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 |
|---|---|---|---|---|
1,405,809 | <p>So we have to find $\lim\limits_{x\to 0} \frac{a^x-1}{x}$ without using any series expansions or the L'Hopital's rule.<br>
I did it using both but I have no idea how to do it. I tried many substitutions but nothing worked. Please point me in the right direction.</p>
| ybtang21c | 285,657 | <p>Firt prove $\lim\limits_{x\to 0}\frac{e^x-1}x=1$. In fact, let $y=e^x-1$, that is $x=\ln(1+y)$, we have
$$\lim\limits_{x\to 0}\frac{e^x-1}x=\lim\limits_{y\to 0}\frac{y}{\ln(1+y)}
=\lim\limits_{y\to 0}\frac1{\frac1y\ln(1+y)}=\frac1{\ln\left[\lim\limits_{y\to 0}
(1+y)^{1/y}\right]}=\frac1{\ln e}=1.$$
Then
$$\lim\limits_{x\to 0}\frac{a^x-1}x=\lim\limits_{x\to 0}\frac{e^{x\ln a}-1}{x}=
\ln a\lim\limits_{x\to 0}\frac{e^{x\ln a-1}}{x\ln a}=\ln a.$$</p>
|
2,263,520 | <p>I'm trying to understand how to simplify summations. My text says that: $$\sum_{i=1}^{\frac{n}{2}} \sum_{j=i}^{n-i} \sum_{k=1}^{j} 1 = \frac{n^3}{8}$$</p>
<p>But does not explain how to get to the right-hand side.</p>
<p>I think the above nested summation evaluates to $\sum_{i=1}^{n/2} \left[\sum_{j=i}^{n-i} j = i + (i + 1) + (i + 2) + .. + (n-i-1) + (n-i)\right]$, but I don't know how to proceed from here.</p>
| Dirk | 379,594 | <p>Hint:
$$\sum_{j=i}^{n-i} j = \sum_{j=1}^{n-i} j - \sum_{j=1}^{i-1} j.$$
The right hand side now should look rather familiar (Gauss...).</p>
<p>There is also a rule for
$$\sum_{i=1}^n i^2,$$
if you look this up and prove it or if you already know it, that should give the final result.</p>
|
2,140,965 | <p>Arithmetic progression
:$a_n=\{a,a+d,a+2d,...,a+nd,...\}$</p>
<p>Geometric progression
:$b_n=\{b,bq,bq^2,...,bq^n,...\}$</p>
<p>if : $a_r,a_s,a_t,$ be a Geometric progression($b_n$)</p>
<p>then : What is q?</p>
<p>my way :
let :$r>s>t$</p>
$$ a_{r} =a +(r-1)d $$
$$ a_{s} =a +(s-1)d $$
$$ a_{t} =a +(t-1)d $$
<p>$$ a_{s}^{2} =a_{r} a_{t} \Rightarrow a^{2} +2ad(s-1)+ d^{2}(s-1)^{2} =a^{2}+ad(r+t-2)+d^{2}(r-1)(t-1)$$</p>
$$ 2a(s-1)+ d(s-1)^{2} =a(r+t-2)+d(r-1)(t-1)=$$
$$ 2as-ar-at=2sd-dt-dr-d s^{2} +drt \Rightarrow $$
$$a= \frac{d(2s-t-r-s^{2} +rt)}{2s-r-t} $$
$$ \frac{a_{r}}{a_{t}} = \frac{a +(r-1)d}{a +(t-1)d} $$
$$ \frac{d(2s-t-r-s^{2} +rt)}{2s-r-t}+ (r-1)d= $$
$$d \frac{2s-t-r-s^{2} +rt+2rs- r^{2}-rt-2s+r+t }{2s-r-t} = $$
$$ d \frac{-s^{2} -d \frac{ (r-s)^{2}}{2s-r-t} +2rs- r^{2}}{2s-r-t}=-d \frac{ (r-s)^{2}}{2s-r-t} $$
<p>$$ -d \frac{ (s-t)^{2}}{2s-r-t} $$</p>
$$ q^{2}=\frac{a_{r}}{a_{t}}= \frac{(r-s)^{2}}{(s-t)^{2}} $$
<p>There is simpler method?</p>
| G Cab | 317,234 | <p>Recall what is the application of the dot product between vectors in physics:
$$
\mathbf{v} \cdot \mathbf{w} = \text{scalar} = \left| {proj_\mathbf{w} \mathbf{v}} \right| \cdot \left| \mathbf{w} \right|
$$
then
$$
\left| {proj_\mathbf{w} \mathbf{v}} \right| = \frac{{\mathbf{v} \cdot \mathbf{w}}}
{{\left| \mathbf{w} \right|}}
$$
and
$$
proj_\mathbf{w} \mathbf{v} = \left| {proj_\mathbf{w} \mathbf{v}} \right|\frac{\mathbf{w}}
{{\left| \mathbf{w} \right|}} = \left( {\frac{{\mathbf{v} \cdot \mathbf{w}}}
{{\left| \mathbf{w} \right|^2 }}} \right)\mathbf{w} = \left( {\frac{{\mathbf{v} \cdot \mathbf{w}}}
{{\mathbf{w} \cdot \mathbf{w}}}} \right)\mathbf{w}
$$</p>
|
411,549 | <p>Here is an modular equation</p>
<p>$$5x \equiv 6 \bmod 4$$</p>
<p>And I can solve it, $x = 2$.</p>
<p>But what if each side of the above equation times <strong>8</strong>, which looks like this</p>
<p>$$40x \equiv 48 \bmod 4$$</p>
<p>Apparently now, $x = 0$. Why is that? Am I not solving the modular equation in a right way, or should I divide both side with their greatest-common-divisor before solving it?</p>
<p>P.S.</p>
<p>To clarify, I was solving a system of modular equations, using <strong>Gaussian Elimination</strong>, and after applying the elimination on the coefficient matrix, the last row of the echelon-form matrix is :</p>
<p>$$0, \dots, 40 | 48$$</p>
<p>but I think each row in the echelon-form should have been divided by its greatest common divisor, that turns it into :</p>
<p>$$0, \dots, 5 | 6$$</p>
<p>But apparently they result into different solution, one is $x = 0,1,2,3....$, the other $x = 2$. And why? Am I applying <strong>Gaussian-Elimination</strong> wrong?</p>
| amWhy | 9,003 | <p>Note that in multiplying each side of the original congruence by $8$, you multiplied the congruence by a <em>multiple of the modulus</em>, $4$, hence, since $4\mid 8$, both sides are thereby divisible by $4$, i.e., each side of the congruence is then a multiple of the modulus, and so congruent, by definition, to $0$. So the second congruence equation has an entirely different solution set.</p>
<p>If you had multiplied both sides of the original congruence by, say $3$, you would have maintained the solution set of $x$, $\text{mod}\;4$</p>
|
4,358,080 | <p>I'm making a pump and I need to make a circular (from the front perspective) hole in a side of a pipe. I can't use a drill and I have to print out a shape that I will stick onto it and cut and file away. Will this circle projected off center onto a cylinder be an ellipse, or is it not an exact ellipse and I have to use a different shape?</p>
<p>To further clarify, it will look something like this picture if you imagine the orange pipe is a boring bit and is not reaching beyond the centerline of the green pipe.
<img src="https://i.stack.imgur.com/eVA8G.jpg" alt="enter image description here" /></p>
| sirous | 346,566 | <p><a href="https://i.stack.imgur.com/qB7xV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qB7xV.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/7gys0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7gys0.jpg" alt="enter image description here" /></a></p>
<p>Hints: Here is three figures that may help you:</p>
<p>Flattened I:The of inserting pipe r is equal to radius of large pipe R and the axis es of two pipes are co-planer.</p>
<p>Flattened II: r<R and axis es of two pipes are co-planer.</p>
<p>Flatten III: r<R and surface of insert pipe and large pipe are co-planer(the axis es of pipes are not co-planer).</p>
<p>The radius of cut of inserting pipe is equal to R in all cases.</p>
|
198,105 | <p>I have runtimes for requests on a webserver. Sometimes events occur that cause the runtimes to skyrocket (we've all seen the occasionaly slow web page before). Sometimes, they plummet, due to terminated connections and other events. I am trying to come up with a consistent method to throw away spurious events so that I can evaluate performance more consistently.</p>
<p>I am trying Chauvenet's Criterion, and I am finding that, in some cases, it claims that all of my data points are outliers. How can this be? Take the following numbers for instance:</p>
<pre><code>[30.0, 38.0, 40.0, 43.0, 45.0, 48.0, 48.0, 51.0, 60.0, 62.0, 69.0, 74.0, 78.0, 80.0, 83.0, 84.0, 86.0, 86.0, 86.0, 87.0, 92.0, 101.0, 103.0, 108.0, 108.0, 109.0, 113.0, 113.0, 114.0, 119.0, 123.0, 127.0, 128.0, 130.0, 131.0, 133.0, 138.0, 139.0, 140.0, 148.0, 149.0, 150.0, 150.0, 164.0, 171.0, 177.0, 180.0, 182.0, 191.0, 200.0, 204.0, 205.0, 208.0, 210.0, 227.0, 238.0, 244.0, 249.0, 279.0, 360.0, 378.0, 394.0, 403.0, 489.0, 532.0, 533.0, 545.0, 569.0, 589.0, 761.0, 794.0, 1014.0, 1393.0]
</code></pre>
<p><code>73</code> values. A mean of <code>222.29</code>, and a standard deviation of <code>236.87</code>. Chauvenet's criterion for the value <code>227</code> would have me calculate the probability according to a normal distribution (<code>0.001684</code> if my math is correct). That number times <code>73</code> is <code>.123</code>, less than <code>.5</code> and thus an outlier. What am I doing wrong here? Is there a better approach that I should be taking?</p>
| Henry | 6,460 | <p>Your data is not close to being normally distributed even if you exclude outliers so the whole procedure is rather dubious. </p>
<p>That being said, if you follow the <a href="http://en.wikipedia.org/wiki/Chauvenet%27s_criterion">procedure set out in Wikipedia</a>, then you would reject the two most extreme points. </p>
<p>For example $1014$ is $(1014-222.29)/238.51 \approx 3.34$ standard deviations above the mean. </p>
<p>The probability of being $3.34$ or more standard deviations above or below the mean on a normal distribution is about $2(1-\Phi(|3.34|)) \approx 0.001$ and multiplying this by the number of data points $73$ give a figure less than $0.5$, which is apparently the criterion. </p>
<p>The data point $1393$ is even more extreme, while the datapoint $794$ gives a result above $0.5$. </p>
|
1,950,466 | <p>Let $x_1, x_2, ...,x_n \in \mathbb{H}$, $\mathbb{H}$ is the Hilbert space. $x_j \neq 0$ for every $j$. If $x_i \perp x_j$ for $i \neq j$ then show that $x_j$ 's are linearly independent. </p>
<p>I think I need to use the fact that $<\alpha x + \beta y , z> = \alpha<x,z> + \beta<y,z>$ but how?</p>
| Community | -1 | <p>Always try a simpler case first when you are lost. Suppose $n=2$ and write $ax_1+bx_2=0$. Try to do the inner products with $x_1$ and $x_2$ on both sides. Then you see $a=b=0$. Can you generalize the argument?</p>
|
789,802 | <p>How do I solve $\displaystyle \lim_{x \to 0} \sqrt{x^2 + x^3} \sin \frac{\pi}{x}$ using squeeze Theorem?</p>
<p>My book only teaches me the simplest use of the Theorem. I have no idea what should I do with a function as complex as this...</p>
<p>I know I have to start with:</p>
<p>$$ -1 \le \sin \frac{\pi}{x} \le 1$$</p>
<p>But what do I do next?</p>
| Adrian Keister | 30,813 | <p>Then you could say
$$-\sqrt{x^{2}+x^{3}}\le \sqrt{x^{2}+x^{3}} \, \sin\left(\frac{\pi}{x}\right)\le \sqrt{x^{2}+x^{3}}.$$
The outermost functions approach the desired limit, and so by the Squeeze Theorem, you get the desired result.</p>
|
2,721,372 | <p>A sequence is defined by $a_1=2$ and $a_n=3a_{n-1}+1 $ .Find the sum $a_1+a_2+\cdots+a_n$</p>
<p>how to find sum $a_1=2,a_2=7,\ldots$</p>
<p>Also i found the value of $a_n=\frac{5}{6}\cdot3^n-\frac{1}{2}$</p>
| fleablood | 280,126 | <p>1)</p>
<p>$A\subset B$ if all elements of $A$ are elements of $B$. As $\emptyset$ has no elements, then all of them are in $B$.</p>
<p>That's vacuously true.</p>
<p>So $\emptyset \subset B$.</p>
<p>2)</p>
<p>$A\subset B$ if any elements not in $B$ are not in $A$ either. As any element that is not in $B$ is not in $\emptyset$ either, $\emptyset \subset B$.</p>
<p>That's true-true; nothing vacuous about it.</p>
<p>3)</p>
<p>$A \subset B$ if $x \in A \implies x \in B$ is true for all $x$. As $x \in \emptyset$ is always false and $FALSE \implies P$ is always true, $x \in \emptyset \implies x \in B$ is always true.</p>
<p>So $\emptyset \subset B$.</p>
<p>===</p>
<p>So for the most part, yes, they are vacuously true statements, or they are a false premise implies anything true statements.</p>
<p>But it's not all smoke and mirrors. A subset is "embedded" in the superset and everything you can pull out of the subset most come directly from the superset, and there is nothing in the subset that isn't in the superset.</p>
<p>All of those are true about an empty set and a set $B$ and, to me at least, the all feel directly true with no semantic slick word play or gimmicks. The empty skein of the the emptyset (with nothing in it) is embedded every where in the ether of existent space. That <em>doesn't</em> seem to me to be a "trick". And because nothing can be pulled out of the emptyset, if we are standing in the general vacinity of $B$ nothing can be pulled out of the empty set that isn't from $B$. That's a direct objective fact.</p>
|
2,317,406 | <p>How many two-digit positive integers $N$ have the property that the sum of $N$ and the number obtained by reversing the order of the digits of $N$ is a perfect square?
Any help would be much appreciated.</p>
| M. Winter | 415,941 | <p>I wrote a program and it gave me this:</p>
<p>$0 + 0 = 0$ (if you consider $00$ as a two digit number)</p>
<p>$92 + 29 = 121$</p>
<p>$83 + 38 = 121$</p>
<p>$74 + 47 = 121$</p>
<p>$65 + 56= 121$</p>
|
198,298 | <p>Is there a tool for editing the spelling dictionary? Now that <em>Mathematica</em> 12 does spellchecking on the fly, I've been using it. And of course, as I was wondering how I might delete a word accidentally added to the dictionary, I accidentally added a misspelled word to the dictionary. I'm sure I can hunt it down and edit it by hand, but it seems safer to use a proper tool if one exists. Is there one? </p>
| ciao | 11,467 | <p>For removing a misspelled word from the correct word list, go to Edit->Preferences, select the Advanced tab, select Open Option Inspector, lookup/search for SpellingDictionaries, click on the edit button for the CorrectWords entry, highlight the incorrect entry and select remove.</p>
<p><a href="https://i.stack.imgur.com/2n4ML.png" rel="noreferrer"><img src="https://i.stack.imgur.com/2n4ML.png" alt="enter image description here"></a></p>
|
4,473,264 | <p>I have part of a circle described by three two dimensional vectors.</p>
<ul>
<li>start point <code>s1</code></li>
<li>center point <code>c1</code></li>
<li>end point <code>e</code></li>
</ul>
<p>I move the start point <code>s1</code> by <code>m1</code>, which is a <strong>known</strong> two dimensional vector. My question is: Can I calculate the new center point <code>c2</code> from the data I have? And if so, how?</p>
<p>Problem</p>
<p><a href="https://i.stack.imgur.com/3wWOr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3wWOr.jpg" alt="enter image description here" /></a></p>
<p>I'm creating a svg-manuipulation-app (drawing-app) in javascript where I want to edit one point of an arc, but keep the shape of the arc intact by appropriately moving the center of the arc.</p>
<p>It only looks like I want to keep the <code>x</code> value the same. Small coincidence I didn't realised. The question should cover any vector <code>m1</code>, no matter where the new center <code>c2</code> would end up.</p>
| prets | 12,880 | <p>The <strong>indefinite</strong> integral <span class="math-container">$\displaystyle \int f(x) \, d x$</span> represents the collection of all functions whose derivatives with respect to <span class="math-container">$x$</span> equals <span class="math-container">$f(x)$</span>---they're antiderivatives.
This is in contrast with the <strong>definite</strong> integral <span class="math-container">$\displaystyle \int_a^b f(x) \, d x$</span> which represents the area under the curve <span class="math-container">$y = f(x)$</span> from <span class="math-container">$x = a$</span> to <span class="math-container">$x = b$</span>.</p>
<p>These objects are related by the <a href="https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus" rel="noreferrer">Fundamental theorem of calculus</a> which says (in one of its many forms) that the area in the second expression can be computed by evaluating any one of the antiderivatives at the endpoints and subtracting.</p>
<p>Note that it does <em>not</em> say that using antiderivatives is the only way to compute the area under the curve, it's just a very convenient way in many situations. There are many situations in which it is impossible to write down a nice expression for the indefinite integral (e.g. in terms of elementary functions) but nevertheless possible to compute the area just fine.
One way of doing this is to exploit the geometry of the shape whose area we are after.</p>
<p>A classic example (mentioned in comments) is the <a href="https://en.wikipedia.org/wiki/Gaussian_integral" rel="noreferrer">Gaussian integral</a>
<span class="math-container">$$ \int_{-\infty}^\infty e^{-x^2} \, d x = \sqrt{\pi}. $$</span>
There are also examples where it is perfectly possible to find a nice antiderivative, but it may be difficult, or at least more difficult than some other approach.
For instance,
<span class="math-container">$$ \int_{-1}^1 \sqrt{1 - x^2} \, d x = \frac{\pi}{2}. $$</span>
We could compute this by finding an antiderivative (which is doable but tricky if we're not familiar with trigonometric substitutions), but it is (probably) easier to notice that the shape we are after is really a semicircle. (You may argue that this is circular, depending on how you derived the formula for the area of a circle in the first place.)</p>
<p>As to why some functions don't have elementary antiderivatives: this is a fairly deep question to do with an area of mathematics called <a href="https://en.wikipedia.org/wiki/Differential_algebra" rel="noreferrer">differential algebra</a>. A reasonably approachable exposition I enjoy is Brian Conrad's <a href="http://math.stanford.edu/%7Econrad/papers/elemint.pdf" rel="noreferrer">Impossibility theorems for elementary integration</a>.</p>
<p>Finally there are powerful methods to <strong>approximate</strong> definite integrals---to estimate the area under the curve <span class="math-container">$y = f(x)$</span>---without having a clue whatsoever how one might find an antiderivative. This is an entire area of mathematics in its own right called <a href="https://en.wikipedia.org/wiki/Numerical_integration" rel="noreferrer">numerical integration</a>. Depending on what one is doing, this may be all one needs.</p>
|
24,361 | <p>Let $X$ be a topological space and let $\mathcal{F}$ and $\mathcal{G}$ be two sheaves over $X$.</p>
<p>Of course, if one has a morphism $f : \mathcal{F} \to \mathcal{G}$ such that for all $x\in X$, $f_x : \mathcal{F}_x \to \mathcal{G}_x$ is an isomorphism, then it is known that $f$ itself is an isomorphism.</p>
<p>My question is the following: if we don't have such a morphism $f$, but if we know that for all $x\in X$, $\mathcal{F}_x$ and $\mathcal{G}_x$ are isomorphic, is it true that $\mathcal{F}$ and $\mathcal{G}$ are isomorphic ?</p>
| Harry Gindi | 1,353 | <p>A conceptual explanation (since the other answers have given counterexamples) for why this is untrue is as follows: A sheaf consists of local data and global data specifying how those local data fit together. Even if all of the local data of two sheaves are isomorphic, there is no reason to believe that those isomorphisms can be fit together in a compatible way. This is why we require that the isomorphisms on stalks arise from a map that is already a morphism of sheaves, since this exactly says that the data fit together in the proper way.</p>
<p>We even encounter the same problems when we work with presheaves, for instance. Since presheaves are functors, a morphism of presheaves must be a natural transformation of functors. However, simply having isomorphisms "pointwise", as it were, is not enough. The isomorphisms must also commute with the restriction maps.</p>
|
1,870,305 | <p>I am trying to calculate the Jacobian of a function that has quaternions and 3D points in it. I refer to quaternions as <span class="math-container">$q$</span> and 3D points as <span class="math-container">$p$</span>
<span class="math-container">$$h_1(q)=A C(q)p $$</span>
<span class="math-container">$$h_2(q)=q_1\otimes q \otimes q_2 $$</span></p>
<p>where <span class="math-container">$A\in R^{3x3}$</span> and <span class="math-container">$C(q)$</span> is <em>Direction cosine matrix</em>.</p>
<p>I am using the Hamilton form for the quaternions.</p>
<p>I would like to calculate the following Jacobians:
<span class="math-container">$$H_1 = \frac{\partial h_1(q)}{\partial q} $$</span>
<span class="math-container">$$H_2 = \frac{\partial h_2(q)}{\partial q} $$</span></p>
<p>Following <a href="http://www.iri.upc.edu/people/jsola/JoanSola/objectes/notes/kinematics.pdf" rel="nofollow noreferrer">Joan Solà's reference</a> eq. 18 what I have is </p>
<p><span class="math-container">$$H_1 = A^TC(q)^T[p]_x $$</span>
<span class="math-container">$$H_2 = [q_1]_L[q_2]_R $$</span></p>
<p>Where <span class="math-container">$[q]_R$</span> and <span class="math-container">$[q]_L$</span> are the right and left handed conversion of quaternion to matrix form as defined in <a href="http://www.iri.upc.edu/people/jsola/JoanSola/objectes/notes/kinematics.pdf" rel="nofollow noreferrer">Joan Solà's reference</a> eq. 18.</p>
<p>All rotations are body centric.</p>
<p>Is this correct?
Is there a better way to do this?
Can the expression be easily simplified?</p>
| Yonatan Simson | 351,327 | <p>I needed to solve an easier problem then what I posted.</p>
<p>If $\mathbf{a}=(0,a_x,a_y,a_z)^T$ and the differential is by
$\bar{\mathbf{a}}=(a_x,a_y,a_z)^T$ the answer simply becomes:</p>
<p>$$ \frac{\partial(\mathbf{q} \otimes \mathbf{a} \otimes \mathbf{q\ast})}{\partial \bar{\mathbf{a}}} =
\frac{\partial(\mathbf{R} \bar{\mathbf{a}})}{\partial \bar{\mathbf{a}}} = \mathbf{R}$$</p>
<p>Where $\mathbf{R}$ is the rotation matrix derived from $\mathbf{q}$</p>
|
1,217,771 | <p>Let $x,y \in R$.
If $0 \leq y < x$ for all $x > 0$, then $y=0$.</p>
<p>Proof by contradiction: </p>
<p>Assume the opposite that is; "If $0 \leq y < x$ for all $x > 0$, then $y\neq0$".
Subtract $x$ from each part of the inequality to get,
$0-x \leq y-x < 0$
Then multiply through by -1 to get,
$x \geq y+ x > 0$
Since $x>0$, this implies a contradiction of the original statement, therefore we conclude that if $0 < y < x$ for all $x > 0$, then $y=0$.
Is my reasoning correct or is there something I can improve upon?</p>
| Marco Cantarini | 171,547 | <p>We have $$\frac{1}{2}=\lim_{x\rightarrow\infty}\frac{\sqrt{x}}{2\sqrt{x}\left(\sqrt{\left(1+\frac{1}{\sqrt{x}}\right)}\right)}\leq\lim_{x\rightarrow\infty}\frac{\sqrt{x}}{\sqrt{x\left(1+\frac{1}{\sqrt{x}}\right)}+\sqrt{x}}\leq\lim_{x\rightarrow\infty}\frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\leq\lim_{x\rightarrow\infty}\frac{1+\sqrt{x}}{2\sqrt{x}}=\frac{1}{2}$$</p>
|
138,866 | <p>I have data in a csv file. The first row has labels, and the first column, too.</p>
<pre><code>Datos = Import["C:\\Users\\jodom\\Desktop\\Data.csv"]
</code></pre>
<p>Tha data in the csv file is that:</p>
<pre><code>{{"No", "Vol", "Vel"}, {1, 500, 45}, {2, 700, 67}, {3, 350, 87}, {4,
123, 23}, {5, 587, 45}, {6, 435, 89}, {7, 896, 65}, {8, 125,
45}, {9, 476, 27}, {10, 987, 80}}
</code></pre>
<p>I put those csv data into a dataset:</p>
<pre><code>B = Dataset[Datos]
</code></pre>
<p>You can check it out as an image here,on how it has seen on wolfram after the import:
<a href="https://drive.google.com/file/d/0B56r_V66BiodQUhUMWNHcHZFOWc/view?usp=sharing" rel="noreferrer">https://drive.google.com/file/d/0B56r_V66BiodQUhUMWNHcHZFOWc/view?usp=sharing</a></p>
<p>Now I want to convert the first row that has the labels, into a head or label of the dataset, and the first column into a label column, so I can get data from this dataset, like </p>
<pre><code>Dataset[labelrow, labelcolumn]
</code></pre>
| GenericAccountName | 38,159 | <p>As of 11.2, I would suggest doing either:</p>
<p><code>Import["test.csv", "Dataset", "HeaderLines" -> 1]</code></p>
<p>or</p>
<p><code>Import["test.csv", "Dataset", "HeaderLines" -> {1, 1}]</code></p>
<p><a href="https://i.stack.imgur.com/FkENo.png" rel="noreferrer"><img src="https://i.stack.imgur.com/FkENo.png" alt="enter image description here"></a></p>
|
150,084 | <p>I need to say whether or not $f_n(x)=n\left(\sqrt{x+\frac{1}{n}}-\sqrt{x}\right)$ is uniformly convergent on $(0,\infty)$.</p>
<p>I've found that the function is locally convergent to $f(x)=\frac{1}{2\sqrt{x}}$ and was trying to find $\sup{|f_n(x)-f(x)|}$.</p>
<p>I got the derivative $f_n'(x)= \frac{2nx\left(x-\sqrt x\sqrt{x+\frac{1}{n}}\right)+\sqrt{x}\sqrt{x+\frac{1}{n}}}{...}$ and could not find $x$ so that $f_n'(x)=0$</p>
<p>Any ideas?</p>
| Mark McClure | 21,361 | <p>In fact, $f_n'(x)<0$ for all $x>0$. Thus, each $f_n$ is continuous, positive, and decreasing on $[0,\infty)$. It follows that
$$\sup\{f(x):x>0\} = f(0) = \sqrt{n}.$$</p>
<p>As $f$ is unbounded, you can't have uniform convergence.</p>
|
4,216,243 | <p>I have to solve the following problem which seems difficult:</p>
<p>Find
<span class="math-container">$$ \iint_S \nabla \times F\ dS $$</span>
where <span class="math-container">$S$</span> is given by</p>
<p><span class="math-container">$$r(t,s)=\left( 9+(\cos t)(\sin s)\left(2+\frac{\sin (5s)}{2}\right), \ \ \ 9+(\cos t)(\cos s)\left(2+\frac{\sin (5s)}{2}\right), \ \ \ 9+\frac{\sin t}{3}\left(2+\frac{\sin (5s)}{2}\right) \right) $$</span></p>
<p>where <span class="math-container">$0\leq t\leq 2\pi$</span>, <span class="math-container">$\ \ $</span> <span class="math-container">$0\leq s\leq \pi$</span>, <span class="math-container">$\ \ \ $</span> and <span class="math-container">$F:=(z,0,y)$</span>.</p>
<p>I'm not sure how to proceed, any help is appreciated.</p>
<p>Should I use Gauss divergence?</p>
<p>When I plotted <span class="math-container">$S$</span> in wolfram (not sure why is different from the answer below)
<a href="https://www.wolframalpha.com/input/?i=%289%2B%28cos+t%29%28sin+s%29%282%2Bsin+%285s%29%2F%282%29%29%2C++9%2B%28cos+t%29%28cos+s%29%282%2Bsin+%285s%29%2F2%29%2C+9%2B%28%28sin+t%29%2F3%29%282%2Bsin+%285s%29%2F2%29+%29%2C+0%3C%3D+s%3C%3D+pi%2C+0%3C%3Dt%3C%3D+2pi" rel="nofollow noreferrer">https://www.wolframalpha.com/input/?i=%289%2B%28cos+t%29%28sin+s%29%282%2Bsin+%285s%29%2F%282%29%29%2C++9%2B%28cos+t%29%28cos+s%29%282%2Bsin+%285s%29%2F2%29%2C+9%2B%28%28sin+t%29%2F3%29%282%2Bsin+%285s%29%2F2%29+%29%2C+0%3C%3D+s%3C%3D+pi%2C+0%3C%3Dt%3C%3D+2pi</a></p>
<p><a href="https://i.stack.imgur.com/XZFNJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XZFNJ.png" alt="enter image description here" /></a></p>
| rebo79 | 814,741 | <p>As far as I can see the surface <span class="math-container">$S$</span> encloses a solid volume (I plotted it in mathematica)
<a href="https://i.stack.imgur.com/X3vLc.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/X3vLc.jpg" alt="enter image description here" /></a>
Then if <span class="math-container">$V$</span> denotes the solid enclosed by <span class="math-container">$S$</span>, you can apply Gauss's theorem to obtain
<span class="math-container">\begin{align*}
\iint_S (\nabla \times F)·dS &= \iiint_V \nabla·(\nabla \times F) dV = 0
\end{align*}</span>
since the divergence of the curl is zero.</p>
<p><strong>PD:</strong> I can't figure it out how to prove analytically from the parametrization that <span class="math-container">$S$</span> is indeed closed. I'll keep thinking about it.</p>
|
509,856 | <p>Let $K_n$ be a complete $n$ graph with a color set $c$ with $c=\{\text{Red}, \text{Blue}\}$. Every edge of the complete $n$ graph is colored either $\text{Red}$ or $\text{Blue}$. Since $R(3, 3)=6$, the $K_6$ graph must contain at least one monochromatic $K_3$ graph. How can I prove that this graph must contain another (different) monochromatic $K_3$ graph. I saw proofs which uses the fact that there are at most $18$ non-monochromatic $K_3$ graphs. Since there are $20$ $K_3$ graphs (how can you calculate this) there are at least 2 monochromatic $K_3$ graphs. Are there other proofs?</p>
| StatisticalMechanic | 514,930 | <p>There is another counting argument that one can come up with. A possible quantity of interest is a $monochromatic$ $angle$. An angle is monochromatic if both its arms are of the same colour.</p>
<p>Using pigeonhole principle, we can see that, at each vertex of $K_6$, there are at least three edges of the same colour (and hence at least 4 monochromatic angles). This leads to the conclusion that there are at least $24$ $(=6$x$4)$ monochromatic angles in $K_6$. Also, we can see that every monochromatic triangle has $3$ monochromatic angles and every other triangle has exactly $1$ monochromatic angle.</p>
<p>We know that there are a total of $20$ triangles in $K_6$. Let there be $x$ monochromatic triangles. Then there must be $(20−x)$ non-monochromatic triangles.</p>
<p>Putting all of this together, we get:</p>
<p>$3x+(20−x)$$≥24$ which gives $x≥2$.</p>
<p>I really like this argument and feel that the idea of monochromatic angles is very understudied. There are more things one can prove using this quantity and I encourage you to do so :) </p>
|
4,096,516 | <p>Calculate <span class="math-container">$\operatorname{tg}( \alpha), $</span> if <span class="math-container">$\frac{\pi}{2} < \alpha<\pi$</span> and <span class="math-container">$\sin( \alpha)= \frac{2\sqrt{29}}{29}$</span>. Please provide a hint.<br />
I know that <span class="math-container">$\operatorname{tg}( \alpha)=\frac{\sin( \alpha)}{\cos( \alpha)}$</span> and <span class="math-container">$\sin^2( \alpha)+\cos^2( \alpha)=1$</span>, but still can't get the answer from there.</p>
| José Carlos Santos | 446,262 | <p>Since <span class="math-container">$\frac\pi2<\alpha<\pi$</span>,<span class="math-container">$$\cos(\alpha)=-\sqrt{1-\sin^2(\alpha)}=-\frac5{\sqrt{29}},$$</span>and therefore<span class="math-container">$$\tan(\alpha)=\frac{\sin(\alpha)}{\cos(\alpha)}=-\frac25.$$</span></p>
|
2,460,195 | <p>I had the following question:</p>
<p>Three actors are to be chosen out of five — Jack, Steve, Elad, Suzy, and Ali. What is the probability that Jack and Steve would be chosen, but Suzy would be left out?</p>
<p>The answer given was:
Total Number of actors = $5$;
Since Jack and Steve need to be in the selection and Suzy is to be left out, only one selection matters.
Number of actors apart from Jack, Steve, and Suzy = $2$;
Probability of choosing 3 actors including Jack and Steve, but not Suzy = $$\frac{C(2,1)}{C(3,5)} = \frac{1}{5}$$</p>
<p>I do not understand the answer. What do they mean by only one selection matters? It looks like they are choosing $1$ person from $2$ combinations? Why? Can anyone please explain this.</p>
<p>Thanks</p>
| Robert Z | 299,698 | <p>More generally, the equation $y+z=2n-2x$ has $2n-2x+1$ non negative solutions for all $x=0,1,\dots, n$. Therefore the number of non negative solutions is
$\sum_{x=0}^n(2n+1-2x)$, that is the sum of all odd numbers in the interval $[1,2n+1]$ which is equal to $(n+1)^2$. </p>
|
677,708 | <p>Haven't done this for a long time, just want to know if this is the right method for a really simple example. Say we have two (obviously equal) sets $$A= \Big\{\begin{bmatrix}a & b\\c & d\end{bmatrix} : a,b,c,d \in \Bbb R, a+b=c \Big\}$$ $$B= \Big\{\begin{bmatrix}e & f\\g & h\end{bmatrix} : e,f,g,h \in \Bbb R, e+f=g \Big\}$$</p>
<p>Prove $A\subseteq B$.</p>
<p>Let $a\in A$, such that $a=\begin{bmatrix}a & b\\c & d\end{bmatrix}$. Choose $a=e$, $b=f$, $c=g$ and $d=h$. Then $a=\begin{bmatrix}e & f\\g & h\end{bmatrix}$ where $e+f=g$, hence $a\in B$ and $A\subseteq B$. </p>
<p>Thanks! </p>
<p>Edit: I chose an obviously equal example on purpose. It was a question about the method of proving rather than the actual example. </p>
| ricardio | 124,869 | <p>Yes that works. In general, you take something from $A$, and show it has whatever property necessary to belong to $B$. </p>
|
4,116,732 | <p>If we have a topological category and the underlying category forgetting the topological structure, are the nerves same. They should be, is what my guess is from the definition of nerve of a category. Then, I have some facts which leads to contradiction (which should not be, thus I am missing something). The facts are below:</p>
<p>(For a category C, classifying space of C is the realization of the nerve of C.)</p>
<ol>
<li><p>If G is a group consider it a category with the only object is * and set of morphisms is G. Then the nerve of this category has realization K(G,1).</p>
</li>
<li><p>If G is a topological group, then the classifying space (for Principal G-bundle) is not always (weakly-)equivalent to K(G,1). For example, take unit circle. They are same (upto weak-equivalence) when G has discrete topology.</p>
</li>
<li><p>In the paper of Graeme Segal (<a href="https://www.maths.ed.ac.uk/%7Ev1ranick/papers/segalclass.pdf" rel="nofollow noreferrer">https://www.maths.ed.ac.uk/~v1ranick/papers/segalclass.pdf</a>) his model of classifying space (in the sense of principal G-bundles) is in terms of realization of the nerve if the category in 1 (assuming say G is a locally finite CW-complex).</p>
<p>Now, any two classifying spaces (in the sense of Principal G-bundle) are weakly equivalent, then 2 gives contradiction to 1 and 3 . I know I am mistaken at some stage. Any help is welcome.</p>
</li>
</ol>
| Zhen Lin | 5,191 | <p>I am sure that Segal does not mean them to be the same.
In the cited paper, notice that Segal deliberately mentions "semi-simplicial spaces" (in modern terminology: "simplicial spaces") and their realisations.
From that one infers that the nerve of a topological category is a simplicial space.
By contrast, the nerve of an ordinary category is a merely a simplicial set, which one may regard as a simplicial space that is degreewise discrete.</p>
|
383,735 | <p>Given this (very) tricky determinant, how can we calculate it easily?</p>
<p>$$\begin{pmatrix} \alpha + \beta & \alpha \beta & 0 & ... & ... & 0 \\ 1 & \alpha + \beta & \alpha \beta & 0 & ... & 0 \\ 0 & 1 & \alpha + \beta & \alpha \beta & ... & ... \\ ... & ... & ... & ... & ... & 0 \\ ... & ... & .... & ... & ... & \alpha \beta \\ 0 & 0 & 0 & ... & 1 & \alpha + \beta \\ \end{pmatrix} \in M_{n\times n}$$</p>
<p>EDIT:</p>
<blockquote>
<p>I have to prove it is equal to $\frac{{\alpha}^{n+1} - {\beta}^{n+1}}{\alpha - \beta}$</p>
</blockquote>
<p>Any help is appreciated, I just could not find a trick to ease it up!</p>
| Riccardo | 74,013 | <p>You can consider factorizing the matrix using L-U factorization.
Let A be your matrix, $L$ the matrix $$\begin{pmatrix}
p_1 & 0 & 0 & ... & ... & 0 \\
1 & p_2 & 0 & 0 & ... & 0 \\
0 & 1 & p_3 & 0 & ... & ... \\
... & ... & ... & ... & ... & 0 \\
... & ... & .... & 1 & p_{n-1} & 0 \\
0 & 0 & 0 & ... & 1 & p_n \\ \end{pmatrix}$$</p>
<p>and U the matrix</p>
<p>$$\begin{pmatrix}
1 & q_1 & 0 & ... & ... & 0 \\
0 & 1 & q_2 & 0 & ... & 0 \\
0 & 0 & 1 & q_3 & ... & ... \\
... & ... & ... & ... & ... & 0 \\
... & ... & .... & 0 & 1 & q_{n-1} \\
0 & 0 & 0 & ... & 0 & 1\\ \end{pmatrix}$$</p>
<p>where </p>
<p>$p_{11} = a_{11}; \\ q_1= a_{12}/p_1; \\ p_2= a_{22} - a_{21}q_1; \\ q_2 = a_{23} / p_2; \\ p_3= a_{33}-a_{32}q_2; \\ \dots \\ \dots \\ q_{n-1}= a_{n-1,n}/p_{n-1}; \\ p_n = a_{nn} - a_{n,n-1}q_{n-1};$</p>
<p>then by Binet formula, you can note that $\det A = \det L \det U= \det L = \prod_{i=1}^n p_i$ </p>
<p>(used the fact that $L,U$ are upper/lower triangular)</p>
<p>then you can use induction.
hope it helps</p>
|
3,938,951 | <p>The problem:</p>
<p>Given a group <span class="math-container">$G$</span> or order <span class="math-container">$n$</span>, and a Cayley embedding <span class="math-container">$\phi \ :\ G\to S_{n}$</span>. Prove that some <span class="math-container">$g\in G$</span> is of order <span class="math-container">$m$</span> iff <span class="math-container">$\phi ( g)$</span> is a multiplication of <span class="math-container">$\frac{n}{m}$</span> disjoint cycles of length <span class="math-container">$m$</span>.</p>
<p>I was able to prove that if <span class="math-container">$\phi ( g) \ $</span> is s a multiplication of <span class="math-container">$\displaystyle \frac{n}{m}$</span> disjoint cycles of order <span class="math-container">$m$</span>, <em>then</em> the order of <span class="math-container">$g$</span> is <span class="math-container">$m$</span>. This was fairly straightforward given the fact that:</p>
<blockquote>
<p>The order of a multiplication of disjoint cycles is the <span class="math-container">$lcm$</span> of the length of the cycles.</p>
</blockquote>
<p>In our case the <span class="math-container">$lcm$</span> is obviously <span class="math-container">$m$</span>, so it was easy to prove from here that the order of <span class="math-container">$g$</span> is <span class="math-container">$m$</span>.</p>
<p>The other direction, meaning if the order of <span class="math-container">$g$</span> is <span class="math-container">$m$</span>, then <span class="math-container">$\phi ( g)$</span> is a multiplication of <span class="math-container">$\frac{n}{m}$</span> disjoint cycles of length <span class="math-container">$m$</span>, I'm struggling to prove it.</p>
<p>I also want to admit that my understanding of Cayley embeddings, and Cayley's theorem in general, is very poor. What I know is simply that it's a homomorphism and also injective, not much more.</p>
<p>Any help?</p>
| Community | -1 | <p>Let's consider the action by left multiplication of <span class="math-container">$\langle g\rangle$</span> on <span class="math-container">$G$</span>. For <span class="math-container">$x\in G$</span>, the stabilizer is given by:</p>
<p><span class="math-container">\begin{alignat}{1}
\operatorname{Stab}(x) &= \{g^k\in\langle g\rangle\mid g^kx=x\} \\
&= \{g^k\in\langle g\rangle\mid g^k=e\} \\
&= \{e\} \\
\tag 1
\end{alignat}</span></p>
<p>Therefore, <span class="math-container">$G$</span> is partitioned into <span class="math-container">$r:=\frac{n}{m}$</span> orbits each of size <span class="math-container">$m$</span> (orbit-stabilizer theorem). This means that there are <span class="math-container">$r$</span> elements <span class="math-container">$\tilde g_i\in G$</span>, <span class="math-container">$i=1,\dots,r$</span>, such that <span class="math-container">$G$</span> reads:</p>
<p><span class="math-container">$$G=\bigsqcup_{i=1}^rO(\tilde g_i) \tag 2$$</span></p>
<p>where:</p>
<p><span class="math-container">$$O(\tilde g_i):=\{g^k\tilde g_i, k=1,\dots,m\} \tag 3$$</span></p>
<p>Let's now take the permutation by left multiplication by <span class="math-container">$g$</span>, <span class="math-container">$\phi_g\in S_G$</span>. By <span class="math-container">$(2)$</span> and <span class="math-container">$(3)$</span>, <span class="math-container">$\forall x\in G, \exists i\in \{1,\dots,r\}, k\in \{1,\dots,m\}$</span> such that <span class="math-container">$x=g^k\tilde g_i$</span>. Therefore, <span class="math-container">$\forall l\in \{1,\dots,m\}$</span>:</p>
<p><span class="math-container">\begin{alignat}{1}
\phi_g^l(x) &\stackrel{(*)}{=}g^lx \\
&=g^{l+k \pmod m}\tilde g_i \\
\end{alignat}</span></p>
<p>[<span class="math-container">$(*)$</span> induction on <span class="math-container">$l$</span>], whence, <span class="math-container">$\forall x\in G$</span>:</p>
<p><span class="math-container">$$x\in O(\tilde g_i) \iff \phi_g^l(x)\in O(\tilde g_i), \space\forall l\in\{1,\dots,m\} \tag 4$$</span></p>
<p>Now, let's define <span class="math-container">$\alpha_i$</span> as the extension by the identity map of the restriction of <span class="math-container">$\phi_g$</span> to the orbit <span class="math-container">$O(\tilde g_i)$</span>, namely:</p>
<p><span class="math-container">\begin{alignat}{2}
&x \in O(\tilde g_i) &&\Longrightarrow \alpha_i(x) :=\phi_g(x) \\
&x \in O(\tilde g_{j\ne i}) &&\Longrightarrow \alpha_i(x) :=x \\
\tag 5
\end{alignat}</span></p>
<p>Firstly, <span class="math-container">$\alpha_i \in S_G$</span> for every <span class="math-container">$i=1,\dots,r$</span>, because <span class="math-container">${\phi_g}_{|O(\tilde g_i)}$</span> is a bijection on <span class="math-container">$O(\tilde g_i)$</span>. Then, by <span class="math-container">$(4)$</span> and <span class="math-container">$(5)$</span>:</p>
<p><span class="math-container">\begin{alignat}{2}
& x \in O(\tilde g_i) &&\Longrightarrow \alpha_i^m(x)=\phi_g^m(x)=x\\
& x \in O(\tilde g_{j\ne i}) &&\Longrightarrow \alpha_i^m(x)=x\\
\tag 6
\end{alignat}</span></p>
<p>and finally, for every <span class="math-container">$i=1,\dots,r$</span>:</p>
<blockquote>
<p><span class="math-container">$$\alpha_i^m=Id_G \tag 7$$</span></p>
</blockquote>
<p>so <strong>all the <span class="math-container">$\alpha_i$</span>'s are <span class="math-container">$m$</span>-cycles</strong>. Moreover, again by <span class="math-container">$(4)$</span> and <span class="math-container">$(5)$</span>, for <span class="math-container">$j\ne i$</span> we get:</p>
<p><span class="math-container">\begin{alignat}{2}
&x \in O(\tilde g_i) &&\Longrightarrow (\alpha_i\alpha_j)(x)=\alpha_i(\alpha_j(x))=\alpha_i(x)=\phi_g(x) \\
&x \in O(\tilde g_j) &&\Longrightarrow (\alpha_i\alpha_j)(x)=\alpha_i(\alpha_j(x))=\alpha_i(\phi_g(x))=\phi_g(x) \\
&x \in O(\tilde g_{l\ne i,j}) &&\Longrightarrow (\alpha_i\alpha_j)(x)=\alpha_i(\alpha_j(x))=\alpha_i(x)=x \\
\tag 8
\end{alignat}</span></p>
<p>or, equivalently:</p>
<p><span class="math-container">\begin{alignat}{2}
&x \in O(\tilde g_i)\sqcup O(\tilde g_j) &&\Longrightarrow (\alpha_i\alpha_j)(x)=\phi_g(x) \\
&x \in O(\tilde g_{l\ne i,j}) &&\Longrightarrow (\alpha_i\alpha_j)(x)=x \\
\tag 9
\end{alignat}</span></p>
<p>By induction on <span class="math-container">$(9)$</span>:</p>
<p><span class="math-container">\begin{alignat}{2}
&x \in O(\tilde g_1)\sqcup\dots\sqcup O(\tilde g_r)=G &&\Longrightarrow (\alpha_1\dots\alpha_r)(x)=\phi_g(x) \\
\tag {10}
\end{alignat}</span></p>
<p>namely:</p>
<blockquote>
<p><span class="math-container">$$\phi_g=\alpha_1\dots\alpha_r \tag {11}$$</span></p>
</blockquote>
<p>Note that <span class="math-container">$\alpha_i(x)\ne x \iff x \in O(\tilde g_i)$</span>, and hence <span class="math-container">$\alpha_i$</span> and <span class="math-container">$\alpha_j$</span> have disjoint supports as soon as <span class="math-container">$i\ne j$</span>. Therefore, <span class="math-container">$(11)$</span> (along with <span class="math-container">$(7)$</span>) reads: <strong><span class="math-container">$\phi_g$</span> is the product (composition) of <span class="math-container">$\frac{n}{m}$</span> disjoint cycles of length <span class="math-container">$m$</span>.</strong></p>
|
183,039 | <p>I'm pretty weak in the field of mathematics, but a strong programmer. I am looking for a mathematical solution that, given two points on a line will give me a curve between them, including those two points within the curve itself.</p>
<p>For instance, if I have a set of points { (0, 3) (1,10) } I'd like a mathematical way to generate points between the two (I believe this is called interpolate) to create a curve that will contain { (0,3) (1,10) }</p>
<p>Will Linear Interpolation give me this? </p>
<p>Thank you</p>
| Matt Calhoun | 1,417 | <p>You probably want to check out <a href="http://en.wikipedia.org/wiki/B%C3%A9zier_curve" rel="nofollow">Bézier Curves</a>. You can find a live demo <a href="http://jsdraw2d.jsfiction.com/demo/curvesbezier.htm" rel="nofollow">here</a>, just be sure to pick 4 points and then click draw bezier. I know Bézier curves are heavily used in computer graphics, mostly due to the fact you can compute them quickly. You need 4 points to define them, but if you keep two of the points fixed, the other two points can be anything, changing them just changes the shape of the curve. By the way the javascript source for jsDraw2d can be found <a href="http://jsdraw2d.jsfiction.com/viewsourcecode.htm" rel="nofollow">here</a>. If that's not enough, <a href="http://processingjs.nihongoresources.com/bezierinfo/" rel="nofollow">this</a> should keep you busy ;)</p>
|
2,965,865 | <blockquote>
<p>Finding the minimum value of <span class="math-container">$\displaystyle \frac{x^2 +y^2}{y}.$</span> where <span class="math-container">$x,y$</span> are
real
numbers satisfying <span class="math-container">$7x^2 + 3xy + 3y^2 = 1$</span></p>
</blockquote>
<p>Try: Equation <span class="math-container">$7x^2+3xy+3y^2=1$</span> represent Ellipse</p>
<p>with center is at origin.</p>
<p>So substitute <span class="math-container">$x=r\cos \alpha $</span> and <span class="math-container">$y=r\sin \alpha$</span> </p>
<p>in <span class="math-container">$7x^2+3xy+3y^2=1$</span></p>
<p><span class="math-container">$$3r^2+4r^2\cos^2 \alpha+3r^2\sin \alpha \cos \alpha =1$$</span></p>
<p><span class="math-container">$$3r^2+2r^2(1+\cos 2 \alpha)+\frac{3r^2}{2}\sin 2 \alpha =1$$</span></p>
<p><span class="math-container">$$8r^2+r^2(4\cos 2 \alpha+3\sin \alpha)=2$$</span></p>
<p>So <span class="math-container">$$r^2=\frac{2}{8+(4\cos 2 \alpha+3\sin \alpha)}$$</span></p>
<p><span class="math-container">$$\frac{2}{8+5}=\frac{2}{13}\leq r^2\leq \frac{2}{8-5}=\frac{2}{3}$$</span></p>
<p>we have to find minimum of <span class="math-container">$$\frac{x^2+y^2}{y}=\frac{r}{\sin \alpha}$$</span></p>
<p>How can i find it, could some help me </p>
| Aleksas Domarkas | 562,074 | <p>Lagrange function is
<span class="math-container">$$L=k\, \left( 3 {{y}^{2}}+3 x y+7 {{x}^{2}}-1\right) +\frac{{{y}^{2}}+{{x}^{2}}}{y}$$</span>
Solve system:
<span class="math-container">$$L'_x=0,\quad L'_y=0,\quad L'_k=0.$$</span></p>
<p>We get two solutions
<span class="math-container">$$k=1/4,y=-2/5,x=-1/5$$</span> <span class="math-container">$$k=-1/4,y=2/5,x=1/5$$</span></p>
<p>Answer:</p>
<p><span class="math-container">$$f_{min}=f\left(-\frac{1}{5},-\frac{2}{5}\right)=-\frac{1}{2}$$</span></p>
<p>with CAS Maxima:</p>
<p><a href="https://i.stack.imgur.com/EFj6j.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EFj6j.png" alt="enter image description here"></a></p>
|
1,953,628 | <p>0 choices for the 1st person.
17 choices for the 2nd person (must exclude 1st and his/her two neighbours) </p>
<p>For 2 of these choices of 2nd person, there is one shared neighbour, so 15 remaining choices. (e.g. if they are numbered 1 to 20 in a circle, 1st person is #1, 2nd is #3, then people 20,1,2,3,4 are excluded).
For the other 15 choices of 2nd person, there are no shared neighbors, so 14 remaining choices. </p>
<p>So if order matters, total is $20 \cdot (2 \cdot 15 + 15 \cdot 14) $
but since order does not matter, divide by $3! = 6$ to account for the permutations in order of the 3 people.
So total = $20 \cdot \frac{2 \cdot 15 + 15 \cdot 14}{6}$</p>
<p>just redid it; does this make any sense?</p>
| Brian M. Scott | 12,042 | <p>There are $20$ ways to choose a block of $3$ consecutive people. There are also $20$ ways to choose a pair of adjacent people, and for each of those pairs there are $16$ ways to choose a third person who is not adjacent to either of them. Those are the only sets of $3$ people that we don’t want. There are altogether $\binom{20}3$ possible sets of $3$ people, so the number of acceptable sets is</p>
<p>$$\binom{20}3-20-20\cdot16=\frac{20\cdot19\cdot18}6-20\cdot17=20(57-17)=800\;.$$</p>
<p><strong>Added:</strong> More generally, suppose that we have $n$ people at the table, and we want to choose $k$ of them, no two of whom are adjacent. First solve the problem when the $n$ people are in a line. Once the $k$ people are chosen, there will be $k+1$ gaps in which the others can be seated: one before the first chosen person, $k-1$ between adjacent chosen people, and one after the last chosen person. We have to fill these gaps with the $n-k$ people whom we didn’t choose. We must be sure to put at least one person in each of the $k-1$ gaps between adjacent chosen people, so we have only $n-k-(k-1)=n-2k+1$ people left to distribute freely amongst the $k+1$ gaps. This can be done in</p>
<p>$$\binom{(n-2k+1)+(k+1)-1}{(k+1)-1}=\binom{n-k+1}k$$</p>
<p>ways, by a standard <a href="https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)" rel="noreferrer">stars and bars</a> calculation.</p>
<p>However, this count includes the arrangements that in which the two people at the ends of the line are chosen, and when we wrap the line back around the table, those two people are adjacent. Thus, we don’t want to count those arrangements. There is one of them for each way of choosing $k-2$ people from a line of $n-2$ people with the requirement that no two be adjacent and that neither end person be chosen. This time we’re distributing $n-k$ unchosen people amongst $k-1$ slots with a requirement that each slot get at least one person, and another stars and bars calculation gives the number of such distributions as</p>
<p>$$\binom{n-k-1}{k-2}\;.$$</p>
<p>The final answer is therefore</p>
<p>$$\binom{n-k+1}k-\binom{n-k-1}{k-2}\;,$$</p>
<p>which in this specific problem is</p>
<p>$$\binom{18}3-\binom{16}1=816-16=800\;.$$</p>
|
2,485,109 | <blockquote>
<p>Proof by induction that $1^2 + 2^2 + 3^2 +.......+ n^2 = \frac{1}{6}\cdot n\cdot (1+n)\cdot (1+2n)$</p>
</blockquote>
<p>I tried showing that </p>
<p>$1^2 + 2^2 + 3^2 +.......+ (k+1)^2 = \frac{1}{6}\cdot (k+1)\cdot (1+(k+1))\cdot (1+2(k+1))$</p>
<p>By using the left side:</p>
<p>$1^2 + 2^2 + 3^2 +.......+ (k+1)^2$ </p>
<p>$= 1^2 + 2^2 + 3^2 +.......+ k^2 +(k+1)^2 $</p>
<p>$= \frac{1}{6}\cdot k\cdot (1+k)\cdot (1+2k) + (k+1)^2$</p>
<p>I tried expanding and making it equal the right side but I was not able to get it.</p>
| Bruce Ikenaga | 62,033 | <p>Since $5 \cdot 3 + 7 \cdot (-2) = 1$, it follows that $5 \cdot 3702 + 7 \cdot (-2468) = 1234$. Thus, $x = 3702$, $y = -2468$ is a particular solution to the equation. The theory of linear Diophantine equations gives the general solution $x = 3702 + 7 t$ and $y = -2468 - 5 t$.</p>
<p>Since we want positive solutions, $3702 + 7 t > 0$ and $-2468 - 5 t> 0$. You can check that the integers satisfying these inequalities are $t = -528, -527, \ldots -494$. Thus, there are $35$ positive integer solutions.</p>
|
1,730,445 | <p>I am checking whether the limit is true or not. $z$ is complex number
\begin{equation}
\lim_{z \rightarrow 0} z\sin(\frac{1}{z})=0
\end{equation}
I found Laurent series of $z\sin(\frac{1}{z})$ which is $\sum_{n=0}^{\infty} (\frac{1}{z})^{2n}\frac{1}{(2n+1)!}(-1)^n$ </p>
<p>the series is not defined $z =0$.</p>
<p>Can the Laurent series gives us any information about the limit whether it is true or not ? Thank you in advance for your help.</p>
| Arthur | 15,500 | <p>It might be more intuitive to do the substitution $z=1/w$ to get
$$
\lim_{|w|\to\infty}\frac{\sin w}{w}
$$
which clearly goes to $0$ along the real axis. But along the imaginary axis, the sine function grows exponentially, which outweighs the division by $w$ to give an infinite limit. Therefore the limit is not defined.</p>
|
3,709,331 | <p>How can I integrate
<span class="math-container">$$
\int \frac{x\,dx}{(a-bx^2)^2}
$$</span> I've tried to use partial fraction decomposition, but I'm getting six equations for four variables, and they don't give uniform answers.</p>
| Harish Chandra Rajpoot | 210,295 | <p><span class="math-container">$$\int \frac{x\,dx}{(a-bx^2)^2}=-\frac{1}{2b}\int \frac{(-2bx)\,dx}{(a-bx^2)^2}=-\frac1{2b}\int \frac{d(a-bx^2)}{(a-bx^2)^2}=\frac{1}{2b(a-bx^2)}+c$$</span> </p>
|
1,193,472 | <p>Let $ X \subseteq \mathbb{R}^n$ be compact and $ f:X \to \mathbb{R} $ be a function such that for all $ t \in \mathbb{R} $, the set $ f^{-1} [t, \infty) $ is closed. Which one of the following statement is correct and why the rest are incorrect? </p>
<p>1) There exists a $ x_0 \in X $ such that $$ f(x_0)= \sup_{x\in X} f(x)<\infty.$$</p>
<p>2) $ f$ is bounded. </p>
<p>3) There exists $ y_0 \in X $ such that $$ f(y_0)= \inf_{x\in X} f(x)> -\infty.$$
4) $f$ can be unbounded. </p>
| Alex Zorn | 73,104 | <p>There is a nice theorem that will help you here: Suppose $X$ is compact, and let $V_0 \supset V_1 \supset V_2 \supset \cdots$ be a decreasing collection of closed subsets such that $\cap_i V_i = \emptyset$. Then there exists some $k$ such that $V_k = \emptyset$. Now lets prove some stuff.</p>
<p><strong>I: $f$ is bounded above.</strong> Let $V_n = f^{-1}([n,\infty))$, $n \in \mathbb{N}$. Then $V_n$ is closed and we have $V_0 \supset V_1 \supset V_2 \supset \cdots$, and $\cap_iV_i = \emptyset$. So using the above theorem, there is some $k$ such that $V_k = \emptyset$. Hence $f^{-1}([k,\infty)) = \emptyset$, and so $k$ is an upper bound for $f$.</p>
<p><strong>II: $f$ achieves its upper bound.</strong> By $I$, $f$ is bounded above. Let $M = \sup \, f(x)$. Let $V_n = f^{-1}([M - 1/n,\infty))$. Then $V_1 \supset V_2 \supset \cdots$ are closed. But we cannot have $V_k = \emptyset$ for any $k$, otherwise $M$ would not be the least upper bound. By the converse of the above theorem, $\cap_i V_i \neq \emptyset$. Then pick $x_0 \in \cap_i V_i$, we must have $f(x_0) = M$.</p>
<p><strong>III: $f$ need not be bounded below.</strong> For instance, define $f:[0,1] \rightarrow \mathbb{R}$ by $f(0) = 0$ and $f(x) = -1/x$ for all $x \neq 0$.</p>
<p>This should answer all your questions I think.</p>
|
806,779 | <p>Prove, without expanding, that
\begin{vmatrix}
1 &a &a^2-bc \\
1 &b &b^2-ca \\
1 &c &c^2-ab
\end{vmatrix} vanishes.</p>
<p>Any hints ?</p>
| Robert Israel | 8,508 | <p>e.g. if your matrix is $A$, consider $(b-c,c-a,a-b) A$</p>
|
2,294,991 | <p>I am trying to find the limit as $x\to 8$ of the following function. What follows is the function and then the work I've done on it. </p>
<p>$$ \lim_{x\to 8}\frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8}$$</p>
<hr>
<p>\begin{align}\frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8} &= \frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8} \times \frac{\frac{1}{\sqrt{x +1}} + \frac{1}{3}}{\frac{1}{\sqrt{x +1}} + \frac{1}{3}} \\\\
& = \frac{\frac{1}{x+1}-\frac{1}{9}}{(x-8)\left(\frac{1}{\sqrt{x +1}} + \frac{1}{3}\right)}\\\\
& = \frac{8-x}{(x-8)\left(\frac{1}{\sqrt{x +1}} + \frac{1}{3}\right)}\\\\
& = \frac {-1}{\frac{1}{\sqrt{x +1}} + \frac{1}{3}}\end{align}</p>
<p>At this point I try direct substitution and get:
$$ = \frac{-1}{\frac{2}{3}}$$</p>
<p>This is not the answer. Could someone please help me figure out where I've gone wrong?</p>
| Zain Patel | 161,779 | <p>We have $$\begin{align}\frac{\frac{1}{x+1} - \frac{1}{9}}{(x-8)((x+1)^{-1/2} + 1/3)} &= \frac{\frac{8-x}{9(x+1)}}{(x-8)((x+1)^{-1/2} + 1/3)} \\&= -\frac{1}{9(x+1)((x+1)^{-1/2} + 1/3))} \\ &\longrightarrow-\frac{1}{9(9)(2/3)} = \color{blue}{-\frac{1}{54}}.\end{align}$$</p>
<p>In particular you forgot to take into account the $9(x+1)$ term you get when simplifying $\frac{1}{x+1} - \frac{1}{9}$.</p>
|
1,620,175 | <p>Given metric space $M = (\mathbb{R}^2, d)$ where $d = \operatorname{max}\{|x_1 - y_1|, |x_2 - y_2|\}$, how can one measure distance from some arbitrary point $X$ to the line $y = 3$, let's say?</p>
<p>How it can be done, by use of the identity
$$\operatorname{max}\{x, y\} = \frac{1}{2}(x + y+ |x - y|)\text{?}$$</p>
| levap | 32,262 | <p>Denote your point by $(x_0,y_0)$ and the line $y = 3$ by $l$. You have</p>
<p>$$ d((x_0,y_0),l) = \inf_{x \in \mathbb{R}} d((x_0,y_0), (x,3)) = \inf_{x \in \mathbb{R}} \max \{ |x_0 - x|, |y_0 - 3| \}. $$</p>
<p>It is clear that $\max \{ |x_0 - x|, |y_0 - 3| \} \geq |y_0 - 3|$ and so $d((x_0,y_0),l) \geq |y_0 - 3|$. By taking $x = x_0$, we see that the infimum is actually attained and $d((x_0,y_0),l) = |y_0 - 3|$. It is worth interpreting this calculation geometrically using the description of the circles in this norm as squares.</p>
|
603,290 | <p>this might be a stupid question, but is any $C^2$ function $f:\mathbb{R}\to\mathbb{C}$ of period $f(x+L)=f(x)$ automatically analytic (and in particular, infinitely often differentiable)?</p>
<p>I learned that for a $L$-periodic $C^k$ function with Fourier coefficients $f_n$, they converge to zero like $|f_n| |n|^k \to 0$ as $|n|\to \infty$. So if $k\geq 2$, that would mean that the Fourier series converges absolutely and uniformly, right? So it would extend to an analytic funtion $f(z) = \sum_{n= -\infty}^{\infty} f_n e^{i n z}$ and therefore, it is also $C^\infty$. Am I wrong here?</p>
<p>edit: ok, this is wrong. I thought the uniform convergence meant that $f(z) = \sum_{n= -\infty}^{\infty} f_n e^{i n z}$ should be holomorphic, but of course that only works when we have uniform convergence on an open subset of $\mathbb{C}$, here it's only on an interval in $\mathbb{R}$.</p>
| user1337 | 62,839 | <p>Consider $f(x)=\left| \sin \frac{x}{L} \right|^5$.</p>
<p>The first derivative is, by the chain rule $$f'(x)=5 \left| \sin \frac{x}{L} \right|^4 \left[ \text{sgn} \sin \frac{x}{L} \right] \left( \cos \frac{x}{L} \right) \frac{1}{L}$$ for $x \neq 0$. The derivative at $x=0$ can be computed from the definition $$f'(0)=\lim_{h \to 0} \frac{f(h)-f(0)}{h}=\lim_{h \to 0} \frac{\left| \sin \frac{h}{L} \right|^5-0}{h}= \lim_{h \to 0} \left| \frac{\sin^5 \frac{h}{L}}{h} \right|=0. $$
This agrees with the formula for $x \neq 0$, since $\sin 0=0$. Differentiating a second time is more involved, but still possible.</p>
|
2,388,613 | <p>1) a right cone has a surface area 12 m${^2}$ and radius 1.3 m</p>
<p><strong>here is my answers:</strong></p>
<p>1) $s$ = 1.28</p>
<p><a href="https://i.stack.imgur.com/a8ovb.jpg" rel="nofollow noreferrer">(photo of how i got my answer)</a></p>
<p><strong>textbook answer:</strong> 1.64 m</p>
<p>how did they get that??</p>
| K Split X | 381,431 | <p>The formula to calculate the surface area of a cone is:</p>
<p>$$SA=\pi r(r+\sqrt{h^2+r^2})$$</p>
<p>We know the $SA$ and $r$. Now solve for $h$.</p>
<p>$$12=\pi(1.3)(1.3+\sqrt{h^2+(1.3)^2})$$</p>
<p>$$\frac{12}{1.3\pi}-1.3=\sqrt{h^2+(1.3)^2}\text{ moved stuff we already know to left side}$$</p>
<p>$$(\frac{12}{1.3\pi}-1.3)^2=h^2\text{ squared both sides}$$</p>
<p>$$\sqrt{\frac{12}{1.3\pi}-1.3}=h\text{ take square root to solve for $h$}$$</p>
<p>$h\approx 1.28$</p>
<p>I'm guessing they made a mistake.</p>
|
87,271 | <p>For a category $C$, let $C-Set$ denote the category of set-valued functors $\delta\colon C\to Set$. Given categories $C$ and $D$, and a functor $F\colon C\to D$, composition with $F$ yields a functor that I'll denote by $$\Delta_F\colon D-Set\longrightarrow C-Set.$$ The functor $\Delta_F$ has both a left adjoint, which I'll denote by $\Sigma_F\colon C-Set\longrightarrow D-Set$, and a right adjoint $\Pi_F\colon C-Set\longrightarrow D-Set$. One then has a monad $M = \Delta_F\Sigma_F$ on $C-Set$ and a monad $N=\Pi_F\Delta_F$ on $D-Set$. Let $M-alg=(C-Set)^M$ denote the category of $M$-algebras on $C-Set$, and similarly, let $N-alg=(D-Set)^N$ denote the category of $N$-algebras on $D-Set$.</p>
<p>I'll say that $\Delta_F$ is monadic if the obvious functor $D-Set\longrightarrow M-alg$ is an equivalence of categories, and I'll say that $\Pi_F$ is monadic if the obvious functor $C-Set\longrightarrow N-alg$ is an equivalence of categories.</p>
<p>Questions: </p>
<ol>
<li><p>Under what conditions on $F$ is $\Delta_F$ monadic?</p></li>
<li><p>Under what conditions on $F$ is $\Pi_F$ monadic?</p></li>
</ol>
<p>Thanks!</p>
| Ricardo Andrade | 21,095 | <p>I will try to give an answer to your first question.</p>
<p>The functor $\Delta_F$ verifies automatically nearly all the conditions of the monadicity theorem:</p>
<ul>
<li>it is a right adjoint;</li>
<li>it is a left adjoint with cocomplete domain, and thus coequalizers exist in the source and are preserved by $\Delta_F$.</li>
</ul>
<p>It remains to see when the functor $\Delta_F$ is conservative, i.e. reflects isomorphisms. This is true, for example, in each of the following two cases:</p>
<ul>
<li>$F$ is essentially surjective;</li>
<li>$\Delta_F$ is full and faithful. This condition holds if $F$ induces an equivalence of Cauchy completions, i.e. $F$ is full and faithful and every object in the codomain is a retract of an object in the image. $\Delta_F$ is also full and faithful when $F$ is a reflection onto a reflective subcategory, or more generally when $F$ formally inverts some arrows (such as the map from a model category to its homotopy category).</li>
</ul>
|
31,992 | <p>I am attempting to generate partially transparent images for PNG <code>Export</code>, but seem to run to the following issue. If I <code>Rasterize</code> a simple piece of <code>Graphics</code> with <code>Background -> None</code> (transparent) it looks worse:</p>
<pre><code>rings = Image[
Rasterize[
Graphics[{Black, Disk[], White, Disk[{1, 1}/64, 1 - Sqrt[2]/64]}],
Background -> #, RasterSize -> 400, ImageSize -> 400]] & /@ {None, White}
</code></pre>
<blockquote>
<p><img src="https://i.stack.imgur.com/ho1BD.png" alt="enter image description here"></p>
</blockquote>
<p>First version appears jagged, as a circle drawn using a too simple polygon would do. I can verify this with <code>ImageDifference</code>:</p>
<pre><code>ImageAdjust[ImageDifference @@ (RemoveAlphaChannel[#, White] & /@ rings)]
</code></pre>
<blockquote>
<p><img src="https://i.stack.imgur.com/JZsO6.png" alt="enter image description here"></p>
</blockquote>
<p>Jagged pattern is clearly visible.</p>
<p>How to use <code>Rasterize</code> with <code>Background -> None</code> (or anything similar generating an alpha-channel image) and achieve good output quality without resorting to generating primitives such as <code>Disk</code> using hand-crafted code?</p>
<p><strong>Clarification:</strong></p>
<p>I want alpha channel on the output to behave as it does on <code>Rasterize[..., Background -> None]</code>. That is, the image having conceptually three regions: transparent background, black outer disk and white (non-transparent) inner disk.</p>
<p>(These screenshots were taken on Mathematica 9.0.1.0 running on OS X 10.8.4.)</p>
| halirutan | 187 | <p>One easy idea is to set the alpha channel directly but whether this satisfy your needs really depends on what exactly you want to have. Btw, it is worth that you use <a href="http://reference.wolfram.com/mathematica/ref/ColorSeparate.html" rel="nofollow noreferrer"><code>ColorSeparate</code></a> to see for yourself what exactly the output in the single channels of <em>your</em> jagged image is; you may be surprised.</p>
<pre><code>SetAlphaChannel[#, ColorNegate[#]] &@
Rasterize[
Graphics[{Black, Disk[], White, Disk[{1, 1}/64, 1 - Sqrt[2]/64]}],
Background -> White, RasterSize -> 400, ImageSize -> 400]
</code></pre>
<blockquote>
<p><img src="https://i.stack.imgur.com/ERztP.png" alt="Mathematica graphics"></p>
</blockquote>
|
31,992 | <p>I am attempting to generate partially transparent images for PNG <code>Export</code>, but seem to run to the following issue. If I <code>Rasterize</code> a simple piece of <code>Graphics</code> with <code>Background -> None</code> (transparent) it looks worse:</p>
<pre><code>rings = Image[
Rasterize[
Graphics[{Black, Disk[], White, Disk[{1, 1}/64, 1 - Sqrt[2]/64]}],
Background -> #, RasterSize -> 400, ImageSize -> 400]] & /@ {None, White}
</code></pre>
<blockquote>
<p><img src="https://i.stack.imgur.com/ho1BD.png" alt="enter image description here"></p>
</blockquote>
<p>First version appears jagged, as a circle drawn using a too simple polygon would do. I can verify this with <code>ImageDifference</code>:</p>
<pre><code>ImageAdjust[ImageDifference @@ (RemoveAlphaChannel[#, White] & /@ rings)]
</code></pre>
<blockquote>
<p><img src="https://i.stack.imgur.com/JZsO6.png" alt="enter image description here"></p>
</blockquote>
<p>Jagged pattern is clearly visible.</p>
<p>How to use <code>Rasterize</code> with <code>Background -> None</code> (or anything similar generating an alpha-channel image) and achieve good output quality without resorting to generating primitives such as <code>Disk</code> using hand-crafted code?</p>
<p><strong>Clarification:</strong></p>
<p>I want alpha channel on the output to behave as it does on <code>Rasterize[..., Background -> None]</code>. That is, the image having conceptually three regions: transparent background, black outer disk and white (non-transparent) inner disk.</p>
<p>(These screenshots were taken on Mathematica 9.0.1.0 running on OS X 10.8.4.)</p>
| ybeltukov | 4,678 | <p>Another easy idea: use difference between white and black backgrounds to set alpha channel</p>
<pre><code>SetAlphaChannel[#2, ColorNegate@ImageSubtract[##]] & @@ (
Rasterize[
Graphics[{Black, Disk[], White, Disk[{1, 1}/64, 1 - Sqrt[2]/64]}],
Background -> #, RasterSize -> 400,
ImageSize -> 400] & /@ {Black, White})
</code></pre>
|
3,503,518 | <blockquote>
<p>Determine if <span class="math-container">$ \sum_{n=2}^{\infty} \frac{(\sin{n})\sum\limits_{k=1}^{n} \frac{1}{k}}{(\log n)^2}$</span> is convergent or divergent.</p>
</blockquote>
<hr>
<p>[My attempt]</p>
<p>It seems like Dirichlet test, so I tried to show that <span class="math-container">$a_n := \frac{\sum\limits_{k=1}^{n} \frac{1}{k}}{(\log n)^2}$</span> is decreasing and converges to zero.</p>
<p>By the <a href="https://en.wikipedia.org/wiki/Integral_test_for_convergence" rel="nofollow noreferrer">integral test proof</a>, I know that
<span class="math-container">$$
\int_1^{n+1}\frac{dx}{x}\leq\sum_{k=1}^n\frac{1}{k}\leq 1+\int_1^{n}\frac{dx}{x}
$$</span>
Since <span class="math-container">$\int\frac{dx}{x}=\ln(x)+C$</span>, I can calculate that <span class="math-container">$a_n$</span> converges to zero by the squeeze theorem.</p>
<p>However, I can't show that <span class="math-container">$a_n$</span> is a monotonic decreasing sequence...</p>
<p>How to solve this?</p>
| Eugene Sirkiza | 469,093 | <p><span class="math-container">$$ \frac{\sum_{k=1}^{n} \frac{1}{k}}{(\log{n})^2} \ \ \lor \ \ \frac{\sum_{k=1}^{n+1} \frac{1}{k}}{(\log{(n+1)})^2}$$</span></p>
<p><span class="math-container">$$ \left(\sum_{k=1}^{n} \frac{1}{k}\right) \log^2(n+1) - \left(\sum_{k=1}^{n+1} \frac{1}{k}\right) \log^2(n) \ \ \lor \ \ 0$$</span></p>
<p><span class="math-container">$$ \left(\sum_{k=1}^{n} \frac{1}{k}\right) \left[\log^2(n+1) - \log^2(n)\right] - \frac{1}{n+1} \log^2(n) > \\
\{\log^2(n+1) - \log^2(n) > 0\ \ \text{for} \ \ n > 1 \} \\
\log(n+1) \left[\log^2(n+1) - \log^2(n)\right] - \frac{1}{n+1} \log^2(n) > 0
$$</span></p>
<p>Last inequality flows from <a href="https://www.wolframalpha.com/input/?i=ln%28x%2B1%29%28ln%28x%2B1%29%5E2+-+ln%28x%29%5E2%29+-+%28ln%28x%29%5E2%29%2F%28x%2B1%29+%3E+0" rel="nofollow noreferrer">Wolfram</a>, but I'm not sure, how to prove it strictly. Any help with it?</p>
|
381,093 | <p>I would like to prove Chebyshev's sum inequality, which states that:</p>
<p>If <span class="math-container">$a_1\geq a_2\geq \cdots \geq a_n$</span> and <span class="math-container">$b_1\geq b_2\geq \cdots \geq b_n$</span>, then<br />
<span class="math-container">$$
\frac{1}{n}\sum_{k=1}^n a_kb_k\geq \left(\frac{1}{n}\sum_{k=1}^n a_k\right)\left(\frac{1}{n}\sum_{k=1}^n b_k\right)
$$</span><br />
I am familiar with the non-probabilistic proof, but I need a probabilistic one.</p>
| Fedor Petrov | 4,312 | <p>I do not know whether this counts, but you may do the following.</p>
<p>At first, you may suppose that <span class="math-container">$a_n>0$</span> (otherwise replace all <span class="math-container">$a_i$</span> to <span class="math-container">$a_i-a_n+1$</span>). Consider the probabilistic distribution <span class="math-container">$\mu$</span> on <span class="math-container">$\{1,\ldots,n\}$</span> such that <span class="math-container">$p_i={\rm prob} (X=i)=a_i/(a_1+\ldots+a_n)$</span>. Then <span class="math-container">$p_1+\ldots+p_n=1$</span>, <span class="math-container">$p_1\geqslant \ldots \geqslant p_n$</span>, and we want to prove that <span class="math-container">$\mathbb{E}_{\mu} b\geqslant \mathbb{E}_{\lambda} b\,\, (*),$</span> where <span class="math-container">$\lambda$</span> is a uniform distribution, and <span class="math-container">$b:i\to b_i$</span> is a decreasing function. For doing this we construct a coupling, that is, a joint distribution <span class="math-container">$\nu$</span> of <span class="math-container">$(X,Y)\in\{1,\ldots,n\}^2$</span> such that <span class="math-container">$X$</span> is distributed as <span class="math-container">$\mu$</span>, <span class="math-container">$Y$</span> as <span class="math-container">$\lambda$</span> and <span class="math-container">$Y\geqslant X$</span>. This gives <span class="math-container">$(*)$</span> since <span class="math-container">$$\mathbb{E}_{\mu} b=\mathbb{E}_\nu b(X)\geqslant \mathbb{E}_\nu b(Y)=\mathbb{E}_\lambda b.$$</span>
For constructing <span class="math-container">$\nu$</span>, partition the semiinterval <span class="math-container">$[0,1)$</span> onto <span class="math-container">$n$</span> semiintervals <span class="math-container">$\Delta_1,\ldots,\Delta_n$</span> (from left to right) of lengths <span class="math-container">$1/n$</span> and semiintervals <span class="math-container">$\delta_1,\ldots,\delta_n$</span> of lengths <span class="math-container">$p_1,\ldots,p_n$</span>. Choose a random point <span class="math-container">$x\in [0,1]$</span> at uniform and denote <span class="math-container">$(X,Y)=(k,i)$</span> if <span class="math-container">$x\in \Delta_i\cap \delta_k$</span>.</p>
<p>It remains to prove that <span class="math-container">$i\geqslant k$</span>. Assume the contrary: <span class="math-container">$k>i$</span>. Then <span class="math-container">$x\geqslant p_1+\ldots+p_i\geqslant ip_i$</span>; <span class="math-container">$1-x\leqslant p_{i+1}+\ldots+p_n\leqslant (n-i)p_i$</span>; <span class="math-container">$x/i\geqslant p_i\geqslant (1-x)/(n-i)$</span>; <span class="math-container">$x\geqslant i/n$</span>; <span class="math-container">$x\notin \Delta_i$</span>. A contradiction.</p>
|
706,514 | <p>I know that the fundamental group of homeomorphic spaces are isomorphic. Is the converse true? I mean, can we say the two spaces with isomorphic fundamental groups are homeomorphic? </p>
| Ayman Hourieh | 4,583 | <p>The closest one can get to your statement is probably <a href="http://en.wikipedia.org/wiki/Whitehead_theorem" rel="nofollow">Whitehead's theorem</a>:</p>
<blockquote>
<p>If a map $f$ between two connected CW complexes $X$, $Y$ induces isomorphisms on all homotopy groups, then $f$ is a homotopy equivalence between $X$ and $Y$.</p>
</blockquote>
<p>Homotopy groups are the higher-dimensional analogue of the fundamental group. Of course, homotopy equivalence is a weaker condition than the existence of a homeomorphism. Also note that it's note enough for homotopy groups to be isomorphic; one map must induces all isomorphisms. The only "exception" is when all homotpy groups of a CW complex $X$ are trivial. In this case, the constant map from $X$ to a point induces isomorphisms on all homotopy groups, and therefore $X$ is contractible (homotopy equivalent to a point).</p>
|
3,921,255 | <p>If a function <span class="math-container">$f:M\to\mathbb{R}$</span> is continuous at point <span class="math-container">$x_0$</span> we know that for an arbitrary <span class="math-container">$\epsilon>0$</span> there exists a <span class="math-container">$\delta>0$</span> such that for all <span class="math-container">$x\in M$</span> and <span class="math-container">$|x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon$</span>. Let's call those <span class="math-container">$\epsilon$</span> and <span class="math-container">$\delta(\epsilon)$</span> a pair, <span class="math-container">$(\epsilon,\delta)$</span>.</p>
<p>What happens if I shrink the <span class="math-container">$\delta$</span>? Does this imply that <span class="math-container">$|f(x)-f(x_0)|$</span> also shrinks?</p>
<p>My intuition says that we can't make any claim on the behaviour of <span class="math-container">$|f(x)-f(x_0)|$</span>. Sure if <span class="math-container">$\delta$</span> attains a value which is very small then it will be smaller than another <span class="math-container">$\delta'$</span> which belongs to a pair <span class="math-container">$(\epsilon',\delta')$</span> where the <span class="math-container">$\epsilon'<\epsilon$</span>. But if I shrink <span class="math-container">$\delta$</span> only a bit what happens then? How do I argue in a formal way?</p>
| user3482749 | 226,174 | <p>For manifolds with boundary: no, you can't take boundary points to non-boundary points.</p>
<p>For manifolds without boundary: yes. For 1-manifolds, this is easy. For <span class="math-container">$n > 1$</span>, take a neighbourhood of a (shortest) path between your points. After removing problems (by taking a subset), that's homeomorphic to a disk, with both points in the interior. Take any homeomorphism of the disk that interchanges the points while fixing the boundary, and pass it through the homeomorphism to the neighbourhood of the path.</p>
|
3,678,908 | <p>Let</p>
<p><span class="math-container">$$
V = \{ax^3+bx^2+cx+d|a,b,c,d \in Z_7\}
$$</span></p>
<p>Let the subspace <span class="math-container">$U$</span> Be a subspace of <span class="math-container">$V$</span>:</p>
<p><span class="math-container">$$
U = \{p(x) \in V | p(3)=p(5)=0\}
$$</span></p>
<p>The answers says that:</p>
<p><span class="math-container">$$
Dim(U) = 2
$$</span></p>
<p>How did they conclude that the dimension of <span class="math-container">$U$</span> Is <span class="math-container">$2$</span>?</p>
| Community | -1 | <p>The LHS is a piecewise linear, continuous function. If you evaluate it at the "corner points", you will know in which interval(s) there are solutions.</p>
<p><span class="math-container">$$\begin{array}{}x&-\infty&-\frac32&1&\infty\\\hline f(x)&\infty&-\frac52&5&\infty\end{array}.$$</span></p>
<p>Hence, with <span class="math-container">$x>1$</span>,</p>
<p><span class="math-container">$$2x+3-(x-1)=6$$</span></p>
<p>and with <span class="math-container">$x<-\frac32$</span>,</p>
<p><span class="math-container">$$-(2x+3)-(1-x)=6.$$</span></p>
|
796,787 | <p>Given a Lipschitz function $g$ (i.e. $|g(x) - g(y)| \leq L |x - y|, \forall x, y \in dom(g)$), and an function $f$ integrable on $[a, b]$, how do we prove $g \circ f$ is integrable on $[a, b]$, preferably using Darboux integrals/sums?</p>
<p>Let us assume $g, f$ have appropriate domains and codomains.</p>
| copper.hat | 27,978 | <p>Note that if $M=\sup_{x \in [a,b]} f(x), m = \inf_{x \in [a,b] } f(x)$, then
\begin{eqnarray}
\sup_{x \in [a,b]} g(x) -\inf_{x \in [a,b] } g(x) &=& \sup_{x,y\in [a,b]}
g(x)-g(y) \\
&=& \sup_{x,y\in [a,b]} |g(x)-g(y)| \\
&\le & L \sup_{x,y\in [a,b]} |f(x)-f(y)| \\
&\le & L \sup_{x,y\in [a,b]} f(x)-f(y) \\
&=& L (M-m)
\end{eqnarray}</p>
<p>Now consider a partition $P$ such that $U(f,P) - L(f,P) < { \epsilon \over L}$
and use the above to estimate $U(g\circ f,P) - L(g \circ f,P)$.</p>
|
154,955 | <blockquote>
<p>Let $M\in M_n(F)$ and define $\phi:M_n(F)\to M_n(F)$ by $\phi(X)=AX$
for all $X\in M_n(F)$. Prove that $\det(\phi)=\det(A)^n$.</p>
</blockquote>
<p>I can prove it by considering the matrix representation with respect to the usual basis, which turns out to be a block diagonal form consisting of $n$ copies of $A$. Nevertheless, I'm looking for a clean (basis-free) approach to this problem.</p>
| Qiaochu Yuan | 232 | <p>Think of $M_n(F)$ as $V \otimes V^{\ast}$ where $V = F^n$ and $V^{\ast}$ is the <a href="http://en.wikipedia.org/wiki/Dual_space" rel="nofollow">dual space</a>. The multiplication map $M_n(F) \times M_n(F) \to M_n(F)$ is a map
$$(V \otimes V^{\ast}) \otimes (V \otimes V^{\ast}) \to V \otimes V^{\ast}$$</p>
<p>which, as it turns out, can be identified with <a href="http://en.wikipedia.org/wiki/Tensor_contraction" rel="nofollow">tensor contraction</a> of the two middle terms. It follows that $M_n(F)$, as a left $M_n(F)$-module, can be identified with the tensor product of $V$ (with the natural structure of an $M_n(F)$-module) and $V^{\ast}$ (regarded just as a vector space, the "multiplicity space" of $V$), which is a coordinate-free way of saying that it consists of a direct sum of $n$ copies of $V$ (which in turn is a reformulation of your statement about block diagonals). </p>
<p>Some <a href="http://en.wikipedia.org/wiki/Module_(mathematics)" rel="nofollow">module theory</a> will put this result in useful context. The above argument shows that $M_n(F)$ is <a href="http://en.wikipedia.org/wiki/Semisimple_module" rel="nofollow">semisimple</a> as a module over itself, hence that it is a <a href="http://en.wikipedia.org/wiki/Semisimple_module#Semisimple_rings" rel="nofollow">semisimple ring</a>, which means any module over $M_n(F)$ is a direct sum of simple modules; moreover, the above argument shows that $F^n$ is the unique simple module of $M_n(F)$ (since any simple module of a ring $R$ appears as a quotient of $R$ regarded as a module over itself), so in fact <strong>any</strong> module over $M_n(F)$ whatsoever breaks up into a direct sum of copies of $F^n$. This reflects the fact that $M_n(F)$ is <a href="http://en.wikipedia.org/wiki/Morita_equivalence" rel="nofollow">Morita equivalent</a> to $F$. </p>
|
2,936,236 | <p>I have two 2D disks, <span class="math-container">$(C_1, r_1)$</span> (blue) and <span class="math-container">$(C_2, r_2)$</span> (red), where the blue disk somehow overlaps the red disk (the figure below shows one example). I’m interested in finding the positive distance along the <span class="math-container">$y$</span> axis the blue disk needs to be moved so that it touches but does not overlap the red disk. If I move the blue disk radially outward from the red disk then it’s trivial to figure out how far to move in order to avoid overlap; but in my case, where I only want to move the blue disk in the positive <span class="math-container">$y$</span> direction, the general formula (for any configuration of the two disks, assuming there is any overlap) seems more difficult. Could someone provide some insight on what a general formula for this would be, if it's feasible?</p>
<p><a href="https://i.stack.imgur.com/CMCrI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CMCrI.png" alt="enter image description here"></a></p>
| davewy | 595,478 | <p>Just to codify @amd 's answer, the formula appears to just require combining the Pythagorean theorem with the <span class="math-container">$y$</span>-offset of the two circles:</p>
<p><span class="math-container">$dy = \sqrt{(r_2-r_1)^2 - (x_2-x_1)^2} - (y_1-y_2)$</span></p>
<p>where <span class="math-container">$C_1 = (x_1,y_1)$</span> and <span class="math-container">$C_2=(x_2,y_2)$</span>.</p>
|
85,228 | <p>I wonder if the periodic paths of a lightray trapped between two nonconcentric circles,
each perfectly reflecting, are known.
The behavior of such rays seems chaotically complicated. For example, left below the highlighted
initial ray has slope <span class="math-container">$\frac{1}{12}$</span>, while that to the right has slope <span class="math-container">$\frac{1}{9}$</span>.
(The green circle has radius <span class="math-container">$\frac{3}{4}$</span>.) After 500 reflections, neither ray has become periodic.
<br />
<img src="https://i.stack.imgur.com/Yak4Y.jpg" alt="Light rays trapped in annulus">
<br />
I suspect the combination of dispersive and focusing reflection
(from the inner and outer circles respectively)
leads to this complex behavior.
But perhaps the periodic paths are known?</p>
<p>I only know the 2-cycles from rays collinear with the circle centers,
and those that Noam Elkies kindly identified in his comment:
"regular <span class="math-container">$n$</span>-gons (and <span class="math-container">$(n/k)$</span>-gons, i.e. stars) that stay so close to the outer circle that they never hit the inner one." </p>
<p>(Related MO question: "<a href="https://mathoverflow.net/questions/38307/">Trapped rays bouncing between two convex bodies</a>.")
<hr />
<b>Update 1</b>. I found one! :-) No doubt among the "simple short cycles" that Noam had in mind:
<br />
<img src="https://i.stack.imgur.com/1tdDg.jpg" alt="3-cycle">
<br />
<hr />
<b>Update 2</b>.
With the benefit of the search terms helpfully provided by Ian Agol and Igor Riven,
I found a useful <em>Physical Review</em> paper by
G. Gouesbet, S. Meunier-Guttin-Cluzel, and G. Grehan,
"Periodic orbits in Hamiltonian chaos of the annular billiard,"
<a href="http://pre.aps.org/abstract/PRE/v65/i1/e016212" rel="nofollow noreferrer">Phys. Rev. E 65, 016212 (2001)</a>. Their approach is more experimental than theoretical:</p>
<blockquote>
<p>Periodic orbits embedded in the phase space are
systematically investigated, with a focus on inclusion-touching periodic orbits, up to symmetrical orbits of
period 6. Candidates for periodic orbits are detected by investigating grayscale distance charts and, afterward,
each candidate is validated (or rejected) by using analytical and/or numerical methods.</p>
</blockquote>
<p>The (unstable) 3-cycle I found they label "2(1)1" in their
(barely discernable) Fig.5 inventory below:
<br />
<img src="https://i.stack.imgur.com/Diet8.jpg" alt="Fig5"></p>
| Ian Agol | 1,345 | <p>I don't know how to answer your question, but I'll make a reformulation of the question. </p>
<p>Consider the unit tangent bundle to the outer circle, and consider
the subset consisting of vectors pointing into the circle. This is
an annulus, parameterized by $(\theta,\varphi)\in A=[0,2\pi]\times [0,\pi]/\{ (0,\varphi)\sim (2\pi,\varphi)\}$, where $\theta$ parameterizes the point on the outer circle, and $\varphi$ gives the angle that the unit vector makes with the tangent vector. One may consider the first-return map under the geodesic flow $F: A\to A$ (flow in the direction of the vector until you hit the outer circle again, then reflect). This is a piecewise smooth map. For each $\theta$, there are angles $0< f_1(\theta) < f_2(\theta) < \pi$ such that $F(\theta,\varphi)= (\theta+2\varphi, \varphi)$ for $0\leq \varphi \leq f_1(\theta)$ and $f_2(\theta)\leq \varphi \leq \pi$ (in particular, $F$ is the identity on the boundary of $A$). For fixed $\theta$ and $f_1(\theta) < \varphi < f_2(\theta)$, $F(\theta,\varphi)$ is a more complicated trigonometric function depending on how the geodesic reflects off the inner circle and bounces back to the outer circle, but it has the property that $\varphi$ is increasing, and $\theta$ is decreasing. There is a natural measure on geodesic flow in the plane, the Liouville measure, which restricts to a measure on $A$. Clearly $F$ preserves this measure. I haven't computed the measure, but it is invariant under rotation, so is independent of $\theta$, and is invariant under reflection $(\theta,\varphi)\to (\theta,\pi-\varphi)$. One could reparameterize the $\varphi$ coordinate in terms of the Liouville measure to get an area-preserving homeomorphism of the annulus. So I would suggest you could do a literature search for results on periodic points for area-preserving homeomorphisms of an annulus. </p>
|
3,335,257 | <p>This is the full question:</p>
<blockquote>
<p>show that if <span class="math-container">$p$</span> is an odd prime then the number of ordered pair solutions of the congruence<span class="math-container">$x^2-y^2 \equiv a \pmod p,$</span> is <span class="math-container">$p-1$</span> unless <span class="math-container">$a \equiv 0 \pmod p,$</span> in which case number of solutions is <span class="math-container">$2p-1$</span>. </p>
</blockquote>
<p>Considering <span class="math-container">$x,y \,\in \mathbb{Z}$</span>. In the second case, since <span class="math-container">$a \equiv 0 \pmod p,$</span> it follows that <span class="math-container">$p\mid (x-y)(x+y)$</span>, but then there will be infinitely many solutions of this congruence relation because there are no bounds mentioned in the question on <span class="math-container">$x$</span> and <span class="math-container">$y$</span>.</p>
<p>So is the question incomplete? or is it implicitly stated that <span class="math-container">$0\leq x,y<p$</span> .For this bound, do we get <span class="math-container">$2p-1$</span> solutions? </p>
| Mike | 544,150 | <p>ETA: For whatever reason I thought the question here was to prove that the statement in the OP's thread in the shaded yellow box, which is what I did below.</p>
<p>You already answered your own question for the case where <span class="math-container">$a$</span> is zero so we now consider the case where <span class="math-container">$a$</span> is nonzero i.e., <span class="math-container">$a \in (\mathbb{F}_p)^{\times}$</span>.</p>
<ol>
<li><p>Let <span class="math-container">$a \in (\mathbb{F}_p)^{\times}$</span>. For every such nonzero <span class="math-container">$b \in (\mathbb{F}_p)^{\times}$</span>, there is exactly one <span class="math-container">$c \in (\mathbb{F}_p)^{\times}$</span> satisfying <span class="math-container">$a=bc$</span>. </p></li>
<li><p>The above equation factors to <span class="math-container">$(x-y)(x+y) = a; x,y \in (\mathbb{F}_p)^{\times}$</span>. By the above, for each nonzero <span class="math-container">$b=(x-y)\in (\mathbb{F}_p)^{\times}$</span>, there is exactly one <span class="math-container">$c=(x+y)\in (\mathbb{F}_p)^{\times}$</span> such that <span class="math-container">$(x-y)(x+y)=a$</span>. </p></li>
<li><p>So for each nonzero <span class="math-container">$b\in (\mathbb{F}_p)^{\times}$</span>, there is exactly one pair <span class="math-container">$(x,y); x,y \in (\mathbb{F}_p)^{\times}$</span> such that both <span class="math-container">$(x-y)=b$</span> and <span class="math-container">$(x-y)(x+y)=a$</span>.</p></li>
</ol>
<p>Can you finish from here.</p>
|
2,279,660 | <p>On $C^1([0,1],\mathbb{R})$ I have two norms
\begin{align*}
N_1(f)&=|f(0)|+\|f'\|_{\infty}\\
N_2(f)&=\|f\|_{\infty}+\|f'\|_{\infty}.
\end{align*}</p>
<p>We have always that $N_1(f)\leq N_2(f)$ but what about the other inequality please?</p>
<p>And please how to see If $N_1$ and $N_2$ are equivalent to $\left\|\cdot\right\|_{\infty}$. </p>
<p>Thank you. </p>
| copper.hat | 27,978 | <p>Note that $f(x) = f(0) + \int_0^x f'(t) dt$ hence
$\|f\|_\infty \le |f(0)| + \|f'\|_\infty$.</p>
<p>Hence $N_2(f) = \|f\|_\infty + \|f'\|_\infty \le |f(0)| + 2\|f'\|_\infty \le 2 N_1(f)$.</p>
<p>For the second part, take $f_n(t) = \sin (2 \pi nt)$. Clearly $\|f_n\|_\infty = 1$, but
$N_2(f_n) = 1+2\pi n$.</p>
|
3,817,092 | <p>How to show that we may replace the property of a norm <span class="math-container">$$||x||=0 \iff x=0$$</span> by <span class="math-container">$$||x||=0 \implies x = 0$$</span> without altering the concept
of a norm.</p>
| Qiaochu Yuan | 232 | <p>Some points.</p>
<ol>
<li><p>By Poincare duality the Euler characteristic of a closed odd-dimensional manifold vanishes so we restrict our attention to even-dimensional manifolds. By the classification of surfaces <span class="math-container">$S^2$</span> is the only example in dimension <span class="math-container">$2$</span>.</p>
</li>
<li><p>Euler characteristic is multiplicative with respect to products (e.g. by the <a href="https://en.wikipedia.org/wiki/K%C3%BCnneth_theorem" rel="noreferrer">Kunneth theorem</a>). So we can find examples of manifolds with positive Euler characteristic by taking products of an even number of manifolds of negative Euler characteristic together with any number of manifolds of positive Euler characteristic. The Euler characteristic is also multiplicative with respect to (nice) <a href="https://en.wikipedia.org/wiki/Euler_characteristic#Fibration_property" rel="noreferrer">fiber bundles</a> so we can consider nontrivial fiber bundles with suitable bases and fibers also.</p>
</li>
<li><p>The Euler characteristic of a connected sum of closed <span class="math-container">$n$</span>-manifolds satisfies the "inclusion-exclusion" formula <span class="math-container">$\chi(M\#N) = \chi(M) + \chi(N) - \chi(S^n)$</span>. This means when <span class="math-container">$n$</span> is even, connected sum with <span class="math-container">$N$</span> increases the Euler characteristic iff <span class="math-container">$\chi(N) \ge 3$</span>. Also the connected sum of simply connected manifolds is simply connected.</p>
</li>
<li><p>Any manifold whose cohomology is concentrated in even degrees (say, over <span class="math-container">$\mathbb{Q}$</span>) has positive Euler characteristic, and there's a large source of examples of such manifolds coming from algebraic geometry: every <a href="https://en.wikipedia.org/wiki/Generalized_flag_variety" rel="noreferrer">generalized flag variety</a> <span class="math-container">$G/P$</span> over <span class="math-container">$\mathbb{C}$</span> has this property. <span class="math-container">$\mathbb{CP}^n$</span> is a special case of this construction but we also have <a href="https://en.wikipedia.org/wiki/Grassmannian" rel="noreferrer">Grassmannians</a> and <a href="https://en.wikipedia.org/wiki/Generalized_flag_variety#Prototype:_the_complete_flag_variety" rel="noreferrer">complete flag varieties</a>, for example, which are also simply connected (probably generalized flag varieties over <span class="math-container">$\mathbb{C}$</span> are always simply connected but I don't know how to prove it).</p>
</li>
<li><p>By Poincare duality, any closed simply connected <span class="math-container">$4$</span>-manifold has cohomology concentrated in even degrees and hence has Euler characteristic at least <span class="math-container">$2$</span>. <a href="https://qchu.wordpress.com/2014/06/16/hypersurfaces-4-manifolds-and-characteristic-classes/" rel="noreferrer">This blog post</a> on hypersurfaces in <span class="math-container">$\mathbb{CP}^3$</span> discusses their topology; in particular they are <a href="http://www.map.mpim-bonn.mpg.de/4-manifolds:_1-connected#Topological_classification" rel="noreferrer">completely classified</a> up to homotopy (Milnor, Whitehead) and even up to homeomorphism (Freedman). Their Euler characteristics can be arbitrarily large: the linked post shows that the Euler characteristic of a smooth degree <span class="math-container">$d$</span> hypersurface in <span class="math-container">$\mathbb{CP}^3$</span> is <span class="math-container">$d^3 - 4d^2 + 6d$</span>. When <span class="math-container">$d = 1$</span> we get <span class="math-container">$\mathbb{CP}^2$</span> which has <span class="math-container">$\chi = 1 - 4 + 6 = 3$</span>, when <span class="math-container">$d = 2$</span> we get <span class="math-container">$\mathbb{CP}^1 \times \mathbb{CP}^1$</span> which has <span class="math-container">$\chi = 8 - 16 + 12 = 4$</span>, and when <span class="math-container">$d = 4$</span> we get a <a href="https://en.wikipedia.org/wiki/K3_surface" rel="noreferrer">K3 surface</a> which has <span class="math-container">$\chi = 64 - 64 + 24 = 24$</span>.</p>
</li>
<li><p>For the classification of closed simply connected <span class="math-container">$6$</span>-manifolds see <a href="http://www.map.mpim-bonn.mpg.de/6-manifolds:_1-connected" rel="noreferrer">here</a>. Wall showed that any such manifold splits as a connected sum of a number of copies of <span class="math-container">$S^3 \times S^3$</span> (<span class="math-container">$\chi(S^3 \times S^3) = 0$</span> so this connected sum lowers the Euler characteristic by <span class="math-container">$2$</span> and removing it increases the Euler characteristic by <span class="math-container">$2$</span>) and a manifold <span class="math-container">$M$</span> with <span class="math-container">$b_3 = 0$</span>, hence <span class="math-container">$M$</span> has rational cohomology concentrated in even degrees and so has positive Euler characteristic. Incidentally, this makes a connected sum of two copies of <span class="math-container">$S^3 \times S^3$</span> the simplest example of a closed simply connected manifold with negative Euler characteristic, so now we can actually carry out the construction I suggested in point 2: the product of two such sums is a closed simply connected <span class="math-container">$12$</span>-manifold with positive Euler characteristic whose rational cohomology is not concentrated in even degree.</p>
</li>
<li><p>The <a href="https://en.wikipedia.org/wiki/Hopf_conjecture" rel="noreferrer">Hopf conjecture</a> claims in part that a closed even-dimensional manifold admitting a metric with <a href="https://en.wikipedia.org/wiki/Sectional_curvature#Manifolds_with_positive_sectional_curvature" rel="noreferrer">positive sectional curvature</a> has positive Euler characteristic. In dimension <span class="math-container">$2$</span> this of course follows from the <a href="https://en.wikipedia.org/wiki/Gauss%E2%80%93Bonnet_theorem" rel="noreferrer">Gauss-Bonnet theorem</a>, and as Wikipedia discusses this also holds in dimension <span class="math-container">$4$</span>.</p>
</li>
<li><p>If I were looking for more examples I might look through lists of <a href="https://en.wikipedia.org/wiki/Symmetric_space" rel="noreferrer">symmetric spaces</a>.</p>
</li>
</ol>
|
3,294,446 | <p>I'm currently working on a definite integral and am hoping to find alternative methods to evaluate. Here I will to address the integral:
<span class="math-container">\begin{equation}
I_n = \int_0^\frac{\pi}{2}\ln^n\left(\tan(x)\right)\:dx
\end{equation}</span>
Where <span class="math-container">$n \in \mathbb{N}$</span>. We first observe that when <span class="math-container">$n = 2k + 1$</span> (<span class="math-container">$k\in \mathbb{Z}, k \geq 0$</span>) that,
<span class="math-container">\begin{equation}
I_{2k + 1} = \int_0^\frac{\pi}{2}\ln^{2k + 1}\left(\tan(x)\right)\:dx = 0
\end{equation}</span>
This can be easily shown by noticing that the integrand is odd over the region of integration about <span class="math-container">$x = \frac{\pi}{4}$</span>. Thus, we need only resolve the cases when <span class="math-container">$n = 2k$</span>, i.e.
<span class="math-container">\begin{equation}
I_{2k} = \int_0^\frac{\pi}{2}\ln^{2k}\left(\tan(x)\right)\:dx
\end{equation}</span>
Here I have isolated two methods.</p>
<hr>
<p>Method 1:</p>
<p>Let <span class="math-container">$u = \tan(x)$</span>:
<span class="math-container">\begin{equation}
I_{2k} = \int_0^\infty\ln^{2k}\left(u\right) \cdot \frac{1}{u^2 + 1}\:du = \int_0^\infty \frac{\ln^{2k}\left(u\right)}{u^2 + 1}\:du
\end{equation}</span>
We note that:
<span class="math-container">\begin{equation}
\ln^{2k}(u) = \frac{d^{2k}}{dy^{2k}}\big[u^y\big]_{y = 0}
\end{equation}</span>
By Leibniz's Integral Rule:
<span class="math-container">\begin{align}
I_{2k} &= \int_0^\infty \frac{\frac{d^{2k}}{dy^{2k}}\big[u^y\big]_{y = 0}}{u^2 + 1}\:du = \frac{d^{2k}}{dy^{2k}} \left[ \int_0^\infty \frac{u^y}{u^2 + 1} \right]_{y = 0} \nonumber \\
&= \frac{d^{2k}}{dy^{2k}} \left[ \frac{1}{2}B\left(1 - \frac{y + 1}{2}, \frac{y + 1}{2} \right) \right]_{y = 0} =\frac{1}{2}\frac{d^{2k}}{dy^{2k}} \left[ \Gamma\left(1 - \frac{y + 1}{2}\right)\Gamma\left( \frac{y + 1}{2} \right) \right]_{y = 0} \nonumber \\
&=\frac{1}{2}\frac{d^{2k}}{dy^{2k}} \left[ \frac{\pi}{\sin\left(\pi\left(\frac{y + 1}{2}\right)\right)} \right]_{y = 0} = \frac{\pi}{2}\frac{d^{2k}}{dy^{2k}} \left[\operatorname{cosec}\left(\frac{\pi}{2}\left(y + 1\right)\right) \right]_{y = 0}
\end{align}</span></p>
<hr>
<p>Method 2:</p>
<p>We first observe that:
<span class="math-container">\begin{align}
\ln^{2k}\left(\tan(x)\right) &= \big[\ln\left(\sin(x)\right) - \ln\left(\cos(x)\right) \big]^{2k} \nonumber \\
&= \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \ln^j\left(\cos(x)\right)\ln^{2k - j}\left(\sin(x)\right)
\end{align}</span>
By the linearity property of proper integrals we observe:
<span class="math-container">\begin{align}
I_{2k} &= \int_0^\frac{\pi}{2} \left[ \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \ln^j\left(\cos(x)\right)\ln^{2k - j}\left(\sin(x)\right) \right]\:dx \nonumber \\
&= \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \int_0^\frac{\pi}{2} \ln^j\left(\cos(x)\right)\ln^{2k - j}\left(\sin(x)\right)\:dx \nonumber \\
& = \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j F_{n,m}(0,0)
\end{align}</span>
Where
<span class="math-container">\begin{equation}
F_{n,m}(a,b) = \int_0^\frac{\pi}{2} \ln^n\left(\cos(x)\right)\ln^{m}\left(\sin(x)\right)\:dx
\end{equation}</span>
Utilising the same identity given before, this becomes:
<span class="math-container">\begin{align}
F_{n,m}(a,b) &= \int_0^\frac{\pi}{2} \frac{d^n}{da^n}\big[\sin^a(x) \big] \cdot \frac{d^m}{db^m}\big[\cos^b(x) \big]\big|\:dx \nonumber \\
&= \frac{\partial^{n + m}}{\partial a^n \partial b^m}\left[ \int_0^\frac{\pi}{2} \sin^a(x)\cos^b(x)\:dx\right] = \frac{\partial^{n + m}}{\partial a^n \partial b^m}\left[\frac{1}{2} B\left(\frac{a + 1}{2}, \frac{b + 1}{2} \right)\right] \nonumber \\
&= \frac{1}{2}\frac{\partial^{n + m}}{\partial a^n \partial b^m}\left[\frac{\Gamma\left(\frac{a + 1}{2}\right)\Gamma\left(\frac{b + 1}{2}\right)}{\Gamma\left(\frac{a + b}{2} + 1\right)}\right]
\end{align}</span>
Thus,
<span class="math-container">\begin{equation}
I_{2k} = \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \frac{1}{2}\frac{\partial^{2k }}{\partial a^j \partial b^{2k - j}}\left[\frac{\Gamma\left(\frac{a + 1}{2}\right)\Gamma\left(\frac{b + 1}{2}\right)}{\Gamma\left(\frac{a + b}{2} + 1\right)}\right]_{(a,b) = (0,0)}
\end{equation}</span></p>
<hr>
<p>So, I'm curious, are there any other Real Based Methods to evaluate this definite integral?</p>
| Martin Gales | 2,323 | <p>Slightly different way - use well known results</p>
<p><span class="math-container">$$\int_0^{\frac{\pi}{2}}\tan^ay\;dy=\frac{\pi}{2}\frac{1}{\sin \frac{\pi}2(a+1) } $$</span></p>
<p>( this integral is considered in this site probably many times.)
<span class="math-container">$$\frac{1}{\sin x}=\frac{1}{x}+\sum _{n=1}^\infty (-1)^n\left ( \frac{1}{x-n\pi}+ \frac{1}{x+n\pi}\right )$$</span></p>
<p>and differentiate with respect to <span class="math-container">$a$</span> as many times as necessary.</p>
|
1,141,412 | <p>Prove that the ring of integers of $\mathbb Q (\sqrt{-5})$ does not have unique factorisation.</p>
<p>Since $-5\equiv 3\pmod 4$, I know that the ring of integers of $\mathbb Q (\sqrt{-5})$ is $\mathbb Z [\sqrt{-5}]$. </p>
<p>I assume the way to prove that this does not have a unique factorisation is to give two different factorisations, but what exactly do I use to factorise? Do I consider the polynomial $x+\sqrt{-5}$?</p>
| Bernard | 202,857 | <p>We can write two decompositions of $6$:
$$6=2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5})$$
All factors are irreducible, since their norms can't be decomposed as the product of two norms $\neq 1$ and the factors on the rhs are not associated with the factors on the lhs, since their norms are different.</p>
<p>This proves $\mathbf Z[\sqrt{-5}]$ does not have unique factorisation.</p>
|
1,424,561 | <p>I know that $\sum_0^\infty \frac{\lambda^x} {x!} = e^\lambda$, but I'm having a really difficult time dealing with the extra $x$.</p>
| Michael Hardy | 11,667 | <p>\begin{align}
\sum_{x=0}^\infty \frac{x\lambda^x}{x!} & = \sum_{x=1}^\infty \frac{x\lambda^x}{x!} & & \text{(since the first term is $0$)} \\[10pt]
& = \sum_{x=1}^\infty \frac{\lambda^x}{(x-1)!} & & \text{(since $\dfrac x {x!} = \dfrac 1 {(x-1)!}$)} \\[10pt]
& = \lambda\sum_{x=1}^\infty \frac{\lambda^{x-1}}{(x-1)!} \\[10pt]
& = \lambda\sum_{y=0}^\infty \frac{\lambda^y}{y!} & & \text{where $y=x-1$} \\
& & & \text{As $x$ goes through $1,2,3,\ldots\,{}$,} \\
& & & \text{then $y$ goes through $0,1,2,3,\ldots\,{}$.}
\\[10pt]
& = \lambda\sum_{x=0}^\infty \frac{\lambda^x}{x!} =\cdots & & \text{(This is an “alphabetic} \\
& & & \phantom{(}\text{variant'' of the previous line.)}
\end{align}</p>
|
17,259 | <p>I am using <em>Mathematica</em> to go through the examples and exercises on the book <em>Modern Control Systems</em>, 6th edition by Dorf. On page 605, there is a table (Table 8.5) with the Bode plot for several transfer functions. In what follows there is a piece of code that attempts to build the very same table.</p>
<p>Here is the code:</p>
<pre><code>With[{ τ1 = 20, τ2 = 2, τ3 = 0.4, τ4 = 0.05, τa = 10, τb = 1, k = 10},
Grid[
Partition[
Table[ BodePlot[ sys, PlotLabel->sys, GridLines -> Automatic], { sys,
{ k/(s τ1 + 1), (k(s τa + 1))/(s(s τ1 + 1)(s τ2 + 1)),
k/((s τ1 + 1)(s τ2 + 1)), k/s^2, k/((s τ1 + 1)(s τ2 + 1)(s τ3 + 1)),
k/(s^2 (s τ1 + 1)), k/s, (k(s τa + 1))/(s^2 (s τ1 + 1)),
k/(s(s τ1 + 1)), k/s^3, k/(s(s τ1 + 1)(s τ2 + 1)),
(k (s τa + 1))/s^3, (k (s τa + 1)(s τb + 1))/s^3,
(k (s τa + 1))/(s^2 (s τ1 + 1)(s τ2 + 1)),
(k (s τa + 1)(s τb + 1))/(s(s τ1 + 1)(s τ2 + 1)(s τ3 + 1)(s τ4 + 1)) }
}
], 2], Frame->All, Spacings->6] ]
</code></pre>
<p><img src="https://i.stack.imgur.com/DyGQX.gif" alt="a bunch of Bode plots"></p>
<p>All the transfer functions with <code>1/s^n</code> <code>( n > 1 )</code> give the wrong result as far as the phase plot is concerned. Is there a simple way to fix this? <em>Wolfram</em> does not have a time line to go through the problem and sort it out.</p>
| Nasser | 70 | <p>To address the above comment by Ed</p>
<blockquote>
<p>However if we change the example. Matlab - <code>bode(tf(10*[10 1],[1 0 0
0]))</code> - phase is negative (-270 to -180). Mathematica - See plots above
- phase from +90 to +180. If instead of <code>10(10s+1)/sˆ3</code>, one uses <code>10(10s+1)/(s+0.0001)ˆ3</code>, the phase is negative</p>
</blockquote>
<p>This below is a direct implementation of the phase plot part of Bode, using <code>ArcTan</code>. I used the following 2 transfer functions to compare with Mathematica BodePlot: <code>10(10s+1)/(s+0.0001)ˆ3</code> and <code>10(10s+1)/(s)ˆ3</code> and the result does show that there is a sudden phase change shift by <code>180</code> which does not show when using straight calculations using <code>ArcTan</code> to find the phase. Conclusion: There seems to be some convention used that causes this change as I would have expected it to match the <code>ArcTan</code> direct method.</p>
<h2>Mathematica BodePlot phase diagram</h2>
<pre><code>Clear[s];
expr1 = (100 s + 10)/(s)^3;
expr2 = (100 s + 10)/(s + 0.0001)^3;
Grid[{{expr1, expr2},
BodePlot[TransferFunctionModel[#, s], GridLines -> Automatic,
ImageSize -> 300, PlotLayout -> "List",
FrameLabel -> {{{"magnitude (db)", None}, {None,"Bode plot"}},
{{"phase(deg)",None}, {"Frequency (rad/sec)", None}}},
ScalingFunctions -> {{"Log10", "dB"}, {"Log10", "Degree"}},
PlotRange -> {{{0.001, 10}, All}, {{0.001, 10}, All}}][[2]] & /@ {expr1, expr2}}]
</code></pre>
<p><img src="https://i.stack.imgur.com/cvJfN.png" alt="Mathematica graphics"></p>
<h2>Direct implementation of the phase plot using <code>ArcTan</code></h2>
<pre><code>Clear[s];
ticks[min_, max_] := Table[{i, Superscript[10, i]}, {i, Ceiling[min], Floor[max], 1}];
makePhasePlot[expr_, s_Symbol] := Module[{ex, w, re, im, data},
ex = expr /. s -> (w I);
re = ComplexExpand[Re[ex]];
im = ComplexExpand[Im[ex]];
data =Table[{Log[10, w], 180/Pi ArcTan[re, im]}, {w, 0.001, 10, 0.001}];
ListPlot[data,
Joined -> True,
PlotRange -> All,
FrameTicks -> {{Automatic, Automatic}, {ticks, Automatic}},
AxesOrigin -> {0, 0},
Frame -> True,
ImageSize -> 300,
FrameLabel -> {{"angle(deg)", None}, {"Frequency (rad/sec)", None}},
Axes -> False]
];
expr1 = (100 s + 10)/s^3;
expr2 = (100 s + 10)/(s + 0.0001)^3;
Grid[{{expr1, expr2},makePhasePlot[#, s] & /@ {expr1, expr2}}, Spacings -> {3, 0}]
</code></pre>
<p><img src="https://i.stack.imgur.com/scfHb.png" alt="Mathematica graphics"></p>
|
17,259 | <p>I am using <em>Mathematica</em> to go through the examples and exercises on the book <em>Modern Control Systems</em>, 6th edition by Dorf. On page 605, there is a table (Table 8.5) with the Bode plot for several transfer functions. In what follows there is a piece of code that attempts to build the very same table.</p>
<p>Here is the code:</p>
<pre><code>With[{ τ1 = 20, τ2 = 2, τ3 = 0.4, τ4 = 0.05, τa = 10, τb = 1, k = 10},
Grid[
Partition[
Table[ BodePlot[ sys, PlotLabel->sys, GridLines -> Automatic], { sys,
{ k/(s τ1 + 1), (k(s τa + 1))/(s(s τ1 + 1)(s τ2 + 1)),
k/((s τ1 + 1)(s τ2 + 1)), k/s^2, k/((s τ1 + 1)(s τ2 + 1)(s τ3 + 1)),
k/(s^2 (s τ1 + 1)), k/s, (k(s τa + 1))/(s^2 (s τ1 + 1)),
k/(s(s τ1 + 1)), k/s^3, k/(s(s τ1 + 1)(s τ2 + 1)),
(k (s τa + 1))/s^3, (k (s τa + 1)(s τb + 1))/s^3,
(k (s τa + 1))/(s^2 (s τ1 + 1)(s τ2 + 1)),
(k (s τa + 1)(s τb + 1))/(s(s τ1 + 1)(s τ2 + 1)(s τ3 + 1)(s τ4 + 1)) }
}
], 2], Frame->All, Spacings->6] ]
</code></pre>
<p><img src="https://i.stack.imgur.com/DyGQX.gif" alt="a bunch of Bode plots"></p>
<p>All the transfer functions with <code>1/s^n</code> <code>( n > 1 )</code> give the wrong result as far as the phase plot is concerned. Is there a simple way to fix this? <em>Wolfram</em> does not have a time line to go through the problem and sort it out.</p>
| Suba Thomas | 5,998 | <p>This is fixed for v10. In v10 there is also a <code>PhaseRange</code> option for <code>BodePlot</code> and <code>NicholsPlot</code> that can be used to override the default range (if you need to wrap it between $-\pi$ and $\pi$, etc).</p>
<pre><code>With[{τ1 = 20, τ2 = 2, τ3 = 0.4, τ4 = 0.05, τa = 10, τb = 1, k = 10},
Grid[Partition[Table[BodePlot[sys, PlotLabel -> sys, GridLines -> Automatic,
PlotLayout -> "Phase"], {sys, {k/(s τ1 + 1),
(k (s τa + 1))/(s (s τ1 + 1) (s τ2 + 1)),
k/((s τ1 + 1) (s τ2 + 1)), k/s^2,
k/((s τ1 + 1) (s τ2 + 1) (s τ3 + 1)),
k/(s^2 (s τ1 + 1)),
k/s, (k (s τa + 1))/(s^2 (s τ1 + 1)),
k/(s (s τ1 + 1)), k/s^3,
k/(s (s τ1 + 1) (s τ2 + 1)), (k (s τa + 1))/
s^3, (k (s τa + 1) (s τb + 1))/
s^3, (k (s τa + 1))/(s^2 (s τ1 + 1) (s τ2 +
1)), (k (s τa + 1) (s τb + 1))/(s (s τ1 +
1) (s τ2 + 1) (s τ3 + 1) (s τ4 + 1))}}], 2],
Frame -> All, Spacings -> 6]]
</code></pre>
<p><img src="https://i.stack.imgur.com/hSuYe.png" alt="enter image description here"></p>
|
2,339,964 | <p>I wondered whether it is possible to define a sequence without relying on sets or natural numbers and tried with this definition.</p>
<ol>
<li>A symbol which is not a comma is a sequence. </li>
<li>If $S$ is sequence and $s$ a symbol which is not a comma:
<ul>
<li>$S, s$ is a sequence, where every symbol occurring in $S$ precedes $\phi$ and $\phi$ is the last symbol of the sequence.</li>
<li>$s, S$ is a sequence, where every symbol occurring in $S$ follows $\phi$ and $\phi$ is the first symbol of the sequence. </li>
</ul></li>
</ol>
<p>So given the string "$a,b,c$", "$a$" is a sequence and "$a,b$" is a sequence too. Therefore "$a,b,c$" is the sequence "$S, c$", where "$S$" is "$a,b$", "$c$" is its last element and "$a,b$" precede it. Similarly "$a$" can be identified as the first element. </p>
<p>Is that correct?</p>
| DanielV | 97,045 | <p>You seem to be attempting a recursive definition of finite sequences. A recursive definition needs 3 things: construction, distinguishing, and exclusion.</p>
<p>For construction:</p>
<blockquote>
<p>A symbol which is not a comma is a sequence.</p>
</blockquote>
<p>It is very very bad style not to distinguish between a singleton and it's element, between $[a]$ and $a$. This also leaves out the zero length sequence.
I suggest simply "$[]$ is a sequence".</p>
<blockquote>
<p>If S is sequence and s a symbol which is not a comma...</p>
</blockquote>
<p>I suggest just writing "If [a] is a sequence, then [a,x] is a sequence".</p>
<p>Distinguishability:</p>
<p>You can't just assume that because 2 things are written differently that they are different, for example the sets $\{a, b\} = \{b, a\} = \{a, a, b, b\}$. So</p>
<p>$$x \ne y \Rightarrow [x] \ne [y]$$
$$[] \ne [a,\dots]$$
$$[a] \ne [b] \Rightarrow [a,x] \ne [b,x]$$</p>
<p>Exclusion</p>
<p>So far you have said what is a set, but you haven't said what isn't a set. For this induction is the most straightforward, especially since you explicitly don't want to use set theory.</p>
|
4,095,361 | <p>Given <span class="math-container">$f : [0, 1] \rightarrow \mathbb{R}$</span> be a bounded function which is continuous on <span class="math-container">$[0,1]$</span> except at <span class="math-container">$1/2$</span>. Let <span class="math-container">$\alpha(x) = x^2$</span>. No further description of function <span class="math-container">$f$</span> is given. How do I show that <span class="math-container">$f$</span> Riemann Stieltjes integrable with respect to <span class="math-container">$\alpha$</span>?</p>
| Rahul Madhavan | 439,353 | <p>Another way to solve this is by ysing the heaviside step function <span class="math-container">$H(x)$</span> and the dirac delta function <span class="math-container">$\delta(x)$</span>:</p>
<p><span class="math-container">\begin{align*}
F(s)=
\begin{cases}
0, & s<0\\
s, & 0\leq s<5/6\\
\frac{5}{6} + \frac{1}{6}\cdot H(x-\frac{5}{6}), & s\geq 5/6
\end{cases}
\end{align*}</span></p>
<p>We can then write the pdf in terms of the dirac delta function, which is<span class="math-container">\begin{align*}f(s)=
\begin{cases}
0, & s<0\\
1, & 0< s<5/6\\
\frac{1}{6}\delta(x-\frac{5}{6}), & s\geq 5/6\\
\end{cases}\end{align*}</span></p>
<p>The expectation is better computed as an integral over
<span class="math-container">\begin{align*}
\mathbb E [X] &= \int\limits_{x=-\infty}^{0} x\cdot \mathbb P[X=x] + \int\limits_{x=0}^{\frac{5}{6}} x\cdot \mathbb P[X=x]+ \int\limits_{x=\frac{5}{6}}^{\infty} x\cdot \mathbb P[X=x]\\
&= \int\limits_{x=-\infty}^{0} x\cdot f[X=x]dx + \int\limits_{x=0}^{\frac{5}{6}} x\cdot f[X=x]dx+ [X=\frac{5}{6}]\cdot\frac{1}{6} \int\limits_{x=\frac{5}{6}}^{\infty}\delta(x-\frac{5}{6})\\
&= \int\limits_{x=-\infty}^{0} x\cdot 0dx + \int\limits_{x=0}^{\frac{5}{6}} x\cdot 1 dx+ \frac{5}{6}\cdot \frac{1}{6}\int\limits_{x=-\infty}^{\infty}\delta(x-\frac{5}{6})\\
&= 0 + \frac{25}{36\cdot 2}+ \frac{5}{6}\cdot \frac{1}{6}\cdot 1\\
&= \frac{35}{72}
\end{align*}</span></p>
<p>Similarly, one can solve <span class="math-container">$\mathbb E [X^2]$</span></p>
<p><span class="math-container">\begin{align*}
\mathbb E [X^2] &= \int\limits_{x=-\infty}^{0} x^2\cdot \mathbb P[X=x] + \int\limits_{x=0}^{\frac{5}{6}} x^2\cdot \mathbb P[X=x]+ \int\limits_{x=\frac{5}{6}}^{\infty} x^2\cdot \mathbb P[X=x]\\
&= 0 + \int\limits_{x=0}^{\frac{5}{6}} x^2\cdot f[X=x]dx+ [X=\frac{5}{6}]^2\cdot\frac{1}{6} \int\limits_{x=\frac{5}{6}}^{\infty}\delta(x-\frac{5}{6})\\
&= \int\limits_{x=0}^{\frac{5}{6}} x^2dx+ \frac{5^2}{6^2}\cdot \frac{1}{6}\int\limits_{x=-\infty}^{\infty}\delta(x-\frac{5}{6})\\
&= 0 + \frac{125}{216\cdot 3}+ \frac{25}{36}\cdot \frac{1}{6}\cdot 1\\
&= \frac{200}{648}=\frac{25}{81}
\end{align*}</span></p>
|
820,490 | <p>I am studying pre-calculus mathematics at the moment, and I need help in verifying if $\sin (\theta)$ and $\cos (\theta)$ are functions? I want to demonstrate that for any angle $\theta$ that there is only one associated value of $\sin (\theta)$ and $\cos (\theta)$. How do I go about showing this?</p>
| Amir Ayupov | 155,158 | <p>It follows from the definition of $\sin(\theta)$ as a ratio of length of $\theta$'s opposite side to hypotenuse (<a href="http://en.wikipedia.org/wiki/Sine" rel="nofollow">Wikipedia</a>).</p>
<p>As long as hypotenuse length is not equal to zero (otherwise the triangle degenerates into line or point), the division is a function which yields only one associated value.</p>
|
549,254 | <blockquote>
<p>$$\frac{1}{3}=.33\bar{3}$$ </p>
</blockquote>
<p>is a rational number, but the $3$ keeps on repeating indefinitely (infinitely?). How is this a ratio if it shows this continuous pattern instead of being a finite ratio? </p>
<p>I understand that $\pi$ is irrational because it extends infinitely <em>without repetition</em>, but I am confused about what makes $1/3=.3333\bar{3}$ rational. It is clearly repeating, but when you apply it to a number, the answers are different: $.33$ and $.3333$ are part of the same concept, $1/3$, yet:</p>
<p>$.33$ and $.3333$ are different numbers: </p>
<p>$.33/2=.165$ and $.3333/2=.16665$, yet they are both part of $1/3$. </p>
<p>How is $1/3=.33\bar{3}$ rational?</p>
| Khosrotash | 104,171 | <p>$$\begin{align}
0.3333333333333\ldots &= 0.3 +0.03 +0.003 +0.0003+ \ldots\\
&=\frac{3}{10} + \frac{3}{100} + \frac{3}{1000}+ \frac{3}{10000} +\ldots
\end{align}$$</p>
<p>If you know the sum of a geometric sequence, that is:
$$a+aq+aq^2+aq^3+\ldots = \frac{a}{1-q} \quad\text{ if $|q| < 1$}$$</p>
<p>you can use it to conclude that for $q = \frac{1}{10}$:</p>
<p>$$\frac{3}{10} + \frac{3}{100} + \frac{3}{1000}+ \frac{3}{10000} +\ldots =\frac{\frac{3}{10}}{1-\frac{1}{10}}=\frac{1}{3}
$$</p>
|
549,254 | <blockquote>
<p>$$\frac{1}{3}=.33\bar{3}$$ </p>
</blockquote>
<p>is a rational number, but the $3$ keeps on repeating indefinitely (infinitely?). How is this a ratio if it shows this continuous pattern instead of being a finite ratio? </p>
<p>I understand that $\pi$ is irrational because it extends infinitely <em>without repetition</em>, but I am confused about what makes $1/3=.3333\bar{3}$ rational. It is clearly repeating, but when you apply it to a number, the answers are different: $.33$ and $.3333$ are part of the same concept, $1/3$, yet:</p>
<p>$.33$ and $.3333$ are different numbers: </p>
<p>$.33/2=.165$ and $.3333/2=.16665$, yet they are both part of $1/3$. </p>
<p>How is $1/3=.33\bar{3}$ rational?</p>
| Michael Hoppe | 93,935 | <p>Eye catcher: $0.123\overline{5678}=\dfrac{12345678-123}{9999000}$.</p>
|
925,558 | <p>How do I prove that the inductive sequence $y_{n+1}= \dfrac {2y_n + 3}{4}$ is bounded? $y(1)=1$</p>
<p><strong>Attempt:</strong> Let us assume that the given sequence is unbounded. : </p>
<p>Then, $y_{n+1} \rightarrow \infty$ either for some finite $n$ or when $n \rightarrow \infty$</p>
<blockquote>
<p>CASE $1 :$ When $|y_n| \rightarrow \infty$ at a finite $n$</p>
</blockquote>
<p>Since : $y_{n}= \dfrac {2y_{n-1} + 3}{4}$, then $y_n \rightarrow \pm \infty \implies y_{n-1} \rightarrow \pm \infty \implies y_{n-2} \rightarrow \pm \infty ~~\cdots$</p>
<p>This ultimately means $y_2 \rightarrow \pm \infty$ which is not true.</p>
<p>Hence, there does not exist a finite $n$ for which $y_n \rightarrow \pm \infty$.</p>
<blockquote>
<p>CASE $2:$ When $|y_n| \rightarrow \infty$ at $n \rightarrow \infty$</p>
</blockquote>
<p>I think we can proceed the same way as we did above, i.e inductively, we proceed like above and deduce that if the above assumption is true, then $y_2 \rightarrow \infty$, which is not true.</p>
<p>Is my attempt correct?</p>
<p>Does there exist a proof without induction as well?
Thank you for your help.</p>
| André Nicolas | 6,312 | <p>It is clear that $y_n\gt 0$ for all $n$. Let us prove by induction that $y_n\lt 4$. This is certainly true if $n=1$. And if $y_n\lt 4$, then $\frac{2y_n+3}{4}\lt 4$. </p>
<p>There are smaller upper bounds than $4$, but to prove boundedness we need not find the least upper bound.</p>
|
530,920 | <p>I want to know if there is a way to find for example $\ln(2)$, without using the calculator ?</p>
<p>Thanks </p>
| Mats Granvik | 8,530 | <p>$$\log (x)=\sum _{n=1}^{\infty } \frac{\left(\frac{x-1}{x}\right)^n}{n}$$
when $x>1$</p>
|
63,064 | <p>If one defines on a $\mathbb{R},\mathbb{C}$-vector space a norm this gives rise to a metric. Why are particularly mappings that satisfy the norm axioms so important that in every book for beginners on linear algebra/functional analysis norms are studied ? </p>
<p>Aren't there also other functions that always give rise to a metric, that are worth studying? </p>
<p>What are the properties that a norm-induced metric has, that makes it so special (except being translation-invariant, $d(x+z,y+z)=d(x,y)$, and compatible with scalar multiplication, $d(\lambda x, \lambda y)= |\lambda | d(x,y)$; because I imagine that there would be also other mappings defined on the vector space that give, by some other rule of definition, rise to a translation-invariant,scalar multiplication compatible metric) ? </p>
<p>(<a href="https://math.stackexchange.com/questions/55934/why-do-we-have-the-notions-of-both-norm-and-metric">This</a> question was similar but not really what I was looking for - in case someone would want to redirect me to that question)</p>
| Jadmrial | 10,957 | <p>The standard notion of a metric is weaker than that of a norm- see the discrete metric, for example. By adding the multiplicative condition, we introduce extra structure on our space that makes it nicer and more intuitive. </p>
<p>Indeed, as Asaf Karagila says, the really nice thing about the standard norms on $\mathbb{R}$ and $\mathbb{C}$ is how well they deal with the linear structure on each of these spaces. One way to think about norms is that they're metrics with the additional guarantee that they're multiplicative.</p>
<p>The answer to the last part of the question is precisely as Matthew Davis has said- any metric that's multiplicative automatically gives a norm. There are certainly spaces with norms which satisfy something more than multiplicativity. For example, the norm on a Banach algebra is submultiplicative, ie $|XY|<|X|\cdot|Y|$ for all $X,Y\in B$.</p>
|
3,693,814 | <p>For <span class="math-container">$x,y,z>0.$</span> Prove<span class="math-container">$:$</span> <span class="math-container">$$P={x}^{4}y+{x}^{4}z+3\,{x}^{3}{y}^{2}-11\,{x}^{3}yz+3\,{x}^{3}{z}^{2}+3
\,{x}^{2}{y}^{3}+3\,{x}^{2}{y}^{2}z+3\,{x}^{2}y{z}^{2}+3\,{x}^{2}{z}^{
3}+x{y}^{4}-11\,x{y}^{3}z+3\,x{y}^{2}{z}^{2}\\-11\,xy{z}^{3}+x{z}^{4}+{y
}^{4}z+3\,{y}^{3}{z}^{2}+3\,{y}^{2}{z}^{3}+y{z}^{4} \geqq 0$$</span>
There are many SOS way for <span class="math-container">$P,$</span> any one can find<span class="math-container">$?$</span></p>
<p>For example<span class="math-container">$,$</span></p>
<p><em>NguyenHuyen</em> use <em>fsos</em> function and gave the following expression<span class="math-container">$:$</span>
<a href="https://i.stack.imgur.com/XYk9o.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XYk9o.png" alt="enter image description here"></a>
And also<span class="math-container">$:$</span> <span class="math-container">$$P=\frac{1}{4} \sum\limits_{cyc} \,z \left( x-y \right) ^{4}+\frac{3}{4} \sum\limits_{cyc} \, \left( {x}^{2}+{y}^{2}+4\,{z}^{2}
\right) \left( x-y \right) ^{2}z$$</span></p>
<p><span class="math-container">$(\ast)$</span> Result by <em>SBM</em><span class="math-container">$:$</span>
<span class="math-container">$$P=\frac{3}{2} \sum\limits_{cyc} \,z \left( xy+2\,{z}^{2} \right) \left( x-y \right) ^{2}+\sum\limits_{cyc} z \left(
x-y \right) ^{4}$$</span></p>
| Michael Rozenberg | 190,319 | <p>There is also the following SOS:
<span class="math-container">$$\sum_{cyc}(x^4y+x^4z+3x^3y^2+x^3z^2-11x^3yz+3x^2y^2z)=$$</span>
<span class="math-container">$$=\sum_{cyc}(x^4y+x^4z+3x^3y^2+3x^3z^2-18x^3yz+10x^2y^2z)+\sum_{cyc}7x^3yz-7x^2y^2z)=$$</span>
<span class="math-container">$$=\sum_{cyc}z(x-y)^2(x+y-3z)^2+7.5xyz\sum_{cyc}(x-y)^2\geq0.$$</span></p>
|
2,277,117 | <p>For $m = 2$, the fraction is $\frac{3}{4}$. for $m=3$, the fraction is$\frac{8}{15}$. I was wondering why numerators and demoninators of $\frac{1}{m} + \frac{1}{m+2}$ show primitive pythagorean triples a and b.</p>
| helloworld112358 | 300,021 | <p>Notice that $$\frac{1}{m}+\frac{1}{m+2}=\frac{2m+2}{m^2+2m}.$$ We can check that $$(m^2+2m)^2+(2m+2)^2=m^4+4m^3+4m^2+4m^2+8m+4=m^4+4m^3+8m^2+8m+4=(m^2+2m+2)^2.$$ Note that this is always a primitive triple if you put the fraction in lowest form, but $2m+2$ and $m^2+2m$ are not coprime in general (and thus the triple $2m+2,m^2+2m, m^2+2m+2$ is not primitive in general). </p>
|
2,277,117 | <p>For $m = 2$, the fraction is $\frac{3}{4}$. for $m=3$, the fraction is$\frac{8}{15}$. I was wondering why numerators and demoninators of $\frac{1}{m} + \frac{1}{m+2}$ show primitive pythagorean triples a and b.</p>
| Jay Zha | 379,853 | <p>$$\frac{1}{m} + \frac{1}{m+2} = \frac{2m+2}{m^2 + 2m}$$</p>
<p>Now notice that</p>
<p>$$(2m+2)^2 + (m^2 + 2m)^2=8m^2+4m+4+m^4+4m^3=(m^2+2m+2)^2$$</p>
|
1,545,019 | <p>I'm sure this is probably an extremely simple problem but I'm stuck figuring this out.<br>
For example: </p>
<p>$(\frac{1}{5})^{x} + (\frac{7}{10})^{x} = 1$</p>
<p>What would be the steps to solve for x?</p>
| Brevan Ellefsen | 269,764 | <p>This is a hard problem. There is no general closed form for $a^x + b^x = c$, and the best way is probably for you to use something like Newton's method to get an approximation. Here is a <a href="https://math.stackexchange.com/questions/1371722/solve-axbx-c-for-x">link</a> to an answer outlining this</p>
|
3,769,222 | <p>Evaluate: <span class="math-container">$\lim_{n\to \infty } \int_0^n (1-\frac{x}{n})^n \cos(\frac{x}{n})dx$</span></p>
<p>For fixed <span class="math-container">$n$</span>, I first rewrite my integral as
<span class="math-container">$$\int_0^n \left(1-\frac{x}{n}\right)^n \cos\left(\frac{x}{n}\right)dx = \int_0^\infty \left(1-\frac{x}{n}\right)^n \cos\left(\frac{x}{n}\right) \chi_{[0,n]}(x) dx,$$</span>
where <span class="math-container">$\chi_{[0,n]}(x)$</span> is the characteristic function.</p>
<p>We then have that <span class="math-container">$$\lim_{n\to \infty}\left(1-\frac{x}{n} \right)^n = e^{-x},$$</span> <span class="math-container">$$\lim_{n\to \infty}\cos\left(\frac{x}{n}\right) = 1,$$</span> <span class="math-container">$$\lim_{n\to \infty}\chi_{[0,n]}(x) = 1.$$</span></p>
<p>So that <span class="math-container">$$\lim_{n\to \infty}\left(1-\frac{x}{n}\right)^n \cos\left(\frac{x}{n}\right) \chi_{[0,n]}(x) = e^{-x}.$$</span></p>
<p>I'd now like to just quote the dominated convergence theorem and integrate <span class="math-container">$e^{-x}$</span> to get the limit of the integral. But I can't think of an integrable function that dominates the sequence of functions.</p>
<p>Any thoughts would be greatly appreciated.</p>
<p>Thanks in advance.</p>
| juancodmw | 517,676 | <p>You can see that <span class="math-container">$\int_{0}^n (1-\frac{x}{n})^n\cos{(\frac{x}{n})}dx = \int_{0}^{1}n(1-t)^{n}\cos{(t)}dt$</span> and you can integrate by parts</p>
|
1,125 | <p>I know volume preserving diffeomorphisms of a $sphere^2$ make a grou'p sdiff($S_2$). I would to know if it is a Lie group, which I assume if it is that makes interpolation easier (like with rotations). </p>
<p>So that is one question, is it a Lie group?</p>
<p>Also is the group path connected? If so, how can I interpolate between two elements in the group?</p>
<p>These are not subjects I know very little about. I apologize if I'm phrasing it in some way that sounds ridiculous. </p>
| Jason DeVito | 331 | <p>First, as others have pointed out, the group of volume preserving diffeomorphisms will be infinite dimensional.</p>
<p>For the second question, there is a beautiful technique known as Moser's trick which answers it. Moser's trick, in fancy language, says that if <span class="math-container">$(M,w)$</span> and <span class="math-container">$(M,w')$</span> are two symplectic structures on the same manifold, and if <span class="math-container">$[w] = [w']$</span> in <span class="math-container">$H^2(M;R)$</span> (de Rham cohomology), then there is a family of diffeomorphism <span class="math-container">$f_t:M \to M$</span> with <span class="math-container">$f_0 = Id$</span> and such that <span class="math-container">$f_1$</span> pulls <span class="math-container">$w'$</span> back to <span class="math-container">$w$</span>.</p>
<p>For a <span class="math-container">$2$</span>-dimensional compact, oriented manifold (like the sphere), we have <span class="math-container">$H^2(M;R) = R$</span> (the real numbers), and a symplectic form is nothing but a nonzero element in <span class="math-container">$R$</span> (which can be interpreted as the total volume). Since in this setting, <span class="math-container">$[w] = [w']$</span> iff they both give the same signed volume, it follows from Moser's trick that the group of (signed) volume preserving maps is connected.</p>
<p>If we consider unsigned volume, there will be <span class="math-container">$2$</span> components to the group diffeomorphisms preserving the unsigned volume. This is because, as others have pointed out, one has the notion of "degree" which shows the map <span class="math-container">$x \to -x$</span> is not homotopic to Id, even through just continuous maps (not neccesarily volume preserving). This shows there are AT LEAST two componenets. There are at most two components because every volume preserving diffeo can be connected to Id or (<span class="math-container">$x \to -x$</span>) by Moser's trick again.</p>
<hr>
<p>Edit: I misspoke a little bit. Moser's trick says that if you have a family w_t of symplectic forms, then there is a family of diffeomorphisms as I described above. It's not clear to me that what I said (that it's enough to have <span class="math-container">$[w] = [w']$</span> in <span class="math-container">$H^2$</span>) is enough to guarantee that there is a family of symplectic forms connecting them. Further, it seems that Moser's trick only guarantees you have a path of diffeos which starts and ends at a volume preserving diffeo, but may not preserve volume for all time.</p>
<p>However, in the case of <span class="math-container">$S^2$</span> (or any closed, orientable <span class="math-container">$2$</span>-manifold), I can patch things up. Given <span class="math-container">$w$</span> and <span class="math-container">$w'$</span>, volume forms, with <span class="math-container">$[w] = [w']$</span> (i.e, they have the same volume), then the form <span class="math-container">$w_t = tw + (1-t)w'$</span> is a path of symplectic forms which connects them. For a fixed <span class="math-container">$t$</span>, the form <span class="math-container">$w_t$</span> is closed since it's a sum of closed forms (or, even easier, because it has top degree), and is nondegenerate because it's a volume form (integration shows the volume given is that of <span class="math-container">$w$</span>). The fact that the volume is constant for each <span class="math-container">$w_t$</span> implies that the path of diffeos preserves volume for all time.</p>
<p>(I wasn't able to immediately convince myself that in general, the convex sum of symplectic forms was nondegenerate, hence my initial hesitation. In fact, I think that it need not be nondegenerate.)</p>
|
271,554 | <p>Is $\chi_{p(\displaystyle\lim_{n\to\infty} f_n)}=\displaystyle\lim_{n\to\infty}\chi_{p(f_n)}$, assuming $\lim f_n$ exists? Here $\chi_{p(f)}$ is $1$ for the set where the proposition $p$ on the function $f$ is true and $0$ for otherwise. Here the functions are $[0,1]\to\mathbb{R}$ and the limits are almost everywhere convergence.</p>
| Did | 6,179 | <p>No. Try $p=$ "being bounded" and $f_n(x)=\min\{n,1/x\}$.</p>
|
74,290 | <p>I have this function $f(x)$ which is continuous and differentiable on $\mathbb R$. Is the following true - without the assumption that $f$ being absolutely continuous!</p>
<p>$$ \int_{a}^{b} f'(x) dx=f(b)-f(a)$$</p>
<p>Edit: $f$ is infinitely differentiable on $\mathbb R$. I think this may change everything!</p>
| kahen | 1,269 | <p>No, the derivative need not even be integrable! See Wikipedia's article on <a href="http://en.wikipedia.org/wiki/Volterra%27s_function" rel="nofollow">Volterra's function</a>.</p>
<p>PS: If $f$ is differentiable, then $f$ is continuous.</p>
|
1,443,166 | <blockquote>
<p>If the proposition ¬p→v is true, then the truth value of the
proposition ¬p∨(p→q), where ¬ is negation, ∨ is inclusive OR and → is
implication, is</p>
<ol>
<li>True</li>
<li>False</li>
<li>Multiple Values</li>
<li>Cannot be determined</li>
</ol>
</blockquote>
<hr>
<p><strong>I try to explain</strong></p>
<hr>
<p>Given ,¬p→v = p+v is valid
now , proposition ¬p∨(p→q) = ¬p+q ,</p>
<p>hence using rule of inference ,
= (p+v)→(¬p+q) = ¬p.¬v + ¬p + q = ¬p+q ,</p>
<p>so it cannot be determined.</p>
<p><strong>Edit :</strong> Is my method correct ?</p>
| Mauro ALLEGRANZA | 108,274 | <p>The formula $¬p→v$ is True in two cases :</p>
<blockquote>
<p>either when $p$ is True (i.e. $\lnot p$ False) or when $v$ is True.</p>
</blockquote>
<p>Thus, the assumption does not force a truth-value for $p$.</p>
<p>Consider now $¬p∨(p→q)$; we have three cases :</p>
<p><em>(i)</em> $q$ True; in this case $p→q$ is True, irrespective of the truth-value of $p$, and so $¬p∨(p→q)$ is <strong>T</strong>.</p>
<p><em>(ii)</em> $q$ False; now two subcases :</p>
<ul>
<li><p><em>(ii-a)</em> $p$ True; in this case $p→q$ is False, and so $¬p∨(p→q)$ is <strong>F</strong>.</p></li>
<li><p><em>(ii-b)</em> $p$ False; in this case $¬p∨(p→q)$ is <strong>T</strong>.</p></li>
</ul>
<hr>
<p><em>Conclusion</em> : you are right, cannot be determined.</p>
<p><em>Note</em> : please, note that the previous explanation is nothing more than the description in words of the truth-table entered for $p,q,v$, observing that in all the six rows with <strong>T</strong> for the first formula, the second one has different values.</p>
<hr>
<p>Regarding your solution : your transformations are correct, but what are they aimed to ? </p>
<p>The $v$ in $\lnot p \to v = p \lor v$ will not disappear, and thus you will not be able to reach the second formula.</p>
<p>But this "impossibility" (quite evident) is not easy to prove that way.</p>
|
76,457 | <p>I have an ellipse centered at $(h,k)$, with semi-major axis $r_x$, semi-minor axis $r_y$, both aligned with the Cartesian plane.</p>
<p>How do I determine if a point $(x,y)$ is within the area bounded by the ellipse? </p>
| Srivatsan | 13,425 | <p>The region (disk) bounded by the ellipse is given by the equation:
$$
\frac{(x-h)^2}{r_x^2} + \frac{(y-k)^2}{r_y^2} \leq 1. \tag{1}
$$
So given a test point $(x,y)$, plug it in $(1)$. If the inequality is satisfied, then it is inside the ellipse; otherwise it is outside the ellipse. Moreover, the point is on the boundary of the region (i.e., on the ellipse) if and only if the inequality is satisfied tightly (i.e., the left hand side evaluates to $1$). </p>
|
76,457 | <p>I have an ellipse centered at $(h,k)$, with semi-major axis $r_x$, semi-minor axis $r_y$, both aligned with the Cartesian plane.</p>
<p>How do I determine if a point $(x,y)$ is within the area bounded by the ellipse? </p>
| Victor Engel | 49,770 | <p>I have another solution. In summary, transform everything so you can test whether a point is within a circle centered at (0,0). Since the ellipse is oriented orthogonally to the Cartesian plane, we can simply scale one of the dimensions by the quotient of the two axes.</p>
<p>First subtract (h,k) from both points.</p>
<p>(h,k) becomes (0,0).
(x,y) becomes (x-h,y-k).</p>
<p>Now we scale the second coordinate, normalizing it to the first.</p>
<p>Scaling (x-h,y-k) we get
$$
(x-h,(y-k) * \frac{r_x}{r_y})
$$</p>
<p>Now we simply need to test if
$$
|(x-h,(y-k) * \frac{r_x}{r_y})| <= r_x
$$</p>
|
211,123 | <p>I am writing some tests and would like to introduce large chunks of random alphabetic text (just characters a-z) to the input. The way I am generating the text now is like this:</p>
<pre><code>RandomString[length_Integer] :=
StringJoin[
Table[
FromLetterNumber[RandomInteger[{1, 26}]],
length
]
]
</code></pre>
<p>This works great but is slow: <code>AbsoluteTiming[RandomString[100000]]</code> shows it running in 6.69763 seconds, which is too slow to run on lots of tests.</p>
<p>Does anyone know of a faster way to generate random text?</p>
| MarcoB | 27,951 | <p>The following will give you a long string of text using common words separated by a space; it is very fast after the first execution (which loads some indices):</p>
<pre><code>StringRiffle@RandomWord["CommonWords", 100]
</code></pre>
<blockquote>
<p>"dexterous calibration ethical nocturnal misfortune ruining commodious refreshing gable arithmetic sacristy doorknocker thread measles pittance disrupted chorister discharge arbitrarily midday vainglory anvil walking spotty philosopher plutonium balboa lynx resignedly distinct python photostat platypus okra predetermination
fanlight knickknack scuttlebutt adios silents forewarn business carbonic pigeonhole motivation diagnostician awareness flushed sentimentally painting shirring impeccable proficiency racketeering snack soulfully excision chaplain writer anthologist psychologically fulfillment televangelist Thursday gibbon broadcasting hideously recombination voucher brownstone absentee blockbuster freshly unpasteurized freethinker womanliness gender bellwether overrating pronoun tarmacadam reachable hypothermia elsewhere knead egress exonerate nuts slanted hand-to-hand pyramidal visualize characterize thriftless strap quite tourney baryon alarm consanguineous"</p>
</blockquote>
<p>Alternatively, you can also use the built in Lorem Ipsum in the example data collection:</p>
<pre><code>ExampleData[{"Text", "LoremIpsum"}]
</code></pre>
|
898,082 | <p>I have the answer for this, but my teacher hadn't taught the whole "when cosine is an even, the value of $-\arccos (-0.7)$ is a solution too." </p>
<p>Please: </p>
<p>-tell me when a $\cos$/$\sin$ function is even/odd </p>
<p>-what happens if its odd? </p>
<p>-how to use the "$±\arccos(-0.7) + 2kπ$" ( don't understand why you add 2kπ) </p>
<p>-and how to find the solutions! </p>
<p>Another example is $$\sec( x )= -3, -π ≤ x < π $$</p>
<p>I really don't understand what happens if the function is "even" or "odd." And how to determine if it is.</p>
| drhab | 75,923 | <p>I restrict to a solution of the title of your question.</p>
<p>In general $\cos x=\cos\alpha$ if and only if $\frac{x-\alpha}{2\pi}$
is an integer or $\frac{x+\alpha}{2\pi}$ is an integer. </p>
<p>Another way
to express that is: if and only if $x=\pm\alpha+2k\pi$ where $k$
stands for an integer. Take notice of sign $\pm$.</p>
<p>If you have equation $\cos x=-0.7$ then this can be used.
To find is some $\alpha$ that satisfies $\cos\alpha=-0.7$ and here
function $\arccos$ comes in: $\cos\alpha=-0.7$ is true for $\alpha=\arccos\left(-0.7\right)$.</p>
<p>This leads to $x=\pm\arccos\left(-0.7\right)+2k\pi$ and as last step
you must find all integers $k$ that satisfy $\arccos\left(-0.7\right)+2k\pi\in[2\pi,4\pi)$
or $-\arccos\left(-0.7\right)+2k\pi\in[2\pi,4\pi)$.</p>
<p>This solves the equation $\cos x=-0.7$ under the extra condition that
$x\in[2\pi,4\pi)$.</p>
|
1,715,324 | <p>I am wondering what is the difference between algebraic sets and algebraic varieties in complex projective space.</p>
<p>It seems that both are zero sets of polynomials, so what is the difference?</p>
| silvascientist | 108,405 | <p>According to <a href="https://en.wikipedia.org/wiki/Algebraic_variety" rel="nofollow">Wikipedia</a>, some sources require that a variety be irreducible, which means that it is not the union of two smaller sets that are closed in the <a href="https://en.wikipedia.org/wiki/Zariski_topology" rel="nofollow">Zariski topology</a>. In this case, a non-irreducible algebraic variety is called an <em>algebraic set</em>. In other sources, this convention is not followed and both cases are simply called <em>variety</em>.</p>
|
102,966 | <p>Let $T^*$ denote upper triangular matrices (of the appropriate size) with positive diagonal entries and $\mathrm{UT}$ upper triangular matrices with all diagonal entries equal to 1.</p>
<blockquote>
<p>Does every (abstract group) embedding $\varphi:\mathrm{UT}(n,\mathbb{R})\to\mathrm{UT}(m,\mathbb{R})$ extend to $\bar{\varphi}:T^*(n,\mathbb{R})\to T^*(m,\mathbb{R})$?</p>
</blockquote>
<p>(In the other direction, any $\bar{\varphi}:T^*(n,\mathbb{R})\to T^*(m,\mathbb{R})$ restricts to a homomorphism $\mathrm{UT}(n,\mathbb{R})\to\mathrm{UT}(m,\mathbb{R})$, since $\mathrm{UT}$ is the derived subgroup of $T^*$.)</p>
<p>An affirmative answer to this question implies a relatively easy affirmative answer to <a href="https://mathoverflow.net/questions/93091/free-affine-actions-of-borel-subgroups">this question</a>. I've recently answered the latter independently, but wonder if there's a general result that could be used, and which I should know about. I asked the current question on Math.stackexchange but without success.</p>
<p>EDIT: Florian Eisele has shown that as stated above the answer to the original question is no. This makes me wonder if there's a reasonably natural reformulation for which the answer is yes. For the sake of asking a concrete question, let me hazard the following.</p>
<blockquote>
<p>Does every embedding $\varphi:\mathrm{UT}(n,\mathbb{Q})\to\mathrm{UT}(m,\mathbb{Q})$ extend to $\bar{\varphi}:T^*(n,\mathbb{Q})\to T^*(m,\mathbb{Q})$?</p>
</blockquote>
| YCor | 14,094 | <p>Here's a counterexample ($m=n=3$) which is a continuous homomorphism and actually probably also works for your second question with $\mathbf{Q}$.</p>
<p>In short: most automorphisms of $UT(3)$ do not extend to $T^*(3)$.</p>
<p>Since I deal with continuous automorphisms, it boils down to a Lie algebra problem. An automorphism of $UT(3)$ induces an automorphism of its abelianized subgroup, giving a $2\times 2$ matrix and every invertible matrix occurs this way. But my claim is that if the automorphism extends to $T^*(3)$ then (some power of) this matrix has to be diagonal.</p>
<p>Sketch of argument: I work upside down by considering an automorphism of $T^1(3)$, the group of determinant 1 matrices in $T^*(3)$ (it's enough and more convenient to deal with it) and describe its restriction to $UT(3)$.</p>
<p>First, because $UT(3)$ is the derived subgroup, it is stable. Write $T^1(3)=D.UT(3)$ where $D$ is the group of diagonal matrices. Since $D$ is a Cartan subgroup in $T^1(3)$ (that is, its Lie algebra is nilpotent and self-normalized, see Bourbaki, Groups and Lie algebras) and is unique up to conjugacy, it is mapped to a conjugate subgroup. So after composing by a conjugation, we can suppose $D$ mapped into itself. Now any automorphism of $T^1(3)$ induces a finite order automorphism of $D$: indeed, any one-parameter group of automorphisms of $T^1(3)$ induces the identity on $D$, as we see using the fact that the derived subgroup of the semidirect product $\mathbf{R}\ltimes T^1(3)$ is nilpotent (hence contained in the nilpotent radical $UT(3)$ of $T^1(3)$). So after conjugation, some power of the automorphism acts on $D$ as the identity and therefore preserves the weight decomposition of the Lie algebra of $UT(3)$ induced by the action of $D$, and thus induces a diagonal automorphism of the derived subgroup of $UT(3)$. </p>
<p>If you don't follow the argument you can also compute the group of continuous automorphisms of $T^*(3)$ and check that no one maps a matrix $\pmatrix{1 & x & * \cr 0 & 1 & y \cr 0 & 0 & 1\cr}$ to a matrix of the form $\pmatrix{1 & x+y & * \cr 0 & 1 & y \cr 0 & 0 & 1\cr}$.</p>
<p>NB: it's unclear from your question if you mean "extends as an automorphism" or "extends as an endomorphism". However, it is not hard to show that an endomorphism of $T^1(3)$ that is injective on $UT(3)$ remains injective on $T^1(3)$, and it easily follows that an automorphism of $UT(3)$ extends as an endomorphism of $T^*(3)$ iff it extends as an automorphism.</p>
|
440,744 | <p>The vector space dimension of the cohomology group of the <span class="math-container">$2$</span>-plane Grassmannian <span class="math-container">$\mathrm{Gr}_{2,n}$</span> is given by the number of tuples <span class="math-container">$(\lambda_1,\lambda_2)$</span> satisfying
<span class="math-container">$$
n - 2 \geq \lambda_1 \geq \lambda_2 \geq 0.
$$</span>
Explicitly this is given by
<span class="math-container">$$
\binom{n}{2}.
$$</span>
This also happens to be the dimension of <span class="math-container">$V_{\pi_2}$</span> the second fundamental representation of <span class="math-container">$\frak{sl}_n$</span>. I am guessing this is not an accident, especially since the <span class="math-container">$2$</span>-plane Grassmannian corresponds (in the usual way) to <span class="math-container">$V_{\pi_2}$</span>.</p>
<p>Does this extend to the general identity
<span class="math-container">$$
\mathrm{dim}(H^{*}(\mathrm{Gr}_{d,n})) = \mathrm{dim}(V_{\pi_d})?
$$</span>
If it does, then what is a conceptual explanation for this?</p>
<p>EDIT: Since <span class="math-container">$V_{\pi_d}$</span> is isomorphic to the exterior power
<span class="math-container">$$
\Lambda^d(V_{\pi_1})
$$</span>
and <span class="math-container">$V_{\pi_1}$</span> is of dimension of <span class="math-container">$n$</span>, we see that the RHS of the claimed identity is the binomial coefficient
<span class="math-container">$$
\binom{n}{d}.
$$</span>
It follows from the general formula given in this <a href="https://mathoverflow.net/questions/196546/hard-lefschetz-theorem-for-the-flag-manifolds">answer</a> that the LHS is the same binomial coefficient. Thus the identity does indeed extend from <span class="math-container">$2$</span>-planes to <span class="math-container">$d$</span>-planes. So the question is if there is a conceptual reason for this . . .</p>
| Vladimir Dotsenko | 1,306 | <p>I accidentally (looking for something else) came across another paper where a very elegant explanation is given:</p>
<p><a href="https://www.jstor.org/stable/24903230" rel="nofollow noreferrer">Dan Laksov, Anders Thorup: Schubert Calculus on Grassmannians and Exterior Powers</p>
<p>Indiana University Mathematics Journal, Vol. 58, No. 1 (2009), pp. 283-300</a></p>
<p>They introduce the "<span class="math-container">$d$</span>-th factorization algebra" that "controls" the ways to factor a polynomial <span class="math-container">$p(x)$</span> of degree <span class="math-container">$n$</span> into a product of factors of degrees <span class="math-container">$d$</span> and <span class="math-container">$n-d$</span>, relate the <span class="math-container">$d$</span>-th exterior power of the quotient by <span class="math-container">$p(x)$</span> to the <span class="math-container">$d$</span>-th factorization algebra (Main Theorem, see Sec.0.6), and also relate the <span class="math-container">$d$</span>-th factorization algebra to the Schubert calculus (Sections 4 and 5).</p>
|
1,110,122 | <p>Prove $$\large\int_{-\pi}^{\pi}\sin (\sin x) \,dx =0$$ without using the fact that $\sin(x)$ is odd.</p>
<p>Computing this in wolfram says that it is uncomputable, which leads me to believe that the only way to find this would be methods for solving definite integrals. I am wondering if it is possible with any other techniques such as DUIS or residues?</p>
| Robert Israel | 8,508 | <p>$$\int_{-\pi}^\pi \sin(\sin(x)) \; dx = \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!}\int_{-\pi}^\pi \sin^{2n+1}(x)\; dx$$ (the interchange of sum and integral justified by uniform absolute convergence)
so it suffices to show that $\int_{-\pi}^\pi \sin^{2n+1}\; dx = 0$ for nonnegative integers $n$.
Now using the substitution $u = \cos(x)$,
$$\int_{-\pi}^\pi \sin^{2n+1}(x)\; dx = \int_{-\pi}^\pi (1 - \cos^2(x))^n \sin(x)\; dx = -\int_{-1}^{-1} (1-u^2)^n \; du = 0 $$</p>
|
1,110,122 | <p>Prove $$\large\int_{-\pi}^{\pi}\sin (\sin x) \,dx =0$$ without using the fact that $\sin(x)$ is odd.</p>
<p>Computing this in wolfram says that it is uncomputable, which leads me to believe that the only way to find this would be methods for solving definite integrals. I am wondering if it is possible with any other techniques such as DUIS or residues?</p>
| mickep | 97,236 | <p>I think this question is a bit funny (and in principal, I think that any calculation leading to the desired conclusion uses that sine is odd in one or another way). </p>
<p>Are we allowed to use $\sin(t\pm\pi)=-\sin(t)$ and that $\sin t=\frac{\exp(it)-\exp(-it)}{2i}$ (I agree that it follows from the second identity directly that sine is odd. But in some sense it is not worse than saying that $\sin t=\text{Im}\,\exp(it)$ and then use that $\overline{\exp(it)}=\exp(-it)$)? </p>
<p>Nevertheless, here are some calculations using these formulas, leading to the desired result:</p>
<p>We divide the integral in two pieces
$$
I=\int_{-\pi}^\pi \sin(\sin x)\,dx = \int_{-\pi}^0\sin(\sin x)\,dx+\int_0^{\pi}\sin(\sin x)\,dx.
$$
Performing the change of variables $t=x+\pi$ and $t=x-\pi$ respectively in these integrals give
$$
\begin{align}
I&=\int_0^{\pi}\sin(\sin(t-\pi))\,dt+\int_{-\pi}^0 \sin(\sin(t+\pi))\,dt\\
&=\int_{-\pi}^{\pi} \sin(-\sin t)\,dt.
\end{align}
$$
Next, we use that $\sin z=\frac{\exp(iz)-\exp(-iz)}{2i}$,
$$
\begin{align}
I & = \int_{-\pi}^\pi \sin(\sin x)\,dx\\
& = \int_{-\pi}^{\pi}\frac{\exp(i\sin x)-\exp(-i\sin x)}{2i}\,dx\\
& = -\int_{-\pi}^{\pi} \frac{\exp(-i\sin x)-\exp(i\sin x)}{2i}\,dx\\
& = -\int_{-\pi}^{\pi} \sin(-\sin x)\,dx\\
& = -I.
\end{align}
$$
Thus $I=0$.</p>
|
1,616,460 | <blockquote>
<p>If $p(x)$ be a quadratic equation with real coefficient satisfying $$x^2-2x+2\leq p(x)\leq 2x^2-4x+3\;\forall x\in \mathbb{R}$$ and $p(11) = 181.$ Then $p(16)= $ </p>
</blockquote>
<p>$\bf{My\; Try::}$ We can Write Inequality as $$(x-1)^2+1\leq p(x)\leq 2(x-1)^2+1$$</p>
<p>Here from above inequality we have seen that vertices's of $p(x)$ must lie on $\bf{1^{st}}$ quadrant and upward parabola is formed</p>
<p>Now I did not Understand how can I solve after that</p>
<p>Help me, Thanks</p>
| Future | 299,525 | <p>Put the maps into the diagram. Now take anything in $\ker \iota$, say $n$. By the diagram in Thm 8, it gets sent to $0$ in $S \otimes N$ which, since $\Phi$ is a homomorphism, then gets mapped to $0$ in $L$. Thus, by commutativity of the diagram, so then does $\varphi (n) = 0$ meaning that $n \in \ker \varphi$. </p>
|
690,822 | <p>Let <span class="math-container">$G$</span> be a graph of order <span class="math-container">$n$</span> and let <span class="math-container">$k$</span> be an integer with <span class="math-container">$1\leq k\leq n-1$</span>. Prove that if <span class="math-container">$\delta(G)\geq (n+k-2)/2$</span>, then <span class="math-container">$G$</span> is <span class="math-container">$k$</span>-connected.</p>
<p>where, <span class="math-container">$\delta(G) = \text{minimum degree of a vertex in } G.$</span></p>
| Calvin Lin | 54,563 | <p><strong>Hint:</strong> Apply Dirac's Theorem.</p>
<p>Let $K\subset G$ be a set of $k-1$ vertices. Consider $G-K$, which has $n-k+1 $ vertices.</p>
<blockquote class="spoiler">
<p> The minimum degree is $ \frac{ n+k-2}{2} - (k-1) = \frac{n-k}{2}.$</p>
</blockquote>
<p>Consider $G-K + v$, where we add a special vertex $v$ that is connected to all vertices of $G-K$. It has $n-k+2$ vertices.</p>
<blockquote class="spoiler">
<p> The minimum degree is $\frac{n-k+2} {2}$.</p>
</blockquote>
<p>Hence, by Dirac's Theorem, a Hamiltonian circuit exists.</p>
<blockquote class="spoiler">
<p> Now delete $v$, and we get a Hamiltonian path. Thus, $G-K$ is connected.</p>
</blockquote>
|
2,367,788 | <p>Need to find the total numbers out of all 6 digit numbers where a digit is repeated exactly 4 times in the number. </p>
<p>Eg. 111122, 111123 is valid
But 111121 is not valid.</p>
| Tony | 258,406 | <p>If you mean any digit combinations as "numbers":</p>
<p>$10\times1\times1\times1\times1\times9\times9+...+10\times9\times9\times1\times1\times1\times1 = 10\times9^2\times \binom64=10\times9^2\times 15=12150$</p>
<p>If the qualities of numbers are meant to be satisfied (leading zeros excluded), simply multiply by $9/10$ as any case that starts with a zero is to be excluded.</p>
|
2,012,080 | <p>Consider the PDE for $v(x,y)$:</p>
<blockquote>
<p>$v_{xx} − v_{xy} − 6v_{yy} − 5v_{x} − 10v_{y} + 25 = 0$</p>
</blockquote>
<p>Taking $\xi = y-2x$ and $\eta=y+3x$ that I found from $b^2-4ac$ I have transformed the equation to the form</p>
<p>$v_{\eta\xi} + v_{\eta}=1$</p>
<p>I have been posed the question:</p>
<blockquote>
<p>Using this result obtain the general solution, casting your answer in terms of
the x and y variables.</p>
</blockquote>
<p>I am not sure how to solve this problem. Could anyone point me in the right direction?</p>
| Mahmoud Hassan | 282,900 | <p>we have :
$$v_{\eta\xi} + v_{\eta}=1$$
Now let $u=v_{\eta} $
$$ u=v_{\eta} \Rightarrow u_{\xi}=v_{\eta\xi}$$
$$\Rightarrow u_{\xi}+u = 1$$
$$\frac{-d_{\xi}}{1}=\frac{-d_{\eta}}{0}=\frac{du}{u-1}$$
which implies :
$$\eta=a \ , \ u-1=be^-{\xi}\Rightarrow u=1+f(\eta)e^-{\xi}$$
Finally :
$$v_{\eta}=1+f(\xi)e^-{\xi}\Rightarrow v=\eta +\int_{\eta}f(s)e^-{\xi}ds+g(\xi) $$
$$ v=\eta+e^-\xi \int_{\eta}f(s)ds+g(\xi) $$
$$ v=\eta +h(\eta)e^-\xi +g(\xi) $$
$$v=x+3y+h(x+3y)e^{-(x-2y)}+g(x-2y)$$</p>
|
3,902,836 | <p>There're similar questions already, but I somewhat struggled to apply their reasoning to this particular statement.</p>
<p>I wanted to ask if my proof is correct (and clarify several things which I've seemingly figured out when writing it here). Here's the proof:</p>
<p><span class="math-container">$$\mbox{(1) }\lim_{x \to 0^-} f\left(\frac1x\right)=l \mbox{ if } \forall\varepsilon>0(\exists\delta>0(\forall x(0<0-x<\delta \longrightarrow|f\left(\frac1x\right)-l|<\varepsilon$$</span></p>
<p><span class="math-container">$$\mbox{(2) }\lim_{x \to -\infty} f(x)=m \mbox{ if } \forall\varepsilon>0(\exists N(\forall x(x<N \longrightarrow|f(x)-m|<\varepsilon$$</span></p>
<p>We need to show that <span class="math-container">$l=m$</span>.</p>
<p>from (1) we have
<span class="math-container">$$-\delta>x>0\longrightarrow|f\left(\frac1x\right)-l|<\varepsilon$$</span></p>
<p>let <span class="math-container">$g(x)=f\left(\frac1x\right)$</span>, then we have
<span class="math-container">$$-\delta>x>0\longrightarrow|g\left(x\right)-l|<\varepsilon$$</span></p>
<p>in (2) we suppose that <span class="math-container">$x<N$</span>. Hence <span class="math-container">$\frac1x>\frac1N$</span>. Let <span class="math-container">$x'=\frac1x$</span>. Furthermore, since <span class="math-container">$-\delta$</span> is negative and we can assume that <span class="math-container">$N$</span> is negative. (If it's not, we can take <span class="math-container">$N$</span> to be -1 or any other negative number, because if the conclusion holds <span class="math-container">$\forall x<N$</span>, it surely holds for a subset <span class="math-container">$(-\infty,-1)$</span>). We can take <span class="math-container">$-\delta=\frac1N$</span>, which gives us</p>
<p><span class="math-container">$$\frac1N=-\delta<x'<0$$</span> and this is the hypothesis from (1). Hence we conclude that</p>
<p><span class="math-container">$$|g\left(x'\right)-l|<\varepsilon$$</span> or <span class="math-container">$$|f\left(\frac1{\frac1x}\right)-l|<\varepsilon$$</span> or <span class="math-container">$$|f\left(x\right)-l|<\varepsilon$$</span></p>
<p>To summarise, <span class="math-container">$x<N\longrightarrow|f\left(x\right)-l|<\varepsilon$</span>. By the definition of the limit <span class="math-container">$$\lim_{x \to -\infty}f(x)=l$$</span> but also <span class="math-container">$$\lim_{x \to -\infty}f(x)=m$$</span> Therefore <span class="math-container">$l=m$</span>.</p>
| WA Don | 542,712 | <p>The proof you give becomes a bit unwieldy and it does not make clear that <span class="math-container">$N$</span> must be negative.</p>
<p>I think it can be shortened as follows:</p>
<hr />
<p>Assume <span class="math-container">$\lim_{x\to0^-} f(1/x)$</span> exists and equals <span class="math-container">$\ell$</span>. Then take any <span class="math-container">$\varepsilon > 0$</span>.</p>
<p>Because the limit above exists, there exists <span class="math-container">$\delta > 0$</span> such that
<span class="math-container">$$\lvert f(1/x) - \ell\rvert \leqslant \varepsilon \text{ for all } -\delta \leqslant x < 0. \tag{1}\label{eq1}$$</span></p>
<p>Now take any <span class="math-container">$y \leqslant -1/\delta$</span>, write <span class="math-container">$x=1/y$</span>. It follows <span class="math-container">$-\delta \leqslant x < 0$</span>. This satisifies the criterion in \eqref{eq1}, so <span class="math-container">$\lvert f(1/x)-\ell \rvert \leqslant \varepsilon$</span>, which is the same as <span class="math-container">$\lvert f(y) - \ell\rvert \leqslant \varepsilon$</span>.</p>
<p>Thus for every <span class="math-container">$\varepsilon > 0$</span> there exists <span class="math-container">$m$</span> (taking <span class="math-container">$m=1/\delta$</span>) such that <span class="math-container">$\lvert f(y)-\ell\rvert \leqslant \varepsilon$</span> whenever <span class="math-container">$y \leqslant -m$</span>. This provbes <span class="math-container">$\lim_{y\to-\infty}f(y)$</span> exists and equals <span class="math-container">$\ell$</span>.</p>
|
251,799 | <p>I have a PDE in the form of
$$
\frac{\partial u}{\partial t} + \frac{\partial u}{\partial x} + u = \delta(x-1),
$$
with initial condition $u(x,0)=100$. I'm trying to solve it numerically, but I have no idea on which method should I use. Most of the examples that I refer to has no delta function in the PDE.</p>
<p>Can anyone guide me on what should I do?</p>
<p>Thank you very much. </p>
| Pragabhava | 19,532 | <p>You can restate the problem by integrating the equation in $x \in (1-\epsilon,1+\epsilon)$ and taking $\epsilon \to 0$.</p>
<p>$$
\int_{1-\epsilon}^{1+\epsilon} \left(\frac{\partial u}{\partial t} + \frac{\partial u}{\partial x} + u\right) dx = \int_{1-\epsilon}^{1+\epsilon} \delta(1-x)dx = 1.
$$</p>
<p>Now we have to make the assumption that we can exchange the temporal derivative with the integral, and then
$$
\frac{\partial}{\partial t} \left(\int_{1-\epsilon}^{1+\epsilon} u dx \right) + \int_{1-\epsilon}^{1+\epsilon}\frac{\partial u}{\partial x} dx + \int_{1-\epsilon}^{1+\epsilon} u dx = 1
$$
Taking the limit as $\epsilon \to 0$ and assuming it can be exchanged with the time differentiation (and $u$ is integrable I think, I'm not so fresh on my analysis courses) we have that
$$
\lim_{\epsilon \to 0} \big[u(1+\epsilon, t) - u(1-\epsilon,t)\big] = 1
$$
or, in other words, that the jump of the function $u$ at the point $x = 1$ is of magnitude $1$.</p>
<p>Finally, solve for $0 < x < 1$, $1 < x < 1000$ and glue the solutions using the jump condition.</p>
<h2>Analytic solution</h2>
<p><strong>For $0 < x < 1$</strong> you have that
$$
\frac{\partial u}{\partial t} + \frac{\partial u}{\partial x} + u = 0
$$
with initial conditions $u(x,0) = 100$.</p>
<p>Using the method of characteristics you have the equivalent system
\begin{align}
\frac{d x}{d \eta} &= 1 &x(\xi)\big|_{\eta = 0} &= \xi\\
\frac{d t}{d \eta} &= 1 &t(\xi)\big|_{\eta = 0} &= 0\\
\frac{d u}{d \eta} &= -u &u(\xi)\big|_{\eta = 0} &= 100
\end{align}
wich can easily be solved, leading to
\begin{align}
x(\xi, \eta) &= \eta + \xi \\
t(\xi,\eta) &= \eta \\
u(\xi,\eta) &= 100e^{-\eta}
\end{align}
and then we can invert for $(\xi,\eta)$ , leading to the solution $u(x,t) = 100e^{-t}$. On $x=1$, $u(1,t) = 100e^{-t}$, the jump condition states that
$$
u(1^+,t) = 1 + 100e^{-t}
$$
and then,</p>
<p><strong>For $1 < x < 1000$</strong> the problem is
$$
\frac{\partial u}{\partial t} + \frac{\partial u}{\partial x} + u = 0
$$
with $u(x,0) = 100$, and $u(1,t) = 1 + 100e^{-t}$. Clearly, there is an inconsistency in $t=0$, $x=1$, derived from $\delta(x-1)$, and the solution is not well defined there. Given that the PDE is a transport equation, the problem will propagate along the characteristic $t = x - 1$, as we will see next.</p>
<p>Case $t < x - 1$.</p>
<p>In this case, the method of characteristics leads to the system
\begin{align}
\frac{d x}{d \eta} &= 1 &x(\xi)\big|_{\eta = 0} &= \xi\\
\frac{d t}{d \eta} &= 1 &t(\xi)\big|_{\eta = 0} &= \eta\\
\frac{d u}{d \eta} &= -u &u(\xi)\big|_{\eta = 0} &= 100
\end{align}
and, as before, the solution is $u(x,t) = 100 e^{-t}$.</p>
<p>Case $x - 1 < t$.</p>
<p>In this case, the method of characteristics leads to the system
\begin{align}
\frac{d x}{d \xi} &= 1 &x(\eta)\big|_{\xi = 0} &= 1\\
\frac{d t}{d \xi} &= 1 &t(\eta)\big|_{\xi = 0} &= \eta\\
\frac{d u}{d \xi} &= -u &u(\eta)\big|_{\xi = 0} &= 1 + 100e^{-\eta}
\end{align}
hence
\begin{align}
x(\xi, \eta) &= \xi + 1 \\
t(\xi,\eta) &= \xi + \eta \\
u(\xi,\eta) &= \left(1 + 100e^{-\eta}\right)e^{-\xi}
\end{align}
then $\xi = x - 1$, $\eta = t - x +1$ and
$$
u(x,t) = e^{1-x} + 100e^{-t}
$$</p>
<p>It helps to draw a $(x,t)$ diagram to see exactly whats going on. In the whole upper plane, there's the solution $100 e^{-t}$ that comes from the initial condition. Starting from $x = 1$, there is a new component to the solution, namely $e^{1-x}$. This arises from the discontinuity of the function at $x = 1$, and propagates along the characteristics $t = x - c$, where $c \le 1$. Since the solution has to propagate with velocity $1$, the discontinuity wont be seen until time $t = x-1$, where the <em>delta generated wave</em> will reach the observer. The solution can be written as
$$
u(x,t) = 100e^{-t} + e^{1-x}\big(H(t - x + 1) - H(x - 1)\big)
$$</p>
|
2,636,188 | <p>How would I go about proving:</p>
<blockquote>
<p>For every non-prime $n \in \mathbb N$ there exists $m \in \mathbb Z / n\mathbb Z \setminus \{0\}$ with $m^{n-1} \not\equiv 1 \mod n$.</p>
</blockquote>
<p>I already proved the other way around (straight forward using Euler-Fermat), but am stuck at this point.</p>
| Dave | 334,366 | <p>I think it makes sense that we need to use the fact that $n$ is composite somehow. So what is the one thing we know about composite numbers: they have nontrivial diviors! So choose $m$ to be a divisor of $n$ different from $1$. Now is it possible for $ma\equiv 1\pmod n$ for any $a\in\Bbb Z/n\Bbb Z$?</p>
|
711,922 | <p>I know there are well-ordered sets that are not countable.</p>
<p>Suppose you are given an uncountable, well-ordered set $S$.</p>
<p>Isn't it possible to provide a bijection $f:\mathbb{N} \rightarrow S$ as following?</p>
<p>$S$ is well-ordered, so it has the smallest element, say $s_1$. $S \setminus$ {$s_1$} is also well-ordered, so there is the next smallest element, $s_2$, Similarly, there is the next smallest element $s_3$, and so on.</p>
<p>Continuing like this, define $f(i) = s_i$. </p>
<p>I know there is something wrong with this, but I cannot really see why...</p>
| user126154 | 126,154 | <p>Even if you made a mistake about surjectivity, as pointed out by 5xum and alex, you did a good thing: you proved that </p>
<p>given any two well-ordered sets, one is always the intial segment of the other! (which is, of course, true. The proof needs transfinite induction in general, but works exactly as your attempt of building the bijection)</p>
|
3,279,503 | <p>I'm relatively new in tensor space theory, and while reading some materials i've came across authors describing a inner product as a <span class="math-container">$(0,2)$</span> tensor. I'm not sure why it is, but i think if i write a map <span class="math-container">$f$</span> as <span class="math-container">$$f: P \times P \rightarrow \mathbb{R} \\(p,q) \mapsto \int_{-T}^{T}p(x)q(x) \mathrm{d}x$$</span> This clearly defines a inner product , and here i'm taking two elements from <span class="math-container">$P$</span> which the integral eats and spits out something in <span class="math-container">$R$</span> , so it's like a <span class="math-container">$(0,2)$</span> tensor, can i think like this ?</p>
<p>But i can't picture it for a linear transform as a (1,1) tensor. For a linear map <span class="math-container">$T$</span> defined between two finite dimensional vector spaces i.e <span class="math-container">$$T:V \rightarrow W$$</span>How is it tensor ? because the target is in <span class="math-container">$W$</span> which is not in <span class="math-container">$\mathbb{R}$</span> and by definition of tensor it eats <span class="math-container">$r$</span> copies of <span class="math-container">$V^{*}$</span> and <span class="math-container">$s$</span> copies of <span class="math-container">$V$</span> and spits out a real number . Also what happens if T is a endomorphism ? i'm having hard time imagining it.</p>
| Lutz Lehmann | 115,115 | <p>The linear map <span class="math-container">$T:V\to W$</span> belongs to the tensor space <span class="math-container">$W\otimes V^*$</span>. As that is not a product of <span class="math-container">$V$</span> and <span class="math-container">$V^*$</span>, it is not really appropriate to call it an <span class="math-container">$(1,1)$</span> tensor. That would apply to linear maps <span class="math-container">$T:V\to V$</span> that belong to <span class="math-container">$V\otimes V^*=T^{(1,1)}V$</span>.</p>
<p>The simple tensor products <span class="math-container">$a\otimes \beta\in W\otimes V^*$</span> get identified with the rank-1 linear maps <span class="math-container">$v\mapsto \beta(v)a$</span>. With any biorthogonal pair of bases <span class="math-container">$e_1,...e_m$</span> and <span class="math-container">$\theta^1,..,\theta^m$</span> of <span class="math-container">$V$</span> and <span class="math-container">$V^*$</span> the tensor to a general linear map <span class="math-container">$T$</span> is <span class="math-container">$\sum_{k=1}^m T(e_k)\otimes \theta^k$</span>.</p>
|
626,837 | <p>If $a+b+c+d = 30$ and $a,b,c,d$ lie between $0$ and $9$. How to find number of solutions of this equation.</p>
| robjohn | 13,854 | <p><strong>Hint:</strong> Consider the coefficient of $x^{30}$ in
$$
\left(\frac{x^{10}-1}{x-1}\right)^4
$$
Since
$$
(1-x)^{-4}=\sum_{n=0}^\infty\binom{n+3}{3}x^n
$$
and
$$
\left(1-x^{10}\right)^4=\sum_{k=0}^4(-1)^k\binom{4}{k}x^{10k}
$$
we get the coefficient of $x^n$ to be
$$
\sum_{k=0}^4(-1)^k\binom{4}{k}\binom{n-10k+3}{n-10k}
$$
Note that $\binom{n-10k+3}{n-10k}=\binom{n-10k+3}{3}$ when $n\ge10k$.</p>
|
3,170,742 | <p>I have a system of recurrence relations in the following form:</p>
<p><span class="math-container">$$
\begin{pmatrix}
f(n+1)\\
g(n+1)\\
\end{pmatrix}
= \textbf{A}
\begin{pmatrix}
f(n)\\
g(n)\\
\end{pmatrix} +\vec{b}
$$</span></p>
<p>which hold for all <span class="math-container">$n \in \{ 0,1,2,...,N\}.$</span> I also have the conditions: <span class="math-container">$f(0) = g(N+1) = 0.$</span> I've been trying to find a way to solve this but I'm really not sure how to proceed. Any help would be appreciated.</p>
| Dr. Sonnhard Graubner | 175,066 | <p>Use that <span class="math-container">$$\sin(x-y)=\sin(x)\cos(y)-\sin(y)\cos(x)$$</span></p>
|
3,663,374 | <p>I asked a question about if we have in general that if <span class="math-container">$G=G_1\times \cdots \times G_n$</span> (where <span class="math-container">$G_i$</span> are characteristic in <span class="math-container">$G$</span> for <span class="math-container">$i=1,\cdots ,n$</span>), then
<span class="math-container">$${\rm Out}(G)\cong {\rm Out}(G_1)\times\cdots\times {\rm Out}(G_n).$$</span></p>
<p><a href="https://math.stackexchange.com/questions/3661034/do-we-have-rm-outg-cong-rm-outg-1-times-cdots-times-rm-outg-n#comment7525007_3661034">A comment</a> suggested that I should use the following two facts:</p>
<ol>
<li>The inner automorphism group of a direct product is the direct product of the inner automorphism groups.</li>
<li>The direct product of quotients is a quotient of the direct product.</li>
</ol>
<p>I can prove the first one. I know how to prove a similar result for <span class="math-container">${\rm Aut}(G)$</span>. In addition, the <span class="math-container">${\rm Aut}(G)$</span> case requires that those <span class="math-container">$G_i$</span> be characteristic, while the <span class="math-container">${\rm Inn}(G)$</span> case only requires normality.</p>
<p>But how to prove the second one? Is it really true in general? Any help is appreciated.</p>
| Community | -1 | <p>It's true if <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are normal in <span class="math-container">$G$</span> and <span class="math-container">$H$</span> respectively. Then <span class="math-container">$A×B\triangleleft G×H$</span>. Define <span class="math-container">$i:(G×H)/(A×B)\to G/A×H/B$</span> by <span class="math-container">$i((g,h)+(A×B))=(g+A,h+B)$</span>. It's straight forward to check that <span class="math-container">$i$</span> is an isomorphism.</p>
|
1,437,961 | <p>Are comparison notations such as <, >, ≤, ≥, =, ≠ valid for sets?</p>
<p>I'm interested in stating size (number of elements) relations between sets.</p>
| Siminore | 29,672 | <p>Equality of course is. I have never seen the notation $A<B$ to say that $A$ has less elements than $B$. Everybody uses a more descriptive notation as $\#A < \# B$, for instance.</p>
|
1,437,961 | <p>Are comparison notations such as <, >, ≤, ≥, =, ≠ valid for sets?</p>
<p>I'm interested in stating size (number of elements) relations between sets.</p>
| Ove Ahlman | 222,450 | <p>$=$ and $\neq$ are defined and valid (see the axiom of extensionality).</p>
<p>$\leq, \geq, <$ and $>$ are more tricky. We may compare the number of elements in two sets, called the cardinality, and write that $|A|> |B|$ meaning that the cardinality of $A$ is greater than B, however this notation is never used directly on sets i.e. without the use of bars around the sets .</p>
<p>On the other hand $\subseteq, \subset,\supseteq $ and $\supset$ (the subset and superset relation) works just like ordering on sets, so one might argue that writing $A<B$ means $A\subseteq B$ as it does so in the subset partial order which is induced between all sets. </p>
<p>In general however $\leq, \geq, <$ and $>$ are not defined explicitly for sets, and If you're reading litterature which uses this notation, you need to look up what they define it as.</p>
|
1,437,961 | <p>Are comparison notations such as <, >, ≤, ≥, =, ≠ valid for sets?</p>
<p>I'm interested in stating size (number of elements) relations between sets.</p>
| Gabor Bakos | 270,432 | <p>You can use the $=$ sign, which has the definition: both sets have the same members.</p>
<p>The other comparison notations are not used unless you are comparing the sizes (cardinality, the number of elements) of the sets:</p>
<p>A={1,2,3,4}, B={1,5}, here $≥$ can be used (|A|≥|B|)</p>
<hr>
<p>Some comparison notations which look similar but mean different things:</p>
<p><strong>$⊆$</strong> (similar to <strong>$≤$</strong>): subset has fewer elements or equal to the set</p>
<p>{9,14,28} $⊆$ {9,14,28}</p>
<p><strong>$⊇$</strong> (similar to <strong>$≥$</strong>): set A has more elements or equal to the set B</p>
<p>{9,14,28} $⊇$ {9,14,28}</p>
<p><strong>$⊊$</strong> (similar to <strong>$<$</strong>): subset has fewer elements than the set</p>
<p>{9,14} $⊂$ {9,14,28}</p>
<p><strong>$\supsetneq$</strong> (similar to <strong>$>$</strong>): set A has more elements than set B</p>
<p>{9,14,28} $⊃$ {9,14}</p>
<p>Note that some people use the symbol ⊂ to mean what you've written as ⊆ (despite the fact that the analogy with < stops working). To be completely unambiguous, use ⊆ and ⊊. – </p>
|
289,757 | <p>I am writing this, as I am a currently an intern at an aircraft manufactur. I am studying a mixture of engineering and applied math. During the semester I focussed on numerical courses and my applied field is CFD. Even though every mathematician would say I have not heard a lot of math, for myself I would say that I get the "most amount of math" you can get while not studying math.</p>
<p>In my courses I have done deep theoretical analysis for numerical concepts and application in CFD. But currently I am starting to wonder, how much the e.g. Calculus of Variation course really helps me in my future career. The theory you learn at university seems to get only a little application in the <em>real word</em>. </p>
<p>Example: In my numerics for PDE class I have spent (wasted?) so many hours on trying to figure out the CFL number of certain schemes, but what I am doing right now has nothing to do with that. <em>Oh your simulation does diverge? Well let's take 2 instead of 4 as our CFL number.</em>
Furthermore, I am not really programming stuff as I hoped I could, but I am rather scripting. Fact is, 99 out of 100 people are not going to program a CFD solver. You rather use the code and apply it to your needs.</p>
<p>I am aware that university always follows a way more theoretical path than industry, but I am actually disappointed how little math I am really doing. Okay you might, say that's due to the fact that I am an intern and of course you are right. But I am in the lucky situation, that my team comes really close to research. Most of the members hold a PhD and studied engineering or math, and the focus is definetely on research ( in this departure of the company). But if the amount of math is that small in such an environment, where are you really able to make use of what you have learned at university.</p>
<p>So here comes my question</p>
<blockquote>
<p>How much math are you actually doing at your job?
And I don't mean, how much math is helping you to understand things, but how often does it happen, that you sit down and really <strong>do math</strong> in your non-academic job?</p>
</blockquote>
<p>Personally I get the impression that I could do the exact work without having heard most of my courses. Don't get me wrong, I really enjoy the theory, but currently I am rather frustrated.</p>
<p>Note: As this is my first Question, I hope I did not screw up completely. I did not found similar questions on this side. And feel free to edit or ask questions if thinks are not clear.</p>
| Mike Spivey | 2,370 | <p>I've spent the past twelve years as a professor. However, for five months last spring I spent part of a sabbatical working in the long-term forecasting group of an investment firm. I used <em>a lot</em> of math in those five months. (Admittedly, it was mostly a research-type position, and I gravitated toward the math-heavy problems.) Here are some problems I tackled on this job that required me to use math. </p>
<ol>
<li><p>We have huge gaps in our set of stock prices because countries change currencies or come into existence or cease to exist or stocks move in and out of major indexes or name your reason. We need to know how these stocks rise and fall (or don't) with each other. How do you calculate a covariance matrix in the presence of missing data? The best solution often results in the matrix becoming singular. This is a big problem because your model requires you to invert it. Do you try one of the other solutions to avoid the singular matrix problem and accept the resulting drawbacks, or do you try to "fix" your matrix somehow? If the latter, what are the best ways to do that? Answering this question required a great deal of understanding of (well, to be honest, learning about) numerical issues in linear algebra. </p></li>
<li><p>We have a model that we're happy with that makes short-term predictions, and we have a model that we're happy with that makes long-term predictions. How about the medium term? How do we smooth our short-term predictions into our long-term predictions? To answer this question for us I had to, among other things, solve a couple of differential equations that resulted from trying variations on the logistic curve as models. </p></li>
<li>Our model is giving weird, erratic results. Why? Does it have some
fundamental economic flaw? It takes me a couple of days to
determine that the answer to this one has to do with the eigenvalues
of the covariance matrix at the core of the model. Linear algebra
once again.</li>
<li>I'm analyzing a set of economic indicators to determine their
predictive value. Are they random walks of some sort, or do recent
values have something to do with slightly less recent values? This
requires time series analysis.</li>
<li>The woman sitting next to me is having trouble with her linear
regression. I can't remember now exactly what the difficulty was,
but I needed some statistics to solve the problem for her.</li>
</ol>
<p>The job wasn't all math; I spent more time coding than anything else. But I did use linear algebra, numerical methods, optimization, statistics, differential equations, and even some calculus on this job. </p>
<p>So don't give up hope yet. Maybe they are in the research arm of a company, or maybe they require an advanced degree, but there are some jobs out there that require a good deal of math. </p>
|
2,466,022 | <p>In Real and Complex Analysis, 3rd Edition, Walter Rudin advances the following:</p>
<p><a href="https://i.stack.imgur.com/XEVwK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XEVwK.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/dTwbg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dTwbg.png" alt="enter image description here"></a></p>
<p>How does $e^z \cdot e^{-z} = 1$ entail $(a)$?</p>
| MathematicianByMistake | 237,785 | <p>If there was a $z\in \mathbb{C}$ such that $e^z=0$ we would have </p>
<p>$e^z\cdot e^{-z}=0\cdot e^{-z} \Rightarrow\\1=0$</p>
<p>which simply is not true.</p>
<p><em>EDIT</em> </p>
<p>Another way to look at it is by using the form $e^z=e^{x+iy}=e^x(\cos y+i\sin y)$</p>
<p>with $x,y \in \mathbb{R}$</p>
<p>Thus $0=e^{x+iy}=e^x(\cos y+i\sin y)\Rightarrow \cos y=-i\sin y$</p>
<p>which cannot hold.</p>
|
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