more rigorous definition, this would be awesome, then maybe there's ways to prove the answer to my question mathematically. </p>
<p>For example, in the picture of <a href="https://i.stack.imgur.com/Ue0Fg.jpg" rel="noreferrer">this cube</a>, you can see 3 faces. This is the maximum you can see for a cube. How can that be proved? </p>
<p>What methods might you use to prove this for a convex solid of any size and shape? Are there ways to do so using only relatively basic math (Multivariable calc, linear algebra, high school geometry)? </p>
| Vincent | 332,815 | <p>Here are some definitions (in dimention 3, but you can easily generalize):</p>
<p><em>Definition</em> : Given a finite number of points with coordinates $ P_1 = (x_1,y_1,z_1), .., P_n = (x_n,y_n,z_n) $, a convex solid is the convex hull of these points, i.e. all the points of the space defined as $\sum_{1\leq i \leq n} \lambda_i P_i $ with $\forall i \in [1,n], \lambda_i \in \mathbb{R}, 0 \leq \lambda_i \leq 1, \sum_{1 \leq i \leq n } \lambda_i = 1 $</p>
<p>For instance, the points (0,0,0); (1,0,0); (0,1,0); (1,1,0); (0,0,1); (1,0,1); (0,1,1); (1,1,1) define a cube. </p>
<p>Rq : If I add the point (0.5,0.5,0.5), I define the same cube.</p>
<p><em>Definition</em> : An oriented plane is 4 real numbers $a, b, c$ and $d$.</p>
<p><em>Definition</em> : An oriented plane is a face of a solid, iff at least three points defining the solid verify $ax+by+cz = d$ (are in the plane), and all points of the solid verify $ax+by+cz \leq d$.</p>
<p>For instance, the plane defined by $a=1, b=0, c=0, d=1$ in the cube described above is a face. The one defined $a=1, b=1, c=0, d=1$ is not, since it fails the second condition of the definition. The one defined by $a=-1, b=0, c=0, d=-1$ is not either. Indeed, the definition specifies $ax+by+cz \leq d$, and it is not the case here. </p>
<p>The definition of a face given here enables to control on which "side" of the plane the solid is. Putting a "-" sign in front of all the coefficients defining a face would mean the solid is on the other side of the plane.</p>
<p><em>Definition</em> : A view direction is a vector $ (x_v,y_v,z_v) $ with $x_v^2+y_v^2+z_v^2=1$.</p>
<p>The direction can be identified at the way you look at. Here, it is assumed you look from a point at an infinite distance.</p>
<p><em>Definition</em> : A face $(a,b,c,d)$ of a solid is visible from a direction $ (x_v,y_v,z_v) $ iff $ax_v+by_v+cz_v < 0 $.</p>
<p>Here is the proof (short version) for the cube (with the points described earlier).</p>
<p>All faces of this cube are defined by the following quadruples (and no others) :</p>
<ul>
<li><p>(1,0,0,1)</p></li>
<li><p>(0,1,0,1)</p></li>
<li><p>(0,0,1,1)</p></li>
<li><p>(-1,0,0,0)</p></li>
<li><p>(0,-1,0,0)</p></li>
<li><p>(0,0,-1,0)</p></li>
</ul>
<p>So, with any viewing direction, one sees at most three faces (the ones corresponding to the sign of your viewing direction).
(you see less if one of the viewing direction coordinates is 0, it would mean we see only some of its edges).</p>
<p>For a different kind of solid, you need its defining points, and then the proof would basically go this way</p>
<ul>
<li><p>first find all faces of this solid</p></li>
<li><p>assume there is another face and discover it is one you already have (so you have proven that you have all of them)</p></li>
<li><p>Assume you have a viewing direction. Then, a lot of cases to look at, like "if my direction is such that $x_v + ... < ...$"</p></li>
<li><p>find, in all the cases, the maximal number of faces</p></li>
</ul>
<p>If you think there is a leaner/more conventional way to define the problem, do not hesitate to mention it.</p>
|
898,683 | <p>Given a pool of 30 balls (5 of each color). When drawing 8 balls without replacement, what is the probability of getting at least one of each color?</p>
<p>Related: <a href="https://math.stackexchange.com/questions/897730/probability-of-drawing-at-least-one-red-and-at-least-one-green-ball">Probability of drawing at least one red and at least one green ball.</a></p>
<p>When drawing more than 2 colors you need to exclude overlapping 'hands'. Thus when finding the probability of drawing no red, you can have a hand made up of blue, green, white, black and grey. But when you are determining the probability of drawing no blue you draw from red, green, white, black, grey. So you need to exclude all green, white, black, grey hands as they have already been counted. And the same for the other colors as well.</p>
<p>The other complexity of the problem is that since there are only 5 of each color, no draw will only include balls of the same color.</p>
| André Nicolas | 6,312 | <p><strong>Edit:</strong> The question has changed. We leave the answer to the original question, which specified $6$ draws. We append a sketch for the new question.</p>
<hr>
<p>We want the probability that the balls all have different colours. Imagine (it makes no difference) that we draw the balls one at a time. </p>
<p>Whatever the first ball was, the probability the second has a different colour is $\frac{25}{29}$.</p>
<p>Given that the second ball had a different colour from the first, the probability we get a new colour on the third is $\frac{20}{28}$.</p>
<p>Thus the probability that the first three balls have different colours is $\frac{25}{29}\cdot \frac{20}{28}$.</p>
<p>Given that the first three are of different colours, the probability of a new colour on the fourth is $\frac{15}{27}$. So the probability the first four balls are of different colours is $\frac{25}{29}\cdot \frac{20}{28}\cdot \frac{15}{27}$.</p>
<p>Continue. We are close to the end. </p>
<hr>
<p><strong>Drawing $8$:</strong> There are $\binom{30}{8}$ equally likely ways to draw $8$ balls. We count the number of <strong>favourable</strong> draws, in which we get all $6$ colours. Since $8$ is not much larger than $6$, a division into cases is efficient. For larger numbers, we would suggest using Inclusion/Exclusion. </p>
<p>These favourables are of two types.</p>
<p>(i) One each of $5$ colours, and $3$ of another. The abundant colour can be chosen in $\binom{6}{1}$ ways, and for each of these the actual balls can be chosen in $\binom{5}{3}$ ways. The rest of the balls can then be chosen in $\binom{5}{1}^5$ ways. Multiply.</p>
<p>(ii) One each of $4$ colours, and $2$ each of $2$ others. The lucky colours can be chosen in $\binom{6}{2}$ ways. For each way, the actual balls can be chosen in $\binom{5}{2}^2$ ways. And the remaining $4$ can be chosen in $\binom{5}{1}^4$ ways.</p>
|
898,683 | <p>Given a pool of 30 balls (5 of each color). When drawing 8 balls without replacement, what is the probability of getting at least one of each color?</p>
<p>Related: <a href="https://math.stackexchange.com/questions/897730/probability-of-drawing-at-least-one-red-and-at-least-one-green-ball">Probability of drawing at least one red and at least one green ball.</a></p>
<p>When drawing more than 2 colors you need to exclude overlapping 'hands'. Thus when finding the probability of drawing no red, you can have a hand made up of blue, green, white, black and grey. But when you are determining the probability of drawing no blue you draw from red, green, white, black, grey. So you need to exclude all green, white, black, grey hands as they have already been counted. And the same for the other colors as well.</p>
<p>The other complexity of the problem is that since there are only 5 of each color, no draw will only include balls of the same color.</p>
| Thomas Andrews | 7,933 | <p>The quickest approach is to count ways to get $6$ colors on $8$ balls. There are essentially two cases, $\langle 3,1,1,1,1,1\rangle$ and $\langle 2,2,1,1,1,1\rangle$. There are $\binom{6}{1}\binom{5}{3}\binom{5}{1}^5=187,500$ ways to get the first case. The $\binom{6}{1}$ counts the number of ways of choosing one color to get three balls, and the rest is the number of ways of choosing three balls from that one color and one for each of the other colors.</p>
<p>The second case has $\binom6 2\binom 5 2^2\binom 5 1^4=937,500$ different ways. There are $\binom 6 2$ ways to choose two colors to receive two balls, and the rest is the number of ways of choosing two balls from each of those colors and one ball from the others.</p>
<p>So the total cases are $1,125,000$, out of $\binom{30}{8}=5,852,925$. That gives a probability of about $0.1922$.</p>
|
973,035 | <p>I'm wondering whether there is an invertible function $f: \mathbb{R} \to \mathbb{R}$ such that $f(-1)=0$, $f(0)=1$ and $f(1)=-1$. I think it's not but I'm missing a real proof.</p>
<p>The easiest would be to show such a function cannot be injective... But I don't see how? I don't see any other way of starting this either! Can you hint me please?</p>
<p>Thank you!</p>
| Tomasz Kania | 17,929 | <p>There is no invertible function like that which is <em>continuous</em>. Indeed, $f(0)=1$ and $f(1)=-1$ so by the <a href="http://en.wikipedia.org/wiki/Intermediate_value_theorem" rel="nofollow">Intermediate Value Theorem</a>, there is $\xi \in (0,1)$ such that $f(\xi)=0 = f(-1)$.</p>
<p>There are discontinuous bijections of $\mathbb{R}$ which satisfy your conditions.</p>
|
172,292 | <p>I am trying to find the residue of the function $$f(z)=(z+1)^2e^{3/z^2}$$
at $z=0$.
I tried the following in Mathematica</p>
<pre><code>Residue[(z+1)^2*Exp[3/z^2],{z,0}]
</code></pre>
<p>which remains unevaluated. Computing this by hand gives the value of $6$. What is going on?</p>
<p>I’ve noticed that Mathematica has a problem with the Laurent series of $e^{3/z^2}$ at $z=0$.</p>
| Akku14 | 34,287 | <p>Or integrate around zero</p>
<pre><code>Integrate[(z + 1)^2 Exp[3/z^2], {z, 1, I, -1, -I,
1}]/(2 Pi I) // Simplify
(* 6 *)
</code></pre>
|
2,347,820 | <p>What is the solution to $\log_{10} x -x=2?$</p>
<p>I have tried to solve it but I couldn't. I've got to $x^x =200$.</p>
| 高田航 | 407,845 | <p>$f(x)=log_{10}(x)-x$ remains completely within the fourth quadrant, so $f(x)$ will never be equal to $2$. </p>
|
1,709,713 | <p>How do you make the jump from:</p>
<p>$$\frac{1-(\frac{4}{25})^{21}}{1-\frac{4}{25}}$$</p>
<p>To:</p>
<p>$$\frac{25^{21}-4^{21}}{25^{21}-4(25^{20})}$$</p>
| Henricus V. | 239,207 | <p>$$ \frac{1-(\frac{4}{25})^{21}}{1-\frac{4}{25}}
= \frac{1-(\frac{4}{25})^{21}}{1-\frac{4}{25}} \frac{25^{21}}{25^{21}}
= \frac{25^{21}-(\frac{4}{25})^{21}25^{21}}{25^{21}-\frac{4}{25}25^{21}}
= \frac{25^{21}-4^{21}}{25^{21}-4 \cdot 25^{20}}
$$</p>
|
1,896,008 | <p>Is the following statement correct: </p>
<blockquote>
<p>If $A$ and $B$ are closed subsets of $[0,\infty)$, then $A+B=\{x+y:x \in A,y \in B\}$ is closed in $[0,\infty)$.</p>
</blockquote>
| Piquito | 219,998 | <p>HINT.-The addition in $\Bbb R$ is continuous (from $\Bbb R^2$ to $\Bbb R$). You have, for example two ways to prove $A+B$ is closed.</p>
<p>1) If $\{z_n\}$ is a sequence convergent to $z$ and contained in $A+B$ then $z\in A+B$.</p>
<p>2) The complement of $A+B$ is easily proved to be open (using also the continuity of the sum of reals).</p>
<p>There are more (for example the adherence of $A+B$ is equal to $A+B$).</p>
|
1,621,269 | <p>I have tried everything in my knowledge and no, I cannot state it.
I have tried a factorizor online which tells me that it is not factorizable hence irreducible. But I cannot reason why.</p>
<p>I looked at Eisenstein's criteria but obviously, there is no prime $q$ that fits the criteria so this is useless.</p>
<p>I then tried reducibility via modulo reduction, and this should give me the options to test irreducibility up to mod $8$ since that is the largest coefficient in the polynomial...yes? But every mod arrives at the polynomial being reducible...so it basically fails to tell me that it is irreducible.
For mod 2, I get 0 as a solution to the reduced polynomial so that means I can factor it out with $x$.
Similarly, mod 3 says 1 is a solution so $(x-1)$ should be a solution.
In a similar fashion, I get mod 4,5,6,7 to have solutions 0,3,4,6.</p>
<p>Am I missing anything? This is the best I can do, nothing more assures me irreducibility at all. Ideas please...?</p>
| Alex | 48,061 | <p>Let $x = y+1$. Then this equation becomes:</p>
<p>$y^5+3y^2+9y+3$. Use Eisenstein to show this is irreducible. Therefore, the original is also irreducible.</p>
|
489,907 | <p>I've only got the following parts of a triangle:</p>
<ul>
<li>Line A to B </li>
<li>Line B to C</li>
</ul>
<p>And optionally the Line from A to C if needed?</p>
<p>I'm trying to get the point X</p>
<p><a href="https://i.stack.imgur.com/f6leO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/f6leO.png" alt="Triangle"></a></p>
<p>Now the problem is, i've got absolutly no idea of trigonometry so I dont even know how to search for it.
Thanks in advance!</p>
| Mufasa | 49,003 | <p>OK, then working off your diagram above and assuming AX is perpendicular to BC, you can proceed as follows:<br/>
Let's call the co-ordinates of point A $(x_a, y_a)$, point B $(x_b, y_b)$, point C $(x_c, y_c)$ and point X $(x, y)$.<br/>
For line BC we can write: $\frac{y-y_c}{x-x_c}=\frac{y_b-y_c}{x_b-x_c}=m$ where $m$ represents the slope of this line. This leads to:<br/>
$y=y_c+mx-mx_c\tag{1}$<br/></p>
<p>Similarly, for line AX we can write: $\frac{y-y_a}{x-x_a}=-\frac{1}{m}$ (since line AX is perpendicular to line BC). This leads to:<br/>
$y=y_a-\frac{x}{m}+\frac{x_a}{m}\tag{2}$<br/></p>
<p>Now just equate equations (1) and (2) and solve for $x$ and then use either of these equations to find $y$. $(x, y)$ will then be the point you desire.<br/></p>
<p>You will have to be careful in dealing with vertical lines as their slopes will turn out to be infinite.</p>
<p>To help you along, if you equate (1) and (2) you'll get:<br/>
$y_c+mx-mx_c=y_a-\frac{x}{m}+\frac{x_a}{m}$<br/>
Then multiply both sides by $m$ to get:<br/>
$my_c+m^2x-m^2x_c=my_a-x+x_a$<br/>
Take all terms involving $x$ to the left-hand-side of the equals and all others to the right-hand-side to get:<br/>
$(m^2+1)x=m(y_a-y_c)+x_a+m^2x_c$<br/>
Then divide both sides by $(m^2+1)$ to get:<br/>
$\large x=\frac{m(y_a-y_c)+x_a+m^2x_c}{m^2+1}\tag{3}$<br/></p>
<p>This will give you a value for the $x$ co-ordinate of point X which you can then use in equation (1) to calculate the corresponding $y$ value for point X.</p>
|
2,670,082 | <p>I have quite an interesting infinite totient sum. My task is to evaluate</p>
<p>$\sum_{n=1}^{\infty} \frac{\phi(n)}{5^n +1}.$</p>
<p>The problem is that I have no idea how to go from here as I have never seen such a problem before. The usual techique of writing $n$ and $\phi(n)$ in terms of the prime factorization of $n$ doesn't work. I get $\frac{\phi(n)}{5^n+1}=\frac{\prod_i p_i^{e_i-1} (p_i-1)}{5^{\prod p_i^{e_i}}+1}$, which turns into a complete crocodile when plugged back into the summation.</p>
<p>I know that the problem has an elegant, closed form solution because it appeared on a problem set that is expected to be solved by mere high school students. Furthermore, the only way to simplify such an infinite sum is to turn it into a numerical answer. Wolfram Alpha is unable to give a closed form solution, so I know that there is some clever trick to demolish this problem that I have not seen before. Does anyone have any hints or ideas on how do deal with this?</p>
<p>My only guess is that 5 is quite a peculiar number, and so it seems reasonable that the sum $\sum_{n=1}^{\infty} \frac{\phi(n)}{a^n +1}$ should also have a closed form solution for any integer $a$.</p>
<p>Update: Summing up to $n=1000$ using Wolfram-Alpha (it seems that my mistake was asking it to sum infinitely many terms) gives $0.22569444444444...$ with no end to the relentless onslaught of the 4's. Therefore I believe that the sum is $\frac{65}{288} = \frac{5 \cdot 13}{2^5 \cdot 3^2}.$ I have split up the possible answer into prime factors so that the clever trick I had earlier alluded to may be hopefully found. However, I do not have the luxury of pulling out a computational device on a math competition, and so I hope to find the actual solution, i.e. the method through one may derive this answer.</p>
<p>Update: We define $f(a) = \sum_{n=1}^{\infty} \frac{\phi(n)}{a^n +1}$. Here are some reasonable values for prime $a$ in case someone may try to use these to find the solution. At this point, I believe the only way to find a rigorous solution is to work backwards from the likely solution given by WA.</p>
<p>$f(2)=22569444444444...=\frac{5 \cdot 13}{2^5 3^2}$</p>
<p>$f(3)=0.$</p>
<p>$f(5)=0.$</p>
<p>$f(7)=0.$</p>
<p>$f(11)=0.09319444444...=\frac{11 \cdot 61}{2^5 3^2 5^2}$</p>
<p>Additional Update: With the help of achille hui, I have found that $$\sum_{n=1}^{\infty} \frac{\phi(n)}{1-(-1)^n q^n} = \frac{q}{(q+1)^2}.$$</p>
<p>This is almost the right sum.</p>
| achille hui | 59,379 | <p>Notice for any prime <span class="math-container">$p$</span> and integer <span class="math-container">$k \ge 0$</span>, we have</p>
<p><span class="math-container">$$\varphi(p^k) = \begin{cases}p^k - p^{k-1}, & k > 0\\ 1,& k = 0\end{cases}
\quad\implies\quad \sum_{\ell=0}^k \varphi(p^\ell) = p^k
$$</span>
For any <span class="math-container">$n \in \mathbb{Z}_{+}$</span>, if we factorize it into a product of primes <span class="math-container">$n = p_1^{e_1}\ldots p_m^{e_m}$</span>, we can use above fact to deduce following identity by Gauss
<span class="math-container">$$\begin{align}\sum_{d|n} \varphi(d)
&=
\sum_{\ell_1=0}^{e_1} \cdots\sum_{\ell_m=0}^{e_m} \varphi(p_1^{e_1} \cdots p_m^{e_m})=
\sum_{\ell_1=0}^{e_1} \cdots\sum_{\ell_m=0}^{e_m} \varphi(p_1^{e_1}) \cdots \varphi(p_m^{e_m})\\
&= \prod_{i=1}^m \left(\sum_{\ell_i=0}^{e_i}\varphi(p_i^{e_i})\right)
= \prod_{i=1}^m p_i^{e_i} = n
\end{align}
$$</span>
As a result, for any <span class="math-container">$|q| < 1$</span>, we have
<span class="math-container">$$F(q) \stackrel{def}{=}\sum_{n=1}^\infty \frac{\varphi(n)q^n}{1-q^n}
= \sum_{n=1}^\infty\sum_{m=1}^\infty \varphi(n)q^{nm}
= \sum_{k=1}^\infty \left(\sum_{d|k} \varphi(d)\right) q^k
= \sum_{k=1}^\infty k q^k = \frac{q}{(1-q)^2}$$</span></p>
<p>As a corollary, for any <span class="math-container">$|a| > 1$</span>, we have</p>
<p><span class="math-container">$$\sum_{n=1}^\infty \frac{\varphi(n)}{a^n-1} = \sum_{n=1}^\infty \frac{\varphi(n)a^{-n}}{1-a^{-n}} = F(a^{-1}) = \frac{a}{(a-1)^2}$$</span></p>
<p>This leads to</p>
<p><span class="math-container">$$\sum_{n=1}^\infty \frac{\varphi(n)}{a^n+1}
=\sum_{n=1}^\infty \varphi(n)\left(\frac{1}{a^n-1} - \frac{2}{a^{2n}-1}\right)
= \frac{a}{(a-1)^2} - \frac{2a^2}{(a^2-1)^2} = \frac{a(a^2+1)}{(a^2-1)^2}
$$</span>
Substitute <span class="math-container">$a = 5$</span>, we get
<span class="math-container">$$\sum_{n=1}^\infty \frac{\varphi(n)}{5^n+1} = \frac{5(5^2+1)}{(5^2-1)^2} = \frac{65}{288}$$</span></p>
|
515,659 | <p>Question is to check :</p>
<p>For any real number $c$, the polynomial $x^3+x+c$ has exactly one real root .</p>
<p>the way in which i have proceeded is :</p>
<p>let $a$ be one real root for $x^3+x+c$ i.e., we have $a^3+a+c=0$</p>
<p>i have seen that $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a$</p>
<p>But, $a^3+a+c=0$. So, $-a^3-a=c$.</p>
<p>so, $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a=x^3+x+c$</p>
<p>i.e, $(x-a)(x^2+ax+(a^2+1))=x^3+x+c$</p>
<p>Now, determinant for quadratic $x^2+ax+(a^2+1)$ is $a^2-4(a^2+1)=-3a^2-4 <0$ for any real number $a$</p>
<p>Thus, quadratic has no real root and so is the cubic $x^3+x+c$</p>
<p>I would like to know if this justification is sufficient and if this can be generalized.</p>
<p>I mean, can i say $x^3+ax^2+bx+c$ for $a,b,c\in \mathbb{Z}$ have exactly one real root. (Conditions apply)</p>
<p>can this be generalized to any odd degree polynomial (at least for some special cases)</p>
| Hagen von Eitzen | 39,174 | <p>You essentially split off the linear factor belnging to a real root and showed per determinant that the remaining quadratic has no real root.
This is fine but does not readily generalize to higher degrees.
Instead, it is probably easier to show that $f$ is injective: Assume $a,b$ are two real roots, i.e. $f(a)=f(b)=0$.
Then by Rolle, there exists $\xi$ between $a$ and $b$ (including the case $a=\xi=b$ if you also want to show that no multiple root - $a=b$ - exists) with $f'(\xi)=0$. But for $f(x)=x^3+x$ we have $f'(x)=3x^2+1\ge1$ for all $x$.</p>
|
515,659 | <p>Question is to check :</p>
<p>For any real number $c$, the polynomial $x^3+x+c$ has exactly one real root .</p>
<p>the way in which i have proceeded is :</p>
<p>let $a$ be one real root for $x^3+x+c$ i.e., we have $a^3+a+c=0$</p>
<p>i have seen that $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a$</p>
<p>But, $a^3+a+c=0$. So, $-a^3-a=c$.</p>
<p>so, $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a=x^3+x+c$</p>
<p>i.e, $(x-a)(x^2+ax+(a^2+1))=x^3+x+c$</p>
<p>Now, determinant for quadratic $x^2+ax+(a^2+1)$ is $a^2-4(a^2+1)=-3a^2-4 <0$ for any real number $a$</p>
<p>Thus, quadratic has no real root and so is the cubic $x^3+x+c$</p>
<p>I would like to know if this justification is sufficient and if this can be generalized.</p>
<p>I mean, can i say $x^3+ax^2+bx+c$ for $a,b,c\in \mathbb{Z}$ have exactly one real root. (Conditions apply)</p>
<p>can this be generalized to any odd degree polynomial (at least for some special cases)</p>
| Mark Bennet | 2,906 | <p>It is a bit unclear what you want to generalise. $x^3+x+c$ is a strictly increasing function of $x$ - a strictly increasing (or decreasing) polynomial of odd order with real coefficients will always have a single real root.</p>
<p>Or you may be saying that the polynomial can be written as $p(x)=(x-a)q(x)$ where $q(x)$ is a polynomial of even degree which has no real roots. Any odd degree polynomial with a single simple zero (ie not a multiple zero) can be written in this way. However this form of factorisation does not imply that $p(x)$ is strictly increasing, simply that it has just one real zero.</p>
<p>For example take $p(x)=x\left((x-3)^2+1\right)$ with $a=0$</p>
<p>We have $p(0)=0, P(1)=5, p(2)=4, p(3)=3, p(4)=8$, so $p(x)$ is not increasing, while $q(x)$, which is a sum of squares, one of which is non-zero, has no real roots.</p>
|
515,659 | <p>Question is to check :</p>
<p>For any real number $c$, the polynomial $x^3+x+c$ has exactly one real root .</p>
<p>the way in which i have proceeded is :</p>
<p>let $a$ be one real root for $x^3+x+c$ i.e., we have $a^3+a+c=0$</p>
<p>i have seen that $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a$</p>
<p>But, $a^3+a+c=0$. So, $-a^3-a=c$.</p>
<p>so, $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a=x^3+x+c$</p>
<p>i.e, $(x-a)(x^2+ax+(a^2+1))=x^3+x+c$</p>
<p>Now, determinant for quadratic $x^2+ax+(a^2+1)$ is $a^2-4(a^2+1)=-3a^2-4 <0$ for any real number $a$</p>
<p>Thus, quadratic has no real root and so is the cubic $x^3+x+c$</p>
<p>I would like to know if this justification is sufficient and if this can be generalized.</p>
<p>I mean, can i say $x^3+ax^2+bx+c$ for $a,b,c\in \mathbb{Z}$ have exactly one real root. (Conditions apply)</p>
<p>can this be generalized to any odd degree polynomial (at least for some special cases)</p>
| Michael Hoppe | 93,935 | <p>$(x^3+x+1)'=3x^2+1>0$ for all real $x$.</p>
|
80,918 | <p>Could anyone please tell me what could be the math function to get the number of zeros in given decimal representation of numbers? I scratched my head on Combination and Permutation but couldn't come up with generic answer. The number length can be up to 1000 digits, so you can represent a number as a String.</p>
<p>For example, if numbers range is $1-100$, the answer should be $11$, for $1-200$, it's $20$ and so on! Now, how would you find total number of zeros between $1-19447494833737292827272\cdots 444$ ( or any big number)?</p>
<p>Thanks.</p>
| Dima | 44,950 | <p>The approach is correct, but the answer is wrong. There are 145 zeros from 1 to 751:
9 zeros from 1 to 99, 120 zeros from 100 to 699 (20 x 6) and 16 zeros from 700 to 759</p>
<p>Total: 9+120+16 = 145.</p>
|
1,100,906 | <p>What would be the highest power of two in the given expression?</p>
<p>$32!+33!+34!+35!+...+87!+88!+89!+90!\ ?$</p>
<p>I know there are 59 terms involved. I also know the powers of two in each term. I found that $32!$ has 31 two's. If we take 32! out of every term the resulting 59 terms has 2 odd terms and 57 even terms. So it is an even number of the form $2K$. So min possible highest power of 2 will be 32. But I don't know how to calculate the exact value. Surely, We cannot go term by term.</p>
<p>Can anyone throw light in this matter?</p>
| Milo Brandt | 174,927 | <p>You've simply stopped calculating one step too soon! You've shown that that expression, with $32!$ divided out is divisible by $2$, but if you just checked whether it was divisible by $4$, you would see that it is indeed not. In particular, the divided expression would be
$$\frac{32!}{32!}+\frac{33!}{32!}+\frac{34!}{32!}+\frac{35!}{32!}+\frac{36!}{32!}+\ldots+\frac{90!}{32!}$$
Now, take this mod $4$. All but the first four terms are eliminated, since $4$ divides each of them (as each has $36$ as a factor). However, the first four terms, mod $4$ are $1,\,1,\,2,\,2$, which sums to $2$ mod $4$. Ergo, $2^{33}$ does not divide the original expression, and $2^{32}$ is thus the maximum power of two dividing the expression.</p>
|
2,280,666 | <p>Let AD be the altitude corresponding to the hypotenuse BC of the right triangle ABC. The circle of diameter AD intersects AB at M and AC at N shown. Prove $\frac{BM}{CN}$= $\bigg(\frac{AB}{AC}\bigg)^{3}$.</p>
<p>So far I have...</p>
<p>The power of B is $BD^{2}=(BM)(BA)$</p>
<p>The power of C is $CD^{2}=(CN)(CA)$</p>
<p>I am stuck after this. Any help would be appreciated!!</p>
| Martín Forsberg Conde | 249,492 | <p><strong>Hint</strong>: What can you say about triangle $BMD$ given that A forms a right angle? Try to express $\frac{BM}{CN}$ in terms of the relations of sides that you know.</p>
<p>__</p>
<p>__</p>
<p>__</p>
<p>__</p>
<p>__</p>
<p><strong>Solution</strong>: The angle in M of $AMD$ is right, since it sees a diameter of the circle. Hence $BMD$ has two angles equal to $ABC$, and is similar to it. The same holds for $AMD$ and $DNC$. Also, the quadrilateral $AMDN$ has four right angles, and therefore it is a rectangle, and $AM=DN$. Now we operate.
$$\frac{BM}{CN} = \frac{BM}{MD}\frac{MD}{AM}\frac{AM}{DN}\frac{DN}{CN} = \frac{AB}{BC}\frac{AB}{BC}1\frac{AB}{BC}$$ so we get the desired result.</p>
|
189,074 | <p>the function f defined by $f(x)=(x^3+1)/3$ has three fixed points say α,β,γ where
$-2<α<-1$, $0<β<1$, $1<γ<2$.
For arbitrarily chosen $x_{1}$, define ${x_{n}}$ by setting $x_{n+1}=f(x_{n})$
If $α<x_{1}<γ$, prove that $x_{n}\rightarrow β$ as $n \rightarrow \infty$</p>
<p>I think I must prove three things, but not sure:</p>
<p>1: if $α<x_{1}<γ$, then $α<f(x)<γ$</p>
<p>2: if $α<x_{1}<β$, then $x_{1}<f(x)<β$</p>
<p>3: if $β<x<γ$, then $β<f(x)<x_{1}$</p>
<p>could you please help me?</p>
| copper.hat | 27,978 | <p>This is not meant to be an answer, but I can't add a picture to comments or another answer...</p>
<p><img src="https://i.stack.imgur.com/WwgrH.png" alt="enter image description here"></p>
|
564,378 | <p>Suppose we start with a rational number $a_0$, and define $a_{n+1}=2a_n^2-1$ for $n\geq 0$. For what $a_0$ will it be the case that $a_i=a_j$ for some $i\neq j$?</p>
<p>We can start with something like $a_0=1$, then $a_1=1$ so $a_0=a_1$.</p>
<p>If $a_0=0$, we get $0, -1, 1, 1, \ldots$</p>
<p>Likewise if $a_0=-1$, we get $-1,1,1,\ldots$.</p>
<p>But how can we find all $a_0$?</p>
| DonAntonio | 31,254 | <p>In general</p>
<p>$$a_{n+1}:=2a_n^2-1=2a_m^2-1=:a_m\iff a_n=\pm a_m$$</p>
<p>If we choose $\;n\;$ to be the minimal index s.t. $\;a_{n+1}=a_{m+1}\;$ , for some $\;m\neq n\;$ , the above means that</p>
<p>$$a_n=-a_m\iff 2a_{n-1}^2-1=-2a_{m-1}^2+1\iff a_{n-1}^2+a_{m-1}^2=1\ldots$$</p>
<p>Try to take it from here.</p>
|
1,550,603 | <p>I got confirmed from a graduate school starting from next year and I will major algebraic geometry.</p>
<p>Until now, I have never thought that I study little things than others with my age. However, I heard that <strong>some</strong> of my colleagues already studied Hartshorne at least once and quite a few of them have read Rudin's RCA when they were undergaduates. It's kinda unbelievable to me, but it seems like if they really <strong>did</strong> study and understood, then they will write absolutely a better Ph.D thesis than mine.</p>
<p>So I'm now very worrying myself. I want to know whether this situation is general. Is it recommenable to study graduate subjects as early as possible? Or are there people here who experienced the same thing too? Was that beneficial?</p>
<p>Between "studying each thing deep and slow" and "skimming many subjects as fast as possible", which one is better?</p>
| paul garrett | 12,291 | <p>Your questions in order:</p>
<p>Yes, it is highly advantageous to <em>be</em> <em>exposed</em> <em>to</em> more sophisticated ("graduate") subjects as early as one can tolerate it.</p>
<p>Not clear that one should "study" them.</p>
<p>Yes, some people do "read ahead". I myself found it very helpful.</p>
<p>I would claim that the widely-believed sense of "deep and slow" versus "skimming ..." is a fake comparison, and is not the question anyone should really ask. That is, studying "slowly" cannot possibly be "deep", because "slowly" also entails maintaining naivete and one sort of shallowness for an unfortunately long time.</p>
<p>Likewise, the "skimming... as fast as possible..." is not any sort of "other" alternative. A more sane "other" is "looking around, not getting bogged down in details, trying to see where things are going".</p>
<p>And, my recommendation would be to <em>both</em> read fairly carefully and also look around to see what's going on. <em>Both</em> lower-level details <em>and</em> some idea of the goals and larger enterprise.</p>
<p>Very specifically: much of "comfort" and "facility" consists of familiarity more than anything else. The psychological obstacle of novelty is surprisingly great, while the psychological ease of (even superficial) familiarity is surprising. Simply hearing the words and a bit of a story a year or two (or more) earlier is extremely beneficial, in all my observation (and my own experience).</p>
|
3,527,919 | <p>I've tried to prove this property of Bessel function but I don't seem to be going anywhere</p>
<p><span class="math-container">$$\sqrt{\frac 12 \pi x} J_\frac 32 (x) = \cfrac{\sin x}{x} - \cos x$$</span></p>
<p>I have tried substituting <span class="math-container">$\frac 32$</span> for <span class="math-container">$J_n (x)$</span> and then manipulating with the product but it doesn't seem to give me something similar with the series on my LHS. I don't know if there is another different approach which I must follow. </p>
| Jean Marie | 305,862 | <p>Three methods :</p>
<p>1) By using the general (recurrence) formula :</p>
<p><span class="math-container">$$J_{\nu-1}(x)+J_{\nu+1}(x)=\frac{2 \nu}{x}J_{\nu}(x)$$</span></p>
<p>(formula 2.4 p. 13 of this excellent <a href="https://www.math.ust.hk/~machiang/150/Intro_bessel_bk_Nov08.pdf" rel="nofollow noreferrer">document</a>)</p>
<p>Taking <span class="math-container">$\nu=\tfrac12$</span>:</p>
<p><span class="math-container">$$J_{3/2}(x)=\frac{1}{x}J_{1/2}(x)-J_{-1/2}(x)$$</span></p>
<p>Knowing that (formulas (2.16) of the document):</p>
<p><span class="math-container">$$\begin{align}
J_{1/2}(x) = \sqrt{\frac{2}{\pi x}} \sin{(x)}\\
J_{-1/2}(x) = \sqrt{\frac{2}{\pi x}} \cos{(x)}
\end{align}$$</span></p>
<p>we get your result. I assume that <span class="math-container">$\phi$</span> is a typo for <span class="math-container">$\pi$</span>... </p>
<p>2) By using Laplace Transform. Let us prove the equivalent formula :</p>
<p><span class="math-container">$$\sqrt{\frac {\pi}{2}} x^{3/2} J_\frac 32 (x) = \sin x - x \cos x \tag{1}$$</span></p>
<p>As indicated for example in this <a href="http://eqworld.ipmnet.ru/en/auxiliary/inttrans/laplace8.pdf" rel="nofollow noreferrer">table</a>, </p>
<p><span class="math-container">$$\mathfrak{L}{(x^{\nu}J_{\nu}(x))}=2^{\nu+1}\dfrac{1}{\sqrt{\pi}}\Gamma(\nu+\tfrac12)(s^2+1)^{-\nu-\tfrac12}$$</span></p>
<p>Knowing the Laplace Transforms:</p>
<p><span class="math-container">$$\mathfrak{L}(\sin x)=\frac{1}{s^2+1} \ \ \text{and} \ \ \mathfrak{L}(x \cos x)=\frac{s^2-1}{(s^2+1)^2},$$</span></p>
<p>it is easy to conclude to the exactness of (1).</p>
<p>3) by taking <span class="math-container">$n=1$</span> in the (rather classical) formula :</p>
<p><span class="math-container">$J_{p+{\frac{1}{2}}}(x)=\sqrt{\dfrac{2}{\pi}}
(-1)^p x^{p+{\frac{1}{2}}}
\left(\frac{1}{x}\frac{\mathrm d}{\mathrm dx}\right)^p\left(\frac{\sin x}{x}\right)\tag{2}$</span></p>
<p>(see <a href="https://math.stackexchange.com/q/417110">this answer</a>, knowing that the <strong>spherical Bessel</strong> function of order <span class="math-container">$n$</span> is defined by :</p>
<p><span class="math-container">$$j_n(x):=\sqrt{\dfrac{\pi}{x}}J_{n+\tfrac12}(x)$$</span></p>
<p>(please note the lowercase <span class="math-container">$j$</span>).</p>
<p><a href="https://i.stack.imgur.com/IMeXR.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IMeXR.jpg" alt="enter image description here"></a></p>
<p><em>Fig. 1 : Blue curve : <span class="math-container">$y=$$\text{sinc}$$(x)$</span>, red curve : <span class="math-container">$y=\cos(x)$</span>, magenta curve : <span class="math-container">$y=\sqrt{\frac12 \pi x} J_\frac 32 (x)$</span></em>
<strong>Remarks :</strong> </p>
<p>1) The roots <span class="math-container">$r_k$</span> of Bessel functions in general are important. Here, the roots of <span class="math-container">$J_{3/2}(x)=0$</span>, are (except <span class="math-container">$0$</span>) the same as the roots of:</p>
<p><span class="math-container">$$\tan x = x \tag{3}$$</span></p>
<p>These roots have a nice property (please note that we deal only with positive roots):</p>
<p><span class="math-container">$$\sum_{k=1}^{\infty} \dfrac{1}{r_k^2}=\dfrac{1}{10}.\tag{4}$$</span></p>
<p>See proofs of (4) <a href="https://math.stackexchange.com/q/75206">here</a> or <a href="https://math.stackexchange.com/q/851035">here</a>.</p>
<p>2) Connected : the second and third integral of <a href="https://math.stackexchange.com/q/3442741">this question</a>.</p>
|
1,023,575 | <p>How would one factor a number, say $9+4\sqrt{2}$ in $\mathbb{Z}[\sqrt{2}]$?</p>
<p>This is what I've attemped to do:
$$(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2}) $$
$$a_1a_2+a_1b_2\sqrt{2}+a_2b_1\sqrt{2}+2b_1b_2$$
Thus,
\begin{eqnarray}
a_1a_2+2b_1b_2&=&9 \\
a_1b_2+a_2b_1 &=& 4.
\end{eqnarray}</p>
<p>But this results in 4 variables and only 2 equations.</p>
| laerne | 192,873 | <p>Not all numbers can be factored in $\mathbb{Z}[\sqrt{2}]$, it has also $\mathbb{Z}[\sqrt{2}]$-prime numbers, like $3$ or $\sqrt{2}$. Besides, other numbers have many factorisation. For example numbers of $\mathbb{z}$ like $60$ which have at least all the $\mathbb{Z}$ like factorisation, like $10\cdot6$ or $12\cdot5$.
So there's no unique solution. There might even be no non-trivial solution.</p>
<p>You have two $\mathbb{Z}$-degree of freedom too much. Try to proceed carefully, and deduce the value of two variables from the value of two others.
You'd probably have to check existence condition to avoid divide by $0$ and since your equations are quadratic, you'll likely end up with square roots.
So basically you're be in a problem of finding whether some formulas can be made to be a integer square.</p>
<p><em>Sidenote</em>: For example if you where in $\mathbb{Z}$, you're basically trying to solve something like $a \cdot b = 17$. It's not always possible to solve non-trivially and sometimes there are many solutions, like in $a \cdot b = 60$.</p>
<p><em>Remark</em>: You may want to do a prime factorization. In which case you have to compute the prime factor first, and would probably be better off asking another question.</p>
|
211,705 | <p>I am given a table of possible <span class="math-container">$X_1$</span> and <span class="math-container">$X_2$</span> values that can be generated in a casino. In the game, both are generated with each turn.</p>
<p><img src="https://i.stack.imgur.com/G0nLn.jpg" alt="enter image description here" /></p>
<blockquote>
<p>The questions asks me to determine the minimum fee that should be charged per turn so that that casino doesn't lose money, if the payouts are:</p>
<p><span class="math-container">$a) \ 8X_1$</span></p>
<p><span class="math-container">$b) \ 4X_1 + 8(X_2)^2-\frac{51}{128}$</span></p>
<p><span class="math-container">$c) \ 8X_1X_2$</span></p>
</blockquote>
<p>For a), I added up the total possible odds for each value <span class="math-container">$X_1$</span> could take, <span class="math-container">$$ 0\cdot\frac{2}{16} + 1\cdot\frac{4}{16}+2\cdot\frac{6}{16} + 3\cdot\frac{4}{16} = \frac{7}{4}$$</span>
and determined that the casino would need to charge at least <span class="math-container">$\$1.75$</span> per turn.</p>
<p>For b) I tried the pretty much the same thing, only breaking it down into individual situations, for example; <span class="math-container">$$P\{X_1=0, X_2 =1\} = \frac{1}{16}$$</span> <span class="math-container">$$P\{X_1=0, X_2 =2\} = \frac{1}{16}$$</span> <span class="math-container">$$etc...$$</span></p>
<blockquote>
<p>My question is, is there a faster, more efficient way to solve these types of problems? What if there were many more possible outcomes?</p>
</blockquote>
| Ross Millikan | 1,827 | <p>For a) you need to multiply by $8$, as you are paying $8X_1$</p>
<p>For b) you are doing the right thing. The only way to simplify it is to find some pattern. For example, in c) if $X_1$ and $X_2$ were independent (they are not) you could just multiply the expectations.</p>
|
3,440,873 | <p>I do not how to solve this, can such equation even exist? For the root to lie on the y intercept, the line would have to pass through origin, which means one root will be 0, breaking down the whole the thing. Am I missing something here?</p>
| Dietrich Burde | 83,966 | <p>The German wikipedia entry <a href="https://de.wikipedia.org/wiki/Intervall_(Mathematik)" rel="nofollow noreferrer">here</a> explains the notations of open, half-open, closed intervals. It agrees with the <a href="https://en.wikipedia.org/wiki/Interval_(mathematics)" rel="nofollow noreferrer">English entry</a>.</p>
<p><span class="math-container">\begin{align}
{\color{Red}(} a,b{\color{Red})} = \mathopen{\color{Red}]}a,b\mathclose{\color{Red}[} &= \{x\in\Bbb R\mid a\mathbin{\color{Red}<}x\mathbin{\color{Red}<}b\}, \\{}
{\color{DarkGreen}[}a,b{\color{Red})} = \mathopen{\color{DarkGreen}[} a,b\mathclose{\color{Red}[} &= \{x\in\Bbb R\mid a\mathbin{\color{DarkGreen}\le} x\mathbin{\color{Red}<}b\}, \\{}
{\color{Red}(} a,b{\color{DarkGreen}]} = \mathopen{\color{Red}]}a,b\mathclose{\color{DarkGreen}]} &= \{x\in\Bbb R\mid a\mathbin{\color{Red}<}x\mathbin{\color{DarkGreen}\le} b\}, \\{}
{\color{DarkGreen}[}a,b{\color{DarkGreen}]} = \mathopen{\color{DarkGreen}[} a,b\mathclose{\color{DarkGreen}]} &= \{x\in\Bbb R\mid a\mathbin{\color{DarkGreen}\le} x\mathbin{\color{DarkGreen}\le} b\}.
\end{align}</span></p>
|
1,159,860 | <p>If $$f:[a,b]\times [c,d] \to \mathbb{R}$$ is continuous and $f_{y}$ is continuous, let $$F(x,y)=\int_{a}^{x} f(t,y)dt.$$ </p>
<ol>
<li>Find $F_x$ and $F_y$.</li>
<li>If $G(x)=\int_{a}^{g(x)}f(t,x)dt$, find $G'(x)$</li>
</ol>
<p>My try: </p>
<p>For (1) $$F(x+h,y)-F(x,y)=\int_{a}^{x+h} f(t,y)dt-\int_{a}^{x}f(t,y)dt=\int_{x}^{x+h}f(t,y)dt$$
Let $\int f(t,y)dt= H(t,y)$. Then $\int_{x}^{x+h}f(t,y)dt=H(x+h,y)-H(x,y)$. Hence $$F_x= \frac {\partial H(x,y)}{\partial x} $$. I am kind of stuck here.</p>
<p>$F_{y}$ is easy. $$F(x,y+h)-F(x,y)=\int_{a}^{x}\left(f(t,y+h)-f(t,y)\right) dt$$
Hence $$F_{y}=\int_{a}^{x}\frac{\partial f}{\partial y} dt $$</p>
<p>For (2) we have $$G(x+h)-G(x)=\int_{a}^{g(x+h)}f(t,x+h)dt-\int_{a}^{g(x)} f(t,x)dt$$
$$=\int_{a}^{g(x+h)}f(t,x+h)dt-\int_{a}^{g(x)} f(t,x+h) dt+ \int_{a}^{g(x)} f(t,x+h) dt-\int_{a}^{g(x)} f(t,x)dt$$
$$=\int_{g(x)}^{g(x+h)} f(t,x+h) dt+\int_{a}^{g(x)} \left(f(t,x+h)-f(t,x)\right)dt$$</p>
<p>If I let $\int f(t,x+h) dt= H(t,x+h)$. Then we have
$$ \frac{G(x+h)-G(x)}{h}=\frac{H(g(x+h),x+h)-H(g(x),x+h)}{g(x+h)-g(x)}. \frac{g(x+h)-g(x)}{h}+\int_{a}^{g(x)}\frac{ \left(f(t,x+h)-f(t,x)\right)}{h}dt$$</p>
<p>Taking limit on both sides we have $$G'(x)=H'(g(x),x+h).g'(x)+\int_{a}^{g(x)}\frac{ \partial f}{\partial x}dt$$</p>
<p>From here How do I get thr result??</p>
<p>Thanks for the help!!</p>
| JMP | 210,189 | <p>$P(A-B)$ means the probability that $A$ happens and $B$ doesn't. As $P(A)=a$ and $P(¬B)=1-b$, the answer is $a(1-b)$.</p>
|
3,668,101 | <p>I know that if <span class="math-container">$n \bmod k \le k-1$</span> then this sum is converge then it has finite sum, I just guess it's <span class="math-container">$\ln(k)$</span> because when <span class="math-container">$k=1$</span> sum is <span class="math-container">$0=ln(1)$</span>. I really don't know how to find it. Please help me.</p>
| HOANXA | 760,747 | <p>My idea is:</p>
<p><span class="math-container">$\sum \frac{nmodk}{n(n+1)}=\frac{1}{1.2}+\frac{2}{2.3}+...+\frac{k-1}{(k-1)k}+...=\frac{1}{1}-\frac{1}{2}+\frac{2}{2}-\frac{2}{3}+...$</span>
<span class="math-container">$=(\frac{\:1}{1\:}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k-1}+\frac{k-1}{k})+(\frac{\:1}{k+1\:}+\frac{1}{k+2}+\frac{1}{k+3}+...+\frac{1}{2k-1}+\frac{k-1}{2k})+...$</span>
<span class="math-container">$=\sum _{n=1}^{\infty }\left(\frac{1}{\left(n-1\right)k+1}+\frac{1}{\left(n-1\right)k+2}+...+\frac{1}{\left(n-1\right)k+k-1}-\frac{k-1}{\left(n-1\right)k+k}\right)$</span></p>
<p>But it still not work</p>
|
2,346,489 | <p>I am studying the Proposition: Let $D$ be a Dedekind domain, $F$ its field of fractions, $E$ a finite dimensional extension field of $F$ and $D'$ the subring of $E$ of $D$ integral elements. Assume that $E/F$ is a finite separable field extension. Then $D'$ is a finitely generated $D$-module.</p>
<p>I have to show that $D'$ is Noetherian. It is clear from the Proposition that $D'$ is a Noetherian $D$ module. Why does it follow that $D'$ is Noetherian as $D'$ module?</p>
<p>Would you help me, please?
Thank you in advance.</p>
| Community | -1 | <p>If I understand correctly your question, here is a sort of answer that you might like.</p>
<p><strong>Hints and facts</strong>:</p>
<p>Because of the situation we have (which is really good in fact since the rings we have very <em>good</em> structures) the extension $D \subset D^{\prime}$ is integral, and $D$ is normal (as Dedekind domain is integrally closed). Hence we have very good correspondence between the prime ideals inside $D$ and primes inside $D^{\prime}$ respectively. Commonly in bibliography are referred as <strong>Going Down, Going up</strong> theorems (if you haven't heart them so far is a good chance to meet them because are one of the basics in commutative algebra, otherwise skip them) and you can find them in <em>any</em> book of commutative algebra, included this online notes <a href="http://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/main.pdf" rel="nofollow noreferrer">http://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/main.pdf</a> (page 89).
So what this correspondence establishes, is that prime ideals in $D^{\prime}$ are giving prime ideals in $D$ and vice versa. In a Dedekind domain every ideal is a product of primes, and in <em>commutative case only</em>, finite product of finitely generated ideals is finitely generated ideal, so in order to prove your question it's enough to restrict our attention to prime ideals in $D^{\prime}$, since the latter is Dedekind domain too (is integrally closed by its definition, Noetherian domain and since our extension $D \subset D^{\prime}$ is integral has Krull dimension $1$ as well). Now from all the above facts your question maybe is more clear now. Instead of writing out explicitly the proof (which is kind of long-scale in fact) it's good I think to work it out alone and if you have any queries please do let me know!</p>
<p>The above result you can find it out as <strong>Krull-Akizuki</strong> theorem in many textbooks. (also a short proof can be found on wiki article too)</p>
|
2,136,411 | <p>According to the Theorem 12.7 of the book Analytic Nymber Theory by Apostol, $$\zeta(1-s) = 2(2\pi)^{-s} \Gamma(s) \cos \big(\frac{\pi s}{2}) \zeta(s)$$ which results in (as the book also says) that $\zeta(-2n) =0$ for $n=1,2,3, \dots$, the so-called trival zeros of $\zeta(s)$. </p>
<p>But how on earth $\zeta(-2n) = \sum_{i=1}^{\infty} \frac{1}{i^{-2n}}= \sum_{i=1}^{\infty} i^{2n}=\infty=0$? </p>
| Start wearing purple | 73,025 | <p>In the same spirit, we have
$$2^0+2^1+2^2+2^3+\ldots =-1.$$
The seeming paradox is that the sum on the left is <em>defined</em> as the analytic continuation of the series $\sum_{k=0}^{\infty}z^k=\frac{1}{1-z}$ outside its original domain of convergence $|z|<1$.</p>
|
3,258,642 | <blockquote>
<p>If the roots of quadratic equation <span class="math-container">$$x^2 − 2ax + a^2 + a – 3 = 0$$</span>
are real and less than <span class="math-container">$3$</span>, find the range of <span class="math-container">$a$</span>.</p>
</blockquote>
<p>The roots are <span class="math-container">$a \pm \sqrt {3 – a}$</span></p>
<p>For the roots to be real, we must have a < 3.</p>
<p>Also, for the roots to be less than 3, we must have <span class="math-container">$\pm \sqrt {3 – a } \lt 3 – a $</span></p>
<p>If squaring both sides is allowable, I will get <span class="math-container">$(a – 2)(a – 3) > 0$</span>. Then the problem is solved.</p>
<p>The question is:- how to convince others that the squaring of both sides of <span class="math-container">$\pm \sqrt {3 - a } \lt 3 - a $</span> is allowable?</p>
| Bernard | 202,857 | <p>You don't have to calculate the roots, nor square them – only use high-school theorems on the sign of quadratic polynomials:</p>
<ol>
<li>The polynomial <span class="math-container">$p(x)=x^2 − 2ax + a^2 + a – 3$</span> has <em>real</em> roots, so its reduced discriminant is non-negative:
<span class="math-container">$$\Delta'=a^2-(a^2+a-3)=3-a \ge 0,\text{ i.e. } a\le 3.$$</span></li>
<li><span class="math-container">$3$</span> has to be <em>outside</em> the interval of the roots <span class="math-container">$x_1, x_2$</span>: this means <span class="math-container">$p(3)=a^2-5a+6>0$</span>. As the roots of <span class="math-container">$a^2-5a+6$</span> are <span class="math-container">$2$</span> and <span class="math-container">$3$</span>, we must have <span class="math-container">$a<2\quad\text{or}\quad a>3$</span>.</li>
<li>From condition 2, we know that either <span class="math-container">$3< x_1<x_2$</span> or <span class="math-container">$x_1<x_2<3$</span>. In other words either both roots are greater than <span class="math-container">$3$</span> or both are less. To ensure that we're in the latter case, it is enough to ensure their arithmetic mean is less than <span class="math-container">$3$</span>. Now
<span class="math-container">$$\frac{x_1+x_2}2=a$$</span>
so the condition is <span class="math-container">$a<3$</span>.</li>
</ol>
<p>Combining all these conditions, we obtain <span class="math-container">$\;a\in \color{red}{(-\infty,2)}$</span>.</p>
|
3,275,732 | <p>How can I solve it without using matrix? I tried it to solve by using systems. But I have no idea how deal with "<span class="math-container">$0$</span>"</p>
| awkward | 76,172 | <p>Since we have the values of the polynomial at successive integers (-1,0,1 and 2), one way to find the interpolating polynomial is to use a table of finite differences:
<span class="math-container">$$\begin{matrix}
-2\\
&3\\
1 & &-2\\
&1 & &8\\
2 & &6\\
&7\\
9
\end{matrix}$$</span>
Each column in the table is derived from the the previous column by applying the forward difference operator <span class="math-container">$\Delta(x) = f(x+1) - f(x)$</span>. We can then plug the differences on the upper diagonal into <a href="http://mathworld.wolfram.com/NewtonsForwardDifferenceFormula.htmlhttp://mathworld.wolfram.com/NewtonsForwardDifferenceFormula.html" rel="nofollow noreferrer">Newton's Forward Difference Formula</a>:
<span class="math-container">$$f(s) = -2 + (3) s + (-2)(1/2!) s(s-1) + 8 (1/3!)s(s-1)(s-2)$$</span>
But the Newton formula assumes the series of <span class="math-container">$s$</span> values starts at <span class="math-container">$0$</span>, whereas our data starts at <span class="math-container">$-1$</span>, so we need to make the substitution <span class="math-container">$s=x+1$</span> in order to fit the data given:
<span class="math-container">$$f(x) = -2 + (3) (x+1) + (-2)(1/2!)(x+1)x + 8(1/3!) (x+1)x(x-1)$$</span></p>
<p>Expanding this polynomial will yield the coefficients <span class="math-container">$a,b,c$</span> and <span class="math-container">$d$</span>.</p>
|
1,860,459 | <blockquote>
<p>Prove that $4k < 2^k$ by induction.</p>
</blockquote>
<p>It holds for $k = 5$. Assume $ k = n + 1 $. Then</p>
<p>$4(n+1) < 2^{(n+1)}$</p>
<p>$4n + 4 < 2^n * 2$</p>
<p>$2n + 2 \leq 2^n$</p>
<p>Now I just need to show that</p>
<p>$2n + 2 \leq 4n$</p>
<p>$n + 1 \leq 2n$</p>
<p>$1 \leq n$</p>
<p>And because I chose $n = 5$ which is greater than $1$, this should prove that the formula holds for $n \geq 5$.</p>
<p>Is this correct?</p>
| Researcher314 | 165,956 | <p>You got the basis step correct by checking that $4n<2^n$ for $n=5$.</p>
<p>Next, you must prove that $4N<2^N \implies 4(N+1) < 2^{N+1}$ for $N\geq5$</p>
<p>When trying to solve this problem, simplify the right-hand-side of this implication to a form that is easily comparable to the left-hand-side.</p>
<p>Notice that $4(N+1) < 2^{N+1}$ holds if and only if </p>
<p>$4N+4 < 2^N *2$</p>
<p>Then you might notice that the inductive hypothesis gives us that </p>
<p>$4N+4 < 2^N + 4$ (since we are assuming $4N < 2^N$).</p>
<p>These two formulae look similar, so we are almost done.</p>
<p>Now we need only show that $2^N + 4 < 2^N * 2 $</p>
<p>Notice $2^N * 2 = 2^N + 2^N $.</p>
<p>Can you now show that $2^N + 4 < 2^N + 2^N$ for $N \geq 5$?</p>
<p>Do that, and you're done! Hooray for you!</p>
|
1,860,459 | <blockquote>
<p>Prove that $4k < 2^k$ by induction.</p>
</blockquote>
<p>It holds for $k = 5$. Assume $ k = n + 1 $. Then</p>
<p>$4(n+1) < 2^{(n+1)}$</p>
<p>$4n + 4 < 2^n * 2$</p>
<p>$2n + 2 \leq 2^n$</p>
<p>Now I just need to show that</p>
<p>$2n + 2 \leq 4n$</p>
<p>$n + 1 \leq 2n$</p>
<p>$1 \leq n$</p>
<p>And because I chose $n = 5$ which is greater than $1$, this should prove that the formula holds for $n \geq 5$.</p>
<p>Is this correct?</p>
| Bill Dubuque | 242 | <p><strong>Hint</strong> $ $ Equivalently we seek to prove that $\,f(n) = 2^n/(4n) > 1\,$ for all $\,n\ge 5.$</p>
<p>Note $\,f(5)>1\,$ and $\,f(n\!+\!1)/f(n) = 2n/(n\!+\!1) \ge 1\,$ for $\,n\ge 5\,$ so $\,\color{#c00}{f(n\!+\!1) \ge f(n)}\,$ </p>
<p>Hence the induction reduces to a trivial one: $ $ an $\rm\color{#c00}{increasing}$ sequence stays $\,\ge\, $ its initial value. From this view, the induction step becomes obvious, boiling down to transitivity of $\,\ge,\,$ i.e.</p>
<p>$$\begin{align} f(n)\ge f(5)\,&\Rightarrow\, \color{#c00}{f(n\!+\!1)\ge f(n)} \ge f(5)\\[0.3em]
{\rm i.e.}\quad P(n)\,&\Rightarrow\,P(n\!+\!1)\end{align}$$ </p>
<p><strong>Remark</strong> $ $ This is not an ad-hoc trick. Rather, it is a special case of a general method of transforming such problems into a simpler form where the inductive step is more obvious. It is a special case of <a href="https://math.stackexchange.com/search?q=user%3A242+telescopy+multiplicative">multiplicative telescopy.</a> Follow the link for many further examples. </p>
<p>Note that once you prove by induction that result about increasing sequences, you can invoke the result as a Lemma for other induction problems of this type (which are quite common, as you can see from the links).</p>
|
1,860,459 | <blockquote>
<p>Prove that $4k < 2^k$ by induction.</p>
</blockquote>
<p>It holds for $k = 5$. Assume $ k = n + 1 $. Then</p>
<p>$4(n+1) < 2^{(n+1)}$</p>
<p>$4n + 4 < 2^n * 2$</p>
<p>$2n + 2 \leq 2^n$</p>
<p>Now I just need to show that</p>
<p>$2n + 2 \leq 4n$</p>
<p>$n + 1 \leq 2n$</p>
<p>$1 \leq n$</p>
<p>And because I chose $n = 5$ which is greater than $1$, this should prove that the formula holds for $n \geq 5$.</p>
<p>Is this correct?</p>
| fleablood | 280,126 | <p>You made lots of little mistakes but your biggest mistake is that you inductions step, doesn't actually do any inducing.</p>
<p>A proper induction step always goes like this:</p>
<p>=======</p>
<p>We assume that something is true for a specific $k = n$</p>
<p>$\implies$</p>
<p>Then something is true for $k = n+1$.</p>
<p>======</p>
<p>Thus, we start with showing it is true for $k= $ base case; thus we can conclude it is true of $k = $ base case $ + 1$; thus we can conclude it is true for $k = $ base case $ + 2$; thus we con conclude it is true for $k = $ base case $ + 3$; the we can....</p>
<p>... and so on. We can conclude it is true for all possible numbers $k$.</p>
<p>But your induction step didn't try to conclude something for $k = n + 1$ from something for $k = n$. </p>
<p>Instead you tried to assume something was true for $k = n+1$ and then show if it's true for $k = n + 1 $ then $n \ge 1$ and therefor it must be true for all numbers.</p>
<p>This is wrong. 1) your logic is backwards. You can't start at a conclusion and get a basic result and assume your conclusion is true. That just isn't logical. But more importantly 2) your induction doesn't build forward step by step. You didn't do <em>anything</em> to do with the statement being true for $k = n$.</p>
<p>THat's your big mistake. Your small mistakes are:</p>
<p>a) In stating that you are trying to prove $4k < 2^k$ you didn't state the <em>essential</em> condition that $k \ge 5$. This just isn't true if $k = 1,2,3,4$.</p>
<p>b)Your logic can't start with a conclusion, get the result $n \ge 1$ and assume you conclusion is true. </p>
<p>Consider this: Assume $25 > 36$. Then as $47 > 25$ I get $47 > 25> 36$ so $47 > 36$. So $11 > 0$. So $1 > 0$. That is a true statement so $25 > 36$.</p>
<p>I hope you see why that is bull#*@%.</p>
<p>c) You claimed $2n+2 \le 2^n$ when you meant $2n+2 < 2^n$.</p>
<p>d) You claimed "all we need to show is $2n + 2 \le 4n$". That isn't true. If $2n+ 2 \le 4n$ and $2n+2 \le 2^n$ we can not conclude anything about how $4n$ relates to $2^n$. Maybe $4n < 2^n$ maybe $4n > 2^n$ or maybe $4n = 2^n$.</p>
<p>To reach that conclusion you have to show that $4n \le 2n + 2$ and therefore as $2n+2 < 2^n$ we can conclude by transitivity that $4n < 2^n$. </p>
<p>But we can't prove $4n \le 2n+2$ ... because it isn't. Or proof was flawed from the beginning because the logic of mistake b) doesn't work.</p>
<p>So let's fix this:</p>
<p>=====</p>
<p>Prove $4k < 2^k$ for all $k \ge 5$.</p>
<p>Well it is true for $k = 5$ as $4*5 = 20 < 32 = 2^5$.</p>
<p>Induction step:</p>
<p>We assume it is true for some $k = n$. (This is okay because we know it is true for $k = n = 5$.)</p>
<p>So $4n < 2^n$</p>
<p>We need to find steps </p>
<p>$\implies.....$</p>
<p>$\implies.....$</p>
<p>$\implies.....$ $4(n+1) < 2^{n+1}$.</p>
<p>Can we do that?</p>
<p>Well,</p>
<p>$4n < 2^n$</p>
<p>$4n + 4 < 2^n + 4$</p>
<p>$4n + 4 < 2^n + 4 \le 2^n + 4n$ (because $n \ge 1$ so $4n \ge 4$)</p>
<p>$4n + 4 < 2^n + 4n < 2^n + 2^n$.</p>
<p>$4(n+1) < 2*2^n = 2^{n+1}$</p>
<p>and we have proven the induction step.</p>
<p>Thus we can conclude as it is true for $k = 5$, it is true for $k = 6$. And as it is true for $k = 6$ it is true for $k = 7$, and so on. By <em>induction</em> we have shown it is true for <em>all</em> $k \ge 5$.</p>
|
4,071,619 | <blockquote>
<p>There are two German couples, two Japanese couples and one unmarried person. If all 9 persons are two be interviewed one by one then the total number of ways of arranging their interviews such that no wife gives an interview before her husband is?</p>
</blockquote>
<p>I tried using the string method, but that will be only counting the cases where the wife speaks just after her husband.</p>
<p>There are too many elements to take care of, unlike a similar question which involved only two people, so by symmetry the answer is half of the total number of ways of arranging those <span class="math-container">$n$</span> people.</p>
<p>Then how is this one solved</p>
| John Omielan | 602,049 | <p>You can use induction to prove</p>
<p><span class="math-container">$$\frac{10^n-1}{9n} \tag{1}\label{eq1A}$$</span></p>
<p>is an integer when <span class="math-container">$n = 3^k$</span>. The base case of <span class="math-container">$k = 0$</span> gives <span class="math-container">$3^k = 1$</span>, with \eqref{eq1A} becoming <span class="math-container">$\frac{10-1}{9} = 1$</span>. Next, assume \eqref{eq1A} is an integer for <span class="math-container">$k = m$</span> for some integer <span class="math-container">$m \ge 0$</span>. Then, for <span class="math-container">$k = m + 1$</span>,</p>
<p><span class="math-container">$$\begin{equation}\begin{aligned}
\frac{10^{3^{m+1}} - 1}{9(3^{m+1})} & = \frac{\left(10^{3^{m}}\right)^3 - 1}{\left(9(3^{m})\right)3} \\
& = \frac{(10^{3^{m}} - 1)(10^{3^{2m}} + 10^{3^{m}} + 1)}{\left(9(3^{m})\right)3} \\
& = \left(\frac{10^{3^{m}} - 1}{9(3^{m})}\right)\left(\frac{10^{3^{2m}} + 10^{3^{m}} + 1}{3}\right)
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$</span></p>
<p>The first factor above is an integer based on the induction hypothesis, i.e., \eqref{eq1A} being an integer for <span class="math-container">$n = 3^{m}$</span>. For the second factor, note</p>
<p><span class="math-container">$$10 \equiv 1 \pmod{3} \implies 10^{3^{2m}} \equiv 10^{3^{m}} \equiv 1 \pmod{3} \tag{3}\label{eq3A}$$</span></p>
<p>This means</p>
<p><span class="math-container">$$10^{3^{2m}} + 10^{3^{m}} + 1 \equiv 3 \equiv 0 \pmod{3} \implies 3 \mid 10^{3^{2m}} + 10^{3^{m}} + 1 \tag{4}\label{eq4A}$$</span></p>
<p>i.e., the second factor is also an integer. With the product of <span class="math-container">$2$</span> integers being an integer, \eqref{eq2A} shows \eqref{eq1A} is an integer for <span class="math-container">$n = 3^{m+1}$</span>. This proves by induction that \eqref{eq1A} is an integer for <span class="math-container">$n = 3^k$</span> for all integers <span class="math-container">$k \ge 0$</span>.</p>
|
551,662 | <p>I am reading "What Is Mathematics? An Elementary Approach to Ideas and Methods"
And I am stuck here, I don't get it. I have posted a screen shot underlining what my doubt is..</p>
<p>I dont get it when the author says while the pythagoras theorem is : $a^2 + b^2 = c^2$
and then he says $x=a/c$ and $y=b/c$
and then the equation should be according to me , $ax+by=c$.. right??
but the author writes y^2 = (1-x)(1+x)</p>
<p>Which I think may have came from something like
$x^2 + y^2 = 1$
$y^2= 1^2 + x^2$
$y^2= (1+x)(1-x)$ (since $a^2 + b^2 = (a+b)(a-b)$ )</p>
<p>but I dont get it where did that x^2 + y^2 = 1 came from ??
Is it that author assumed that x=a/c and y=b/c and then stoped talking about pythagoras theorem and started talking on x^2 + y^2 = 1???</p>
<p>and then further he introduces a number t.. i don't get it how it got converted into y=t(1+x) and (1-x)=ty ???</p>
<p>Can please someone help?</p>
<p>also , while i was writing the question it clicked me that if:
x=a/c (i.e. opposite upon hypotenuse means x is <strong><em>sin</em></strong> )
y=b/c (i.e. adjacent upon hypotenuse means y is <strong><em>cos</em></strong>)
therefore x^2 + y^2 = 1 (sin^2 + cos^2 = 1)</p>
<p>but then if i assume i am right about the sin cos thing then how come the last step is derived ?? (the one in blue color underline and box)
also that as per me x is sin , but the formula in that book is of cos2x? when t=tanx??</p>
<p>so m lil confused .. if you want any more clean way of me asking my doubt then do tell me i will rephrase the entire question .. :)</p>
<p><img src="https://i.stack.imgur.com/QQIVM.png" alt="Image of my question asked above,, screenshot of the book"></p>
| Hagen von Eitzen | 39,174 | <p>You replaced one of the two $a$s making up $a^2$ with $cx$. Try replacing both (and similarly for the two $b$'s in $b^2$ with $cy$)</p>
|
3,997,968 | <p>I'm trying to figure out how to get the point x = 3 :
What's given here are the points S and G .
(Assuming the 2 angles are equal)
<a href="https://i.stack.imgur.com/COFMn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/COFMn.png" alt="enter image description here" /></a></p>
<p>Apparently, we can assume that the ball does not bounce off the x-axis. Instead, we can change the target with respect to the symmetrical point on the x-axis so that we can now look at the intersection of the x-axis and the line connecting the current and target point.</p>
<p>The Formula to get x=3 is below :</p>
<p><span class="math-container">$$\frac{S_x.G_y + G_x.S_y}{S_y+G_y}$$</span></p>
<p>Is there any explanation for this formula ?</p>
| Narasimham | 95,860 | <p>Slightly different notation for co-ordinates.</p>
<p>To arrive at that equation equate slopes of straight lines <span class="math-container">$(SI, IG2)$</span> because after reflection the three points now are in a straight line.</p>
<p>( Reflection law requires that incidence and reflected angles be equal).</p>
<p>It can be also derived by total time minimization using Fermat's Law for given light speed.</p>
<p><a href="https://i.stack.imgur.com/GLy41.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GLy41.png" alt="enter image description here" /></a></p>
|
147,742 | <p>Is this statement true?</p>
<blockquote>
<p>$L:V\to V$ is a linear map with eigenvalue (not necessarily the only one) $a$. Suppose $(L-aI)^{m+1}(v)=0$ where $m$ is the power of the $(x-aI)$ term in the minimal polynomial of $L$. Then $(L-aI)^m(v)=0$ also.</p>
</blockquote>
<p>Some thoughts: </p>
<p>So this is essentially saying that $(L-aI)[(L-aI)^m(v)]=0\implies (L-aI)^m(v)=0$. In other words the nullspace of $ (L-aI)$ is $\{0\}$. Therefore I don't think the statement is necessarily true. Is there a way of constructing an explicit example to disprove this?</p>
<p>Ah wait, I haven't used the fact that $m$ is the power of the term in the minimal polynomial. So this may yet be true...</p>
| DonAntonio | 31,254 | <p>But $\,L-aI\,$ cannot have nullspace equal to zero as it is a singular map: $$\exists\,0\neq v\in V\,\,s.t.\,\,Lv=av\Longleftrightarrow (L-aI)(v)=0\,$$ and this seems to disprove your idea.</p>
<p>Now, if $\,Tv=0\,$ for some lin. transf., then $\,T^n(v)=0\,,\,\forall n\in\mathbb{N}\,$ , so both $\,(L-aI)^m\,\,,\,\,(L-aI)^{m+1}\,$ vanish at $\,v\,$, but not that from one we can deduce the other.</p>
|
147,742 | <p>Is this statement true?</p>
<blockquote>
<p>$L:V\to V$ is a linear map with eigenvalue (not necessarily the only one) $a$. Suppose $(L-aI)^{m+1}(v)=0$ where $m$ is the power of the $(x-aI)$ term in the minimal polynomial of $L$. Then $(L-aI)^m(v)=0$ also.</p>
</blockquote>
<p>Some thoughts: </p>
<p>So this is essentially saying that $(L-aI)[(L-aI)^m(v)]=0\implies (L-aI)^m(v)=0$. In other words the nullspace of $ (L-aI)$ is $\{0\}$. Therefore I don't think the statement is necessarily true. Is there a way of constructing an explicit example to disprove this?</p>
<p>Ah wait, I haven't used the fact that $m$ is the power of the term in the minimal polynomial. So this may yet be true...</p>
| Simon Markett | 30,357 | <p>The statement is true. In general you have
$$Ker(L-aI)\subseteq Ker(L-aI)^2\subseteq\cdots $$
This sequence stabilises, i.e. there is an $m$ such that
$$Ker(L-aI)^i=Ker(L-aI)^{i+1}$$
for all $i\geq m$. This $m$ is precisely the $m$ which is the exponent of $x-a$ in the minimal polynomial. In particular you have
$$Ker(L-aI)^m=Ker(L-aI)^{m+1}.$$</p>
<p>You write that the kernel of $L-aI$ <em>restricted to</em> $Ker(L-aI)^m$ is zero. Indeed it is, but for trivial reasons, since the whole source is the zero space.</p>
<p>Edit: The sequence stabilises since the dimension is bounded from above by the dimension of $V$. To see that the $m$ for which this happens is precisely the power in the minimal polynomial, one has to carefully analise the proof of the Caley Hamilton Theorem. This is done in any basic book on linear algebra. The key idea is that there is a squence of spaces with decreasing dimension which finally will be the image of the characteristic polynomial evaluated at $L$. This shows that the image is trivial. One observes carefully that we may reduce the power of the linear coefficients precisely to $m$ to obtain the similar result for the minimal polynomial.</p>
|
1,229,227 | <p>Does it make sense to write $\frac{d}{dx}u(x,t)$ or can one only write $\frac{\partial}{\partial x}u(x,t)$?</p>
| hyperkahler | 188,593 | <p>Actually, the $\frac{dy}{dx}$ maybe considered as a quotient of two differentials, since $dy= y' \cdot dx$, whereas $\frac{\partial y}{\partial x}$ is an indivisible symbol.</p>
<p>According to this, it's convinient to write such type of equations $\frac{d(f(g(x))}{dx}= \frac{\partial{f}}{\partial{g}} \cdot \frac{\partial{g}}{\partial{x}}$</p>
|
1,229,227 | <p>Does it make sense to write $\frac{d}{dx}u(x,t)$ or can one only write $\frac{\partial}{\partial x}u(x,t)$?</p>
| GPerez | 118,574 | <p>For convention's sake, I'll pretend you asked about $\frac{d}{dt}$ and $\frac{\partial}{\partial t}$. It's the same exact question this way, and you'll see the reason for the change in notation afterwards.</p>
<p>Getting to the point, say you have a function of two variables $$u:\Bbb R^2\to\Bbb R^k\atop (x,t)\mapsto u(x,t)$$</p>
<p>For example, let $k=1$ and we have a scalar field (which we can think of as dependent on "position" $x$ and time $t$, if we wish). Now consider a real function $x(t)$ (forgive me for using the same letter $x$ for a completely new term). This determines a path through $\Bbb R^2: \gamma (t) =(x(t),t)$. We can compose the two functions to get a real function of $t$: $u\circ\gamma$. Now, this is a sort of liberal use of notation, but we can "identify" $u$ with $u\circ\gamma$. Now the derivative of $u$ is $$\frac{d}{dt}u = \frac{d}{dt}u(x(t),t) = \frac{\partial u }{\partial x}x'(t) + \frac{\partial u}{\partial t}$$</p>
<p>You can see for yourself that both $\frac{\partial}{\partial t}$ and $\frac{d}{dt}$ are applied to $u$, to mean different things. Again, this is artificious notation (at least, I haven't seen it formalized), but it's usually just comfortable to write this way, so many people do.</p>
|
503,358 | <p>I remember I saw this question somewhere in Lang's undergraduate real analysis.</p>
<blockquote>
<p>Given any real number $\ge0$, show that it has a square root.</p>
</blockquote>
| user66733 | 66,733 | <p>It depends on how you define a real number. If you use Dedekind cuts, then you should show that the set $\{ x \in \mathbb{Q}^+: x^2<2\}$ is a Dedekind cut. If you use Cauchy sequences to define a real number, you can prove that the the sequence that is obtained from Newton's method is Cauchy:</p>
<p>$$ a_{n+1} = \frac{a_n}{2} + \frac{1}{a_n}$$</p>
|
982,780 | <p>I have the following system of <span class="math-container">$M$</span> linear equations in <span class="math-container">$N$</span> unknowns.</p>
<p><span class="math-container">$$
\begin{bmatrix}
3 & 0 & 1 & 0 & -1 & -3 & 2\\
1 & 2 & 0 & 4 & 0 & 0 & -1\\
1 & 1 & 0 & 0 & -1 & -1 & -2\\
0 & 0 & 1 & 0 & -3 & -1 & 1 \\
\end{bmatrix}
\begin{bmatrix}
x_{1}\\
x_{2}\\
x_{3}\\
x_{4} \\
x_{5} \\
x_{6} \\
x_{7} \\
\end{bmatrix} =
\begin{bmatrix}
1\\
0\\
0\\
-1\\
\end{bmatrix}$$</span></p>
<p>Is there any algorithm for finding answers of this equations that <span class="math-container">${x_{i} \ge 0}$</span>?</p>
<p><strong>Comment</strong>: I just want that <span class="math-container">$x_i \ge 0$</span>.</p>
<p>It can change to</p>
<p><span class="math-container">$$
\begin{bmatrix}
1 & 0 & 0 & 0 & 2/3 & -2/3 & 1/3 & 2/3\\
0 & 1 & 0 & 0 & -5/3 & -1/3 & -7/3 & -2/3 \\
0 & 0 & 1 & 0 & -3 & -1 & 1 & -1 \\
0 & 0 & 0 & 1 & 2/3 & 1/3 & 5/6 & 1/6 \\
\end{bmatrix}
$$</span></p>
| Harald Hanche-Olsen | 23,290 | <p>I would try this: Given a family of closed subsets where any finite intersection is nonempty, add all such finite intersections to the family, so you have a family of nonempty, closed subsets which is closed under finite intersections. Any $F$ in the family has a smallest and a largest element. Consider the supremum of all the $\min F$, or the infimum of all the $\max F$. Either should be in the intersection of all $F$.</p>
|
982,780 | <p>I have the following system of <span class="math-container">$M$</span> linear equations in <span class="math-container">$N$</span> unknowns.</p>
<p><span class="math-container">$$
\begin{bmatrix}
3 & 0 & 1 & 0 & -1 & -3 & 2\\
1 & 2 & 0 & 4 & 0 & 0 & -1\\
1 & 1 & 0 & 0 & -1 & -1 & -2\\
0 & 0 & 1 & 0 & -3 & -1 & 1 \\
\end{bmatrix}
\begin{bmatrix}
x_{1}\\
x_{2}\\
x_{3}\\
x_{4} \\
x_{5} \\
x_{6} \\
x_{7} \\
\end{bmatrix} =
\begin{bmatrix}
1\\
0\\
0\\
-1\\
\end{bmatrix}$$</span></p>
<p>Is there any algorithm for finding answers of this equations that <span class="math-container">${x_{i} \ge 0}$</span>?</p>
<p><strong>Comment</strong>: I just want that <span class="math-container">$x_i \ge 0$</span>.</p>
<p>It can change to</p>
<p><span class="math-container">$$
\begin{bmatrix}
1 & 0 & 0 & 0 & 2/3 & -2/3 & 1/3 & 2/3\\
0 & 1 & 0 & 0 & -5/3 & -1/3 & -7/3 & -2/3 \\
0 & 0 & 1 & 0 & -3 & -1 & 1 & -1 \\
0 & 0 & 0 & 1 & 2/3 & 1/3 & 5/6 & 1/6 \\
\end{bmatrix}
$$</span></p>
| Andreas Blass | 48,510 | <p>If the conclusion fails, then there is, in $X$, a first element $\beta$ such that $]\leftarrow,\beta]$ is not compact. Let $C$ be an open cover of $]\leftarrow,\beta]$ with no finite subcover, and let $U$ be an element of $C$ that contains $\beta$. Of course, $U$ doesn't include all of $]\leftarrow,\beta]$, so, by definition of the topology, it includes $]\gamma,\beta]$ for some $\gamma<\beta$. By the minimality of $\beta$, finitely many elements of $C$ cover $]\leftarrow,\gamma]$. These finitely many together with $U$ cover $]\leftarrow,\beta]$; contradiction.</p>
|
1,282,419 | <p>Let $\Delta\subset\Bbb C$ be the open unitary disk. Let $\varphi:\Delta\to\Bbb R$ defined as follows: $\varphi(z)=1$ if $\Re z\ge0$, $\varphi(z)=0$ otherwise. So $\varphi$ is upper semicontinous.</p>
<p>In order to prove $\varphi$ is NOT subharmonic, I've to find a compact subset $K\Subset\Delta$ and a real valued function $h\in\mathcal{C}^0(K)\cap\mathcal H(\stackrel{\circ}{K})$ such that</p>
<ul>
<li>$\varphi\leq h$ on $\partial K$</li>
<li>$\varphi\nleq h$ on $\stackrel{\circ}{K}$</li>
</ul>
<p>Here $\mathcal H(\stackrel{\circ}{K})$ is the set of harmonic functions on the interior of $K$.</p>
<p>I tried for hours, but I did't find an answer.</p>
<p>The only thing which gave me a chance for one second was $h(z):=\log\left(\frac{|z|}{a}\right)+1$ on $K:=\bar\Delta_{0,a}$ which is the closure of the disk centered in $0$ of radius $a\in]0,1[$. This function would be right, the problem is it's not defined in $0$. So I tried cutting out a small circle around zero, but we get other obvious problems in satisfying the given conditions.</p>
| zhw. | 228,045 | <p>Another approach is through the mean value inequality: If $\varphi$ is subharmonic, then</p>
<p>$$1=\varphi(1/4) \le (1/2\pi)\int_0^{2\pi}\varphi(1/4 + (1/2)e^{it})\,dt.$$</p>
<p>But it's clear the last integral is $<1,$ contradiction.</p>
|
2,476,973 | <p>A fair six-sided die carries $1$ on one face, $2$ on two of its faces, and<br>
$3$ on the remaining three faces. </p>
<p>Suppose the die is rolled twice, and let $X$ be the random variable ’total score'. Find the probability distribution of $X$.</p>
| Graham Kemp | 135,106 | <p>The six-sided die has <code>1</code> on one face, <code>2</code> on two faces, and <code>3</code> on three faces. That gives the support and priobabilities for an individual roll.</p>
<p>The die is rolled twice, and the result of each roll added to give $T$. Thus let $T=T_1+T_2$, with $T_1,T_2$ being the individual die rolls, which are independent and identically distributed as above.</p>
<p>$$\therefore\qquad \mathsf P(T{=}t) ~=~\sum_{s=\max\{1,\,t-3\}}^{\min\{3,\,t-1\}} \mathsf P(T_1{=}s)\,\mathsf P(T_2{=}t{-}s) ~\mathbf 1_{t\in\{2,3,4,5,6\}}$$</p>
|
214,766 | <p>Is there an efficient way to check a number x and remove all prime factors in the number which are less than some n? For example for n = 200:</p>
<pre><code>x=88984589931961415442566827779929187431222364934742868664124547963532933
FactorInteger[x]
{{29, 2}, {31, 1}, {37, 2}, {269, 1}, {271,
1}, {34200471605536976187361939984030218061598132568100785528233,
1}}
</code></pre>
<p>After removing all prime factors < n from x gives:</p>
<pre><code>2493180179572040027082498062895818866472442266081979164222657467
</code></pre>
<p>I'd like to use as large n as possible and then use PrimeQ to check the remaining number, which is faster than checking for large prime factors.</p>
<p>I made this code which works but may be slow:</p>
<pre><code>x=2*53*6571*18313*31259
n=20000;
n=PrimePi[n];
listWithSmallPrimeFactorsRemoved={};
AppendTo[listWithSmallPrimeFactorsRemoved,x];
For[i=1,i<=n,i++,
z=Last[listWithSmallPrimeFactorsRemoved];
a=IntegerExponent[z,Prime[i]];
z=z/(Prime[i]^a);
AppendTo[listWithSmallPrimeFactorsRemoved,z];
]
CountDistinct[listWithSmallPrimeFactorsRemoved]-1 (*count of how many prime factors were removed*)
Last[listWithSmallPrimeFactorsRemoved] (*the remaining number after removing prime factors \[LessEqual] n*)
</code></pre>
<p>cheers,
Jamie</p>
| kglr | 125 | <p>The first list:</p>
<pre><code>listsa = GatherBy[Join @@ (Thread /@ list), Last];
Column[listsa]
</code></pre>
<p><a href="https://i.stack.imgur.com/b5H55.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/b5H55.png" alt="enter image description here"></a></p>
<p>The second list:</p>
<pre><code>listsb1 = Join @@@
GatherBy[Thread[Thread /@ list, List, {3, 5}] /. {{a_, b_}} :> {a, b}, #[[-1, -1]] &];
</code></pre>
<p>or</p>
<pre><code>listsb2 = TemporalData[Transpose@PadLeft[#[[All, -1]]], {#[[All, 1]]}]["Paths"] &@list;
listsb1 == listsb2
</code></pre>
<blockquote>
<p>True</p>
</blockquote>
<pre><code>Column[listsb1]
</code></pre>
<p><a href="https://i.stack.imgur.com/n2zbP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/n2zbP.png" alt="enter image description here"></a></p>
<p>The third:</p>
<pre><code>listsc = Join[##, 2] & @@ Map[List, Thread /@ list, {-2}];
Column[listsc]
</code></pre>
<p><a href="https://i.stack.imgur.com/HztQB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HztQB.png" alt="enter image description here"></a></p>
|
3,078,097 | <blockquote>
<p>Why it is impossible to split the natural numbers into two sets <span class="math-container">$A$</span> and <span class="math-container">$B$</span> such that for distinct elements <span class="math-container">$m, n \in A$</span> we have <span class="math-container">$m + n \in B$</span> and vice-versa?</p>
</blockquote>
<p>Also, does vice-versa means that there are distinct elements such that <span class="math-container">$x + y \in A$</span>? </p>
<p>How do I show the proof?</p>
| Julián C. Cano | 731,543 | <p><strong>Theorem [Schur, 1916]</strong>. For any partition of the set of positive integers in finite pieces, there are integers <span class="math-container">$x$</span> and <span class="math-container">$y$</span> with <span class="math-container">$x \neq y$</span> such that the set <span class="math-container">$\{x,y,x+y\}$</span> is contained in the
same piece.</p>
<p><em>Proof</em>. We define a <em><span class="math-container">$\Delta-$</span>set</em> as a set of natural numbers of the form <span class="math-container">$\{a_{i}-a_{j} \, | \, j<i<\omega\}$</span>, where <span class="math-container">$\{ a_{i} \}_{i<\omega}$</span> is an increasing sequence of natural numbers. </p>
<p>Fact: The set of positive integers is a <span class="math-container">$\Delta-$</span>set. </p>
<p>Let <span class="math-container">$A=\{a_{i}-a_{j} \, | \, j<i<\omega\}$</span> be any fixed <span class="math-container">$\Delta-$</span>set and let <span class="math-container">$c: A \rightarrow r$</span> be a finite coloration of <span class="math-container">$A$</span>. Now, let <span class="math-container">$\pi: [\mathbb{N}]^{2} \rightarrow r$</span> be the finite coloration of <span class="math-container">$[\mathbb{N}]^{2}$</span> given by <span class="math-container">$\pi (\{i,j\}) = c (a_{i}-a_{j})$</span> when <span class="math-container">$j<i$</span>. </p>
<p>Thus, the Ramsey's theorem implies that there exist <span class="math-container">$H \in [\mathbb{N}]^{\omega}$</span> such that <span class="math-container">$\pi$</span> is constant on <span class="math-container">$[H]^{2}$</span>. Therefore, for all <span class="math-container">$i,j,k \in H$</span> with <span class="math-container">$k<j<i$</span> we have that <span class="math-container">$\{i,j,k\}$</span> is monochromatic for <span class="math-container">$\pi$</span>, hence if <span class="math-container">$x=a_{i}-a_{j}$</span>, <span class="math-container">$y=a_{j}-a_{k}$</span> and <span class="math-container">$z=a_{i}-a_{k}$</span> then <span class="math-container">$x,y,z \in A$</span> are such that <span class="math-container">$z=x+y$</span> and <span class="math-container">$\{x,y,x+y\}$</span> is monochromatic for <span class="math-container">$c$</span>. <span class="math-container">$\; \; \;\square$</span></p>
<p><strong>Corollary</strong>. There are not sets <span class="math-container">$A \neq \emptyset$</span> and <span class="math-container">$B \neq \emptyset$</span> of positive integers, with <span class="math-container">$A \cap B = \emptyset$</span> and <span class="math-container">$A \cup B = \mathbb{N} - \{0\}$</span>, such that either <span class="math-container">$x+y\in B$</span> for all <span class="math-container">$x,y \in A$</span> with <span class="math-container">$x\neq y$</span> or <span class="math-container">$x+y\in A$</span> for all <span class="math-container">$x,y \in B$</span> with <span class="math-container">$x\neq y$</span>.</p>
|
25,853 | <p>With regard to an undergraduate statistics course, I am developing a standardized list of point deductions with the TAs (doctoral students) so that graders are consistent in what they are taking off intermediate points for. For example, most problems are 10 points total, and my proposed point deductions for intermediate math errors are (for example):</p>
<ul>
<li>-2 pts, erroneous +, - , *, /</li>
<li>-2 pts, erroneous sign, e.g. 3.02 instead of -3.02</li>
<li>-3 pts, failed to square, e.g. (x) instead of (x)^2</li>
<li>-3 pts, failed to take square root, e.g. (x) instead of sqrt(x)</li>
</ul>
<p>If, after grading, you discover on a particular exam that the final answers for five 10-point questions are incorrect because of only making a minor -2 point intermediate error, the student could conceivably obtain a score of 80% on an exam if they missed only -2 points per question (40/50).</p>
<p>However, in statistics, there is a contextual element to every question, not just solving for a numerical answer -- that is, in addition to the worked problem, students need to write a text-based response for the following:</p>
<ul>
<li>(2 pts) state whether the hypothesis test is significant or not</li>
<li>(2 pts) state whether the null hypothesis is rejected or accepted</li>
<li>(2 pts) state whether the p-value is less than 0.05 or not.</li>
</ul>
<p>So if there was only one minor (-2 point) intermediate error made, causing an incorrect final numerical answer, the student will also incorrectly respond to the final text-based answers (above) as well.</p>
<p>Thus, would you also take off e.g. -2 points for an incorrect final numerical answer, as well as -6 points for missing the final text-based sub-items listed above?</p>
<p>In other words, would you only deduct -2 points for a complex (multi-step) algebra or calculus question if only a minor intermediate step was erroneous, or would you also deduct for having an incorrect final numerical answer as well?</p>
<p>Maybe I could propose to the TAs to augment the point deduction list with:</p>
<ul>
<li>-1 pt, incorrect final numerical answer</li>
<li>-1 pt, state whether the hypothesis test is significant or not</li>
<li>-1 pt, state whether the null hypothesis is rejected or accepted</li>
<li>-1 pt, state whether the p-value is less than 0.05 or not.</li>
</ul>
| guest | 20,674 | <p>I advise being less intricate and put less load on the graders. I personally would go with all, half or zero credit for every question.</p>
<p>*All is correct answer (and some reasonable explication, not an essay, but also not a bare number.</p>
<p>*Half shows some decent knowledge of the process, but founders partway. Or has a "dumb mistake" on the algebra/arithmetic.</p>
<p>*And zero is for a mess.</p>
<p>After all there is more than one question on a test. More than one test in the course. And more than one course that the students take. Life is statistical after all, so these things even out.</p>
<p>Don't invest too much time in intricate grading. The cost/benefit is not worth it.</p>
|
25,853 | <p>With regard to an undergraduate statistics course, I am developing a standardized list of point deductions with the TAs (doctoral students) so that graders are consistent in what they are taking off intermediate points for. For example, most problems are 10 points total, and my proposed point deductions for intermediate math errors are (for example):</p>
<ul>
<li>-2 pts, erroneous +, - , *, /</li>
<li>-2 pts, erroneous sign, e.g. 3.02 instead of -3.02</li>
<li>-3 pts, failed to square, e.g. (x) instead of (x)^2</li>
<li>-3 pts, failed to take square root, e.g. (x) instead of sqrt(x)</li>
</ul>
<p>If, after grading, you discover on a particular exam that the final answers for five 10-point questions are incorrect because of only making a minor -2 point intermediate error, the student could conceivably obtain a score of 80% on an exam if they missed only -2 points per question (40/50).</p>
<p>However, in statistics, there is a contextual element to every question, not just solving for a numerical answer -- that is, in addition to the worked problem, students need to write a text-based response for the following:</p>
<ul>
<li>(2 pts) state whether the hypothesis test is significant or not</li>
<li>(2 pts) state whether the null hypothesis is rejected or accepted</li>
<li>(2 pts) state whether the p-value is less than 0.05 or not.</li>
</ul>
<p>So if there was only one minor (-2 point) intermediate error made, causing an incorrect final numerical answer, the student will also incorrectly respond to the final text-based answers (above) as well.</p>
<p>Thus, would you also take off e.g. -2 points for an incorrect final numerical answer, as well as -6 points for missing the final text-based sub-items listed above?</p>
<p>In other words, would you only deduct -2 points for a complex (multi-step) algebra or calculus question if only a minor intermediate step was erroneous, or would you also deduct for having an incorrect final numerical answer as well?</p>
<p>Maybe I could propose to the TAs to augment the point deduction list with:</p>
<ul>
<li>-1 pt, incorrect final numerical answer</li>
<li>-1 pt, state whether the hypothesis test is significant or not</li>
<li>-1 pt, state whether the null hypothesis is rejected or accepted</li>
<li>-1 pt, state whether the p-value is less than 0.05 or not.</li>
</ul>
| TomKern | 15,671 | <p>When designing a grading system for a question on an exam, I would identify the key skills being tested on that question. This can involve looking at other questions on the exam: if there are other questions that test a skill, I like to weigh it less unless it's a vital skill. A good exam is actually created starting from a list of skills you want to test.</p>
<p>Then allot some points for each key skill being tested on the problem. For each skill, I would suggest either two possible scores (all or nothing) or maybe an additional "demonstrated a significant partial understanding of the skill" score.</p>
<p>I advise avoiding putting emphasis on skills which are not taught/reviewed in the course. So for instance, I won't deduct points for an arithmetic mistake since arithmetic is not a skill taught in the course. However there are similar skills that are taught: using the right formula, knowing how to use that formula (what plugs in where), and checking to make sure the answer makes intuitive sense.</p>
<p>What this means for multi-step problems is: if the student got step 1 wrong, they get points for step 2 if:</p>
<ul>
<li>Their answer would have been correct if the answer to step 1 had been what they said</li>
<li>They demonstrated the skill in question</li>
</ul>
<p>So, for instance, if they get the wrong p-value on a question, but interpret that incorrect p-value correctly to get the corresponding (but incorrect) interpretation, they get full points for demonstrating the "interpret p-values" skill. If they interpret their incorrect p-value incorrectly to get the interpretation of the correct answer, they don't get points for demonstrating the "interpret p-values skill".</p>
<p>Of course the best test of a grading system is to actually look at graded work and see if it gives reasonable scores. Try whatever you choose on some sample student work before locking it in.</p>
<p>Also, depending on the number of exams, questions, and TAs, consider having each TA grade a specific problem across all sections. This improves uniformity and distances the grader from the student.</p>
|
564,360 | <p>Lets take the example, if we take the expression $\frac{X!}{y_1!\cdot y_2!\cdots y_n!} $as long as summation $S=y_1+y_2+...y_n$ is less than or equals $X$, the remainder is always $0$. Thats How the permutation of $X$ things where there is $y_1$ things same , $y_2$ things same works. My question is, why does this happen, what is the mathematical explanation behind this?
when its like $\frac{100!}{49!\cdot49!}$ that still works? Here the first $49$ consecutive digits already divided, but how the second consecutive $1..49$ also divides by $50...100$? </p>
| Arthur | 15,500 | <p>Not a full answer, but an outline of how I convinced myself of this fact:</p>
<p>Prime factorization is the key word. The key result is that if $y_1 + y_2 + \cdots y_n \leq x$ then for any prime $p$ the power of $p$ in the factorization of $x!$ is at least as high as in $y_1!y_2!\cdots y_n!$.</p>
<p>Once you've convinced yourself of this, you can (pretend to) write down a prime factorization of the numerator and denominator and see that the denominator can be simplified away completely. Thus you end up with just a product of primes (whatever's left of the numerator after the simplification), which is trivially an integer.</p>
|
782,507 | <p>So here's a somewhat incoherent question.</p>
<p>To define characteristic classes in the Chern–Weil way, one takes a curvature form $\Omega$ on a vector bundle $E \to M$ and an invariant polynomial $f$ on $\mathrm{GL}(\mathrm{rk }(E),\mathbb R)$, and then forms the cohomology class $c = \Big[f\big(\!\frac 1{2\pi}\Omega\big)\Big]$.</p>
<p>Why do we divide by $2\pi$? I understand why in the sense that "it works": if we want an integral class, so that $\langle c, [M] \rangle = \int_M f\big(\!\frac 1{2\pi}\Omega\big) \in \mathbb Z$, and agreeing with other standard definitions of these classes, dividing by $2\pi$ works, and not doing it doesn't.</p>
<p>But <em>why</em> does it work? "Morally," why is this the right thing to do?</p>
<p>I suppose this is analogous to asking why one always divides by $2\pi i$ in complex analysis, but there I feel I have some grasp on the answer: Cauchy's theorem holds, the only power of $z$ whose antiderivative isn't well-defined everywhere a power is $1/z$, and $t \mapsto z_0 + re^{it}$ describes one loop around a point $z_0$ as $t$ ranges from $0$ to $2\pi$.</p>
<p>I don't have even that clear an understanding what's going on in the case of the Chern–Weil homomorphism.</p>
| Mikhail Katz | 72,694 | <p>The point is to think of dual integer lattices. This can be seen already in the case of the unit circle. The class in cohomology dual to the fundamental homology class will be $d\theta$ divided by $2\pi$ since the circle has length $2\pi$. On the unit 2-sphere the curvature is $1$ but the total area is $4\pi$, which is why the fundamental cohomology class has to be normalized in a similar way to the circle to get an integer class. Hope this helps.</p>
|
782,507 | <p>So here's a somewhat incoherent question.</p>
<p>To define characteristic classes in the Chern–Weil way, one takes a curvature form $\Omega$ on a vector bundle $E \to M$ and an invariant polynomial $f$ on $\mathrm{GL}(\mathrm{rk }(E),\mathbb R)$, and then forms the cohomology class $c = \Big[f\big(\!\frac 1{2\pi}\Omega\big)\Big]$.</p>
<p>Why do we divide by $2\pi$? I understand why in the sense that "it works": if we want an integral class, so that $\langle c, [M] \rangle = \int_M f\big(\!\frac 1{2\pi}\Omega\big) \in \mathbb Z$, and agreeing with other standard definitions of these classes, dividing by $2\pi$ works, and not doing it doesn't.</p>
<p>But <em>why</em> does it work? "Morally," why is this the right thing to do?</p>
<p>I suppose this is analogous to asking why one always divides by $2\pi i$ in complex analysis, but there I feel I have some grasp on the answer: Cauchy's theorem holds, the only power of $z$ whose antiderivative isn't well-defined everywhere a power is $1/z$, and $t \mapsto z_0 + re^{it}$ describes one loop around a point $z_0$ as $t$ ranges from $0$ to $2\pi$.</p>
<p>I don't have even that clear an understanding what's going on in the case of the Chern–Weil homomorphism.</p>
| R.S. | 151,441 | <p>The inclusion $\mathbb{Z}\hookrightarrow\mathbb{R}$ induces a homomorphism $\check{H}^2(M;\mathbb{Z})\longrightarrow \check{H}^2(M;\mathbb{R})$ and under the identification $\check{H}^2(M;\mathbb{R})\cong H_{dR}^2(M;\mathbb{R})$ one has a homomorphism $\check{H}^2(M;\mathbb{Z})\longrightarrow H_{dR}^2(M;\mathbb{R})$.</p>
<p>I think that the problem lies in the use of THE for $\mathbb{Z}\hookrightarrow\mathbb{R}$. This is related to user72694's answer about the unit circle. If one is willing to define $S^1$ by $\mathbb{R}/2\pi\mathbb{Z}$, one can also define the inclusion $\mathbb{Z}\hookrightarrow\mathbb{R}$ by $\mathbb{Z}\ni k\mapsto 2\pi k\in\mathbb{R}$. This choice can wash $2\pi$'s from some formulae, and makes them appear elsewhere.</p>
|
8,658 | <p>$f(x) = \frac{1}{\cos x}$</p>
<p>$f'(x) = \frac{\sin(x)}{\cos^2(x)}$</p>
<p>$f''(x) = \frac{2\sin^2(x)+\cos^2(x)}{\cos^3(x)}$</p>
<p>$f^{(3)}(x) = \frac{6\sin^3(x)+5\cos^2(x)\sin(x)}{cos^4(x)}$</p>
<p>$\vdots$</p>
<p>$f^{(n)}(x) = \frac{ ?}{cos^{n+1}(x)}$</p>
<p>Some of these are easy: <a href="http://darkwing.uoregon.edu/~jcomes/251exn.pdf" rel="nofollow">http://darkwing.uoregon.edu/~jcomes/251exn.pdf</a>
Others are not. Why?</p>
| Qiaochu Yuan | 232 | <p>That depends on what you mean by "easy." As far as the examples in the .pdf you link to, I claim that the following are true for any reasonable definition of "easy":</p>
<ul>
<li>It is easy to compute the iterated derivatives of powers and logarithms. </li>
<li>It is easy to compute the iterated derivatives of solutions to homogeneous linear ODEs.</li>
<li>It is easy to compute the iterated derivatives of $x^n f(x)$ if it is easy to compute the iterated derivatives of $f(x)$.</li>
<li>It is easy to compute the iterated derivatives $f(x) + g(x)$ if it is easy to compute the iterated derivatives of $f(x)$ and of $g(x)$.</li>
</ul>
<p>I think that covers all of them. Otherwise, there is no general reason to expect that it should be easy to compute iterated derivatives. Even at $x = 0$ they may define complicated and highly nontrivial sequences, for example the <a href="http://en.wikipedia.org/wiki/Bernoulli_number" rel="nofollow">Bernoulli numbers</a> (which are related to this problem).</p>
|
8,658 | <p>$f(x) = \frac{1}{\cos x}$</p>
<p>$f'(x) = \frac{\sin(x)}{\cos^2(x)}$</p>
<p>$f''(x) = \frac{2\sin^2(x)+\cos^2(x)}{\cos^3(x)}$</p>
<p>$f^{(3)}(x) = \frac{6\sin^3(x)+5\cos^2(x)\sin(x)}{cos^4(x)}$</p>
<p>$\vdots$</p>
<p>$f^{(n)}(x) = \frac{ ?}{cos^{n+1}(x)}$</p>
<p>Some of these are easy: <a href="http://darkwing.uoregon.edu/~jcomes/251exn.pdf" rel="nofollow">http://darkwing.uoregon.edu/~jcomes/251exn.pdf</a>
Others are not. Why?</p>
| MarkV | 2,515 | <p>You might want to check out Faa di Bruno's formula: <a href="http://mathworld.wolfram.com/FaadiBrunosFormula.html" rel="nofollow">http://mathworld.wolfram.com/FaadiBrunosFormula.html</a> </p>
<p>In your case you have $f(x) = g(h(x))$ where $g(x) = 1/x$ and $h(x) = \cos(x)$. The answer is not simple unfortunately, and involves some combinatorics. </p>
<p>Alternatively, notice $f(x) \cos(x) = 1$, and thus by taking the $m$th derivative of both sides we get</p>
<p>$$ \sum_{k=0}^m {m \choose k} f^{(k)}(x) \left( \frac{d^{m-k}}{dx^{m-k}} \cos(x) \right) = 0 $$</p>
<p>if $m \geq 1$. Thus, you can view the vector $(f(x),f'(x),\dots,f^{(n)}(x))$ as the solution to the system of linear equations above with $m=0,1,\dots,n$. You can invert this matrix numerically, if that if your interest. A closed form might be possible too. </p>
|
33,543 | <p>Let $M$ be a filtered module over a filtered algebra $A$, and suppose $gr(M)$ is flat over $gr(A)$, where $gr$ means the associated graded module and algebra, respectively.</p>
<p>What can one say in general about the flatness of $M$ over $A$, or with relevant assumptions (for instance in the above, we should assume both filtrations are complete to avoid dumb counterexamples)? Are there good references for this sort of question? I have played with the various definitions of flatness trying to find an obvious relationship, but I find flatness proofs confusing.</p>
<p>The particular examples I have in mind are comparing $U(\mathfrak{g})$-modules to $S(\mathfrak{g})$-modules, and $D(X)$-modules to $O(T^*X)$-modules for affine varieties $X$, if it helps. I suspect the answer doesn't depend on any of the details though.</p>
| Emerton | 2,874 | <p>Dear David,</p>
<p>See Prop. 1.2 of <a href="http://arxiv.org/abs/math/0206056" rel="nofollow">Algebras of p-adic distributions and admissible representations</a> by Peter Schneider and Jeremy Teitelbaum, for one such result:</p>
<p>Proposition 1.2. Suppose that $gr^{\bullet} R$ and $gr^{\bullet} A$ are left noetherian and that $gr^{\bullet} A$ as a right $gr^{\bullet}R$-module (via $gr^{\bullet}\phi$) is flat; then $A$ is flat as a right
$R$-module (via $\phi$).</p>
<p>There are some hypotheses which you can find in the preamble to section 1 of the paper,
as well as useful references to literature on this kind of question. </p>
|
320,348 | <p>I would like to prove inductively that $${2n\choose n}=\sum_{i=0}^n{n\choose i}^2.$$</p>
<p>I know a couple of non-inductive proofs, but I can't do it this way. The inductive step eludes me. I tried naively things like $${2n+2\choose n+1}={2n+2\over n+1}{2n+1\choose n}=2\cdot {2n+1\over n+1}{2n\choose n},$$</p>
<p>But I don't think it can lead me anywhere. I would like the proof to be as simple as possible.</p>
| Bartek | 23,371 | <p>After Martin Sleziak and Marc van Leeuwen's comments, I've found <a href="https://math.stackexchange.com/a/219938/23371">this inductive proof</a> of Vandermonde's identity. (On this very site.)</p>
|
1,250,258 | <p>It's been 10 years since my last math class so I'm very rusty. How would I go about proving
$$3^n < n!$$
where $n \geq 7$?</p>
<p>I understand that factorials grow faster than set values with a variable exponent. Just not sure how to start proving it mathematically.</p>
| Belgi | 21,335 | <p>Since $2187=3^{7}<7!=5040$ then for $n>7$
$$
3^{n}=3^{7}\cdot3^{n-7}<7!\cdot8\cdot9\cdot\ldots\cdot n=n!
$$</p>
<p>since $8,9,\dots n>3$ </p>
|
122,770 | <p>Given that $A$ is an open set in $\mathbb R^n$ and $f:A \to \mathbb R^n$ is differentiable, and its derivative is non-singular at every point in $A$, prove that $f(A)$ is open in $\mathbb R^n$</p>
<p>Note $f$ is differentiable, <em>not</em> continuously differentiable. </p>
| azarel | 20,998 | <p>By the inverse function theorem for each $x\in A$ there exists open sets $x\in U$ and $f(x)\in V$ so that $f|_U:U\to V$ is a diffeomorphism. So in particular $f(U)=V$ hence $f(x)\in V\subset f(A)$. </p>
|
2,014,067 | <p>Please advise how to integrate this function. I think I need to use parts to do this, but I can't seem to get to the right answer. I know what the final solution is supposed to be, but I can't figure out where I'm going wrong in my effort to get there.</p>
<blockquote>
<p>$$\int \ln (1+2x)\:dx$$</p>
</blockquote>
<p>Thanks in advance.
Marcia</p>
| parsiad | 64,601 | <p><strong>Hint</strong>: what happens when you substitute $u \equiv 1+2x$? </p>
|
2,014,067 | <p>Please advise how to integrate this function. I think I need to use parts to do this, but I can't seem to get to the right answer. I know what the final solution is supposed to be, but I can't figure out where I'm going wrong in my effort to get there.</p>
<blockquote>
<p>$$\int \ln (1+2x)\:dx$$</p>
</blockquote>
<p>Thanks in advance.
Marcia</p>
| Olivier Oloa | 118,798 | <p><strong>Hint</strong>. One may integrate by parts as follows
$$
\int \ln(1+2x)\:dx=\frac{1+2x}2\:\ln(1+2x)-\int \frac{1+2x}2\cdot\frac{(1+2x)'}{(1+2x)}\:dx=\:?
$$</p>
|
223,162 | <p>I have a modulus function that looks like this $f(x) = 2x+3 \bmod b$, and i have to show that $x = y$ to prove that the function is $1-1.$</p>
<p>I know that mod functions can't be algebraically manipulated like regular function, so I was wondering if it was even possible or if I was just wasting my time scratching my head.</p>
| André Nicolas | 6,312 | <p>The number $b$ you are given must be <strong>odd</strong>. </p>
<p>We have $2x+3 \equiv 2y+3 \pmod {b}$ iff $(2x+3)-(2y+3)$ is divisible by $b$ iff $2(x-y)$ is divisible by $b$. </p>
<p>If $b$ is odd, this is true iff $x-y$ is divisible by $b$, that is, iff $x\equiv y \pmod{b}$.</p>
<p>Thus if $b$ is odd, our function is one to one. This need not be true if $x$ is even. For example. $(2)(1)+3 \equiv 2(5)+3\pmod {8}$ but $1\not\equiv 5\pmod{8}$.</p>
<p><strong>Remark:</strong> Actually, what is remarkable is the degree to which mod expressions <em>can</em> be manipulated like regular functions.</p>
|
223,162 | <p>I have a modulus function that looks like this $f(x) = 2x+3 \bmod b$, and i have to show that $x = y$ to prove that the function is $1-1.$</p>
<p>I know that mod functions can't be algebraically manipulated like regular function, so I was wondering if it was even possible or if I was just wasting my time scratching my head.</p>
| Bill Dubuque | 242 | <p>It is much more insightful to consider the following generalization.</p>
<p><strong>Theorem</strong> $\ $ The following are equivalent for integers $\rm\:c,\, m\,$.</p>
<p>$(1)\rm\ \ \ gcd(c,m) = 1$<br>
$(2)\rm\ \ \ c\:$ is invertible $\rm\,(mod\ m)$<br>
$(3)\rm\ \ \ x\to cx+d\:$ is $\:1$-$1\:$ $\rm\,(mod\ m)$<br>
$(4)\rm\ \ \ x\to cx+d\:$ is onto $\rm\,(mod\ m)$ </p>
<p><strong>Proof</strong> $\ \ (1\Rightarrow 2)\ \ $ By <a href="http://en.wikipedia.org/wiki/Bezout%27s_identity" rel="nofollow">Bezout,</a> $\rm\: gcd(c,m) = 1\:\Rightarrow\: cd+km = 1\:$ for $\rm\:d,k\in\Bbb Z\:$ $\rm\Rightarrow\:cd\equiv 1\pmod m$<br>
$(2\Rightarrow 3)\ \ \ \rm cx+d \equiv cy+d\:\Rightarrow\:c(x-y)\equiv 0\:\Rightarrow\:x-y\equiv 0\:$ by multiplying by $\rm\:c^{-1}$<br>
$(3\Rightarrow 4)\ \ $ Every $1$-$1$ function on a finite set is onto.<br>
$(4\Rightarrow 1)\ \ \ \rm x\to cx\:$ is onto, so $\rm\:cd\equiv 1,\:$ some $\rm\,d,\:$ i.e. $\rm\: cd+km = 1,\:$ some $\rm\,k,\:$ so $\rm\:gcd(c,m)=1$</p>
|
2,055,878 | <p>If$$ x^2+x+1=0$$ find the value of $$8x^{282}+1799x^{183}+87x^{51}+124x^{-3}+1$$</p>
<p>Solving this equation gives imaginary solutions. </p>
<p>Is there an easy way to do this ?</p>
| DonAntonio | 31,254 | <p>Hint:</p>
<p>Observe that such a $\;x\;$ fufills $\;x^3=1\;$ ...</p>
|
1,410,586 | <blockquote>
<p>For certain pairs $ (m,n)$ of positive integers with $ m\ge n$ there are exactly $ 50$ distinct positive integers $ k$ such that $ |\log m - \log k| < \log n$. Find the sum of all possible values of the product $ mn$.</p>
</blockquote>
<p><strong>HINTS ONLY!</strong></p>
<p>Obviously, converting it into a simpler form is good idea.</p>
<p>$$\frac{1}{n} < \frac{m}{k} < n \implies k < mn < n^2k.$$</p>
<p>Okay, so we already have a pretty simplified form, I broke it into two cases.</p>
<p>Case 1: $m = n$ then saw:</p>
<p>$\implies k < n^2 < kn^2$, which is true for ALL $k \ge 2$ and any $n$. </p>
<p>So $m= n$ is an impossible case.</p>
<p>The remaining case left is $m > n$. </p>
<p>This is actually getting quite difficult. </p>
<p>There must be exactly $50$ values of $k$. </p>
<p>How should I go about this?</p>
<p><strong>SMALL - HINTS ONLY! Please don't give it away!</strong></p>
| user137794 | 137,794 | <p>Perhaps it will help to see</p>
<p>$$|\log m-\log k| < \log n \implies \log m-\log n < \log k < \log m+\log n \implies \log\frac{m}{n} < \log k < \log mn$$</p>
<p>Where $\log$ is monotonic.</p>
|
2,206,247 | <p><strong>Question:</strong> Consider the following non linear recurrence relation defined for $n \in \mathbb{N}$:</p>
<p>$$a_1=1, \ \ \ a_{n}=na_0+(n-1)a_1+(n-2)a_2+\cdots+2a_{n-2}+a_{n-1}$$</p>
<p>a) Calculate $a_1,a_2,a_3,a_4.$</p>
<p>b) Use induction to prove for all positive integers that:</p>
<p>$$a_n=\dfrac{1}{\sqrt{5}}\left[\left(\dfrac{3+\sqrt{5}}{2}\right)^n-\left(\dfrac{3-\sqrt{5}}{2}\right)^n\right]$$
Hi all! I'm having trouble solving this problem. I have no problem with part (a), but I'm having lots of troubles with part (b). I proved the base case (which is quite trivial), but I'm having trouble for the inductive step (proving k->k+1).</p>
<p><a href="https://i.stack.imgur.com/JgGfv.png" rel="nofollow noreferrer">Attempt</a></p>
<p>I don't know what to do from this point. Thank you! </p>
| Stefano | 387,021 | <p>Since $a$ and $b$ are non negative, this is equivalent to</p>
<p>$$\sqrt{x^2+y^2} \le |x|+|y|, $$</p>
<p>which has a nice geometric interpretation if you think of right triangles.</p>
|
64,643 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/4467/a1-2-is-either-an-integer-or-an-irrational-number">$a^{1/2}$ is either an integer or an irrational number</a> </p>
</blockquote>
<p>I know how to prove $\sqrt 2$ is an irrational number. Who can tell me that why $\sqrt 3$ is a an irrational number?</p>
| Aryabhata | 1,102 | <p>A continued fraction proof of the irrationality of $x = \sqrt{3} - 1$, from which the irrationality of $\sqrt{3}$ follows. (A continued fraction proof of $\sqrt{2}$ can be found here: <a href="https://math.stackexchange.com/questions/5/how-can-you-prove-that-the-square-root-of-two-is-irrational/16526#16526">How can you prove that the square root of two is irrational?</a> )</p>
<p>Notice that $x = \sqrt{3} - 1$ is a root of the equation $x^2 + 2x - 2 = 0$</p>
<p>This can be re-written as</p>
<p>$$x(3+x) = 2+x$$</p>
<p>$$x = \frac{2+x}{3+x} = \cfrac{1}{1 + \cfrac{1}{2 + x}}$$</p>
<p>And thus</p>
<p>$$x = [1,2,1,2,\dots]$$</p>
<p>and so is irrational.</p>
|
3,244,866 | <p>How can i prove that <span class="math-container">$$2^n\not \in O(n^2)$$</span> by formal definition and not using limits?</p>
<p>With:</p>
<p><a href="https://i.stack.imgur.com/id9gx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/id9gx.png" alt="enter image description here"></a></p>
| auscrypt | 675,509 | <p>Like you noted, <span class="math-container">$n$</span> odd doesn't work.</p>
<p>For <span class="math-container">$n$</span> even, note that looking only at non-negative integers is sufficient. We have, for <span class="math-container">$n\ge 4$</span>,</p>
<p><span class="math-container">$$\left(n^3+\frac{n}{2}\right)^2=n^6+n^4+\frac{n^2}{4}> {\bf n^6+n^4+1}> n^6+n^4-2n^3+\frac{n^2}{4}-n+1=\left(n^3+\frac{n}{2}-1\right)^2$$</span></p>
<p>And thus since your expression lies between two consecutive squares, it cannot be a square.</p>
|
3,244,866 | <p>How can i prove that <span class="math-container">$$2^n\not \in O(n^2)$$</span> by formal definition and not using limits?</p>
<p>With:</p>
<p><a href="https://i.stack.imgur.com/id9gx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/id9gx.png" alt="enter image description here"></a></p>
| Mike | 544,150 | <p>The only value that works is <span class="math-container">$n=2$</span>.</p>
<p>Indeed, if there is another value that works, then there exists integers <span class="math-container">$n > 2$</span> and <span class="math-container">$k>0$</span> such that</p>
<p><span class="math-container">$$n^6 + n^4 +1 = (n^3+k)^2$$</span></p>
<p>which implies that there exists exists integers <span class="math-container">$n > 2$</span> and <span class="math-container">$k>0$</span> such that</p>
<p><span class="math-container">$$n^4+1 = 2n^3+k^2$$</span></p>
<p>However, note that on the one hand, for <span class="math-container">$k=n/2$</span> that <span class="math-container">$2n^3+k^2 = n^4+n^2/4$</span> which is strictly greater than <span class="math-container">$n^4+1$</span> for <span class="math-container">$n > 2$</span> [and equal to <span class="math-container">$n^4+1$</span> for <span class="math-container">$n=2$</span>]. On the other hand, or <span class="math-container">$k=n/2-1$</span>, note that
<span class="math-container">$2n^3+k^2 < n^4-n^3+n^2/4 < n^4+1$</span> for <span class="math-container">$n > 2$</span>.</p>
<p>Can you see how this implies no integral <span class="math-container">$k$</span> will solve the equation <span class="math-container">$n^4+1 = 2n^3+k^2$</span>, and then in turn <span class="math-container">$n^4+1 = 2n^3+k^2$</span> for integral <span class="math-container">$n >2$</span>?</p>
|
695,648 | <p>How can I count the numbers of $5$ digits such that at least one of the digits appears more than one time? </p>
<p>My thoughts are:<br>
I count all the possible numbers of $5$ digits: $10^5 = 100000$. Then, I subtract the numbers that don't have repeated digits, which I calculate this way: $10*9*8*7*6$ $= 30240 $. Thus, I have $100000 - 30240 = 69760 $ numbers that have at least one digit repeated more than one time. </p>
<p>Is this correct?</p>
| user76568 | 74,917 | <p>It is helpful to look at the negation of your requirement.<br>
No digit is ever repeated for $10 \cdot 9 \cdot 8 \cdot 7 \cdot 6=30240$ numbers.<br>
The total number of $5$ digit numbers is $10^5=100000$. </p>
<p>So the number of $5$ digits numbers with at least $1$ digit repeating more than once is:
$$10^5 - 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = 69760$$
So, yes you are correct. <strong>NOW</strong> I noticed I have just repeated your correct arguments. Down vote for it to disappear! :-)</p>
|
695,648 | <p>How can I count the numbers of $5$ digits such that at least one of the digits appears more than one time? </p>
<p>My thoughts are:<br>
I count all the possible numbers of $5$ digits: $10^5 = 100000$. Then, I subtract the numbers that don't have repeated digits, which I calculate this way: $10*9*8*7*6$ $= 30240 $. Thus, I have $100000 - 30240 = 69760 $ numbers that have at least one digit repeated more than one time. </p>
<p>Is this correct?</p>
| Community | -1 | <p><strong>Helpful question:</strong> In how many ways can we make or arrange five digits so that there is no repetition?</p>
<p>We know that there are 10 possible digits to choose from for the first digit, 0 - 10. However, we can't have the first digit be a 0, else we get a four-digit number. So, we're down to 9. For each of the 9 digits, there are 9 digits to choose from for the second digit to avoid repetition and so on. </p>
<p>This gives <span class="math-container">$9 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = 27,216$</span> ways. </p>
<p><strong>Actual question:</strong> How many five-digit numbers have at least one digit that occurs more than once?</p>
<p>We also know that there are <span class="math-container">$9 \cdot 10 \cdot 10 \cdot 10 \cdot 10 = 90,000$</span> possible five-digit numbers. At this point, it should be clear why the first digit can only be selected in 9 ways. </p>
<p>In order to get the number of five-digit numbers that have at least one digit repeated, we simply subtract the number of possible numbers without repeated digits from the total number of five-digit numbers, which gives us <span class="math-container">$90,000 - 27,216 = 62,784$</span></p>
<p>My answer is probably redundant at this point. Nevertheless, I'll just leave it here. :)</p>
|
4,519,106 | <p>After I learned about the existence of such a concept as a contrapositive, I always try to translate any statements into a contrapositive. And every time I fail. I haven't found a general technique for this yet. I think that if I know the statement and its contrapositive form, it will give me a better understanding.</p>
<p>The statement :<span class="math-container">$$A\subset B \implies \min A \geq \min B.$$</span></p>
<p>Its contrapositive form must look like <span class="math-container">$$\neg (\min A\geq \min B)\implies \neg (A\subset B).$$</span></p>
<p>The first thing which comes to mind is <span class="math-container">$$\min A<\min B \implies B\subseteq A.$$</span></p>
<p>On the other hand I think I must write down what I mean by <span class="math-container">$A\subset B$</span> and I get</p>
<p><span class="math-container">$$\neg (\min A\geq \min B)\implies \neg (x\in A \implies x\in B ).$$</span></p>
<p>And then I get <span class="math-container">$$\min A<\min B \implies x\in A, x\notin B$$</span> which looks very strange.</p>
<p><strong>Question</strong>: What is the correct way to do this?</p>
| Nitin Uniyal | 246,221 | <p><span class="math-container">$\sim(A\subset B)\equiv A\not\subset B$</span> rather than <span class="math-container">$B\subseteq A$</span>.</p>
|
79,869 | <p>Let <span class="math-container">$(X,\mu,\mathcal{F})$</span> be a probability space. The paper <em><a href="http://projecteuclid.org/euclid.aoms/1177693405" rel="nofollow noreferrer">Equiconvergence of Martingales</a></em> by Edward Boylan introduced a pseudometric on sub-<span class="math-container">$\sigma$</span>-fields (sub-<span class="math-container">$\sigma$</span>-algebras) of <span class="math-container">$\mathcal{F}$</span> as follows:</p>
<p><span class="math-container">$\rho(\mathcal{G},\mathcal{H})
:= \sup_{A\in \mathcal{G}} \inf_{B\in \mathcal{H}} \mu(A \triangle B) + \sup_{B\in \mathcal{H}} \inf_{A\in \mathcal{G}} \mu(A \triangle B)$</span></p>
<p>where <span class="math-container">$A \triangle B$</span> is symmetric difference.</p>
<p>It seems to be called the Hausdorff pseudometric on <span class="math-container">$\sigma$</span>-fields in later papers. (Does anyone know why?) Further, if we only consider a <span class="math-container">$\mu$</span>-complete <span class="math-container">$\sigma$</span>-fields then <span class="math-container">$\rho$</span> is a metric. Also, the paper shows <span class="math-container">$\rho$</span> is complete.</p>
<blockquote>
<p>Is this metric <span class="math-container">$\rho$</span>
separable---assuming, say, <span class="math-container">$X=[0,1]$</span>
and <span class="math-container">$\mu$</span> is the Lebesgue measure?</p>
</blockquote>
<p>My guess is that it is not, but I cannot off-hand come up with a witnessing set to show this. Considering the paper is 40 years old, I imagine this might be well-known. And if it is not separable, then my follow up question is this?</p>
<blockquote>
<p>Is there a known separable, complete
metric on the space of
<span class="math-container">$\mu$</span>-complete sub-<span class="math-container">$\sigma$</span>-fields?</p>
</blockquote>
<p>For reference, I found the following list <a href="https://groups.google.com/forum/#!searchin/sci.math/%22Hausdorff-metric<span class="math-container">$20of$</span>20sigma-fields%22%7Csort:date/sci.math/iz249jCUvEU/rkKOei1NV5YJ" rel="nofollow noreferrer">online</a>, compiled by Dave L. Renfro, of papers dealing with metrics on <span class="math-container">$\sigma$</span>-fields (listed in Chronological order). I quickly looked though these papers and didn't find what I was looking for, but maybe I missed something.</p>
<ol>
<li><p>Edward S. Boylan, "Equiconvergence of martingales",<br>
Annals of Mathematical Statistics 42
(1971), 552-559. [MR 44 #7603; Zbl 218.60049]</p></li>
<li><p>Jacques Neveu, "Note on the tightness of the metric on the
set of complete sub sigma-algebras of a probability space",
Annals of Mathematical Statistics 43 (1972), 1369-1371.
[MR 48 #5133; Zbl 241.60036]</p></li>
<li><p>Hirokichi Kudo, "A note on the strong convergence of
sigma-algebras", Annals of Probability 2 (1974), 76-83.
[MR 51 #6900; Zbl 275.60007]</p></li>
<li><p>Lothar Rogge, "Uniform inequalities for conditional
expectations", Annals of Probability 2 (1974), 486-489.
[MR 50 #14858; Zbl 285.28010]</p></li>
<li><p>Louis H. Blake, "Some further results concerning
equiconvergence of martingales", Revue Roumaine de
Mathématiques Pures et Appliquées 28 (1983), 927-932.
[MR 86i:60130; Zbl 524.60029]</p></li>
<li><p>Hari G. Mukerjee, "Almost sure equiconvergence of
conditional expectations", Annals of Probability 12
(1984), 733-741. [MR 86c:28012; Zbl 557.28001]</p></li>
<li><p>Beth Allen, "Convergence of sigma-fields and applications
to mathematical economics", pp. 161-174 in Gerald Hammer
and Diethard Pallaschke (editors), SELECTED TOPICS IN
OPERATIONS RESEARCH AND MATHEMATICAL ECONOMICS (Proceedings,
Karlsruhe, West Germany, 22-25 August 1983), Lecture Notes
in Economics and Mathematical Systems #226, Springer-Verlag, 1984.
[MR 86f:90029; Zbl 547.28001]</p></li>
<li><p>Dieter Landers and Lothar Rogge, "An inequality for the
Hausdorff-metric of sigma-fields", Annals of Probability
14 (1986), 724-730. [MR 87h:60006; Zbl 597.60003]</p></li>
<li><p>Abdallah M. Al-Rashed, "On countable unions of sigma
algebras", Journal of Karachi Mathematical Association
8 (1986), 57-63. [MR 88f:28001; Zbl 639.28001]</p></li>
<li><p>Maxwell B. Stinchcombe, "A further note on Bayesian
information topologies", Journal of Mathematical Economics
22 (1993), 189-193. [MR 93k:60011; Zbl 773.90016]</p></li>
<li><p>Timothy Van Zandt, "The Hausdorff metric of sigma-fields
and the value of information", Annals of Probability 21
(1993), 161-167. [MR 94d:62012; Zbl 777.62007]</p></li>
<li><p>Xikui Wang, "Completeness of the set of sub-sigma-algebras",
International Journal of Mathematics and Mathematical
Sciences 16 (1993), 511-514. [MR 94f:28002; Zbl 782.28001]</p></li>
<li><p>Zvi Artstein, "Compact convergence of sigma-fields and
relaxed conditional expectation", Probability Theory and
Related Fields [= Zeitschrift für Wahrscheinlichkeits-
theorie] 120 (2001), 369-394. [MR 2002g:28003; Zbl 992.28001]</p></li>
</ol>
| Bill Johnson | 2,554 | <p>Take a sequence $A_n$ of independent sets of measure $1/2$. Given two different subsets $B$ and $C$ of natural numbers, suppose WLOG that there is an $n$ in $B\sim C$. Now $\mu(A_n\Delta A) = 1/2$ for all sets $A$ which are independent of $A_n$, so the distance from the sigma algebra generated by $(A_n)_{n\in B}$ to the sigma algebra generated by
$(A_n)_{n\in C}$ is at least $1/2$. This shows that the density character of your space is at least the continuum.</p>
|
1,918,408 | <p>Let $C(B)$ be a $\infty$-order polynomial: $$ C(B) = \sum_{k=0}^\infty \alpha_k B^k$$</p>
<p>Show that $$C(B) = C(1) + (1-B)C^*(B)$$ where $C^*(B)$ is a another $\infty$-order polynomial.</p>
<p>This comes from the prove of the Engle-Granger Representation Theorem in their <a href="http://www.uta.edu/faculty/crowder/papers/Engle_Granger_1987.pdf" rel="nofollow">original paper</a></p>
<p>Here, the polynomials $C(B), C^*(B)$ are moving-average polynomials in time series analysis.</p>
<p>I can't seem to understand how its derived... any help?</p>
| ajotatxe | 132,456 | <p>When $\inf B<0$, $|\inf B|\ge \sup B$, $\sup A>0$ and $\sup A\ge |\inf A|$.</p>
|
1,807,479 | <blockquote>
<p>I recently took a test and was confused about a question. I feel that
the answer is B. Could anyone please elucidate it. Thanks!</p>
</blockquote>
<p>The point $(−4, 3)$ is on the terminal side of angle $\theta$ as sketched below. Find $\cos\theta$.</p>
<p><a href="https://i.stack.imgur.com/BiOiI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BiOiI.png" alt="terminal_side_of_angle"></a></p>
<p>A. $-(4/5)$</p>
<p>B. $-(3/4)$</p>
<p>C. $-(\sqrt{2}/2)$</p>
<p>D. $(\sqrt{3}/2)$</p>
<p>E. $(4/5)$</p>
| Bernard | 202,857 | <p>The correct answer is A. The hypotenuse in the following figure has length $5$ by Pythagoras:</p>
<p><a href="https://i.stack.imgur.com/xlJyj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xlJyj.png" alt="enter image description here"></a></p>
|
2,464,890 | <p>Here is link to some limit questions:</p>
<p><a href="https://i.stack.imgur.com/2rM9f.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2rM9f.png" alt="Example" /></a>
Can anyone explain how has answers were derived? In (a), how can we cancel out <span class="math-container">$(x-2)$</span>? And how can answer be 0? When <span class="math-container">$x\to 2$</span>, <span class="math-container">$x-2\to 0$</span> and the answer should be infinity. Similarly in (b) the answer should be infinity. Can anyone explain?</p>
| Bernard | 202,857 | <p>They simply use the following rule to simplify fractions when there's a common factor in the numerator and the denominator:
$$\frac{A\color{blue}\not\! C}{B\color{blue}\not\! C}=\frac AB,$$
and they determine the limit of the simplified fraction.</p>
|
108,010 | <p>It is not necessarily true that the closure of an open ball $B_{r}(x)$ is equal
to the closed ball of the same radius $r$ centered at the same point $x$. For a quick example, take $X$ to be any set and define a metric
$$
d(x,y)=
\begin{cases}
0\qquad&\text{if and only if $x=y$}\\
1&\text{otherwise}
\end{cases}
$$
The open unit ball of radius $1$ around any point $x$ is the singleton set $\{x\}$. Its closure is also the singleton set. However, the closed unit ball of radius $1$ is everything. </p>
<p>I like this example (even though it is quite artificial) because it can show that this often-assumed falsehood can fail in catastrophic ways. My question is: are there necessary and sufficient conditions that can be placed on the metric space $(X,d)$ which would force the balls to be equal? </p>
| Mathecm | 544,686 | <blockquote>
<p>More general, we can prove that for any normed space <span class="math-container">$(V,\|\cdot\|)$</span>, if <span class="math-container">$x_0\in V$</span> and <span class="math-container">$R>0$</span>, then <span class="math-container">$$\overline{B(x_0;R)}^{\|\cdot\|}=B[x_0;R],$$</span> where <span class="math-container">$\overline{B(x_0;R)}^{\|\cdot\|}$</span> is the closure of the open ball <span class="math-container">$B(x_0;R)$</span> in topology induced by <span class="math-container">$\|\cdot\|$</span>.</p>
</blockquote>
<p>The inclusion <span class="math-container">$\overline{B(x_0;R)}^{\|\cdot\|}\subseteq B[x_0;R]$</span> is trivial, because <span class="math-container">$B[x_0;R]$</span> is a closed set that contains <span class="math-container">$B(x_0;R)$</span> and <span class="math-container">$\overline{B(x_0;R)}^{\|\cdot\|}$</span> is the smallest closed set that contains <span class="math-container">$B(x_0;R)$</span>.</p>
<p>To show that <span class="math-container">$B[x_0;R]\subseteq\overline{B(x_0;R)}^{\|\cdot\|}$</span>, we can prove that any element of <span class="math-container">$B[x_0;R]$</span> is the limit of some sequence in <span class="math-container">$B(x_0;R)$</span>. Let <span class="math-container">$x\in B[x_0;R]$</span>, then of course <span class="math-container">$$\|x-x_0\|\leqslant R.$$</span> Now defines, for each <span class="math-container">$n\in\Bbb{N}$</span>, <span class="math-container">$$x_n:=\left(1-\frac{1}{n}\right)x+\frac{1}{n}x_0.$$</span></p>
<p><strong>Claim 1</strong>: The sequence <span class="math-container">$(x_n)$</span> converges to <span class="math-container">$x$</span> in the norm <span class="math-container">$\|\cdot\|$</span>. In fact, <span class="math-container">\begin{eqnarray}
\|x_n-x\| & = & \left\|\left(1-\frac{1}{n}\right)x+\frac{1}{n}x_0-x\right\| \\
& = & \frac{1}{n}\underbrace{\|x-x_0\|}_{\leqslant R} \\
& \leqslant & \frac{R}{n}\longrightarrow0,\quad\text{when}\ n\to\infty,
\end{eqnarray}</span> which proves that <span class="math-container">$x_n\to x$</span> in <span class="math-container">$\|\cdot\|$</span>.</p>
<p><strong>Claim 2</strong>: For all <span class="math-container">$n\in\Bbb{N}$</span>, <span class="math-container">$x_n\in B(x_0;R)$</span>, that is, <span class="math-container">$(x_n)$</span> is a sequence in the set <span class="math-container">$B(x_0;R)$</span>. This is obviously by the construction of <span class="math-container">$x_n$</span>, since <span class="math-container">\begin{eqnarray}
\|x_n-x_0\| & = & \left\|\left(1-\frac{1}{n}\right)x+\frac{1}{n}x_0-x_0\right\| \\
& = & \left\|\left(1-\frac{1}{n}\right)x-x_0\left(1-\frac{1}{n}\right)\right\| \\
& = & \left\|\left(1-\frac{1}{n}\right)(x-x_0)\right\| \\
& = & \left(1-\frac{1}{n}\right)\underbrace{\|x-x_0\|}_{\leqslant R} \\
& \leqslant & \underbrace{\left(1-\frac{1}{n}\right)}_{<1}R<R,
\end{eqnarray}</span> which gives us <span class="math-container">$x_n\in B(x_0;R)$</span>.</p>
<p>So we conclude that any element of <span class="math-container">$B[x_0;R]$</span> is the limit of a sequence in <span class="math-container">$B(x_0;R)$</span>, hence <span class="math-container">$B[x_0;R]\subseteq\overline{B(x_0;R)}^{\|\cdot\|}$</span>.</p>
|
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