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 |
|---|---|---|---|---|
2,419,057 | <p>Trying to assimilate the meaning of the differential I have looked for different examples of functions which:</p>
<ol>
<li>Admits all directional derivatives but are not continuous $(f: \mathbb{R}^2 \rightarrow \mathbb{R} \quad ,\quad (x,y) \mapsto
\begin{cases}
0 & \text{for } (x,y)=(0,0) \\
\frac{xy^4}{x^4+y^8} & \text{for } (x,y) \neq (0,0)
\end{cases})$</li>
<li>Admits all directional derivatives and are continuous but not differentiable:<a href="https://math.stackexchange.com/questions/372070/f-not-differentiable-at-0-0-but-all-directional-derivatives-exist?rq=1">$f$ not differentiable at $(0,0)$ but all directional derivatives exist</a></li>
</ol>
<p>For what I am looking for now I will use:</p>
<p>Suppose we have a function $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$</p>
<p>that in a point, let say $x_0$, admits all direccional derivatives. Then, we can consider the map</p>
<p>$\phi_{x_{0}}: \mathbb{R}^n \rightarrow \mathbb{R}^m \quad$ $v \mapsto D_{v}f(x_{0}) \quad$ , which sends a vector to its directional derivative in $x_0$.</p>
<p>If I am not mistaken, if this map is linear, it will be the differential of $f$.
What I am searching now is:</p>
<blockquote>
<p>A function $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$, continuos at $x_0$, which admits all directional derivatives at this point, but in which $\phi_{x_{0}}$ is not continuous.</p>
</blockquote>
<p>I don´t know if this would be possible.
In the example 2, the function $\phi_{(0,0)}$ is smooth (continuous, although non linear) and I can´t imagine how you could "break" the smoothness without breaking the continuity of $f$ too (that happens in the example 1).
I would appreciate if the example is in 2 dimensions ($f: \mathbb{R}^2 \rightarrow \mathbb{R}$) for it to be possible to visualize.</p>
<p>Thanks in advance</p>
| Sangchul Lee | 9,340 | <p>Define the function $f:\mathbb{R}^2 \to \mathbb{R}$ such that $f(0) = 0$ and</p>
<p>$$ f(r\cos\theta, r\sin\theta) = r \left| \cos\theta \right|^{1/r}.$$</p>
<p>Then $f$ is continuous everywhere and satisfies</p>
<p>$$ D_v f(0) = \begin{cases}
|v|, & \text{if } v = (v_1, 0) \\
0, & \text{otherwise}
\end{cases} $$</p>
<p>This behavior is transparent from its graph:</p>
<p>$\hspace{5.5em}$<a href="https://i.stack.imgur.com/buFPS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/buFPS.png" alt="Graph of $f$"></a></p>
|
1,717,149 | <p>Is it true or false that if $V$ is a vector space and $T:V \to W$ is a linear transformation such that $T^2 = 0$, then $Im(T) \subseteq Ker(T)$ ?<br>
I don't understand it that much. It doesn't seem related... I can have a vector $v$ from $V$ that its power by 2 equals zero but $T(v) \neq 0_{v}$ </p>
| Community | -1 | <p>I recommend the Marcus du Sautoy's book: <a href="http://rads.stackoverflow.com/amzn/click/0007214626" rel="nofollow noreferrer"><strong>Finding Moonshine</strong> : Mathematician's Journey Through Symmetry</a><br>
<a href="https://i.stack.imgur.com/wTUyG.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wTUyG.jpg" alt="enter image description here"></a> </p>
<p>He is also the author of <a href="http://rads.stackoverflow.com/amzn/click/0062064010" rel="nofollow noreferrer">The Music of the Primes</a>.</p>
|
1,717,149 | <p>Is it true or false that if $V$ is a vector space and $T:V \to W$ is a linear transformation such that $T^2 = 0$, then $Im(T) \subseteq Ker(T)$ ?<br>
I don't understand it that much. It doesn't seem related... I can have a vector $v$ from $V$ that its power by 2 equals zero but $T(v) \neq 0_{v}$ </p>
| Guru | 743 | <p>I would suggest to read, Simon Singh's Fermat's Last Theorem. <a href="http://simonsingh.net/books/fermats-last-theorem/" rel="nofollow">http://simonsingh.net/books/fermats-last-theorem/</a></p>
<p>It starts with the introduction to simple problem, solutions. And entire history as it evolves to solve the problem surrounding group theory. And then into advanced theory. </p>
|
468 | <p>Textbook writers are blessed with only solving problems with neat answers. Numerical coefficients are small integers, many terms cancel, polynomials split into simple factors, angles have trigonometric functions with known values. Pure bliss.</p>
<p>The "real life" is different (as any of us knows).</p>
<p>Giving such questions for homework runs the risk that the student knows she is wrong when some crooked formula/value shows up. On the other hand, checking messy results (unless perhaps directly numerical values) is harder.</p>
<p>What do you think about posing questions which don't have neat derivations/results? I presume the answer could depend on the subject matter, the level of the students, and perhaps on exactly what the question should teach.</p>
| quid | 143 | <p>I agree that such exercises have a built in risk of giving the students the wrong idea what is the generic situation and what is quite a special case. </p>
<p>However, "the risk that the student knows she is wrong when some crooked formula/value shows up" you mention is more a feature in my opinion. It is not a bad thing to do heuristic checks that suggest one <em>might</em> have made an error and to then double-check. </p>
<p>Also, for some things it is more or less unavoidable to present constructed exercises. A polynomial will not have a multiple root just so, you need to take care it has one. </p>
<p>Or, in linear algebra one has to choose the matrices somewhat carefully if one wants to illustrate certain things. And, it is a real pain to calculate even with moderately sized matrices if they involve radical expressions, and if you go for decimal approximations you 'lose' the clear notion of singularity of a matrix and would actually need to do something more advanced. So, somehow one is stuck with matrices whose eigenvalues are "too nice" (and this even leaving the issue of factoring the polynomial aside.) </p>
<p>Personally, I mostly tend to give exercises where things work out nicely, yet I do talk with students about the fact that this is what happens and that this does not capture what would be a "typical" situation were one to choose things "randomly." </p>
<p>That is, I think it is important to raise awareness of the issue of "artificially nice" problems with students, but as mentioned above consider it as unavoidable to use them in certain cases. </p>
|
201,807 | <p>I heard this problem, so I might be missing pieces. Imagine there are two cities separated by a very long road. The road has only one lane, so cars cannot overtake each other. $N$ cars are released from one of the cities, the cars travel at constant speeds $V$ chosen at random and independently from a probability distribution $P(V)$. What is the expected number of groups of cars arriving simultaneously at the other city? </p>
<p>P.S.: Supposedly, this was a Princeton physics qualifier problem, if that makes a difference. </p>
| joriki | 6,622 | <p>There are already two answers that show that under a certain interpretation of the question the answer is the $N$-th harmonic number. This can be seen more directly by noting that the $k$-th car is the "leader" of a group iff it is the slowest of the first $k$ cars, which occurs with probability $1/k$. Thus the expected number of leaders of groups is the $N$-th harmonic number, and of course there are as many groups as there are leaders of groups.</p>
|
1,377,412 | <p>I am brand new to ODE's, and have been having difficulties with this practice problem. Find a 1-parameter solution to the homogenous ODE:$$2xy \, dx+(x^2+y^2) \, dy = 0$$assuming the coefficient of $dy \ne 0$
The textbook would like me to use the subsitution $x = yu$ and $dx=y \, du + u \, dy,\ y \ne 0$
Rewriting the equation with the subsititution:
$$2uy^2(y \, du + u \, dy)+(x^2+y^2) \, dy = 0$$
divide by $y^2$
$$2u(y \, du + u \, dy) + (u^2+1 ) \, dy=0$$
but after further simplification I end up getting ${dy \over y}$ which would mean I would get a logarithm after integrating, and the answer is given as $$3x^2y+y^3 = c$$
Could I get some help/hints as to how this answer was obtained? </p>
| Michael Hardy | 11,667 | <p>The variables $y$ and $u$ can be separated from each other:
$$
2u(y \, du + u \, dy) + (u^2+1 ) \, dy=0
$$
$$
2u\left( du + u \, \frac{dy} y \right) + (u^2+1 ) \, \frac{dy} y = 0
$$
$$
2u\left( \frac{du} u + \frac{dy} y \right) + \left( u + \frac 1 u \right) \, \frac{dy} y = 0
$$
$$
2 \left( \frac{du} u + \frac{dy} y \right) + \left( 1 + \frac 1 {u^2} \right) \, \frac{dy} y = 0
$$
$$
2 \frac{du} u + 2 \frac{dy} y + \left( 1 + \frac 1 {u^2} \right) \, \frac{dy} y = 0
$$
$$
2 \frac{du} u + \left( 3 + \frac 1 {u^2} \right) \, \frac{dy} y = 0
$$
$$
\left( 3 + \frac 1 {u^2} \right) \, \frac{dy} y = -2 \frac{du} u
$$
$$
\frac{dy} y = \frac{-2 \, du/u}{3 + \frac 1 {u^2}}
$$
$$
\frac{dy} y = \frac{-2 u\, du}{3u^2 + 1}
$$
$$
\log|y| = - \frac 1 3 \log|3u^2 + 1| + \text{constant}
$$
$$
y = (3u^2+1)^{-1/3}\cdot\text{constant}
$$
$$
y^3 = \frac{\text{constant}}{3u^2+1}
$$</p>
|
2,432,213 | <p>I am having a really hard time understanding this problem. I know that for uniqueness we need that the derivative is continuous and that the partial derivative is continuous. I also know that the lipschitz condition gives continuity. I can't figure out what to do with this problem though. </p>
<p><a href="https://i.stack.imgur.com/xktm5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xktm5.png" alt="enter image description here"></a></p>
| Hans H | 481,395 | <p>Here is a hint. Consider the operator $T$ defined below that maps continuous functions into continuous functions. For any $h\in C(-\infty,r]$, let</p>
<p>$$
(Th)(t) := \phi(0) + \int_0^t f(\tau, h(\tau), h(g(\tau)))\ d\tau
$$
when $t>0$ and let $(Th)(t)= \phi(t)$ when $t\leq0$. Notice that $T: C(-\infty,r] \rightarrow C(-\infty,r]$. </p>
<p>Think about the fixed points of $T$. If $h(t)$ is a fixed point of $T$, what differential equation would it satisfy?</p>
<p>To show that $T$ is a contraction for some value of $r>0$, you need to show that there exists an $r>0$ and a corresponding $\alpha\in (0,1)$ such that for any two functions $h_1$ and $h_2$ in $C(-\infty,r]$,
$$
d(Th_1, Th_2) \leq \alpha \,d(h_1, h_2).
$$</p>
<p>That proof could be broken down into a few steps:</p>
<p>1) $d(Th_1, Th_2) := \sup_{t\in (-\infty, r]} | Th_1(t) - Th_2(t)|$.</p>
<p>2) $d(Th_1, Th_2) = \sup_{t\in (0, r]} \left| \int_0^t f(\tau, h_1(\tau), h_1(g(\tau))) - f(\tau, h_2(\tau), h_2(g(\tau)))\, d\tau \right|$.</p>
<p>3) $d(Th_1, Th_2) \leq \sup_{t\in (0, r]} \int_0^t |\,f(\tau, h_1(\tau), h_1(g(\tau))) - f(\tau, h_2(\tau), h_2(g(\tau)))|\, d\tau$.</p>
<p>4) $d(Th_1, Th_2) \leq 2 r L\, d(h_1, h_2)$.</p>
<p>5) $T$ is a contraction if $0\leq r < 1/(2 L)$.</p>
<p>Proving each of those statements would be most of the complete proof. I have left some of the details for you to fill in. Once you have done a dozen similar proofs, the whole process will seem much easier.</p>
|
4,617,031 | <p>How would I order <span class="math-container">$x = \sqrt{3}-1, y = \sqrt{5}-\sqrt{2}, z = 1+\sqrt{2} \ $</span> without approximating the irrational numbers? In fact, I would be interested in knowing a general way to solve such questions if there is one.</p>
<p>What I tried to so far, because they are all positive numbers, is to square <span class="math-container">$x,y,z$</span> but, obviously, the rational parts will not be equal so I cannot compare the radicals. Proving that <span class="math-container">$x<z$</span> is easy and so is <span class="math-container">$y<z$</span>, but I'm stuck at <span class="math-container">$x < y \text{ or } x>y$</span>.</p>
| GohP.iHan | 151,481 | <p>Suppose <span class="math-container">$ x \geqslant y $</span>, then</p>
<p><span class="math-container">$$ \begin{array} { r c l }
\sqrt3 + \sqrt2 &\geqslant& \sqrt5 + 1 \\
(\sqrt3 + \sqrt2)^2 &\geqslant& (\sqrt5 + 1)^2 \\
5 + 2 \sqrt6 &\geqslant& 6 + 2 \sqrt5 \\
2( \sqrt6 - \sqrt5) &\geqslant& 1 \\
2( \sqrt6 - \sqrt5)( \sqrt6 + \sqrt5) &\geqslant& ( \sqrt6 + \sqrt5) \\
2(6 - 5) &\geqslant& \sqrt6 + \sqrt5 \\
2 &\geqslant& \sqrt6 + \sqrt5 > \sqrt4 + \sqrt4 = 4 \\
\end{array}
$$</span></p>
<p>A contradiction.</p>
|
1,614,274 | <p><br>
I need to find the splitting field of $\; x^2+1 \in \mathbb Z_7 [x] \;$ over $\mathbb Z_7 $.<br><br> The roots of the polynomial are $-i \;$ and $i$. Therefore I would conclude that the splitting field is $\mathbb Z_7(i)$ but that would be more like a guess.<br><br> I don´t really understand splitting field over $ \mathbb Z_7$. Can seomeone give me a hint, on how to find the splitting field ?</p>
| Hagen von Eitzen | 39,174 | <p>The smallest field extension of $\Bbb F_7$ is $\Bbb F_{49}$. As $\Bbb F_{49}^\times $ is a cyclic group of order $48$, it has an element of order $4$, let's call that $i$. Then $i^4= 1$ but $i^2\ne 1$, meaning that $i^2=-1$, as desired. </p>
<p>Note however that even though I introduced the name $i$ for an element of $\Bbb F_{49}$, it is not "the" $i$ you know from $\Bbb C$; actually $0\in \Bbb F_7$ is not "the" $0$ you know from $\Bbb C$, either.</p>
|
1,614,274 | <p><br>
I need to find the splitting field of $\; x^2+1 \in \mathbb Z_7 [x] \;$ over $\mathbb Z_7 $.<br><br> The roots of the polynomial are $-i \;$ and $i$. Therefore I would conclude that the splitting field is $\mathbb Z_7(i)$ but that would be more like a guess.<br><br> I don´t really understand splitting field over $ \mathbb Z_7$. Can seomeone give me a hint, on how to find the splitting field ?</p>
| AdLibitum | 210,743 | <p>Since $x^2+1$ has no roots in $\Bbb Z_7$ and the <em>polynomial is quadratic</em> the splitting field of $X^2+1$ is the quadratic extension
$$
\Bbb Z_7[X]/(X^2+1)
$$
which you can concretely model as the degree $\leq1$ polynomials $aX+b$ with coefficients in $\Bbb Z_7$ with the multiplication rule given (unsurprisingly) by $X^2=-1$.</p>
<p>It follows from the general theory of finite fields that this is <strong>the</strong> quadratic extension of $\Bbb Z_7$, i.e. the <strong>unique</strong> (up to isomorphism) field with $7^2=49$ elements, usually denoted $\Bbb F_{49}$.</p>
|
19,373 | <p>I posted this question earlier today on the Mathematics site (<a href="https://math.stackexchange.com/q/3988907/96384">https://math.stackexchange.com/q/3988907/96384</a>), but was advised it would be better here.</p>
<p>I had a heated argument with someone online who claimed to be a school mathematics teacher of many years standing. The question which spurred this discussion was something along the lines of:</p>
<p>"A horseman was travelling from (location A) along a path through a forest to (location B) during the American War of Independence. The journey was of 22 miles. How far was it in kilometres?"</p>
<p>To my mind, the answer is trivially obtained by multiplying 22 by 1.6 to get 35.2 km, which can be rounded appropriately to 35 km.</p>
<p>I was roundly scolded by this ancient mathematics teacher for a) not using the official conversion factor of 1.60934 km per mile and b) not reporting the correct value as 35.405598 km.</p>
<p>Now I have serious difficulties with this analysis. My argument is: this is a man riding on horseback through a forest in a pre-industrial age. It would be impractical and impossible to measure such a distance to any greater precision than (at best) to the nearest 20 metres or so, even in this day and age. Yet the answer demanded was accurate to the nearest millimetre.</p>
<p>But when I argued this, I was told that it was not my business to round the numbers. I was to perform the conversion task given the numbers I was quoted, and report the result for the person asking the question to decide how accurately the numbers are to be interpreted.</p>
<p>Is that the way of things in school? As a trained engineer, my attitude is that it is part of the purview of anybody studying mathematics to be able to estimate and report appropriate limits of accuracy, otherwise you get laughably ridiculous results like this one.</p>
<p>I confess I have never had a good relationship with teachers, apart from my A-level physics teacher whom I adored, so I expect I will be given a hard time over my inability to understand the basics of what I have failed to learn during the course of the above.</p>
| Jan | 15,298 | <p>When I was in school, I once got an answer marked as error for having too many digits. IIRC it was in trigonometry and I had just written down as many digits as the calculator displayed. (I was able to discuss it away, but was told to avoid unreasonable amounts of digits in the future)</p>
<p>That was in the 1990s in continental Europe, but I think it is still good enough for s counterexample: Not all teachers are like that.</p>
|
747,519 | <p><img src="https://i.stack.imgur.com/jYzfz.png" alt="enter image description here"></p>
<p>I tried this problem on my own, but got 1 out of 5. Now we are supposed to find someone to help us. Here is what I did:</p>
<p>Let $f:[a,b] \rightarrow \mathbb{R}$ be continuous on a closed interval $I$ with $a,b \in I$, $a \leq b$</p>
<p>If $f(a), f(b) \in f(I)$ let $f(a)\leq y \leq f(b)$. Then by IVT there exists $x$, $a\leq x \leq b$ where $f(x)=y$ $Rightarrow$ The image is also an interval. </p>
<p>Show closed: Let m be the lowest upper bound and M the greatest lower bound of the image interval. $I=[a,b]$ must be a subset of $[M,m]$ and the function attains its bounds,
$m,M\in f(I)$. so $f(I)$ is a subset of $[M,m]$, thus is closed. </p>
<p>Can anyone provide a proof of this statement? Thanks!</p>
| Faqir Chand | 812,679 | <p>Closed intervals are compact in <span class="math-container">$ \mathbb{R}$</span>. <span class="math-container">$f$</span> continuous implies image of <span class="math-container">$f$</span> is compact. By Heine-Borel, we know that compact sets are closed and bounded in <span class="math-container">$\mathbb{R}$</span>, in particular.
Since a closed set contains its sup and inf use that to show that they form the end points of the interval and use the IVP to conclude that it is a closed interval.</p>
|
102,721 | <p>This is probably a very simple question, but I couldn't find a duplicate.</p>
<p>As everybody knows, <code>{x, y} + v</code> gives <code>{x + v, y + v}</code>. But if I intend <code>v</code> to represent a vector, for example if I am going to substitute <code>v -> {vx, vy}</code> in the future, then the result <code>{x + v, y + v}</code> is meaningless.</p>
<p>How I can indicate to Mathematica that <code>v</code> is not a scalar and functions like <code>Plus</code> should not treat it as one? I tried setting <code>$Assumptions = v ∈ Vectors[2]</code> but that didn't help.</p>
| Edmund | 19,542 | <p>You may use <code>Indexed</code> to represent <code>v</code> as its components in the calculation.</p>
<pre><code>{x, y} + (Indexed[v, #] & /@ Range[2])
(* {x + Indexed[v, {1}], y + Indexed[v, {2}]} *)
</code></pre>
<p>Later when <code>v</code> is assigned the <code>Indexed</code> components will become <code>Part</code> values of <code>v</code>.</p>
<pre><code>v = {1, 2};
{x, y} + (Indexed[v, #] & /@ Range[2])
(* {1 + x, 2 + y} *)
</code></pre>
<p>As a function</p>
<pre><code>f[t_] := {x, y} + (Indexed[t, #] & /@ Range[2])
</code></pre>
<p>Hope this helps.</p>
|
102,721 | <p>This is probably a very simple question, but I couldn't find a duplicate.</p>
<p>As everybody knows, <code>{x, y} + v</code> gives <code>{x + v, y + v}</code>. But if I intend <code>v</code> to represent a vector, for example if I am going to substitute <code>v -> {vx, vy}</code> in the future, then the result <code>{x + v, y + v}</code> is meaningless.</p>
<p>How I can indicate to Mathematica that <code>v</code> is not a scalar and functions like <code>Plus</code> should not treat it as one? I tried setting <code>$Assumptions = v ∈ Vectors[2]</code> but that didn't help.</p>
| John Doty | 27,989 | <p>I like to wrap lists in some suitable function to block this evaluation. <code>MatrixForm</code> is good because it looks pretty, but an undefined function would also work.</p>
<pre><code>a = MatrixForm[{x, y}] + MatrixForm[v]
</code></pre>
<p>Then, once you have concrete definitions for your arrays, take away the <code>MatrixForm</code>:</p>
<pre><code>b = a /. v -> {1, 2}
b /. MatrixForm -> Identity
(* {1 + x, 2 + y} *)
</code></pre>
|
167,575 | <p>I have 6 sets of 4D points. Here is an example of one set :</p>
<pre><code>{{30., 5., 111.925, 113.569}, {30., 7.5, 114.7, 158.286}, {30., 10., 115.625, 206.023},
{30., 12.5, 115.625, 257.528}, {30., 15., 117.475, 294.663}, {30., 17.5, 119.325, 328.03},
{30., 20., 121.175, 357.982}, {30., 22.5, 122.1, 393.646}, {30., 25., 122.1, 437.384},
{30., 27.5, 122.1, 481.123}}
</code></pre>
<p>I want to plot the x,y coordinates of the points on the 2D plane and use the z coordinate to define the size of the symbol (bubble radius or area) and the last coordinate to define a color for that bubble. So the color will be different depending on the fourth coordinate. Any help would be appreciated !</p>
<p>I would like to have a 4D graphic like that :
<a href="https://i.stack.imgur.com/wk30Z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wk30Z.png" alt="Bubble chart 4D"></a></p>
| J. M.'s persistent exhaustion | 50 | <p>Using the given data:</p>
<pre><code>data = {{30., 5., 111.925, 113.569}, {30., 7.5, 114.7, 158.286},
{30., 10., 115.625, 206.023}, {30., 12.5, 115.625, 257.528},
{30., 15., 117.475, 294.663}, {30., 17.5, 119.325, 328.03},
{30., 20., 121.175, 357.982}, {30., 22.5, 122.1, 393.646},
{30., 25., 122.1, 437.384}, {30., 27.5, 122.1, 481.123}};
sc = {"ThermometerColors", MinMax[data[[All, -1]]]}; cf = ColorData[sc];
Legended[Graphics[{cf[#4], Disk[{#, #2}, #3/100]} & @@@ data, Frame -> True],
BarLegend[sc]]
</code></pre>
<p><img src="https://i.stack.imgur.com/hWS2a.png" alt="diagram"></p>
|
3,167,571 | <p>Let consider a square <span class="math-container">$10\times 10$</span> and write in the every unit square the numbers from <span class="math-container">$1$</span> to <span class="math-container">$100$</span> such that every two consecutive numbers are in squares which has a common edge. Then there are two perfect squares on the same line or column. Can you give me an hint? How to start?</p>
| Cesareo | 397,348 | <p>With the help of the slack variables <span class="math-container">$\epsilon_i$</span> and calling</p>
<p><span class="math-container">$$
L(p,\mu,\epsilon,\lambda)= p_1 p_2 p_3 p_4 p_5+\mu _5 \left(p_1-p_2-\epsilon _5^2\right)+\mu _4 \left(p_2-p_3-\epsilon _4^2\right)+\mu _3 \left(p_3-p_4-\epsilon _3^2\right)+\mu _2
\left(p_4-p_5-\epsilon _2^2\right)+\mu _1 \left(p_5-\epsilon _1^2\right)+\lambda \left(p_1 x_1+p_2 \left(x_2-x_1\right)+p_3
\left(x_3-x_2\right)+p_4 \left(x_4-x_3\right)+p_5 \left(x_5-x_4\right)-1\right)
$$</span></p>
<p>and solving the sationary conditions</p>
<p><span class="math-container">$$
\nabla L = 0
$$</span></p>
<p>we obtain a set of solutions jointly with a set of conditions <span class="math-container">$\epsilon_i^2\ge 0$</span> to qualify those solutions. To be feasible the solution requires that <span class="math-container">$\epsilon_i^2\ge 0$</span> Also when <span class="math-container">$\epsilon_i^2 = 0$</span> it means that the corresponding restriction is active.</p>
<p>Due to the length of the symbolic response, we leave a script in MATHEMATICA that summarizes the results. There are sixteen non trivial solutions in <strong>res0</strong> with structure <span class="math-container">$\{p_i,\epsilon_i^2,p_1p_2p_3p_4p_5\}$</span></p>
<pre><code>n = 5;
X = Table[Subscript[x, k], {k, 1, n}];
P = Table[Subscript[p, k], {k, 1, n}];
EE = Table[Subscript[epsilon, k], {k, 1, n}];
M = Table[Subscript[mu, k], {k, 1, n}];
vars = Join[Join[Join[P, M], EE], {lambda}]
prod = Product[P[[k]], {k, 1, n}]
L = prod
L += lambda (Sum[P[[k]] (X[[k]] - X[[k - 1]]), {k, 2, n}] + P[[1]] X[[1]] - 1)
L += Sum[M[[k]] (P[[n - k + 1]] - P[[n - k + 2]] + EE[[k]]^2), {k, 2, n}] + M[[1]] (P[[n]] - EE[[1]]^2)
grad = Grad[L, vars]
equs = Thread[grad == 0];
E2 = EE^2;
results = Join[Join[P, E2], {prod}];
sols = Quiet[Solve[equs, vars]];
results0 = results /. sols // FullSimplify;
For[i = 1; res = {}, i <= Length[results0], i++,
If[NumberQ[results0[[i]][[2 n + 1]]] == False, AppendTo[res,results0[[i]]]]
]
res0 = Union[res];
res0 // MatrixForm
</code></pre>
<p>Results for <span class="math-container">$n = 3$</span></p>
<p><span class="math-container">$$
\left[
\begin{array}{ccccccc}
p_1&p_2&p_3&\epsilon_1^2&\epsilon_2^2&\epsilon_3^2&p_1p_2p_3\\
\frac{1}{3 x_1} & \frac{1}{3 \left(x_2-x_1\right)} & \frac{1}{3 \left(x_3-x_2\right)} & \frac{1}{3 \left(x_3-x_2\right)} & \frac{1}{3 x_2-3
x_3}+\frac{1}{3 \left(x_2-x_1\right)} & \frac{1}{3} \left(\frac{1}{x_1-x_2}+\frac{1}{x_1}\right) & \frac{1}{27 \left(x_1^2-x_1 x_2\right)
\left(x_2-x_3\right)} \\
\frac{1}{3 x_1} & \frac{2}{3 \left(x_3-x_1\right)} & \frac{2}{3 \left(x_3-x_1\right)} & \frac{2}{3 \left(x_3-x_1\right)} & 0 & \frac{1}{3}
\left(\frac{2}{x_1-x_3}+\frac{1}{x_1}\right) & \frac{4}{27 x_1 \left(x_1-x_3\right){}^2} \\
\frac{2}{3 x_2} & \frac{2}{3 x_2} & \frac{1}{3 \left(x_3-x_2\right)} & \frac{1}{3 \left(x_3-x_2\right)} & \frac{1}{3 x_2-3 x_3}+\frac{2}{3
x_2} & 0 & \frac{4}{27 x_2^2 \left(x_3-x_2\right)} \\
\frac{1}{x_3} & \frac{1}{x_3} & \frac{1}{x_3} & \frac{1}{x_3} & 0 & 0 & \frac{1}{x_3^3} \\
\end{array}
\right]
$$</span></p>
<p>Results for <span class="math-container">$n = 4$</span></p>
<p><span class="math-container">$$
\left[
\begin{array}{ccccccccc}
p_1 & p_2 & p_3 & p_4 & \epsilon_1^2&\epsilon_2^2&\epsilon_3^2&\epsilon_4^2&p_1p_2p_3p_4\\
\frac{1}{4 x_1} & \frac{1}{4 \left(x_2-x_1\right)} & \frac{1}{4 \left(x_3-x_2\right)} & \frac{1}{4 \left(x_4-x_3\right)} & \frac{1}{4
\left(x_4-x_3\right)} & \frac{1}{4 x_3-4 x_4}+\frac{1}{4 \left(x_3-x_2\right)} & \frac{1}{4 x_2-4 x_3}+\frac{1}{4 \left(x_2-x_1\right)} &
\frac{1}{4} \left(\frac{1}{x_1-x_2}+\frac{1}{x_1}\right) & -\frac{1}{256 \left(x_1^2-x_1 x_2\right) \left(x_2-x_3\right)
\left(x_3-x_4\right)} \\
\frac{1}{4 x_1} & \frac{1}{4 \left(x_2-x_1\right)} & \frac{1}{2 \left(x_4-x_2\right)} & \frac{1}{2 \left(x_4-x_2\right)} & \frac{1}{2
\left(x_4-x_2\right)} & 0 & \frac{1}{2 x_2-2 x_4}+\frac{1}{4 \left(x_2-x_1\right)} & \frac{1}{4}
\left(\frac{1}{x_1-x_2}+\frac{1}{x_1}\right) & -\frac{1}{64 x_1 \left(x_1-x_2\right) \left(x_2-x_4\right){}^2} \\
\frac{1}{4 x_1} & \frac{1}{2 \left(x_3-x_1\right)} & \frac{1}{2 \left(x_3-x_1\right)} & \frac{1}{4 \left(x_4-x_3\right)} & \frac{1}{4
\left(x_4-x_3\right)} & \frac{1}{4 x_3-4 x_4}+\frac{1}{2 \left(x_3-x_1\right)} & 0 & \frac{1}{4}
\left(\frac{2}{x_1-x_3}+\frac{1}{x_1}\right) & \frac{1}{64 x_1 \left(x_1-x_3\right){}^2 \left(x_4-x_3\right)} \\
\frac{1}{4 x_1} & \frac{3}{4 \left(x_4-x_1\right)} & \frac{3}{4 \left(x_4-x_1\right)} & \frac{3}{4 \left(x_4-x_1\right)} & \frac{3}{4
\left(x_4-x_1\right)} & 0 & 0 & \frac{1}{4} \left(\frac{3}{x_1-x_4}+\frac{1}{x_1}\right) & \frac{27}{256 x_1 \left(x_4-x_1\right){}^3} \\
\frac{1}{2 x_2} & \frac{1}{2 x_2} & \frac{1}{4 \left(x_3-x_2\right)} & \frac{1}{4 \left(x_4-x_3\right)} & \frac{1}{4 \left(x_4-x_3\right)} &
\frac{1}{4 x_3-4 x_4}+\frac{1}{4 \left(x_3-x_2\right)} & \frac{1}{4 x_2-4 x_3}+\frac{1}{2 x_2} & 0 & \frac{1}{64 x_2^2
\left(x_2-x_3\right) \left(x_3-x_4\right)} \\
\frac{1}{2 x_2} & \frac{1}{2 x_2} & \frac{1}{2 \left(x_4-x_2\right)} & \frac{1}{2 \left(x_4-x_2\right)} & \frac{1}{2 \left(x_4-x_2\right)} &
0 & \frac{1}{2} \left(\frac{1}{x_2-x_4}+\frac{1}{x_2}\right) & 0 & \frac{1}{16 x_2^2 \left(x_2-x_4\right){}^2} \\
\frac{3}{4 x_3} & \frac{3}{4 x_3} & \frac{3}{4 x_3} & \frac{1}{4 \left(x_4-x_3\right)} & \frac{1}{4 \left(x_4-x_3\right)} & \frac{1}{4 x_3-4
x_4}+\frac{3}{4 x_3} & 0 & 0 & \frac{27}{256 x_3^3 \left(x_4-x_3\right)} \\
\frac{1}{x_4} & \frac{1}{x_4} & \frac{1}{x_4} & \frac{1}{x_4} & \frac{1}{x_4} & 0 & 0 & 0 & \frac{1}{x_4^4} \\\end{array}
\right]
$$</span></p>
<p>and for <span class="math-container">$n = 5$</span></p>
<p><span class="math-container">$$
\left[
\begin{array}{ccccccccccc}
p_1&p_2&p_3&p_4&p_5&\epsilon_1^2&\epsilon_2^2&\epsilon_3^2&\epsilon_4^2&\epsilon_5^2&p_1p_2p_3p_4p_5\\
\frac{1}{5 x_1} & -\frac{1}{5 \left(x_1-x_2\right)} & -\frac{1}{5 \left(x_2-x_3\right)} & -\frac{1}{5 \left(x_3-x_4\right)} & -\frac{1}{5
\left(x_4-x_5\right)} & -\frac{1}{5 \left(x_4-x_5\right)} & \frac{x_3-2 x_4+x_5}{5 \left(x_3-x_4\right) \left(x_4-x_5\right)} &
\frac{x_2-2 x_3+x_4}{5 \left(x_2-x_3\right) \left(x_3-x_4\right)} & \frac{x_1-2 x_2+x_3}{5 \left(x_1-x_2\right) \left(x_2-x_3\right)} &
\frac{2 x_1-x_2}{5 x_1 \left(x_1-x_2\right)} & \frac{1}{3125 x_1 \left(x_1-x_2\right) \left(x_2-x_3\right) \left(x_3-x_4\right)
\left(x_4-x_5\right)} \\
\frac{1}{5 x_1} & -\frac{1}{5 \left(x_1-x_2\right)} & -\frac{1}{5 \left(x_2-x_3\right)} & -\frac{2}{5 \left(x_3-x_5\right)} & -\frac{2}{5
\left(x_3-x_5\right)} & -\frac{2}{5 \left(x_3-x_5\right)} & 0 & \frac{2 x_2-3 x_3+x_5}{5 \left(x_2-x_3\right) \left(x_3-x_5\right)} &
\frac{x_1-2 x_2+x_3}{5 \left(x_1-x_2\right) \left(x_2-x_3\right)} & \frac{2 x_1-x_2}{5 x_1 \left(x_1-x_2\right)} & \frac{4}{3125 x_1
\left(x_1-x_2\right) \left(x_2-x_3\right) \left(x_3-x_5\right){}^2} \\
\frac{1}{5 x_1} & -\frac{1}{5 \left(x_1-x_2\right)} & -\frac{2}{5 \left(x_2-x_4\right)} & -\frac{2}{5 \left(x_2-x_4\right)} & -\frac{1}{5
\left(x_4-x_5\right)} & -\frac{1}{5 \left(x_4-x_5\right)} & \frac{x_2-3 x_4+2 x_5}{5 \left(x_2-x_4\right) \left(x_4-x_5\right)} & 0 &
\frac{2 x_1-3 x_2+x_4}{5 \left(x_1-x_2\right) \left(x_2-x_4\right)} & \frac{2 x_1-x_2}{5 x_1 \left(x_1-x_2\right)} & \frac{4}{3125 x_1
\left(x_1-x_2\right) \left(x_2-x_4\right){}^2 \left(x_4-x_5\right)} \\
\frac{1}{5 x_1} & -\frac{1}{5 \left(x_1-x_2\right)} & -\frac{3}{5 \left(x_2-x_5\right)} & -\frac{3}{5 \left(x_2-x_5\right)} & -\frac{3}{5
\left(x_2-x_5\right)} & -\frac{3}{5 \left(x_2-x_5\right)} & 0 & 0 & \frac{3 x_1-4 x_2+x_5}{5 \left(x_1-x_2\right) \left(x_2-x_5\right)} &
\frac{2 x_1-x_2}{5 x_1 \left(x_1-x_2\right)} & \frac{27}{3125 x_1 \left(x_1-x_2\right) \left(x_2-x_5\right){}^3} \\
\frac{1}{5 x_1} & -\frac{2}{5 \left(x_1-x_3\right)} & -\frac{2}{5 \left(x_1-x_3\right)} & -\frac{1}{5 \left(x_3-x_4\right)} & -\frac{1}{5
\left(x_4-x_5\right)} & -\frac{1}{5 \left(x_4-x_5\right)} & \frac{x_3-2 x_4+x_5}{5 \left(x_3-x_4\right) \left(x_4-x_5\right)} &
\frac{x_1-3 x_3+2 x_4}{5 \left(x_1-x_3\right) \left(x_3-x_4\right)} & 0 & \frac{3 x_1-x_3}{5 x_1 \left(x_1-x_3\right)} & \frac{4}{3125 x_1
\left(x_1-x_3\right){}^2 \left(x_3-x_4\right) \left(x_4-x_5\right)} \\
\frac{1}{5 x_1} & -\frac{2}{5 \left(x_1-x_3\right)} & -\frac{2}{5 \left(x_1-x_3\right)} & -\frac{2}{5 \left(x_3-x_5\right)} & -\frac{2}{5
\left(x_3-x_5\right)} & -\frac{2}{5 \left(x_3-x_5\right)} & 0 & \frac{2 \left(x_1-2 x_3+x_5\right)}{5 \left(x_1-x_3\right)
\left(x_3-x_5\right)} & 0 & \frac{3 x_1-x_3}{5 x_1 \left(x_1-x_3\right)} & \frac{16}{3125 x_1 \left(x_1-x_3\right){}^2
\left(x_3-x_5\right){}^2} \\
\frac{1}{5 x_1} & -\frac{3}{5 \left(x_1-x_4\right)} & -\frac{3}{5 \left(x_1-x_4\right)} & -\frac{3}{5 \left(x_1-x_4\right)} & -\frac{1}{5
\left(x_4-x_5\right)} & -\frac{1}{5 \left(x_4-x_5\right)} & \frac{x_1-4 x_4+3 x_5}{5 \left(x_1-x_4\right) \left(x_4-x_5\right)} & 0 & 0 &
\frac{4 x_1-x_4}{5 x_1 \left(x_1-x_4\right)} & \frac{27}{3125 x_1 \left(x_1-x_4\right){}^3 \left(x_4-x_5\right)} \\
\frac{1}{5 x_1} & -\frac{4}{5 \left(x_1-x_5\right)} & -\frac{4}{5 \left(x_1-x_5\right)} & -\frac{4}{5 \left(x_1-x_5\right)} & -\frac{4}{5
\left(x_1-x_5\right)} & -\frac{4}{5 \left(x_1-x_5\right)} & 0 & 0 & 0 & \frac{5 x_1-x_5}{5 x_1 \left(x_1-x_5\right)} & \frac{256}{3125 x_1
\left(x_1-x_5\right){}^4} \\
\frac{2}{5 x_2} & \frac{2}{5 x_2} & -\frac{1}{5 \left(x_2-x_3\right)} & -\frac{1}{5 \left(x_3-x_4\right)} & -\frac{1}{5
\left(x_4-x_5\right)} & -\frac{1}{5 \left(x_4-x_5\right)} & \frac{x_3-2 x_4+x_5}{5 \left(x_3-x_4\right) \left(x_4-x_5\right)} &
\frac{x_2-2 x_3+x_4}{5 \left(x_2-x_3\right) \left(x_3-x_4\right)} & \frac{3 x_2-2 x_3}{5 x_2 \left(x_2-x_3\right)} & 0 & -\frac{4}{3125
x_2^2 \left(x_2-x_3\right) \left(x_3-x_4\right) \left(x_4-x_5\right)} \\
\frac{2}{5 x_2} & \frac{2}{5 x_2} & -\frac{1}{5 \left(x_2-x_3\right)} & -\frac{2}{5 \left(x_3-x_5\right)} & -\frac{2}{5
\left(x_3-x_5\right)} & -\frac{2}{5 \left(x_3-x_5\right)} & 0 & \frac{2 x_2-3 x_3+x_5}{5 \left(x_2-x_3\right) \left(x_3-x_5\right)} &
\frac{3 x_2-2 x_3}{5 x_2 \left(x_2-x_3\right)} & 0 & -\frac{16}{3125 x_2^2 \left(x_2-x_3\right) \left(x_3-x_5\right){}^2} \\
\frac{2}{5 x_2} & \frac{2}{5 x_2} & -\frac{2}{5 \left(x_2-x_4\right)} & -\frac{2}{5 \left(x_2-x_4\right)} & -\frac{1}{5
\left(x_4-x_5\right)} & -\frac{1}{5 \left(x_4-x_5\right)} & \frac{x_2-3 x_4+2 x_5}{5 \left(x_2-x_4\right) \left(x_4-x_5\right)} & 0 &
\frac{2 \left(2 x_2-x_4\right)}{5 x_2 \left(x_2-x_4\right)} & 0 & -\frac{16}{3125 x_2^2 \left(x_2-x_4\right){}^2 \left(x_4-x_5\right)} \\
\frac{2}{5 x_2} & \frac{2}{5 x_2} & -\frac{3}{5 \left(x_2-x_5\right)} & -\frac{3}{5 \left(x_2-x_5\right)} & -\frac{3}{5
\left(x_2-x_5\right)} & -\frac{3}{5 \left(x_2-x_5\right)} & 0 & 0 & \frac{5 x_2-2 x_5}{5 x_2 \left(x_2-x_5\right)} & 0 & -\frac{108}{3125
x_2^2 \left(x_2-x_5\right){}^3} \\
\frac{3}{5 x_3} & \frac{3}{5 x_3} & \frac{3}{5 x_3} & -\frac{1}{5 \left(x_3-x_4\right)} & -\frac{1}{5 \left(x_4-x_5\right)} & -\frac{1}{5
\left(x_4-x_5\right)} & \frac{x_3-2 x_4+x_5}{5 \left(x_3-x_4\right) \left(x_4-x_5\right)} & \frac{4 x_3-3 x_4}{5 x_3 \left(x_3-x_4\right)}
& 0 & 0 & \frac{27}{3125 x_3^3 \left(x_3-x_4\right) \left(x_4-x_5\right)} \\
\frac{3}{5 x_3} & \frac{3}{5 x_3} & \frac{3}{5 x_3} & -\frac{2}{5 \left(x_3-x_5\right)} & -\frac{2}{5 \left(x_3-x_5\right)} & -\frac{2}{5
\left(x_3-x_5\right)} & 0 & \frac{5 x_3-3 x_5}{5 x_3 \left(x_3-x_5\right)} & 0 & 0 & \frac{108}{3125 x_3^3 \left(x_3-x_5\right){}^2} \\
\frac{4}{5 x_4} & \frac{4}{5 x_4} & \frac{4}{5 x_4} & \frac{4}{5 x_4} & -\frac{1}{5 \left(x_4-x_5\right)} & -\frac{1}{5
\left(x_4-x_5\right)} & \frac{5 x_4-4 x_5}{5 x_4 \left(x_4-x_5\right)} & 0 & 0 & 0 & -\frac{256}{3125 x_4^4 \left(x_4-x_5\right)} \\
\frac{1}{x_5} & \frac{1}{x_5} & \frac{1}{x_5} & \frac{1}{x_5} & \frac{1}{x_5} & \frac{1}{x_5} & 0 & 0 & 0 & 0 & \frac{1}{x_5^5} \\
\end{array}
\right]
$$</span></p>
|
209,869 | <p>I am very interested in the maximum number of triangles could a connected graph with $n$ vertices and $m$ edges have. For example, if $m\leq n−1$, this number is $0$, if $m=n$, this number is $1$, if $m=n+1$, this number is $2$, and if $m=n+2$, this number is $4$. </p>
| Shahrooz | 19,885 | <p>It is a bound and since it is very long, I wrote it an answer, may be it can be helpful.</p>
<p>Let $G$ be a connected graph with $n$ vertices and $m$ edges. Suppose the eigenvalues of this graph are $\lambda_1\geq \lambda_2\geq\ldots\geq\lambda_n$. We know that $\sum{\lambda_i^3}=6\Delta_G$, where $\Delta_G$ counts the total number of triangles of the graph $G$.</p>
<p>Also,we have:</p>
<p>$$\lambda_1\leq\sqrt{2m-\delta(n-1)+\Delta(\delta-1)}.$$</p>
<p>Since your graph is connected, we can set $\delta=1$ and obtain:
$$\lambda_1\leq\sqrt{2m-n+1}.$$</p>
<p>So we have:</p>
<p>$$\Delta_G\leq\frac{n}{6}(2m-n+1)^{\frac{3}{2}},$$</p>
<p>as you wanted in your comments.</p>
<p>Actually, you can get more information from this method since we exactly know when the upper inequalities which I used are equality for which graphs. You can search for "SHARP UPPER BOUNDS OF SPECTRAL RADIUS OF GRAPHS" or similar keywords.</p>
|
65,059 | <p>I have two points ($P_1$ & $P_2$) with their coordinates given in two different frames of reference ($A$ & $B$). Given these, what I'd like to do is derive the transformation to be able to transform any point $P$ ssfrom one to the other.</p>
<p>There is no third point, but there <em>is</em> an extra constraint, which is that the y axis of Frame $B$ is parallel to the $X$-$Y$ plane of Frame $A$ (see sketch below). I <em>believe</em> that is enough information to be able to do the transformation.</p>
<p><img src="https://i.stack.imgur.com/2d6QH.png" alt="Two frames"></p>
<p>Also:</p>
<ul>
<li>The points are the same distance apart in both frames (no scaling).</li>
<li>The points don't coincide.</li>
<li>The origins don't necessarily coincide.</li>
</ul>
<p>As you may have gathered, I'm <em>not</em> a mathematician (ultimately this will end up as code), so please be gentle...</p>
<p><sub>I've seen this question (<a href="https://math.stackexchange.com/questions/23197/finding-a-rotation-transformation-from-two-coordinate-frames-in-3-space">Finding a Rotation Transformation from two Coordinate Frames in 3-Space</a>), but it's not <em>quite</em> the same as my problem, and unfortunately I'm not good enough at math to extrapolate from that case to mine.</sub></p>
<p><strong>EDIT</strong> I've updated the diagram, which makes it a bit cluttered, but (I hope) shows all the 'bits': $P3_B$ is what I'm trying to calculate...</p>
| davidlowryduda | 9,754 | <p>So given two points, an initial frame of reference, and the fact that the 'new y axis', which I will designate $y'$, is in the $xy$ plane, we do not have enough information to determine a unique new frame of reference around the new point. Even if $y'$ were parallel to $y$, we still don't quite have enough information. We need one more constraint, somewhere.</p>
<p>But let's say for a moment that you know both the new and the old frame of reference. By this, I mean that you know the vectors of the three mutually perpendicular basis elements (the vectors along the axes). Then, you are in business. As long as there is no scaling (vectors in the first frame don't get shorter or longer in the second), this is a simple process. I don't know what math you do or don't know, so I'll say it in some sort of general way and if it doesn't make sense to you, I can edit in more.</p>
<p>I assume throughout that the first frame of reference is at the origin (this is not unreasonable, and if we can't keep it then that's okay too). Assume for a moment that the two frames of reference are actually at the origin (i.e. that the second frame is at the origin too, but only for a moment). Then your task is to find the unique matrix transformation that rotates the original basis to the new basis. There are a few ways to do this, but the big idea is to remember that the columns of a matrix take one basis element to another. Good. Call this matrix $T$, so I can refer to it later.</p>
<p>Now find the difference between the two points. If $x_1$ is in the old frame, and $x_2$ is in the new frame, then find the vector $d = x_2 - x_1$. Then any element, given by the vector $v$ in the old frame will be given by the new vector $v' = [T][v] - d$, where I used brackets to emphasize the fact that there is a matrix multiplication.</p>
<p>Does that make any sense?</p>
|
1,319,767 | <p>If we know that $\frac{2^n}{n!}>0$ for every $n\in \mathbb{N}$ and $$\frac{2^n}{n!}=\frac{2}{1}\frac{2}{2}...\frac{2}{n}$$ how to bound this sequence above?</p>
| Timbuc | 118,527 | <p>$$a_n:=\frac{2^n}{n!}\implies\frac{a_{n+1}}{a_n}=\frac{2^{n+1}}{(n+1)!}\frac{n!}{2^n}=\frac2{n+1}\xrightarrow[n\to\infty]{}0$$</p>
<p>so by the ratio test (d'Alembert's), we get that the series</p>
<p>$$\sum_{n=1}^\infty\frac{2^n}{n!}\;\;\;\text{converges}\;\;\;\implies\;\;\;\lim_{n\to 0}\frac{2^n}{n!}=0$$</p>
|
2,115,484 | <p>I am not too sure how to prove that a hyperplane in $\mathbb{R}^{n}$ is convex? So far I know the definition of what convex is, but how do we add that hyperplane in $\mathbb{R}^{n}$ is convex?</p>
<p>Thanks in advance!</p>
| Kuifje | 273,220 | <p>Roughly speaking, you need to show that any two points of the hyperplane of $\mathbb{R}^n$ can be joined by a line segment.</p>
<p>First, lets define what this hyperplane $H$ is:
$$
H=\left\{ \pmatrix{x_1\\ \vdots\\ x_n} \in \mathbb{R}^n \;|\;
a_1x_1+\cdots+a_n x_n = c \right\}
$$
where $a_1,\cdots,a_n \neq 0$ and $c\in \mathbb{R}$.</p>
<p>So you need to show that
$$
\forall X,Y \in H, \; \lambda X+(1-\lambda)Y \in H, \; 0\le \lambda \le 1
$$</p>
<p>Lets do it:</p>
<p>$$
\lambda X+(1-\lambda)Y =\pmatrix{\color{red}{\lambda x_1+(1-\lambda)y_1}\\ \vdots \\ \color{red}{\lambda x_n+(1-\lambda)y_n}}
$$
Since $X,Y \in H$, we can write
$$
\begin{cases}
a_1x_1+\cdots+a_n x_n = c \\
a_1y_1+\cdots+a_n y_n = c
\end{cases}
$$
and so $\forall \lambda \in [0,1]$:
$$
\begin{cases}
\lambda a_1x_1+\cdots+\lambda a_n x_n = \lambda c \\
(1-\lambda)a_1y_1+\cdots+(1-\lambda)a_n y_n = (1-\lambda)c
\end{cases}
$$
Summing these two equations yields
\begin{align}
&\lambda a_1x_1+(1-\lambda)a_1y_1+\cdots+\lambda a_n x_n+(1-\lambda)a_n y_n = \lambda c+(1-\lambda)c \\
\Rightarrow \quad &a_1\left(\color{red}{\lambda x_1+(1-\lambda)y_1}\right)+\cdots+a_n\left(\color{red}{\lambda x_n+(1-\lambda)y_n}\right)= c \\
\Rightarrow \quad &\lambda X+(1-\lambda)Y \in H
\end{align}</p>
|
1,802,020 | <p>Let $S$ be a set such that if $A,B\in S$ then $A\cap B,A\triangle B\in S,$ where $\triangle$ denotes the symmetric difference operator. I would like to show that if $S$ contains $A$ and $B$, then it also contains $A\cup B, A\setminus B$.</p>
<p>The difference was easy to find, but I am not succeeding with the union. I was able to show that each of the sets $$\emptyset,A\setminus B, B\setminus A,(A\cap B)\cup(A\setminus B),(A\cap B)\cup(B\setminus A),A\cup(A\triangle B),B\cup(A\triangle B),\\(A\cap B)\cup(B\setminus A)\cup(A\triangle B)$$ is an element of $S$. Combining these I could go on producing other elements, and probably I would eventually find $A\cup B$. But I have already spent too long on the problem, there must be a cleverer approach, right?</p>
| Bérénice | 317,086 | <p>$$A\cup B=((A \triangle B) \cap A)\triangle B$$</p>
|
2,124,068 | <p>I came across the following problem in a book I was reading on continuous probability distributions:-</p>
<p>$Q.$ Let $Y$ be uniformly distributed on $(0,1)$. Find a function $\phi$ such that $\phi(Y )$ has the gamma density $\Gamma(\frac12,\frac12)$.</p>
<p>I know that the probability density represented by $\Gamma(\frac12,\frac12)$ is the following:-</p>
<p>$$\Gamma\left(\frac12,\frac12\right)=\begin{cases}\frac1{\sqrt{2\pi x}}.e^{-\frac x2} &&&& x \ge 0 \\ 0 &&&& x <0\end{cases}$$</p>
<p>I don't have any idea what to do after this. I would also like to have insight on similar questions.</p>
| David G. Stork | 210,401 | <pre><code>Manipulate[
Column[
{TextGrid[{{"x", "y", "z"},
v = {Cos[tt], Sin[tt], tt/10}},
Frame -> All],
Show[
ParametricPlot3D[{Cos[t], Sin[t], t/10}, {t, 0, 30}],
Graphics3D[{Red, PointSize[0.05], Point[v]}]]},
Alignment->Center],
{tt, 0, 30}]
</code></pre>
|
729,054 | <p>Let $f$ be continuous and $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$.</p>
<p>Suppose $|f(x)-x| \leq 2$ holds for all $x$. Is $f$ surjective?</p>
| Thomas Andrews | 7,933 | <p>A homotopy proof (outline.) Not sure how to do it with only analysis.</p>
<p>Assume $x_0\in\mathbb R^2$ is not in the range of $f$. consider $S=\{x\in\mathbb R^2: \left|x-x_0\right|=3\}$ be the circle of radius $3$ around $x_0$. </p>
<p>Then $f:S\to\mathbb R^2\setminus\{x_0\}$ is homotopic with the inclusion map $S\to\mathbb R^2\setminus\{x_0\}$, but the map: $h(x,t) = tx + (1-t)f(x)$, which never hits $x_0$ by the rule $|f(x)-x|\leq 2$.</p>
<p>But the function $f$ extends to the ball $B=\{x:|x-x_0|\leq 3\}$, which means that $f_{|S}$ retracts to a constant function as a map to $\mathbb R^2\setminus\{x_0\}$. But the inclusion map on $S$ does not retract to a point, which is a contradiction.</p>
<p>So our assumption is false.</p>
<p>Basically, this is a property that is a multi-dimensional intermediate value theorem. If a continuous $f:\mathbb R^2\to\mathbb R^2$ sends some circle to a path that "winds" a non-zero total number of times around some point $y_0$, then $y_0$ is in the range of the function. There is an $n$-dimensional version of this, too. I'm just not sure how to even define this concept of "wind" without some homotopy.</p>
|
422,948 | <p>How could I/is it possible to take a fourier transform of text? i.e. What domain would/does text exist in? Any help would be great.</p>
<p>NOTE: I do not mean text as an image. I understand it's value, but I'm wondering if it is possible to map text to some domain and transform text on the basis of letters. This is in hopes of performing frequency filtering on said text.</p>
| Seth | 229,250 | <p>Not quite in the frequency domain, but there is a way to look for periodic structures in text -- the Index of Coincidence. For normal text the IoC will be pretty much flat. But for text encrypted with, say, an 8-letter key and the Vigenère cipher, the IoC will show a pattern of 7 low values and a spike every 8th.</p>
<p>That tells you to take every 8th character and look for a key for that position, then try the next position. This also works with XOR and other ciphers.</p>
|
4,106,273 | <p>In how many ways can a committee of four be formed from 10 men (including Richard) <br>
and 12 women (including Isabel and Kathleen) if it is to have two men and two women <br></p>
<p>a) Isabel refuses to serve with Richard,</p>
<p>b) Isabel will serve only if Kathleen does, too</p>
<p>My Thoughts : <br>
a) Total number of ways to select 4 people = 22C4 now for part a) I just need to deduct the <br>
pairs where Isabel and Richard are both in the committe which = 20C2
<br>
I am not sure how to proceed with part (b) and if part (a) is entirely correct</p>
| IanJ | 854,548 | <p>You could divide the problem into all the committees without Isabel and all those with her. <span class="math-container">${10 \choose 2} {11 \choose 2}$</span> for those without her. And <span class="math-container">${9 \choose 2} * 1$</span> for those with her.</p>
|
1,255,376 | <blockquote>
<p>For all $q \in \Bbb Q$ there exists $n \in \Bbb Z$ so that $q + n = 271$.</p>
</blockquote>
<p>This is true? Because both $q$ and $n$ are rational numbers and $271$ is an integer thus it's a rational number? </p>
<p>Also,</p>
<blockquote>
<p>For all $n \in \Bbb Z$ there exists $q \in \Bbb Q$ so that $q + n = 271$. </p>
</blockquote>
<p>Isn't it the same thing? One should be true and one should be false. Any help? </p>
| Newb | 98,587 | <blockquote>
<p>For all $q \in \Bbb Q$ there exists $n \in \Bbb Z$ such that $q + n = 271$.</p>
</blockquote>
<p>False. </p>
<p>$\Bbb Q$ is the set of rationals, i.e. fractions: $\Bbb Q = \{\ldots, -\frac 1 3,- \frac 2 2,-\frac 1 2, 0, \frac 1 2, \frac 2 2, \frac 1 3, \ldots\}$.</p>
<p>$\Bbb Z$ is the set of integers: $\{\ldots, -3, -2, -1, 0, 1,2,3,\ldots\}$.</p>
<p>Take $q = \frac 1 2$ for an easy counterexample. Generally, any $q \in \Bbb Q$ such that $q \notin \Bbb Z$ yields a counterexample to the claim.</p>
<blockquote>
<p>For all $n \in \Bbb Z$ there exists $q \in \Bbb Q$ such that $q + n = 271$.</p>
</blockquote>
<p>True. It's easy to see by example why this is true, but there's a more general reason.</p>
<p>From the axioms of arithmetic, we know that for any $n \in \Bbb Z$ there exists an $m \in \Bbb Z$ such that $n+m = 271$. We also know that $\Bbb Z \subset \Bbb Q$, i.e. that the integers are a subset of the rationals. So when we go to pick the second integer $m$, that integer $m$ is also necessarily contained in the rationals (as $\frac m 1$).</p>
|
4,253,598 | <p>My textbook states that if <span class="math-container">$f(x) \to 0$</span> as <span class="math-container">$x \to 0$</span> <span class="math-container">$$\lim_{x \to 0} (1+f(x))^\frac{1}{g(x)} = e^l$$</span> where <span class="math-container">$$l=\lim_{x \to 0} \frac{f(x)}{g(x)}$$</span><br />
I try to do this as follows and I get a different result<br />
<span class="math-container">$$\lim_{x \to 0} (1+f(x))^\frac{1}{g(x)}= \lim_{x \to 0} ((1+f(x))^\frac{1}{f(x)})^\frac{f(x)}{g(x)}$$</span><br />
And now we take the exponent out of the limit to get<br />
<span class="math-container">$$[\lim_{x \to 0} (1+f(x))^\frac{1}{f(x)}]^\frac{f(x)}{g(x)}=e^l$$</span><br />
Where <span class="math-container">$$l=\frac{f(x)}{g(x)}$$</span>
Am I doing something wrong here because of which my <span class="math-container">$l$</span> is different?</p>
| Alessio K | 702,692 | <p>You are taking the limit with respect to <span class="math-container">$x$</span>, so the RHS should not depend on <span class="math-container">$x$</span>. Then you arrive at <span class="math-container">$\lim_{x \to 0} ((1+f(x))^\frac{1}{f(x)})^\frac{f(x)}{g(x)}$</span> and took the limit inside the bracket, which is not correct, since the limit will only act on the function inside the bracket. Note you can only write the limit inside a function if it exists and the function is continuous.</p>
<p>For example we know that <span class="math-container">$\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n}=e,$</span> but <span class="math-container">$\left(\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)\right)^{n}=1^{n}$</span> since the limit acts only inside the brackets.</p>
<hr />
<p>Let</p>
<p><span class="math-container">$$L=\lim_{x \to 0} (1+f(x))^\frac{1}{g(x)}$$</span></p>
<p>then <span class="math-container">$$\ln L = \lim_{x\rightarrow 0}\ln\left((1+f(x))^{\frac{1}{g(x)}}\right) \quad(1)$$</span>
<span class="math-container">$$\ln L = \lim_{x\rightarrow 0}\frac{\ln(1+f(x))}{g(x)}$$</span></p>
<p><span class="math-container">$$\ln L = \lim_{x\rightarrow 0}\frac{\ln(1+f(x))}{f(x)}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}$$</span>
<span class="math-container">$$\implies L=e^{\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}}$$</span></p>
<p>provided <span class="math-container">$\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}$</span> exist, and where I used the fact that <span class="math-container">$\lim_{x\rightarrow 0}\frac{\ln(1+f(x))}{f(x)}=1$</span>, which can be shown by expanding in power series. Indeed <span class="math-container">$\lim_{t\rightarrow 0}\frac{\ln(1+t)}{t}=1$</span>. Also in <span class="math-container">$(1)$</span> the continuity of <span class="math-container">$\ln$</span> and the fact that the limit inside this function exists have been used.</p>
|
1,382,479 | <p>I would like to know why $a^p \equiv a \pmod p$ is the same as $a^{p-1} \equiv 1 \pmod p$, and also how Fermat's little theorem can be used to derive Euler's theorem, or vice versa. </p>
<p>Please keep in mind that I have little background in math, and I am trying to understand these theorems to understand the math behind RSA encryption, so I would appreciate it if the explanation could be as simple as possible (i.e. not using too much mathematical notation). I have had trouble finding resources online to explain these theorems that explain it in simple enough terms that I can understand.</p>
| almahed | 258,478 | <p>They are <strong>not</strong> the same, if $\gcd(a,p)>1$, that is, if $p\mid a$, then $a^p\equiv a\pmod p$ is true while $a^{p-1}\equiv 1\pmod p$ is false. It follows that the first is always true for any $a\in\mathbb{Z}$, but the second only in the case where $\gcd(a,p)=1$.</p>
<p>As for the second question, assume $\gcd(a,p)=1$, thus provided that Fermat's little theorem is true as you ask, we know that $a^{\phi(m)}\equiv 1\pmod m$ when $m=p$, since the definition of Euler's totient function $\phi(x)$ says $\phi(p)=p-1$. In particular, $\phi(p^b)=(p-1)p^{b-1}$.</p>
<p>Thus, write $m=2^kp_1^{b_1}\cdot\ldots\cdot p_n^{b_n}$ for given $k,b_i>0$. Since</p>
<p>$$\phi(m)=2^{k-1}(p_1-1)p_1^{b_1-1}\cdot\ldots\cdot(p_n-1) p_n^{b_n-1}$$</p>
<p>you know that</p>
<p>\begin{array}{ll}
a^{\phi(m)}&\equiv 1\pmod 2\\
a^{\phi(m)}&\equiv 1\pmod{p_1}\\
&\vdots\\
a^{\phi(m)}&\equiv 1\pmod{p_n}\\
\end{array}</p>
<p>because for each case $p_i-1$ is in the exponent for a given $i-th$ prime (for 2 it is trivial).</p>
<p>The last step is to show that $a^{(p-1)p^{b-1}}\equiv 1\pmod{p^b}$. Notice that by showing this we can conclude that</p>
<p>\begin{array}{ll}
a^{\phi(m)}&\equiv 1\pmod 2\\
a^{\phi(m)}&\equiv 1\pmod{p_1^{b_1}}\\
&\vdots\\
a^{\phi(m)}&\equiv 1\pmod{p_n^{b_n}}\\
\end{array}</p>
<p>and from the fact that all divide $a^{\phi(m)}-1$ it would be safe to conclude that $a^{\phi(m)}\equiv 1\pmod m$ for all $m\in\mathbb{N}_{>0}$.</p>
<p>So, to prove that $a^{(p-1)p^{b-1}}\equiv 1\pmod{p^b}$ is the tricky part. We know that $a^{p-1}\equiv 1\pmod{p}$, but that a power $p^b$ must divide $a^{(p-1)p^{b-1}}-1$ is not trivial. To prove this fact, I use an old friend called Lifting the Exponent Lemma -- LTE (link: <a href="http://www.taharut.org/imo/LTE.pdf" rel="nofollow">Lifting the Exponent Lemma</a>).</p>
<p>It says: let $\gcd(x, p)=\gcd(y, p)=1$ for a prime $p$ and given whole numbers $x,\ y$, where $v_p$ is the maximum exponent $b$ such that $p^b\mid\ell$, for $\ell\in\mathbb{Z}$, then $v_p(x^n\pm y^n)=v_p(x\pm y)+v_p(n)$ for $n\in\mathbb{N}$.</p>
<p>From that we conclude that $v_p(a^{(p-1)p^{b-1}}-1^{(p-1)p^{b-1}})=v_p(a^{p-1}-1^{p-1})+v_p(p^{b-1})$, so clearly $a^{(p-1)p^{b-1}}\equiv 1\pmod{p^b}$.</p>
<p>I should say that the proof given by Euler is more economic and elegant. We don't need to actually <em>use</em> Fermat's little theorem to get Euler's theorem only because one is a particular case of the other.</p>
|
24,550 | <p>Let <span class="math-container">$H$</span> and <span class="math-container">$K$</span> be affine group schemes over a field <span class="math-container">$k$</span> of characteristic zero. Let <span class="math-container">$\varphi:H\to Aut(K)$</span> be a group action. Then we can form the semi-direct product <span class="math-container">$G = K\ltimes H$</span>.</p>
<p><strong>Problem</strong>: Describe the tannakian category <span class="math-container">$Rep_k(G)$</span> in terms of <span class="math-container">$Rep_k(K)$</span>, <span class="math-container">$Rep_k(H)$</span> and <span class="math-container">$\varphi$</span>.</p>
<p>If <span class="math-container">$\varphi$</span> is the trivial action, <span class="math-container">$G$</span> is the direct product <span class="math-container">$K\times H$</span>. In this case, <span class="math-container">$Rep_k(G) = Rep_k(K) \boxtimes Rep_k(G)$</span> is just the Deligne tensor product of the tannakian categories.</p>
<p><strong>Question</strong>: What happens in the opposite scenario where the action <span class="math-container">$\varphi$</span> is faithful? Can one characterize this situation in terms of Ext groups of simple objects in <span class="math-container">$Rep_k(K)$</span> and <span class="math-container">$Rep_k(H)$</span>?</p>
<p>I'm mostly insterested in the case where <span class="math-container">$K$</span> and <span class="math-container">$H$</span> are extensions of <span class="math-container">$\mathbb{G}_m$</span> by a pro-unipotent group but I have no idea where to start.</p>
| David Ben-Zvi | 582 | <p>The algebraic stack BG is the quotient of BK by the action of H, induced by its action on K. So we can describe coherent sheaves on it (aka reps of G) via descent from BK - i.e. reps of G are H-equivariant sheaves on BK, or H-equivariant objects in Rep K. Now this is not yet the answer you want, since it involves Rep K and H, rather than Rep H. </p>
<p>One indirect (and maybe slightly imprecise) answer is to use the Morita equivalence between Rep H-module categories and
categories with H action: the Rep H-module category Rep G corresponds under this equivalence to Rep K as a category with H action. One direction takes a category with H
action to its H-equivariant objects, which carry a natural Rep H action. The other
direction takes a Rep H category to the category of its eigenobjects.
Here an eigenobject is an object $M\in C$ with a functorial identification
$$V\ast M \simeq \underline{V}\otimes M$$ of the module action of $V\in Rep K$ with the naive tensor product by the underlying vector space of $V$. (see e.g. <a href="http://arxiv.org/abs/math/0010270" rel="noreferrer">this paper</a>) -- again this is just playing with monadic formalism.. So I would then characterize Rep G as the Rep H-module (via induction of representations) whose eigenobject category is Rep K as an H-category.. </p>
<p>This Morita equivalence for finite groups appears in papers of Mueger and Ostrik cited in <a href="http://arxiv.org/abs/0805.0157" rel="noreferrer">here</a>. For affine group schemes I proved a derived version of this result with Nadler and Francis, but it's not available sadly. I haven't thought this through in the usual Tannakian setting, so maybe I'm missing something obvious, but it should be a fairly straightforward (though 2-categorical!) application of Barr-Beck I would think: we're simply claiming that categories over BH are described via descent from a point, and the descent data is an action of H.</p>
|
416,253 | <p>I am playing around a bit with <span class="math-container">$W^*$</span>-algebras, and I'm trying to come up with a definition for the <span class="math-container">$W^*$</span>-algebraic tensor product. Here is my first attempt:</p>
<p><a href="https://i.stack.imgur.com/WE797.png" rel="noreferrer"><img src="https://i.stack.imgur.com/WE797.png" alt="enter image description here" /></a></p>
<p>It is easy to show that such an object exists. Simply represent the <span class="math-container">$W^*$</span>-algebras as concrete von Neumann algebras on Hilbert spaces and consider the usual von Neumann algebra tensor product (i.e. the weak closure of the algebraic tensor product <span class="math-container">$M \odot N$</span>).</p>
<p>However, I fail to show that the above definition determines the <span class="math-container">$W^*$</span>-algebra <span class="math-container">$M \overline{\otimes} N$</span> uniquely (up to (normal) <span class="math-container">$*$</span>-isomorphism). Is this true? If not, can we still modify the above universal property to get a well-working definition?</p>
<p>Basically, I want my universal property to convey the intuition that the <span class="math-container">$W^*$</span>-tensor product is simply the usual von Neumann tensor product when we appropriately represent the spaces involved.</p>
<p>Thanks in advance for any help!</p>
| Matthew Daws | 406 | <p>Something similar was studied by Wiersma in <a href="https://arxiv.org/abs/1506.01671" rel="noreferrer">arXiv:1506.01671 [math.OA]</a>. However, that paper takes a different definition of the universal property: it wants the individual maps <span class="math-container">$\sigma$</span> and <span class="math-container">$\tau$</span> to be weak<span class="math-container">$^*$</span>-homeomorphisms onto their ranges, and the map <span class="math-container">$\iota$</span> needs to have this property as well.</p>
<p>However, even here you do not recover the usual von Neumann tensor product, but rather something larger.</p>
|
3,644,870 | <p><strong>Give an example or argue that it is impossible.</strong></p>
<p>I argue that it is impossible because, if <span class="math-container">$(x_n)_n$</span> is a sequence which converges to 0, then <span class="math-container">$(x_n)_n$</span> must be bounded above or below by 0. As <span class="math-container">$(x_n)_n$</span> is in K, then the set K must have the element 0 which is false as K is in the open interval of <span class="math-container">$(0,∞)$</span>. </p>
<p>This is my argument. I am not sure if this is the right argument or the right way to express it. </p>
| Raoul | 761,097 | <p>Compactness is not needed only the fact that <span class="math-container">$K$</span> is closed: if <span class="math-container">$(x_n)$</span> is a sequence in a closed set that converges to <span class="math-container">$\ell$</span>, then <span class="math-container">$\ell$</span> belongs to said closed set.</p>
|
1,477,871 | <p>If the space $X$ is banach , then I want to show that any linear map $T:X \to X$ is continuous iff the null space is closed. I could show that if $T$ is continuous then the null space is closed. But I am unable to prove the converse. Any hints are appreciated. Thanks</p>
| Arpit Kansal | 175,006 | <p>This is not the solution of your problem but its a interesting result in case of $f$ is linear functional.</p>
<p>Here is a sketch of more general result.Try to fill the gaps in the proof.</p>
<p><strong>Lemma:</strong> Let $f$ be a linear functional on $X$ .A hyperplane $H= \{ x\in X : f(x) = \alpha\}$ is closed iff $f$ is continuous.</p>
<p><strong>Proof:</strong> Clearly if $f$ is continuous then $H$ is closed.Conversely,Assume $H$ is closed therefore $H^c$ is open and since $f$ is nonzero its nonempty.(Why?)</p>
<p>WLOG,assume that $x_0 \in H^c$ is such that $f(x_0) < \alpha$.Since $H^c$ is open there exists $r>0$ such that $B(x_0,2r) \subset H^c$.Now for all $x \in B(x_0,2r)$,we have $f(x) < \alpha$ (Why?)</p>
<p>Thus for any $z \in V$ such that $\|z\| \leq 1$,we get $f(x_0 +rz) \leq \alpha$ or equivalently $$ f(z) \leq \frac { \alpha -f(x_0)}{r}$$.Thus the image of the unit ball is bounded and hence $f$ is bounded (Why?) therefore continuous.</p>
|
25,778 | <p>I am going to teach a Calculus 1 course next semester, and I have 15 weeks for the course material. The class meets MWF for 50 minutes each. I have taught this class before using the same syllabus, but my colleague shared concerns that my pacing is too fast:</p>
<p>Week 1: Review of Functions</p>
<p>Week 2: Limits and Continuity; Infinite Limits</p>
<p>Week 3: Derivative (Limit Definition); Differentiation Rules; Transcendental Functions</p>
<p>Week 4: Implicit Differentiation; Related Rates; Linear Approximation and Differentials</p>
<p>Week 5: Extrema; Curve Sketching; L'Hopital's Rule</p>
<p>Week 6: Optimization; <em>Newton's Method</em>; Antiderivatives</p>
<p>Week 7: Integrals; Fundamental Theorem of Calculus</p>
<p>Week 8: Applications of Integrals: Work; Areas; Volumes of Solids</p>
<p>Week 9: Integration by Parts; Partial Fraction Decomposition</p>
<p>Week 10: Trig Substitution; <em>Approximate Integration</em></p>
<p>Week 11: Arc Length; Surface Area</p>
<p>Week 12: First-Order Differential Equations</p>
<p>Week 13: Parametric Equations and Polar Coordinates</p>
<p>Week 14: Introduction to Sequences and Series</p>
<p>Week 15: Review for Final</p>
<p>This is the syllabus I used last Spring, and I didn't have any problems with running out of time. I get straight to the point with my lectures, and my student grades have been above average compared to other instructors. However, the exams I use from my department only cover up to the fundamental theorem. So my students end up with a lot of excess information, since I cover up towards the end of a traditional Calculus 2 course. This means they're more than prepared for Calculus 2. However, I don't feel like going slower if I do not have to. I mean, if I were to get rid of the first week review, I could theoretically cover all of Calculus 1 and 2 in one course. Not sure why my colleagues take so long in lecturing. I sat in on a class and it took my colleague the entire 50 minutes to teach about power rule and product rule, when in the same time frame I can cover all differentiation rules plus transcendentals. Student evaluations seemed to be good. In a class of 33 students, 26 got "A's" and 5 got "B's" and 2 got "C's". No one failed.</p>
| David E Speyer | 51 | <p>As fedja says, if your students are doing as well as you say, and you believe the students are similar this year, there isn't a reason to change.</p>
<p>It seems hard to believe, though. I've taught calculus at U Michigan and UC Berkeley, which are generally considered to be good schools, and in 14 weeks with 4 hours a week, we generally get to your week 7. To give very rough numbers, about 25% of our students skip over first semester calculus and take a more advanced class, and about 50% never take it; so this course is targeting the 50th-75th percentile of math preparation. A bit more than half of them have had some high school calculus. Of course, we also have honors courses that cover much more; I'm talking about the non-honors section.</p>
<p>The reason is that the students who take these classes don't have a strong understanding about how to think about functions, graphing and formulae. So that each concept has to be presented many ways before they can use it. Some reasons you might be seeing different results:</p>
<ol>
<li><p>You are an amazing instructor.</p>
</li>
<li><p>You are at an extremely selective college. (But then why do your colleagues not get the same results?)</p>
</li>
<li><p>For some reason, you have unusually strong students in your section. Do your students come from a different major than the norm, or are a large number of them transfer students from a different educational system?</p>
</li>
<li><p>You are pushing out the students who would fail before they get to the final exam. How many students drop the course, or change out of your section, before they get to the final?</p>
</li>
<li><p>Your exams are way too easy.</p>
</li>
<li><p>There is a large amount of cheating on your exams. (Were they administered online?)</p>
</li>
</ol>
<p>If 1,2 or 3 applies and will apply next term then what you are doing is working so there is no reason to change. If 4 applies, then I believe you should work to keep and help those students, although it gets complicated. Obviously, if 5 or 6 applies you should fix your exams!</p>
<p>And, if 1 applies, then congratulations!</p>
|
1,186,516 | <p>Please the highlighted part in the image below. I don't understand why w(c2) must be larger than s(c1, c2) considering s(c1, c2) is counting the position where c1 + c2 = 0, c1 != 0 and c2 != 0 while w(c2) is only counting position where w(c2) != 0.</p>
<p>Should s(c1, c2) be larger than w(c2)?</p>
<p>Thanks for helping out!</p>
<p><img src="https://i.stack.imgur.com/MJzcm.png" alt="This is the question and its answer key"></p>
| Surb | 154,545 | <p>Since $|h|\leq H$ we have $\frac{|h|}{H}\leq 1$ and thus $$\frac{|h|^k}{H^k}\leq \frac{|h|^2}{H^2}$$ for every $k\geq 2$. Moreover adding two nonnegative terms (since $0\leq H$) in the sum will also make it larger so that</p>
<p>$$\sum_{k=2}^n\binom{n}{k}|x|^{n-k}\frac{|h|^k}{H^k}H^k \leq \sum_{k=2}^n\binom{n}{k}|x|^{n-k}\frac{|h|^2}{H^2}H^k=\frac{|h|^2}{H^2}\sum_{k=2}^n\binom{n}{k}|x|^{n-k}H^k\\ \leq \frac{|h|^2}{H^2}\left(\sum_{k=2}^n\binom{n}{k}|x|^{n-k}H^k +\binom{n}{1}|x|^{n-1}H^1+\binom{n}{0}|x|^{n-0}H^0\right) = \frac{|h|^2}{H^2}\sum_{k=0}^n\binom{n}{k}|x|^{n-k}H^k$$</p>
|
4,380,748 | <p>I suppose by contradiction that <span class="math-container">$x+a$</span> is a factor of <span class="math-container">$x^n-a^n$</span> for all odd <span class="math-container">$n$</span>. In particular for <span class="math-container">$n=1$</span>, we have that <span class="math-container">$x+a$</span> is a factor of <span class="math-container">$x-a$</span>, but that is not possible. So that would be a contradiction. If my proof is correct?</p>
| Brendan Connery | 1,082,137 | <p>Humans always tell the truth
Werewolves always lie.</p>
<p>Each option is a statement identifying how many of them are liars in the group. meaning, that each statement is contradictory to one another, so only one of them can be true. There can only be 1 2 or 3 werewolves, but a werewolf cannot tell the truth. This immediately eliminates the possibility of 3 werewolves, because one of the statements HAS to be true, and each werewolf is a liar. The same goes for 3 humans. For a statement to be true, there must be at least a liar, but only a human can truthfully identify the number of liars.</p>
<p>Therefore only one statement can be true, and a human must be the one to tell you the truth. BECAUSE the statements contradict one another, it is impossible to have two humans, only one of the statements can be true at once, which would make one of them statements false.</p>
<p>There are 2 werewolves, and the human tells you that. the other 2 werewolves lie and tell you there are 1 and 3 werewolves.</p>
|
2,859,463 | <blockquote>
<p>Prove or disprove. All four vertices of every regular tetrahedron in $ \mathbb{R}^3$ have at least two irrational coordinates.</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/hYrWv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hYrWv.png" alt="enter image description here"></a></p>
<p>This question arose from my inability to construct a tetrahedron in $\mathbb{R}^3$ with all the coordinates
$$
M_x,M_y,M_z,N_x,N_y,N_z,P_x,P_y,P_z,Q_x,Q_y \;\mathrm
{and}\; Q_z,
$$
of its vertices $M$, $N$, $P$ and $Q$ being rational numbers. </p>
| Christian Blatter | 1,303 | <p>Choose four suitable vertices of the unit cube $[0,1]^3$, and you have a regular tetrahedron.</p>
|
3,757,222 | <p>Let <span class="math-container">$n_{1}, n_{2}, ... n_{k} $</span> be a sequence of k consecutive odd integers. If <span class="math-container">$n_{1} + n_{2} + n_{3} = p^3$</span> and <span class="math-container">$n_{k} + n_{k-1} + n_{k-2} + n_{k-3} + n_{k-4} = q^4$</span> where both p and q are prime, what is k?</p>
<p>I am struggling with this question. I know the first sum can be written as <span class="math-container">$3n_{1} + 6 = p^3$</span> and the second sum can be written as <span class="math-container">$5n_{k} - 20 = q^4$</span>. I believe the second sum is also <span class="math-container">$5n_{1} +10k - 30 = q^4$</span>. However rearranging these I get no workable equations.</p>
| quasi | 400,434 | <p>Hints:</p>
<p>The sum of <span class="math-container">$3$</span> consecutive odd integers is always a multiple of <span class="math-container">$3$</span>.</p>
<p>What does that imply about <span class="math-container">$p$</span>?</p>
<p>Can you then find <span class="math-container">$n_1,n_2,n_3$</span>?</p>
<p>The sum of <span class="math-container">$4$</span> odd numbers is always a multiple of <span class="math-container">$2$</span>.</p>
<p>
What does that imply about <span class="math-container">$q$</span>?
<p>
Can you finish?
<p>
<b>Note:</b><span class="math-container">$\;$</span>The wording of the problem is a little sneaky. The phrase "consecutive odd integers" doesn't automatically mean consecutive <i>increasing</i> (although by default, it usually does). Keep that in mind.
|
3,757,222 | <p>Let <span class="math-container">$n_{1}, n_{2}, ... n_{k} $</span> be a sequence of k consecutive odd integers. If <span class="math-container">$n_{1} + n_{2} + n_{3} = p^3$</span> and <span class="math-container">$n_{k} + n_{k-1} + n_{k-2} + n_{k-3} + n_{k-4} = q^4$</span> where both p and q are prime, what is k?</p>
<p>I am struggling with this question. I know the first sum can be written as <span class="math-container">$3n_{1} + 6 = p^3$</span> and the second sum can be written as <span class="math-container">$5n_{k} - 20 = q^4$</span>. I believe the second sum is also <span class="math-container">$5n_{1} +10k - 30 = q^4$</span>. However rearranging these I get no workable equations.</p>
| salfaris | 453,441 | <p>The idea is to figure out <span class="math-container">$p$</span>, then <span class="math-container">$n_1$</span>, then <span class="math-container">$q$</span> and finally <span class="math-container">$k$</span>.</p>
<p><span class="math-container">$3n_1 + 6 = 3(n_1 + 2) = p^3$</span> implies that <span class="math-container">$p$</span> divides <span class="math-container">$3(n_1 + 2)$</span>. Since <span class="math-container">$p$</span> is prime, then either <span class="math-container">$p = 3$</span> or <span class="math-container">$p \mid (n_1 + 2)$</span>. If <span class="math-container">$p \neq 3$</span>, then we can run a descent argument to get a contradiction. An explicit run of this argument is given below.</p>
<blockquote>
<p>Explicitly: if <span class="math-container">$p \neq 3$</span>, then <span class="math-container">$p \mid (n_1+2)$</span>, which implies that
<span class="math-container">$n_1 + 2 = p \ell$</span> for some <span class="math-container">$\ell \in \mathbb{Z}$</span>. Substituting in the
original equation, we get <span class="math-container">$3p\ell = p^3$</span> or <span class="math-container">$3 \ell = p^2$</span>. But <span class="math-container">$p$</span>
is prime and is not <span class="math-container">$3$</span>, so <span class="math-container">$p \mid \ell$</span> and <span class="math-container">$\ell = p \ell'$</span>.
Substituting, we get <span class="math-container">$3p^2 \ell' = p^3$</span> or <span class="math-container">$3\ell' = p$</span>.
Again, <span class="math-container">$p$</span> is prime but is not <span class="math-container">$3$</span> so we can write <span class="math-container">$\ell' = p \ell''$</span>, and substituting
yields <span class="math-container">$3p^3 \ell'' = p^3$</span> or <span class="math-container">$3 \ell'' = 1$</span> which is a contradiction
as <span class="math-container">$\ell''$</span> is an integer and <span class="math-container">$3$</span> is not a unit in <span class="math-container">$\mathbb{Z}$</span>.</p>
</blockquote>
<p>Thus, <span class="math-container">$p = 3$</span> and so <span class="math-container">$n_1 = 7$</span>.</p>
<p>How about <span class="math-container">$q$</span>? As you suggested, we have <span class="math-container">$5(n_1 + 2k - 6) = q^4$</span>. Using <span class="math-container">$n_1 = 7$</span>, we get <span class="math-container">$5(1 + 2k) = q^4$</span>. Since <span class="math-container">$q$</span> is prime, then either <span class="math-container">$q = 5$</span> or <span class="math-container">$q \mid (1 + 2k)$</span>. If <span class="math-container">$q \neq 5$</span>, we can run a descent argument again to arrive at a contradiction.</p>
<p>So <span class="math-container">$q = 5$</span> and <span class="math-container">$q^4 = 625$</span>. This implies that <span class="math-container">$k = 62$</span>.</p>
|
584,171 | <blockquote>
<p>Show that every graph can be embedded in $\mathbb{R}^3$ with all
edges straight. </p>
</blockquote>
<p>(Hint: Embed the vertices inductively, where should you
not put the new vertex?)</p>
<p>I've also received a tip about using the curve ${(t, t^2 , t^3 : t \in \mathbb{R} )}$ but I'm not sure how to do that and how I should go about proving this rigorously but without using any pure topology. </p>
<p>Any hints or useful suggestions?</p>
<p>Thanks a lot!</p>
| alejopelaez | 1,318 | <p><strong>Hint:</strong> To expand on the hint you were given. If you are able to show that you can put the vertices in such a way that no four of them are coplanar then you are done (Can you see why?).</p>
<p>Now, place the first three vertices, in light of my previous comment, where should you not place the fourth? and the fifth? etc.</p>
|
120,992 | <p>An algorithm book <a href="http://rads.stackoverflow.com/amzn/click/1849967202" rel="nofollow">Algorithm Design Manual</a> has given an description:</p>
<blockquote>
<p>Consider a graph that represents the street map of Manhattan in New York City. Every junction of two streets will be a vertex of the graph. Neighboring junctions are connected by edges. <strong>How big is this graph? Manhattan is basically a grid of 15 avenues each crossing roughly 200 streets. This gives us about 3,000 vertices and 6,000 edges</strong>, since each vertex neighbors four other vertices and each edge is shared between two vertices.</p>
</blockquote>
<p>If it says "The graph is a grid of 15 avenues each crossing roughly 200 streets", how can I calculate the number of vertices and edges? Although the description above has given the answers, but I just can't understand.</p>
<p>Can anyone explain the calculation more easily?</p>
<p>Thanks</p>
| El'endia Starman | 10,537 | <p>Every junction between an avenue and a street is a vertex. As there are $15$ avenues and (about) $200$ streets, there are (about) $15*200=3000$ vertices. Furthermore, every vertex has an edge along an avenue and an edge along a street that connect it to two other vertices. Hence, there are (about) $2*3000 = 6000$ edges<sup>1</sup>. Does that answer your question?
<hr />
<sup>1</sup> With regards to edges, a visual way to imagine it would be to imagine that the avenues are going north-south and the streets are going east-west. Start with the junction/vertex in the northwesternmost corner. It is adjacent to two other vertices: one south along the avenue and one east along the street. Similarly, every vertex has a vertex to the south and a vertex to the east (as well as north and west for most of them, but those are irrelevant for this). Hence, there are two edges for every vertex.</p>
|
120,992 | <p>An algorithm book <a href="http://rads.stackoverflow.com/amzn/click/1849967202" rel="nofollow">Algorithm Design Manual</a> has given an description:</p>
<blockquote>
<p>Consider a graph that represents the street map of Manhattan in New York City. Every junction of two streets will be a vertex of the graph. Neighboring junctions are connected by edges. <strong>How big is this graph? Manhattan is basically a grid of 15 avenues each crossing roughly 200 streets. This gives us about 3,000 vertices and 6,000 edges</strong>, since each vertex neighbors four other vertices and each edge is shared between two vertices.</p>
</blockquote>
<p>If it says "The graph is a grid of 15 avenues each crossing roughly 200 streets", how can I calculate the number of vertices and edges? Although the description above has given the answers, but I just can't understand.</p>
<p>Can anyone explain the calculation more easily?</p>
<p>Thanks</p>
| FangyuanJ | 27,056 | <p>This is so simple, I mean 15 avenues crossing 200 streets, which means there are 15 * 200 = 3000 crossings, i.e. 3000 nodes. </p>
<p>each nodes have upper, lower, left and right neighbors, so for each node, there are 4 edges connecting to the neighbors. However, each edge has been counted twice since node 1 has an edge connected with node 2, and the node 2 has the same edge connected to node 1</p>
<p>So that is totally 3000*4/2 = 6000 edges</p>
|
546,123 | <p>Let <span class="math-container">$X$</span> be any uncountable set with the cofinite topology. Is this space first countable?</p>
<p>I don't think so because it seems that there must be an uncountable number of neighborhoods for each <span class="math-container">$ x \in X$</span>. But I am not sure if this is true.</p>
| Brian M. Scott | 12,042 | <p>You are correct: it is not first countable. However, this is not because each point of $X$ has uncountably many nbhds: each point of $\Bbb R$ also has uncountably many nbhds, but $\Bbb R$, being a metric space, is certainly first countable.</p>
<p>To prove that $X$ is not first countable, you must show that some point of $X$ does not have a countable local base. All points of $X$ ‘look alike’ in the cofinite topology, so it doesn’t matter what point we pick, so let $x\in X$ be any point. Suppose that $\mathscr{B}=\{B_n:n\in\Bbb N\}$ is a countable local base of open sets at the point $x$, meaning that if $U$ is any open nbhd of $x$, then $x\in B_n\subseteq U$ for some $n\in\Bbb N$. For each $n\in\Bbb N$ let $F_n=X\setminus B_n$: $B_n$ is open, so by definition $F_n$ is finite. Let $F=\{x\}\cup\bigcup_{n\in\Bbb N}F_n$; $F$ is the union of countably many finite sets, so $F$ is countable. $X$ is uncountable, so there is some $y\in X\setminus F$. Let $U=X\setminus\{y\}$.</p>
<ul>
<li>Is $U$ an open nbhd of $x$? </li>
<li>Is there any $n\in\Bbb N$ such that $x\in B_n\subseteq U$?</li>
</ul>
|
1,282,545 | <p>If $G$ is a infinite group, then $G$ must have an element of infinite order.</p>
<p>Is this true?</p>
<p>I know that if $G$ is infinite cyclic, then it's isomorphic to $\mathbb Z$. </p>
<p>(I guess fact is irrelevant now)</p>
| Hagen von Eitzen | 39,174 | <p>Consider $(\mathbb Z/2\mathbb Z)^{\mathbb N}$ or $\mathbb Q/\mathbb Z$.</p>
|
1,282,545 | <p>If $G$ is a infinite group, then $G$ must have an element of infinite order.</p>
<p>Is this true?</p>
<p>I know that if $G$ is infinite cyclic, then it's isomorphic to $\mathbb Z$. </p>
<p>(I guess fact is irrelevant now)</p>
| David Holden | 79,543 | <p>in an abelian group the elements of finite order form its <i>torsion</i> subgroup. a torsion group is infinite iff it is not finitely generated. $\Bbb Q / \Bbb Z$, cited by Hagen in his answer is a particularly clear example. </p>
|
1,369,076 | <p>Are there any good "analysis through problems" type books? I've tried reading analysis books but I literally get bored to death, and, until I manage to concoct a way of transforming a normal textbook into a problem book (maybe by trying to prove all the theorems myself, but that probably requires more math maturity on my behalf and I don't have that yet I think), I am really interested in an analysis through problems book. I know there exist good ones for number theory (Burn's pathway into number theory), linear algebra (halmos' problem book), abstract algebra (clarke's abstract algebra), geometry (prasolov), etc. Any for analysis?</p>
<p>Thanks</p>
<p>Edit: New title - I think it expresses "analysis through problems" better.</p>
| Danny | 76,017 | <p>One book that you can get for free online is <em>Introductory Single Variable Real Analysis: A Learning Approach Through Problem Solving</em> by Marcel Finan.</p>
<p>One book that I'd particularly recommend if you're looking for really unique and interesting analysis problems is <em>Real Mathematical Analysis</em> by Charles Pugh. To give you a taste, here's an exercise from the topology chapter:</p>
<p><em>Prove that there is no way to place uncountably many copies of the letter T disjointly in the plane.</em></p>
|
1,724,419 | <p>I can create a large collection of normalized real valued $n$-dimensional vectors from some random process which I hypothesis should be equidistributed on the unit sphere. I would like to test this hypothesis.</p>
<ul>
<li>What is a good way numerically to test if vectors are equidistributed on the unit sphere? I am writing computer code so I will be testing that way</li>
<li>Is there some way to visualise the distribution given that my vectors are in $n$ dimensions?</li>
</ul>
| Raaja_is_at_topanswers.xyz | 286,483 | <p>In the whole vector space defined by your normalised vectors in $\mathbb{R}^n$, you can try to find the inner product of the vectors (in $\mathcal{L}_2$ space) and with the output, you can decide whether it is equidistributed on the unit sphere (n-dimension).</p>
<p>This is one of the numerically reliable method.</p>
|
3,203,678 | <blockquote>
<p>Find all prime numbers <span class="math-container">$p$</span>, for which there are positive integers <span class="math-container">$m$</span> and <span class="math-container">$n$</span> such that <span class="math-container">$p=m^2+n^2$</span> and <span class="math-container">$p \mid m^3+n^3-4$</span>.</p>
</blockquote>
<p>I simplified this a little bit.</p>
<p><span class="math-container">$$m^2+n^2 \mid m^3+n^3-4 =(m+n)(m^2+n^2-mn)-4 \\ \Longrightarrow m^2+n^2 \mid (m+n)mn+4 $$</span></p>
<p>The only case that <span class="math-container">$m$</span> and <span class="math-container">$n$</span> can both be odd is <span class="math-container">$m=n=1$</span>, which leads to <span class="math-container">$p=2$</span>. If one of <span class="math-container">$m$</span> and <span class="math-container">$n$</span> is <span class="math-container">$1$</span> (For example <span class="math-container">$n=1$</span>), then <span class="math-container">$p=m^2+1$</span> and </p>
<p><span class="math-container">$$m^2+1 \mid m^2+m+4 \Longrightarrow m^2+1 \mid m+3 \Longrightarrow m=2$$</span>
Which gives <span class="math-container">$p=5$</span>.</p>
<p>For <span class="math-container">$m,n>1$</span> with different parity, I could not find a strong argument. Can you guys give it a try?</p>
| TBTD | 175,165 | <p>This problem is from 2004 Silkroad Mathematical Competition. By the way, if you are into this problems, I think you should really visit AoPS forums, audience here at stack exchange are not very experienced with olympiadic problems.</p>
<p>Here is a short reasoning (similar to above). Check that, <span class="math-container">$p=2,5$</span> indeed works. Suppose now <span class="math-container">$p>5$</span>. Note that, <span class="math-container">$m^3+n^3=(m+n)(m^2-mn+n^2)$</span>. Notice that, <span class="math-container">$m^2-mn+n^2\equiv -mn\pmod{p}$</span>. This yields, <span class="math-container">$p\mid mn(m+n)+4$</span>. Next, note also that, <span class="math-container">$p=m^2+n^2$</span> implies, <span class="math-container">$(m+n)^2\equiv 2mn\pmod{p}$</span>, and thus, <span class="math-container">$mn\equiv \frac{(m+n)^2}{2}\pmod{p}$</span>. This yields, <span class="math-container">$p\mid (m+n)^3+8=(m+n+2)((m+n)^2+2(m+n)+4)$</span>. Now, if <span class="math-container">$p\mid m+n$</span>, we have <span class="math-container">$m^2+n^2\mid m+n$</span> with <span class="math-container">$m,n$</span> not simultaneously one, which is not sound. Similarly, if <span class="math-container">$p\mid m^2+2mn+n^2+2(m+n)+4$</span>, then note that, <span class="math-container">$p\mid 2mn+2(m+n)+4$</span>, or equivalently, <span class="math-container">$p\mid mn+m+n+2$</span>, since <span class="math-container">$p>2$</span>. This yields <span class="math-container">$m^2+n^2\leq mn+m+n+2$</span>. Now, using AM-GM inequality, we also have <span class="math-container">$m^2+n^2\geq 2mn$</span>, which implies, <span class="math-container">$mn-m-n+1\leq 3$</span>, that is, <span class="math-container">$(m-1)(n-1)\leq 3$</span>. In particular, either <span class="math-container">$(m-1)(n-1)=1$</span>, which is not possible, or <span class="math-container">$(m-1)(n-1)=2$</span>, yielding <span class="math-container">$m=3,n=2$</span>, and <span class="math-container">$p=13$</span>, for which the condition does not hold, or <span class="math-container">$(m-1)(n-1)=3$</span>, yielding <span class="math-container">$m=4,n=2$</span>, for which, again, <span class="math-container">$m^2+n^2$</span> is not prime.</p>
|
3,772,534 | <p>Tangents to a circumference of center O, drawn by an outer point C, touch the circle at points A and B. Let S be any point on the circle. The lines SA, SB and SC cut the diameter perpendicular to OS at points A ', B' and C ', respectively. Prove that C 'is the midpoint of A'B'.</p>
<p><a href="https://i.stack.imgur.com/Ggwjj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ggwjj.jpg" alt="My draw (geogebra)" /></a></p>
<p>I saw a solution by Projective Geometry. I want to know if there is a solution by euclidean geometry. I think that is possible to do with Menelaus Theorem, but I don't know wich triangles I have to use. Thanks for attention.</p>
| Mikael Helin | 418,258 | <p>Use <span class="math-container">$(-1)^{\frac{1}{4}}$</span> to be able to use the series, so you get
<span class="math-container">$$\frac{\cos (\frac{(1+i)x}{\sqrt 2})+\cosh (\frac{(1+i)x}{\sqrt 2})}{2}=\sum_{n=0}^\infty\frac{(-1)^nx^{4n}}{(4n)!}
$$</span>
Then use
<span class="math-container">$$
\cos(a+ib)=\cos a\cosh b-i\sin a\sinh b
$$</span>
and
<span class="math-container">$$
\cosh(a+ib)=\cos b\cosh a+i\sin b\sinh a.
$$</span>
In our case <span class="math-container">$a=b$</span> so <span class="math-container">$\cos(a+ia)+\cosh(a+ib)=2\cos a\cosh a$</span>. Set <span class="math-container">$a=\frac{x}{\sqrt 2}$</span> and we get the solution
<span class="math-container">$$
\sum_{n=0}^\infty\frac{(-1)^nx^{4n}}{(4n)!}=\cos \frac{x}{\sqrt 2}\cosh \frac{x}{\sqrt 2}.
$$</span></p>
|
1,972,927 | <p>What's the function determined by the series $1+\sin(x)\cos(x) + \sin(x)^2 \cos(x)^2 + \cdot \cdot \cdot$?</p>
<hr>
<p>Note, the series converges uniformly.</p>
| lab bhattacharjee | 33,337 | <p>As $\sin x\cos x=\dfrac{\sin2x}2$ and $-1\le\sin2x\le1$ for real $x$</p>
<p>$$\sum_{r=0}^\infty ar^n=\dfrac a{1-r}$$ for $|r|<1$</p>
|
1,690,346 | <p>How can we solve this integral?
$\int_0^1\:\frac{\ln(x)\:\Big[1+x^{-\frac{1}{3}}\Big]}{(1-x)\sqrt[3]{x}}\:dx$</p>
| Ron Gordon | 53,268 | <p>Subbing $x=u^3$ is very natural here and leads to a simple sum...</p>
<p>$$I = \int_0^1 dx \frac{\left (1+x^{-1/3} \right ) \log{x}}{x^{1/3} (1-x)} = 9 \int_0^1 du \, \frac{(1+u) \log{u}}{1-u^3}$$</p>
<p>Expand the denominator...</p>
<p>$$\begin{align}I &= 9 \sum_{k=0}^{\infty} \int_0^1 du \, (1+u) u^{3 k} \log{u}\\ &= -9 \sum_{k=0}^{\infty} \left (\frac1{(3 k+1)^2} + \frac1{(3 k+2)^2} \right ) \\ &= -9 \frac{\pi^2}{6} + 9 \sum_{k=0}^{\infty} \frac1{(3 k+3)^2}\\ &= - \frac{9\pi^2}{6} + \frac{\pi^2}{6}\\ &= -\frac{4 \pi^2}{3}\end{align}$$</p>
|
2,522,342 | <p>So far I have only got 9 from just guess and check. I am thinking of using Vieta's Formula, but I am struggling over the algebra. Can someone give me the first few steps?</p>
| Ovi | 64,460 | <p>If $x$ is odd, there we can apply the factoring formula of $a^x + b^x$ to show that $2^x+1$ is divisible by $3$. If $x$ is even, you can factor $2^x-1$ and see that $x$ must be $2$.</p>
|
171,364 | <p>So I'm looking for a function that takes in the degree of the polynomial and the range of coefficients from -c to c, and outputs a list of all the monic polynomials of that degree and with coefficients in that range.</p>
<p>I already have code to numerically compute the roots and plot in the complex plane, I just need a way to compute this list. I haven't been able to find previously posted code to do this task on stackexchange. </p>
| ulvi | 1,714 | <p><code>toMonicpol[lis_] :=
x^(Length[lis]) + Dot[lis, Table[x^r, {r, 0, Length[lis] - 1}]]</code></p>
<p><code>pols[deg_, c_] := Map[toMonicpol[#] &, Tuples[Range[-c, c], deg]]</code></p>
<p><code>pols[2, 3]</code></p>
<p><code>{-3 - 3 x + x^2, -3 - 2 x + x^2, -3 - x + x^2, -3 + x^2, -3 +
x + x^2, -3 + 2 x + x^2, -3 + 3 x + x^2, -2 - 3 x + x^2, -2 - 2 x +
x^2, -2 - x + x^2, -2 + x^2, -2 + x + x^2, -2 + 2 x + x^2, -2 +
3 x + x^2, -1 - 3 x + x^2, -1 - 2 x + x^2, -1 - x + x^2, -1 +
x^2, -1 + x + x^2, -1 + 2 x + x^2, -1 + 3 x + x^2, -3 x +
x^2, -2 x + x^2, -x + x^2, x^2, x + x^2, 2 x + x^2, 3 x + x^2,
1 - 3 x + x^2, 1 - 2 x + x^2, 1 - x + x^2, 1 + x^2, 1 + x + x^2,
1 + 2 x + x^2, 1 + 3 x + x^2, 2 - 3 x + x^2, 2 - 2 x + x^2,
2 - x + x^2, 2 + x^2, 2 + x + x^2, 2 + 2 x + x^2, 2 + 3 x + x^2,
3 - 3 x + x^2, 3 - 2 x + x^2, 3 - x + x^2, 3 + x^2, 3 + x + x^2,
3 + 2 x + x^2, 3 + 3 x + x^2}</code></p>
|
1,111,041 | <p>Given: $y=\log(1+x)$</p>
<p>Show that $y≈x$ if $x$ gets small (less than 1).</p>
<p>I don't think we're supposed to use Taylor series (because they were never formally introduced in class), but I do think we have to differentiate and show that the derivative of $\log(1+x)$ is approximately equal to $\log(1+x)$ on the interval $0$ to $1$. How should I show this?</p>
| Barry Cipra | 86,747 | <p>By definition of the (natural) logarithm,</p>
<p>$$\log(1+x)=\int_1^{1+x}{du\over u}$$</p>
<p>If $x\approx0$, then ${1\over u}\approx1$ for $1\le u\le 1+x$, in which case</p>
<p>$$\log(1+x)\approx\int_1^{1+x}du=u\Big|_1^{1+x}=(1+x)-1=x$$</p>
<p>(Remark: I wrote $1\le u\le 1+x$ with $x\gt0$ in mind. A more precise version would be $1-|x|\le u\le1+|x|$.)</p>
|
2,935,693 | <p>I am trying to prove that </p>
<p><span class="math-container">$(a\to(b\to c))\to((a\to b)\to(a\to c))$</span></p>
<p>holds in natural deduction, in particular when I work backwards from a Fitch style proof I can only get so far:</p>
<p><a href="https://i.stack.imgur.com/w2BYf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/w2BYf.png" alt="enter image description here"></a></p>
<p>How can I prove it?</p>
| Mauro ALLEGRANZA | 108,274 | <p>You have only to remove the unnecessary LEM application : lines 3,9,10,11, and you will get the correct derivation :</p>
<p>1) <span class="math-container">$a \to (b \to c)$</span> --- assumed [a]</p>
<p>2) <span class="math-container">$a \to b$</span> --- assumed [b]</p>
<p>3) <span class="math-container">$a$</span> --- assumed [c]</p>
<p>4) <span class="math-container">$b$</span> --- from 2) and 3), by <span class="math-container">$\to$</span>-elim</p>
<p>5) <span class="math-container">$b \to c$</span> --- from 1) and 3), by <span class="math-container">$\to$</span>-elim</p>
<p>6) <span class="math-container">$c$</span> --- from 5) and 4), by <span class="math-container">$\to$</span>-elim</p>
<p>7) <span class="math-container">$a \to c$</span> --- from 3) and 6), by <span class="math-container">$\to$</span>-intro, discharging [c]</p>
<p>8) <span class="math-container">$(a \to b) \to (a \to c)$</span> --- from 2) and 7) by <span class="math-container">$\to$</span>-intro, discharging [b]</p>
<blockquote>
<p>9) <span class="math-container">$\vdash (a \to (b \to c)) \to ((a \to b) \to (a \to c))$</span> --- from 1) and 8) by <span class="math-container">$\to$</span>-intro, discharging [a].</p>
</blockquote>
|
1,025,321 | <p>$\ln(1+xy) = xy$</p>
<p>When I try to implicitly differentiate this I get</p>
<p>$\frac{1}{1+xy}(y + xy')$ = (y + xy')</p>
<p>At which point the $(y + xy')$ terms cancel out, leaving no $y'$ to solve for.</p>
<p>However, the answer to this is $-\frac{y}{x}$... How do you get this?</p>
| orangeskid | 168,051 | <p>Let $\phi \colon [0,1) \to [0, \infty)$ a diffeomorphism with inverse $\psi$. Some possible choices: $t \mapsto \frac{t}{1-t}$, $t \mapsto \tan (\frac{\pi}{2}\cdot t)$.</p>
<p>The map
$$x \mapsto \phi(||x||) \cdot \frac{x}{||x||}$$</p>
<p>is a diffeomorphism from $B^n$ to $ \mathbb{R}^n$ with inverse
$$y \mapsto \psi(||y||) \cdot \frac{y}{||y||}$$</p>
<p>$\bf{Added:}$ It turns out that the choice of the diffeomorphism from $[0,1)$ to $[0,\infty)$ matters a lot, since $x \mapsto ||x||$ is not smooth at $0$. This was brought to my attention by @Freeze_S and I thank him a lot! One can check that the map obtained for $\phi(t) = \frac{t}{1-t}$ is only $C^1$ at $0$... However, we can use the map so kindly suggested by @Jesus RS: ( big thanks! ) $\phi(t) = \frac{t}{\sqrt{1-t^2}}$ with inverse
$\psi(s) = \frac{s}{\sqrt{1+s^2}}$ and it will work just fine. The diffeomorphisms are, as written by @Jesus RS:
$$x \mapsto \frac{x}{\sqrt{1-||x||^2}} \\
y \mapsto \frac{y}{\sqrt{1+||y||^2}}$$</p>
<p>In fact, as long as $\phi(t)$ is an odd function of $t$ things will work OK. So, another example is </p>
<p>$$x \mapsto \frac{\tan (\frac{\pi}{2} \cdot ||x|| )}{||x||} \cdot x $$</p>
|
1,025,321 | <p>$\ln(1+xy) = xy$</p>
<p>When I try to implicitly differentiate this I get</p>
<p>$\frac{1}{1+xy}(y + xy')$ = (y + xy')</p>
<p>At which point the $(y + xy')$ terms cancel out, leaving no $y'$ to solve for.</p>
<p>However, the answer to this is $-\frac{y}{x}$... How do you get this?</p>
| Jesus RS | 203,197 | <p>Or you can try $f(x)=x/\sqrt{1-|x|^2}$ for $x\in B^n$.</p>
|
2,010,434 | <p>Say we are given n piles of stones.
Sizes are $s_{1}, s_{2}, .. , s_{n}$, they can be any positive integer numbers.
The game is played by two players, they alternate their moves.
The allowed moves are:
1. Take exactly 1 stone from 1 pile.
2. Take all stones from 1 pile.</p>
<p>Wins the player who mades the last move.
Both play optimally.</p>
<p>Is it possible to find the winner for some state?
I was thinking about Grundy numbers, but stuck with definitions and do not know can it be applied here. </p>
| Reese Johnston | 351,805 | <p>Assuming we only care about positive integers for the time being, notice that if $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ is an integer then it must be at least $1$. If three numbers add up to at least $1$, then at least one of them must be at least $\frac{1}{3}$ - so at least one of $a$, $b$, and $c$ must be no more than $3$.</p>
<p>If none of them is $2$ or less, then $\frac{1}{a}$, $\frac{1}{b}$, and $\frac{1}{c}$ are all no more than a third. But if they add up to at least $1$, then they must all then be exactly $\frac{1}{3}$. So $a = b = c = 3$.</p>
<p>Say the smallest number we have is $2$. Then we can't have two of them (because if $\frac{1}{2} + \frac{1}{2} + \frac{1}{c}$ is a whole number, then $c = 1$ and that's smaller than $2$). But $\frac{1}{2} + \frac{1}{b} + \frac{1}{c}$ can only be at least $1$ if $\frac{1}{b} + \frac{1}{c}$ is at least $\frac{1}{2}$, so we need $\frac{1}{b},\frac{1}{c}$ to be at least $\frac{1}{4}$. Therefore at least one of $b$ and $c$ is at most $4$. So we have either $\frac{1}{2} + \frac{1}{3} + \frac{1}{c}$ (in which case $c = 6$) or $\frac{1}{2} + \frac{1}{4} + \frac{1}{c}$ (in which case $c = 4$).</p>
<p>So far we have $(3, 3, 3)$, $(2, 3, 6)$, and $(2, 4, 4)$, and we've found all of the ones that don't involve a $1$. I'll leave it to you to try to apply this approach to the case when we <em>do</em> have a $1$.</p>
|
355,489 | <p>What are suggestions for reducing the transmission rate of the current epidemics?</p>
<p>In summary, my best one so far is (once we are down to the stay home rule) to discretize time, i.e., to introduce the following rule for the general populace not directly involved in necessary services:</p>
<p><em>If members of your household go to public services on a certain day, the whole household should not use any public service for 2 weeks after. That way you still can get infected but cannot infect without knowing it.</em></p>
<p>Do you have some suggestions? Good models to look at? No predictions please, just advices what to do.</p>
<p>Edit: In more detail:</p>
<p>Model (the simplest version to make things as clear as possible): There are several categories of people <span class="math-container">$C_i$</span> that constitute portion <span class="math-container">$p_i$</span> of the population and have a certain matrix <span class="math-container">$A$</span> of interactions per day. Then, if <span class="math-container">$x_i(t)$</span> is the number of ever infected people in category <span class="math-container">$C_i$</span> by the time <span class="math-container">$t$</span>, the driving ODE is
<span class="math-container">$$
\dot x(t)=\alpha A[x(t)-x(t-\tau)]
$$</span>
where <span class="math-container">$\tau$</span> is the ("typical") time after which the sick person is removed from the population and <span class="math-container">$\alpha$</span> is the transmission probability. In this model the exponential growth is unsustainable if <span class="math-container">$\alpha\lambda(A)\tau<1$</span> where <span class="math-container">$\lambda$</span> is the largest eigenvalue of <span class="math-container">$A$</span>. We do not know <span class="math-container">$\alpha$</span> (though we can try to make suggestions how to reduce it, most such suggestions are already made by the government). The government can modify <span class="math-container">$A$</span> by issuing orders. Some orders merely reduce <span class="math-container">$a_{ij}$</span> to <span class="math-container">$0$</span>, but the government cannot shut essential public services completely this way.</p>
<p>Questions: What is <span class="math-container">$A$</span>, which entries <span class="math-container">$a_{ij}$</span> are most important to reduce, how to issue a sensible order that will modify them, and by how much they will reduce the eigenvalue? </p>
<p>First suggestion for these 4 answers: There are two categories of people: ordinary population that only goes to public services and<br>
public servants that both provide services and go to them. There is only one ("averaged") type of service involving a dangerous client-server interaction and all infection goes there. The portion of public servants in the population is <span class="math-container">$p$</span>. The server sees <span class="math-container">$M$</span> clients a day. Then the current social interaction matrix (say, for the grocery store I've seen yesterday) is <span class="math-container">$A=\begin{bmatrix}0 & M(1-p)\\ Mp & 2Mp\end{bmatrix}$</span> (ordinary population does not transmit to ordinary population, servers transmit to clients who can be both servers and clients, clients transmits to servers. The largest eigenvalue is <span class="math-container">$M(p+\sqrt p)$</span>. The lion's share comes from <span class="math-container">$\sqrt p$</span>, which is driven by the off-diagonal entries, The order should be issued as above, the effect that ordinary people never come to the service infected, which will remove the left bottom corner and drop the largest eigenvalue to <span class="math-container">$2Mp$</span>. Assuming <span class="math-container">$p=1/9$</span> (not too unrealistic), the drop will be two-fold even if you leave the service organization as it is. </p>
<p>That ends the solution I propose in mathematical language.
In layman terms, the public will completely do this part (you cannot ask for more) and still have some life, and we can concentrate on the models of how servers should be organized.</p>
<p>Edit:</p>
<p>Time to remove the non-relevant part and add some relevant thoughts about what else we can help with plus the response to JCK.</p>
<p>First of all, <em>It is very hard to formulate the orders correctly</em>. The stay home rule really means "avoid all close contacts outside your household except the necessary interactions with public servants providing vital services to you" (and even that version is, probably flawed). It is not about dogs, etc., as the Ohio version reads now. If everybody understood and implemented that meaning, my suggestion could be formulated as I said. However the intended meaning really is</p>
<p><em>When going to public services, minimize the probability that you can infect others as much as feasible and consider it to be <span class="math-container">$0$</span> if in the last two weeks nobody from your household had a contact with a stranger and nobody in the household had any symptoms</em>. </p>
<p>Now it is more to the point, but also more complicated. And if a professional mathematician like myself is so inept, imagine the difficulties of other people. </p>
<p>So, within that model, what would be the best formulation of the order to give?</p>
<p>Second, the set of questions I asked is clearly incomplete.
One has to add for instance "What assumption can be wrong and what effect that will have on the outcome <em>under the condition that the order is given in the currently stated form</em>. I have never seen a book that teaches the influence of the order formulation on the possible model behavior and that may be a crucial thing now. The interaction between the formal logic and differential equations within a given scenario is a non-existing science (or am I just ignorant of something? <em>That</em> reference would really be useful).</p>
<p>Third, if we have a particular question (say, how much to reduce and how organize the public transportation, which is NY and Tel Aviv headache now), what would be a good mathematical model for just that and what would be the corresponding order statement under this model? </p>
<p>The questions like that are endless and if there were ready answers in textbooks, the governments would just implement them already instead of having 7-hour meetings. So I can fairly safely conclude that they are not there.</p>
<p>What I tried with my model example was, in particular, to show that there may be some non-trivial moves in even seemingly optimal situations (strict stay home order and running only the absolutely vital services at the minimal rate that still allows to serve the population) that also make common sense and can be used by everyone right now and right here. Finding such moves can really help now. The main real life question now is "What can I (as government, business, or individual) do to reduce the largest eigenvalue of the social interaction matrix?" Now show me the textbook that teaches that and I'll stop the "ballspitting" and apologize for the wasted time of the people reading all this. </p>
| Steven Landsburg | 10,503 | <p>This is just a slight expansion of my comment.</p>
<p>When the environment changes, behavioral parameters (that is, parameters like <span class="math-container">$M$</span> that describe people's behavior --- in this case the behavior of public servants deciding how many clients to serve each day) are going to change. Therefore such parameters should not be taken as constants; they should be determined within the model.</p>
<p>This means we need to be able to predict how <span class="math-container">$M$</span> will change in an unprecedented circumstance. Fortunately, we have a lot of relevant data. For example, consider the function <span class="math-container">$f(p)$</span> that tells you how much income a person is willing to forgo in order to avoid a probability <span class="math-container">$p$</span> of death. At least in the United States, we know (somewhat roughly and with various caveats) that when <span class="math-container">$p$</span> is small, <span class="math-container">$f(p)\approx \$10,000,000\times p$</span>. (Theory predicts, and evidence seems to confirm, that <span class="math-container">$f(p)$</span> is linear for small <span class="math-container">$p$</span>.) We infer this, for example, from the premiums you have to pay people in order to get them to take on dangerous jobs, or from the amount people are willing to pay for safety devices. I'm not sure whether <span class="math-container">$p\approx 1\%$</span> counts as small for this purpose, but there are data available that will help decide that.</p>
<p>(This, incidentally, is precisely what economists mean when they say that "In the United States, the value of a life is about $10,000,000".)</p>
<p>The best way to account for all this is to assume that people are maximizing some functionn <span class="math-container">$U$</span> which takes as arguments things like income, social interaction, time spent being sick, and probability of death. Try to estimate the function <span class="math-container">$U$</span> by observing the choices people make in a great variety of ordinary circumstances. In other words, observe their behavioral parameters in ordinary times, assume that those behavioral parameters are the solutions to some maximization problem, and try to infer what's being maximized.</p>
<p>Now when the pandemic comes along, you've got to assume that <i>something</i> is unchanged; otherwise you have no basis to make any predictions whatsoever. The idea is to assume that what's unchanged is the maximand <span class="math-container">$U$</span>, and that the pandemic represents a change in the constraints subject to which people are trying to maximize. Having estimated <span class="math-container">$U$</span>, having assumed it's fixed, and writing down the new constraints, you can calculate the new behavioral parameters that result from the new maximization problem. </p>
<p>Now your model is essentially a fixed point problem: Behavioral parameters (that is, the solutions to the maximization problem) cause changes in behavior, which cause changes in the way the pandemic spreads (that is, the constraints on the maximization problem), which cause changes in behavioral parameters. The solution to the model is a fixed point of that process. </p>
<p>You can also estimate how the fixed point will change if you add additional constraints, such as penalties for going outside, prohibitions on meetings, etc.</p>
<p>This sort of modeling is what economists try to do all the time. I expect, but do not know, that one could say the same of epidemiologists. No model is perfect, but economists have learned the painful lesson that some models are a lot less perfect than others, and that fixed behavioral parameters are generally a hallmark of such models. </p>
|
355,489 | <p>What are suggestions for reducing the transmission rate of the current epidemics?</p>
<p>In summary, my best one so far is (once we are down to the stay home rule) to discretize time, i.e., to introduce the following rule for the general populace not directly involved in necessary services:</p>
<p><em>If members of your household go to public services on a certain day, the whole household should not use any public service for 2 weeks after. That way you still can get infected but cannot infect without knowing it.</em></p>
<p>Do you have some suggestions? Good models to look at? No predictions please, just advices what to do.</p>
<p>Edit: In more detail:</p>
<p>Model (the simplest version to make things as clear as possible): There are several categories of people <span class="math-container">$C_i$</span> that constitute portion <span class="math-container">$p_i$</span> of the population and have a certain matrix <span class="math-container">$A$</span> of interactions per day. Then, if <span class="math-container">$x_i(t)$</span> is the number of ever infected people in category <span class="math-container">$C_i$</span> by the time <span class="math-container">$t$</span>, the driving ODE is
<span class="math-container">$$
\dot x(t)=\alpha A[x(t)-x(t-\tau)]
$$</span>
where <span class="math-container">$\tau$</span> is the ("typical") time after which the sick person is removed from the population and <span class="math-container">$\alpha$</span> is the transmission probability. In this model the exponential growth is unsustainable if <span class="math-container">$\alpha\lambda(A)\tau<1$</span> where <span class="math-container">$\lambda$</span> is the largest eigenvalue of <span class="math-container">$A$</span>. We do not know <span class="math-container">$\alpha$</span> (though we can try to make suggestions how to reduce it, most such suggestions are already made by the government). The government can modify <span class="math-container">$A$</span> by issuing orders. Some orders merely reduce <span class="math-container">$a_{ij}$</span> to <span class="math-container">$0$</span>, but the government cannot shut essential public services completely this way.</p>
<p>Questions: What is <span class="math-container">$A$</span>, which entries <span class="math-container">$a_{ij}$</span> are most important to reduce, how to issue a sensible order that will modify them, and by how much they will reduce the eigenvalue? </p>
<p>First suggestion for these 4 answers: There are two categories of people: ordinary population that only goes to public services and<br>
public servants that both provide services and go to them. There is only one ("averaged") type of service involving a dangerous client-server interaction and all infection goes there. The portion of public servants in the population is <span class="math-container">$p$</span>. The server sees <span class="math-container">$M$</span> clients a day. Then the current social interaction matrix (say, for the grocery store I've seen yesterday) is <span class="math-container">$A=\begin{bmatrix}0 & M(1-p)\\ Mp & 2Mp\end{bmatrix}$</span> (ordinary population does not transmit to ordinary population, servers transmit to clients who can be both servers and clients, clients transmits to servers. The largest eigenvalue is <span class="math-container">$M(p+\sqrt p)$</span>. The lion's share comes from <span class="math-container">$\sqrt p$</span>, which is driven by the off-diagonal entries, The order should be issued as above, the effect that ordinary people never come to the service infected, which will remove the left bottom corner and drop the largest eigenvalue to <span class="math-container">$2Mp$</span>. Assuming <span class="math-container">$p=1/9$</span> (not too unrealistic), the drop will be two-fold even if you leave the service organization as it is. </p>
<p>That ends the solution I propose in mathematical language.
In layman terms, the public will completely do this part (you cannot ask for more) and still have some life, and we can concentrate on the models of how servers should be organized.</p>
<p>Edit:</p>
<p>Time to remove the non-relevant part and add some relevant thoughts about what else we can help with plus the response to JCK.</p>
<p>First of all, <em>It is very hard to formulate the orders correctly</em>. The stay home rule really means "avoid all close contacts outside your household except the necessary interactions with public servants providing vital services to you" (and even that version is, probably flawed). It is not about dogs, etc., as the Ohio version reads now. If everybody understood and implemented that meaning, my suggestion could be formulated as I said. However the intended meaning really is</p>
<p><em>When going to public services, minimize the probability that you can infect others as much as feasible and consider it to be <span class="math-container">$0$</span> if in the last two weeks nobody from your household had a contact with a stranger and nobody in the household had any symptoms</em>. </p>
<p>Now it is more to the point, but also more complicated. And if a professional mathematician like myself is so inept, imagine the difficulties of other people. </p>
<p>So, within that model, what would be the best formulation of the order to give?</p>
<p>Second, the set of questions I asked is clearly incomplete.
One has to add for instance "What assumption can be wrong and what effect that will have on the outcome <em>under the condition that the order is given in the currently stated form</em>. I have never seen a book that teaches the influence of the order formulation on the possible model behavior and that may be a crucial thing now. The interaction between the formal logic and differential equations within a given scenario is a non-existing science (or am I just ignorant of something? <em>That</em> reference would really be useful).</p>
<p>Third, if we have a particular question (say, how much to reduce and how organize the public transportation, which is NY and Tel Aviv headache now), what would be a good mathematical model for just that and what would be the corresponding order statement under this model? </p>
<p>The questions like that are endless and if there were ready answers in textbooks, the governments would just implement them already instead of having 7-hour meetings. So I can fairly safely conclude that they are not there.</p>
<p>What I tried with my model example was, in particular, to show that there may be some non-trivial moves in even seemingly optimal situations (strict stay home order and running only the absolutely vital services at the minimal rate that still allows to serve the population) that also make common sense and can be used by everyone right now and right here. Finding such moves can really help now. The main real life question now is "What can I (as government, business, or individual) do to reduce the largest eigenvalue of the social interaction matrix?" Now show me the textbook that teaches that and I'll stop the "ballspitting" and apologize for the wasted time of the people reading all this. </p>
| Gerhard Paseman | 3,402 | <p>Here is a suggestion that may get downvoted. I make it not because it is a good idea, but because some modification may lead to a good idea.</p>
<p>Get everyone sick to get better.</p>
<p>If you hunt down my WordPress blog (grpaseman) you will see a mild expansion of this idea. The crux is to interrupt the replication of SARS-CoV-2 by introducing a different virus whose systemic effects are known , are not fatal, but use the same resources that SARS-CoV-2 would use to replicate. This might delay replication inside a host long enough for the immune system to manufacture a response.</p>
<p>The idea has problems and needs thinking through. If it suggests a better path to interrupt replication at a different level, it will be worth all the down votes.</p>
<p>Gerhard "Getting Downvotes To Inspire Others" Paseman, 2020.03.26.</p>
|
808,144 | <p>Here is a fun looking one some may enjoy. </p>
<p>Show that:</p>
<p>$$\int_{0}^{1}\log\left(\frac{x^{2}+2x\cos(a)+1}{x^{2}-2x\cos(a)+1}\right)\cdot \frac{1}{x}dx=\frac{\pi^{2}}{2}-\pi a$$</p>
| Felix Marin | 85,343 | <p>$\newcommand{\+}{^{\dagger}}
\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\down}{\downarrow}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\fermi}{\,{\rm f}}
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\isdiv}{\,\left.\right\vert\,}
\newcommand{\ket}[1]{\left\vert #1\right\rangle}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
\newcommand{\sgn}{\,{\rm sgn}}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
\newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\int_{0}^{1}\ln\pars{x^{2} + 2x\cos\pars{a} + 1
\over x^{2} - 2x\cos\pars{a} + 1}\,{\dd x \over x}:\ {\large ?}}$</p>
<blockquote>
<p>\begin{align}
&x^{2} + 2\cos\pars{a}x + 1\quad\mbox{has roots}\quad
-p\quad\mbox{and}\quad -p^{*}\quad\mbox{where}\quad p \equiv \expo{\ic a}
\\[3mm]&\mbox{Similarly,}\quad
x^{2} - 2\cos\pars{a}x + 1\quad\mbox{has roots at}\quad p\quad\mbox{and}\quad p^{*}.\qquad
\mbox{Note that}\quad pp^{*} = 1.
\end{align}</p>
</blockquote>
<p>Then, with $\ds{0 < \epsilon < 1}$:
\begin{align}
&\int_{\epsilon}^{1}\ln\pars{x^{2} + 2x\cos\pars{a} + 1
\over x^{2} - 2x\cos\pars{a} + 1}\,{\dd x \over x}
\\[3mm]&=
\int_{\epsilon}^{1}{\ln\pars{x + p} \over x}\,\dd x
+\int_{\epsilon}^{1}{\ln\pars{x + p^{*}} \over x}\,\dd x
-\int_{\epsilon}^{1}{\ln\pars{x - p} \over x}\,\dd x
-\int_{\epsilon}^{1}{\ln\pars{x - p^{*}} \over x}\,\dd x
\end{align}</p>
<blockquote>
<p>However,
\begin{align}
&\int_{\epsilon}^{1}{\ln\pars{x + b} \over x}\,\dd x
=-\ln\pars{b}\ln\pars{\epsilon}
+\int_{\epsilon}^{1}{\ln\pars{x/b + 1} \over x}\,\dd x
\\[3mm]&=-\ln\pars{b}\ln\pars{\epsilon}
+\int_{\epsilon/b}^{1/b}{\ln\pars{x+ 1} \over x}\,\dd x
=-\ln\pars{b}\ln\pars{\epsilon}
+\int_{-\epsilon/b}^{-1/b}{\ln\pars{1 - x} \over x}\,\dd x
\\[3mm]&=-\ln\pars{b}\ln\pars{\epsilon}
-\int_{-\epsilon/b}^{-1/b}{{\rm Li}_{1}\pars{x} \over x}\,\dd x
=-\ln\pars{b}\ln\pars{\epsilon}
-\int_{-\epsilon/b}^{-1/b}{\rm Li}_{2}'\pars{x}\,\dd x
\end{align}
$$
\begin{array}{|c|}\hline\\ \quad
\int_{\epsilon}^{1}{\ln\pars{x + b} \over x}\,\dd x
=-\ln\pars{b}\ln\pars{\epsilon} + {\rm Li}_{2}\pars{-\,{\epsilon \over b}}
-{\rm Li}_{2}\pars{-\,{1 \over b}}
\quad\\ \\ \hline
\end{array}
$$</p>
</blockquote>
<p>Since $\ds{pp^{*} = 1}$, we get in the limit $\ds{\epsilon \to 0^{+}}$:
\begin{align}
&\int_{0}^{1}\ln\pars{x^{2} + 2x\cos\pars{a} + 1
\over x^{2} - 2x\cos\pars{a} + 1}\,{\dd x \over x}
\\[3mm]&=-{\rm Li}_{2}\pars{-\,{1 \over p}}
-{\rm Li}_{2}\pars{-\,{1 \over p^{*}}}
+{\rm Li}_{2}\pars{1 \over p}
+{\rm Li}_{2}\pars{1 \over p^{*}}
\\[3mm]&=-\bracks{{\rm Li}_{2}\pars{-p} + {\rm Li}_{2}\pars{-\,{1 \over p}}}
+\bracks{{\rm Li}_{2}\pars{p} + {\rm Li}_{2}\pars{1 \over p}}
\end{align}</p>
<blockquote>
<p>\begin{align}
&\int_{0}^{1}\ln\pars{x^{2} + 2x\cos\pars{a} + 1
\over x^{2} - 2x\cos\pars{a} + 1}\,{\dd x \over x}
\\[3mm]&=-\bracks{{\rm Li}_{2}\pars{-\expo{\ic a}}
+{\rm Li}_{2}\pars{-\expo{-\ic a}}}
+\bracks{{\rm Li}_{2}\pars{\expo{\ic a}} + {\rm Li}_{2}\pars{\expo{-\ic a}}}
\end{align}
This is the general solution for $\ds{a \in {\mathbb R}}$. The OP proposed solution is found when $\ds{0 \leq a \leq \pi}$. A little later, I'll explain how to handle the general case by means of the DiLogarithm Inversion Formula.</p>
</blockquote>
|
2,467,228 | <p>I attempted to start with the $L_p$ norm and raise it to the power of $p$ but got stuck because I realized that I have no idea how to eliminate the integrand.</p>
<hr>
<p><strong>$L_p$ norm:</strong>
$||f||_p = ||f||_{L_p[a,b]} = (\int_{a}^{b}~|f(x)|^p~~dx)^{\frac{1}{p}}$</p>
<p>$\\$</p>
<hr>
<p><strong>C-norm:</strong>
<a href="https://i.stack.imgur.com/OiHHj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OiHHj.png" alt="enter image description here"></a></p>
| Fred | 380,717 | <p>We have $|f_n(t)-f(t)| \le ||f_n-f||$ for all $f,f_n \in C$ , all $n \in \mathbb N$ and all $t \in [a,b]$. Hence</p>
<p>$|f_n(t)-f(t)|^p \le ||f_n-f||^p$ for all $f,f_n \in C$ , all $n \in \mathbb N$ and all $t \in [a,b]$.</p>
<p>This gives</p>
<p>$ \int_a^b|f_n(t)-f(t)|^p dt \le \int_a^b ||f_n-f||^p dt =(b-a)||f_n-f||^p$.</p>
<p>Therefore</p>
<p>$||f_n-f||_p \le (b-a)^{1/p}||f_n-f||$.</p>
<p>Conclusion: $||f_n-f|| \to 0$ implies $||f_n-f||_p \to 0$ .</p>
|
119,552 | <p>I'm trying to create a function that does the cross product between two lists of the same dimensions, element by element. Each entry of the list is a 3D vector. Something like this:</p>
<pre><code>a = {{a1, a2, a3}, {b1, b2, b3}};
b = {{c1, c2, c3}, {d1, d2, d3}};
listCross[a, b] = {Cross[{a1, a2, a3}, {c1, c2, c3}], Cross[{b1, b2, b3}, {d1, d2, d3}]};
</code></pre>
<p>All my lists have 3 dimensions, so something like <code>Dimensions[a] = {32, 32, 32, 3}</code> would be typical, and I have the following code that works:</p>
<pre><code>listCross[list1_, list2_] := Module[{vec11, vec12, vec13, result},
vec11 = -list1[[All, All, All, 3]]list2[[All, All, All, 2]] + list1[[All, All, All, 2]]list2[[All, All, All, 3]];
vec12 = +list1[[All, All, All, 3]]list2[[All, All, All, 1]] - list1[[All, All, All, 1]]list2[[All, All, All, 3]];
vec13 = -list1[[All, All, All, 2]]list2[[All, All, All, 1]] + list1[[All, All, All, 1]]list2[[All, All, All, 2]];
result = Partition[Partition[MapThread[{#1, #2, #3}&, {Flatten[vec11], Flatten[vec12], Flatten[vec13]}], Dimensions[list1][[3]]], Dimensions[list1][[2]]]
];
</code></pre>
<p>It exploits the result of <code>Cross[{a1, a2, a3}, {b1, b2, b3}]</code> and generalizes to lists with list sums. In the end it uses <code>Partition</code> and <code>MapThread</code> to join the x, y and z components into a new list with the same dimensions as the input lists.</p>
<p>However this is not as fast as I would like it to be, and I haven't been able to come up with anything better. I also tried doing the cross product in it's matrix-vector product form, creating a block diagonal sparse matrix from the first list that I would multiply with the <code>Flatten</code> of the second list, but I couldn't create the matrix fast enough.</p>
<p>Does anyone have any thoughts on this? I also know nothing about <code>Compile</code>, so no idea if it would be of use here. Any help will be greatly appreciated.</p>
| Simon Woods | 862 | <p>How about this</p>
<pre><code>furious[a_, b_] := Module[{a1, a2, a3, b1, b2, b3, c},
{a1, a2, a3} = Transpose[a, {2, 3, 4, 1}];
{b1, b2, b3} = Transpose[b, {2, 3, 4, 1}];
c = {-a3 b2 + a2 b3, a3 b1 - a1 b3, -a2 b1 + a1 b2};
Transpose[c, {4, 1, 2, 3}]]
</code></pre>
<p>Timing results (from march's answer) for version 10.4.1</p>
<pre><code>list1 = RandomReal[{-1, 1}, {32, 32, 32, 3}];
list2 = RandomReal[{-1, 1}, {32, 32, 32, 3}];
l1 = listCrossMarch1[list1, list2]; // RepeatedTiming // First
l2 = listCrossMarch2[list1, list2]; // RepeatedTiming // First
l3 = listCross[list1, list2]; // RepeatedTiming // First
l4 = furious[list1, list2]; // RepeatedTiming // First
2.67
0.6064
0.0386
0.0015
</code></pre>
|
119,552 | <p>I'm trying to create a function that does the cross product between two lists of the same dimensions, element by element. Each entry of the list is a 3D vector. Something like this:</p>
<pre><code>a = {{a1, a2, a3}, {b1, b2, b3}};
b = {{c1, c2, c3}, {d1, d2, d3}};
listCross[a, b] = {Cross[{a1, a2, a3}, {c1, c2, c3}], Cross[{b1, b2, b3}, {d1, d2, d3}]};
</code></pre>
<p>All my lists have 3 dimensions, so something like <code>Dimensions[a] = {32, 32, 32, 3}</code> would be typical, and I have the following code that works:</p>
<pre><code>listCross[list1_, list2_] := Module[{vec11, vec12, vec13, result},
vec11 = -list1[[All, All, All, 3]]list2[[All, All, All, 2]] + list1[[All, All, All, 2]]list2[[All, All, All, 3]];
vec12 = +list1[[All, All, All, 3]]list2[[All, All, All, 1]] - list1[[All, All, All, 1]]list2[[All, All, All, 3]];
vec13 = -list1[[All, All, All, 2]]list2[[All, All, All, 1]] + list1[[All, All, All, 1]]list2[[All, All, All, 2]];
result = Partition[Partition[MapThread[{#1, #2, #3}&, {Flatten[vec11], Flatten[vec12], Flatten[vec13]}], Dimensions[list1][[3]]], Dimensions[list1][[2]]]
];
</code></pre>
<p>It exploits the result of <code>Cross[{a1, a2, a3}, {b1, b2, b3}]</code> and generalizes to lists with list sums. In the end it uses <code>Partition</code> and <code>MapThread</code> to join the x, y and z components into a new list with the same dimensions as the input lists.</p>
<p>However this is not as fast as I would like it to be, and I haven't been able to come up with anything better. I also tried doing the cross product in it's matrix-vector product form, creating a block diagonal sparse matrix from the first list that I would multiply with the <code>Flatten</code> of the second list, but I couldn't create the matrix fast enough.</p>
<p>Does anyone have any thoughts on this? I also know nothing about <code>Compile</code>, so no idea if it would be of use here. Any help will be greatly appreciated.</p>
| march | 29,734 | <p>Interestingly enough, <code>MapThread</code>ing <code>Cross</code> works but is much slower:</p>
<p>Using sample lists:</p>
<pre><code>list1 = Array[c, {20, 20, 20, 3}];
list2 = Array[d, {20, 20, 20, 3}];
</code></pre>
<p>We can perform this operation in the following two ways, using <code>MapThread</code>:</p>
<pre><code>listCrossMarch1[list1_, list2_] := MapThread[Cross, {list1, list2}, 3]
listCrossMarch2[list1_, list2_] := MapThread[{#1[[2]] #2[[3]] - #1[[3]] #2[[2]], #1[[3]] #2[[1]] - #1[[1]] #2[[3]], #1[[1]] #2[[2]] - #1[[2]] #2[[1]]} &, {list1, list2}, 3]
</code></pre>
<p>As we can see below, there is a lot of overhead associated with <code>Cross</code> apparently, since that version is <em>much</em> slower than coding the cross-product explicitly. The <code>MapThread</code> version with the explicit cross-product (rather than <code>Cross</code>) is almost as fast as the OP's version and much cleaner to write down. <a href="https://mathematica.stackexchange.com/users/862/simon-woods">Simon Woods</a>' <a href="https://mathematica.stackexchange.com/a/119553/29734">answer</a> is the fastest (and also very clean).</p>
<pre><code>l1 = listCrossMarch1[list1, list2]; // AbsoluteTiming // First
l2 = listCrossMarch2[list1, list2]; // AbsoluteTiming // First
l3 = listCross[list1, list2]; // AbsoluteTiming // First
l4 = furious[list1, list2]; // AbsoluteTiming // First
l1 === l2 === l3 === l4
(* 1.436369 *)
(* 0.190366 *)
(* 0.120962 *)
(* 0.084740 *)
(* True *)
</code></pre>
<hr>
<p>As suggested by Simon Woods, here are <code>Timing</code>s with packed arrays of real numbers. Using</p>
<pre><code>list1 = RandomReal[{-1, 1}, {20, 20, 20, 3}];
list2 = RandomReal[{-1, 1}, {20, 20, 20, 3}];
</code></pre>
<p>we do</p>
<pre><code>l1 = listCrossMarch1[list1, list2]; // AbsoluteTiming // First
l2 = listCrossMarch2[list1, list2]; // AbsoluteTiming // First
l3 = listCross[list1, list2]; // AbsoluteTiming // First
l4 = furious[list1, list2]; // AbsoluteTiming // First
(* 0.429275 *)
(* 0.094540 *)
(* 0.008337 *)
(* 0.028592 *)
</code></pre>
<p>The OP's messy version is significantly the fastest here! Simon Woods' answer still does a great job, and <code>Cross</code> still has lots of overhead.</p>
|
28,104 | <p>It occured to me that the Sieve of Eratosthenes eventually generates the same prime numbers, independently of the ones chosen at the beginning. (We generally start with 2, but we could chose any finite set of integers >= 2, and it would still end up generating the same primes, after a "recalibrating" phase).</p>
<p>If I take 3 and 4 as my first primes, starting at 5:</p>
<ul>
<li>5 is prime,</li>
<li>6 is not (twice 3),</li>
<li>7 is prime,</li>
<li>8 is not,</li>
<li>9 is not,</li>
<li>10 is not,</li>
<li>etc.</li>
</ul>
<p>Eventually, I will find all the primes as if I had started with 2 only.</p>
<p>To me, it means that we can generate big prime numbers without any knowledge of their predecessors (until a certain point). If I want primes higher than 1000000, then I can generate them without any knowledge of primes under, say, 1000. (It may not be as effective computationnally, but I find this philosophically interesting.)</p>
<p>Is this result already known ? </p>
<p>Are there any known implications ?</p>
<p><strong>Edit</strong></p>
<p>The number from which we are assured to get the right primes again is after the square of the last missing natural prime.</p>
<p>That is, if I start with a seed set containing only the number 12, the highest missing natural prime is 11, so I'll end up having 121 as my last not-natural prime.
Non-natural primes found are 12,14,15,16,18,20,21,22,25,27,33,35,49,55,77 and 121.</p>
<p>This is a bit better than what I thought at first (namely, as stated below, somewhere under the square of the highest seed).</p>
| Noldorin | 602 | <p>By missing out only the first prime (2), your statement is indeed correct, but it is a special case. This is because all even numbers > 4 are guaranteed to have a divisor > 2 (i.e. in the set of tested numbers). Likewise, all odd numbers that are not primes will have a divisor within your set, since you start with the lowest odd number > 1.</p>
<p>You claim:</p>
<blockquote>
<p>We generally start with 2, but we could chose any finite set of integers >= 2, and it would still end up generating the same primes, after a "recalibrating" phase</p>
</blockquote>
<p>Yet if you consider the simple case of starting with 4 and 5 now, you immediately run into problems. Given your method, 6 and 7 would be resolved as primes, since there are no divisors within your set!</p>
<p>I believe I've understood your statement correctly, and hopefully this should clarify why it actually fails in general. Let me know if you think I've missed any point, however.</p>
|
117,432 | <p>All varieties are over $\mathbb{C}$. Notions related to weights etc. refer to mixed Hodge structures (say rational, but I would be grateful if the experts would point out any differences in the real setting).</p>
<p>I am trying to get some intuition for the geometric meaning of/when to expect the weight filtration on the cohomology groups $H^i(X)$ of a variety to split. By the weight filtration splitting, I mean that the Hodge structure on each $H^i(X)$ is a direct sum of pure Hodge structures.</p>
<p>The simplest situation in which this happens is the classical one of smooth projective varieties. The next simplest situation I think of is the "smooth" being weakened to "mild singularities" (for instance, rationally smooth). So at least in the projective case I think of the "mixed" as encoding singularities. This also gels well with the construction of these Hodge structures using resolution of singularities. Are there other helpful perspectives?</p>
<p>Instead of fiddling with the "smooth" one can make the variety non-compact (but still smooth say). Examples: affine $n$-space, tori. Here I don't know how to think of or when to expect the weight filtration to split. Any intuition would be appreciated. The only rough picture I have is that this encodes information about the complement in a good compactification. But I don't find this particularly illuminating.</p>
<p>Generalizing affine space and tori is the situation of toric varieties for which the weight filtration always splits (thanks to a lift of Frobenius to characteristic $0$). In general when should one expect a splitting of the weight filtration to be given "geometrically" by a morphism (or say correspondence in the context of Borel-Moore homology)?</p>
<p>Related is the following: when should one expect the Hodge structure on each $H^i(X)$ to be pure (not necessarily of weight $i$). Here I am again more interested in weakening the "projective" rather than the "smooth".</p>
<p>At the risk of being even more vague, let me add some motivation from left field. There are several situations in representation theory where one expects/knows that the weight filtration on some cohomology groups splits (and is even of Hodge-Tate type). For instance, the cohomology of intersections of Schubert cells with opposite Schubert cells. However, the reasoning/heuristic has, a priori, nothing to do with geometry but more with the philosophy of "graded representation theory" (ala Soergel, see for instance his ICM94 address) I would love to have a geometric reason/heuristic for this. Pertinent to this is also the question of when should one expect the canonical Hodge structure on extensions between perverse sheaves of geometric origin to be split Tate? The only examples I know come from representation theory (see Section 4 of Beilinson-Ginzburg-Soergel's "Koszul duality patterns in representation theory").</p>
| Dan Petersen | 1,310 | <p>I don't have a general answer, but let me add some more examples.</p>
<ol>
<li><p>For your second question, examples of smooth varieties with $H^i$ pure of the 'wrong' weight, a good example is the complement of an affine arrangement of hyperplanes in $\mathbf C^n$. In this case, the Hodge structure on $H^i$ is pure of type $(i,i)$. One way to see this is that the cohomology is generated as an algebra by $H^1$, and $H^1$ is spanned by logarithmic forms $$ \omega_H = \frac 1 {2\pi i} \mathrm d \log H,$$
where $H=0$ is the defining equation of one of the hyperplanes; the class of $\omega_H$ is of type $(1,1)$. A reference for this is Brieskorn's "Sur les groupes des tresses [d'après V.I. Arnol'd]". See also the simple and "motivic" proof in "Weights in cohomology groups arising from hyperplane arrangements" by Minhyong Kim, as well as near-simultaneous papers by Boris Shapiro and Gus Lehrer which present basically the same result.</p></li>
<li><p>Another example is when $X$ is an abelian variety, and $F(X,n)$ is the configuration space of $n$ points in $X$, i.e. the complement of the "big diagonal" in $X^n$. Then the mixed Hodge structure on $H^i(F(X,n))$ is always a direct sum of pure Hodge structures. A reference for this is Gorinov's preprint "Rational cohomology of the moduli spaces of pointed genus 1 curves" (on his webpage), Section 3. </p></li>
<li><p>Let $Y(N)$ be the open modular curve parametrizing full level $N$ structures on elliptic curves. Then $H^1(Y(N))$ is a sum of pure Hodge structures, by the criterion in Donu Arapura's answer and the theorem of Drinfel'd and Manin.</p></li>
</ol>
|
71,822 | <p><em>I have moved this question here from MSE, because I did not receive any answers as of yet over there.</em></p>
<p>I know that there are statements that are neither provable nor disprovable within some set of axioms, and I also know that such statements are called undecidable. Please allow me to call these statements to be undecidable to the first order. These statements belong to $U_1$.</p>
<p>I was wondering if there is some kind of generalization of this concept. </p>
<blockquote>
<p>Question 1. Are there any conjectures/statements of which we can prove that we cannot prove whether it is decidable or not? Related: how would such a proof look like?</p>
</blockquote>
<p>Such a statement/conjecture would be undecidable to the second order, or belong to $U_2$.
Generalizing even further
(Question 2):</p>
<blockquote>
<p>What about statements that are in $U_\infty$ ?</p>
</blockquote>
| Emil Jeřábek | 12,705 | <p>Let $T$ be a fixed theory (recursively axiomatized, extending $I\Delta_0+\mathrm{EXP}$, sound). I read your first question as:</p>
<blockquote>
<p>Q1: Is there a sentence $A$ such that $T$ proves that “$T$ does not prove ‘$T$ proves $A$ or $T$ proves $\neg A$’ and $T$ does not prove ‘$T$ does not prove $A$ and $T$ does not prove $\neg A$’”?</p>
</blockquote>
<p>The answer is no, by Gödel’s theorem: if $T$ does not prove some formula, then in particular $T$ is consistent, hence a positive answer would imply that $T$ proves its own consistency. In light of this argument, we can make the question more sensible by putting enough consistency of $T$ in the assumptions:</p>
<blockquote>
<p>Q1’: Is there a sentence $A$ such that $T$ proves that “If $T$ + ‘$T$ is consistent’ is consistent, then $T$ does not prove ‘$T$ proves $A$ or $T$ proves $\neg A$’ and $T$ does not prove ‘if $T$ is consistent, then $T$ does not prove $A$ and $T$ does not prove $\neg A$’”?</p>
</blockquote>
<p>The answer is yes, and the convenient way to solve this and similar question is to use <a href="http://en.wikipedia.org/wiki/Provability_logic" rel="noreferrer">provability logic</a>.</p>
<p>The basic provability logic works with the language of propositional modal logic: we have propositional variables $p_0$, $p_1$, $p_2$, ..., Boolean connectives such as $\land$, $\lor$ and $\neg$ (including the constants $\bot$ and $\top$ for falsity and truth), and the unary modal connective $\Box$. We also define $\Diamond A=\neg\Box\neg A$. An arithmetical interpretation of this language is an assignment $*$ of an arithmetical sentence $p^*$ to every propositional variable $p$, which is extended to all modal formulas by making it commute with Boolean connectives, and putting $(\Box A)^*=\Pr_T(\ulcorner A^*\urcorner)$, where $\Pr_T$ is the formalized provability predicate for $T$. That is, we read $\Box A$ as “$A$ is provable in $T$”, and $\Diamond A$ as “$A$ is consistent with $T$”. In particular, $\Diamond\top$ translates to “$T$ is consistent”.</p>
<p>$\DeclareMathOperator\prl{PRL}$
Then, the provability logic of $T$ with respect to a metatheory $S$, denoted $\prl_S(T)$, is the set of all modal formulas $A$ such that $S\vdash A^*$ for every arithmetical interpretation $*$. The two most important special cases are the ordinary provability logic of $T$, which is $\prl(T):=\prl_T(T)$, and the true provability logic of $T$, which is $\prl^+(T):=\prl_{\mathrm{Th}(\mathbb N)}(T)$ (here, $\mathrm{Th}(\mathbb N)$ is the true arithmetic, i.e., the set of all arithmetical sentences true in the standard model).</p>
<p>The modal logics $\prl_S(T)$ have been completely characterized for all sufficiently strong theories $T,S$. In particular, under our assumptions, $\prl(T)$ is the so-called Gödel–Löb logic GL, and $\prl^+(T)$ is Solovay’s logic S. Both logics are decidable, and they have transparent Kripke semantics.</p>
<p>Now, a moment’s reflection shows that Q1’ is equivalent to:</p>
<blockquote>
<p>Q1’’: the formula $B:=\neg\Box[\Diamond\Diamond\top\to\neg\Box(\Box p\lor\Box\neg p)\land\neg\Box(\Diamond\top\to\neg\Box p\land\neg\Box\neg p)]$ is not in $\prl^+(T)$.</p>
</blockquote>
<p>One can easily produce a Kripke model of S where $B$ fails, hence the answer is indeed positive. One can solve all sorts of similar questions about unprovability like this, by expressing it in provability logic and then finding a suitable Kripke model. You can learn more about provability logic from references given in the Wikipedia article.</p>
|
2,138,241 | <p>I tried to prove $$\lim_{x\to \infty}\frac 1x = 0$$
I started as thus
$$\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac x{x^2}$$
Applying <a href="https://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule" rel="nofollow noreferrer">L'Hospital's Rule</a>
$$\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac x{x^2}=\lim_{x\to \infty}\frac 1{2x}=\frac12\lim_{x\to \infty}\frac 1x$$
Thus,
$$\frac12\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac 1x$$
which therefore implies
$$\lim_{x\to \infty}\frac 1x = 0$$
QED.</p>
| Theorem | 346,898 | <p>This is incorrect, as you can only use L'Hospital's Rule when you know the limit of the derivative ratio exists.</p>
|
2,138,241 | <p>I tried to prove $$\lim_{x\to \infty}\frac 1x = 0$$
I started as thus
$$\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac x{x^2}$$
Applying <a href="https://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule" rel="nofollow noreferrer">L'Hospital's Rule</a>
$$\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac x{x^2}=\lim_{x\to \infty}\frac 1{2x}=\frac12\lim_{x\to \infty}\frac 1x$$
Thus,
$$\frac12\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac 1x$$
which therefore implies
$$\lim_{x\to \infty}\frac 1x = 0$$
QED.</p>
| Ovi | 64,460 | <p><a href="https://i.stack.imgur.com/sKQRt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sKQRt.png" alt=""></a></p>
<p>The horizontal lines in the picture are $y = \pm \dfrac 12$. As you can see, after $P = 3$ on the $x$ axis, the values of $f(x)$ are contained on the interval $\left(-\dfrac 12, \dfrac 12 \right)$ on the $y$ axis. In informal terms, the rigorous definition of $\lim_{x \to \infty} \dfrac 1x = 0$ is simply the assertion that that you can do exactly what I did above for any horizontal lines $y = \pm \epsilon$, no matter what (positive) $\epsilon$ you pick. That is, for any positive number $\epsilon$, you can always find some point $P$ somewhere on the $x$ axis such that for every $x$ larger than $P$, its $f(x)$ value is on the interval $(-\epsilon, \epsilon)$.</p>
|
2,138,241 | <p>I tried to prove $$\lim_{x\to \infty}\frac 1x = 0$$
I started as thus
$$\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac x{x^2}$$
Applying <a href="https://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule" rel="nofollow noreferrer">L'Hospital's Rule</a>
$$\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac x{x^2}=\lim_{x\to \infty}\frac 1{2x}=\frac12\lim_{x\to \infty}\frac 1x$$
Thus,
$$\frac12\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac 1x$$
which therefore implies
$$\lim_{x\to \infty}\frac 1x = 0$$
QED.</p>
| Jones | 414,995 | <p>Your expression In other words:</p>
<p>As x approaches infinity, then 1/x approaches 0 so its answer is 0
Try to think in that way...</p>
<p>Your method is wrong as you can only use L'Hospital's Rule when you know the limit of the derivative ratio exists.</p>
|
3,890,064 | <p>When Max is 8 m from a lamp post which is 6 m high his shadow is 2 m long. When Max is 3 m from the lamp post, what is the length of his shadow?</p>
| Toby Mak | 285,313 | <p>Let the top of the lamp post be at <span class="math-container">$A = (0,6)$</span>, and Max be at <span class="math-container">$B = (8,0)$</span>, with his head being at height <span class="math-container">$h$</span> or <span class="math-container">$H = (8, h)$</span>.</p>
<p>The gradient of line <span class="math-container">$AH$</span> is <span class="math-container">$\frac{h-6}{8-0} = \frac{h-6}{8}$</span>. When Max moves to the point <span class="math-container">$B' = (3,0)$</span>, with his head at <span class="math-container">$H' = (3,h)$</span>, the gradient of line <span class="math-container">$AH'$</span> is <span class="math-container">$\frac{h-6}{3}$</span>.</p>
<p>Now <span class="math-container">$\text{gradient} = \frac{\text{change in } y}{\text{change in } x}$</span>. Since Max's height does not change, the change in <span class="math-container">$y$</span> is the same for both lines. Therefore, for both lines, gradient <span class="math-container">$\times$</span> change in <span class="math-container">$x$</span> are equal, and the change in <span class="math-container">$x$</span> is the length of Max's shadow (<span class="math-container">$s$</span>):</p>
<p><span class="math-container">$$\frac{h-6}{8} \cdot 2 = \frac{h-6}{3} \cdot s\Rightarrow \frac{1}{8} \cdot 2=\frac{1}{3}s \Rightarrow s = \boxed{\frac{3}{4}}.$$</span></p>
<p>and here is a diagram (look at triangles <span class="math-container">$H'B'E$</span> and <span class="math-container">$HBD$</span>):</p>
<p><a href="https://i.stack.imgur.com/9ezsHm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9ezsHm.png" alt="enter image description here" /></a></p>
<p>Note that we don't need to know Max's height.</p>
|
1,388,565 | <p>Given for example $\omega_1$ coin tosses (i.e. a mapping from the elements of $\omega_1$ to $\{H,T\}$ with independent probabilities half), what is the probability that there is an infinite <del>subsequence</del> subinterval [<em>corrected following comments</em>] consisting only of heads?</p>
<p>Is this question even well defined? Does it depend on the set theory axiomatisation?</p>
<p>For a countable sequence of tosses (i.e. $<\omega_1$) the answer is presumably zero, as an infinite subinterval of heads is as likely as any of the uncountably many other possible infinite subintervals.</p>
| Eric Wofsey | 86,856 | <p>Taking "subsequence" to mean "subinterval", the set you are describing is not measurable, and in fact has outer measure $1$ and inner measure $0$. Indeed, let $S\subset\{0,1\}^{\omega_1}$ be the set of sequences which are constant with value $1$ on some infinite interval in $\omega_1$. Suppose $B\subset S$ is a Borel set. Then there is some $\alpha<\omega_1$ and some $A\subseteq \{0,1\}^\alpha$ such that $B=A\times\{0,1\}^{[\alpha,\omega_1)}$. It is then clear that every element of $A$ must be constant with value $1$ on some infinite interval in $\alpha$. There are only countably many such infinite intervals, and the set of sequences that are constant on each one has measure zero. So $A$ must have measure zero, and hence $B$ has measure zero. This shows $\omega_1$ has inner measure zero.</p>
<p>Now suppose $B\supset S$ is Borel, and let $B=A\times\{0,1\}^{[\alpha,\omega_1)}$ as before. An element of $S$ can have any restriction to $\alpha$ (since its infinite interval of $1$s could come after $\alpha$), so $A$ must be all of $\{0,1\}^\alpha$. Thus $B=\{0,1\}^{\omega_1}$ and has measure 1. Hence $S$ has outer measure $1$. </p>
<p>Since the inner measure and outer measure of $S$ do not agree, $S$ is not measurable. That is, the probability of the event you are describing happening is undefined, at least for the standard definition of "probability".</p>
|
2,784,784 | <p>Let $f(x)=$$x-1 |x \in \mathbb{Q} \brace 5-x| x \in \mathbb{Q}^c$</p>
<p>Show that $\lim_{x \to a}f(x)$ does not exists for any $a \not= 3$</p>
<p>I first showed that $lim_{x \to 3}f(x)=2$. </p>
<p>I don't know how to approach this part. Can anyone please guide? I was thinking of using density theorem at first to show that there will exist sequences of rationals and irrationals that will approach a but their limits would not equal but that is problematic as I don't really know the limit.</p>
<p>thank you</p>
| Community | -1 | <p>As $(x-1)-(5-x)=2x-6$, you will always find values in the neighborhood of $x$ that have their images $2x-6$ apart so that not all $\epsilon$ can be met. Unless $x=3$.</p>
|
2,495,555 | <p>I want to use Konig's theorem to show that the pictured graph $G$ has no perfect matching. By this theorem it suffices to find a vertex cover of size $|G|/2-1= 20$, but so far I have only been able to find vertex covers of size 21. I'm just doing this by inspection as opposed to using any algorithms, and it's not immediately obvious to me how to find a cover of size 20.</p>
<p><a href="https://i.stack.imgur.com/3gsoV.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/3gsoV.jpg" alt="enter image description here"></a></p>
| Donald Splutterwit | 404,247 | <p>Must write at least $30$ charcters.</p>
<p><a href="https://i.stack.imgur.com/kXcag.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/kXcag.jpg" alt="enter image description here"></a></p>
|
2,908,993 | <p>There are $6967$ notes labelled $1,2,3,4,...,6967$. If you choose $K$ notes at random, what is the smallest number $K$ that would guarantee that you pick <strong>two notes labelled by consecutive numbers</strong>? Use the pigeonhole principle to explain</p>
<p>I'm not quite sure in this question what the pigeons and pigeonholes are. If someone could explain that would be great</p>
| Rushabh Mehta | 537,349 | <p>The <a href="https://en.wikipedia.org/wiki/Pigeonhole_principle" rel="nofollow noreferrer">pigeonhole principle</a> is summarized as the following</p>
<blockquote>
<p>If $n$ items are put into $m$ containers, with $n>m$, then at least one container must contain more than one item.</p>
</blockquote>
<p>It's quite a simple concept, but a powerful one. Let's think about how to apply it in your problem.</p>
<p>We realize immediately that there are two "pigeonholes" for us to consider in this problem: even and odd integers. It's obvious that no even numbers can be consecutive, and neither can any odds. So, we try to fill in as many numbers from one "pigeonhole" as we can (in this case, we can fill in the $3484$ odd integers) and then add one more to force there to be two consecutive integers (since the last number must be even.) Hence, the answer is $\color{red}{3485}$.</p>
|
2,908,993 | <p>There are $6967$ notes labelled $1,2,3,4,...,6967$. If you choose $K$ notes at random, what is the smallest number $K$ that would guarantee that you pick <strong>two notes labelled by consecutive numbers</strong>? Use the pigeonhole principle to explain</p>
<p>I'm not quite sure in this question what the pigeons and pigeonholes are. If someone could explain that would be great</p>
| hartkp | 23,399 | <p>Picking all the odd numbers gives a maximal set without neighbours and $3484$ elements.
Divide the set in $\{1,2\}$, $\{3,4\}$, ..., $\{6965,6966\}$, and $\{6967\}$; that's $3484$ pigeon holes, so if you pick $3485$ numbers at least one pigeon hole gets picked from twice.</p>
|
4,116,252 | <p>I'm trying to prove (or disprove) the following:</p>
<p><span class="math-container">$$ \sum_{i=1}^{N} \sum_{j=1}^{N} c_i c_j K_{ij} \geq 0$$</span>
where <span class="math-container">$c \in \mathbb{R}^N$</span>, and <span class="math-container">$K_{ij}$</span> is referring to a <a href="https://en.wikipedia.org/wiki/Kernel_method" rel="noreferrer">kernel matrix</a>:</p>
<p><span class="math-container">$$K_{ij} = K(x_i,x_j) = \frac{\sum_{k=1}^{N} \min(x_{ik}, x_{jk})}{\sum_{k=1}^{N} \max(x_{ik}, x_{jk})}$$</span>
Here, <span class="math-container">$x \in \mathbb{R}^N \geq 0$</span>.</p>
<p>I'm basically trying to prove that <span class="math-container">$K_{ij}$</span> is a positive definite matrix, so I can use it as a Kernel, but I'm really stuck trying to work with <span class="math-container">$\max$</span></p>
<p>Edit: the function I'm refering to is:</p>
<p><span class="math-container">$$K(u,v) = \frac{\sum_{k=1}^{N} \min(u_{k}, v_{k})}{\sum_{k=1}^{N} \max(u_{k}, v_{k})}$$</span>
where <span class="math-container">$u, v \in \mathbb{R}^N \geq 0$</span></p>
| Gunnar Þór Magnússon | 3,225 | <p>I have some comments in a different direction from g g. Maybe they'll be useful to someone.</p>
<p>First, like g g noted, <span class="math-container">$K \geq 0$</span> if <span class="math-container">$n = 1$</span>, so the kernel matrix is semipositive there. For <span class="math-container">$n = 2$</span>, the kernel matrix we get is of the form
<span class="math-container">$$
B = \begin{pmatrix} 1 & a \\ a & 1 \end{pmatrix}
$$</span>
where <span class="math-container">$0 \leq a \leq 1$</span>. By inspection, the eigenvectors of this matrix are <span class="math-container">$(1,1)$</span> and <span class="math-container">$(1,-1)$</span> with eigenvalues <span class="math-container">$1 + a$</span> and <span class="math-container">$1-a$</span>, which are nonnegative, so the kernel matrix is semipositive.</p>
<p>In general,
we have <span class="math-container">$\min(x,y) = \frac12(x + y - |x-y|)$</span> and <span class="math-container">$\max(x,y) = \frac12(x + y + |x-y|)$</span>. Then
<span class="math-container">$$
K(x,y)
= \frac{\sum_j \min(x_j,y_j)}{\sum_j \max(x_j,y_j)}
= \frac{|x+y|_1 - |x-y|_1}{|x+y|_1 + |x-y|_1},
$$</span>
where <span class="math-container">$|\,\cdot\,|_1$</span> is the <span class="math-container">$1$</span>-norm. I haven't been able to do anything with this so far.</p>
<p>This is close to some trigonometric identities: If we write <span class="math-container">$t(x,y) := \sqrt{|x-y|_1/|x+y|_1}$</span> then <span class="math-container">$K = (1-t^2)/(1+t^2)$</span> so if we set <span class="math-container">$t(x,y) =: \tan \theta(x,y)$</span> then <span class="math-container">$K = \cos 2\theta(x,y)$</span>. But I don't know how to do anything useful with that in the quadratic form the kernel matrix <span class="math-container">$B$</span> defines.</p>
<p>As I said in a comment, we can have numpy crunch some randomly sampled examples:</p>
<p><a href="https://pastebin.com/UGKrvxSK" rel="nofollow noreferrer">https://pastebin.com/UGKrvxSK</a></p>
<p>This has only given me positive-definite matrices so far for <span class="math-container">$K$</span>, while randomly sampling matrices with entries between <span class="math-container">$0$</span> and <span class="math-container">$1$</span> and <span class="math-container">$1$</span>s on the diagonal yields matrices that are almost never positive-definite as <span class="math-container">$n$</span> grows, which seems like some kind of evidence for the kernel matrix being positive-definite in general.</p>
|
4,150,776 | <p>Let <span class="math-container">$(a_n)_{n=1}^\infty$</span> Let be a positive, increasing, and unbounded sequence. Prove that the series:</p>
<p><span class="math-container">$$\sum_{n=1}^\infty\left(\frac{1}{a_{2n-1}}-\frac{1}{a_{2n}}\right)$$</span></p>
<p>convergent.</p>
<hr />
<p>We know that since <span class="math-container">$a_n$</span> is increasing and unbounded, than <span class="math-container">$\lim_{n \to \infty}a_n=\infty$</span>, so I want to apply that to say that, <span class="math-container">$\sum_{n=1}^\infty(\frac{1}{a_{2n-1}}-\frac{1}{a_{2n}})$</span> is decreasing and its limit will be <span class="math-container">$0$</span>.</p>
<p>My problem is that every time I get confused while it says <span class="math-container">$a_{2n}$</span> or <span class="math-container">$a_{2n-1}$</span>, it is less intuitive for me than just "normal" <span class="math-container">$a_n$</span>...</p>
<p>Appreciate your help!</p>
<p>Thanks a lot!</p>
| Martin R | 42,969 | <p>The fact that <span class="math-container">$(a_n)$</span> is unbounded is actually not needed, only that the sequence is positive and increasing.</p>
<p>First note that all terms <span class="math-container">$b_n = \frac{1}{a_{2n-1}}-\frac{1}{a_{2n}}$</span> are non-negative. Then verify that
<span class="math-container">$$
\sum_{n=1}^N b_n = \frac{1}{a_1} + \underbrace{\left(-\frac{1}{a_2}+\frac{1}{a_3}\right)}_{\le 0}+ \cdots
+\underbrace{\left(-\frac{1}{a_{2N-2}}+\frac{1}{a_{2N-1}}\right)}_{\le 0}
-\frac{1}{a_{2N}} \le \frac{1}{a_1} \, .
$$</span>
So the partial sums of <span class="math-container">$\sum_{n=1}^\infty b_n$</span> are increasing and bounded above. It follows that the series is convergent.</p>
|
172,124 | <p>$$\int \frac{1}{x^{10} + x}dx$$</p>
<p>My solution :</p>
<p>$$\begin{align*}
\int\frac{1}{x^{10}+x}\,dx&=\int\left(\frac{x^9+1}{x^{10}+x}-\frac{x^9}{x^{10}+x}\right)\,dx\\
&=\int\left(\frac{1}{x}-\frac{x^8}{x^9+1}\right)\,dx\\
&=\ln|x|-\frac{1}{9}\ln|x^9+1|+C
\end{align*}$$</p>
<p>Is there completely different way to solve it ?</p>
| Generic Human | 26,855 | <p>Not really different, but even simpler:
$$\begin{align}
\int\frac{1}{x^{10}+x} dx=&\int\frac{x^{-10}}{1+x^{-9}} dx
=-\frac 1 9 \log |1+x^{-9}| + C
\end{align}$$</p>
|
2,730,407 | <p>How many different numbers must be selected from the first 25 positive integers to be certain that at least one of them will be twice the other ?</p>
| saulspatz | 235,128 | <p>Here's how I did it. Group the integers we get by doubling</p>
<p>A:$1,2,4,8,16$ </p>
<p>B:$3,6,12,24$</p>
<p>C:$5,10,20$</p>
<p>D:$7,14$</p>
<p>E:$9,18$</p>
<p>F:$11,22$</p>
<p>We have $7$ odd numbers from $13$ to $25$ not included and we can take them all. We can take $1$ each from groups $F$, $E$, and $D$, $2$ each from groups $C$ and $B$ and $3$ from group $A$ giving $17$ at most with no doubles, so if we choose 18, we are certain of having a double. </p>
|
4,576,868 | <p>I was provided the following generating function, and was unsure how to use it. I have never seen an example where the function “involved” itself.
The generating function is
<span class="math-container">$F(z)^8$</span>
Where
<span class="math-container">$$F(z)=z+z^6 F(z)^5+z^{11} F(z)^{10}+z^{16} F(z)^{15}+z^{21} F(z)^{20}$$</span></p>
<p>Any help appreciated</p>
| Alexander Burstein | 499,816 | <p>Use Lagrange inversion. From the functional equation that <span class="math-container">$F(z)$</span> satisfies, we get, by multiplying through by <span class="math-container">$z$</span>,
<span class="math-container">$$
zF(z)=z^2\left(1+(zF(z))^5+(zF(z))^{10}+(zF(z))^{15}+(zF(z))^{20}\right).
$$</span>
Thus,
<span class="math-container">$$
\frac{z}{1+z^5+z^{10}+z^{15}+z^{20}}\circ(zF(z))=z^2,
$$</span>
so
<span class="math-container">$$
zF(z)=\left(\frac{z}{1+z^5+z^{10}+z^{15}+z^{20}}\right)^{\langle-1\rangle}\circ z^2,
$$</span>
where "<span class="math-container">$\langle-1\rangle$</span>" denotes the compositional inverse. Thus, <span class="math-container">$zF(z)$</span> is an even function, i.e. <span class="math-container">$zF(z)=H(z^2)$</span> for the invertible power series <span class="math-container">$H(z)=\left(\frac{z}{1+z^5+z^{10}+z^{15}+z^{20}}\right)^{\langle-1\rangle}$</span>. This means that <span class="math-container">$H=H(z)$</span> satisfies the functional equation <span class="math-container">$H=z\varphi(H)$</span> for <span class="math-container">$\varphi(z)=1+z^5+z^{10}+z^{15}+z^{20}$</span>. Note that <span class="math-container">$\varphi(0)\ne 0$</span>. We need to find the coefficients of
<span class="math-container">$$
F(z)^8=\frac{H(z^2)^8}{z^8}=\frac{H(z)^8}{z^4}\circ z^2.
$$</span>
By Lagrange inversion,
<span class="math-container">$$
[z^n]f(H(z))=\frac{1}{n}[z^{n-1}]\!\left(\varphi(z)^n f'(z)\right),
$$</span>
where <span class="math-container">$[z^n]$</span> means the coefficient at <span class="math-container">$z^n$</span>.
Here, <span class="math-container">$f(z)=z^8$</span>, so <span class="math-container">$f'(z)=8z^7$</span>, and we need essentially
<span class="math-container">\begin{multline*}
[z^n]\frac{H(z)^8}{z^4}=[z^{n+4}]H(z)^8=\frac{1}{n+4}[z^{n+3}]\!\left(8z^7\varphi(z)^{n+4}\right)=\\=\frac{8}{n+4}[z^{n-4}] (1+z^5+z^{10}+z^{15}+z^{20})^{n+4}.
\end{multline*}</span>
And since <span class="math-container">$1+z^5+z^{10}+z^{15}+z^{20}=(1+z+z^2+z^3+z^4)\circ z^5$</span>, only the terms where <span class="math-container">$n\equiv 4\pmod 5$</span> would be nonzero in the <span class="math-container">$[z^{n-4}] (1+z^5+z^{10}+z^{15}+z^{20})^{n+4}$</span>. And since we are also composing with <span class="math-container">$z^2$</span> at the end, <span class="math-container">$F(z)^8$</span> will only have terms with powers congruent to <span class="math-container">$8\!\!\!\mod\!\!10$</span>.</p>
<p>Finally, here is the <a href="https://oeis.org/A035343" rel="nofollow noreferrer">pentanomial coefficient triangle</a>. Scroll down to example, it's easier to see there.</p>
|
3,003,672 | <p>Say I have an infinte 2D grid (ex. a procedurally generated world) and I want to get a unique number for each integer coordinate pair. How would I accomplish this?</p>
<p>My idea is to use a square spiral, but I cant find a way to make a formula for the unique number other than an algorythm that just goes in a square spiral and stops at the wanted coords.</p>
<p>The application for this converstion could be for example a way to save an n dimensional shape to a file where each line represents a chunk of the shape (by using <span class="math-container">$u(x, y, z) = u(u(x, y), u(y, z))$</span> ), or have a very unique random seed for each integer point (ex. a way to hash an integer vector to a data point in an n dimensional array)</p>
| Michael Stachowsky | 337,044 | <p>This is an interesting question. I will provide a method to simplify your algorithm, but not necessarily a formula just yet (I'm sure that what I'm about to show you will lead to a formula...probably).</p>
<p>Let's begin with a point in the center. That is, we are not starting at the top corner of a semi-infinite plane, we are instead assuming the plane is infinite. The point in the center is assigned the number 1, and we call it the <span class="math-container">$i = 1$</span> point. We surround this square with a border of squares. This border has 8 such squares. We repeat the process and get 16 squares. In general, each "border" has <span class="math-container">$2(2i-1) + 2(2i-3) = 8i - 8 = 8(i-1)$</span> squares. </p>
<p>We now want to form a sum of the first <span class="math-container">$N$</span> such squares:</p>
<p><span class="math-container">$$S_N = \bigg(\sum_{i=2}^{N}8(i-1)\bigg) + 1$$</span></p>
<p>Now, you want to simplify your life, so let's simplify that sum. We would end up with:</p>
<p><span class="math-container">$$S_N = 8\bigg(\sum_{i=2}^{N}(i-1)\bigg) + 1$$</span></p>
<p><span class="math-container">$$S_N = 8\frac{N(N+1)}{2}-1 - (8N - 8) + 1$$</span> (I'll leave you to simplify) and check my algebra. I'm 99% sure it's accurate.</p>
<p>Now you want to unpack this. This involves several steps. Say you have the number <span class="math-container">$M$</span>. You need to find the largest <span class="math-container">$N$</span> such that <span class="math-container">$S_N \le M \lt S_{N+1}$</span>. Other than a strict search, I'm afraid I don't know how to do that, sorry.</p>
<p>Once you know the <span class="math-container">$N$</span>, then you need to compute the quantity <span class="math-container">$M - S_N$</span>. This quantity tells you how many squares to "walk" from some starting square on border <span class="math-container">$N+1$</span> to where you want to be. Since you know where the border squares start, and you know how far you've walked, you know where on the border you are and therefore what the coordinates are.</p>
<p>The method needs some cleanup, but should do it. Good luck doing that in N-D.</p>
|
2,482,341 | <p>I have tried to solve $\frac{\mathrm{d}}{\mathrm{dx}}\int_{0}^{x^2}e^{x+t}\mathrm{dt}$ by two different ways and I'm getting two answers. Please let me know the mistake: </p>
<p><strong>Method One</strong><br>
Let $F(t)$ be the antiderivative of $e^{x+t}$.<br>
Thus $F^{'}(t)=e^{x+t}$ </p>
<p>So </p>
<p>\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\int_{0}^{x^2}e^{x+t}\mathrm{dt} & = \frac{\mathrm{d }}{\mathrm{dx}}(F(x^2)-F(0) \nonumber \\
& = F^{'}(x^2)\cdot2x -F^{'}(0) \nonumber \\
& = e^{x+x^2}\cdot2x-e^{x} \nonumber
\end{align}
But consider the following<br>
<strong>Method Two:</strong><br>
\begin{align}
\text{Let } I &=\int_{0}^{x^2}e^{x+t}\mathrm{dt} \\
&=e^{x}\int_{0}^{x^2}e^{t}\mathrm{dt} \\
&=e^{x}(e^{x^2}-1)\\
&=e^{x+x^2}-e^{x}
\end{align}</p>
<p>\begin{align}
\text{Thus } \frac{\mathrm{d}I}{\mathrm{d}x} &=(2x+1)e^{x+x^2}-e^{x} \\
\end{align}</p>
<p>Thank you in advance</p>
| Koshinder | 493,809 | <p>you can also use Heron's formula.
Heron's formula states that the area of a triangle whose sides have lengths a, b, and c is :-
<a href="https://i.stack.imgur.com/1dN0i.png" rel="nofollow noreferrer">Put the upper limits of a=2,b=3,c=4</a></p>
|
2,273,506 | <p>I was able to simplify a boolean expression $$(\neg a*\neg b*c)+(a*\neg b*\neg c)+(a*b*\neg c)+(a*b*c)$$into the form $$\neg b*(a\oplus c)+a*b$$ where $*$ is the logical and, $+$ is the logical or, and $\oplus$ is the logical XOR.</p>
<p>Apparently, from Wolfram Alpha, this expression can be simplified to $$\left(a\oplus c\right)\oplus(b*c)$$</p>
<p>I would like to know how Wolfram Alpha was able to simplify this.</p>
| G Tony Jacobs | 92,129 | <p>That expression you got from WA isn't the same as your original. It includes, for example, $(\neg a * b * \neg c)$, which is not included in your original expression. I figured that out by looking at a Venn diagram.</p>
<hr>
<p>Edit: With that correction, WA's solution actually matches up. As for how they obtained it, I don't know what specific algorithm they use, but we can see that their expression is indeed equivalent to the original. It consists of two disjoint pieces: $(a\oplus c)*\neg(b*c)$, and $\neg(a\oplus c)*(b*c)$. The first piece is equivalent to $(a*\neg b*\neg c)+(a*b*\neg c)+(\neg a*\neg b*c)$, while the second is the same as $(a*b*c)$.</p>
<p>Again, this is probably easiest to see by drawing a Venn diagram.</p>
|
2,984,918 | <p>How can I prove this? </p>
<blockquote>
<p>Prove that for any two positive integers <span class="math-container">$a,b$</span> there are two positive integers <span class="math-container">$x,y$</span> satisfying the following equation:
<span class="math-container">$$\binom{x+y}{2}=ax+by$$</span></p>
</blockquote>
<p>My idea was that <span class="math-container">$\binom{x+y}{2}=\dfrac{x+2y-1}{2}+\dfrac{y(y-1)}{2}$</span> and choose <span class="math-container">$x,y$</span>, such that <span class="math-container">$2a=x+2y-1, 2b=y(y-1)$</span>, but using this idea, <span class="math-container">$x,y$</span> won’t be always positive. </p>
| nonuser | 463,553 | <p>Because of simmetry we can assume WLOG that <span class="math-container">$a\geq b$</span></p>
<ul>
<li>If <span class="math-container">$a>b$</span>:</li>
</ul>
<p>Let <span class="math-container">$m=2a+1$</span> and <span class="math-container">$n=2b+1$</span> and <span class="math-container">$k= m-n$</span>. Since <span class="math-container">$k$</span> is even <span class="math-container">$k$</span> does not divide odd numbers <span class="math-container">$m$</span> and <span class="math-container">$n$</span>. Since <span class="math-container">$${m\over k}-{n\over k} = {m-n\over m-n} = 1$$</span> we see exists natural number <span class="math-container">$p$</span> in <span class="math-container">$\displaystyle\Big({n\over k},{m\over k}\Big)$</span>. Now let
<span class="math-container">$$x=p(pk-n)$$</span> and <span class="math-container">$$y=p(m-pk)$$</span> We can easly see that <span class="math-container">$x,y$</span> are good and we are done. </p>
<ul>
<li>If <span class="math-container">$a=b$</span> then let <span class="math-container">$z=x+y$</span> and we have <span class="math-container">$z(z-1)=2az $</span> so <span class="math-container">$z=2a+1$</span>. Clearly then <span class="math-container">$x=a+1$</span> and <span class="math-container">$y=a$</span> works.</li>
</ul>
|
2,984,918 | <p>How can I prove this? </p>
<blockquote>
<p>Prove that for any two positive integers <span class="math-container">$a,b$</span> there are two positive integers <span class="math-container">$x,y$</span> satisfying the following equation:
<span class="math-container">$$\binom{x+y}{2}=ax+by$$</span></p>
</blockquote>
<p>My idea was that <span class="math-container">$\binom{x+y}{2}=\dfrac{x+2y-1}{2}+\dfrac{y(y-1)}{2}$</span> and choose <span class="math-container">$x,y$</span>, such that <span class="math-container">$2a=x+2y-1, 2b=y(y-1)$</span>, but using this idea, <span class="math-container">$x,y$</span> won’t be always positive. </p>
| Community | -1 | <p>If <span class="math-container">$a=b$</span> then let <span class="math-container">$(x,y)=(a,a+1)$</span>. </p>
<p>Otherwise, w.l.g. suppose <span class="math-container">$a>b$</span> and let <span class="math-container">$x+y=2t(a-b)$</span> for some positive integer <span class="math-container">$t$</span>. Then
<span class="math-container">$$t(a-b)\Big(2t(a-b)-1\Big) =ax+by=(a-b)x+2bt(a-b)$$</span>
<span class="math-container">$\text{Therefore } x=t\Big(2t(a-b)-1\Big)-2bt=t\Big(2t(a-b)-(2b+1)\Big)$</span>.</p>
<p><span class="math-container">$x$</span> will be a positive integer providing <span class="math-container">$t>\frac{2b+1}{2(a-b)}.$</span></p>
<p><span class="math-container">$y=2t(a-b)-x$</span> will be a positive integer providing <span class="math-container">$2(a-b)>2t(a-b)-(2b+1)$</span> i.e. <span class="math-container">$\frac{2b+1}{2(a-b)}>t-1$</span>.</p>
<p><span class="math-container">$\frac{2b+1}{2(a-b)}$</span> is positive but not an integer and so, precisely as required, there is a positive integer <span class="math-container">$t$</span> such that
<span class="math-container">$$t>\frac{2b+1}{2(a-b)}>t-1.$$</span></p>
|
90,812 | <p>How do I use the Edmonds-Karp algorithm to calculate the maximum flow? I don't understand this algorithm $100\%$. What I need to know is about flow with minus arrow. Here is my graph: </p>
<p><img src="https://i.stack.imgur.com/nep5F.jpg" alt="the graph">. </p>
<p>Our $1-6-11-12$, the flow is $4$. On the next iteration $1-2-4-11-6-7-9-12$, the flow on $6-11$ decrease on $3$, on other $+3$
on how do next? $1-3-5-11-6-8-10-12$? What will be with $11-6$? We must take $-3$, we will get negative flow on $6-11$ or what? Help me please.</p>
| Jesko Hüttenhain | 11,653 | <p>You augment</p>
<ol>
<li>$1-6-11-12$ by $4$</li>
<li>$1-2-4-11-6-7-9-12$ by $3$</li>
<li>$1-3-5-11-6-8-10-12$ by $ 1$</li>
</ol>
<p>and you are done: $12$ is no longer reachable from $1$ in the residual graph. You have already <em>found</em> the maximal flow.</p>
|
3,310,193 | <p>I want to show:
<span class="math-container">$$\mu(E)=0 \rightarrow \int_E f d\mu=0$$</span> </p>
<p><span class="math-container">$f:X \rightarrow [0, \infty] $</span> is measurable and <span class="math-container">$E \in \mathcal{A} $</span></p>
<p>Consider a step function <span class="math-container">$s=\sum_i a_i \chi_{A_i}$</span></p>
<p>Then I get: <span class="math-container">$\int_E s d \mu = \sum_i a_i \mu(A_i\cap E) $</span></p>
<p>Can I conclude that <span class="math-container">$ \mu(A_i \cap E) \leq \mu(E)=0 $</span>?</p>
| Tsemo Aristide | 280,301 | <p>The map <span class="math-container">$H_t(x)=tF(x), t\in [0,1]$</span> restricted to the unit ball defines a deformation retract of the unit ball to a point.</p>
|
4,498,296 | <p>Is there any subtle way to compute the following integral?</p>
<p><span class="math-container">$$\int \frac{\sqrt{u^2+1}}{u^2-1}~ \mathrm{d}u$$</span></p>
<p>The solution i had in mind was substituting <span class="math-container">$u=\tan (\theta)$</span>,then after a few calculations the integral became <span class="math-container">$$\int \sec (\theta)
~ \mathrm{d}\theta+2\int \frac{\sec (\theta)}{\sec^2 (\theta) -2} ~\mathrm{d}\theta$$</span> I think we can formulate the last integral as <span class="math-container">$$\frac{1}{2}\int \left(\frac{1}{\sec (\theta) -\sqrt{2}}+\frac{1}{\sec (\theta)+\sqrt{2}} \right)\mathrm{d}\theta$$</span> But it still seems to be a daunting task and i think it will require further substitutions.</p>
<p>So could anyone please provide a <em>out of the blue</em> kind of solution or a clever approach to this?Or is it possible to go along my approach shortening the calculations?</p>
| Quanto | 686,284 | <p>Note that <span class="math-container">$\int \frac1{\sqrt{u^2+1}}du=\sinh^{-1}u$</span>. Then <span class="math-container">\begin{align}
&\int \frac{\sqrt{u^2+1}}{u^2-1}du
- \int \frac1{\sqrt{u^2+1}}du\\
=&\int \frac2{(u^2-1)\sqrt{u^2+1}} \ du
=2\int \frac{d(\frac u{\sqrt{1+u^2}})}{\frac{2u^2}{u^2+1}-1}
= -\sqrt2\tanh^{-1}\frac{\sqrt2 u}{\sqrt{u^2+1}}
\end{align}</span></p>
|
3,861,319 | <p>I have 3 features <code>age</code>, <code>income</code> and <code>rating</code>.</p>
<p>In case of age I have 3 buckets.</p>
<p>for income I have 4 buckets.</p>
<p>and for rating I have 2 buckets.</p>
<p>If one could filter data where a person could select 1 or more than 1 bucket from each feature what would be the total number of combinations that can be selected?</p>
<p>E.g</p>
<pre><code>Age = [less than 18, 18 < age <= 30, greater than 30]
Rating = [Low, High]
Combination_1 = [less than 18, Low]
Combination_2 = [18 < age <= 30, Low]
Combination_3 = [greater than 30, Low]
Combination_4 = [less than 18, High]
Combination_5 = [18 < age <= 30, High]
Combination_6 = [greater than 30, High]
Combination_7 = [less than 18, None]
Combination_8 = [18 < age <= 30, None]
Combination_9 = [greater than 30, None]
Combination_10 = [None, Low]
Combination_11 = [None, High]
Combination_12 = [None,None]
</code></pre>
<p>In this example there will be 12 combinations. What would be a generalized formula to achieve this?
Here <code>None</code> represents nothing is filtered from the feature.</p>
<p>Edit:
<code>None</code> here is not a feature but just a representation to tell it can be possible that nothing is selected.</p>
<pre><code>Combination_1 = [less than 18, Low]
Combination_2 = [18 < age <= 30, Low]
Combination_3 = [greater than 30, Low]
Combination_4 = [less than 18, High]
Combination_5 = [18 < age <= 30, High]
Combination_6 = [greater than 30, High]
Combination_7 = [less than 18, None]
Combination_8 = [18 < age <= 30, None]
Combination_9 = [greater than 30, None]
Combination_10 = [None, Low]
Combination_11 = [None, High]
Combination_12 = [None,None]
</code></pre>
<p>As one could select more than 1 bucket from each feature following combinations are also possible.</p>
<pre><code>Combination_13 = [[less than 18,18 < age <= 30], Low]
Combination_14 = [[less than 18,18 < age <= 30], high]
Combination_15 = [[less than 18,18 < age <= 30], None]
Combination_16 = [[less than 18,18 < age <= 30, greater than 30], Low]
Combination_17 = [[less than 18,18 < age <= 30, greater than 30], high]
Combination_18 = [[less than 18,18 < age <= 30, greater than 30], None]
Combination_19 = [less than 18, [High,Low]]
Combination_20 = [18 < age <= 30, [High,Low]]
Combination_21 = [greater than 30, [High,Low]]
Combination_22 = [[less than 18,18 < age <= 30], [High,Low]]
Combination_23 = [[less than 18,18 < age <= 30], [High,Low]]
Combination_24 = [[less than 18,18 < age <= 30], [High,Low]]
Combination_25 = [[less than 18,18 < age <= 30, greater than 30], [High,Low]]
Combination_26 = [[less than 18,18 < age <= 30, greater than 30], [High,Low]]
Combination_27 = [[less than 18,18 < age <= 30, greater than 30], [High,Low]]
</code></pre>
| Prasiortle | 554,727 | <p>Write the expression as <span class="math-container">$4(x+y)+x$</span>, so we can see that it is even precisely when <span class="math-container">$x$</span> is even. Thus the statement becomes "there exists <span class="math-container">$x$</span> such that for all <span class="math-container">$y$</span>, <span class="math-container">$x$</span> is even", which is obviously true: just take any even <span class="math-container">$x$</span>.</p>
|
1,548,159 | <p>This is a question asked in India's CAT exam: <a href="http://iimcat.blogspot.in/2013/08/number-theory-questions-and-solutions.html" rel="nofollow noreferrer">http://iimcat.blogspot.in/2013/08/number-theory-questions-and-solutions.html</a> </p>
<blockquote>
<p>How many numbers with distinct digits are possible product of whose digits is 28?</p>
<p>A. 6</p>
<p>B. 4</p>
<p>C. 8</p>
<p>D. 12 </p>
</blockquote>
<p>Firstly, I couldn't even understand the question because the English seems grammatically incorrect. </p>
<p>Secondly, I couldn't understand how the answer was arrived at either.</p>
<pre><code>Two digit numbers; The two digits can be 4 and 7: Two possibilities 47 and 74
Three-digit numbers: The three digits can be 1, 4 and 7: 3! Or 6 possibilities.
We cannot have three digits as (2, 2, 7) as the digits have to be distinct.
We cannot have numbers with 4 digits or more without repeating the digits.
So, there are totally 8 numbers.
</code></pre>
<p>If you can't have three digits, then how can four digits even be considered? And how on earth did they eventually reach 8 numbers? What does this even mean? </p>
<p>ps: I considered asking on <a href="https://puzzling.stackexchange.com/">puzzling.stackexchange</a>, but felt that it'd be more appropriate in a math forum.</p>
| Bharathwaj | 561,197 | <p>Two digit numbers; The two digits can be 4 and 7: Two possibilities 47 and 74.</p>
<p>Three-digit numbers: The three digits can be 1, 4 and 7: 3! Or 6 possibilities.</p>
<p>We cannot have three digits as (2, 2, 7) as the digits have to be distinct.</p>
<p>We cannot have numbers with 4 digits or more without repeating the digits.</p>
<p>So, there are totally 8 numbers.</p>
<p>Correct Answer: 8</p>
<p><a href="http://iim-cat-questions-answers.2iim.com/quant/number-system/other/number-system_1.shtml" rel="nofollow noreferrer">http://iim-cat-questions-answers.2iim.com/quant/number-system/other/number-system_1.shtml</a></p>
|
654,198 | <p>$6x^3 -11x^2 + 6x + 5 \equiv (Ax-1)(Bx - 1)(x - 1) + c$</p>
<p>Find the value of A, B and C.</p>
<p>I started it like this: </p>
<p>$6x^3 -11x^2 + 6x + 5 \equiv (Ax-1)(Bx - 1)(x - 1) + c$</p>
<p>Solving the right hand side:</p>
<p>$ (ABx^2 - Ax - Bx + 1)(x - 1) + C$</p>
<p>$ ABx^3 - ABx^2 - Ax^2 + Ax - Bx^2 + Bx + x - 1 + C$</p>
<p>$ABx^3 - (AB + A + B)x^2 + (A + B + 1)x - 1 + C$</p>
<p>Comparing the coefficients: </p>
<p>$AB = 6$</p>
<p>$A = \frac6 B$</p>
<p>$AB + A + B = 11$</p>
<p>Then substitute the value of A in the above equation...is this right? Is there any error?</p>
| Bill Dubuque | 242 | <p><strong>Hint</strong> $\ x=1\,\Rightarrow\,c = 6.\,$ Cancelling $\,x-1\,$ yields $6x^2-5x+1 = (Ax-1)(Bx-1)\ $ so $\,\ldots$</p>
|
2,065,639 | <p>$\displaystyle \int_a^b (x-a)(x-b)\,dx=-\frac{1}{6}(b-a)^3$</p>
<p>$\displaystyle \int_a^{(a+b)/2} (x-a)(x-\frac{a+b}{2})(x-b)\, dx=\frac{1}{64}(b-a)^4$ </p>
<p>Instead of expanding the integrand, or doing integration by part, is there any faster way to compute this kind of integral?</p>
| Community | -1 | <p>You can first get rid of the integration bounds by the linear transform $a+(b-a)t$:</p>
<p>$$\int_a^b (x-a)(x-b)\,dx=(b-a)^3\int_0^1t(t-1)\,dt.$$</p>
<p>Mentally expanding the polynomial, the integral is $\frac13-\frac12=-\frac16$.</p>
<p>For the other case, $a+(b-a)t/2$:
$$ \int_a^{(a+b)/2} (x-a)(x-\frac{a+b}{2})(x-b)\, dx=\frac{(b-a)^4}{16}\int_0^1t\left(t-1\right)(t-2)\,dt.$$</p>
<p>Also mentally, $(\frac14-1+1)$$\frac1{16}=\frac1{64}$.</p>
<hr>
<p>Also, using a remark by @mickep, Newton's formula applies and</p>
<p>$$\int_0^1t(t-1)\,dt=\frac16\left(0-4\frac12\frac12+0\right)=-1,$$</p>
<p>$$\int_0^1t(t-1)(t-2)\,dt=\frac16\left(0+4\frac12\frac12\frac32+0\right)=\frac14.$$</p>
|
947,626 | <p>What are the conditions under which the center of a group will have a cyclic subgroup? (with proof, of course)</p>
| orangeskid | 168,051 | <p>Every nontrivial subgroup has a nontrivial cyclic subgroup. The condition is : the center is nontrivial. </p>
|
478,713 | <p>I have this logic statement:</p>
<pre><code> (A and x) or (B and y) or (not (A and B) and z)
</code></pre>
<p>The problem is that accessing A and B are rather expensive. Therefore I'd like to access them only once each. I can do this with an if-then-else construct:</p>
<pre><code>if A then
if x then
true
endif
else
if B then
if y then
true
endif
else
if z then
true
endif
endif
endif
</code></pre>
<p>Is there a way to express this as a boolean expression? I have "and", "or" (both short-circuit) but no "xor".</p>
<p>I thought this would work:</p>
<pre><code>X and (A or (Y and (B xor Z)))
</code></pre>
<p>But my test program (<a href="http://pastebin.com/EjURvpM4" rel="nofollow">http://pastebin.com/EjURvpM4</a>) shows it doesn't.</p>
| Argon | 27,624 | <p>$x$-interscepts are when the function intersects the $x$ axis, i.e. when $f(x)=0$. Thus</p>
<p>$$0=(x-3)^2 \implies 0=x-3\implies {x=3}$$</p>
|
1,476,982 | <p>I'm trying to understand why the volume of a parallelepiped whos sides are $s,u,w$ is $ V = s \cdot(u \times w)$.</p>
<p>Even the units of measurement don't add up. The length of the vectors $s,u,w$ is measured in centimeters, the volume is measured in cubic cm.</p>
<p>$u\times w$ is a vector. It is a vector that is orthogonal to $u$ and $w$, but still a vector, so its length is again measure in cms. So overall $V=s \cdot(u \times w)$ means that $V$ is equal to the product of $2$ vectors, so the unit of measurement for $V$ is squared centimeters, not cubed.</p>
<p>I'm struggling to understand how can $|u\times w|$ be equal to the area of a parallelogram. That is equivalent to saying "The time it takes for me to solve a problem is the distance between New York and London."</p>
| Community | -1 | <p>The norm of the vector $u\times v$ is <a href="https://math.stackexchange.com/questions/1395970/what-is-the-logic-rationale-behind-the-vector-cross-product/1471129#1471129"><em>defined</em> as the area of the parallelogram</a> (scroll down to <em>Geometric Definition</em> under <em>The Cross Product</em> if you click that link)with sides $u$ and $v$. Also, as both $u$ and $v$ have units of cm, their product will have units of cm$^2$ -- regardless of the fact that $u\times v$ is a vector. Vectors don't have to have units of length -- they can have whatever units we like.</p>
<p>So if $\|u\times v\|$ is the area of a parallelogram, then the area of the parallelopiped will just be this area times the height of the parallelepiped ("bases times height" is the formula we use here). So because $\|s\|\cos(\theta)$ is the height of the parallelepiped (draw a picture to confirm this for yourself), the volume will just be $\|s\|\cos(\theta)\|u\times v\| = \|s\|\|u\times v\|\cos(\theta)$. But that's just the dot product of $s$ and $u\times v$.</p>
|
4,173 | <p>I asked this question on mathoverflow, but it was deemed too simple, so I'm posting here instead -- </p>
<p>Is there a nice way to characterize an orthonormal basis of eigenvectors of the following $d\times d$ matrix?</p>
<p>$$\mathbf{I}-\frac{1}{d} \mathbf{v}\mathbf{v}'$$</p>
<p>Where $\mathbf{v}$ is a $d\times 1$ vector of 1's. This is similar to the <a href="http://en.wikipedia.org/wiki/Householder_transformation" rel="nofollow">Householder matrix</a>, except the $v's$ are not normalized. One eigenvector is $\mathbf{v}$ with corresponding eigenvalue 0, remaining eigenvalues should be 1. I'm looking for an expression in terms of unknown d.</p>
<p>Motivation: this is covariance matrix of uniform multinomial distribution, so expression for orthonormal basis produces a linear transformation that will make variables uncorrelated for large n</p>
<p>Example: below are 5 orthonormal eigenvectors vectors I get from Gram-Schmidt for d=5...what is the expression for general d? An even bigger example -- columns of <a href="http://yaroslavvb.com/upload/bigmatrix.png" rel="nofollow">this</a> form orthonormal basis for d=20</p>
<p>$$-\frac{1}{\sqrt{2}},0,0,0,\frac{1}{\sqrt{2}}$$</p>
<p>$$-\frac{1}{\sqrt{6}},0,0,\sqrt{\frac{2}{3}},-\frac{1}{\sqrt{6}}$$</p>
<p>$$-\frac{1}{2 \sqrt{3}},0,\frac{\sqrt{3}}{2},-\frac{1}{2 \sqrt{3}},-\frac{1}{2 \sqrt{3}}$$</p>
<p>$$-\frac{1}{2 \sqrt{5}},\frac{2}{\sqrt{5}},-\frac{1}{2 \sqrt{5}},-\frac{1}{2 \sqrt{5}},-\frac{1}{2 \sqrt{5}}$$</p>
<p>$$\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}}$$</p>
<p><b>Update 09/08</b>
I came across another interesting characterization, when d=2^k, for some k, then Walsh Functions form orthogonal basis for this matrix. In particular, let {$\mathbf{x_i}$} represent the list of vectors of binary expansion of integers 1 to d, ie {(0,0,0),(0,0,1),(0,1,0)...}. Then, rows (and columns) of $M$ define the orthonormal basis of matrix in question, where</p>
<p>$$M_{ij}=(-1)^{x_i \cdot x_j}$$</p>
| Agustí Roig | 664 | <p>If you write $P= I - \frac{1}{d}vv^t$ (I assume your $v'$ means the transpose of $v$), then you certainly have a symmetric matrix</p>
<p>$$
P^t = \left( I- \frac{1}{d}vv^t\right)^t = I -\frac{1}{d}v^{tt}v^t = I - \frac{1}{d}vv^t = P \ .
$$</p>
<p>So it diagonalizes and has an orthonormal basis of eigenvectors. That is, there exists an orthogonal matrix $S$, ($S^{-1} = S^t$) and a diagonal matrix $D $ such that $S^tP S = D$.</p>
<p>Moreover, since $P^2 = P$, </p>
<p>\begin{align}
P^2 &= \left( I- \frac{1}{d}vv^t\right) \left( I- \frac{1}{d}vv^t\right) \\
&= I - 2\frac{1}{d}vv^t + \frac{1}{d^2}vv^tvv^t \\
&= I - 2\frac{1}{d}vv^t + \frac{1}{d}vv^t \\
&=I- \frac{1}{d}vv^t = P
\end{align}</p>
<p>and $P \neq 0,I$, its minimal polynomial is $x^2-x = x(x-1)$. So the only eigenvalues of $P$ are $0,1$.</p>
<p>As for eigenvectors, as you say, $v$ is one of them, of $0$ eigenvalue:</p>
<p>$$
Pv = \left( I- \frac{1}{d}vv^t\right)v = v - \frac{1}{d}vv^tv = v - v = 0 \ .
$$</p>
<p>Hence, you can take $\frac{v}{\|v\|} = \frac{1}{\sqrt{d}} (1, \dots ,1)$ as the first vector of your orthonormal basis. The other vectors are an orthonormal basis of the orthogonal complement of $v$, $[v]^\bot$:</p>
<p>\begin{align}
\left( I- \frac{1}{d}vv^t\right)w = w &\quad\Longleftrightarrow \quad w - \frac{1}{d}vv^tw = w \\
&\quad\Longleftrightarrow \quad vv^tw = 0 \\
&\quad\Longleftrightarrow \quad v^tw = 0
\end{align}</p>
<p>So, you only need to compute a basis for $[v]^\bot$ and orthonormalize it. In coordinates you must find the solutions of the linear equation</p>
<p>$$
x_1 + \dots + x_d = 0 \ .
$$</p>
<p>For instance,</p>
<p>$$
(-1, 1, 0, \dots ,0), (-1, 0, 1, 0, \dots ,0), \dots , (-1, 0, \dots , 0,1) \ .
$$</p>
<p>And now you apply the Gram-Schmidt process <a href="http://en.wikipedia.org/wiki/Gram_schmidt" rel="nofollow">http://en.wikipedia.org/wiki/Gram_schmidt</a> to these vectors.</p>
<hr>
<p>EDIT. I think Unkz intuition is right. Before normalizing, the general pattern looks as follows. You have $d-1$ vectors of the form</p>
<p>$$
(-1, 0,\dots,0, i, -1,\dots, -1), \qquad \text{for} \quad i=1,\dots,d-1 \ ,
$$</p>
<p>where $i$ is in the $d-i+1$ coordinate (so there are $i-1$ coordinates with $-1$ after it) and one more last vector</p>
<p>$$
(d,\dots, d) \ .
$$</p>
<p>Let's check that:</p>
<p>(a) Those first $d-1$ vectors belong to $\ker (P-I) = [(1,\dots, 1)]^\bot$ (so they are eigenvectors corresponding to the eigenvalue $1$):</p>
<p>$$
(1,\dots, 1)\cdot (-1, 0,\dots,0, i, -1,\dots, -1) = -1 + i + (i-1)(-1)= 0 \ .
$$</p>
<p>(b) Those first $d-1$ are mutually orthogonal vectors. We may assume $i>j$, for instance:</p>
<p>$$
(-1, 0,\dots,0, i, -1,\dots, -1) \cdot (-1, 0,\dots,0, j, -1,\dots, -1) = 1 - i +(i-1)(-1)^2 = 0 \ .
$$</p>
<p>Now, you quotient out their norms</p>
<p>$$
\sqrt{1 + i^2 + (i-1)} = \sqrt{i^2 + i} = \sqrt{i (i+1)} \qquad \text{for} \quad i= 1, \dots, d-1
$$</p>
<p>and </p>
<p>$$
\sqrt{d^2d}
$$</p>
<p>and you are done.</p>
|
4,173 | <p>I asked this question on mathoverflow, but it was deemed too simple, so I'm posting here instead -- </p>
<p>Is there a nice way to characterize an orthonormal basis of eigenvectors of the following $d\times d$ matrix?</p>
<p>$$\mathbf{I}-\frac{1}{d} \mathbf{v}\mathbf{v}'$$</p>
<p>Where $\mathbf{v}$ is a $d\times 1$ vector of 1's. This is similar to the <a href="http://en.wikipedia.org/wiki/Householder_transformation" rel="nofollow">Householder matrix</a>, except the $v's$ are not normalized. One eigenvector is $\mathbf{v}$ with corresponding eigenvalue 0, remaining eigenvalues should be 1. I'm looking for an expression in terms of unknown d.</p>
<p>Motivation: this is covariance matrix of uniform multinomial distribution, so expression for orthonormal basis produces a linear transformation that will make variables uncorrelated for large n</p>
<p>Example: below are 5 orthonormal eigenvectors vectors I get from Gram-Schmidt for d=5...what is the expression for general d? An even bigger example -- columns of <a href="http://yaroslavvb.com/upload/bigmatrix.png" rel="nofollow">this</a> form orthonormal basis for d=20</p>
<p>$$-\frac{1}{\sqrt{2}},0,0,0,\frac{1}{\sqrt{2}}$$</p>
<p>$$-\frac{1}{\sqrt{6}},0,0,\sqrt{\frac{2}{3}},-\frac{1}{\sqrt{6}}$$</p>
<p>$$-\frac{1}{2 \sqrt{3}},0,\frac{\sqrt{3}}{2},-\frac{1}{2 \sqrt{3}},-\frac{1}{2 \sqrt{3}}$$</p>
<p>$$-\frac{1}{2 \sqrt{5}},\frac{2}{\sqrt{5}},-\frac{1}{2 \sqrt{5}},-\frac{1}{2 \sqrt{5}},-\frac{1}{2 \sqrt{5}}$$</p>
<p>$$\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}}$$</p>
<p><b>Update 09/08</b>
I came across another interesting characterization, when d=2^k, for some k, then Walsh Functions form orthogonal basis for this matrix. In particular, let {$\mathbf{x_i}$} represent the list of vectors of binary expansion of integers 1 to d, ie {(0,0,0),(0,0,1),(0,1,0)...}. Then, rows (and columns) of $M$ define the orthonormal basis of matrix in question, where</p>
<p>$$M_{ij}=(-1)^{x_i \cdot x_j}$$</p>
| Unkz | 326 | <p>While I haven't taken the time to prove it, if you look at the numbers I think you'll see a very nice pattern if you multiply out all your irrational denominators.</p>
<p>(-1,0,0,0,1) norm=$\sqrt{1 \cdot 2}$</p>
<p>(-1,0,0,2,-1) norm=$\sqrt{2 \cdot 3}$</p>
<p>(-1,0,3,-1,-1) norm=$\sqrt{3 \cdot 4}$</p>
<p>(-1,4,-1,-1,-1) norm=$\sqrt{4 \cdot 5}$</p>
<p>(5,5,5,5,5) norm=$\sqrt{5^2 \cdot 5}$</p>
<p>This appears to work for a few other random choices of d, where you have -1 in the first column, (1..d) down the backwards diagonal, and -1 under the backwards diagonal, along with 1,1,...,1 on the bottom row.</p>
|
2,000,013 | <p>On an NFA, how can the empty set ∅ and {ϵ} be considered regular languages? Does it make sense that a machine that accepts no symbols or a machine that takes the empty symbol exist? I could think of a machine (laptop) that is in off mode, where no entries (symbols) are accepted, but, is that of interest? or could it still be considered as a machine? the same model could apply for a rock for example.</p>
<p>thanks!</p>
| MJD | 25,554 | <p>We will play a game. There is a machine with buttons and a green lamp. Your job is to press the right sequence of buttons that make the green lamp light up.</p>
<p>For example, perhaps the machine has only one button, labeled <code>a</code>. When you press it, the lamp lights up! Then you press the button again and the lamp turns off. Then you press a third time and the lamp turns on again. You find that pressing the button always turns the lamp on if it was off, and off if it was on. To light up the lamp and win the game all you need to do is press the button an <em>odd</em> number of times. We say that the machine accepts the language <code>a(aa)*</code>, which is a regular expression that represents the set of all strings of <code>a</code>s of odd length—all the possible winning sequences of button presses.</p>
<p>Now imagine an even simpler machine: the green lamp is already lit! And there is a sign over the button that says DON'T PRESS THE BUTTON. Indeed, if you do press the button the lamp goes off and no amount of pressing makes it turn on again. Don't you wish you had followed the sign's advice? This machine accepts only the string $\epsilon$, which is the string of zero button presses. The regular expression $\epsilon$ represents the set that contains this one string and nothing else.</p>
<p>Now imagine the simplest machine of all: the green lamp never lights up no matter what buttons you press. Here you can't win the game. The set of strings accepted by the machine is empty, which we write $\varnothing$.</p>
|
2,109,197 | <p><strong>Update:</strong><br>
(because of the length of the question, I put an update at the top)<br>
I appreciate recommendations regarding the alternative proofs. However, the main emphasis of my question is about the correctness of the reasoning in the 8th case of the provided proof (with a diagram).</p>
<p><strong>Original question:</strong> </p>
<p>I would like to know, whether the following proof, is a valid way to prove that $a^2 + b^2 \neq 3c^2$ for all $a, b, c \in Z$ (except the trivial case, when $a=b=c=0$). More formally, we have to prove the correctness of the following statement:</p>
<p>$$P: (\forall a,b,c \in Z, a^2 + b^2 \neq 3c^2 \lor (a=b=c=0))$$</p>
<p><strong>Proof.</strong> <em>(by contradiction)</em><br>
For the sake of contradiction let's assume, that there exist such $a, b, c \in Z$, that $a^2 + b^2 = 3c^2$ (and the combination of $a,b,c$ is not a trivial case). More, formally, let's assume that $\neg P$ is <em>true</em>:</p>
<p>$$\neg P: (\exists a,b,c \in Z, a^2 + b^2 = 3c^2 \land \neg (a=b=c=0))$$</p>
<p>There are $2^3$ possible combinations of different parities of $a,b,c$ (8 disjoint cases, which cover entire $Z^3$). So, in order to prove the original statement, we have to consider each case, and show that the <em>true</em>-ness of the $\neg P$ always leads to some sort of contradiction.</p>
<p>Let's consider 8 possible cases (7 of which are simple, <strong>whereas the 8th case looks a bit intricate, and I am not sure regarding its correctness</strong>):</p>
<p><strong>Case 1)</strong> $a$ is odd, $b$ is odd, $c$ is odd<br>
Thus:<br>
$a = (2x + 1)$, $b = (2y + 1)$, $c = (2z + 1)$ for some $x, y, z \in Z$<br>
So:<br>
$$
a^2 + b^2 = 3c^2 \\
\implies (2x + 1) ^2 + (2y + 1)^2 = 3 \cdot (2z + 1)^2 \\
\implies 2 \cdot (2x^2 + 2x + 2y^2 + 2y + 1) = 2 \cdot (6z^2 + 6z + 1) + 1 \\
\implies even\ number = odd\ number \\
$$</p>
<p>However, the derived result contradicts to the fact that <em>odd numbers</em> and <em>even numbers</em> can't be equal.
Hence: $(even\ number = odd\ number) \land (even\ number \neq odd\ number)$, or equivalently: $(even\ number = odd\ number) \land \neg (even\ number = odd\ number)$. Contradiction.</p>
<p><strong>Case 2)</strong> $a$ is odd, $b$ is odd, $c$ is even<br>
Thus:<br>
$a = (2x + 1)$, $b = (2y + 1)$, $c = 2z$ for some $x, y, z \in Z$<br>
So:<br>
$$
a^2 + b^2 = 3c^2 \\
\implies (2x + 1) ^2 + (2y + 1)^2 = 3 \cdot (2z)^2 \\
\implies 2 \cdot (2x^2 + 2x + 2y^2 + 2y + 1) = 12z^2 \\
\implies 2 \cdot (x^2 + x + y^2 + y) + 1 = 6z^2 \\
\implies odd\ number = even\ number
$$
Contradiction.</p>
<p><strong>Case 3)</strong> $a$ is odd, $b$ is even, $c$ is odd<br>
Thus:<br>
$a = (2x + 1)$, $b = 2y$, $c = (2z + 1)$ for some $x, y, z \in Z$<br>
So:<br>
$$
a^2 + b^2 = 3c^2 \\
\implies (2x + 1) ^2 + (2y)^2 = 3 \cdot (2z + 1)^2 \\
\implies 4x^2 + 4x + 1 + 4y^2 = 12z^2 + 12z + 3 \\
\implies 4\cdot(x^2 + x + y^2) = 2 \cdot (6z^2 + 6z + 1) \\
\implies 2\cdot(x^2 + x + y^2) = 6z^2 + 6z + 1 \\
\implies even\ number = odd\ number
$$
Contradiction.</p>
<p><strong>Case 4)</strong> $a$ is odd, $b$ is even, $c$ is even<br>
The square of an odd number is odd (so, $a^2$ is odd).<br>
The square of an even number is even (so, $b^2$ and $3c^2$ are even).<br>
Fact: <em>the sum of an even number and an odd number is odd</em>.<br>
However, equality: $a^2 + b^2 = 3c^2$ leads to the conclusion, that: $odd\ number + even\ number = even\ number$<br>
Contradiction.</p>
<p><strong>Case 5)</strong> $a$ is even, $b$ is odd, $c$ is odd<br>
Symmetric to the Case 3 (because $a$ and $b$ are mutually exchangeable), which shows the contradiction.</p>
<p><strong>Case 6)</strong> $a$ is even, $b$ is odd, $c$ is even<br>
Symmetric to the Case 4, which shows the contradiction.</p>
<p><strong>Case 7)</strong> $a$ is even, $b$ is even, $c$ is odd<br>
Thus:<br>
$a = 2x$, $b = 2y$, $c = (2z + 1)$ for some $x, y, z \in Z$<br>
So:<br>
$$
a^2 + b^2 = 3c^2 \\
\implies 4x^2 + 4y^2 = 12z^2 + 12z + 3 \\
\implies even\ number = odd\ number
$$
Contradiction.</p>
<p><strong>Case 8)</strong> $a$ is even, $b$ is even, $c$ is even<br>
Thus:<br>
$a = 2x$, $b = 2y$, $c = 2z$ for some $x, y, z \in Z$<br>
So:
$$
a^2 + b^2 = 3c^2 \\
\implies 4x^2 + 4y^2 = 3 \cdot 4z^2 \\
\implies x^2 + y^2 = 3z^2
$$</p>
<p>Now, we are faced with the similar instance of the problem, <strong>however, the size of the problem is strictly smaller</strong> ($x = {a \over 2}$, $y = {b \over 2}$, $z = {c \over 2}$).<br>
At first glance, it seems that we have to consider again the eight possible parities of $x, y, z$. However, if we analyze all dependencies between the cases of the problem, we will notice that the only possible outcomes are either <em>contradiction</em> or <em>the trivial case</em>:</p>
<p><a href="https://i.stack.imgur.com/KVWUL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KVWUL.png" alt="enter image description here"></a></p>
<p>We have shown the contradiction in all cases, hence we have subsequently proved the original statement.
$\blacksquare$</p>
<p>So, I would like to know, if there is any problem with reasoning in the 8th case?</p>
| Nosrati | 108,128 | <p>The basic concept for this answer is the area of a triangle. The area of a triangle which is made on vectors $z$ and $w$ is $A=\dfrac12{\bf Im}(\bar{z}w)$. for three points in complex plane, we make a triangle on them with sides $z_1-z_2$, $z_2-z_3$ and $z_3-z_1$, we have
\begin{eqnarray}
\dfrac12{\bf Im}\overline{(z_1-z_2)}(z_2-z_3)&=&0\\
{\bf Im}\overline{z_1}z_2+\overline{z_2}z_3-\overline{z_2}z_2-\overline{z_1}z_3&=&0\\
{\bf Im}\overline{z_1}z_2+\overline{z_2}z_3+{\bf Im}\overline{z_3}z_1&=&{\bf Im}\overline{z_2}z_2+\overline{z_1}z_3+{\bf Im}\overline{z_3}z_1\\
{\bf Im}\overline{z_1}z_2+\overline{z_2}z_3+\overline{z_3}z_1&=&0
\end{eqnarray}</p>
<p>$z_1$, $z_2$ and $z_3$ are collinear, if and only if, the area of triangle made on them is zero.</p>
|
1,247,185 | <p>I already know, and so ask NOT about, the proof of: <a href="https://math.stackexchange.com/a/463407/53259">$A$ only if $B$ = $A \Longrightarrow B$</a>.<br>
Because I ask only for intuition, please do NOT prove this or use truth tables. </p>
<p><strong>My problem:</strong> I try to avoid memorisation. So whenever I see this statement, I always need to pause for 5 minutes to remember my linked post above, in order to determine the meaning.<br>
This pause reveals deficiency in my knowledge that stifles me. So please help me dig deeper. How can this statement be naturalised? </p>
| Matt Samuel | 187,867 | <p>There have been some comments about this requiring memorization or it being different from the way the word is used in normal conversation, but this simply isn't true. It's English.</p>
<p>You can walk the dog only if you have a leash. Therefore, if you can walk the dog, it follows logically that you must have a leash. Can walk the dog implies have a leash.</p>
<p>You can legally drive a car only if you have a license. Therefore, if you can legally drive a car, it follows that you must have a license. Can legally drive a car implies have a license.</p>
<p>Etc.</p>
|
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