qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
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3,051,207 | <p>In the very first chapter in my class "mathematical analysis 1" I've seen something called <strong>the triangle inequality</strong>, which is <span class="math-container">$||a| - |b|| \leq |a \pm b| \leq |a| + |b|$</span>. Now the thing is that I do understand why this is true, but i fail to see what this actually has to do with triangles. It was never explained nor proven. I'd like to know how this inequality relates to triangles to get a better understanding of it, because it's also used a lot further on. Would someone be able to explain this?</p>
| Lee Mosher | 26,501 | <p>In Euclidean space with Cartesian coordinates, let <span class="math-container">$+$</span> denote vector addition, and let <span class="math-container">$O$</span> denote the origin. Choose points <span class="math-container">$a,b$</span>, and form a triangle with vertices <span class="math-container">$a,b,O$</span>. One side of this triangle has side length <span class="math-container">$|a|$</span>, another has side length <span class="math-container">$|b|$</span>, and the third has side length <span class="math-container">$|a-b|$</span>. Now as we all know, the shortest path between two points is a straight line. So, the shortest path between the two points <span class="math-container">$a,b$</span> has length <span class="math-container">$|a-b|$</span>. There is a longer path one can take which goes around the other two sides, the length of that path is <span class="math-container">$|a| + |b|$</span>. So, <span class="math-container">$|a-b| \le |a| + |b|$</span>. </p>
|
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)$</span> <br /><br /> So i am getting positive value in 1st case whereas in 4th quadrant my answer is Negative. Why is so ? am i making some mistake when considering 4th quadrant ? or both are acceptable answer ? also in MIT lecture they said whatever acceptable quadrant you choose you'll get the same answer. so why i m getting Negative answer ?</p>
| Mark Saving | 798,694 | <p>Take a basis <span class="math-container">$B \subseteq W$</span>. For each <span class="math-container">$b \in B$</span>, pick <span class="math-container">$s_b \in V$</span> such that <span class="math-container">$T(s_b) = b$</span>.</p>
<p>Then <span class="math-container">$S = \{s_b \mid b \in B\}$</span> must span all of <span class="math-container">$V$</span>, since <span class="math-container">$T(S) = \{T(s_b) \mid b \in B\} = B$</span> spans all of <span class="math-container">$W$</span>.</p>
<p>Now suppose we have <span class="math-container">$T(x) = 0$</span>. Since <span class="math-container">$S$</span> spans <span class="math-container">$T$</span>, write <span class="math-container">$x = \sum\limits_{i = 1}^n c_i s_{b_i}$</span>, where <span class="math-container">$c_i$</span> is a scalar and <span class="math-container">$b_i \in B$</span> for all <span class="math-container">$i$</span>. Then we have <span class="math-container">$T(x) = 0 = \sum\limits_{i = 1}^n c_i b_i$</span>.</p>
<p>Since <span class="math-container">$B$</span> is a basis, we see that for all <span class="math-container">$i$</span>, we must have <span class="math-container">$c_i = 0$</span>. Therefore, <span class="math-container">$x = 0$</span>.</p>
<p>Thus, <span class="math-container">$T$</span> is injective.</p>
<p>I don't see a way to do this without invoking the axiom of choice at some point, which is rather unfortunate.</p>
|
4,245,475 | <p>Evaluate <span class="math-container">$ \, \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)$</span> <br /><br /> So i am getting positive value in 1st case whereas in 4th quadrant my answer is Negative. Why is so ? am i making some mistake when considering 4th quadrant ? or both are acceptable answer ? also in MIT lecture they said whatever acceptable quadrant you choose you'll get the same answer. so why i m getting Negative answer ?</p>
| baharampuri | 50,080 | <p>Let <span class="math-container">$v \neq 0$</span> be such that <span class="math-container">$T(v)=0$</span>. Extend this to a basis of <span class="math-container">$V$</span>. Let's call that basis <span class="math-container">$B$</span>. Now as <span class="math-container">$T$</span> is surjective <span class="math-container">$T(B) $</span> spans <span class="math-container">$W$</span>. But <span class="math-container">$T(v) =0$</span> so <span class="math-container">$T(B \setminus v) $</span> spans <span class="math-container">$W$</span> which means <span class="math-container">$B\setminus v$</span> spans <span class="math-container">$V$</span> a contradiction as <span class="math-container">$B$</span> is a basis. Thus our assumption <span class="math-container">$v\neq 0$</span> is false.</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>
| Natalio | 499,705 | <p>I know that this question is quite old <a href="https://math.stackexchange.com/questions/2712583/find-the-distribution-of-r-sqrtx2y2-where-x-y-is-uniform-on-the-un/2717764#2717764">but I asked the same some days ago</a> and now I think I have an answer which involves more calculus. Since this question is more popular than mine, I'm gonna copy-paste my answer here so that other people in the futere can see it:
Consider the following function:
$$ g:\mathbb{R}^2\to \mathbb{R}^2 \quad g(x,y)=\left(\sqrt{x^2+y^2},\arctan\left(\frac{x}{y}\right)\right) $$
It is inyective and has inverse $ g^{-1}(r,s) =(r\sin(s),r\cos(s)) $ then
$$ f_{g(X,Y)}(r,s)=f_{(X,Y)}(g^{-1}(r,s))|\det(J_{g^{-1}})(r,s)| =\frac{r}{\pi}1_{[0,1]}(r^2)=\frac{r}{\pi}1_{[-1,1]}(r) $$
Recall that we want to know $f_R $ which is the marginal distribution of $ f_{g(X,Y)}$:
$$ f_R(r)= \int_{\mathbb{R}} f_{g(X,Y)}(r,s) \; ds = \int_{0}^{2\pi} \frac{r}{\pi}1_{[-1,1]}(r) \; ds =2r 1_{[-1,1]}(r) $$</p>
|
253,359 | <p>I'm trying to prove by induction the following statement without success:<br>
$$\forall n \ge 2, \;\forall d \ge 2 : d \mid n(n+1)(n+2)...(n+d-1) $$</p>
<p>For the base case: $n = 2$, $d = 2$<br>
$2\mid 2(2+1)$ which is true.<br></p>
<p>Now, the confusion begins! I assume I would need to use the second induction principle to proof this because $P(n)$ and $P(n+1)$ are not related at all. It is also the first time I am dealing with more than one variable so it makes it harder for me.</p>
<p>I tried the following:<br>
- Trying to prove by simple induction. I did not go very far.<br>
- Trying to split my induction step in 2 parts: If $d\mid n$, I'm done. <br></p>
<p>If $d$ does not divide $n$, then I would need to do a second proof and this is where I'm blocked.</p>
<p>Anyone could tell me what's wrong in the approach I take to solve this problem?<br></p>
<p>Any help would be appreciated!</p>
| Hagen von Eitzen | 39,174 | <p>As abbreviation, define $f(n,d)=n\cdot (n+1)\cdot\ldots\cdot (n+d-1)$.</p>
<p>Base case $n=2$:
$$\forall d\ge 2\colon d|f(2,d) = 2\cdot 3\cdot\ldots \cdot d$$
This is true because $d$ occurs among the factors on the rihght.</p>
<p>$n\to n+1$:</p>
<p>Assume that we know about $n$ that
$$\tag1\forall d\ge 2\colon d|f(n,d)$$
Claim: then also
$$\tag2\forall d\ge 2\colon d|f(n+1,d)$$
Indeed, let $d\ge 2$ be arbitrary.
Then by$(1)$ we know that $d|f(n,d)$, say $f(n,d)=d a$.
Note that $$n\cdot f(n+1,d) = f(n,d)\cdot (n+d)=n\cdot f(n,d)+d\cdot f(n,d)$$
Since $n$ divides $f(n,d)$ (as first factor), write $f(n,d)=n b$.
Then we obtain
$$f(n+1,d) = \frac1n(n\cdot f(n,d)+d\cdot f(n,d))=d (a+b).$$
SInce $d\ge 2$ was arbitrary, this shows $(2)$ and thus our induction step $n\to n+1$.</p>
|
2,455,408 | <p>I am encountering questions like this below.</p>
<p>$$\frac{dP}{dt}=(a-b\cos t ) \left(P+ \frac{P^2}{M}\right)$$</p>
<p>Then there is information stating $M$ is a positive integer and $a$ and $b$ are positive. That when $t=0$, $P=P_0$.</p>
<p>It wants me to solve the differential equation and show the process.</p>
<p>How am I supposed to derive this? I suspect it is a separation of variables question but I don't know how to proceed. Can anyone point me in the direction of an exercise or explanation video.</p>
| Mathemagical | 446,771 | <p>The area you need is 8 times the shaded area. Observe the equations carefully to see the symmetries. <a href="https://i.stack.imgur.com/549TS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/549TS.png" alt="enter image description here"></a></p>
<p>How to find it? Well you can see that (0,0) is where the curves meet. And the right extreme of the interval is at $x=\frac{\pi}{2}$. So the total area is $$8\int_0^{\pi/2}(x- \sin x) dx = 8\left( \frac{\pi^2}{8}-1\right)=\pi^2-8$$</p>
<p>Note: A zoom out to see that no other area is finite.<a href="https://i.stack.imgur.com/vlebM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vlebM.png" alt="enter image description here"></a></p>
|
499,476 | <p>Use mathematical induction to prove that the derivative of $f(x)=\sin(ax+b)$ is given by</p>
<p>$f^{(n)}(x)= (-1)^ka^n\sin(ax+b)$ if $n=2k$, and $(-1)^ka^n\cos(ax+b)$ if $n=2k+1$</p>
<p>for a number $k=0,1,2,3,...$</p>
<p>I have done som proofs by induction, but I seem to struggle as soon as trig functions appear.</p>
| Cameron Buie | 28,900 | <p>You'll need to use the Chain Rule, together with the fact that $$\frac{d}{du}\sin(u)=\cos(u)$$ and $$\frac{d}{du}\cos(u)=-\sin(u).$$</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>
| BLAZE | 144,533 | <p>It might help you see what's going on if you write them as two story fractions: $$\frac{x^{-6}}{-x^{-4}}=\frac{\left(\dfrac{1}{x^6}\right)}{\left(-\dfrac{1}{x^4}\right)}=\underbrace{\color{red}{\frac{1}{x^6}\times -\frac{x^4}{1}}}_{\large \color{blue}{\text{by the reciprocal rule}}}=-\frac{1}{x^2}=\frac{1}{-x^2}$$ So you are correct, and the "answer" is wrong.</p>
|
315,844 | <p>What is the probability P(X>Y) given that X,Y are Uniformly distributed between [0,1]?</p>
| Seyhmus Güngören | 29,940 | <p>Assuming $X$ and $Y$ are <em>independent</em>, Let $Z=X-Y$, Then were looking for $P(Z>0)$. To obtain the distribution of $Z$, you can convolve $X$ by $-Y$. The result will be a triange in the range $-1<x<1$. Then it it is obvious to see that $P(Z>0)=0.5$.</p>
|
866,654 | <p>Reading various betting forum I came across different threads claiming <strong><em>betting multiple is worse than betting on single events</em></strong>.</p>
<p>Could you explain why?</p>
<p>[Clairification for the ones not familiar with betting:
Betting on a single event: predict the outcome of a single match.
Betting on multiple events: predict the outcome of multiple matches, you win only if you guess correctly the outcome of all the events, even if you only make one mistake you win nothing.]</p>
<p>I think that the key point is the commission taken by bookmakers on each match (they take something like 5% on every bet).</p>
<p>For example if you bet 100dollars on a tennis match where each player has P(win) = 0.5, You will win something like 195 instead of 200, due to commissions.</p>
| Thanos Darkadakis | 105,049 | <p>I'm afraid you're wrong. Betting on multiple events is worse. This happens because of the commissions. I will try to explain it with numbers and not with formulas, because it's easier to be understood.</p>
<p>Assume you want to bet on 2 events with P(win)=0.5. Initially you have €100.</p>
<p><strong>Case 1: No commissions:</strong></p>
<p><strong>Case 1a</strong>: Betting on single events:</p>
<p>1st event: You bet €100. You win. You win €100. You get €100.You have €200</p>
<p>2nt event: You bet €200. You win. You win €200. You get €200.You have €400</p>
<p><strong>Case 1b</strong>: Betting on multiple events:</p>
<p>You bet €100. You win. You win €300. You get €300. You have €400.</p>
<p>The result is the same.</p>
<p><strong>Case 2: With commissions:</strong> (assuming commission is 5%)</p>
<p>First of all, you have to know that the commissions are a percentage of your <strong>winnings</strong>. This means if you bet €1000 and you win €10, you will be charged the same as if you had bet €1 and you had won €10.</p>
<p><strong>Case 2a</strong>: Bettins on single events:</p>
<p>1st event: You bet €100. You win. You win €100. Commission is 100*0.05=€5.You get €95. You have €195.</p>
<p>2st event: You bet €195. You win. You win €195. Commission is 195*0.05=€9,75.You get €185,25. You have €380,25.</p>
<p><strong>Case 2b</strong>: Betting on multiple events:</p>
<p>You bet €100. You win. You win €300. Commission is 300*0.05=€15.You get €285. You have €385.</p>
|
2,406,043 | <p>Let the triangle $\triangle ABC$ have sides $a,b,c$ and be inscribed in a circle with radius $R$. If $p=\frac{a+b+c}{2}$ The radius of the circle can be expressed as</p>
<p>a) $$R=\frac{\sqrt{p(p-a)(p-b)(p-c)}}{4abc}$$</p>
<p>b) $$R=\frac{4\sqrt{p(p-a)(p-b)(p-c)}}{abc}$$</p>
<p>c) $$R=\frac{abc}{4\sqrt{p(p-a)(p-b)(p-c)}}$$</p>
<p>d) $$R=\frac{4abc}{\sqrt{p(p-a)(p-b)(p-c)}}$$</p>
<hr>
<p>So clearly Heron's formula is being used here. I know that the radical is an expression of the area of the triangle. Let's denote $A_T=\sqrt{p(p-a)(p-b)(p-c)}$ and solve for it in all four cases:</p>
<p>a) $A_T=4Rabc$</p>
<p>b) $A_T=\frac{Rabc}{4}$</p>
<p>c) $A_T=\frac{abc}{4R}$</p>
<p>d) $A_T=\frac{4abc}{R}$</p>
<p>None of the RHS's even closely resembles to the area of the triangle of the form $A=\frac{base \cdot height}{2}.$ How should I do this? Keep in mind that this is from a test where one is to have about 3 minutes per question, so no complicated solution should be needed.</p>
| Michael Rozenberg | 190,319 | <p>Also, we can use SOS here:
$$a^3+b^3+c^3-3abc=\sum_{cyc}(a^3-abc)=\sum_{cyc}a(a^2-bc)=$$
$$=\frac{1}{2}\sum_{cyc}a((a-b)(a+c)-(c-a)(a+b))=$$
$$=\frac{1}{2}\sum_{cyc}(a-b)(a(a+c)-b(b+c))=$$
$$=\frac{1}{2}\sum_{cyc}(a-b)(a^2-b^2+ac-bc)=\frac{1}{2}\sum_{cyc}(a-b)^2(a+b+c)\geq0.$$</p>
|
3,636,667 | <blockquote>
<p>Evaluate
<span class="math-container">$$\lim_{n\to\infty}\frac{1}{n^{p+1}}\cdot \sum_ \limits{i=1}^{n} \frac{(p+i)!}{i!} , p \in N$$</span> </p>
</blockquote>
<p>Now, I found this problem while doing some practice and I am curious on how to solve it . I have no good ideas yet, so I will appreciate any hints !</p>
| bernat tobella | 745,671 | <p>Did you try by factoring the inside of summatory? Also, you can solve it for different values of p. (this should be a comment but i don't have enough reputation)</p>
|
3,636,667 | <blockquote>
<p>Evaluate
<span class="math-container">$$\lim_{n\to\infty}\frac{1}{n^{p+1}}\cdot \sum_ \limits{i=1}^{n} \frac{(p+i)!}{i!} , p \in N$$</span> </p>
</blockquote>
<p>Now, I found this problem while doing some practice and I am curious on how to solve it . I have no good ideas yet, so I will appreciate any hints !</p>
| CHAMSI | 758,100 | <p>First of all, <span class="math-container">$ \left(\forall x\in\mathbb{R}_{+}\right),\ \mathrm{e}^{x}-1=x\int_{0}^{1}{\mathrm{e}^{xy}\,\mathrm{d}y}\leq x\, \mathrm{e}^{x} \cdot $</span></p>
<p>Let <span class="math-container">$ n,p $</span> be positive integers, we have the following : <span class="math-container">\begin{aligned} \left|\frac{1}{n^{p+1}}\sum_{i=1}^{n}{\frac{\left(p+i\right)!}{i!}}-\frac{1}{n^{p+1}}\sum_{i=1}^{n}{i^{p}}\right|&=\frac{1}{n^{p+1}}\sum_{i=1}^{n}{i^{p}\left(\prod_{j=1}^{p}{\left(1+\frac{j}{i}\right)}-1\right)}\\ &\leq\frac{1}{n^{p+1}}\sum_{i=1}^{n}{i^{p}\left(\prod_{j=1}^{p}{\exp{\left(\frac{j}{i}\right)}}-1\right)}\\&\leq\frac{1}{n^{p+1}}\sum_{i=1}^{n}{i^{p}\left(\exp{\left(\frac{p\left(p+1\right)}{2i}\right)}-1\right)}\\ &\leq\frac{1}{2n^{p+1}}\sum_{i=1}^{n}{i^{p-1}\exp{\left(\frac{p\left(p+1\right)}{2i}\right)}}\\ \left|\frac{1}{n^{p+1}}\sum_{i=1}^{n}{\frac{\left(p+i\right)!}{i!}}-\frac{1}{n^{p+1}}\sum_{i=1}^{n}{i^{p}}\right|&\leq\frac{\mathrm{e}^{\frac{p\left(p+1\right)}{2}}}{2n}\times\frac{1}{n}\sum_{i=1}^{n}{\left(\frac{i}{n}\right)^{p-1}}\underset{n\to +\infty}{\longrightarrow}0\times\int_{0}^{1}{x^{p-1}\,\mathrm{d}x}=0 \end{aligned}</span></p>
<p>Thus <span class="math-container">$$ \lim_{n\to +\infty}{\frac{1}{n^{p+1}}\sum_{i=1}^{n}{\frac{\left(p+i\right)!}{i!}}}=\lim_{n\to +\infty}{\frac{1}{n^{p+1}}\sum_{i=1}^{n}{i^{p}}}=\int_{0}^{1}{x^{p}\,\mathrm{d}x}=\frac{1}{p+1} $$</span></p>
|
4,274,600 | <p><span class="math-container">$(x^3+x+1)^{-1} \mod (x^4+x+1)$</span> over <span class="math-container">$\text{GF}(2)$</span></p>
<p>I understand well how to solve the equation without inverse but don't know how to solve it with inverse.</p>
| Bernard | 202,857 | <p>Just perform the <em>extended Euclidean algorithm</em> in <span class="math-container">$\mathbf F_2$</span> to obtain a Bézout's relation between <span class="math-container">$X^4+X+1$</span> and <span class="math-container">$X^3+X+1$</span>:
<span class="math-container">\begin{array}{r| lll}
r(X) & u & v & q \\ \hline
X^4+X+1 & 0 & 1 \\
X^3+X+1&1 & 0 & X \\ \hline
X^2+1 & X & 1 & X \\
1 & 1 +X^2 & X \\
\hline
\end{array}</span></p>
<p>Therefore, <span class="math-container">$(X^2+1)(X^3+X+1)+X(X^4+X+1)=0 $</span>.</p>
|
3,682,987 | <blockquote>
<p>Let <span class="math-container">$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>
| Anas A. Ibrahim | 650,028 | <p>Well, by Cauchy-Schwarz,
<span class="math-container">$$((a+b)+(a+c)+(b+c))\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right) \ge (1+1+1)^2=9$$</span>
<span class="math-container">$$\iff (a+b+c)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right) \ge \frac{9}{2} $$</span></p>
|
1,309,670 | <p>Suppose $D \subset \mathbb{R}$ is open, $f : D \to \mathbb{R}$ is a smooth (not necessarily real analytic) function, $x_0 \in D$, and $T_n$ is the degree $n$ Taylor polynomial of $f$ centered at $x_0$. Let $S=\{ x \in D : f(x)=T_n(x) \}$. It is not hard to see that $S$ is closed and contains $x_0$. What else can be said about it (possibly with additional hypotheses)? I suppose the most general possible question is: can $S$ be an arbitrary closed subset of $D$?</p>
<p>I know that $S$ need not be finite; for instance, the constant approximation of $\sin$ at $\pi/2$ has $S=\pi/2 + 2\pi \mathbb{Z}$. I also know that $S$ need not be discrete, at least if the domain is not compact; for instance, if the domain is $(0,1]$ and $f(x)=\sin(1/x)$ and we take the constant approximation at any point, then $S$ has a limit point at $0$.</p>
<p>Major edit: evidently $D \setminus S$ can at least be any open interval, because $f$ could be a bump function supported interval $I$ and take a point of expansion outside $I$, in which case $T_n$ will be zero and $T_n - f$ will be zero exactly outside $I$. Can we use this to prove that $S$ can be any closed set containing $x_0$?</p>
| Robert Israel | 8,508 | <p>Outside a neighbourhood of $x_0$, $T_n - f$ is an arbitrary smooth function.
So essentially you're asking what can be said about the zero set of an arbitrary smooth function. That can be any closed subset of $\mathbb R$.</p>
|
997,602 | <blockquote>
<p>Prove that the function <span class="math-container">$x \mapsto \dfrac 1{1+ x^2}$</span> is uniformly continuous on <span class="math-container">$\mathbb{R}$</span>.</p>
</blockquote>
<p>Attempt: By definition a function <span class="math-container">$f: E →\Bbb R$</span> is uniformly continuous iff for every <span class="math-container">$ε > 0$</span>, there is a <span class="math-container">$δ > 0$</span> such that <span class="math-container">$|x-a| < δ$</span> and <span class="math-container">$x,a$</span> are elements of <span class="math-container">$E$</span> implies <span class="math-container">$|f(x) - f(a)| < ε.$</span></p>
<p>Then suppose <span class="math-container">$x, a$</span> are elements of <span class="math-container">$\Bbb R. $</span>
Now
<span class="math-container">\begin{align}
|f(x) - f(a)|
&= \left|\frac1{1 + x^2} - \frac1{1 + a^2}\right|
\\&= \left| \frac{a^2 - x^2}{(1 + x^2)(1 + a^2)}\right|
\\&= |x - a| \frac{|x + a|}{(1 + x^2)(1 + a^2)}
\\&≤ |x - a| \frac{|x| + |a|}{(1 + x^2)(1 + a^2)}
\\&= |x - a| \left[\frac{|x|}{(1 + x^2)(1 + a^2)} + \frac{|a|}{(1 + x^2)(1 + a^2)}\right]
\end{align}</span></p>
<p>I don't know how to simplify more. Can someone please help me finish? Thank very much.</p>
| Michael Hardy | 11,667 | <p>According to the mean value theorem,
$$
\left|\frac1{1 + x^2} - \frac1{1 + a^2}\right| = f'(c)|x-a|
$$
where $f(x)=\dfrac 1 {1+x^2}$ and $c$ is somewhere between $x$ and $a$. But $|f'(c)|\le\max |f'|$, the absolute maximum value of $|f'|$. In order for this to make sense, you need to show that $|f'|$ does have an absolute maximum value. But that is not hard. So you have
$$
|f(x)-f(a)|\le M|x-a| \text{ for ALL values of $x$ and $a$},
$$
(where $M$ is the absolute maximum of $|f'|$). So $f$ is Lipschitz-continuous and therefore uniformly continuous.</p>
|
86,800 | <p>I am curious about how the Heegaard genus changes after a finite covering. </p>
<p>Is there anyone constructing an closed hyperbolic 3-manifold $N$ such that </p>
<p>the Heegaard genus of a finite covering of $N$ is less than the Heegaard genus of $N$? </p>
<p>Thank you!</p>
<p>Note: Heegaard genus of a 3-manifold means the minimal genus of all Heegaard splittings.</p>
| Michael Siler | 5,413 | <p>There are examples like this. Check out section 4.5 of Shalen's paper "Hyperbolic volume, Heegaard genus and ranks of groups." It's here: <a href="http://arxiv.org/abs/0904.0191">http://arxiv.org/abs/0904.0191</a></p>
<p>He gives a reference for a genus 3 example by Alan Reid and a sketch of a technique for producing examples by Hyam Rubinstein. Shalen also conjectures that the genus can drop by at most 1 in a finite cover of a closed hyperbolic 3-manifold.</p>
|
2,449,443 | <p>Set of numbers $\ x_1, \ldots, x_m , y_1, \ldots, y_n $ where $\ x_i=0 $ for $i = 1,\ldots, m$ and $\ y_i=1 $ for $i = 1,\ldots, n$</p>
<p>Show that mean $M$ of this set is given by $\frac{n}{m+n}$ and the standard deviation $S$ by $\frac{ \sqrt{mn}} {m+n} $</p>
<p>I know the definitions of the mean and standard deviation and how to get them but Im really stuck at that question</p>
| Donald Splutterwit | 404,247 | <p>Calculate the zero, first and second moments
\begin{eqnarray*}
\sum 1 = ? \\
\sum z_j = ? \\
\sum z_j^2 = ? \\
\end{eqnarray*}
Then use the formulea
\begin{eqnarray*}
\mu&=&\frac{\sum z}{\sum 1 } \\
\sigma^2 &=& \frac{\sum z^2}{\sum 1 }- \frac{(\sum z)^2}{(\sum 1)^2 }.
\end{eqnarray*}</p>
<blockquote class="spoiler">
<p> $\sum 1 = n+m \,\,\, \sum z_j = n \,\,\, \sum z_j^2 = n $</p>
</blockquote>
|
801,081 | <p>I was doing some school work and got bored so I started messing with k-gonal numbers. I started with the triangular numbers, square numbers and looked for patterns. I noticed something.</p>
<p>Let $n^{(k)}$ denote the $n$-th $k$-gonal number. For example, $3^{(3)}$ is the third triangular number, 6.</p>
<p>I found that there was an easy way to compute the formula for each $k$-gonal and noticed that the
$$n^{(k)}=n^{(k-1)}+(n-1)^{(3)}$$</p>
<p>So to find the formula for the $n$-th pentagonal number,
$$n^{(5)}=n^{(4)}+(n-1)^{(3)}$$
$$n^{(5)}=n^2+\frac{n(n-1)}{2}$$
$$n^{(5)}=\frac{n(3n-1)}{2}$$</p>
<p>So after doing this a bunch of times, I think I found the pattern...</p>
<p>Let $f:\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}$ such that
$$f(n,k)=\frac{n[(k-2)n-(k-4)]}{2}$$
Is this the formula for the $n$-th $k$-gonal number? Are there any other intersting formulas that come out of the 2 x 2 array of these numbers like the one I derived above?</p>
| Janaka Rodrigo | 1,043,137 | <p>Method- (ii) <br/>
3-gonal numbers
1,3,6,10,15,..
4-gonal numbers
1,4,9,16,25,...
5-gonal numbers <br/>
1,5,12,22,35,... <br/>
6-gonal numbers <br/>
1,6,15,28,45,... <br/>
Let u(k,n) be the n th k-gonal number <br/>
In above sequences second difference is a constant, 3-gonalsequence it is 1, 4- gonal sequence it is 2, 5-gonal sequence it is 3
therefore it is k -2 in the case of k-gonal sequence <br/>
u(k,n) need to be a quadratic polynomial of n <br/>
u(k,n) = an^2 + bn + c <br/>
In all above sequences first term is 1. <br/>
By observing the patterns second term of the k- gonal number sequence is given by adding k-1 to the first term and the third term is given by adding (k-1) + (k-2) that is 2k-3 to the second term. <br/>
Therfore first three terms of k-gonal sequence is 1,k,3k-3 <br/>
By substituting n =1,2,3 for u(k,n) <br/>
a+b+c =1 <br/>
4a+2b+c = k <br/>
9a+3b+c = 3k-3 <br/>
By solving these equations <br/>
a= (k-2)/2 , b= (4-k)/2 , c =0 <br/>
Therefore <br/>
u(k,n) = {(k-2)/2}n^2 + {(4-k)/2}n <br/>
u(k,n) =( n/2){ (n-1)k-2n +4 } <br/>
Importance of this approach is now you can deduce the relationship <br/>
u(k,n) = u(k-1,n) + u(3,n-1)</p>
|
2,199,222 | <p>I have the feeling of being stuck or missing something trying to prove
$$ \lim_{N\to\infty}\sum_{k=1}^{N} \frac{1}{N+k} =\int_{1}^{2} \frac{1}{x} dx = ln(2)$$</p>
<p>Using Riemann-Sums I have shown that $$\int_{1}^{a} \frac{1}{x} dx=\lim_{N\to\infty}\sum_{k=1}^{N} (a^{1/N}-1)=\lim_{N\to\infty}N(a^{1/N}-1)=\lim_{h\to 0}\frac{a^h-1}{h}=ln(a)$$</p>
<p>So I would have to show that </p>
<p>$$\lim_{N\to\infty}\sum_{k=1}^{N} \frac{1}{N+k}=\lim_{N\to\infty}\sum_{k=1}^{N} (2^{1/N}-1)$$</p>
<p>However the summands are not equal. How does one prove this equality?</p>
| Riley | 320,172 | <p>I believe you are thinking of a skew-symmetric matrix, but this requires the diagonal to be 0 as well. It is skew-Hermitian if you require the diagonal to be imaginary and all other entries to be real. If the diagonal must have real values, then I don't believe there is an appropriate term for this, but you can call it the sum of a skew-symmetric and a diagonal matrix.</p>
|
1,618,753 | <p>Trying to expand $f(x)=\cot(x)$ to Taylor series (Maclaurin, actually).
But I keep "adding up" infinities when using the formula. (Because of $\cot(0)=\infty$) Could you perhaps give me a hint on how to proceed?</p>
| latorrefabian | 306,438 | <p>The correct answer is that x = 0 is not in the domain of cot(x). Continuity is a property of elements of the domain of the function.</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>
| shardulc | 140,607 | <p>As @almagest has pointed out, the absolute maximum number of faces you can see of a polyhedron with $n$ faces is $n-1$. This is achieved in the case of a right pyramid with a base and $n-1$ sides; if you view the pyramid from above the apex, you can see all the sides except the base. This is perhaps true for non-right pyramids and other shapes as well.</p>
<p>The absolute minimum number of faces you can see is 1, as you said: just place yourself (or the camera) arbitrarily close to any one face. As the polyhedron is convex, none of the faces will 'tower over' any one shape and you will see only one shape.</p>
<p>Both these bounds, however, are quite obvious and useless. As in the answer above, half the faces of a regular polyhedron with a center of symmetry can be seen from sufficiently far away. I would extend this to say that roughly half of the faces of a roughly regular polyhedron can be seen from sufficiently far away, where 'roughly' is an appropriate tolerance constant. I think it is illustrative to think of the sphere that completely circumscribes the polyhedron, of which you can obviously see exactly half. Maybe that half can be 'mapped' on to the polyhedron's faces.</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>
| G Cab | 317,234 | <p>Not only for a pyramid, but also for a "hemi-spheric diamond" cut with any number of flat faces, when looked from far enough, will show $n-1$ faces.<br>
If you consider regular polyhedron only, than the answer may vary.<br>
Coming back to a general method, for a general convex polyhedron and a "common" camera, then consider the camera visual cone (something less than $
{2\pi }$ steradians), with the vertex placed at the camera position. It will collect all the rays within that cone. Now consider the polyhedron as light emitting. You may consider that each face radiate from its centre in all the directions radiating outwards. If the polyhedron is convex, each face should freely radiated in the outward hemisphere. Exclude rays at $90°$ from vertical if you do not account for face=line. Check if there is a radiated ray that can reach the camera, and if it is within its view cone. Or invert the ray path.<br>
With a proper math description of the polyhedron, the SW task is not too difficult.</p>
|
1,690,715 | <p>I have this space $E=\mathcal{C}([0,1],\mathbb{R})$ and the inner product $d(f,g)=\int_0^1 |f(x)-g(x)|\,{\rm d}x$.</p>
<p>Who have an idea about a simple sequence $\{f_n\}_{n=1}^\infty$ which is Cauchy but not convergent in $(E,d)$?</p>
| Community | -1 | <p>Hint: consider a sequence of functions like:</p>
<p>$$f_n(x)=\begin{cases} 0 \text{ for } x\in [0,\frac{1}{2}-\frac{1}{n}]\\
\text{linear} \text{ for } x\in [\frac{1}{2}-\frac{1}{n},\frac{1}{2}+\frac1n]\\
1 \text{ for } x\in [\frac{1}{2}+\frac{1}{n},1]
\end{cases}$$</p>
|
266,832 | <p>Let $p(x) = \sum_{k \geq 0} a_k x^k$ where the $a_k$'s are IID random variables taken from a mean-zero random variable taking finitely many values in $\mathbb{R}$; it clearly converges for $-1<x<1$. Is it a.s. true that the sign of $p(x)$ oscillates infinitely often as $x \rightarrow 1^-$? That is, is it the case (with probability 1) that there exist $x_1 < x_2 < x_3 < \dots$ in (0,1) such that $p(x_k)$ has the same sign as $(-1)^k$?</p>
<p>I imagine that this is well-known for $P(1) = P(-1) = 1/2$ (where $P(r)$ denotes the probability that $a_k = r$); the case that most interests me is $P(1) = P(-1) = 1/4$ and $P(0) = 1/2$, but I'm sure that the same technique settles both cases.</p>
<p>I am also interested in knowing about the magnitude of the oscillations.</p>
| Igor Rivin | 11,142 | <p><em>Edelman, Alan; Kostlan, Eric</em>, <a href="http://dx.doi.org/10.1090/S0273-0979-1995-00571-9" rel="nofollow noreferrer"><strong>How many zeros of a random polynomial are real?</strong></a>, Bull. Am. Math. Soc., New Ser. 32, No.1, 1-37 (1995). <a href="https://zbmath.org/?q=an:0820.34038" rel="nofollow noreferrer">ZBL0820.34038</a>.</p>
<p>Yes, this follows immediately from the Kac-Rice formula, see Edelman-Kostlan </p>
|
4,510,795 | <p>I need to find the % difference between two numbers. One person told me to use <span class="math-container">$\frac{x-y} {x} $</span>, another told me to use <span class="math-container">$\frac x y$</span> <span class="math-container">$- 1$</span> . Who is right?</p>
<p>Example:
Today's price: <span class="math-container">$23892$</span>
Yesterday's price: <span class="math-container">$23941$</span></p>
<p>Using <span class="math-container">$\frac{x-y} {x} $</span>: <span class="math-container">$-0.21%.$</span>
Using <span class="math-container">$\frac x y$</span> <span class="math-container">$- 1$</span> : <span class="math-container">$-0.20%$</span></p>
<p>It's almost the same result, but not quite. How to intuitively understand why?</p>
| Yaroslav Nikitenko | 186,740 | <p>The original equation is equivalent to a system with <span class="math-container">$n+1$</span> unknowns:</p>
<p><span class="math-container">$$ \eqalign{
\sum a_i r_i & = 0, \cr
r_1^2 - f_1(x) & = 0, \cr
&... \cr
r_n^2 - f_n(x) & = 0,
}
$$</span></p>
<p>with additional requirements all <span class="math-container">$r_i \geq 0$</span>.</p>
<p>I was able to solve such system with sagemath (UPD: only for the simplest case, not for the updated question). Please add answers/comments, if there is some more powerful method than this or if I missed something!</p>
|
3,631,648 | <p>Suppose <span class="math-container">$X_1, ..., X_n \stackrel{iid}{\sim}$</span> Exponential(rate = <span class="math-container">$\lambda$</span>) independent of <span class="math-container">$Y_1, ..., Y_n \stackrel{iid}{\sim}$</span> Exponential<span class="math-container">$(1)$</span>. </p>
<p>Define <span class="math-container">$Z_i \equiv \min\{X_i, Y_i\}$</span></p>
<p>I want to find the maximum likelihood estimator for <span class="math-container">$\lambda$</span> in the following scenario: I observe <span class="math-container">$Z_1, ..., Z_n$</span> and <span class="math-container">$Y_1, ..., Y_n$</span> but NOT any of the <span class="math-container">$X_i$</span>.</p>
<p>First I need to determine the likelihood and then maximize it over <span class="math-container">$\theta > 0$</span>, but I'm not really sure of the right approach. I calculate the joint cdf as follows:</p>
<p><span class="math-container">$$P(Z_i \leq z, Y_i \leq y) = \begin{cases} P(Y_i \leq y), & y \leq z \\ P(Y_i \leq z, Y_i \leq X_i) + P(Y_i \leq y, X_i \leq z, X_i < Y_i), & y > z\end{cases} \\
= \begin{cases} 1- e^{-y}, & y \leq z \\
1-e^{-z} + (e^{-z}-e^{-y})(1-e^{-\lambda z}), & y > z \end{cases}$$</span></p>
<p>This is because <span class="math-container">$Z_i \leq Y_i$</span> always. Would the likelihood function therefore be:</p>
<p><span class="math-container">$$L(\lambda |Y_i, Z_i, i \in \{1,...n\}) = \prod_{\{i : Y_i = Z_i\}} (1-e^{-Y_i}) \prod_{\{i:Y_i > Z_i\}} \lambda e^{-Y_i}e^{-\lambda Z_i}$$</span></p>
<p>splitting into the "discrete" and "continuous" parts? Or am I getting this wrong? Or should I be doing something like <a href="https://math.stackexchange.com/questions/3512290/exam-prep-maximum-likelihood-estimator">here</a> or <a href="https://math.stackexchange.com/questions/3156467/how-to-find-the-mle-of-these-parameters-given-distribution?noredirect=1&lq=1">here</a>? I should note my scenario is different than theirs, as intuitively at least, observing the magnitude of the difference between the minimum and the maximum (in the cases where <span class="math-container">$Z_i$</span> and <span class="math-container">$Y_i$</span> differ) should give us more information about <span class="math-container">$\lambda$</span>, right?</p>
| qp212223 | 428,839 | <p>Would this be <span class="math-container">$$\prod_{\{i: Y_i = Z_i\}} \frac{1}{\lambda +1} \prod_{\{i: Y_i > Z_i\}} e^{-Y_i}\lambda e^{-\lambda Z_i} $$</span></p>
<p>where we just have the point mass/probability of equality contributing when <span class="math-container">$Y_i = Z_i$</span> and the joint density contributing otherwise. Can someone please provide some insight?</p>
|
3,631,648 | <p>Suppose <span class="math-container">$X_1, ..., X_n \stackrel{iid}{\sim}$</span> Exponential(rate = <span class="math-container">$\lambda$</span>) independent of <span class="math-container">$Y_1, ..., Y_n \stackrel{iid}{\sim}$</span> Exponential<span class="math-container">$(1)$</span>. </p>
<p>Define <span class="math-container">$Z_i \equiv \min\{X_i, Y_i\}$</span></p>
<p>I want to find the maximum likelihood estimator for <span class="math-container">$\lambda$</span> in the following scenario: I observe <span class="math-container">$Z_1, ..., Z_n$</span> and <span class="math-container">$Y_1, ..., Y_n$</span> but NOT any of the <span class="math-container">$X_i$</span>.</p>
<p>First I need to determine the likelihood and then maximize it over <span class="math-container">$\theta > 0$</span>, but I'm not really sure of the right approach. I calculate the joint cdf as follows:</p>
<p><span class="math-container">$$P(Z_i \leq z, Y_i \leq y) = \begin{cases} P(Y_i \leq y), & y \leq z \\ P(Y_i \leq z, Y_i \leq X_i) + P(Y_i \leq y, X_i \leq z, X_i < Y_i), & y > z\end{cases} \\
= \begin{cases} 1- e^{-y}, & y \leq z \\
1-e^{-z} + (e^{-z}-e^{-y})(1-e^{-\lambda z}), & y > z \end{cases}$$</span></p>
<p>This is because <span class="math-container">$Z_i \leq Y_i$</span> always. Would the likelihood function therefore be:</p>
<p><span class="math-container">$$L(\lambda |Y_i, Z_i, i \in \{1,...n\}) = \prod_{\{i : Y_i = Z_i\}} (1-e^{-Y_i}) \prod_{\{i:Y_i > Z_i\}} \lambda e^{-Y_i}e^{-\lambda Z_i}$$</span></p>
<p>splitting into the "discrete" and "continuous" parts? Or am I getting this wrong? Or should I be doing something like <a href="https://math.stackexchange.com/questions/3512290/exam-prep-maximum-likelihood-estimator">here</a> or <a href="https://math.stackexchange.com/questions/3156467/how-to-find-the-mle-of-these-parameters-given-distribution?noredirect=1&lq=1">here</a>? I should note my scenario is different than theirs, as intuitively at least, observing the magnitude of the difference between the minimum and the maximum (in the cases where <span class="math-container">$Z_i$</span> and <span class="math-container">$Y_i$</span> differ) should give us more information about <span class="math-container">$\lambda$</span>, right?</p>
| Henry | 6,460 | <p>I would guess that the useful information is in the values of <span class="math-container">$Z_i$</span> and how often <span class="math-container">$Y_i=Z_i$</span> or not (perhaps call this <span class="math-container">$Q$</span>); the actual values of <span class="math-container">$Y_i$</span> may not help beyond this. </p>
<p>I think you could show <span class="math-container">$Z_1, ..., Z_n \stackrel{iid}{\sim} \text{ Exponential(rate }= \lambda+1)$</span> and independently <span class="math-container">$Q \sim \text{ Binomial}\left(n,\frac{1}{\lambda+1}\right)$</span>. In that case the useful likelihood of observing <span class="math-container">$z_1,\ldots,z_n$</span> and <span class="math-container">$q$</span> (so ignoring parts related to <span class="math-container">$Y_i-Z_i$</span> when that is positive) would be proportional to </p>
<p><span class="math-container">$$(\lambda+1)^ne^{-\sum(\lambda+1) z_i} {n \choose q}\frac{\lambda^{n-q}}{(\lambda+1)^n}={n \choose q} \lambda^{n-q} e^{-(\lambda+1)\sum z_i}$$</span></p>
<p>with logarithm a constant plus <span class="math-container">$$(n-q) \log(\lambda) -(\lambda+1)\sum z_i$$</span></p>
<p>and derivative of the logarithm with respect to <span class="math-container">$\lambda$</span> <span class="math-container">$$\frac{n-q}{\lambda} - \sum z_i$$</span> </p>
<p>and the maximum likelihood estimator <span class="math-container">$$\hat \lambda = \frac{n-q}{\sum z_i}$$</span></p>
|
208,802 | <p>Is there a continuous increasing function $ f : [0, \pi] \to [0, e] $ such that $ f(0) = 0, f(\pi) = e $ and $ f (q ) \in \mathbb{Q} $ for $ q \in \mathbb{Q} $ and $ f (q ) \in \mathbb{Q}^c $ for $ q \in \mathbb{Q}^c $? I think there should be, but I am unable to construct one. </p>
| Ross Millikan | 1,827 | <p>The <a href="http://en.wikipedia.org/wiki/Back-and-forth_method" rel="nofollow">back-and-forth method</a> that shows the isomorphism of dense countable linear orders gets you there. Let $a_i$ be an enumeration of the rationals in $(0,\pi)$ and $b_j$ be an enumeration of the rationals in $(0,e)$. Set $f(a_1)$ to the first $b_j$ that has not been used and is not prohibited by the order. In this case it will be $b_1$. Then set $f^{-1}(b_2)$ to the lowest $a_i$ that is not used yet and acceptable. Alternate back and forth, assigning $f(a_i)$'s on the odd steps and $f^{-1}(b_j)$'s on the even steps. As the rationals are dense, there will always be one available. As they are countable, each has only finitely many predecessors and we will always get to it. Now for the irrationals, use continuity.</p>
|
3,484,052 | <p>Let's say you have a series that looks like <span class="math-container">$\sum^\infty_{n=N}f(n)$</span>, where <span class="math-container">$f(n)$</span> is some <span class="math-container">$n$</span>-dependent thing. If you take the limit of this series as <span class="math-container">$N$</span> approaches infinity, what kind of stuff can you use to figure out what the limit is? For instance, I read somewhere else <span class="math-container">$\sum^\infty_{n=N}\frac{1}{n^2} \rightarrow 0$</span> as <span class="math-container">$N \rightarrow \infty$</span>, but why?</p>
| user284331 | 284,331 | <p>Because of the Cauchy-criterion of the convergent sequence. For the existence of <span class="math-container">$\displaystyle\sum_{n=1}^{\infty}a_{n}$</span>, it simply means that the sequence <span class="math-container">$s_{n}=\displaystyle\sum_{k=1}^{n}a_{k}$</span> is convergent, then it is Cauchy. Then for each <span class="math-container">$\epsilon>0$</span>, we can find an <span class="math-container">$N$</span> such that <span class="math-container">$n,m\geq N$</span> implies that <span class="math-container">$|s_{n}-s_{m}|<\epsilon$</span>. Suppose without loss of generality that <span class="math-container">$m>n$</span>, so <span class="math-container">$|s_{n}-s_{m}|=\left|\displaystyle\sum_{k=n+1}^{m}a_{k}\right|<\epsilon$</span>. Now we take <span class="math-container">$m\rightarrow\infty$</span> to get <span class="math-container">$\left|\displaystyle\sum_{k=n+1}^{\infty}a_{k}\right|\leq\epsilon$</span> for all such <span class="math-container">$n\geq N$</span>, this means that <span class="math-container">$\displaystyle\sum_{k=n}^{\infty}a_{k}\rightarrow 0$</span> as <span class="math-container">$n\rightarrow\infty$</span>.</p>
|
4,302,213 | <blockquote>
<p>Let <span class="math-container">$R,S$</span> be rings and <span class="math-container">$\varphi : R\to S$</span> be a ring homomorphism. Verify that</p>
<ol>
<li><span class="math-container">$\varphi(na) = n\varphi(a)$</span> for all <span class="math-container">$n\in\mathbb Z$</span> and <span class="math-container">$a\in R$</span>.</li>
<li><span class="math-container">$\varphi(a^n) = (\varphi(a))^n$</span> for all <span class="math-container">$n\in\mathbb Z^+$</span> and all <span class="math-container">$a\in R$</span>.</li>
<li>If <span class="math-container">$A$</span> is a subring of <span class="math-container">$R$</span>, then <span class="math-container">$\varphi(A) = \{\varphi(a):a\in A\}$</span> is a subring of <span class="math-container">$S$</span>.</li>
</ol>
</blockquote>
<p>For (1) I have the following:
<span class="math-container">$$
\begin{split}
\varphi(na) &= \varphi((n-1)a+a) \\
&= \varphi((n-1)a)+\varphi(a) \\
&= (n-1)\varphi(a)+\varphi(a)\\
&= n\varphi(a).
\end{split}
$$</span></p>
<p>For (2): For <span class="math-container">$n=1$</span>, we have <span class="math-container">$\varphi(a) = \varphi(a)$</span>. Suppose for some <span class="math-container">$n\in\mathbb Z^+$</span> that <span class="math-container">$\varphi(a^n)=(\varphi(a))^n$</span>. Observe:
<span class="math-container">$$
\begin{split}
\varphi(a^n) & =(\varphi(a))^n \\
\varphi(a)\cdot \varphi(a^n) & =\varphi(a)\cdot (\varphi(a))^n \\
\varphi(a \cdot a^n) & =(\varphi(a))^{n+1} \\
\varphi( a^{n+1}) & =(\varphi(a))^{n+1} \\
\end{split}
$$</span></p>
<p>For (3): I verify for <span class="math-container">$a,b\in\varphi(A)$</span> then <span class="math-container">$a-b\in\varphi(A)$</span> and <span class="math-container">$a\cdot b\in \varphi(A)$</span> (for brevity).</p>
<ul>
<li>Well, <span class="math-container">$0\in A \implies 0 \in \varphi(A) $</span></li>
<li>For <span class="math-container">$a,b \in \varphi(A)$</span> we have <span class="math-container">$\varphi(a) - \varphi(b) = \varphi(a-b)$</span>, thus, verified, because <span class="math-container">$a-b\in A$</span>.</li>
<li>For <span class="math-container">$a,b \in \varphi(A)$</span>, we have <span class="math-container">$\varphi(a) \cdot \varphi(b) = \varphi (a\cdot b)$</span>, thus, verified, because <span class="math-container">$a\cdot b \in A$</span>.</li>
</ul>
<p>Did I do these right? Feedback appreciated!</p>
| Matt E. | 948,077 | <p>I agree with Wuestenfux's answer, and have some feedback to give. There is some error in your answer to part <span class="math-container">$(1)$</span>, which is similar to one in part <span class="math-container">$(2)$</span> as well. How do you know that <span class="math-container">$\phi((n-1)a)=(n-1)\phi(a)$</span>? For <span class="math-container">$(2)$</span> you suppose that there exists some <span class="math-container">$n\in\mathbb{Z}^+$</span> satisfies <span class="math-container">$\phi(a^n) = (\phi(a))^n$</span>, but it could be the case that
<span class="math-container">$\{n\in\mathbb{Z}^+|\phi(a^n) = (\phi(a))^n\}$</span> is empty. For both <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span>, one can use induction after showing the properties hold with base case <span class="math-container">$n=2$</span>.</p>
<p>For problem <span class="math-container">$(3)$</span>, remember that if <span class="math-container">$b = \phi(a)$</span> for some <span class="math-container">$a\in A$</span> then <span class="math-container">$b\in \phi(A)$</span>. It is only once we write an element <span class="math-container">$b$</span> in the form <span class="math-container">$\phi(a)$</span> that the containment <span class="math-container">$b\in \phi(A)$</span> becomes clear. For example, given <span class="math-container">$0\in A$</span>, we know by properties of ring homomorphisms that
<span class="math-container">$$\phi(0) =0.$$</span>
Hence <span class="math-container">$0 = \phi(0) \in \phi(A)$</span>.</p>
<p>"For <span class="math-container">$a,b\in \phi(A)$</span>, we have <span class="math-container">$\phi(a) - \phi(b) = \phi(a-b)$</span>." This does not show that <span class="math-container">$a-b \in \phi(A)$</span>, because <span class="math-container">$a \neq \phi(a)$</span> and <span class="math-container">$b\neq \phi(a)$</span>. Again, to say that <span class="math-container">$a\in \phi(A)$</span> means there exists some <span class="math-container">$x\in A$</span> for which <span class="math-container">$\phi(x) = a$</span>.</p>
|
1,722,964 | <p>Expression :$$(p\rightarrow q)\leftrightarrow(\neg q\rightarrow \neg p)$$
What does the symbol $\leftrightarrow$ mean ? Please explain by drawing the truth table for this expression and also with other examples if possible. <strong>I'm in a desperate situation so I'd really appreciate a quick response !</strong></p>
| Paul Evans | 190,670 | <p>The expression $A\iff B$ means <em>if and only if</em>. </p>
<p>For $\iff$, we get the following truth table
$$\begin{array}{c|c|c|}
& B\text{ is true}& B\text{ is false}\\\hline
A\text{ is true} & \text{true} & \text{false}\\\hline
A\text{ is false} & \text{false} & \text{true}\\\hline
\end{array}$$</p>
|
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>
| Integral | 33,688 | <p>Solving $\log_{10}x - x = 2$ is equivalent to solving $x = 10^{x+2}$. Now note that $x = 0 < 10^{0+2}$ and that the function $10^{x+2}$ grows faster than the function $x$ for $x \geq 0$. From this we can conclude that $x < 10^{x+2}$ for all $x \geq 0$. Therefore, the equality can't be valid.</p>
<p>Another way to see this is to plot the functions.</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>
| user146925 | 146,925 | <p>Have
$$\sum_{k=0}^{20}(\frac{4}{25})^k=\frac{1-(\frac{4}{25})^{21}}{1-\frac{4}{25}}=\frac{\frac{1}{25}\cdot(\frac{1}{25})^{20}(25^{21}-4^{21})}{\frac{1}{25}(25-4)}=\frac{25^{21}-4^{21}}{21\cdot25^{20}}$$</p>
|
997,587 | <p>The first sequence given is 3, 7, 16, 41, 77,....
I really am quite stuck on this because I can't seem to find any relationship between one term and the terms prior to it. I first noticed that it seemed like we were adding a perfect square to each one, since 3+4=7, 7+9=16, etc. But we skipped over adding the perfect square of 16 to anything so that must not be a useful idea.</p>
<p>The next problem is a recursive sequence given by $3s_{n-1}+2$ and I need to write it as a closed-form formula. I believe I should use something involving $3^n$ and I have tried subtracting $2^n$ but it only works for the first two terms.</p>
| Community | -1 | <p>For the second problem,</p>
<p>$s_n=3s_{n-1}+2 \implies s_{n-1}=3s_{n-2}+2$</p>
<p>$\therefore s_n-4s_{n-1}+3s_{n-2}=0$</p>
<p>Therefore the <a href="http://en.wikipedia.org/wiki/Recurrence_relation#Theorem" rel="nofollow">characteristic equation</a> is $t^2-4t+3=0$. Can you take it from here?</p>
|
2,153,340 | <p>Let $G$ be a group, and $C$ a set of proper subgroups of $G$.</p>
<p>Each subgroup in $C$ is normal subgroup of $G$.</p>
<p>For $G_1 , G_2\in C$, if $G_1 \ne G_2$ then $G_1\cap G_2=\{e_G\}$</p>
<p>$\bigcup\limits_{H\in C}H= G$.</p>
<p>Need to prove that G is Abelian group, hint someone?</p>
| Andreas Caranti | 58,401 | <p>First note that elements from two different normal subgroups in the family $\mathcal{C}$ commute. If $a \in A$, $b \in B$, with $A, B$ <em>different</em> normal subgroups in $\mathcal{C}$, we have
$$
[a, b] = a^{-1} b^{-1} a b = a^{-1} a^{b} = (b^{-1})^{a} b \in A \cap B = 1.
$$</p>
<p>Now let $x, y \in A$, with $A \in \mathcal{C}$. Since the subgroups in $\mathcal{C}$ are proper, there is $z \notin A$. So $z$ is contained in a subgroup of the family $\mathcal{C}$ other than $A$, and thus by the above $[x, z] = 1$. Clearly $z y \notin A$, and then again
$$
1 = [x, z y] = [x, y] [x, z]^{y} = [x, y].
$$</p>
|
4,269,898 | <p>I've a question concerning inverse limits, since I don't usually work with them this extensively.</p>
<p>I'm considering the inverse limit of the following "bi-inverse system" of <span class="math-container">$R$</span>-modules and black arrows <span class="math-container">$f_{\bullet,\bullet}$</span>, and <span class="math-container">$g_{\bullet,\bullet}$</span>. Since inverse limits commutes among themselves we can define <span class="math-container">$$A := \varprojlim_{i,j}A_{i,j}$$</span> regardless of the order we take them. Let <span class="math-container">$$\widetilde{A}:=\varprojlim_k A_{k,k}$$</span> be the limit of the diagonal of this inverse system.</p>
<p><a href="https://i.stack.imgur.com/UUsPK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UUsPK.png" alt="enter image description here" /></a></p>
<p>It's easy to see we have a unique map <span class="math-container">$$\widetilde{A} \to A$$</span> that makes everything commute (induced by the obvious maps in the system).</p>
<blockquote>
<p>I was wondering if it's "easy to see" if there's a map going in the other direction and then prove that <span class="math-container">$A\cong \widetilde{A}$</span> maybe.</p>
</blockquote>
| Kevin Arlin | 31,228 | <p>Just for the sake of argument, it is indeed easy to give a more concrete approach here. The limit of your double system <span class="math-container">$A$</span> is a submodule <span class="math-container">$\prod_{i,j} A_{i,j}$</span> defined by those tuples <span class="math-container">$(a_{i,j})$</span> such that <span class="math-container">$a_{i-1,j}=f_{i,j}(a_{i,j})$</span> and similarly for the <span class="math-container">$g$</span>'s. The canonical map <span class="math-container">$A\to \widetilde A$</span> simply sends <span class="math-container">$(a_{i,j})$</span> to <span class="math-container">$(a_{k,k}).$</span></p>
|
88,565 | <p>Today I had an argument with my math teacher at school. We were answering some simple True/False questions and one of the questions was the following:</p>
<p><span class="math-container">$$x^2\ne x\implies x\ne 1$$</span></p>
<p>I immediately answered true, but for some reason, everyone (including my classmates and math teacher) is disagreeing with me. According to them, when <span class="math-container">$x^2$</span> is not equal to <span class="math-container">$x$</span>, <span class="math-container">$x$</span> also can't be <span class="math-container">$0$</span> and because <span class="math-container">$0$</span> isn't excluded as a possible value of <span class="math-container">$x$</span>, the sentence is false. After hours, I am still unable to understand this ridiculously simple implication. I can't believe I'm stuck with something so simple.<br><br>
<strong>Why I think the logical sentence above is true:</strong><br>
My understanding of the implication symbol <span class="math-container">$\implies$</span> is the following:
If the left part is true, then the right part must be also true. If the left part is false, then nothing is said about the right part. In the right part of this specific implication nothing is said about whether <span class="math-container">$x$</span> can be <span class="math-container">$0$</span>. Maybe <span class="math-container">$x$</span> can't be <span class="math-container">$-\pi i$</span> too, but as I see it, it doesn't really matter, as long as <span class="math-container">$x \ne 1$</span> holds. And it always holds when <span class="math-container">$x^2 \ne x$</span>, therefore the sentence is true.</p>
<h3>TL;DR:</h3>
<p><strong><span class="math-container">$x^2 \ne x \implies x \ne 1$</span>: Is this sentence true or false, and why?</strong></p>
<p>Sorry for bothering such an amazing community with such a simple question, but I had to ask someone.</p>
| Samuel Muldoon | 697,677 | <h2>Introduction</h2>
<p>You asked whether the following statement is true or not:</p>
<blockquote>
<p><span class="math-container">$x^2 \ne x \implies x \ne 1$</span></p>
</blockquote>
<p>Another way to write the above is as:</p>
<blockquote>
<p>if <span class="math-container">$(x^2 \ne x)$</span> then <span class="math-container">$(x \ne 1)$</span></p>
</blockquote>
<p>The arrow symbol (<span class="math-container">$\implies$</span>) is similar, but not quite the same, as the English phrase <code>if....then....</code></p>
<h2>The contra-positive of <span class="math-container">$(P \implies Q)$</span></h2>
<p>Okay now: suppose that you have the statement of the form "If <span class="math-container">$P$</span> then <span class="math-container">$Q$</span>"<br />
<span class="math-container">$P$</span> and <span class="math-container">$Q$</span> can be any true or false statements of your choice.<br />
For example, <span class="math-container">$P$</span> could be the statement "<code>pillows are soft</code>"<br />
"If <span class="math-container">$P$</span> then <span class="math-container">$Q$</span>" is also sometimes written as <span class="math-container">$(P \implies Q)$</span></p>
<p>There is a well known theorem in logic which states the following:</p>
<blockquote>
<p>For any <span class="math-container">$P$</span> and <span class="math-container">$Q$</span> taken from the set <span class="math-container">$\{$</span> true, false <span class="math-container">$\}$</span>, the following is true:<br />
<span class="math-container">$(\text{not } P \implies \text{not } Q)$</span> if and only if <span class="math-container">$(Q \implies P)$</span></p>
</blockquote>
<p>That means that the following two statements are equivalent:</p>
<blockquote>
<p>if <span class="math-container">$(x^2 \ne x)$</span> then <span class="math-container">$(x \ne 1)$</span><br />
if <span class="math-container">$(x = 1)$</span> then <span class="math-container">$(x^2 = x)$</span></p>
</blockquote>
<p>In some sense, the statement says that <span class="math-container">$1^2 = 1$</span>, which is true enough.</p>
<h2>Sets</h2>
<p>It considered to be very <em>bad</em> math to write things like the following:</p>
<blockquote>
<p>Is it true or false that <span class="math-container">$(x^2 \ne x \implies x \ne 1)$</span>?</p>
<p>Find <span class="math-container">$x$</span> such that <span class="math-container">$3 + (x-5)^{2} = 0$</span></p>
</blockquote>
<p>This is because the above examples fail to explain what <span class="math-container">$x$</span> is.</p>
<ul>
<li>Is <span class="math-container">$x$</span> a <strong>whole number</strong>, such as <span class="math-container">$1, 2, 3, \dots, 98, 99, 100$</span>?</li>
<li>Is <span class="math-container">$x$</span> a <strong>decimal number</strong>, such as <span class="math-container">$\sqrt{2}$</span></li>
<li>Is <span class="math-container">$x$</span> a <strong>non-real complex number</strong>, such as <span class="math-container">$20 + 5*i$</span>?</li>
</ul>
<p>You are supposed to write things like this instead:</p>
<blockquote>
<p>Is the following true or false?</p>
<ul>
<li>For every <strong>real number</strong> <span class="math-container">$x, (x^2 \ne x \implies x \ne 1)$</span></li>
</ul>
<p>Find every <strong>complex non-real number</strong> <span class="math-container">$x$</span> such that:</p>
<ul>
<li><span class="math-container">$3 + (x-5)^{2} = 0$</span></li>
</ul>
</blockquote>
<p>My last example uses the non-real number <span class="math-container">$i$</span><br />
<span class="math-container">$i*i = -1$</span></p>
<p>I think if you think it will help if you start using "<strong>sets</strong>."</p>
<p>The following is an example of a <em><strong>set</strong></em>:</p>
<blockquote>
<p><strong>my_set</strong> <span class="math-container">$= \{1, 3, 6, 7, 22\}$</span></p>
</blockquote>
<p>A <strong>set</strong> is like a suitcase full of clothing
A set is also like a cookie jar, or cardboard box.<br />
A set is a <em>container</em>.<br />
A suitcase might contain a t-shirt.
Well, <strong>my_set</strong> contains the numbers <span class="math-container">$1, 3, 6, 7,$</span> and <span class="math-container">$22$</span>.</p>
<p><a href="https://i.stack.imgur.com/DOT50.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DOT50.png" alt="PICTURE OF A SUITCASE" /></a></p>
<p>The number <span class="math-container">$3$</span> is like a t-shirt in the sense that the number <span class="math-container">$3$</span> is inside the suitcase.</p>
<p>I went to public school in the United States.<br />
I did not see sets until I was in college.</p>
<p>Sets are basic, basic math.
sets are more basic than knowing how to compute <span class="math-container">$4.5/0.3$</span></p>
<p>You are not allowed to write math like <span class="math-container">$x^2 \ne x \implies x \ne 1$</span> unless you first tell the reader what <strong>set</strong> <span class="math-container">$x$</span> comes from.</p>
<p>The following is a very ugly formula for a function named <span class="math-container">$WEIRD$</span>:</p>
<blockquote>
<p><span class="math-container">$\text{WEIRD_FUNC}(X) = [1-w(x)]*\text{LEFT_PIECE}(X) + [w(x)]*\text{RIGHT_PIECE}(X)$</span></p>
<p><span class="math-container">$\text{LEFT_PIECE}(X) = (x+10)*(x+5)$</span></p>
<p><span class="math-container">$\text{RIGHT_PIECE}(X) = 3 + (x-7)^{2}$</span></p>
<p><span class="math-container">$W(x) = \frac{tanh(10)}{2} + \frac{tanh(x)}{2}$</span></p>
</blockquote>
<p>A plot of the weird function is shown below:</p>
<p><a href="https://i.stack.imgur.com/awBR1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/awBR1.png" alt="PLOT OF WEIRD LOOKING FUNCTION" /></a></p>
<p>Suppose I asked you,</p>
<blockquote>
<p>"Find all <span class="math-container">$x$</span> such that <span class="math-container">$WEIRD(x) = 0$</span>"</p>
</blockquote>
<p>The answers vary depending on which set <span class="math-container">$x$</span> is taken from.</p>
<p>The set of all integers <span class="math-container">$x$</span> such that <span class="math-container">$WEIRD(X) = 0$</span> is <span class="math-container">$\{-10\}$</span><br />
The set of all real numbers <span class="math-container">$x$</span> such that <span class="math-container">$WEIRD(X) = 0$</span> is approximately <span class="math-container">$\{-10, -5.01\}$</span><br />
The set of all complex numbers <span class="math-container">$x$</span> such that <span class="math-container">$WEIRD(X) = 0$</span> is approximately <span class="math-container">$\{-10, -5.01, 7+i*\sqrt{3}, 7-i*\sqrt{3}\}$</span></p>
<h3>Analyzing your classmates argument regarding the number zero</h3>
<p>Is the following statement true or not?</p>
<blockquote>
<p><span class="math-container">$x^2 \ne x \implies x \ne 1$</span></p>
</blockquote>
<p>The following is your description of your teacher, and/or classmates, reasoning:</p>
<blockquote>
<ol>
<li>When <span class="math-container">$x^2$</span> is not equal to <span class="math-container">$x$</span>, <span class="math-container">$x$</span> also can't be <span class="math-container">$0$</span>.</li>
<li><span class="math-container">$0$</span> is not excluded as a possible value of <span class="math-container">$x$</span></li>
<li>some sentence, or another, is false.</li>
</ol>
</blockquote>
<p>That description is very difficult to understand.<br />
I will say that the following three statements are equivalent to each-other.<br />
Also, all statements are false:</p>
<blockquote>
<p><span class="math-container">$\forall x \in \mathbb{R}, x^2 \ne x$</span>
for any decimal number <span class="math-container">$x$</span>, <span class="math-container">$x^2 \ne x$</span>
if <span class="math-container">$x$</span> is a real number, then <span class="math-container">$x^2 \ne x$</span></p>
</blockquote>
<p>Here is a proof:</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th>Line No.</th>
<th>statement</th>
<th>justification</th>
</tr>
</thead>
<tbody>
<tr>
<td>0</td>
<td><span class="math-container">$0$</span> exists</td>
<td>axiom</td>
</tr>
<tr>
<td>1</td>
<td><span class="math-container">$0$</span> is a decimal number.</td>
<td>axiom</td>
</tr>
<tr>
<td>2</td>
<td><span class="math-container">$0^{2} = 0$</span></td>
<td>axiom</td>
</tr>
<tr>
<td>3</td>
<td>not <span class="math-container">$(0^2 \neq 0)$</span></td>
<td>from line 2</td>
</tr>
<tr>
<td>4</td>
<td>there exists a decimal number <span class="math-container">$x$</span> such that not <span class="math-container">$(x^2 \neq x)$</span></td>
<td>from lines 0,1,3</td>
</tr>
<tr>
<td>5</td>
<td>not for every decimal number <span class="math-container">$x$</span>, <span class="math-container">$(x^2 \neq x)$</span></td>
<td>from line 4</td>
</tr>
</tbody>
</table>
</div>
<p>Notice that your teacher said, "<em><strong>possible</strong></em> value of <span class="math-container">$x$</span>"</p>
<p>In mathematics, the symbol <strong>◇</strong> is sometimes used as short-hand notation for the word "<strong>possible</strong>"</p>
<p>All of the following are logically equivalent:</p>
<blockquote>
<ul>
<li>it is not possible for me to see a movie this weekend</li>
<li><span class="math-container">$NOT$</span> <strong>◇</strong> for me to see a movie this weekend</li>
<li>It is necessary for me to <strong>NOT</strong> see a movie this weekend</li>
<li><strong>▢</strong> for me to <strong>NOT</strong> see a movie this weekend</li>
</ul>
</blockquote>
<p>The statement "<span class="math-container">$0$</span> is not excluded as a possible value of <span class="math-container">$x$</span>" can be written as:</p>
<blockquote>
<ul>
<li>not(not <strong>◇</strong> <span class="math-container">$x = 0$</span>)</li>
<li>It is not the case that it is not possible that <span class="math-container">$x$</span> is zero.</li>
</ul>
</blockquote>
<p>The above is equivalent to "it is possible that <span class="math-container">$x = 0$</span>"</p>
<p>I think that I can write a proof outline of what your professor and/or classmates were trying to say</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th>Line No.</th>
<th>statement</th>
<th>justification</th>
</tr>
</thead>
<tbody>
<tr>
<td><span class="math-container">$0$</span></td>
<td><strong>◇</strong> <span class="math-container">$x = 0$</span></td>
<td>axiom</td>
</tr>
<tr>
<td><span class="math-container">$1$</span></td>
<td><strong>▢</strong><span class="math-container">$(x^2 \ne x) \implies$</span> not <strong>◇</strong> <span class="math-container">$x = 0$</span></td>
<td>axiom</td>
</tr>
<tr>
<td><span class="math-container">$2$</span></td>
<td><strong>▢</strong> <span class="math-container">$x^2 \ne x$</span></td>
<td>axiom</td>
</tr>
<tr>
<td><span class="math-container">$3$</span></td>
<td>not <strong>◇</strong> <span class="math-container">$x = 0$</span></td>
<td>1,2 modus ponens</td>
</tr>
<tr>
<td>4</td>
<td><strong>◇</strong> <span class="math-container">$x = 0$</span> and not <strong>◇</strong> <span class="math-container">$x = 0$</span></td>
<td>lines 0, 3 conjunction</td>
</tr>
<tr>
<td>5</td>
<td><span class="math-container">$NOT$</span> <strong>▢</strong><span class="math-container">$(x^2 \ne x)$</span></td>
<td>from line 4, reject 2</td>
</tr>
</tbody>
</table>
</div>
<p>I think that your teacher was saying that if <span class="math-container">$x = 0$</span> is possible, then it is not necessary that <span class="math-container">$x^2 \ne x$</span></p>
<p><span class="math-container">$0 \in S \implies NOT(\forall x \in S, x^2 \ne x)$</span></p>
<h3>One last Note</h3>
<p>All of the following are equivalent:</p>
<blockquote>
<ul>
<li>(it is necessary that <span class="math-container">$x^2 \neq x$</span>) implies that (it is necessary that <span class="math-container">$x \neq 1$</span>)</li>
<li>(it is <strong>NOT</strong> necessary that <span class="math-container">$x \neq 1$</span>) implies that (it is <strong>NOT</strong> necessary that <span class="math-container">$x^2 \neq x$</span>)</li>
<li>(it is possible that <span class="math-container">$x = 1$</span>) implies that (it is possible that <span class="math-container">$x^2 = x$</span>)</li>
<li>For any set <span class="math-container">$S$</span>, [(there exists <span class="math-container">$x$</span> in <span class="math-container">$S$</span> such that <span class="math-container">$x = 1$</span>) implies that (there exists <span class="math-container">$x$</span> in <span class="math-container">$S$</span> such that <span class="math-container">$x^2 = x$</span>)]</li>
</ul>
</blockquote>
|
3,085,181 | <p>I have to cope with a constraint of the form (1) in the following problem: </p>
<p><span class="math-container">$$\begin{align}\max\quad& x+y\\
\text{s.t.}\quad&
x + y \leq \max \{x,y\} &(1)\\
&0 \leq x \leq U_x&(2)\\
&0 \leq y \leq U_y&(3)\\
\end{align}$$</span></p>
<p>In the following link you can find an approach but I don't understand it.</p>
<p><a href="https://www.leandro-coelho.com/how-to-linearize-max-min-and-abs-functions/" rel="nofollow noreferrer">https://www.leandro-coelho.com/how-to-linearize-max-min-and-abs-functions/</a></p>
<p>I don't understand: what is <span class="math-container">$S^+$</span>, <span class="math-container">$S^-$</span> and how would a penalization look like?
(I refer to the text: "The max function can be linearized as follows: ..." in the reference).</p>
<p>I would be grateful if somebody could help. </p>
<hr>
<p><a href="https://i.stack.imgur.com/NLyjI.jpg" rel="nofollow noreferrer">The linked figure shows the problem in LP Format and the solution.</a></p>
| Jorge Cordova | 966,421 | <p>Hello I hope it is not too late; the penalty could be representative like this:</p>
<blockquote>
<p>max x + y<br />
x <=s1 + s2<br />
y <=s1 + s2<br />
s1 >= x - y<br />
s2 >= y - x<br />
s1 - bigM<em>b <= 0<br />
s2 - bigM</em>(1-b) <= 0<br />
0 <= x <= Ux<br />
0 <= y <= Uy</p>
</blockquote>
<p>where b is binary variable</p>
|
3,051,480 | <p>Now we have the equation
<span class="math-container">$$\sum_{i}(x_i-\hat x_i)^2,$$</span>
where <span class="math-container">$x_i$</span> is the observed value of a data sample <span class="math-container">$S$</span>. Here is the question:</p>
<blockquote>
<p>Why does this expression get its minimum value when <span class="math-container">$\hat x_i$</span> is the average of the data sample <span class="math-container">$S$</span> ?</p>
</blockquote>
<p>I tried to take the derivatives of that equation and make it to zero, but it seems there's something wrong, because <span class="math-container">$\hat x_i$</span> is kind of multi-variable.
Can anyone help me out? Thanks a lot!</p>
| Noble Mushtak | 307,483 | <p>Let's take the function:</p>
<p><span class="math-container">$$f(\hat x)=\sum_{i=1}^n (x_i-\hat x)^2$$</span></p>
<p>Here, we want to find the value of <span class="math-container">$\hat x$</span> which minimizes <span class="math-container">$f(\hat x)$</span>. Now, even though there are multiple variables of this function because of <span class="math-container">$x_i$</span>, we can just treat these variables as constants since they are independent from the <span class="math-container">$\hat x$</span>, which essentially changes this to a single-variable calculus problem. Now, let's take the derivative of <span class="math-container">$f$</span> with respect to <span class="math-container">$\hat x$</span>.</p>
<p><span class="math-container">$$f'(\hat x)=\sum_{i=1}^n 2(\hat x-x_i)$$</span></p>
<p>From here, can you find the value of <span class="math-container">$\hat x$</span> satisfying <span class="math-container">$f(\hat x)=0$</span>? Once you solve that equation, use second-derivative test to show that it is indeed an absolute minimum.</p>
|
2,251,998 | <p>Here is a question that I am working on:</p>
<blockquote>
<p>If $G$ is a group such that <em>every</em> non-identity element has order $2$, show that $G$ is abelian (commutative).</p>
</blockquote>
<p><strong>My attempt</strong></p>
<p>Suppose that for all $a \in G$, we have $$a^2 = Id$$</p>
<p>My goal is to show that $ab = ba$</p>
<p>Since $(ab)^2 = abab$ = Id, we multiply both left sides by $ba$ and we have</p>
<p>$(ba)abab = baId$</p>
<p>$\Rightarrow ba^2bab = ba$</p>
<p>$\Rightarrow bIdbab = ba$</p>
<p>$\Rightarrow b^2ab = ba$</p>
<p>$\Rightarrow Idab = ba$</p>
<p>$\Rightarrow ab = ba$</p>
<p>This condition will hold if I multiply $(ba)$ on the right side as well.</p>
<p>How does this look? </p>
| Carl Schildkraut | 253,966 | <p>As far as I know there isn't a system like OEIS, but as vrugtehagel said, WolframAlpha is pretty good. Also, if you know it's algebraic, it might be feasible to search for a minimal polynomial with some computer algebra system.</p>
|
1,162,161 | <p>A patient would like to take a test to determine if he has a nasty disease. Let the variable A denote that
the patient has the disease and the variable B denote a positive test. The following assumptions apply:
• The probability that the test is positive given the patient has the disease is 99%.
• The probability that the test is positive given the patient does not have the disease is 5%.
• The rate of occurrence of the disease in the general population is 0.1%.</p>
<p>Consider a second test for the same disease. A positive result for this test is denoted by C. The following
assumptions apply:
• The probability that the test is positive given the patient has the disease is 80%.
• The probability that the test is positive given the patient does not have the disease is 0.01%.</p>
<p>Now that both tests have been taken, assume conditional independence of these tests.
Both tests give a positive answer. What is the probability that the patient has the disease?</p>
<p>so my question is do i apply P(A|BnC) ?if so how do i deduce this</p>
| cactus314 | 4,997 | <p>Using geometric series we know exactly what your series should be</p>
<p>$$ \sum_{n=1}^\infty n x^{n-1} = \frac{1}{(1-x)^2}$$</p>
<p>Uniform convergence asked about the remaining terms and you can try to give uniform estimate</p>
<p>$$ \sum_{n=N+1}^\infty n x^{n-1} = x^N \left( \frac{1}{(1-x)^2}
+ \frac{N}{1-x} \right) \leq (\tfrac{9}{10})^N \left( \frac{1}{(1-\frac{9}{10})^2}
+ \frac{N}{1-\frac{9}{10}} \right) \to 0 $$</p>
<p>for all $0 \leq x \leq \frac{9}{10}$.</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>
| Robert Israel | 8,508 | <p>Hints: Since there are only those three fixed points, we either have $f(x) < x$ for all $x$ with $\alpha < x < \beta$ or $f(x) > x$ for all those $x$. Check one point $x$ in that interval to see which it is. Similarly for $\beta < x < \gamma$. </p>
<p>Since $f$ is an increasing function, if $x > \alpha$ then $f(x) > f(\alpha) = \alpha$, and similarly ....</p>
<p>You will also want to use the fact that an increasing sequence that is bounded above, or a decreasing sequence that is bounded below, has a limit. </p>
|
3,495,852 | <p>I couldn't find an example or explanation why the following sentence is correct </p>
<p>If a Transformation is linear, and vectors <span class="math-container">$u_1$</span>,<span class="math-container">$u_2$</span>,<span class="math-container">$u_3$</span> are dependent then
<span class="math-container">$T(u_1)$</span>,<span class="math-container">$T(u_2)$</span>,<span class="math-container">$T(u_3)$</span>
must also be dependent </p>
<p><strong>but</strong></p>
<p>If a Transformation is linear, and <span class="math-container">$T(u_1)$</span>,<span class="math-container">$T(u_2)$</span>,<span class="math-container">$T(u_3)$</span> are dependent , that doesn't mean vectors <span class="math-container">$u_1$</span>,<span class="math-container">$u_2$</span>,<span class="math-container">$u_3$</span> are dependent </p>
<p>Couldn't think of an example that would justify the 2nd phrase </p>
| Kavi Rama Murthy | 142,385 | <p>It is not assumed that <span class="math-container">$(\theta_n)$</span>'s exist on the same space. It is rarely necessary to construct random variables on a particular space to prove theorems in Probability Theory. What Kallenberg wants is some probability space on which random variables <span class="math-container">$\xi_n'$</span>'s and <span class="math-container">$\theta_n$</span>'s exist such that <span class="math-container">$(\xi_n')$</span> has the same distribution as <span class="math-container">$(\xi_n)$</span>, <span class="math-container">$(\theta_n)$</span> has the stated properties and <span class="math-container">$\xi_n'$</span>'s are independent of <span class="math-container">$\theta_n$</span>'s. Existence of such random variables is guaranteed by Kolmogorov's Existence Theorem (also known as Kolmogorov's Consistency Theorem). </p>
|
70,603 | <p>We were shown in class this next calculation: (Here, $V_n(RB^n)$ is the volume of an $n$ dimensional ball of radius $R$, likewise $S_{n-1}$ is the surface area of the $n$ dimensional sphere in $\mathbb{R}^n$. $rS^{n-1}$ denotes the $n$ dimensional sphere of radius $r$ and integrating $d\textbf{S}$ means a surface integral.)
$$V_n(RB^n)=\int_{RB^n}1dx=\int_0^R\int_{rS^{n-1}}1d\textbf{S}dr=\int_0^R\int_{S^{n-1}}r^{n-1}d\textbf{S}dr=$$$$=\int_0^Rr^{n-1}\int_{S^{n-1}}1d\textbf{S}dr=\int_0^Rr^{n-1}S_{n-1}dr=\frac{R^n}{n}S_{n-1}$$
and finally $V_n=\frac{1}{n}S_{n-1}$ since $V_n(RB^n)=R^nV_n$. My problem is with the 3rd equality. The first is obvious and the second is the coarea formula. I assume the third equality is a result of a change of variables, but since this is taking place in $\mathbb{R}^n$ I'd expect the change of variables to be $x\mapsto rx$ which gives the Jacobian of $r^n$ - not the $r^{n-1}$ we see after the third equality.</p>
<p>It'd be easier for me to assume the teacher had a mistake here, had she not used this result later on in her lectures... So my question is, was she wrong in the change of variables there or am I missing something about surface integrals?</p>
| Ross Millikan | 1,827 | <p>As an example, if we let $n=3$, the area of a two-sphere is proportional to $r^2$, not $r^3$. You haven't changed the $dr$ integral, which still goes from $0$ to $R$.</p>
|
475,863 | <p>Let $f,P,Q$ three analytic functions. Here $P$ is a polynomial.</p>
<p>I want to solve this equation: $$f(s)=P(s)\exp(Q(s)).$$</p>
<p>The unknown here are $P, Q$ and $f$ is known. </p>
| Daron | 53,993 | <p>$P(s)$ is an arbitrary polynomial. Then $Q(s) = \log(\frac{f(s)}{P(s)})$.</p>
|
1,146,802 | <p>Often seen similar systems of equations. Usually consider such systems in which decisions no. Such as there. <a href="https://math.stackexchange.com/questions/1146460/is-there-a-b-c-d-in-mathbb-n-so-that-a2b2-c2-b2c2-d2">Is there $a,b,c,d\in \mathbb N$ so that $a^2+b^2=c^2$, $b^2+c^2=d^2$?</a></p>
<p>I think it would be more interesting to solve the system in which there are solutions. For example to find out whether such a system solution?</p>
<p>$$\left\{\begin{aligned}&a^2+b^2+c^2=q^2\\&c^2+q^2=k^2\end{aligned}\right.$$</p>
<p>What is the right approach? And how to solve it?</p>
| individ | 128,505 | <p>If you solve the system of equations:</p>
<p>$$\left\{\begin{aligned}&a^2+b^2+c^2=q^2\\&c^2+q^2=w^2\end{aligned}\right.$$</p>
<p>When the standard approach solution and using a replacement.</p>
<p>$$p=9t^2-10tk+5k^2$$</p>
<p>$$s=5t^2-10tk+9k^2$$</p>
<p>$$x=7t^2-10tk+7k^2$$</p>
<p>$$y=4(t^2-k^2)$$</p>
<p>$$z=5t^2-14tk+5k^2$$</p>
<p>Then the solution can be written as :</p>
<p>$$a=2zx$$</p>
<p>$$b=z^2+y^2-x^2$$</p>
<p>$$c=2yx$$</p>
<p>$$q=z^2+y^2+x^2$$</p>
<p>$$w=p^2+s^2$$</p>
<p>$t,k$ - integers which we ask.</p>
|
1,535,731 | <p>I am currently self-learning about matrix exponential and found that determining the matrix of a diagonalizable matrix is pretty straight forward :).</p>
<p>I do not, however, know how to find the exponential matrix of a non-diagonalizable matrix.</p>
<p>For example, consider the matrix
$$\begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}$$</p>
<p>Is there any process of finding the exponential matrix of a non-diagonalizable matrix? If so, can someone please show me an example of the process? :). I am not looking for an answer of the above mentioned matrix (since I just made it up), but rather I'm interested in the actual method of finding the matrix exponential to apply to other examples :)</p>
| Dr. Sundar | 1,040,807 | <p>Laplace Transforms approach is an useful for finding <span class="math-container">$e^{A t}$</span>, whether <span class="math-container">$A$</span> is diagonalizable or non-diagonalizable. It is a very useful method in Electrical
Engineering.</p>
<p>Basically, we use the formula
<span class="math-container">$$
e^{A t} = \mathcal{L}^{-1}\left[ (s I - A)^{-1} \right]
$$</span></p>
<p>As an illustration, consider the defective matrix
<span class="math-container">$$
A = \left[ \begin{array}{cc}
1 & 0 \\
1 & 1 \\
\end{array} \right]
$$</span></p>
<p>We find
<span class="math-container">$$
s I - A = \left[ \begin{array}{cc}
s - 1 & 0 \\
-1 & s - 1 \\
\end{array} \right]
$$</span>
where <span class="math-container">$s$</span> is a complex variable.</p>
<p>We find that
<span class="math-container">$$
(s I - A)^{-1} = \left[ \begin{array}{cc}
{1 \over s - 1} & 0 \\[2mm]
{1 \over (s - 1)^2} & {1 \over s - 1} \\[2mm]
\end{array} \right]
$$</span></p>
<p>The inverse Laplace transform of <span class="math-container">$(s I - A)^{-1}$</span> is the state transition matrix, which is also the matrix exponential, <span class="math-container">$e^{A t}$</span>.</p>
<p>Hence, we get
<span class="math-container">$$
e^{A t} = \mathcal{L}^{-1}\left[ (s I - A)^{-1} \right]
= \left[ \begin{array}{cc}
\mathcal{L}^{-1}\left( {1 \over s - 1} \right) & \mathcal{L}^{-1}(0) \\[2mm]
\mathcal{L}^{-1}\left( {1 \over (s - 1)^2} \right) & \mathcal{L}^{-1}\left( {1 \over s - 1} \right) \\[2mm]
\end{array} \right] = \left[ \begin{array}{cc}
e^{t} & 0 \\[2mm]
t e^t & e^t \\[2mm]
\end{array} \right]
$$</span></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>
| xpaul | 66,420 | <p>Note that <span class="math-container">$n=mk+r$</span>, <span class="math-container">$r=0,1,\cdots k-1$</span>. So
<span class="math-container">\begin{eqnarray}
&&\sum_{n=1}^{\infty }\frac{(n)\mod(k)}{n(n+1)}=\sum_{m=0}^{\infty }\sum_{r=1}^{k-1}\frac{(mk+r)\mod(k)}{(mk+r)(mk+r+1)}\\
&=&\sum_{m=0}^{\infty }\sum_{r=1}^{k-1}\frac{r}{(mk+r)(mk+r+1)}=\sum_{r=1}^{k-1}r\sum_{m=0}^{\infty }\frac{1}{(mk+r)(mk+r+1)}\\
&=&\sum_{r=1}^{k-1}r\sum_{m=0}^{\infty }\bigg(\frac{1}{mk+r}-\frac{1}{mk+r+1}\bigg)=\sum_{r=1}^{k-1}\frac{r}{k}\sum_{m=0}^{\infty }\bigg(\frac{1}{m+\frac{r}{k}}-\frac{1}{m+\frac{r+1}{k}}\bigg)\\
&=&\sum_{r=1}^{k-1}\frac{r}{k}\sum_{m=1}^{\infty }\bigg(\frac{1}{m+\frac{r}{k}-1}-\frac{1}{m+\frac{r+1}{k}-1}\bigg)\\
&=&\sum_{r=1}^{k-1}\frac{r}{k}\sum_{m=1}^{\infty }\bigg[\bigg(\frac1m-\frac{1}{m+\frac{r+1}{k}-1}\bigg)-\bigg(\frac{1}{m}-\frac{1}{m+\frac{r}{k}-1}\bigg)\bigg]\\
&=&\sum_{r=1}^{k-1}\frac{r}{k}\bigg[\psi ^{(0)}\left(\frac{r+1}{k}\right)-\psi ^{(0)}\left(\frac{r}{k}\right)\bigg].
\end{eqnarray}</span>
Here the Digamma function
<span class="math-container">$$ \psi^{(0)}(z+1)=-\gamma+\sum_{n=1}^\infty\bigg(\frac1n-\frac{1}{n+z}\bigg) $$</span>
is used from <a href="https://en.wikipedia.org/wiki/Digamma_function" rel="nofollow noreferrer">here</a>. </p>
<p>Update: Now @Gary proves
<span class="math-container">$$ \sum_{r=1}^{k-1}\frac{r}{k}\bigg[\psi ^{(0)}\left(\frac{r+1}{k}\right)-\psi ^{(0)}\left(\frac{r}{k}\right)\bigg]=\ln k. $$</span>
Thank you, Gary.</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>
| Gary | 83,800 | <p>Continuing xpaul's answer
<span class="math-container">\begin{align*}
& \sum\limits_{r = 1}^{k - 1} {\frac{r}{k}\left[ {\psi ^{(0)} \left( {\frac{{r + 1}}{k}} \right) - \psi ^{(0)} \left( {\frac{r}{k}} \right)} \right]} = \sum\limits_{r = 1}^{k - 1} {\frac{r}{k}\psi ^{(0)} \left( {\frac{{r + 1}}{k}} \right)} - \sum\limits_{r = 1}^{k - 1} {\frac{r}{k}\psi ^{(0)} \left( {\frac{r}{k}} \right)}
\\ &
= \sum\limits_{r = 1}^{k - 1} {\frac{{r + 1}}{k}\psi ^{(0)} \left( {\frac{{r + 1}}{k}} \right)} - \sum\limits_{r = 1}^{k - 1} {\frac{r}{k}\psi ^{(0)} \left( {\frac{r}{k}} \right)} - \sum\limits_{r = 1}^{k - 1} {\frac{1}{k}\psi ^{(0)} \left( {\frac{{r + 1}}{k}} \right)}
\\ &
= \sum\limits_{r = 2}^k {\frac{r}{k}\psi ^{(0)} \left( {\frac{r}{k}} \right)} - \sum\limits_{r = 1}^{k - 1} {\frac{r}{k}\psi ^{(0)} \left( {\frac{r}{k}} \right)} - \frac{1}{k}\sum\limits_{r = 2}^k {\psi ^{(0)} \left( {\frac{r}{k}} \right)}
\\ &
= \psi ^{(0)} (1) - \frac{1}{k}\sum\limits_{r = 1}^k {\psi ^{(0)} \left( {\frac{r}{k}} \right)} = - \gamma + \gamma + \log k = \log k.
\end{align*}</span></p>
|
1,951 | <p>In <a href="https://matheducators.stackexchange.com/a/1949/704">this answer</a>, user <a href="https://matheducators.stackexchange.com/users/942/robert-talbert">Robert Talbert</a> stated that</p>
<blockquote>
<p>There are some amazing things you can do pedagogically with clickers.</p>
</blockquote>
<p>I'd like to see some examples. (Not that I'm eager to try that myself, but I'm just curious.) As I stated in one of the comments to the linked question,</p>
<blockquote>
<p>I can't see much point in using such devices, at least in the context of teaching.</p>
</blockquote>
<p>– but I'm probably mistaken. Please enlighten me, and at the same time give MESE users some neat ideas! :)</p>
| Brian S | 328 | <p>Several times in my high school Computer Science classes, my teacher used devices like this to administer quizzes <em>in parallel</em> with a lecture. That is, as he covered material, he would put up quiz questions on a projector, and we would answer with the clickers. The questions in general tested comprehension of what the teacher was saying, rather than parroting the words that had just come out of his mouth.</p>
<p>The specific devices we had were basically IR remotes, however, making their usage difficult. With one receiver and a couple dozen remotes, we had to fight over each others' signals to get our responses in. The fact that the remotes needed to be pointed fairly accurately towards the receiver was also annoying. These problems would likely be solved by different physical technology being put to use, though.</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 $</span> is allowable?</p>
| nonuser | 463,553 | <p>Since <span class="math-container">$x_1,x_2<3$</span> we have <span class="math-container">$3-x_i>0$</span> so their product is positive:<span class="math-container">$$ 0<9-3(x_1+x_2)+x_1x_2$$</span>
thus <span class="math-container">$$0<9-6a+a^2+a-3 = a^2-5a+6=(a-3)(a-2)$$</span></p>
<p>So <span class="math-container">$a\in (-\infty ,2)\cup (3,\infty)$</span>. But since the discriminat must be <span class="math-container">$\geq 0$</span> we get <span class="math-container">$a\leq 3$</span> so we have <span class="math-container">$a\in (-\infty ,2)$</span></p>
<p>Now let us prove that all <span class="math-container">$a<2$</span> are good.</p>
<p>We have <span class="math-container">$$x_1= a-\sqrt{3-a}\leq a<3$$</span> We are left if <span class="math-container">$x_2= a+\sqrt{3-a}$</span> is smaller than <span class="math-container">$3$</span> if <span class="math-container">$a<2$</span> i.e. <span class="math-container">$$\sqrt{3-a} <3-a$$</span> which is true since <span class="math-container">$3-a>1$</span></p>
|
459,374 | <p>Let $X$ be the random variable which denotes the number of times a die has been rolled till each side has appeared. The order does not matter.
We are trying to find $E[X]$.</p>
<p>Let $X_i$ be a random variable which denotes how many times a die has to be rolled till side i has appeared.</p>
<p>So,</p>
<p>$$E[X]= E[X1+X2+X3+X4+X5+X6] = E[X1]+E[X2]+E[X3]+E[X4]+E[X5]+E[X6]$$</p>
<p>$$E[X1]=E[X2]=E[X3]=E[X4]=E[X5]=E[X6]=6$$</p>
<p>$$E[X]=36$$?</p>
<p>Why is this solution wrong?</p>
| ronno | 32,766 | <p>Let the vertices be <span class="math-container">$a_1,a_2,a_3$</span>. Then any <span class="math-container">$x,y \in T$</span> can be written as convex combinations of these, so let <span class="math-container">$x = \sum \lambda_i a_i$</span> and <span class="math-container">$y = \sum \mu_i a_i$</span> with <span class="math-container">$\sum \lambda_i = \sum \mu_i = 1$</span>. Then by the triangle inequality
<span class="math-container">$$\newcommand{\norm}[1]{\lVert #1 \rVert}
\norm{x - y} \le \sum_{i, j} \lambda_i \mu_j \norm{a_i - a_j} \le \left(\sum \lambda_i \mu_j\right) \left(\max_{i, j}\ \norm{a_i - a_j}\right) = \max_{i, j}\ \norm{a_i - a_j}
$$</span>
Thus, the diameter <span class="math-container">$\sup_{x, y} \norm{x - y} \le \max_{i, j} \norm{a_i - a_j}$</span>. The reverse inequality holds by taking <span class="math-container">$x = a_i$</span>, <span class="math-container">$y = a_j$</span>.</p>
<p>So the same proof works for any convex polytope in <span class="math-container">$\mathbb{R}^n$</span> once you replace edge by "segment between two vertices".</p>
|
459,374 | <p>Let $X$ be the random variable which denotes the number of times a die has been rolled till each side has appeared. The order does not matter.
We are trying to find $E[X]$.</p>
<p>Let $X_i$ be a random variable which denotes how many times a die has to be rolled till side i has appeared.</p>
<p>So,</p>
<p>$$E[X]= E[X1+X2+X3+X4+X5+X6] = E[X1]+E[X2]+E[X3]+E[X4]+E[X5]+E[X6]$$</p>
<p>$$E[X1]=E[X2]=E[X3]=E[X4]=E[X5]=E[X6]=6$$</p>
<p>$$E[X]=36$$?</p>
<p>Why is this solution wrong?</p>
| achille hui | 59,379 | <p>Given any triangle $T$, in fact any bounded closed convex subset of $\mathbb{R}^2$.
The map</p>
<p>$$T^2 \ni (x,y) \mapsto |x-y|^2 \in \mathbb{R}$$</p>
<p>is a continuous function on $T^2$ bounded from above. Since $T^2$ is compact, the map achieves its maximum on some $(u, v) \in T^2$. i.e.</p>
<p>$$\sup \{\;|x-y|^2 : x, y \in T\;\} = | u - v |^2.$$</p>
<p>If either $u$ or $v$ is not an <a href="http://en.wikipedia.org/wiki/Extreme_point" rel="noreferrer">extremal point</a>
of $T$, say $u = \frac12 ( u_1 + u_2 )$ where $u_1, u_2 \in T$, then by substituting $x_1$ by $u - v$ and $x_2$ by $\frac12 ( u_1 - u_2 )$ into
<a href="http://en.wikipedia.org/wiki/Parallelogram_law" rel="noreferrer">parallelogram identity</a>:
$$| x_1 + x_2 |^2 + |x_1 - x_2|^2 = 2 ( |x_1|^2 + |x_2|^2 )$$
one find at least one of $|u_1 - v|^2$ and $|u_2 - v|^2$ is greater than $|u-v|^2$. This contradicts with the role of $u, v$ that maximize $|u - v|^2$. </p>
<p>As a result, the $u, v$ that maximize $|u-v|^2$ are both extremal points and</p>
<p>$$\sup\{\;|x-y|^2 : x,y \in T\;\} = \max\{\;|u-v|^2 : u, v \in T, \text{ both extremal}\;\}$$</p>
<p>A triangle has 3 extremal points, i.e. its 3 vertices, and hence its diameter is the length of its longest edge. </p>
|
2,558,870 | <p>Suppose $f:[0,1]\to \mathbb{R}$ is uniformly continuous, and $(p_n)_{n\in\mathbb{N}}$ is a sequence of polynomial functions converging uniformly to $f$.</p>
<p>Does it follow that $\mathcal{F}=\{p_n\mid n\in\mathbb{N}\}\cup \{f\}$ is equicontinuous?</p>
<p>Also, if $C_n$ are the Lipschitz constants of the polynomials $p_n$, does it follow that $C_n<\infty$ for all $n$, and $\lim_{n\to\infty} C_n=\infty?$</p>
<p>I'm preparing for a test, but I'm not sure how to go about answering these two question. Any hints or tips as to what to look for would be appreciated. </p>
| QED | 91,884 | <p>If $(A-pI)=-p(I-A/p)$ is invertible then $[-p(I-A/p)]^{-1}=-\frac{1}{p}\left[I-\frac{A}{p}+\frac{A^2}{p^2}+\frac{A^3}{p^3}+\cdots\right]$. If $v$ is an eigenvector of $A$ corresponding to the eigenvalue $q$, then $(A-pI)^{-1}v=-\frac{1}{p}\left[1-\frac{q}{p}+\frac{q^2}{p^2}+\frac{q^3}{p^3}+\cdots\right]v=(q-p)^{-1}v$</p>
|
847 | <p>Apologies in advance if this is obvious.</p>
| moonface | 513 | <p>Not a satisfying argument: We can, first of all, find a basis in which the entries lie in some algebraic number field <span class="math-container">$K$</span>. Let <span class="math-container">$\mathcal{O}$</span> be the ring of integers of <span class="math-container">$K$</span>.
Then there is a locally free <span class="math-container">$\mathcal{O}$</span>-module <span class="math-container">$M$</span> of rank <span class="math-container">$n$</span> preserved by <span class="math-container">$G$</span>: add up all the translates of <span class="math-container">$\mathcal{O}^n$</span> under <span class="math-container">$G$</span>. Now, <span class="math-container">$M$</span> need not itself be free, but it is isomorphic
as an <span class="math-container">$\mathcal{O}$</span>-module to the sum of various ideals of <span class="math-container">$\mathcal{O}$</span>. Now pass to an extension <span class="math-container">$L/K$</span> so that every ideal class of <span class="math-container">$K$</span> trivializes in <span class="math-container">$L$</span>, e.g. the Hilbert class field; then <span class="math-container">$G$</span> preserves a free rank <span class="math-container">$n$</span> module
for the ring of integers of <span class="math-container">$L$</span>. Sorry!</p>
|
2,569,557 | <p>I'm still confused by the use of $\Rightarrow$ in (ε,δ)-definition of limit. <br/>
Take for example the definition of $\underset{x\rightarrow x_{0}}{\lim}f\left(x\right)=l$ :<br/></p>
<blockquote>
<p>$$\forall\varepsilon>0,\;\exists\delta>0\quad\mathrm{such\:that\quad}\forall x\in\mathrm{dom}\,f,\;0<\left|x-x_{0}\right|<\delta\;\Rightarrow\;\left|f\left(x\right)-l\right|<\varepsilon$$</p>
</blockquote>
<p>My questions are:</p>
<p>Why is $\left|f\left(x\right)-l\right|<\varepsilon$ not a sufficient condition for $0<\left|x-x_{0}\right|<\delta\;$? <br/></p>
<p>Or, stated in another way, shouldn't $\left|f\left(x\right)-l\right|<\varepsilon\;\Rightarrow\;0<\left|x-x_{0}\right|<\delta\;$ also be true ?
If $f\left(x\right)$ becomes arbitrarily close to $l$, doesn't $x$ becomes arbitrarily close to $x_0$?</p>
| user | 505,767 | <p>You can think about it in this way:</p>
<blockquote>
<p>first set <span class="math-container">$\epsilon$</span> and then you have to find <span class="math-container">$\delta$</span> such that the inequality:</p>
<p><span class="math-container">$$\left|f\left(x\right)-l\right|<\varepsilon$$</span></p>
<p>is satisfied.</p>
</blockquote>
<p>If you can do it for every <span class="math-container">$\epsilon>0$</span> then the limit exists.</p>
|
2,669,292 | <p>$g_n(x) = \frac{\ln(1+x/n)}{n}$ on $\mathbb{R}$.
Don't they all converge to 0?</p>
| Ian | 83,396 | <p>Sure (though the domain can't be all of $\mathbb{R}$, presumably you intend $[0,\infty)$ or maybe $(-n,\infty)$). But for any fixed $n$, the whole thing still goes to infinity as $x$ goes to infinity, so $\| g_n - g \|_\infty=+\infty \not \to 0$. </p>
<p>On the other hand the convergence is uniform on compact subsets, which is generally the best that one should hope for on unbounded domains.</p>
|
4,481,314 | <p>This is an exercise in Tristan Needham's <em>Visual Differential Geometry and Forms</em>. He uses the term <em>ultimate equality</em> to mean roughly the same thing as first order approximation, which he says is motivated by Newton's Principia. The book is dedicated to Needham's longtime personal friend Roger Penrose, and is worthy of the dedication.</p>
<p>The first part, using calculus is pretty straight forward. Divide the triangle into a rectangle of area <span class="math-container">$ab,$</span> an upper triangle of height <span class="math-container">$a\tan{\theta}$</span> and a lower triangle of base <span class="math-container">$b\cot{\theta}.$</span> Add the resulting areas to get <span class="math-container">$\mathcal{A}.$</span> Set the derivative equal to zero. Put the resulting value for <span class="math-container">$\tan\theta$</span> into the expression for area.</p>
<p><span class="math-container">\begin{align*}
\mathcal{A}= & \frac{1}{2}\left(2ab+a^{2}\tan\theta+b^{2}\cot\theta\right)\\
\mathcal{A}^{\prime}= & \frac{1}{2}\left(\frac{a^{2}}{\cos^{2}\theta}-\frac{b^{2}}{\sin^{2}\theta}\right)=0\\
\tan\theta= & \frac{b}{a}\implies\mathcal{A}=2ab
\end{align*}</span></p>
<p>But I haven't figured out the "trick" intended by the second part. See the text in bold-face. The solution involves drawing a picture something like my first drawing. The "ultimate equality" expressions will be the kinds physicists write, and mathematicians say "you can't do that."</p>
<blockquote>
<p>Let <span class="math-container">$L$</span> be a general line through the point <span class="math-container">$\left\{ a,b\right\} $</span>
in the first quadrant of <span class="math-container">$\mathbb{R}^{2}$</span>, and let <span class="math-container">$\mathcal{A}$</span>
be the area of the triangle bounded by the <span class="math-container">$x$</span>-axis, the <span class="math-container">$y$</span>-axis and
<span class="math-container">$L$</span>.</p>
</blockquote>
<blockquote>
<p>(i) Use ordinary calculus to find the position of <span class="math-container">$L$</span> that minimizes
<span class="math-container">$\mathcal{A}$</span>, and show that <span class="math-container">$\mathcal{A_{\min}}=2ab.$</span></p>
</blockquote>
<blockquote>
<p>(ii) Use Newtonian reasoning to solve the problem <em>instantly</em>, without
calculation! (Hints: Let <span class="math-container">$\delta\mathcal{A}$</span> be the change in the area resulting
from a small (ultimately vanishing) rotation <span class="math-container">$\delta\theta$</span> of <span class="math-container">$L$</span>.
<strong>By drawing <span class="math-container">$\delta\mathcal{A}$</span> in the form of two triangles, and
observing that each triangle is ultimately equal to a sector of a
circle, write down an ultimate equality <span class="math-container">$\delta\mathcal{A}$</span> in terms
of <span class="math-container">$\delta\theta$</span>. Now set <span class="math-container">$\delta\theta=0.$</span>)</strong></p>
</blockquote>
<p>The drawings represent two attempts to produce the "immediate" solution. But neither approach seems to give a simple, and obvious formulation of <span class="math-container">$\delta\mathcal{A}$</span> that leads directly to the equation <span class="math-container">$\mathcal{A}=2ab.$</span></p>
<p>The red line is the correct solution. The black (or green) line is the result of rotating through <span class="math-container">$\delta\theta$</span>. I've added another image with a greater difference between <span class="math-container">$a$</span> and <span class="math-container">$b$</span> to show more clearly that the light blue triangles are not equal.</p>
<p><strong>How should the approach described in the "hints" be depicted?</strong></p>
<p><a href="https://i.stack.imgur.com/BVWea.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BVWea.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/HZKtu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HZKtu.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/B535z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B535z.png" alt="enter image description here" /></a></p>
| Z Ahmed | 671,540 | <p><span class="math-container">$\lim_{x\rightarrow 0 } \sin\left(\dfrac{\pi}{x}\right)$</span></p>
<p>Think of two sequences <span class="math-container">$x_n=\frac{1}{(2n+1/2)}$</span> and <span class="math-container">$x'_n=\frac{1}{(n+1)}$</span>
such that when <span class="math-container">$n \rightarrow \infty$</span> bot <span class="math-container">$x_n,x'_n$</span> tend to 0. As <span class="math-container">$f(x)=\sin(\pi/x)$</span>, then <span class="math-container">$f(x_n)=1$</span> and <span class="math-container">$f(x'_n)=0$</span> being unequal, the said limit does not exist.</p>
|
3,995,492 | <p>I have no clue how to do this, I manage to get I get that <span class="math-container">$11^{36} \equiv 1 \hspace{0.1cm} \text{mod} (13)$</span> but I can't get anywhere from there.</p>
| Dietrich Burde | 83,966 | <p>If <span class="math-container">$11^{36}=1$</span> in <span class="math-container">$\Bbb Z/13$</span>, then <span class="math-container">$11^{35}=11^{-1}$</span>. But since <span class="math-container">$6\cdot 11=66=1$</span> in <span class="math-container">$\Bbb Z/13$</span>, we have <span class="math-container">$11^{-1}=6$</span>.</p>
<p>Here I just write <span class="math-container">$a=b$</span> in the ring <span class="math-container">$\Bbb Z/n$</span> for <span class="math-container">$a\equiv b\bmod n$</span>, so that we see it a bit easier.
For <span class="math-container">$n=13$</span>, <span class="math-container">$\Bbb Z/n$</span> is a field and all nonzero elements have an inverse.</p>
|
3,995,492 | <p>I have no clue how to do this, I manage to get I get that <span class="math-container">$11^{36} \equiv 1 \hspace{0.1cm} \text{mod} (13)$</span> but I can't get anywhere from there.</p>
| jacopoburelli | 530,398 | <p>What were your efforts ? However, if you proved that <span class="math-container">$11^{36} \equiv 1 \hspace{0.1cm} (13)$</span> then you have that <span class="math-container">$11^{35}\cdot 11 \equiv 1 \hspace{0.1cm}(13)$</span>. In this line it is written that <span class="math-container">$11^{35}$</span> is the inverse of <span class="math-container">$11$</span> by definition, by uniqueness you only have to prove that <span class="math-container">$6$</span> does the same job, i.e <span class="math-container">$6 \cdot 11 \overset{?}{\equiv} 1 \hspace{0.1cm}(13)$</span>, which is true, and you have the desired result.</p>
|
3,695,127 | <p>Before the moderators close my question, I cant think of any starting approach to the question. </p>
<p>Another question of the similar type I am having trouble with is: 12 balls are distributed at random among 3 boxes. What is the probability that the first box will contain 3 balls?
For the second question I can figure out the exhaustive number of outcomes will be 3 raised to the power 12 since each ball has 12 options.</p>
| E. KOW | 443,898 | <p><span class="math-container">$S^1 \times S^1$</span> is homeomorphic to <span class="math-container">$\left[-1, 1\right]^2 / \sim$</span> where <span class="math-container">$\left(x,-1\right) \sim \left(x, 1\right), \left(-1,y\right) \sim \left(1, y\right)$</span>. There is a deformation retract of <span class="math-container">$\left[-1,1\right]^2-\left(0,0\right)$</span> onto <span class="math-container">$\partial\left[-1,1\right]$</span> that fixes the boundary, and in particular well defined also after quotiening by <span class="math-container">$\sim$</span>. This will give you the desired deformation retract.</p>
|
366,249 | <p>$3xy^2dx+2x^3dy$
where is the boundary of the region between the circles $x^2+y^2=25$ and $x^2+y^2=64$ having positive orientation.</p>
<p>Not quite sure how to evaluate this...</p>
| Mikasa | 8,581 | <p>Hints:</p>
<p>First of all set $(x+1)=t$ to have $t^2y''+ty'-y=0, ~~t\neq -1$ instead. Then, solve the auxiliary equation $$am^2+(b-a)m+c=0$$ wherein </p>
<p>$a=1$ (the coffecient of $t^2$), </p>
<p>$b=+1$ (the cofficient of $t$ above) and $c=-1$</p>
<p>for finding the possible $m$'s.</p>
<p>If $m_1,m_2$ are distinct solutions so the general solution of your Cauchy-Euler ODE will be as $$y_c=C_1t^{m_1}+C_2t^{m_2}$$ If you have $m_1=m_2=m=\frac{a-b}{2a}$ then $y_c=C_1t^m+C_2t^m\ln(t)$ and finally if you have $m=\alpha\pm i\beta$ then $$y_c=t^{\alpha}(C_1\cos(\beta\ln t)+C_2\sin(\beta\ln t))$$ where $C_1,C_2$ are constants. Note that you assume $x+1=t$ before and ofcourse $t\in(0,+\infty)$.</p>
|
1,274,317 | <blockquote>
<p>Let $f:R \longrightarrow S$ a surjective ring homomorphism. Is the inverse image $f^{-1}(M)$ a maximal left ideal of $R$ for any maximal
left ideal $M$ of $S$?</p>
</blockquote>
<p><strong>Comments:</strong> I tied something like this: if $M$ is maximal then</p>
<p>$M \neq S$ and if $J$ is a left ideal such that $M \varsubsetneq J \subseteq S \Rightarrow J = S$. </p>
<p>I want to show that:
$f^{-1}(M) \neq R$ and if $I$ is a left ideal such that $f^{-1}(M) \varsubsetneq I \subseteq R \Rightarrow I = R$.</p>
<p>The first statement follows from the fact that $f$ is surjective. I am unable to prove the second.</p>
| user 1 | 133,030 | <p>Hint.<br>
Let $K:=ker f$. Then $R/ K \cong S$. So maximal ideals of $S$ correspond to maximal ideals of $R/ K$. On the other hand maximal ideals of $R/K$ correspond to <em>those maximal ideals of $R$ which contain $K$</em>.</p>
|
4,309,812 | <p>Recently I knew about <a href="https://en.m.wikipedia.org/wiki/Heron%27s_formula" rel="nofollow noreferrer">Heron's formula</a> for the area of some triangle, and its generalizations to quadrilaterals by Bretschneider's formula. According to Wikipedia there are also generalizations for pentagons and hexagons inscribed in a circle.</p>
<p>My question: has someone researched on the possibility to generalize this formula for <span class="math-container">$n$</span>-gons? Is there any underlying reason for this to be a difficult (or impossible) task?</p>
<p>Thanks!</p>
| Sidvhid Hsinynjad | 866,977 | <p>Hint:</p>
<p>You may use the following identity</p>
<p><span class="math-container">$$(y+x)^2-(y-x)^2=4xy$$</span></p>
|
134,987 | <blockquote>
<p>$$3x^2 + 2y^4 = z^4$$</p>
</blockquote>
<p><em>How do I solve this??</em> I would like to use so-called "elementary number theory", not abstract algebra (e.g. $\mathbb{Z} ( \sqrt d)$) or elliptic curves.</p>
<p>Note: I'm not asking <em>what</em> the solutions are, but rather <em>how</em> to find them.</p>
<p>My instincts are: </p>
<ul>
<li>search the internet (I compared this equation with the ~280 here on MSE, and tried a variety of similar searches on uniquation.com ...)</li>
<li>search the 3 number theory books that I have</li>
<li>try to find solutions "by inspection" (possibly after reducing the order of the variables)</li>
<li>do some magic with modular arithmetic </li>
<li>use <a href="http://www.alpertron.com.ar/QUAD.HTM" rel="nofollow">Alpern's solver</a> - which seemed to indicate that there are no solutions (though I might have made an illegal substitution, so to speak) </li>
</ul>
<p>I was able to identify $A = 6, B = 3, C = 6$ as solutions of $ \ 3A + 2B \ ^2 = C \ ^2$, but those aren't squares!</p>
<blockquote>
<p>What is the number-theoretic approach to such problems? Is there a general method?</p>
</blockquote>
| Dayo Adeyemi | 29,565 | <p>Supposing we did have a solution lets consider the equation modulo $3$, since a square (hence a fourth power) must be congruent to $0$ or $1$ so the LHS is congruent to $0$ or $2$ and the RHS is $0$ or $1$ we see $3$ must divide both $y$ and $z$ thus $3^3$ must divide $x^2$ so $3^2$ divides $x$ hence $3^4$ divides the entire equation and dividing through leaves the same equation as we started with so we fall into an infinite decent which is absurd, hence we can have no solution in integers.</p>
|
134,987 | <blockquote>
<p>$$3x^2 + 2y^4 = z^4$$</p>
</blockquote>
<p><em>How do I solve this??</em> I would like to use so-called "elementary number theory", not abstract algebra (e.g. $\mathbb{Z} ( \sqrt d)$) or elliptic curves.</p>
<p>Note: I'm not asking <em>what</em> the solutions are, but rather <em>how</em> to find them.</p>
<p>My instincts are: </p>
<ul>
<li>search the internet (I compared this equation with the ~280 here on MSE, and tried a variety of similar searches on uniquation.com ...)</li>
<li>search the 3 number theory books that I have</li>
<li>try to find solutions "by inspection" (possibly after reducing the order of the variables)</li>
<li>do some magic with modular arithmetic </li>
<li>use <a href="http://www.alpertron.com.ar/QUAD.HTM" rel="nofollow">Alpern's solver</a> - which seemed to indicate that there are no solutions (though I might have made an illegal substitution, so to speak) </li>
</ul>
<p>I was able to identify $A = 6, B = 3, C = 6$ as solutions of $ \ 3A + 2B \ ^2 = C \ ^2$, but those aren't squares!</p>
<blockquote>
<p>What is the number-theoretic approach to such problems? Is there a general method?</p>
</blockquote>
| user55514 | 55,514 | <p>The Chaz said:</p>
<p>""- use <a href="http://www.alpertron.com.ar/QUAD.HTM" rel="nofollow">Alpern's solver</a> - which seemed to indicate that there are no solutions (though I might have made an illegal substitution, so to speak) </p>
<p>I was able to identify $A = 6, B = 3, C = 6$ as solutions of $ \ 3A + 2B \ ^2 = C \ ^2$, but those aren't squares!""</p>
<p>You forgot that the useful and nice Alpern solver solves only in <em>two</em> variables; and here we have <em>three</em> variables.</p>
<p>I am surprised Gerry and Will did not correct you, Gerry is too busy correcting spelling or the aesthetics of the writing representation of mathematics; in particular not allowing a simpler and less "beautiful", but correct, ASCII representation of them. Very nice, i do certainly agree, the Dayo proof of no solutions to the equation. A short and simple and elegant proof at a time.</p>
<p>Post Scriptum: I do not dislike editing myself when i make a typographic error and maths have also to be well written and well expressed. They are ( Do maths have a plural in English like in Spanish and French ?) also a kind of literature where things have to be said well.
But today i wrote an answer to somebody in which i said his question was also sometimes named the Polignac conjecture. Later someone edited me and gave a link to a wikipedia article on this conjecture. I like Wikipedia and can say i have learnt, as an amateur, a few good things with it. The matter and still slight problem is i did not put any link, somebody put it for me without my consent. What will be next, change the style i (we) use to write; the verbs i (we) use ? I mean there must be a limit of some sort to the editing mania.</p>
|
4,187,932 | <p>Is there a general <strong>algebraic</strong> form to the integral <span class="math-container">$$\int_{k_1}^{k_2} x^2 e^{-\alpha x^2}dx?$$</span> I know that if this integral is an improper one, then the integral can be calculated quite easily (i.e. is a well known result). However, when these bounds are not imposed, I am getting results in the form of the error function which turns out to be another integral. So I am confused if this can even be solved by hand or the only way to do so is to plug it in through a calculator.</p>
<p>The reason I am asking is because this integral is highly related to physics as in that it is the proportional to the Maxwell Boltzmann distribution. So if I want to know the number of particles between two velocities, I have to calculate this. I am sorry if this is a basic question. I just could not find anything similar online.</p>
<p>I would also be interested in the results if only the upper bound is <span class="math-container">$\infty$</span> and the lower bound is <span class="math-container">$k$</span>.</p>
| Kyle Miller | 172,988 | <p>Expanding @Intelligenti pauca's comment, the power series for <span class="math-container">$\sqrt{x^2+y^2}$</span> in terms of <span class="math-container">$y$</span> centered at <span class="math-container">$0$</span> starts as
<span class="math-container">$$\lvert x\rvert + \frac{y^2}{2\lvert x\rvert} + \frac{y^4}{8 \lvert x\rvert^3} + \frac{y^6}{16\lvert x\rvert^5} +\cdots$$</span>
So, to first order with respect to <span class="math-container">$y$</span>, <span class="math-container">$\sqrt{x^2+y^2}\approx \lvert x\rvert$</span>.</p>
<p>Before seeing how this interacts with calculating <span class="math-container">$\frac{d\lvert \mathrm{v}\rvert}{dt}$</span>, let's calculate the derivative directly so we can believe it's zero at least mathematically. Write <span class="math-container">$v_x$</span> and <span class="math-container">$v_y$</span> for the two components of <span class="math-container">$\mathrm{v}$</span>. Taking derivatives, we get
<span class="math-container">$$\frac{d\lvert\mathrm{v}\rvert}{dt} = \frac{d\sqrt{v_x^2+v_y^2}}{dt} = \frac{1}{\sqrt{v_x^2+v_y^2}}\left(v_x\frac{d v_x}{dt} + v_y\frac{d v_y}{dt}\right) = \frac{1}{\lvert \mathrm{v}\rvert}\left(v_x\frac{d v_x}{dt} + v_y\frac{d v_y}{dt}\right)$$</span>
When we evaluate at the <span class="math-container">$t$</span> where <span class="math-container">$v_x=v$</span> and <span class="math-container">$v_y=0$</span>, which by physics is when <span class="math-container">$\frac{d v_x}{dt}=0$</span> (non-rigorously, this is that "<span class="math-container">$\mathrm{v}$</span> and <span class="math-container">$d\mathrm{v}$</span> are orthogonal"), then we can see the derivative evaluates to <span class="math-container">$0$</span>.</p>
<p>Now let's use the power series to compute the derivative. Substituting in <span class="math-container">$v_x$</span> and <span class="math-container">$v_y$</span>, we get
<span class="math-container">$$\lvert\mathrm{v}\rvert = \lvert v_x\rvert + \frac{v_y^2}{2\lvert v_x\rvert} + \cdots$$</span>
Taking the derivative with respect to <span class="math-container">$t$</span>, we get
<span class="math-container">$$ \frac{d\lvert\mathrm{v}\rvert}{dt} = \frac{d\lvert v_x\rvert}{dt}+\left(-\frac{v_y^2}{2v_x^2}\frac{d\lvert v_x\rvert}{dt} + \frac{v_y}{\lvert v_x\rvert}\frac{d v_y}{dt} \right) + \cdots
$$</span>
This is where we can see that the fact that using <span class="math-container">$\sqrt{x^2+y^2}$</span> to first order is all what matters: since <span class="math-container">$v_y=0$</span>, everything from the second term onward is zero and we are left with <span class="math-container">$\frac{d\lvert\mathrm{v}\rvert}{dt} = \frac{d\lvert v_x\rvert}{dt}$</span>.</p>
<p><strong>In short:</strong> when in doubt, try to eliminate differentials and instead use derivatives directly. Most of the time statements like "<span class="math-container">$dv^2=0$</span>" represent manipulation of a power series, in this case that second-order terms and above don't contribute to the first derivative.</p>
|
2,746,153 | <p>Assume $m\ \mathrm{and}\ n\ \mathrm{are\ two\ relative\ prime\ positive\ integers.}$</p>
<p>Given $x \equiv a\ \pmod m$ and $x \equiv a\ \pmod n$.</p>
<p>Prove that $x \equiv a\ \pmod {mn}\ \mathrm{by\ using\ Chinese\ Remainder\ Theorem}.$<br/></p>
<p>And I did the following:
<br>
$$ \mathrm {M_1 = }\ n\ \ and\ \ \mathrm {M_2 = }\ m\ \\
\mathrm {y_1 = }\ n’\ \ and\ \ \mathrm {y_2 = }\ m’ \\
\mathrm{where}\ n\cdot n’\equiv 1\ \mathrm{(mod}\ m) \ \ and\ \ m\cdot m’\equiv1\ \mathrm{(mod}\ n) \\
Then\ x\equiv\ (a\cdot n\cdot n’\ +a\cdot m\cdot m’ )\pmod{mn} $$
<br>
But how could I conclude
“$x \equiv a\ (\mathrm {mod}\ mn)$” from the last statement or I did it wrongly? I would be grateful for your help :)</p>
| Joffan | 206,402 | <p>We have:<br>
$n \mid (x-a)$, and<br>
$m \mid (x-a)$</p>
<p>and $n$ and $m$ have no common factors, so<br>
$nm \mid (x-a)$</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>
| Leandro | 633 | <p>The answer is given in this excellent post by Terence Tao </p>
<p><a href="https://terrytao.wordpress.com/2011/09/12/the-inverse-function-theorem-for-everywhere-differentiable-maps/" rel="nofollow">https://terrytao.wordpress.com/2011/09/12/the-inverse-function-theorem-for-everywhere-differentiable-maps/</a></p>
|
3,045,677 | <p>I am trying to create a program for a school project where I need to plot points of a triangle given all 3 side lengths (a=10, b=20, c=30).
I tried the solution from the other topic and it didn't work since the result produced was C(20,0) and that cant be right since one of the sides is already placed on the axis and its not a right triangle.</p>
<p>Is there a formula I can plug into my program that will produce point C, given that I place A and B on x or y axis and will work with any triangle?</p>
<p>Thank you.</p>
| fleablood | 280,126 | <blockquote>
<p>I am trying to create a program for a school project where I need to plot points of a triangle given all 3 side lengths (a=10, b=20, c=30).</p>
</blockquote>
<p>By the triangle inequality <span class="math-container">$a + b > c$</span>. In this case you have <span class="math-container">$a + b =c$</span>. So this will be a straight line where <span class="math-container">$C$</span> is in a line in between <span class="math-container">$A$</span> and <span class="math-container">$B$</span>.</p>
<blockquote>
<p>I tried the solution from the other topic</p>
</blockquote>
<p><em>WHAT</em> other topic?</p>
<blockquote>
<p>and it didn't work since the result produced was C(20,0)</p>
</blockquote>
<p>That is correct. <span class="math-container">$A = (0,0)$</span> and <span class="math-container">$B = (30, 0)$</span> and <span class="math-container">$C = (20,0)$</span> then <span class="math-container">$a = BC = \sqrt{(30-20)^2 - (0-0)^2} = \sqrt{(10)^2} = 10$</span> and <span class="math-container">$b = AC = \sqrt{(0-20)^2 - (0-0)^2} = \sqrt{(-20)^2} = 20$</span> and <span class="math-container">$c = AB = \sqrt{(0-30)^2 - (0-0)^2} = \sqrt{ (-30)^2} = 30$</span>.</p>
<blockquote>
<p>and that cant be right since one of the sides is already placed on the axis and its not a right triangle.</p>
</blockquote>
<p>Who said it was a right triangle?</p>
<p><span class="math-container">$a^2 + b^2 = 10^2 + 20^2 = 100 + 400 = 500$</span> but <span class="math-container">$c^2 = 30^2 = 900$</span> and <span class="math-container">$500 \ne 900$</span>. It isn't a right triangle.</p>
<blockquote>
<p>Is there a formula I can plug into my program that will produce point C, given that I place A and B on x or y axis and will work with any triangle?</p>
</blockquote>
<p>So if <span class="math-container">$A= (0,0)$</span> and <span class="math-container">$B= (c,0)$</span> then <span class="math-container">$C = (x,y)$</span> where</p>
<p><span class="math-container">$b^2 = AC^2 = x^2 +y^2$</span></p>
<p><span class="math-container">$a^2 = BC^2 = (x-c)^2 + y^2$</span></p>
<p>So <span class="math-container">$a^2 - b^2 = (x-c)^2 - x^2$</span> or </p>
<p><span class="math-container">$a^2 - b^2 = x^2 - 2cx + c^2 - y^2 = -2cx + c^2$</span> or</p>
<p><span class="math-container">$x = \frac {c^2 + b^2 -a^2}{2c}$</span></p>
<p>And so <span class="math-container">$x^2 +y^2 = b^2$</span> so</p>
<p><span class="math-container">$y = \sqrt{ b^2 - x^2}$</span></p>
<hr>
<p>Plugging in <span class="math-container">$a = 10; b=20; c= 30$</span> we get</p>
<p><span class="math-container">$x =\frac {c^2 + b^2 -a^2}{2c}= \frac {30^2 + 20^2 -10^2}{2*30}= \frac {900+ 400 - 100}{60} = \frac {1200}{60} = 20$</span></p>
<p>And <span class="math-container">$x =\sqrt{ b^2 - x^2}= \sqrt {20^2 - 20^2} = 0$</span>.</p>
<p>So <span class="math-container">$C = (20, 0)$</span></p>
<hr>
<p>For a more proper triangle sa <span class="math-container">$a = 20; b= 15; c=25$</span> (an actual right triangle.</p>
<p>You get <span class="math-container">$A=(0,0)$</span> and <span class="math-container">$B = (25,0)$</span> and <span class="math-container">$C = (x,y)$</span> where</p>
<p><span class="math-container">$x = \frac {c^2 + b^2 -a^2}{2c}= \frac {25^2 + 15^2-20^2}{50}= 9 $</span></p>
<p>and
<span class="math-container">$y = \sqrt{ b^2 - x^2}= \sqrt{15^2 - 9^2} = 12$</span></p>
<p>So <span class="math-container">$C = (9,12)$</span>.</p>
<p>======</p>
<p>Now for an example where that fails: Say <span class="math-container">$a = 20; b=30; c = 60$</span>. That's impossible because <span class="math-container">$a+b < c$</span>. No such triangle exists.</p>
<p>If <span class="math-container">$A=(0,0)$</span> and <span class="math-container">$B=(60,0)$</span> then <span class="math-container">$C = (x,y)$</span> with</p>
<p><span class="math-container">$x = \frac {c^2 + b^2 -a^2}{2c}= \frac {60^2 + 30^2-20^2}{60}= 10\sqrt{41}$</span></p>
<p>and
<span class="math-container">$y = \sqrt{ b^2 - x^2}= \sqrt{20^2 - (10\sqrt{41}^2} = \sqrt{400-410} = \sqrt{-10}$</span></p>
<p>So <span class="math-container">$C = (10\sqrt{41},\sqrt{-10})$</span> is not possible.</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>
| Dietrich Burde | 83,966 | <p>With $x^3-1=(x-1)(x^2+x+1)$ we can use $x^{3}=1$ and then reduce the polynomial to $f(x)=8+1799+87+124+1=2019$. This is almost $3$ years ahead, though. </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>
| Deepak | 151,732 | <p>Yes it's true, and it's easily proven by squaring both sides. This operation is allowed (without worrying about the direction of the inequality) because all terms are non-negative.</p>
|
1,238,210 | <p>How we can solve that $\lim _{_{x\rightarrow \infty }}\int _0^x\:e^{t^2}dt$ ?</p>
<p>P.S: This is my method as I thought:
$\int _0^x\:\:e^{t^2}dt>\int _1^x\:e^tdt=e^x-e$ which is divergent, so all your answers, helped me to think otherwise, maybe my method help something else :D</p>
| davidlowryduda | 9,754 | <p>This function diverges extremely fast. Notably, $e^{t^2}$ is monotone increasing with limit $\infty$ as $t \to \infty$. Thus your integral diverges (and it get very, very large very, very quickly).</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>
| lhf | 589 | <p>A well-known variant of the usual proof may be clearer. If $\sqrt{3}=\frac{a}{b}$, then $a^2=3b^2$. Recall the Fundamental Theorem of Arithmetic and consider the exponent of $3$ in the prime factorization of both sides. On the left you have an even exponent. On the right you have an odd exponent, contradiction. This approach goes one to prove that $\sqrt m$ is rational iff $m$ is 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>
| J. W. Tanner | 615,567 | <p><strong>Hint:</strong></p>
<p>for <span class="math-container">$n>2,$</span></p>
<p><span class="math-container">$
n^3+\dfrac{n-1}2<\sqrt{n^6+n^4+1}<n^3+\dfrac n2.$</span></p>
|
4,249,573 | <p>I'm studying Set Theory in my own, with Goldrei's textbook. The chapter I'm reading is on order-isomorphism and well-ordering. One exercise asks (i) to argue that, in general, a collection of well-ordered sets order-isomorphic to a given well-ordered set is a proper class (rather than a set). The proof, IMHO, is fairly straightforward and applicable to any isomorphism class. Then the same exercises asks (ii) if there is a well-ordered set, X say, s.t. the collection of all well-ordered set, order-isomorphic to X, is a set (rather than a proper class). I stumbled on that one.</p>
| Community | -1 | <p>The class of the empty well-order is the only such class that is also a set. For every other class <span class="math-container">$[(X,\le)]$</span> the class-function sending <span class="math-container">$(Y,\preceq)\mapsto Y$</span> would be onto the class of sets that are in bijection with <span class="math-container">$X$</span>, which is proper as soon as <span class="math-container">$X\ne\emptyset$</span>.</p>
|
208,615 | <p>Maybe I'm not using the best programming practices in Mathematica, but my notebooks usually contain a mixture of "definitions" and "computations" with these definitions. Say,</p>
<pre><code>f[x_] := x^2 (* Definition *)
f[10] (* Computation *)
g[x_] := f[x] - 1/f[x] (* Definition *)
g[100] (* Computation *)
</code></pre>
<p>Then each time I open the notebook anew, to ensure that all definitions are in place the easiest way is to evaluate everything. But some computations can be costly, and I do not need their results. </p>
<p>So, is there a way to only evaluate "definitions" or assignments in the notebook, and omit the "computations"? In the code sample above only 1st and 3rd lines must be evaluated, while <code>f[10]</code> and <code>g[100]</code> must be ignored.</p>
<p>Or, if the question does not make much sense, what would be good solution to the described problem? Keeping assignments and computations separately does not make much sense to me, commenting out all the computations make notebook look weird and if I do need them some time later, I'll have to do a lot of typographic work removing them.</p>
| Carl Woll | 45,431 | <p><strong>Update</strong></p>
<p><em>(I added a gray background when in the "Initialization" screen style to remind the user to switch back after evaluating code)</em></p>
<p>Here's a stylesheet solution. The basic idea is to add an "Initialization" screen style environment (like "Working", "SlideShow" and "Presentation"). When the screen style environment is "Initialization" the style definition for "Input" cells looks at the the contents of the cell, and if doesn't match a <a href="http://reference.wolfram.com/language/ref/Set" rel="nofollow noreferrer"><code>Set</code></a> or <a href="http://reference.wolfram.com/language/ref/SetDelayed" rel="nofollow noreferrer"><code>SetDelayed</code></a> expression, the <a href="http://reference.wolfram.com/language/ref/CellEvaluationFunction" rel="nofollow noreferrer"><code>CellEvaluationFunction</code></a> is neutered. Finally, key shortcuts are used to switch between the "Working" and "Initialization" screen style environments. Here is the stylesheet:</p>
<pre><code>SetOptions[
EvaluationNotebook[],
StyleDefinitions -> Notebook[
{
Cell[StyleData[StyleDefinitions->"Default.nb"]],
Cell[StyleData["Input","Initialization"],
CellProlog :> If[
!MatchQ[
NotebookRead[EvaluationCell[]],
Cell[BoxData[RowBox[{_,"="|":=",__}]],__]
],
CurrentValue[EvaluationCell[],CellEvaluationFunction]=Null&
],
CellEpilog :> (CurrentValue[EvaluationCell[], CellEvaluationFunction]=Inherited)
],
Cell[StyleData[All, "Initialization"],
MenuCommandKey -> "=",
Background -> GrayLevel[.92]
],
Cell[StyleData[All, "Working"],
MenuCommandKey -> "-"
]
},
StyleDefinitions->"PrivateStylesheetFormatting.nb"
]
]
</code></pre>
<p>Use <kbd>Cmd</kbd>+<kbd>=</kbd> to switch to the "Initialization" screen style environment. Then, only "Input" cells that are <a href="http://reference.wolfram.com/language/ref/Set" rel="nofollow noreferrer"><code>Set</code></a> or <a href="http://reference.wolfram.com/language/ref/SetDelayed" rel="nofollow noreferrer"><code>SetDelayed</code></a> cells will evaluate. Use <kbd>Cmd</kbd>+<kbd>-</kbd> to switch back to the "Working" screen style environment, and all "Input" cells will evaluate normally.</p>
|
308,329 | <p>I need help with writing $\sin^4 \theta$ in terms of $\cos \theta, \cos 2\theta,\cos3\theta, \cos4\theta$.</p>
<p>My attempts so far has been unsuccessful and I constantly get developments that are way to cumbersome and not elegant at all. What is the best way to approach this problem?</p>
<p>I know that the answer should be:</p>
<p>$\sin^4 \theta =\frac{3}{8}-\frac{1}{2}\cos2\theta+\frac{1}{8}\cos4\theta$</p>
<p>Please explain how to do this.</p>
<p>Thank you!</p>
| Ron Gordon | 53,268 | <p>Write</p>
<p>$$\sin^4{\theta} = \left ( \frac{e^{i \theta} - e^{-i \theta}}{2 i} \right )^4$$</p>
<p>and use the binomial theorem.</p>
<p>$$\begin{align}\left ( \frac{e^{i \theta} - e^{-i \theta}}{2 i} \right )^4 &= \frac{1}{16} (e^{i 4 \theta} - 4 e^{i 2 \theta} + 6 - 4 e^{-i 2 \theta} + e^{-i 4 \theta}) \\ &= \frac{1}{8} (\cos{4 \theta} - 4 \cos{2 \theta} + 3)\end{align}$$</p>
<p>Item of curiosity: the <a href="http://en.wikipedia.org/wiki/Chebyshev_polynomials">Chebyshev polynomials</a> are defined such that </p>
<p>$$T_n(\cos{\theta}) = \cos{n \theta}$$</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>
| Community | -1 | <p>For instance, when $\inf A>0$ and $\sup B<0$. Since all the products will be negative, you want to minimise the absolute values of each factor.</p>
|
39,423 | <ul>
<li><p>case1</p>
<pre><code>Options[f] = {"t" -> "0"};
f[___, OptionsPattern[]] := StringReplace["content", "t" :> OptionValue["t"]]
f[]
(*
con0en0
*)
</code></pre></li>
<li><p>case2</p>
<pre><code>rule = {"t" -> OptionValue["t1"]};
Options[gg] = {"t1" -> "T1", "t2" -> "1"};
gg[___, OptionsPattern[]] := StringReplace["content", rule]
gg[1]
(*
con~~OptionValue[t1]~~en~~OptionValue[t1]
*)
</code></pre></li>
</ul>
<p>Here <code>OptionValue</code> couldn't get the value of "t1"
So, how to make case 2 works like case 1? </p>
<hr>
<p>I found one solution is </p>
<pre><code>Options[gg]={"t1"->"T1","t2"->"1"};
gg[___,OptionsPattern[]]:=Hold[StringReplace]["content",rule]//ReleaseHold//Evaluate
</code></pre>
<p>Any simpler methods?</p>
| Chris Degnen | 363 | <p>One solution is to keep <code>OptionValue</code> inside the function :-</p>
<pre><code>rule = {"t" -> "t1"};
Options[gg] = {"t1" -> "T1", "t2" -> "1"};
gg[___, OptionsPattern[]] := StringReplace["content", #1 -> OptionValue[#2] & @@@ rule]
gg[1]
</code></pre>
<blockquote>
<p>"conT1enT1"</p>
</blockquote>
|
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>
| Vladhagen | 79,934 | <p>It is pretty easy to show (and might even be given in most textbooks) that
$$\cos(\pi - \alpha) = -\cos(\alpha)$$ for any angle $\alpha$. </p>
<p>Note that the angle opposite $\theta$ in your diagram is found by computing $\pi-\theta$. Since it is pretty easy to get that $\cos(\pi-\theta) = \frac{-4}{5}$ from the diagram, we have the following:</p>
<p>$$-\cos(\theta) = \cos(\pi-\theta) = -\frac{4}{5}$$
thus $\cos(\theta) = -(-\frac{4}{5}) = \frac{4}{5}.$</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>
| Lyapunov | 25,683 | <blockquote>
<p>Let $(X,\|\cdot\|)$ be a normed linear space. Then $\overline{B_1(0)}=\bar{B}_1(0)$.</p>
</blockquote>
<p>Proof. Observe that $\overline{B_1(0)}$ is the smallest closed set containing $B_1(0)$ and $B_1(0)\subset \bar{B}_1(0)$, so trivially $\overline{B_1(0)}\subset\bar{B}_1(0)$. Now to show
$\bar{B}_1(0)\subset \overline{B_1(0)}$. Observe that, $\bar{B}_1(0)=B_1(0)\cup \partial B_1(0)$, i.e., for all $x\in \partial B_1(0), \, \exists x_n\in B_1(0)$ such that $\|x_n-x\|\to 0$: for any given $x\in \partial B_1(0),$ let $x_n=(1-\frac{1}{n})x, \, n\in \mathbb{N}.$ Then show $x_n\in B_1(0)$ and $\|x_n-x\|\to 0$. </p>
|
1,987,317 | <blockquote>
<p>Prove that the system $$A^T A x = A^T b$$ always has a solution. The matrices and vectors are all real. The matrix $A$ is $m \times n$. </p>
</blockquote>
<p>I think it makes sense intuitively but I can't prove it formally.</p>
| Taha Akbari | 331,405 | <p>Sorry for bumping but I had a solution which I liked to share.</p>
<p>Using the <a href="https://en.wikipedia.org/wiki/Fundamental_theorem_of_linear_algebra" rel="nofollow noreferrer">fundamental theorem of linear algebra</a> We can decompose our matrix <span class="math-container">$b$</span> into two matrices <span class="math-container">$b_1$</span> and <span class="math-container">$b_2$</span> such that <span class="math-container">$b_1$</span> is in the column space of <span class="math-container">$A$</span> and <span class="math-container">$b_2$</span> is in the null-space of <span class="math-container">$A^T$</span>.</p>
<p>Now let <span class="math-container">$x^*$</span> be a solution to <span class="math-container">$Ax^* = b_1$</span> (It always has a solution since <span class="math-container">$b_1$</span> is in the column space of <span class="math-container">$A$</span>. We claim that <span class="math-container">$x$</span> is also a solution to <span class="math-container">$A^TAx = A^Tb$</span>.</p>
<p><span class="math-container">$$A^T(Ax* - b) = A^T(b_1 - (b_1 + b_2)) = A^T(-b_2) = - A^T(b_2) = 0$$</span></p>
<p>The last equality being a consequence of <span class="math-container">$b_2$</span> being in the null-space of <span class="math-container">$A^T$</span>.</p>
|
3,832,484 | <p>Title's all there is to say.
I'm very new to linear algebra and haven't wrapped my head around determinant rules yet.
Any help would be appreciated.</p>
| jacopoburelli | 530,398 | <p>Solution : <span class="math-container">$\det(A + A) = \det (2A) = 2^3\det(A)$</span> since <span class="math-container">$A$</span> is <span class="math-container">$3\times3$</span>, so <span class="math-container">$\det(A + A) = 2^3 4 = 32.$</span></p>
|
200,322 | <p>Is there a compact topological space $(X,\tau)$ such that for no cardinal $\kappa$ there is a surjective continuous map $e:\{0,1\}^\kappa \to X$? </p>
<p>(We assume that $\{0,1\}$ is endowed with the discrete topology, and $\{0,1\}^\kappa$ has the product topology.)</p>
| Włodzimierz Holsztyński | 8,385 | <p>I'd say that the simplest counter-example is the $1$-point compactification of a non-countable discrete space (of course any compactification of any non-countable discrete space would do).</p>
|
65,223 | <p>The least rational number greater than $\sqrt{2}$ that can be written as a ratio of integers $x/y$ with $y\le10^{100}$ can be found in a moment using a little Python program. Can anyone write a program that finds, in hours rather than centuries, the least rational greater than $\sqrt{2}$ of the form $x/y^2$ with $y^2\le 10^{100}$? </p>
<p>More generally, my question is whether the following computation is known to be feasible or not feasible:</p>
<p>Given $N$, find the least rational greater than $\sqrt{2}$ of the form $x/y^2$,
with $x$ and $y$ integers and $y^2\le N$. For definiteness, let's say that the output should be the required rational written in lowest form.</p>
<p>By a feasible computation I mean one that can be done in $O((\log N)^k)$ bit operations for some constant $k$. </p>
<p>Of course the square root of 2 is not essential here. Any irrational would do, as long as comparisons with rationals are feasible. I don't know of any such irrational for which I can answer the question I've posed.</p>
| Victor Miller | 2,784 | <p>You should try the algorithms in Elkies' paper (from 2000) "Rational points near curves ..." <a href="http://arxiv.org/abs/math/0005139" rel="nofollow">http://arxiv.org/abs/math/0005139</a> . His idea is to cover the curve with a bunch of small rectangles, and use lattice basis reduction within each such region. He proves a result which either says that there are small number of solutions or all the solutions lie on a line.</p>
|
658,078 | <p>I'm embarrassed to ask this question, but my child has the following homework question:</p>
<p>"Use absolute value to describe the relationship between a negative credit card balance and the amount owed."</p>
<p>I'm not sure for what it is they're looking. Clearly a <code>-$25</code> balance means you have <code>$25</code> credit. However, the absolute value of <code>-$25</code> is <code>$25</code>, and positive balances are money you owe.</p>
<p>Is there a simple formula describing this relationship? </p>
| John Habert | 123,636 | <p>You have identified the identity element $e=-5$. To show that you have inverses, you need to prove that for every $a \in \mathbb{Z}$ there is some $b \in \mathbb{Z}$ such that $a*b = b*a = e$. Assuming you proved $*$ is commutative as mentioned in the comments, then it suffices to show there is a $b$ such that $a*b=e$. </p>
<p>I claim that $b=-a-10$ fits the bill. If $a \in \mathbb{Z}$, clearly $-a-10\in \mathbb{Z}$. Then $a*(-a-10) = a + (-a-10) + 5 = a - a - 10 + 5 = -5 = e$. Thus $-a-10$ is the inverse of $a$. </p>
|
1,234,093 | <p>Given that $ e= \frac{a^2-b^2}{b^2} $ , and $L$ is the length of the perimeter, which equals $4aE(e, \pi/2)$, find the length of the perimeter up to $e^2$ in terms of $a$ and $b$.</p>
<p>How does one begin this?</p>
| Christian Blatter | 1,303 | <p>Note that the eccentricity of the ellipse is defined by $e:=\sqrt{a^2-b^2}/a$, whereby $a$ is the longer semiaxis.</p>
<p>A beginning: Use the parametrization
$$x=a\cos \phi,\quad y=b\sin\phi\qquad(0\leq\phi\leq2\pi)$$
in order to set up the integral for $L$. In the resulting expression substitute $b^2:=a^2(1-e^2)$, whereupon $a$ can be factored out. Now expand the integrand into a power series with respect to $e$ and keep as many terms as needed for the desired accuracy.</p>
|
2,138,916 | <p>My question read: </p>
<p>Show that $S_{10}$ contains elements of orders $10,20,30$. Does it contain an element of order $40$? </p>
<p>I am not too sure what the question is asking. Would I have to explicitly write out all the permutations in $S_{10}$ first and then find the orders for all of them? </p>
<p>Update: I understand I need to only show a few examples of disjoint cycles, but I am not sure how to show if order 40 is possible.</p>
| Chappers | 221,811 | <p>It's easy to see that your integral is the same as
$$ I(a) = \int_0^{\infty} \frac{\log{(x^2+a^2)}}{x^2+b^2} \, dx $$
Now, we can do the case $a=0$ fairly easily, by setting $x=b^2/y$:
$$ \begin{align}
I(0) &= 2\int_0^{\infty} \frac{\log{x}}{x^2+b^2} \, dx \\
&= 2\int_0^{\infty} \frac{\log{(b^2/y)}}{b^2/y^2+b^2} \frac{b}{y^2} \, dy \\
&= 2\int_0^{\infty} \frac{2\log{b}-\log{y}}{y^2+b^2} \, dy \\
&= 4\log{b} \int_0^{\infty} \frac{dy}{y^2+b^2} -I(0),
\end{align} $$
so
$$ I(0) = \frac{\pi}{b}\log{b}. $$
To get from here to nonzero $a$, differentiate under the integral sign:
$$ I'(a) = \int_0^{\infty}\frac{\partial}{\partial a} \frac{\log{(x^2+a^2)}}{x^2+b^2} \, dx = \int_0^{\infty} \frac{2a \, dx}{(x^2+a^2)(x^2+b^2)} $$
But this is easy to calculate using partial fractions: we find
$$ I'(a) = \frac{2a\pi}{2(b^2-a^2)} \left( \frac{1}{a} - \frac{1}{b} \right) = \frac{\pi}{b(a+b)} $$
Now
$$ I(a) = I(0) + \int_0^{a} \frac{\pi}{b(A+b)}\, dA = \frac{\pi}{b}(\log{(b+a)}-\log{b}+\log{b}) = \frac{\pi}{b}\log{(a+b)}, $$
as desired.</p>
<hr>
<p>A complex analysis method will work in the same way as that given in <a href="https://math.stackexchange.com/a/371327/221811">this answer</a>, although the pole is in a different place from the branch point in your case, rather than coincident.</p>
|
59,567 | <p>I am looking for a way to add a legend showing the identity of various atoms (with different colours) to this picture. Any Clues?</p>
<pre><code>Import["ExampleData/1PPT.pdb", "Rendering" -> "BallAndStick"]
</code></pre>
<p><img src="https://i.stack.imgur.com/FSFoH.png" alt="enter image description here"></p>
| Jason B. | 9,490 | <p>To show the colors associated with atoms, use the <code>Automatic</code> setting for the PlotLegends option:</p>
<pre><code>MoleculePlot3D[
Molecule @ "1-(amino-dimethyl-silyl)-2,3,4,5,6-pentafluoro-benzene",
PlotLegends -> Automatic
]
</code></pre>
<p><a href="https://i.stack.imgur.com/8ugWk.png" rel="noreferrer"><img src="https://i.stack.imgur.com/8ugWk.png" alt="enter image description here"></a></p>
<p>The example in the OP does not work yet, but the PDB importer will be updated in a future release.</p>
|
3,878,723 | <blockquote>
<p>Find the value of <span class="math-container">$k$</span> if the curve <span class="math-container">$y = x^2 - 2x$</span> is tangent to the line <span class="math-container">$y = 4x + k$</span></p>
</blockquote>
<p>I have looked at the solution to this question and the first step is the "equate the two functions":<br />
<span class="math-container">$ x^2 - 2x = 4x + k$</span></p>
<p>Why? How does that help solve the equation? And how can I use what I get from equating the two functions to find the solution?</p>
| Bernard | 202,857 | <p>It's rather simple: equating the functions yields the equation for the abscissæ of the intersection points of the parabola and the line, and the curve & the line are tangent if and only if this equation has a double root, i.e., as we have a quadratic equation, if and only if its (reduced) discriminant is <span class="math-container">$0$</span>.</p>
|
317,547 | <p>Let $X$ and $Y$ have joint mass function</p>
<p>$f(j,k)=\frac {c(j+k)a^{j+k}}{j!k!}$, $j,k\geq 0$</p>
<p>where $a$ is a constant. Find $c$</p>
<p>This sum seems hard to to. How to complete this sum?</p>
| Ross Millikan | 1,827 | <p>Hint: It looks like $\frac c{j!k!} \frac d{da} a^{j+k+1}$ so binomial identities have a lot to say.</p>
|
145,785 | <p>Let $V$ be a $\mathbb{C}$-vector space of finite dimension. Denote its $d$-th symmetric power by $V^{\odot d}$. I am looking for a proof that $V^{\odot d}$ is generated by the elements $v^{\odot d}$ for $v\in V$. </p>
<p>A different way to look at it is the following: Consider the polynomial ring $R=\mathbb{C}[x_1,\ldots,x_n]$ and $f$ a homogeneous polynomial of degree $d$: Then, I want to show that there are linear polynomials $h_1,\ldots,h_k$ such that $f$ is a linear combination of the $d$-th powers $h_i^d$.</p>
<p>In the case $d=2$, this follows from $2xy = (x+y)^2 - x^2 - y^2$. For higher $d$, I recall seeing a proof involving <a href="http://en.wikipedia.org/wiki/Multinomial_coefficient">multinomial coefficients</a> once, but I do not remember the details. I have tried to work it out again, but it seems a bit cumbersome, so I am asking whether you know any textbook where this result is proved. If you know an easy proof, I'd be very happy if you could outline it, though.</p>
| anon | 11,763 | <p>In <a href="https://math.stackexchange.com/a/137911/11763">my answer here</a> I note that symmetric tensors, as multilinear functionals, descend to linear maps on the symmetric power of the underlying vector space. I then reason that if we could show that $\mathrm{Sym}^n V$ is generated by $n$th powers of elements from $V$ the question on tensors would then be answered in its general form decisively. I remark that this is formally equivalent to the elementary symmetric polynomials $e_n$ being expressible as sums of $n$th powers of homogeneous polynomials.</p>
<p>This was the subject of <a href="https://math.stackexchange.com/questions/137912/can-e-n-always-be-written-as-a-linear-combination-of-n-th-powers-of-linear-p">my question here</a>, which received a <a href="https://math.stackexchange.com/a/138411/11763">correct answer</a> (containing a proof of the claim) from user m_l. It was very combinatorial and indeed involved multinomial coefficients, though I'm not sure how related it is to what you've seen before. (Unfortunately, at this point in time I am the only person to have upvoted poor m_l.) It requires the characteristic of the base field be greater than the power $n$ in question (or zero, of course).</p>
|
3,583,117 | <p>I would like to understand clearly why the following equality is true</p>
<p><span class="math-container">$P[X+Y \leq z] = E_Y[P[X+Y] \leq z | Y]]$</span></p>
<p>I wrote the left part of the equation as follows:</p>
<p><span class="math-container">$E_Y[P[X+Y] \leq z | Y]] = \sum_y y P[X+y \leq z]P(y)$</span></p>
<p>and I have tried with a toy example where <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are two <span class="math-container">$r.v$</span> that model the throw of a die and it works, but I would like to clearly understand why is it true, I know that is linked with the law of total probability right?</p>
| drhab | 75,923 | <p>In situations like this where a probability is involved it is often handsome to convert these probabilities to expectations by means of: <span class="math-container">$$P[A]=\mathbb E[1_A]$$</span></p>
<p>Doing so we find by means of the general rule <span class="math-container">$\mathbb E[Z]=\mathbb E[\mathbb E[Z\mid Y]]$</span> that:<span class="math-container">$$P\left[X+Y\leq z\right]=\mathbb{E}\mathbf{1}_{\left(-\infty.z\right]}\left(X+Y\right)=\mathbb{E}\left[\mathbb{E}\left[\mathbf{1}_{\left(-\infty.z\right]}\left(X+Y\right)\mid Y\right]\right]=$$</span><span class="math-container">$$\mathbb{E}\left[P\left[X+Y\leq z\mid Y\right]\right]$$</span></p>
<p>Here <span class="math-container">$P\left[X+Y\leq z\mid Y\right]$</span> is a random variable that is
measurable wrt <span class="math-container">$\sigma\left(Y\right)$</span> hence takes the shape <span class="math-container">$f\left(Y\right)$</span>
for some function <span class="math-container">$f:\mathbb{R}\to\mathbb{R}$</span> that is Borel-measurable.</p>
<p>This function is determined by:
<span class="math-container">$$f\left(y\right)=P\left[X+Y\leq z\mid Y=y\right]=P\left[X+y\leq z\mid Y=y\right]=P\left[X\leq z-y\mid Y=y\right]$$</span></p>
<p>If moreover <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are independent then we can proceed with:<span class="math-container">$$\cdots=P[X\leq z-y]$$</span></p>
<p>This leads to the observation that <span class="math-container">$$P[X+Y\leq z]=\mathbb Ef(Y)$$</span>where <span class="math-container">$f(y)=P(X\leq z-y)$</span></p>
|
24,186 | <p>Consider the code below:</p>
<pre><code>s = Solve[(3 - Cos[4*x])*(Sin[x] - Cos[x]) == 2, x, InverseFunctions -> True];
Select[s[[All, 1, 2]], Element[#, Reals] &]
</code></pre>
<p>In MMA 8.0, I get </p>
<pre><code>{-\[Pi], \[Pi]/2, \[Pi]}
</code></pre>
<p>but in MMA 9.0, I get an empty set { }</p>
<p>Assuming the solution by MMA 8.0 is correct, can someone show me a work around for MMA 9.0? </p>
| bill s | 1,783 | <p>I know for me, I spent years using Matlab (or should I say, a toolbox-based computational system), where there is a trick called vectorization: you turn almost everything (ifs, ands, sums, products...) into simple vector commands. Doing this with your function is pretty natural since you've already defined the entries in terms of two vectors. You take the <code>Abs</code> of each entry, then subtract the two vectors, then take the <code>Norm</code>, hence</p>
<pre><code>Norm[Abs[u] - Abs[v]]
</code></pre>
<p>If you happen (like rm -r) to know the function <code>EuclideanDistance</code>, this can also be used in place of <code>Norm</code></p>
<pre><code>EuclideanDistance[Abs[u], Abs[v]]
</code></pre>
<p>I guess the general strategy (or maybe it's really a tactic) is to keep things in vector (or should I say "List") form as long as possible, and try to avoid breaking them into individual elements. Recognizing that certain operations recur again and again (in this case we have <code>Abs</code> and <code>Norm</code>) is also key, though it's hard to know how to make this into a rule of thumb.</p>
|
1,579,521 | <p>Find the value of
<span class="math-container">$$ \iint_{\Sigma} \langle x, y^3, -z\rangle. d\vec{S} $$</span>
where <span class="math-container">$ \Sigma $</span> is the sphere <span class="math-container">$ x^2 + y^2 + z^2 = 1 $</span> oriented outward by using the divergence theorem.</p>
<p>So I calculate <span class="math-container">$\operatorname{div}\vec{F} = 3y^2 $</span> and then I convert <span class="math-container">$ x, y, z $</span> into <span class="math-container">$ x = p\sin \phi \cos \theta, y = p\sin \phi \sin \theta, z = p\cos \phi $</span> but then I got stuck from that point.</p>
| N74 | 288,459 | <p>You need to evaluate: $$\int_0^1 3 p^4 \ dp\int_0^{2\pi} \sin^2\theta \ d\theta \int_0^\pi \sin^3\phi \ d\phi $$</p>
|
3,043,039 | <p>Let <span class="math-container">$f:(0,1) \to \mathbb{R}$</span> be a given function. Explain how the following definition is not equivalent to the definition of the limit</p>
<p><span class="math-container">$\lim\limits_{x \to x_0} f(x) = L$</span></p>
<p>of <span class="math-container">$f$</span> at <span class="math-container">$x_0 \in [0,1]$</span> . </p>
<p>For any <span class="math-container">$\epsilon \gt 0$</span>,for any <span class="math-container">$\delta \gt 0$</span> such that for all <span class="math-container">$x \in (0,1)$</span> and <span class="math-container">$0 \lt|x-x_0| \lt \delta$</span>, one has <span class="math-container">$|f(x) - L| \lt \epsilon$</span> .</p>
<p>This definition is incorrect because for any <span class="math-container">$\epsilon \gt 0$</span> there exists some <span class="math-container">$\delta \gt 0$</span> that is small enough. It can't be any delta. Is this the only reason why this definition is not valid?</p>
| hamam_Abdallah | 369,188 | <p>Your definition implies the known definition but the converse is not true.</p>
<p>Take <span class="math-container">$f$</span> defined by :</p>
<p><span class="math-container">$$f(x)=0 \;\; \text{ if } \;\; x<\frac 12$$</span>
and
<span class="math-container">$$f(x)=\color{red}{4}\;\; \text{ if } \;\; x\ge \frac 12.$$</span></p>
<p>then we have <span class="math-container">$$\lim_{x\to 0^+}f(x)=0$$</span>
but</p>
<p>if we take <span class="math-container">$\epsilon = \color{red}{3}$</span></p>
<p>then we do not have
<span class="math-container">$$\forall \eta>0 \;\; \forall x\in(0,1)$$</span>
<span class="math-container">$$|x-0|<\eta \implies \;\; |f(x)-0|<3$$</span></p>
<p>for example <span class="math-container">$f(\frac{9}{10})=\color{red}{4}>\epsilon$</span></p>
|
2,030,739 | <p>Find <span class="math-container">$\frac{d^2y}{dx^2}$</span> of:</p>
<blockquote>
<p><span class="math-container">$$4y^2+2=3x^2$$</span></p>
</blockquote>
<h2>My Attempt</h2>
<p>I attempted the probelm my first solving for the first derivative:</p>
<blockquote>
<p><span class="math-container">$8y*y'=6x$</span><br>
<span class="math-container">$y'=\frac{3x}{4y}$</span></p>
</blockquote>
<p>Then I tried it again; however I was a bit confused, and ended up getting</p>
<blockquote>
<p><span class="math-container">$y''=\frac{6y-3x(2*y')}{16y^2}$</span></p>
</blockquote>
<p>Would I substitute in the first derivative back in to get:</p>
<blockquote>
<p><span class="math-container">$y''=\frac{6y-6x(\frac{3x}{4y})}{4y^2}$</span></p>
</blockquote>
<p>and then finally</p>
<h2>Final Answer</h2>
<p><span class="math-container">$$y''=\frac{6y^2-9x^2}{4y^3}$$</span></p>
<p>Is my final answer correct, if not what is the correct answer please?</p>
| Senex Ægypti Parvi | 89,020 | <p>For future reference,<br>
$F(x,y)=0$<br>
$\frac{d^2[F(x,y)]}{dx^2}=-\frac
{\frac{\partial^2 F}{\partial x^2}\left(\frac{\partial F}{\partial y}\right)^2
-2·\frac{\partial^2 F}{\partial x\partial y}·\frac{\partial F}{\partial y}·\frac{\partial F}{\partial x}
+\frac{\partial^2 F}{\partial y^2}\left(\frac{\partial F}{\partial x}\right)^2}
{\left(\frac{\partial F}{\partial y}\right)^3}$</p>
|
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