qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
1,368,815 | <p>Let $U\subset\mathbb{R}^n$ be an open set. I proved that </p>
<p>$$
H^0_{DR}(U):=\frac{\text{closed forms}}{\text{exact forms}}=\{f\in C^{\infty}(U):\,f\,\,\text{is locally constant} \}
$$</p>
<p>I have to show that $\dim H^0_{DR}=\text{number of connected components of}\,\, U$.</p>
<p>Here is my incomplete proof.</p>
<p>Let's write $U=\bigcup_i U_i$ where $U_i$ are the connected components of $U$.</p>
<p>I have to prove that $f$ is constant on each $U_i$. How?</p>
<p>Then it is sufficient to consider the set
$$
\mathcal{B}=\{\chi_{U_i}\}_i
$$
which is a basis for $H^0_{DR}(U)$.</p>
| Andrew D. Hwang | 86,418 | <p>Hint: Suppose $X$ is a non-empty connected open set, and that $f:X \to \mathbf{R}$ is locally constant. Pick an arbitrary point $x_{0}$ in $X$, and let $U = \{x \in X : f(x) = f(x_{0})\}$.</p>
<p>Use local constancy of $f$ to prove $U$ is open and closed in $X$.</p>
|
52,194 | <p>Assuming you have a set of nodes, how do you determine how many connections are needed to connect every node to every other node in the set?</p>
<p>Example input and output:</p>
<pre><code>In Out
<=1 0
2 1
3 3
4 6
5 10
6 15
</code></pre>
| Jonas Meyer | 1,424 | <p>If there are $n$ nodes, then this is called "$n$ choose $2$", and is equal to the number of $2$-element subsets of a set of $n$ elements. The Wikipedia article on <a href="http://en.wikipedia.org/wiki/Binomial_coefficient">binomial coefficients</a> includes this and generalizations. </p>
<p>Since I started writing you discovered the correct formula. However, if you ever have a similar problem where you are trying to figure out a general form for the terms in a sequence from some initial values, a good tool is <a href="http://oeis.org/">The On-Line Encyclopedia of Integer Sequences</a>. In this case, entering <code>0,1,3,6,10,15</code> brings up a <a href="http://oeis.org/A000217">useful entry</a> in which you can find the general form and references.</p>
|
2,679,525 | <p>Find the $\gcd(x^3-6x^2+14x-15, x^3-8x^2+21x-18)$ over $\mathbb{Q}[x]$. Then find two polynomials $a(x),b(x) \in \mathbb{Q}[x]$ such that, $$a(x)(x^3-6x^2+14x-15) + b(x)(x^3-8x^2+21x-18)=\gcd(x^3-6x^2+14x-15, x^3-8x^2+21x-18)$$ </p>
<p>I have managed to find,
$$x^3-6x^2+14x-15=(x-3)(x^2-3x+5)$$
$$x^3-8x^2+21x-18=(x-3)(x-3)(x-2)$$</p>
<p>Now since $x^2-3x+5$ is irreducible over $\mathbb{Q}[x]$ and so the greatest common divisor is $(x-3)$. Now to find $a(x)$ and $b(x)$ I have no clue how to do that. I have looked online and it seems there is extended euclidean algorithm for polynomials but I haven't formally learned it in my class yet, so I was wondering if there is another efficient way to find these polynomials. Any help is appreciated, thanks!</p>
| Will Jagy | 10,400 | <p>This is just the Extended Euclidean Algorithm. Instead of back-substitution, I have always preferred to write the construction steps in the style of continued fractions. Furthermore, I have always depended on the kindness of strangers.</p>
<p>$$ \left( x^{3} - 6 x^{2} + 14 x - 15 \right) $$ </p>
<p>$$ \left( x^{3} - 8 x^{2} + 21 x - 18 \right) $$ </p>
<p>$$ \left( x^{3} - 6 x^{2} + 14 x - 15 \right) = \left( x^{3} - 8 x^{2} + 21 x - 18 \right) \cdot \color{magenta}{ \left( 1 \right) } + \left( 2 x^{2} - 7 x + 3 \right) $$
$$ \left( x^{3} - 8 x^{2} + 21 x - 18 \right) = \left( 2 x^{2} - 7 x + 3 \right) \cdot \color{magenta}{ \left( \frac{ 2 x - 9 }{ 4 } \right) } + \left( \frac{ 15 x - 45 }{ 4 } \right) $$
$$ \left( 2 x^{2} - 7 x + 3 \right) = \left( \frac{ 15 x - 45 }{ 4 } \right) \cdot \color{magenta}{ \left( \frac{ 8 x - 4 }{ 15 } \right) } + \left( 0 \right) $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{ \left( 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( 1 \right) }{ \left( 1 \right) } $$
$$ \color{magenta}{ \left( \frac{ 2 x - 9 }{ 4 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 2 x - 5 }{ 4 } \right) }{ \left( \frac{ 2 x - 9 }{ 4 } \right) } $$
$$ \color{magenta}{ \left( \frac{ 8 x - 4 }{ 15 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 4 x^{2} - 12 x + 20 }{ 15 } \right) }{ \left( \frac{ 4 x^{2} - 20 x + 24 }{ 15 } \right) } $$
$$ \left( x^{2} - 3 x + 5 \right) \left( \frac{ 2 x - 9 }{ 15 } \right) - \left( x^{2} - 5 x + 6 \right) \left( \frac{ 2 x - 5 }{ 15 } \right) = \left( -1 \right) $$
$$ \left( x^{3} - 6 x^{2} + 14 x - 15 \right) = \left( x^{2} - 3 x + 5 \right) \cdot \color{magenta}{ \left( x - 3 \right) } + \left( 0 \right) $$
$$ \left( x^{3} - 8 x^{2} + 21 x - 18 \right) = \left( x^{2} - 5 x + 6 \right) \cdot \color{magenta}{ \left( x - 3 \right) } + \left( 0 \right) $$
$$ \mbox{GCD} = \color{magenta}{ \left( x - 3 \right) } $$
$$ \left( x^{3} - 6 x^{2} + 14 x - 15 \right) \left( \frac{ 2 x - 9 }{ 15 } \right) - \left( x^{3} - 8 x^{2} + 21 x - 18 \right) \left( \frac{ 2 x - 5 }{ 15 } \right) = \left( - x + 3 \right) $$ </p>
<p>............</p>
|
2,866,389 | <p>Let $\gamma\colon\left[0,1\right]\to\mathbb{R}^{2}$ be a curve such
that $\gamma\in\mathcal{C}^{1}$ (Continuously Differentiable). I
need to show that $\gamma\left(\left[0,1\right]\right)$ has mesure
zero, that is to show there are generalized rectangles $R_{1},R_{2},\ldots$
such that $\gamma\left(\left[0,1\right]\right)\subseteq\bigcup_{i=1}^{\infty}R_{i}$
and $\sum_{i=1}^{\infty}V\left(R_{i}\right)<\varepsilon$ where $V$
is the volume function.</p>
<p>Now I know how to show that the graph of a continuous function $f\colon\mathbb{R}^{n}\to\mathbb{R}$
has measure zero by using the uniform continuity, but that argument
doesn't work here.</p>
<p>Any help?</p>
| Adayah | 149,178 | <p>Hint: A $\mathcal{C}^1$ curve $\gamma : [0, 1] \to \mathbb{R}^2$ is Lipschitz with some constant $L$. For each $n \in \mathbb{N}$ and $1 \leqslant j \leqslant n$ let $I_j = \left[ \frac{j-1}{n}, \frac{j}{n} \right]$ so that $I_j$ are of length $\frac{1}{n}$ and cover $[0, 1]$. Now let $m_j$ be the center of $I_j$ so that for each $x \in I_j$ we have that $|x - m_j| \leqslant \frac{1}{2n}$.</p>
<p>Can you cover $\gamma( [0, 1] )$ with a union of $n$ balls of a small radius?</p>
|
13,478 | <p>This may be a silly question - but are there interesting results about the invariant: the minimal size of an open affine cover? For example, can it be expressed in a nice way? Maybe under some additional hypotheses?</p>
| Dmitri Panov | 943 | <p>Consider for simplicity smooth projective varieties defined over $\mathbb C$. In this case, the minimal size equals $n+1$ where $n$ is the dimension of the variety. </p>
<p>Proof. Let $M^n$ be such a variety. Take $n+1$ generic very ample divisors $D_0,...,D_n$. Then such divisors don't have a common intersection. At the same time for any $i$ $M^n\setminus D_i$ is affine.
In order to show that you do need $n+1$ open affine sub-varieties, notice that the complement to an open affine sub-variety is a closed variety of dimension $n-1$. Now proceed by induction.</p>
<p>For projective irreducible varieties exactly the same reasoning should hold, and it can be generalised further I guess </p>
|
52,338 | <p>I am interested in finding explicit formulae for (better yet characterizing) conformal functions from various domains onto the open unit disc $\mathbb{D}\subset\mathbb{C}$, and in understanding the key ideas necessary to establish such functions.</p>
<p>Specifically, what can $f$ look like when $f:G\to\mathbb{D}$ is conformal and</p>
<p>(1) $G=\{x+iy~|~x,y>0\}$ is the open first quadrant.</p>
<p>(2) $G=\{x+iy~|~x>0,~0<y<1\}$ is an open horizontal strip in the first quadrant.</p>
<p>(3) $G=\{z\in\mathbb{C}~|~\frac{1}{2}<|z|<1\}$ is an annulus.</p>
<p>(4) $G=\mathbb{D}\cap\{|z-\frac{1}{2}|>\frac{1}{2}\}$ is something else (torus?).</p>
| t.b. | 5,363 | <p>Since you didn't show too many own thoughts, here are some hints only. By <em>conformal</em> I understand <em>biholomorphic</em>.</p>
<ol>
<li><p>First take $f(z) = z^2$ to map the quadrant biholomorphically onto the upper half-plane, then compose with the <a href="http://en.wikipedia.org/wiki/Cayley_transform#Conformal_map">Cayley transform</a> $\kappa(z) = \frac{z-i}{z+i}$ to get $\kappa(f(z)) = \frac{z^2-i}{z^2+i}$.</p></li>
<li><p>Look at $\cos{(z)}$ and modify appropriately.</p></li>
<li><p>Impossible, since $G$ is not <a href="http://en.wikipedia.org/wiki/Simply_connected_space">simply connected</a>.</p></li>
<li><p>Map the region $G$ to the strip between two parallel lines using a Möbius transformation sending $1$ to infinity (e.g. using the inverse Cayley transformation). Then use the exponential function. </p></li>
</ol>
<p>This should be enough to figure the solutions out.</p>
<p>For the precise relationship between "conformal" and "analytic", as well as for explanations on how to find such maps, I refer you to <a href="http://books.google.com/books?id=2MRuus-5GGoC">Ahlfors</a> or (probably—I never really read it) <a href="http://books.google.com/books?id=ogz5FjmiqlQC">Needham</a> or any decent text on complex analysis treating conformal mapping.</p>
<p>The characterization of biholomorphisms between simply connected regions is essentially the content of the <a href="http://en.wikipedia.org/wiki/Riemann_mapping_theorem">Riemann mapping theorem</a>.</p>
<p>Sometimes biholomorhic mappings between polygonal regions and the unit disk can be computed via the <a href="http://en.wikipedia.org/wiki/Schwarz%5DChristoffel_mapping">Schwarz-Christoffel formula</a>, but usually it leads to <a href="http://en.wikipedia.org/wiki/Elliptic_integral">elliptic integrals</a> that can't be solved explicitly in elementary terms.</p>
<hr>
<p><strong>Added:</strong></p>
<p>Since the solution of 4. is a bit trickier, here's a rather detailed outline:</p>
<p>First note that $G$ is the region enclosed between the circles $\{|z| = 1\}$ and $\{|z - \frac{1}{2}| = \frac{1}{2}\}$. Applying the Möbius transformation (= the inverse Cayley transform) $\kappa^{-1}(z) = i\frac{1+z}{1-z}$ sends $G$ to the horizontal strip $\{0 \lt \operatorname{Im}{z} \lt 1\}$. To see this, look at this picture <a href="http://en.wikipedia.org/wiki/Cayley_transform#Conformal_map">from Wikipedia</a> illustrating the Cayley transform:</p>
<p><img src="https://i.stack.imgur.com/VIY2O.png" alt="Cayley transform"></p>
<p>Finally, the exponential function $g(z) = \exp{(\pi z)}$ sends this strip to the upper half plane. Composing this with the Cayley transform we get the biholomorphic map $h = \kappa \circ g \circ \kappa^{-1}: G \to \mathbb{D}$. </p>
|
52,338 | <p>I am interested in finding explicit formulae for (better yet characterizing) conformal functions from various domains onto the open unit disc $\mathbb{D}\subset\mathbb{C}$, and in understanding the key ideas necessary to establish such functions.</p>
<p>Specifically, what can $f$ look like when $f:G\to\mathbb{D}$ is conformal and</p>
<p>(1) $G=\{x+iy~|~x,y>0\}$ is the open first quadrant.</p>
<p>(2) $G=\{x+iy~|~x>0,~0<y<1\}$ is an open horizontal strip in the first quadrant.</p>
<p>(3) $G=\{z\in\mathbb{C}~|~\frac{1}{2}<|z|<1\}$ is an annulus.</p>
<p>(4) $G=\mathbb{D}\cap\{|z-\frac{1}{2}|>\frac{1}{2}\}$ is something else (torus?).</p>
| user786 | 786 | <p>Here is a sketch:</p>
<p>For (1) we want to map the open first quadrant onto the unit disk. What we can do is first map the open first quadrant onto the upper half plane then the upper half plane onto the unit disk. the mapping $z \mapsto z^2$ maps the first quadrant to the upper half plane. Then the upper half plane can be mapped to the unit disk by the mapping $z \mapsto \frac{z-i}{z+i}$. What remains to do is to compose.</p>
<p>For (2) $z \mapsto cosh(\pi z)$ maps the open half strip of width 1 to the upper half plane. You can now compose with the mapping in (1) which maps the upper half plane to the unit disk.</p>
<p><strong>EDIT: As noted by Theo, these two last examples are false.</strong></p>
<p>For (3) I am not so sure. $z \mapsto lnz$ maps an annulus onto a rectangle. A rectangle in the plane is simply connected so by the Riemann Mapping Theorem one can find a unique conformal mapping between the rectangle and the unit disk. However I don't know which one.</p>
<p>For (4) It sound like the region described in the interior of a parabola. In which the case the mapping onto the unit disk would be $tan^2 \frac{\pi}{4} \sqrt{\frac{z}{p}} $, $p$ being one fourth of the height of the segment on the $y$-axis formed by the intersection of the parabola with the $y$-axis. I am really unsure about this last one. Maybe someone else could help.</p>
|
1,576,235 | <p>Take, for example: $3x + y + 2z = 6$.</p>
<p>Parameterized as: $ui + (6 -3u - 2v)j + vk$
restricted to the first octant</p>
| Slade | 33,433 | <p>No, everywhere it meets $x=0$, $y=0$, or $z=0$, is a boundary point.</p>
<p>One way to see this: in a manifold $M$ with boundary, $x\in M$ is a boundary point if and only if $x$ has a neighborhood $U$ such that $U\setminus \{x\}$ is contractible. But in your example, $M\setminus \{x\}$ is convex, hence contractible, for any corner point of $M$.</p>
|
342,684 | <p>It would be useful to me to have a result of the following kind (which I would need to generalize, but this case is already interesting). Let <span class="math-container">$r<n$</span> be positive integers and let <span class="math-container">$\delta>0$</span> be a fixed constant such as 1/100. Does there exist a subspace <span class="math-container">$V$</span> of <span class="math-container">$\mathbb F_2^n$</span> that is a <span class="math-container">$\delta r$</span>-separated <span class="math-container">$r$</span>-net? That is, I would like every vector to have Hamming distance at most <span class="math-container">$r$</span> from some <span class="math-container">$v\in V$</span> and any two distinct vectors in <span class="math-container">$V$</span> to have distance at least <span class="math-container">$\delta r$</span> from each other. </p>
<p>This feels like the kind of thing that coding theorists ought to know about, but what comes up when I do a Google search is more to do with the dimension of <span class="math-container">$V$</span> than with whether it is a good net. But perhaps it comes into the "known to be hard" category.</p>
<p>I am interested in more or less the full range of <span class="math-container">$r$</span> for given <span class="math-container">$n$</span>: the existence of such a linear code for some specific <span class="math-container">$r$</span> would be interesting but not enough for my eventual purposes.</p>
| Ilya Bogdanov | 17,581 | <p>Any maximal <span class="math-container">$r$</span>-separated set is an <span class="math-container">$r$</span>-net, otherwise you can augment it with a non-covered point.</p>
<p>But the same holds for linear codes! If a subspace <span class="math-container">$V$</span> is <span class="math-container">$r$</span>-separated but not an <span class="math-container">$r$</span>-net, you may take a far point <span class="math-container">$u$</span>: then <span class="math-container">$\langle V,u\rangle$</span> is still <span class="math-container">$r$</span>-separated. So a maximal <span class="math-container">$V$</span> fits.</p>
|
2,181,633 | <p>Given a real polynomial $P(x)$, find another real polynomial $Q(x)$ such that the powers of $x$ in $P(x)Q(x)$ are multiples of an integer $n$. Is it always possible to find such a polynomial?</p>
<p>For e.g if $P(x) = 1+3x+2x^2$ I can find $Q(x)=(2 x - 1) (x - 1) (4 x^2 + 1) (x^2 + 1)$ such that $P(x)Q(x) = 16 x^8 - 17 x^4 + 1$ has powers which are multiples of 4. </p>
<p>I'm bumping the question because I'm not satisfied of the original answer.</p>
| Community | -1 | <p>Here are three lemmas that underlie one approach to the problem: the first underlies the main idea, and the other two in dealing with the complication of real coefficients instead of complex ones.</p>
<p>Let $\zeta$ be a primitive $n$-th root of unity, and let Let $\bar{R}(x)$ be the polynomial whose coefficients are the complex conjugates of the coefficients of $R(x)$.</p>
<p><strong>Lemma:</strong> $R(x) = R(\zeta x)$ if and only if the only monomials in $R(x)$ are of the form $x^{an}$.</p>
<p><strong>Lemma:</strong> $R(x)$ is a real polynomial if and only if $R(x) = \bar{R}(x)$</p>
<p><strong>Lemma:</strong> If $R(x) = S(\alpha x)$, then $\bar{R}(x) = \bar{S}(\bar{\alpha} x)$.</p>
<p>I think you've already spotted the trick to the problem when $n=2$: take $Q(x) = P(-x)$, so that the product is $P(x) P(-x)$....</p>
|
1,324,824 | <p>Compute the flux integral with vector field,
$\vec F(x,y,z)= <2x,y,3z>$
and a sphere of radius 36 centered at the point $(1,2,-1)$</p>
<p>so what I did was I used divergence theorem which is</p>
<p>$\int\int \vec F \cdot \vec n dS = \int \int \int (div \vec F) dV$</p>
<p>I used the right hand side and so</p>
<p>$\int \int \int 6dV$</p>
<p>From here I can see the answer is 6*(volume of sphere), and hence my answer doesn't change even if the origin of sphere is different. So my question is, does the flux integral answer changes when the origin of sphere change?(assuming volume does not change)</p>
| Raman Singh | 166,287 | <p>I read about flux in my electostatics class . So I may be wrong but :- </p>
<p>Consider flow of a liquid with velocity <strong>v</strong>, through a small
flat surface <strong>dS</strong>, in a direction normal to the surface. The
rate of flow of liquid is given by the volume crossing the
area per unit time <strong>v . dS</strong> and represents the flux of liquid
flowing across the plane. </p>
<p>Now for a sphere it doesn't matter where you place it , the Surface Area remains same and as it is a sphere , changing the angle doesn't make a difference. So flux remains the same.</p>
|
3,653,016 | <p>I am solving this equation: </p>
<p><img src="https://i.stack.imgur.com/WN7ce.png" alt="Image of the equation"><br>
My issue arises on line 2 where we have (n + 1 - 2 + 1)(n + 1 + 2)/2. This is what i understand. We have the formula n! = n(n+1)/2. Subbing values into the equation yields us with 10[(n+1)(n+1 + 1)/2] however we need to account for the fact we are starting at j = 2. I would then do this by 10[(n+1-2)(n+1+1)/2]. What I dont understand is why in the solution they have given </p>
<ul>
<li>(n+1 - 2 <strong>- 1 +1</strong>) the -1 + 1</li>
<li>(n + 1 <strong>+ 2</strong>) the + 2</li>
</ul>
<p>How did they make that logical jump and what were the steps/thought process in doing so.</p>
| JMP | 210,189 | <p><span class="math-container">$n+1-2+1$</span> gives the number of terms, which is the number of integers in the inclusive range <span class="math-container">$[2,n+1]$</span>.</p>
<p><span class="math-container">$\dfrac{n+1+2}{2}$</span> gives the average term, which is the largest term (<span class="math-container">$n+1$</span>) plus the smallest term (<span class="math-container">$2$</span>) divided by <span class="math-container">$2$</span>. </p>
|
723,355 | <p>To win the lottery you must pick the one winning ticket.
Given the option of drawing 1 out of 10 tickets or 10 out of 100 tickets to win which is the better option? Are they both 1 out of 10 or is it better to pick 1 out of 100 first, then 1 out of 99, then 1 out of 98 (assuming you don't pick the winner) up to your 10 picks. Both should be 10% but I don't know why the 10 separate picks out of 100 doesn't add up to more than 10%. Can someone explain the probability equation for this? </p>
| Thanos Darkadakis | 105,049 | <p>Why do you think it's not the same?</p>
<p>It's easier to count the probability that you will NOT win.</p>
<p>In the first case it is $P_1=\frac9{10}$</p>
<p>In the second case it is: </p>
<p>$P_2=\frac{99}{100}\times\frac{98}{99}\times\frac{97}{98}\times\frac{96}{97}\times\frac{95}{96}\times\frac{94}{95}\times\frac{93}{94}\times\frac{92}{93}\times\frac{91}{92}\times\frac{90}{91}=\frac{90}{100}=P_1$</p>
|
247,315 | <p>Please see my Edited version at the end of the post.</p>
<p>================================</p>
<p><a href="http://en.m.wikipedia.org/wiki/Cantor_set" rel="nofollow">http://en.m.wikipedia.org/wiki/Cantor_set</a></p>
<p>My definition of Cantor Set is just like that of wikipedia.</p>
<p>That is, $C=[0,1]\setminus \bigcup_{i=1}^{\infty} \bigcup_{k=0}^{3^{i-1}-1} (\frac{3k+1}{3^i}, \frac{3k+2}{3^i})$.</p>
<p>With this definition, i have shown that $C$ is compact, perfect, equipotent with $2^{\aleph_0}$ and contains no openset. (i.e. Basic properties of Cantor set)</p>
<p>I preferred this definition to another since this definition is simple and strictly written in first-order logic.</p>
<p>Let $C_n = [0,1]\setminus \bigcup_{i=1}^n \bigcup_{k=0}^{3^{i-1}-1} (\frac{3k+1}{3^i},\frac{3k+2}{3^i})$.</p>
<p>Then $\bigcap_{n\in \mathbb{N}} C_n = C$.</p>
<p>Here, how do i prove that $C_n$ is a disjoint union of $2^n$ intervals, each of length $3^{-n}$?</p>
<p>(To make it clear, intervals here refer to closed connected sets)</p>
<p>===========================
EDIT:</p>
<p>This is not actually i meant, but this is exactly the same as what i wanted to prove anyway..</p>
<p>Let $A_0=B_0=[0,1]$.
Define $\{A_n\}$ recursively such as;
$A_{n+1}=\frac{A_n}{3} \cup (\frac{2}{3} + \frac{A_n}{3})$.</p>
<p>Now, define $B_n=[0,1]\setminus \bigcup_{i=1}^n \bigcup_{k=0}^{3^{i-1}-1} (\frac{3k+1}{3^i},\frac{3k+2}{3^i})$, $\forall n\in \mathbb{Z}^+$.</p>
<p>How do i prove $A_n=B_n$?</p>
| Stephen Nand-Lal | 78,546 | <p>In $\mathbb{R}^2$, for any two Cantor sets $A$ and $B$ we can find a homeomorphism $h \colon \mathbb{R}^2 \to \mathbb{R}^2$ which carries $h(A)$ onto $B$. This essentially says they're equivalently embedded. I'm not sure this directly answers your question but in $\mathbb{R}^2$, talk of `the' Cantor set isn't necessarily correct, however upto homeomorphism of the ambient space we do only have one equivalence class of Cantor set.</p>
|
4,253,485 | <p>I understand that if you roll <span class="math-container">$10$</span> <span class="math-container">$6$</span>-faced dice (D6) the different outcomes should be <span class="math-container">$60,466,176$</span> right? (meaning <span class="math-container">$1/60,466,176$</span> probability)</p>
<p>but if you are playing a board-game, you don't care about that right? what matters is the sum of the numbers in the end, so it should anyway be <span class="math-container">$1/60$</span> instead of whatever right? because its a sum not <span class="math-container">$10$</span> individual rolls.</p>
<p>so a <span class="math-container">$60$</span> face die (D60) should be the same as rolling all the dice right? if not can someone please explain? I've been looking all over the net for a compelling answer but there is none. I've seen a D60 can be used as <span class="math-container">$2$</span> D6 but also there is no explanation for this.</p>
<p>if not, what would be the number of faces needed for an equivalent single die that can replace <span class="math-container">$10$</span> D6. I know so far the D120 is the last possible die that can be made, but it doesn't matter, I just want to know too...</p>
<p>EDIT: Sorry I forgot to mention in some tabletop games Dice can have <span class="math-container">$0$</span>, so you'd treat all <span class="math-container">$6$</span> sided dice as going from <span class="math-container">$0$</span> to <span class="math-container">$5$</span>, making minimum value of <span class="math-container">$0$</span> possible in all <span class="math-container">$10$</span> rolls. But still that changes the maximum in the same way minimum used to be, now the max you get in 6 sided dice is <span class="math-container">$50$</span>, while on the 60 sided die it would be <span class="math-container">$59$</span> if we apply a <span class="math-container">$-1$</span> to all results... it gets messy, and this detail doesn't really change much</p>
| Rezha Adrian Tanuharja | 751,970 | <p>A simple argument is you can roll a <span class="math-container">$1$</span> using one <span class="math-container">$60$</span> sided die but you cannot using ten <span class="math-container">$6$</span> sided dice.</p>
|
4,253,485 | <p>I understand that if you roll <span class="math-container">$10$</span> <span class="math-container">$6$</span>-faced dice (D6) the different outcomes should be <span class="math-container">$60,466,176$</span> right? (meaning <span class="math-container">$1/60,466,176$</span> probability)</p>
<p>but if you are playing a board-game, you don't care about that right? what matters is the sum of the numbers in the end, so it should anyway be <span class="math-container">$1/60$</span> instead of whatever right? because its a sum not <span class="math-container">$10$</span> individual rolls.</p>
<p>so a <span class="math-container">$60$</span> face die (D60) should be the same as rolling all the dice right? if not can someone please explain? I've been looking all over the net for a compelling answer but there is none. I've seen a D60 can be used as <span class="math-container">$2$</span> D6 but also there is no explanation for this.</p>
<p>if not, what would be the number of faces needed for an equivalent single die that can replace <span class="math-container">$10$</span> D6. I know so far the D120 is the last possible die that can be made, but it doesn't matter, I just want to know too...</p>
<p>EDIT: Sorry I forgot to mention in some tabletop games Dice can have <span class="math-container">$0$</span>, so you'd treat all <span class="math-container">$6$</span> sided dice as going from <span class="math-container">$0$</span> to <span class="math-container">$5$</span>, making minimum value of <span class="math-container">$0$</span> possible in all <span class="math-container">$10$</span> rolls. But still that changes the maximum in the same way minimum used to be, now the max you get in 6 sided dice is <span class="math-container">$50$</span>, while on the 60 sided die it would be <span class="math-container">$59$</span> if we apply a <span class="math-container">$-1$</span> to all results... it gets messy, and this detail doesn't really change much</p>
| Adam Rubinson | 29,156 | <p>Adding onto other answers…</p>
<p>Rolling ten six-sided dice, the most likely sum you are to get is <span class="math-container">$\frac{1+6}{2}*10=35.$</span> Then, getting a sum of <span class="math-container">$34$</span> is slightly less likely than getting a sum of <span class="math-container">$35$</span>. Etc… all the way down to getting a sum of <span class="math-container">$10$</span>, which has probability just <span class="math-container">$\frac{1}{6^{10}}.$</span></p>
<p>But with a <span class="math-container">$60-$</span> sided die, the probability of getting a <span class="math-container">$10$</span> is just <span class="math-container">$\frac{1}{60}.$</span></p>
|
2,198,462 | <p>Let $\mathbf{f} : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be given by:
$$\mathbf{f}(u, v) = (u − v,\sin(u) \sin(v)).$$</p>
<p>Let $(y, z) = (0, 3/4).$ Find the point $(a, b)$ with $0 < a < \frac{\pi}{2}$ such that $\mathbf{f}(a, b)=(y, z)$.</p>
<p>Then prove that the function $\mathbf{f}^{−1}$ such that $\mathbf{f}^{−1}(y, z) = (a, b)$ exists and is differentiable in some neighbourhood of $(y, z).$</p>
<p>Im really not sure on both questions so any help will be appreciated.</p>
| Jaideep Khare | 421,580 | <p><strong>Answer to Part I</strong></p>
<p>You want $$\mathbf{f}(a, b)=(y, z)=(0, 3/4)$$</p>
<p>Using definition :</p>
<p>$$\mathbf{f}(a, b) = (a − b,\sin(a) \sin(b))=(0,3/4)$$</p>
<p>i.e. $a-b=0 \implies a=b$ </p>
<p>and ,
$$\sin a \sin b =\sin^2a=3/4 \implies \sin a =+\frac{\sqrt{3}}{2}$$ (Since $0< a< \pi/2$, $\sin a$ is positive)$ \implies a= \pi/3$</p>
<p>$$\text{Therefore} ~ a=b= \frac{\pi}{3}$$</p>
|
168,926 | <p>How much of ring theory arises simply from applying group-theoretic results to the additive structure, or semigroup/monoid-theoretic results to the multiplicative structure? To be more precise, can anyone describe <em>how much</em> rigidity is imposed on the structure of rings by the distributive axiom which "connects" addition and multiplication? What are some examples of results, say, where distributivity really plays a key role?</p>
| William | 13,579 | <p>One example of where the distributivity axiom adds additional properties to ring-like structures is the following: </p>
<p>The ring axioms <em>with</em> the multiplicative identity axiom and <em>without</em> the commutativity of addition axiom can already prove the commutativity of addition. </p>
<p>This requires the distributive property. </p>
<p>(1 + 1)(x + y) = (1 + 1)(x) + (1 + 1)(y) = x + x + y + y</p>
<p>(1 + 1)(x + y) = 1(x + y) + 1(x + y) = x + y + x + y</p>
<p>so $x + y = y + x$. </p>
<p>So distributivity forces the commutativity of addition in the case that the ring has a multiplicative identity. </p>
|
219,763 | <p><strong>The function $f$ is defined on $\mathbb{R}$ such that for every $\delta\gt0$, $|f(y)-f(x)|\lt\delta^2$ for all $x,y\in\mathbb{R}$ and $|y-x|\lt\delta$. Prove that $f$ is a constant function.</strong> </p>
<p>So, what I know, is that I need to show that $f(a)=f(b)$ for all points $a,b\in\mathbb{R}$. Or for every $\epsilon\gt0$, $|f(a)-f(b)|\lt\epsilon$. </p>
<p>I'm at a loss here. I got a hint to divide the interval $[a, b]$ into $n$ smaller intervals. But I don't understand why, and how this would help me. </p>
<p>Hints and preferably a proof are very much appreciated. Thanks in advance. </p>
| Berci | 41,488 | <p><strong>Hint:</strong> Show that $f$ is <em>differentiable</em> in each point and the same time that the derivative is $0$.</p>
|
219,763 | <p><strong>The function $f$ is defined on $\mathbb{R}$ such that for every $\delta\gt0$, $|f(y)-f(x)|\lt\delta^2$ for all $x,y\in\mathbb{R}$ and $|y-x|\lt\delta$. Prove that $f$ is a constant function.</strong> </p>
<p>So, what I know, is that I need to show that $f(a)=f(b)$ for all points $a,b\in\mathbb{R}$. Or for every $\epsilon\gt0$, $|f(a)-f(b)|\lt\epsilon$. </p>
<p>I'm at a loss here. I got a hint to divide the interval $[a, b]$ into $n$ smaller intervals. But I don't understand why, and how this would help me. </p>
<p>Hints and preferably a proof are very much appreciated. Thanks in advance. </p>
| Sangchul Lee | 9,340 | <p>It is natural to think that the given condition implies the inequality
$$\left| f(y) - f(x) \right| \leq C \left| y - x \right|^2. \tag{1}$$
But this implication requires some justification. So here is a proof.</p>
<p>Let $\epsilon > 0$ be any positive real number. To prove $(1)$, we decompose the set $\{ 0 < \left| y - x \right| < \epsilon \}$ in a dyadic manner as follow:</p>
<p>$$ \{ 0 < \left| y - x \right| < \epsilon \} = \bigcup_{n=1}^{\infty} \big\{ 2^{-n} \epsilon \leq \left|y - x\right| < 2^{-(n-1)}\epsilon \big\}. $$</p>
<p>Then whenever $n \geq 1$ and $2^{-n} \epsilon \leq \left|y - x\right| < 2^{-(n-1)}\epsilon$, we have</p>
<p>$$ \left|f(y) - f(x)\right| < 2^{-2(n-1)} \epsilon^2 \leq 4 \left| y - x \right|^2. $$</p>
<p>Thus whenever $0 < \left| y - x \right| < \epsilon$, we have $2^{-n} \epsilon \leq \left|y - x\right| < 2^{-(n-1)}\epsilon$ for some $n \geq 1$ and hence</p>
<p>$$ \left|f(y) - f(x)\right| \leq 4 \left| y - x \right|^2. $$</p>
<p>Then it follows that</p>
<p>$$ \left|\frac{f(y) - f(x)}{y - x}\right| \leq 4 \left| y - x \right|, $$</p>
<p>proving $(1)$ with $C = 4$.</p>
<p>Now the rest follows in various ways as many people pointed out. For example, taking the limit as $y \to x$, we find that $f$ is differentiable at any point $x$ with vanishing derivative. Therefore $f$ is constant.</p>
|
396,363 | <p>I have two random variables $X,Y$ which are independent and uniformly distributed on $(\frac{1}{2},1]$. Then I consider two more random variables, $D=|X-Y|$ and $Z=\log\frac{X}{Y}$. I would like to calculate both, the disitrbution functions $F_D(t), F_Z(t)$ and the the density functions $f_D(t),f_Z(t)$</p>
<p>To do that I think the first thing we need to do is to evaluate the density of the common distribution of $X$ and $Y$, but I do not know how to do that.</p>
<p>The only thing which is clear to me is the density and distribution function of $X$ and $Y$ because we know that they are uniform.</p>
<p><strong>EDIT</strong>: Please read my own answer to this question. I need someone who can show me my claculation mistakes.</p>
| wolfies | 74,360 | <p>My goodness ... what a lot of work. The nice thing about using a computer algebra system to solve this is: a) it's basically a one-liner, and b) if it is for a uni course, you will still need to know yourself how to get there, but at least you will know if your workings are correct. Here I am using the mathStatica add-on to Mathematica to do the grunt work.</p>
<p>By independence, the joint pdf of <span class="math-container">$(X,Y)$</span> is say <span class="math-container">$f(x,y)$</span>:</p>
<p><a href="https://i.stack.imgur.com/C9eRZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C9eRZ.png" alt="" /></a><br />
<sub>(source: <a href="http://www.tri.org.au/se/input2.png" rel="nofollow noreferrer">tri.org.au</a>)</sub></p>
<p>Let <span class="math-container">$Z$</span> = Abs<span class="math-container">$[X-Y]$</span>. The cdf of <span class="math-container">$Z$</span> is <span class="math-container">$P(Z<z)$</span> = <span class="math-container">$P($</span>Abs<span class="math-container">$[X-Y] < z)$</span>:</p>
<p><a href="https://i.stack.imgur.com/eGrq0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eGrq0.png" alt="" /></a><br />
<sub>(source: <a href="http://www.tri.org.au/se/output3.png" rel="nofollow noreferrer">tri.org.au</a>)</sub></p>
<p>Let <span class="math-container">$Z = Log[X/Y]$</span>. The cdf of <span class="math-container">$Z$</span> is <span class="math-container">$P(Z<z) = P(Log[X/Y] < z)$</span>:</p>
<p><a href="https://i.stack.imgur.com/uWnyH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uWnyH.png" alt="" /></a><br />
<sub>(source: <a href="http://www.tri.org.au/se/output4.png" rel="nofollow noreferrer">tri.org.au</a>)</sub></p>
<p>All done. Differentiate to get the pdf ...</p>
|
396,363 | <p>I have two random variables $X,Y$ which are independent and uniformly distributed on $(\frac{1}{2},1]$. Then I consider two more random variables, $D=|X-Y|$ and $Z=\log\frac{X}{Y}$. I would like to calculate both, the disitrbution functions $F_D(t), F_Z(t)$ and the the density functions $f_D(t),f_Z(t)$</p>
<p>To do that I think the first thing we need to do is to evaluate the density of the common distribution of $X$ and $Y$, but I do not know how to do that.</p>
<p>The only thing which is clear to me is the density and distribution function of $X$ and $Y$ because we know that they are uniform.</p>
<p><strong>EDIT</strong>: Please read my own answer to this question. I need someone who can show me my claculation mistakes.</p>
| Montaigne | 45,170 | <p>I thought it makes more sense to write an answer myself, but I still need someones help to complete the calculations. Here is my summary:</p>
<p>For $F_D(t)=\int_{\frac{1}{2}}^1\int_{-t+y}^{t+y}f_X(x) dx f_Y(y) dy=\int_{\frac{1}{2}}^1\int_{-t+y}^{t+y} 2*2 dx dy=4t$ for $x\in(1/2,1]$</p>
<p>For $f_D(t)=P(D=t)=\int_{\frac{1}{2}}^1 \int_{\frac{1}{2}}^1 f_X f_Ydx dy=4*1/2*1/2=1$</p>
<p>For $F_Z(t)=\int_\mathbb R\int_0^{y e^t}f_X dx f_Y dy=\int_{\frac{1}{2}}^1\int_0^{y e^t} 2 *2 dx dy=e^t $</p>
<p>For $f_Z(t)=P(Z=t)=\int\int_{\{(x,y): x/y=e^t\}} f_X*f_Y dx dy=?$</p>
<p><strong>Question</strong>: May you help me not doing some calculation mistakes, because I think the bounds from the integrals are false. I understood the idea of the example, but still have difficulties getting the correct numbers, especially in the second case (where does $\log(2)$ appears in wolfies answer?)</p>
|
2,709,532 | <p>(True/False) If $S$ is a non-empty subset of $\mathbb{N}$, then there exists an element $m \in S$ such that $m\ge k$ for all $k \in S$.</p>
<p>So my reasoning is that the above statement is false. Since $S$ is a non-empty subset of $\mathbb{N}$, it may also be the case that $S=\mathbb{N}$, and so there may not be an $m \in S$ such that $m \ge k$ for all $k \in S$. I wanted to know if I'm on the right track, and if not, if someone could provide a hint. </p>
| Maggie Myers | 545,995 | <p>You are correct. You found a counterexample. Every nonempty subset S has an m such that m≤k for all k in S (a minimum but not necessarily a maximum.)</p>
|
856,556 | <p>Let $f_n(x)=\frac{1}{n}\chi_{[0,n]}(x)$, $x\in\mathbb{R}$, $n\in\mathbb{N}$ and $\chi$ is the characteristic/indicator function. Now it is clear that $f_n\rightarrow 0$, but in the text I am using it says we can't apply the Dominated Convergence theorem as there is no function to dominate $f_n$</p>
<p>I have trouble seeing that. To me the $f_n$'s are dominated by the constant function $1$. So I think can apply the Dominated Convergence theorem in some cases which depends on the choice measure where the constant functions are integrable. So if we are using the Lebesgue measure (on $\mathbb{R}$) then we can't use the Dominated Convergence theorem (and the text is right), but if are using a finite measure then will it be valid to use the Dominated Convergence theorem and we will then have $\lim\limits_n\int f_n\rightarrow0$. Also if our space was just a bounded interval and we have the Lebesgue measure (since it is now a finite measure on the internal), then will it be fine to use the Dominated Convergence theorem on the $f_n$'s ? Is this right or wrong? </p>
| StrangerLoop | 161,646 | <p>Yes, if we are working only on the space $[0,1]$, for example, then $f_n$ restricted to this interval is dominated by $1$, which is integrable on $[0,1]$. So the DCT applies, and tells us that the limit of the integral is the integral of the limit, which is $0$. This is easily verified, since $f_n$ integrates to $1/n$ in this case, and $1/n \to 0$.</p>
<p>But as you said, if we are working on an unbounded interval and using Lebesgue measure, then $1$ is NOT integrable, and cannot be used as a dominating function.</p>
|
973,101 | <p>I am trying to find a way to generate random points uniformly distributed on the surface of an ellipsoid.</p>
<p>If it was a sphere there is a neat way of doing it: Generate three $N(0,1)$ variables $\{x_1,x_2,x_3\}$, calculate the distance from the origin</p>
<p>$$d=\sqrt{x_1^2+x_2^2+x_3^2}$$</p>
<p>and calculate the point </p>
<p>$$\mathbf{y}=(x_1,x_2,x_3)/d.$$</p>
<p>It can then be shown that the points $\mathbf{y}$ lie on the surface of the sphere and are uniformly distributed on the sphere surface, and the argument that proves it is just one word, "isotropy". No prefered direction.</p>
<p>Suppose now we have an ellipsoid</p>
<p>$$\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}+\frac{x_3^2}{c^2}=1$$</p>
<p>How about generating three $N(0,1)$ variables as above, calculate</p>
<p>$$d=\sqrt{\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}+\frac{x_3^2}{c^2}}$$</p>
<p>and then using $\mathbf{y}=(x_1,x_2,x_3)/d$ as above. That way we get points guaranteed on the surface of the ellipsoid but will they be uniformly distributed? How can we check that?</p>
<p>Any help greatly appreciated, thanks.</p>
<p>PS I am looking for a solution without accepting/rejecting points, which is kind of trivial.</p>
<p>EDIT:</p>
<p>Switching to polar coordinates, the surface element is $dS=F(\theta,\phi)\ d\theta\ d\phi$ where $F$ is expressed as
$$\frac{1}{4} \sqrt{r^2 \left(16 \sin ^2(\theta ) \left(a^2 \sin ^2(\phi )+b^2 \cos
^2(\phi )+c^2\right)+16 \cos ^2(\theta ) \left(a^2 \cos ^2(\phi )+b^2 \sin
^2(\phi )\right)-r^2 \left(a^2-b^2\right)^2 \sin ^2(2 \theta ) \sin ^2(2 \phi
)\right)}$$</p>
| Max | 654,677 | <p>I have been working on this problem this week and I have found the following approach effective.</p>
<p>The ellipsoid is parameterized by</p>
<p><span class="math-container">$$\vec f(t, u) = \begin{bmatrix}a \cos u \cos t \\ b \cos u \sin t \\ c \sin u\end{bmatrix}, \quad
\begin{align}t &\in [0, 2 \pi] \\ u &\in [0, \pi]\end{align}$$</span></p>
<p>Naively, we might pick <span class="math-container">$T \sim U(0, 2 \pi)$</span> and <span class="math-container">$U \sim U(0, \pi)$</span> and choose <span class="math-container">$\vec f(T, U)$</span> as our random point, but this is equivalent to stretching the sphere, which creates distortion as others have pointed out.</p>
<p>Instead, define the regions</p>
<p><span class="math-container">$$\begin{align}
D_{t'} :\quad & t \in [0, t'] , && u \in [0, \pi] \\
D_{u'} :\quad & t \in [0, 2 \pi] , && u \in [0, u']
\end{align}$$</span></p>
<p>And the functions</p>
<p><span class="math-container">$$\begin{align}
S_t(t') \equiv & \iint_{D_{t'}} dS \\
S_u(u') \equiv & \iint_{D_{u'}} dS
\end{align}$$</span></p>
<p>That is, <span class="math-container">$S_t$</span> yields the cumulative surface area on <span class="math-container">$t\in [0,t']$</span> and for all <span class="math-container">$u$</span> , while <span class="math-container">$S_u$</span> does vice-versa.</p>
<p>Next, pick two numbers <span class="math-container">$X_1, X_2 \sim U(0,A)$</span> where <span class="math-container">$A$</span> is the total surface area <span class="math-container">$$A = S_t(2\pi) = S_u(\pi)$$</span></p>
<p>Since <span class="math-container">$S_t$</span> and <span class="math-container">$S_u$</span> are monotonically increasing over the parameter space, they can be evaluated numerically and inverted through interpolation.</p>
<p>Our random parameters are then</p>
<p><span class="math-container">$$\begin{align}
T = S_t^{-1}(X_1) \\
U = S_u^{-1}(X_2) \\
\end{align}$$</span></p>
<p>And a random point on the surface is <span class="math-container">$$\vec f(T, U)$$</span></p>
<p>The advantage of this method is that because we don't have to discard any points, we don't need to compute a multiplicative factor particular to this ellipse (as another answerer does). In fact, <strong>this method picks points uniformly distributed on <em>any</em> parametric surface whose parameter region is square.</strong></p>
<p>It is also computationally efficient, because we can use the same data (namely, a grid of parameters and the values of the function at those parameters) to determine the overall surface area and, by interpolation, the cumulative surface area functions and their inverses.</p>
<p>I've developed a <a href="https://github.com/maxkapur/param_tools" rel="nofollow noreferrer">small Python module</a> that implements this approach as <code>r_surface()</code>. Example 3 in the module uses an ellipsoid:</p>
<p><a href="https://i.stack.imgur.com/8qzHJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8qzHJ.png" alt="enter image description here" /></a></p>
<p>This implementation is indebted to the responses to these SE questions:</p>
<ul>
<li><a href="https://math.stackexchange.com/questions/3710402/generate-random-points-on-perimeter-of-ellipse/3718774#3718774">Generate random points on perimeter of ellipse</a> (Math SE)</li>
<li><a href="https://codereview.stackexchange.com/questions/243590/generate-random-points-on-perimeter-of-ellipse/243697?noredirect=1#comment478326_243697">Generate random points on perimeter of ellipse</a> (Code Review SE)</li>
</ul>
|
3,930,373 | <p>Hi so im working on a question about finding the <span class="math-container">$rank(A)$</span> and the <span class="math-container">$dim(Ker(A))$</span> of a 7x5 Matrix. Without being given an actual matrix to work from.</p>
<p>I have been told that that the homogeneous equation <span class="math-container">$A\vec x=\vec0$</span> has general solution <span class="math-container">$\vec x=\lambda \vec v$</span> for some non zero <span class="math-container">$\vec v$</span> in <span class="math-container">$R^{5}$</span>.</p>
<p>So my thinking so far is that I know for an <span class="math-container">$m*n$</span> matrix we know that:</p>
<p><span class="math-container">$rk(A)+dimker(A)=n$</span>
which must mean that <span class="math-container">$rk(A)+dimker(A)=5$</span></p>
<p>but this is where I get stuck and dont know how to proceed.</p>
<p>Any help is greately appreciated.</p>
<p><a href="https://i.stack.imgur.com/2kLWC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2kLWC.png" alt="enter image description here" /></a>
.</p>
<p>This is the exact question for the person who asked.</p>
| Adam Rubinson | 29,156 | <p>The function under consideration is <span class="math-container">$f:x \to \sqrt{-x^2} $</span> with domain {<span class="math-container">$0$</span>}. In other words, your function is, <span class="math-container">$f:$</span> {<span class="math-container">$0$</span>} <span class="math-container">$\to $</span>{<span class="math-container">$0$</span>} with <span class="math-container">$f(0)=0.$</span> Your question is, "what is <span class="math-container">$\lim_{x \to 0}f(x)$</span>?"</p>
<p>Now, let's talk about the domain. <span class="math-container">$0$</span> is not a limit point of {<span class="math-container">$0$</span>} because every neighbourhood of <span class="math-container">$0$</span> in {<span class="math-container">$0$</span>} is {<span class="math-container">$0$</span>}, so there are <em>no</em> neighbourhoods of <span class="math-container">$0$</span> that contains a point <span class="math-container">$q \neq 0$</span> such that <span class="math-container">$q \in $</span>{<span class="math-container">$0$</span>}. Note that this is stronger than we needed: we only needed to show that there exists a neighbourhood of <span class="math-container">$0$</span> that does not contain a point <span class="math-container">$q \neq 0$</span> such that <span class="math-container">$q \in $</span>{<span class="math-container">$0$</span>}.</p>
<p>Now, in order for us to consider the <em>limit</em> <span class="math-container">$\ \lim_{x \to 0}f(x)$</span> (e.g. going by Rudin's definition), <span class="math-container">$\ 0$</span> must be a <em>limit point</em> of the domain which we have just seen it is not.</p>
<p>Therefore using the conventional definition of limit of a function at a point in the domain does not tell us anything about <span class="math-container">$\ \lim_{x \to 0}f(x),\ $</span> and in fact, I am not sure what the definition or meaning of "<span class="math-container">$\lim_{x \to 0}f(x)\ $</span>" is for the above function, so let's stop talking about things that haven't been defined.</p>
|
2,972,611 | <p><strong>Q</strong>: Prove the given inequalities for positive a,b,c:<br><span class="math-container">$(i) \left[\frac{bc+ca+ab}{a+b+c}\right]^{a+b+c}>\sqrt{(bc)^a.(ca)^b.(ab)^c}$</span><br><span class="math-container">$(ii) \left(\frac{a+b+c}{3} \right)^{a+b+c}<a^ab^bc^c<\left(\frac{a^2+b^2+c^2}{a+b+c}\right)^{a+b+c} $</span><br>I know that G.M<span class="math-container">$\le$</span>A.M and somehow i guess it must be used.But i really struggling to prove this kind of inequality.Any hints or solution will be appreciated.And i do apologize if this question is very basic.<br>Thanks in advance.</p>
| Batominovski | 72,152 | <p>First of all, you should have <span class="math-container">$\geq $</span> or <span class="math-container">$\leq$</span> instead of <span class="math-container">$>$</span> and <span class="math-container">$<$</span> in your inequalities. In each of them, the equality occurs if and only if <span class="math-container">$a=b=c$</span>.</p>
<p>(i) Use the Weighted AM-GM Inequality to show that
<span class="math-container">$$\frac{bc+ca+ab}{a+b+c}=\frac{b}{a+b+c}c+\frac{c}{a+b+c}a+\frac{a}{a+b+c}b\geq c^{\frac{b}{a+b+c}}a^{\frac{c}{a+b+c}}b^{\frac{a}{a+b+c}}\,.$$</span>
Similarly,
<span class="math-container">$$\frac{bc+ca+ab}{a+b+c}=\frac{c}{a+b+c}b+\frac{a}{a+b+c}c+\frac{b}{a+b+c}a\geq b^{\frac{c}{a+b+c}}c^{\frac{a}{a+b+c}}a^{\frac{b}{a+b+c}}\,.$$</span>
Multiplying these two inequalities to get
<span class="math-container">$$\left(\frac{bc+ca+ab}{a+b+c}\right)^2\geq \left((bc)^a(ca)^b(ab)^c\right)^{\frac{1}{a+b+c}}\,,$$</span>
which is equivalent to the required inequality.</p>
<p>(ii) For the inequality on the right, note that
<span class="math-container">$$\frac{a^2+b^2+c^2}{a+b+c}=\frac{a}{a+b+c}a+\frac{b}{a+b+c}b+\frac{c}{a+b+c}c\geq a^{\frac{a}{a+b+c}}b^{\frac{b}{a+b+c}}c^{\frac{c}{a+b+c}}=\left(a^ab^bc^c\right)^{\frac{1}{a+b+c}}$$</span>
by the Weighted AM-GM Inequality. Thus,
<span class="math-container">$$\left(\frac{a^2+b^2+c^2}{a+b+c}\right)^{a+b+c}\geq a^ab^bc^c$$</span>
as desired.</p>
<p>For the inequality on the left, observe that
<span class="math-container">$$\frac{3}{a+b+c}=\frac{a}{a+b+c}\left(\frac{1}{a}\right)+\frac{b}{a+b+c}\left(\frac1b\right)+\frac{c}{a+b+c}\left(\frac1c\right)\,.$$</span>
Thus, by the Weighted AM-GM Inequality,
<span class="math-container">$$\frac{3}{a+b+c} \geq \left(\frac1a\right)^{\frac{a}{a+b+c}}\left(\frac1b\right)^{\frac{b}{a+b+c}}\left(\frac{1}{c}\right)^{\frac{c}{a+b+c}}=\frac{1}{\left(a^ab^bc^c\right)^{\frac{1}{a+b+c}}}\,.$$</span>
This is equivalent to the inequality to be proven.</p>
|
195,065 | <p>Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a nondecreasing function.</p>
<p>Let $a<f(a)$ and $f(b)<b$. Prove that there is a $a<c<b$ such that $f(c)=c$. </p>
<p>My attempt at a proof is as follows. Let $c:=\sup\{x:a\leq x\leq b\text{, }x\leq f(x)\}$.</p>
<p>This is where I'm stuck. Since I can't use more powerful theorem such as the IVT I find this problem far more complex.</p>
| Brian M. Scott | 12,042 | <p>HINT: Let $L=\{x\in[a,b]:x<f(x)\}$. Is $L$ non-empty? Is $L$ bounded above?</p>
<p><strong>Added:</strong> In other words, your idea is reasonable, though I used a slightly different set from yours. $L$ <strong>is</strong> non-empty and bounded (why?), so we can let $c=\sup L$. What do you know about $f(x)$ for $x\in[c,b]$? What happens if $f(c)\ne c$? Remember, $f$ is non-decreasing.</p>
|
4,361,063 | <blockquote>
<p>Suppose that <span class="math-container">$(X, \tau_1)$</span> is compact and <span class="math-container">$\tau_1 \subset \tau_2$</span>. Is the space <span class="math-container">$(X, \tau_2)$</span> compact? Does the converse hold i.e if <span class="math-container">$(X, \tau_1)$</span> is compact and <span class="math-container">$\tau_2 \subset \tau_1$</span>?</p>
</blockquote>
<p>The first one shouldn't hold since if <span class="math-container">$X= [0,1]$</span> and <span class="math-container">$\tau_1$</span> is the usual topology of <span class="math-container">$\Bbb R$</span>, then I think that if <span class="math-container">$\tau_2$</span> is the lower limit topology we have <span class="math-container">$\tau_1 \subset \tau_2$</span> and <span class="math-container">$X$</span> wouldn' t be compact?</p>
<p>The second one also doesn't seem true. If <span class="math-container">$(X, \tau_1)$</span> is compact and <span class="math-container">$\tau_2 \subset \tau_1$</span>, then every open cover of <span class="math-container">$X$</span> has a finite subcover, but I don't think why this would hold for the coarser topology <span class="math-container">$\tau_2$</span>? I think it could be that <span class="math-container">$\tau_2$</span> doesn't have "enough" elements to satisfy this.</p>
| Sourav Ghosh | 977,780 | <p>Suppose, <span class="math-container">$card(X) \ge \aleph_{0}$</span></p>
<p>Consider two topology on <span class="math-container">$X$</span></p>
<p><span class="math-container">$\tau_{indiscrete}=\{\emptyset, X\}$</span> and</p>
<p><span class="math-container">$\tau_{discrete}={\scr{P}}(X) $</span></p>
<p>Then, <span class="math-container">$\tau_{indiscrete}\subset\tau_{discrete}$</span></p>
<blockquote>
<p>Claim :<span class="math-container">$(X, \tau_{indiscrete}) $</span> is compact but <span class="math-container">$(X,\tau_{discrete}) $</span>
is not compact.</p>
</blockquote>
<p><span class="math-container">$X\subset (X,\tau_{indiscrete}) $</span></p>
<p>Suppose, <span class="math-container">$X\subset \cup_{\alpha} U_{\alpha}$</span> , then <span class="math-container">$U_{\alpha} =X$</span> for at least one <span class="math-container">$\alpha$</span>.</p>
<p><span class="math-container">$X\subset (X, \tau_{indiscrete}) $</span></p>
<p>Then, <span class="math-container">$X\subset \cup_{x\in X} \{x\}$</span></p>
<p>has no finite subcover otherwise it would contradict cardinality of <span class="math-container">$X$</span>.</p>
<p>For your second question,</p>
<p>If <span class="math-container">$X$</span> is compact in a topology then it is also compact in any of it's coarser topology.</p>
<p><span class="math-container">$\tau_{1}\subset \tau_{2}$</span></p>
<p><span class="math-container">$(X, \tau_{2}) $</span> is compact.</p>
<p>Suppose, <span class="math-container">$X\subset \cup_{\alpha} U_{\alpha}$</span></p>
<p>Where, <span class="math-container">$U_{\alpha}\in \tau_{1} \subset \tau_{2}$</span></p>
<p>Then there exists a finite subcover i.e a <span class="math-container">$I$</span> finite index set such that</p>
<p><span class="math-container">$X\subset\cup_{i\in I}U_{\alpha_{i}}$</span></p>
<p>Where, <span class="math-container">$U_{\alpha_{i}}\in \tau_{2}$</span></p>
<p>Then, <span class="math-container">$\{U_{\alpha_{i}}:i\in I\}$</span> is a finite subcover of <span class="math-container">$\{U_{\alpha}\}$</span>.</p>
<p>Hence, <span class="math-container">$(X, \tau_{1}) $</span> is also compact.</p>
|
258,893 | <p>I have a set of three moderately simple simultaneous equations that I'd like to simplify and eliminate a set of variables for (this is a simple example of a more general class of problem - but I'd like to get the simple example working before starting on the bigger ones). Asking Mathematica to <code>Eliminate</code> two of the three variables I'd like to remove and then simplifying the result gets me to the answer fairly quickly: however asking Mathematica to <code>Eliminate</code> all three of the variables at once hangs. Are there any tricks I can use to help <code>Eliminate</code> with this task, or generalisations of it?</p>
<p>My example has three matrices (adjoint representation of so(3) for the interested)</p>
<pre><code>b1 = {{0, 0, 0}, {0, 0, 1}, {0, -1, 0}};
b2 = {{0, 0, -1}, {0, 0, 0}, {1, 0, 0}};
b3 = {{0, 1, 0}, {-1, 0, 0}, {0, 0, 0}};
</code></pre>
<p>and then the equations are</p>
<pre><code>{xt1, xt2, xt3} == {x1, x2, x3}.MatrixExp[t1 b1].MatrixExp[t2 b2].MatrixExp[t3 b3]
</code></pre>
<p>where I would like to eliminate the <code>t1</code>, <code>t2</code>, and <code>t3</code> variables.</p>
<p>Running this to eliminate <code>t1</code> and <code>t2</code> gives</p>
<pre><code>Timing@Simplify@Eliminate[{xt1, xt2, xt3} == {x1, x2, x3}.MatrixExp[t1 b1].MatrixExp[t2 b2].MatrixExp[t3 b3], {t1, t2}]
</code></pre>
<blockquote>
<p>{40.375, x1^2 + x2^2 + x3^2 == xt1^2 + xt2^2 + xt3^2}</p>
</blockquote>
<p>which is the correct answer. However,</p>
<pre><code>Timing@Eliminate[{xt1, xt2, xt3} == {x1, x2, x3}.MatrixExp[t1 b1].MatrixExp[t2 b2].MatrixExp[t3 b3], {t1, t2, t3}]
</code></pre>
<p>runs forever (at least four hours on my laptop). How can I make <code>Eliminate</code>s life easier, or are there other tools I could try to solve this system of equations? (Extra information if helpful: the functional form of the reduced solution(s) to members of this class of equations will always have the same functional form in the <code>xt</code> variables and the <code>x</code> variables, although the actual functional form is unknown - I'm not sure if this can be leveraged within Mathematica to help?)</p>
| Akku14 | 34,287 | <p>The quite fastest way is to use Solve, which is not sensitive if some variables are not independent of each other, add identities for Sin, Cos and eliminate all Sin und Cos at the same time.</p>
<pre><code>Solve[{xt1 ==
Cos[t3] (x1 Cos[t2] + (x3 Cos[t1] + x2 Sin[t1]) Sin[t2]) - (x2 Cos[
t1] - x3 Sin[t1]) Sin[t3],
xt2 == Cos[
t3] (x2 Cos[t1] -
x3 Sin[t1]) + (x1 Cos[t2] + (x3 Cos[t1] + x2 Sin[t1]) Sin[
t2]) Sin[t3],
xt3 == Cos[t2] (x3 Cos[t1] + x2 Sin[t1]) - x1 Sin[t2],
Sin[t1]^2 + Cos[t1]^2 == 1, Sin[t2]^2 + Cos[t2]^2 == 1,
Sin[t3]^2 + Cos[t3]^2 == 1}, {xt1, xt2, xt3}, {Sin[t1], Sin[t2],
Cos[t1], Cos[t2]}]
(* Solve::svars: Equations may not give solutions for all "solve" variables. >>
{{xt3 -> -Sqrt[x1^2 + x2^2 + x3^2 - xt1^2 - xt2^2]},
{xt3 -> Sqrt[x1^2 + x2^2 + x3^2 - xt1^2 - xt2^2]}} *)
</code></pre>
<p>And the same result for all ti with</p>
<pre><code>Solve[{xt1 ==
Cos[t3] (x1 Cos[t2] + (x3 Cos[t1] + x2 Sin[t1]) Sin[t2]) - (x2 Cos[
t1] - x3 Sin[t1]) Sin[t3],
xt2 == Cos[
t3] (x2 Cos[t1] -
x3 Sin[t1]) + (x1 Cos[t2] + (x3 Cos[t1] + x2 Sin[t1]) Sin[
t2]) Sin[t3],
xt3 == Cos[t2] (x3 Cos[t1] + x2 Sin[t1]) - x1 Sin[t2],
Sin[t1]^2 + Cos[t1]^2 == 1, Sin[t2]^2 + Cos[t2]^2 == 1,
Sin[t3]^2 + Cos[t3]^2 == 1}, {xt1, xt2, xt3}, {Sin[t1], Sin[t2],
Sin[t3], Cos[t1], Cos[t2], Cos[t3]}]
</code></pre>
|
258,893 | <p>I have a set of three moderately simple simultaneous equations that I'd like to simplify and eliminate a set of variables for (this is a simple example of a more general class of problem - but I'd like to get the simple example working before starting on the bigger ones). Asking Mathematica to <code>Eliminate</code> two of the three variables I'd like to remove and then simplifying the result gets me to the answer fairly quickly: however asking Mathematica to <code>Eliminate</code> all three of the variables at once hangs. Are there any tricks I can use to help <code>Eliminate</code> with this task, or generalisations of it?</p>
<p>My example has three matrices (adjoint representation of so(3) for the interested)</p>
<pre><code>b1 = {{0, 0, 0}, {0, 0, 1}, {0, -1, 0}};
b2 = {{0, 0, -1}, {0, 0, 0}, {1, 0, 0}};
b3 = {{0, 1, 0}, {-1, 0, 0}, {0, 0, 0}};
</code></pre>
<p>and then the equations are</p>
<pre><code>{xt1, xt2, xt3} == {x1, x2, x3}.MatrixExp[t1 b1].MatrixExp[t2 b2].MatrixExp[t3 b3]
</code></pre>
<p>where I would like to eliminate the <code>t1</code>, <code>t2</code>, and <code>t3</code> variables.</p>
<p>Running this to eliminate <code>t1</code> and <code>t2</code> gives</p>
<pre><code>Timing@Simplify@Eliminate[{xt1, xt2, xt3} == {x1, x2, x3}.MatrixExp[t1 b1].MatrixExp[t2 b2].MatrixExp[t3 b3], {t1, t2}]
</code></pre>
<blockquote>
<p>{40.375, x1^2 + x2^2 + x3^2 == xt1^2 + xt2^2 + xt3^2}</p>
</blockquote>
<p>which is the correct answer. However,</p>
<pre><code>Timing@Eliminate[{xt1, xt2, xt3} == {x1, x2, x3}.MatrixExp[t1 b1].MatrixExp[t2 b2].MatrixExp[t3 b3], {t1, t2, t3}]
</code></pre>
<p>runs forever (at least four hours on my laptop). How can I make <code>Eliminate</code>s life easier, or are there other tools I could try to solve this system of equations? (Extra information if helpful: the functional form of the reduced solution(s) to members of this class of equations will always have the same functional form in the <code>xt</code> variables and the <code>x</code> variables, although the actual functional form is unknown - I'm not sure if this can be leveraged within Mathematica to help?)</p>
| J. M.'s persistent exhaustion | 50 | <p>If I use the <a href="https://en.wikipedia.org/wiki/Weierstrass_substitution" rel="nofollow noreferrer">Weierstrass substitution</a> along with <code>TrigExpand[]</code>, <code>GroebnerBasis[]</code> accomplishes the elimination pretty quickly:</p>
<pre><code>GroebnerBasis[TrigExpand[Thread[{xt1, xt2, xt3} == {x1, x2, x3} . MatrixExp[t1 b1] .
MatrixExp[t2 b2] . MatrixExp[t3 b3]] /.
Thread[{t1, t2, t3} -> {2 ArcTan[u1], 2 ArcTan[u2], 2 ArcTan[u3]}]],
{xt1, xt2, xt3, x1, x2, x3}, {u1, u2, u3}]
{-x1^2 - x2^2 - x3^2 + xt1^2 + xt2^2 + xt3^2}
</code></pre>
<p>which is equivalent to the answer Michael gave.</p>
|
3,276,958 | <p>Given two points <span class="math-container">$P_1(x_1,y_1)$</span> and <span class="math-container">$P_2(x_2,y_2)$</span> with <span class="math-container">$y_1>0$</span> and <span class="math-container">$y_2>0$</span> I need to find the parameters <span class="math-container">$a$</span> and <span class="math-container">$b$</span> of an exponential function having the form <span class="math-container">$y=a*b^x$</span>.</p>
<p>How I can solve this problem in a "generic" way getting the formulas to find <span class="math-container">$a$</span> and <span class="math-container">$b$</span> from known <span class="math-container">$P_1$</span> and <span class="math-container">$P_2$</span>? I tried to find the solution by myself for two hours but I keep getting the same, wrong formulas.</p>
| Martin R | 42,969 | <p>Taking the logarithms of the desired identities <span class="math-container">$y_j = a b^{x_j}$</span> gives a <em>linear equation system</em>
<span class="math-container">$$
\begin{align}
1 \cdot \log a + x_1 \cdot \log b = \log y_1 \\
1 \cdot\log a + x_2 \cdot\log b = \log y_2
\end{align}
$$</span>
which can be solved for <span class="math-container">$(\log a,\log b)$</span> with the usual methods. The determinant of the coefficient matrix is <span class="math-container">$x_2 - x_1$</span>, so that a unique solution exists if <span class="math-container">$x_1 \ne x_2$</span>:
<span class="math-container">$$
\log a = \frac{\begin{vmatrix} \log y_1 & x_1 \\ \log y_2 & x_2\end{vmatrix}}{x_2 - x_1}
= \frac{x_2 \log y_1 - x_1 \log y_2}{x_2 - x_1}
$$</span>
and
<span class="math-container">$$
\log b = \frac{\begin{vmatrix} 1 & \log y_1 \\ 1 & \log y_2 \end{vmatrix}}{x_2 - x_1}
= \frac{\log y_2 - \log y_1}{x_2 - x_1}
$$</span></p>
|
1,401,116 | <p>An aeroplane is observed by two persons travelling at $60 \frac{km}{hr} $ in two vehicles moving in opposite directions on a straight road. To an observer in one vehicle the plane appears to cross the road track at right angles while to the observer in the other vehicle the angle appears to be $45°$. At what angle does the plane actually cross the road track and what is its speed relative to the ground.<br>
My attempt :<br>
Now for vehicle 1 $$ \vec{V_{p1}} =\vec{V_{p}}-\vec{V_{1}} $$ where $ \vec{V_{p}}$ is velocity of plane.<br>
Similarly for vehicle 2
$$ \vec{V_{p2}} =\vec{V_{p}}-\vec{V_{2}} $$
How should I proceed further? I am stuck at this step.</p>
| triple_sec | 87,778 | <p>Found it. It is the “discrete version” of Liouville’s theorem. I was wrong about Alaoglu: the functional-analytic approach uses the Krein–Milman theorem (although proofs based on martingale theory are available, too). See more <a href="https://math.stackexchange.com/questions/849284/looking-for-different-proofs-of-discrete-liouvilles-theorem">here</a>.</p>
|
1,124,417 | <p>I was working my way through some Propositional Logic and had the following doubt :</p>
<blockquote>
<p><strong>Why is this true :</strong></p>
<p>((p <span class="math-container">$\Rightarrow$</span> r) <span class="math-container">$\land$</span> (q <span class="math-container">$\Rightarrow$</span> r)) <span class="math-container">$\equiv$</span> ((p <span class="math-container">$\lor$</span> q) <span class="math-container">$\Rightarrow$</span> r)</p>
</blockquote>
<p><em><strong>Please provide an intuitive explanation and not one that uses a truth table or logic identities to simplify the expression . I have already done both of them :)</strong></em></p>
| mercio | 17,445 | <p>$2$ and $3$ are invertible, so we can look at
$6F(x,y,z) = 3(2x+y-az)^2 + (3y-az)^2 + 2(a^2-3b^2)z^2$.<br>
After an invertible change of basis, the equation $F(x,y,z) = 0$ has the same number of solutions as $3X^2+Y^2 + 2(a^2-3b^2)Z^2 = 0$.</p>
<p>According to wether $3$ and $2(a^2-3b^2)$ (which are nonzero) are squares or not, we can again simplify the equation (up to linear change of variable, and multiplication by a nonsquare if necessary) to get $X^2 + Y^2 + cZ^2 = 0$ for some nonzero $c$.</p>
<p>Now, as Asal Beag Dubh points out, these curves can have either $1$ or $q^2$ points. (and actually it can only depend on wether $c$ is a square or not)</p>
<p>But let's add up the total number of solutions for all the curves with nonzero $c$.<br>
If $x^2+y^2 \neq 0$ then for every $z \neq 0$ there is a unique $c$ for which $(x,y,z)$ is on the curve ; and $(x,y,0)$ is on no curve.<br>
If $x^2+y^2 = 0$ then $(x,y,0)$ is on every curve and $(x,y,z)$ is on no curve for any $z \neq 0$.</p>
<p>If we let $r$ be the number of solutions to $x^2+y^2 = 0$ we get a total number of $(q^2-r)(q-1) + r(q-1) = q^2(q-1)$ points, so each of the $q-1$ curves $X^2+Y^2+cZ^2 = 0$ has exactly $q^2$ points.</p>
|
2,369,133 | <p>I just found out that
if you want to get 1 with the fraction: $$\frac{5}{2}$$
Then you multiply it with: $$ \frac{2}{5} $$
Does anyone have a good way to think about this? </p>
| Bill Dubuque | 242 | <p>$x = a/b\iff bx = a.\,$ If there is a number $x'$ with $\,\color{#c00}{xx' = 1}$ then scaling this by $x'$ yields the</p>
<p>equation $\ b= b\color{#c00}{xx'} = ax',\,$ i.e $\,x' = b/a$</p>
<hr>
<p>More generally if $\,r\,$ is a root of a polynomial $f(x)$ then its reciprocal $\,1/r\,$ is a root of the <a href="http://en.wikipedia.org/wiki/Reciprocal_polynomial" rel="nofollow noreferrer">reciprocal polynomial</a> $\,\bar f(x) = x^n f(1/x),\ n = \deg f,\, $ obtained by <em>reversing</em> the coefficients of $f(x)$. </p>
<p>In the linear case we have $\ r = a/b\ $ is a root of $\,f(x)\, =\, \color{#c00}bx \color{#0a0}{- a}\,$ </p>
<p>so $\,1/r\,$ is a root of $\,\bar f(x) = xf(1/x) = x(b/x-a) = \color{#0a0}{-a}x + \color{#c00}b,\,$ i.e. $\,1/r = b/a$</p>
<p>Note how the coeff list has been reversed $\,f = (\color{#c00}b,\color{#0a0}{-a})\, \mapsto\, \bar f = (\color{#0a0}{-a},\color{#c00}b).\,$ This is one way to explain why inverting a fraction swaps (reverses) the numerator and denominator.</p>
|
2,922,494 | <p>I came across the following problem today.</p>
<blockquote>
<p>Flip four coins. For every head, you get <span class="math-container">$\$1$</span>. You may reflip one coin after the four flips. Calculate the expected returns.</p>
</blockquote>
<p>I know that the expected value without the extra flip is <span class="math-container">$\$2$</span>. However, I am unsure of how to condition on the extra flips. I am tempted to claim that having the reflip simply adds <span class="math-container">$\$\frac{1}{2}$</span> to each case with tails since the only thing which affects the reflip is whether there are tails or not, but my gut tells me this is wrong. I am also told the correct returns is <span class="math-container">$\$\frac{79}{32}$</span> and I have no idea where this comes from.</p>
| William Wong | 589,804 | <p>Expectation of first 4 flips is $2.</p>
<p>Expectation of the fifth flip, condition on there is a tail in the first 4 flips, is 0.5.</p>
<p>Expectation of the fifth flip, condition on there is no tail, is 0.</p>
<p>Probability there is a tail in the first 4 flips is 15/16.</p>
<p>Total expectation = 2 + 0.5*(15/16) + 0*(1/16) = 79/32</p>
|
1,203,116 | <p>I'm trying to prove that in such triangle, AB>AC, but I can't suceed. I tried to use some properties of isósceles triangle since the sides are equal, but couldn't find something nice. Also, whats the criteria needed to prove that a side in a triangle is greater than the other? Thank you so much!</p>
| Joffan | 206,402 | <p>Since $|AC|=|AD|, \triangle ACD$ is isosceles and $\angle ADB$ is obtuse.</p>
<p><img src="https://i.stack.imgur.com/kdtD9.png" alt="enter image description here"></p>
<p>Therefore $|AB| > |BD| = |AC|$</p>
|
4,541,392 | <p>I have an integral that looks like the following:
<span class="math-container">$$
\int_a^b \frac{1}{\left(1 + cx^2\right)^{3/2}} \mathrm{d}x
$$</span>
I have seen a method of solving it being to substitute <span class="math-container">$x = \frac{\mathrm{tan}(u)}{\sqrt{c}}$</span>; however, this seems somewhat sloppy to me. Is there perhaps a better way of tackling this integral?</p>
| Lai | 732,917 | <p><span class="math-container">$$
\begin{aligned}
&\text { Let } y=\frac{1}{x^2} \text {, then } d x=-\frac{1}{2y^{\frac{3}{2}}} d y\\
I&=\int_{\frac{1}{a^2}}^{\frac{1}{b^2}} \frac{1}{\left(1+\frac{c}{y}\right)^{\frac{3}{2}}}\left(-\frac{d y}{2y \frac{3}{2}}\right)\\
&=\frac{1}{2} \int_{\frac{1}{b^2}}^{\frac{1}{a^2}} \frac{d y}{(y+c)^{\frac{3}{2}}}\\
&=\left[-\frac{1}{\sqrt{y+c}}\right]_{\frac{1}{b^2}}^{\frac{1}{a^2}}\\
&=\frac{b}{\sqrt{1+b^2 c}}-\frac{a}{\sqrt{1+a^2 c}}
\end{aligned}
$$</span></p>
|
163,654 | <p>Do You know any successful applications of the geometric group theory in the number theory? GTG is my main field of interest and I would love to use it to prove new facts in the number theory.</p>
| jacob | 4,181 | <p>To expand on my answer above, consider the sequence $\dots\subset Y_2\subset Y_1$
of spaces where $Y_i=X_i$ for $i\leq n$ and $Y_i=X_n$ for $i>n$.</p>
<p>Milnors proof proceeds by taking $M_i$ to be the mapping cylinder of $X_{i+1}\rightarrow X_i$ and $M_0$ the cone over $X_0$ with vertex $t$. Then one takes the union $F=\cup M_i$ with $M_i$ glued to $M_{i+1}$ along $X_{i+1}$. Then one makes a space $T=F\cup X$ by adding $X$ `at infinity'. The upshot is that $T$ can be contracted onto $t$ by collapsing everything down to $X_1$ and then using the cone $M_0$ to get down to $t$.</p>
<p>Now, once this is done, Milnor sets $F_1$ to be the disjoint union of the odd $M_i$
and $F_0$ the disjoint union of the even $M_i$. Then $F_0\cap F_1$ is just the disjoint union of all the $X_i$, and $F_0\cup F_1 = F$.</p>
<p>The proof now follows quickly by looking at the long exact sequence i homology corresponding to the triple $(T,X\cup t,t)$ together with Mayer-Vietoris for $(F_0,F_1)$.</p>
<p>Now carry out the same constructions for $Y$, denoting the relevant spaces by $M_i', t', T', F',F_0',F_1'$.</p>
<p>The point is that there are natural maps $M_i'\rightarrow M_i$ inducing maps $T'\rightarrow T$ and $F'\rightarrow F$, and the induced maps on cohomology are evidently the ones gotten by inclusion. </p>
<p>Now, $Y=\cap_m Y_m = X_n$ and so the map Milne gets $H_q(X)\rightarrow \lim H_q(X_i)$ commutes with the inclusion maps $H_q(X)\rightarrow H_q(X_n)$, which gives what you want.</p>
|
163,654 | <p>Do You know any successful applications of the geometric group theory in the number theory? GTG is my main field of interest and I would love to use it to prove new facts in the number theory.</p>
| Włodzimierz Holsztyński | 8,385 | <p>There is no reason for a homological surjection. And it's false all the time, so-to-speak. Thus let's assume that we are asking either about a homological injection or about the cohomological surjection which goes in the other direction (naturally--as a categorist would say).</p>
<p>Now the answer is YES under some reasonable assumptions. Consider an inverse system of Hausdorff compact spaces. They need not be metrizable, and they need not be <em>nice</em>. Also, the projections can be arbitrary--they need not be injective nor surjective. Just assume that the limit space $X$ is <strong>ANR</strong> (i.e. for any compact pair $(Y\ B)$, and for every continuous map $f:B\rightarrow X$ there is a neighborhood $G\subseteq Y$ of $B$, and a continuous $g:G\rightarrow X$ such that $g$ extends $f$, i.e.$f=g|B$). Then <strong>YES</strong>, when the approximating space $X_t$ has a sufficiently advanced index $t$ then we get the respective homological injection and cohomological surjection.</p>
<p>Indeed, we have a much stronger statement: for any advanced index $\ t\ $ the homotopy class of $\ p:X\rightarrow X_t\ $ is a homotopy $\ell$-map, i.e. it admits a continuous map $q_t:X_t\rightarrow X$ such that $\ q_t\circ p_t:X\rightarrow X\ $ is homotopic to identity. That's what it really is about.</p>
|
100,957 | <p>I have this picture of small particles in a polymer film. I want to count how many particles in the figure, so that I can have a rough estimation of the particle density. But the image quality is poor, I had a hard time to do it.</p>
<p><a href="https://i.stack.imgur.com/ez50R.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/ez50R.jpg" alt="particle in film - grayscale"></a>, <a href="https://i.stack.imgur.com/0OJd7.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/0OJd7.jpg" alt="particle in film - color"></a></p>
<p>I have tried several ways to do it, but failed. below is the code.
The first method I tried is:</p>
<pre><code>`SetDirectory["C:\\Users\\mayao\\documents"]
image = Import["Picture3.jpg"];
imag2 = Binarize[image, {0.0, 0.8}];
cells = SelectComponents[DeleteBorderComponents[imag2], "Count", -400];
circles = ComponentMeasurements[ImageMultiply[image,cells],"Centroid", "EquivalentDiskRadius"}][[All, 2]]; Show[image, Graphics[{Red, Thick, Circle @@ # & /@ circles}]]
</code></pre>
<p>Here is what I got:
<a href="https://i.stack.imgur.com/f7u8C.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/f7u8C.jpg" alt="enter image description here"></a></p>
<p>So it does not count all the particle. Plus, it sometimes take several particle as one.</p>
<p>I read another method from a thread here, the code is:</p>
<pre><code>obl[transit_Image] := (SelectComponents[
MorphologicalComponents[
DeleteSmallComponents@
ChanVeseBinarize[#, "TargetColor" -> Black],
Method -> "ConvexHull"], {"Count", "SemiAxes"},
Abs[Times @@ #2 Pi - #1] < #1/100 &]) &@
transit; GraphicsGrid[{#, obl@# // Colorize,
ImageMultiply[#,
Image@Unitize@
obl@#]} & /@ (Import /@ ("C:\\Users\\mayao\\documents\\" <> # \
& /@ {"Picture1.jpg", "Picture2.jpg", "Picture3.jpg",
"Picture1.jpg"}))]
</code></pre>
<p>But it does not recognize the single particles:</p>
<p><a href="https://i.stack.imgur.com/zav3Y.png" rel="noreferrer"><img src="https://i.stack.imgur.com/zav3Y.png" alt="enter image description here"></a></p>
<p>Is there any other method to do this task?
Thanks a lot for any suggestions.</p>
| Graumagier | 25,421 | <p>You could try something like this</p>
<pre><code>HighlightImage[
ImageTake[img, {100, 400}, {100, 400}],
ImageCorners[
ImageTake[img, {100, 400}, {100, 400}],
1.0,
0.0002,
10
]
]
</code></pre>
<p><a href="https://i.stack.imgur.com/N6qB7.png" rel="noreferrer"><img src="https://i.stack.imgur.com/N6qB7.png" alt="Output of ImageCorners combined with ImageTake."></a></p>
<p>for visualization and then</p>
<pre><code>Length@
ImageCorners[
ImageTake[img, {100, 400}, {100, 400}],
1.0,
0.0002,
10
]
</code></pre>
<p>to get an estimate of the numbers of particles. In any case I would start by selecting a subsection of the image where the lighting is as homogenous as possible. As you can see at the above example there's still a number of false positives and false negatives, and you'll also have to guesstimate the actual area your image is covering in order to calculate the density.</p>
|
3,954,010 | <p>The case <span class="math-container">$n=3$</span> is from <a href="https://usamts.org/Tests/Problems_31_1.pdf" rel="nofollow noreferrer">here</a>. It's straightforward to prove it's true:</p>
<p>First we notice that if any two of <span class="math-container">$x_1, x_2, x_3$</span> are equal then all must be equal.</p>
<p>Denote <span class="math-container">$a=x_1^{x_2}=x_2^{x_3}=x_3^{x_1}$</span> then
<span class="math-container">$$x_1 = a^{1/x_2}, x_2=a^{1/x_3}, x_3=a^{1/x_1}$$</span>
WLOG assume <span class="math-container">$x_1 \ge x_2, x_3$</span>. There are two cases:</p>
<ul>
<li><p><span class="math-container">$x_1 \ge x_2 \ge x_3 \implies \frac{1}{x_2} \ge \frac{1}{x_3} \ge \frac{1}{x_1} \implies x_1 \ge x_3 \ge x_2 \implies x_2=x_3 \implies x_1=x_2=x_3$</span>.</p>
</li>
<li><p><span class="math-container">$x_1 \ge x_3 \ge x_2 \implies \frac{1}{x_2} \ge \frac{1}{x_1} \ge \frac{1}{x_3} \implies x_3 \ge x_1 \ge x_2 \implies x_1=x_3 \implies x_1=x_2=x_3$</span>.</p>
</li>
</ul>
<hr />
<p>If <span class="math-container">$n$</span> is even, then <span class="math-container">$x_{2i-1} = 2, x_{2i}=4$</span> is a counterexample.</p>
<p>When <span class="math-container">$n=5$</span> the above method will need to examine <span class="math-container">$4!=24$</span> cases. Basically we map <span class="math-container">$x_i$</span> to <span class="math-container">$x_{(i \mod 5) +1}$</span> and reverse the order. In many of cases we can deduce <span class="math-container">$4$</span> or all <span class="math-container">$5$</span> of the <span class="math-container">$x_i$</span>'s are equal. For example, if <span class="math-container">$$x_1 \ge x_3 \ge x_2 \ge x_5 \ge x_4 \tag 1$$</span> then
<span class="math-container">$$
x_5 \ge x_1 \ge x_3 \ge x_4 \ge x_2 \tag 2
$$</span>
Since the order of <span class="math-container">$x_2$</span> and <span class="math-container">$x_5$</span> reversed from <span class="math-container">$(1)$</span> to <span class="math-container">$(2)$</span>, they must be equal and so are everything in between them from both <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span> hence all <span class="math-container">$x_i$</span>'s are equal.</p>
<p>There are other cases that are different.</p>
<p><strong>Example 2:</strong> <span class="math-container">$$x_1 \ge x_3 \ge x_4 \ge x_2 \ge x_5 \tag 3$$</span> then
<span class="math-container">$$x_1 \ge x_3 \ge x_5 \ge x_4 \ge x_2 \implies x_1 \ge x_3 \ge x_2=x_4=x_5 \tag 4$$</span></p>
<p><strong>Example 3:</strong> <span class="math-container">$$x_1 \ge x_3 \ge x_4 \ge x_5 \ge x_2 \tag 5$$</span> then
<span class="math-container">$$x_3 \ge x_1 \ge x_5 \ge x_4 \ge x_2 \implies x_1=x_3 \ge x_2=x_4=x_5 \tag 6$$</span></p>
<p>But they all lead to the conclusion that <span class="math-container">$x_1=x_2=x_3=x_4=x_5$</span>.</p>
<hr />
<p>Now my questions:</p>
<p><strong>Question #1:</strong> Is it always true if <span class="math-container">$n>1$</span> and <span class="math-container">$n$</span> is odd?</p>
<p><strong>Question #2:</strong> What if we allow <span class="math-container">$x_i>0$</span> instead of <span class="math-container">$x_i>1$</span>?</p>
| Andreas | 317,854 | <p>Consider the question as a sequence. It starts with two values and has the recursive definition <span class="math-container">$x(n) = x(n-1)\frac{\log x(n-2)}{\log x(n-1)}$</span>. For <span class="math-container">$x_1^{x_2} = x_2^{x_3} = \cdots = x_{n-1}^{x_n} = x_n^{x_1} $</span> to hold, this sequence has to be periodic with some period length. This gives rise to the hypothesis (to be proved) that the sequence will indeed be monotonously rising / falling for odd/even <span class="math-container">$n$</span>, hence they cannot be periodic. If this can be established, indeed <span class="math-container">$x_1 = x_2= \cdots = x_n$</span> is the only solution.</p>
<p>Let <span class="math-container">$x(1) = z > 1$</span> and <span class="math-container">$x(2) = z^q$</span>. Then <span class="math-container">$x(3) = \frac{z^q}{q}$</span>. For <span class="math-container">$x(3) = x(1)$</span> to hold we require that <span class="math-container">$z = q^{\frac{1}{q-1}}$</span>. One can for example choose <span class="math-container">$q=2$</span> which gives <span class="math-container">$z=2$</span>, hence we obtain <span class="math-container">$x(1)= 2$</span> and <span class="math-container">$x(2) =4$</span> which gives the series <span class="math-container">$2,4,2,4, \cdots$</span> which OP already quoted. Infinite other choices of <span class="math-container">$q$</span> are possible as well. For <span class="math-container">$q=3$</span> we would have <span class="math-container">$z = \sqrt 3$</span> and hence <span class="math-container">$x(1)= \sqrt 3$</span> and <span class="math-container">$x(2) =\sqrt{27}$</span> which gives the series <span class="math-container">$\sqrt 3, \sqrt{27}, \sqrt 3, \sqrt{27}, \cdots$</span>. All of the series constructed this way are periodic with (even) period length 2 or integer multiples thereof.</p>
<p>Now let's prove that there are no series possible with odd period length. Start with the situation that the initially (for <span class="math-container">$x(1)$</span> and <span class="math-container">$x(2)$</span>) chosen <span class="math-container">$z$</span> and <span class="math-container">$q$</span> with <span class="math-container">$z \ne q^{\frac{1}{q-1}}$</span>. Then the first case is that <span class="math-container">$x(3) > x(1)$</span> and hence we have <span class="math-container">$x(4) = x(2)\frac{\log x(1)}{\log x(3)} < x(2)$</span>. This gives <span class="math-container">$x(5) = x(3)\frac{\log x(2)}{\log x(4)} > x(3)$</span>. Continuing like this, the odd sequence is rising whereas the even sequence is falling. If initially <span class="math-container">$x(1) > x(2)$</span>, as the odd sequence values will always be larger than <span class="math-container">$x(1)$</span>, and the even sequence values will always be smaller than <span class="math-container">$x(2)$</span>, hence no periodicity will be possible. If initially <span class="math-container">$x(1) < x(2)$</span>, hence the odd sequence (rising) and the even sequence (falling) will converge to some <span class="math-container">$x^*$</span> in between <span class="math-container">$x(1)$</span> and <span class="math-container">$x(2)$</span>, which means no periodicity will be possible.</p>
<p>In the second case, <span class="math-container">$x(3) < x(1)$</span>. With the same reasoning as above, <span class="math-container">$x(4) = x(2)\frac{\log x(1)}{\log x(3)} > x(2)$</span> and <span class="math-container">$x(5) = x(3)\frac{\log x(2)}{\log x(4)} < x(3)$</span>. Continuing like this, the odd sequence is falling whereas the even sequence is rising. This again means that the sequences either diverge from each other or converge to some <span class="math-container">$x^*$</span> in between <span class="math-container">$x(1)$</span> and <span class="math-container">$x(2)$</span>. In both cases, no periodicity will be possible.</p>
<p>Note that the condition <span class="math-container">$x_i >1$</span> is not guaranteed with all initial conditions. The lower values can fall rapidly, consider <span class="math-container">$x(1) = 1.1, x(2) = 2, x(3) = 0.275007, \cdots$</span></p>
<p>In summary, in the cases where <span class="math-container">$x(1) = z > 1$</span> and <span class="math-container">$x(2) = z^q$</span> with <span class="math-container">$z = q^{\frac{1}{q-1}}$</span>, the period is <span class="math-container">$2$</span> or any integer multiple of that, hence even. In all other cases, there is no periodicity, hence a constant <span class="math-container">$x(i) =z$</span> is the only solution.</p>
|
1,227,330 | <p>I have the following hypotheses: $\alpha_n \to 0, \beta_n \to 0, \alpha_n < 0 < \beta_n$. I need to show that the following two limits exist <strong>so that I may add them</strong>:</p>
<p>$$\lim \limits_{n \to \infty} \frac{-\alpha_n}{\beta_n -\alpha_n}, \lim \limits_{n \to \infty} \frac{\beta_n}{\beta_n - \alpha_n}$$</p>
<p>Well, I know that each limit is between $0$ and $1$. But I'm not sure how to show that these limits actually converge and don't fluctuate? (Or, it's also a possibility that I don't need the fact that the limits exist to add them, but I'm not sure if that's true)</p>
<p>This question is motivated by Exercise 5.19 in Baby Rudin.</p>
<hr>
<p>Edit: So either I'm doing something wrong by wanting to add the limits, or <strong>we don't need existence of the limits</strong> because I found a counter example: take $\beta_n = \frac{1}{n} \cos^2 n, \alpha_n = - \frac{1}{n} \sin^2 n$. Then $ \beta_n < 0 < \alpha_n$ (strict because $\pi$ is irrational), and the quotients do not converge (they are equal to $\cos^2 n$ and $\sin^2 n$).</p>
<p>Thus, here is my whole solution so someone can point out where I'm going wrong. I need to prove that for $D_n = \frac{f(\beta_n) - f(\alpha_n)}{\beta_n - \alpha_n}$, $f$ defined in $(-1,1)$, $f'$ exists at $0$, and $-1 < \alpha_n < 0 < \beta_n < 1$, $\alpha_n, \beta_n \to 0$, we have $\lim D_n = f'(0)$.</p>
<p>\begin{align}
\lim D_n &= \lim \frac{f(\beta_n) - f(0) + f(0) - f(\alpha_n)}{\beta_n - \alpha_n}\\
&= \lim \frac{f(\beta_n) - f(0)}{\beta_n - \alpha_n} + \lim \frac{f(0) - f(\alpha_n)}{\beta_n - \alpha_n}\\
&= \lim \frac{f(\beta_n) - f(0)}{\beta_n - 0} \lim \frac{\beta_n}{\beta_n - \alpha_n} + \lim \frac{f(0) - f(\alpha_n)}{0 - \alpha_n} \lim \frac{-\alpha_n}{\beta_n - \alpha_n}\\
&= \lim f'(0) \lim \frac{\beta_n - \alpha_n}{\beta_n - \alpha_n} = f'(0)
\end{align}</p>
| Rolf Hoyer | 228,612 | <p>Note that in the usual proof of the limit $\lim_{h\to 0} \frac{\sin(h)}{h} = 1$, it is shown that the inequality $$\cos(h) \le \frac{\sin(h)}{h} \le 1$$ holds for all $h$ in an interval around $0$. Assuming this result, we have $$\cos(h) - 1 \le \frac{\sin{h}}{h} - 1 \le 0$$ in this interval. Taking absolute value and multiplying by $|1/h|$ then yields
$$0 \le \left|\frac{\sin{h}-h}{h^2}\right| \le \left|\frac{\cos(h)-1}{h}\right|$$
for all $h$ in an interval around $0$.
The desired limit is then $0$ by the squeeze theorem and the standard trig limit $\lim_{x\to 0} \frac{\cos(x)-1}{x} = 0$.</p>
|
228,105 | <p>I want the last number after the "/". How can I do that?</p>
<pre><code>"http://arxiv.org/abs/math/0208009v1"
</code></pre>
<p>I just want <strong>0208009v1</strong></p>
<p>I want to import multiple links from the following:</p>
<pre><code>{{"http://arxiv.org/abs/math/0208009v1"}, \
{"http://arxiv.org/abs/0905.0227v1"}, \
{"http://arxiv.org/abs/0907.5143v2"}, \
{"http://arxiv.org/abs/math/0509348v1"}, \
{"http://arxiv.org/abs/math/0608711v2"}, \
{"http://arxiv.org/abs/math-ph/0002018v2"}}
</code></pre>
<p>I just want the last number in the following links. How would that work?</p>
| user1066 | 106 | <p>In addition, using a modification of this <a href="https://stackoverflow.com/a/8798197/499167">regex</a>:</p>
<pre><code>(StringCases[#, RegularExpression["[^/]+$"]]&/@list)//Flatten
</code></pre>
<blockquote>
<p>{0208009v1, 0905.0227v1, 0907.5143v2, 0509348v1, 0608711v2, 0002018v2}</p>
</blockquote>
<p>where</p>
<pre><code>list = {{"http://arxiv.org/abs/math/0208009v1"},
{"http://arxiv.org/abs/0905.0227v1"},
{"http://arxiv.org/abs/0907.5143v2"},
{"http://arxiv.org/abs/math/0509348v1"},
{"http://arxiv.org/abs/math/0608711v2"},
{"http://arxiv.org/abs/math-ph/0002018v2"}};
</code></pre>
|
1,235,973 | <p>Im reaching a point in programing where I need to create basic shapes which I simply cant since my math skills are very bad. After finding out that the skills required are trigonometry I read a few books. The result was that I couldnt understand any word at all.</p>
<p>What is required in order to understand basic 2D and 3D trigonometry? Calculus? Algebra? Also how long in general would it take for an average person to learn the material. It would be great if you could give me a guide for someone like me who only knows adding, multiplication, division and substracting.</p>
| Michael Hardy | 11,667 | <p>You certainly do not need calculus. You need some basic algebra. There are a few things in basic <b>geometry</b> that you show be thoroughly aware of:</p>
<ul>
<li>The number $\pi$ is the ratio of circumference to diameter of a circle. For example, a circle whose diamater is $1$ foot has a circumference of $\pi$ feet, i.e. about $3.14159\ldots$ feet. And $2\pi$ is the ratio of circumference to radius, so if the radius is $1$ foot (and thus the diameter is $2$ feet) then the circumference is $2\pi$ feet.</li>
<li>There are $90^\circ$ in a right angle, and $180^\circ$ in a straight angle.</li>
<li>The sum of the angles in every triangle is $180^\circ$. There are easy geometric arguments showing why that is so. You show learn to understand those arguments.</li>
<li>An isoceles triangle is one in which two sides have equal lengths. You should know that that happens if, and only if, the measures of the angles opposite those two sides are equal.</li>
<li>In particular, an isoceles right triangle, i.e. an a triangle with one right angle and two smaller angles whose measures are equal to each other, must have two $45^\circ$ angles. That that must be so is a logical consequence of things said above, and you should learn to understand why it is their logical consequence.</li>
<li>Also as a consequence of other things above, a triangle is equilateral, i.e. its three sides all have equal lengths, if, and only if, its three angles are equal. You should understand how the points above logically entail that and how they entail that in that case, the angles must be $60^\circ$ each.</li>
<li>You should know how to explain what the Pythagorean theorem says <b>without</b> saying anything that sounds like "A squared plus B squared equals C squared". It says: <b>The sum of the areas of the squares on the legs of a right triangle equals the area of the square on the hypotenuse.</b> It's about areas of squares, not just about multiplying each of three numbers by itself. Learn how to prove that and how to use it.</li>
</ul>
|
3,148,623 | <p>I want to find all prime ideals in <span class="math-container">$\mathbb{Z}/n\mathbb{Z}$</span> where <span class="math-container">$n>1$</span>.</p>
<p>I think I have to use the following theorem (because they asked me to prove it right before this exercise, which wasn't too complicated):</p>
<blockquote>
<p>Let R be a ring and I ⊂ R an ideal. Let φ : R → R/I
be the natural residue class homomorphism. Let I ⊂ J be another ideal.
Then J is a prime ideal in R if and only if φ(J) is a prime ideal in R/I.</p>
</blockquote>
<p>And</p>
<blockquote>
<p>Let R be a principal ideal domain and <span class="math-container">$I\subset R, I\neq(0)$</span> an ideal. Then every ideal generated by a irreducible element is a prime ideal</p>
</blockquote>
<p>An irreducible element in <span class="math-container">$\mathbb{Z}$</span> are exactly the prime numbers. I also know that <span class="math-container">$\mathbb{Z}$</span> is a principal ideal domain.</p>
<p>I tried supposing that <span class="math-container">$J=(p)$</span> is an ideal of <span class="math-container">$\mathbb{Z}$</span>, and when <span class="math-container">$p$</span> prime, it is a prime ideal. Then if <span class="math-container">$J$</span> is a superset of <span class="math-container">$n\mathbb{Z}$</span>, we have that <span class="math-container">$\phi(J)$</span> is a prime ideal of <span class="math-container">$\mathbb{Z}/n\mathbb{Z}$</span>. Is this correct? For what <span class="math-container">$p$</span> values does this hold? What if <span class="math-container">$J$</span> is not a superset of <span class="math-container">$n\mathbb{Z}$</span> as the theorem requires?</p>
| Bernard | 202,857 | <p>Another theorem says the prime ideals of <span class="math-container">$R/I$</span> correspond bijectively to the prime ideals of <span class="math-container">$R$</span> containing <span class="math-container">$I$</span>.</p>
<p>In the case of <span class="math-container">$\mathbf Z/n\mathbf Z$</span>, this means its prime ideals are generated by the congruence classes of the prime divisors of <span class="math-container">$n$</span>.</p>
<p>If <span class="math-container">$J=(m)$</span> does not contain <span class="math-container">$n$</span>, i.e. if <span class="math-container">$m$</span> does not divide <span class="math-container">$m$</span>, the image of <span class="math-container">$J$</span> in <span class="math-container">$\mathbf Z/n\mathbf Z$</span> is the ideal
<span class="math-container">$$J\cdot \mathbf Z/n\mathbf Z=(m)\cdot \mathbf Z/n\mathbf Z=(m,n)/(n)=(\gcd(m,n))/(n).$$</span></p>
|
3,148,623 | <p>I want to find all prime ideals in <span class="math-container">$\mathbb{Z}/n\mathbb{Z}$</span> where <span class="math-container">$n>1$</span>.</p>
<p>I think I have to use the following theorem (because they asked me to prove it right before this exercise, which wasn't too complicated):</p>
<blockquote>
<p>Let R be a ring and I ⊂ R an ideal. Let φ : R → R/I
be the natural residue class homomorphism. Let I ⊂ J be another ideal.
Then J is a prime ideal in R if and only if φ(J) is a prime ideal in R/I.</p>
</blockquote>
<p>And</p>
<blockquote>
<p>Let R be a principal ideal domain and <span class="math-container">$I\subset R, I\neq(0)$</span> an ideal. Then every ideal generated by a irreducible element is a prime ideal</p>
</blockquote>
<p>An irreducible element in <span class="math-container">$\mathbb{Z}$</span> are exactly the prime numbers. I also know that <span class="math-container">$\mathbb{Z}$</span> is a principal ideal domain.</p>
<p>I tried supposing that <span class="math-container">$J=(p)$</span> is an ideal of <span class="math-container">$\mathbb{Z}$</span>, and when <span class="math-container">$p$</span> prime, it is a prime ideal. Then if <span class="math-container">$J$</span> is a superset of <span class="math-container">$n\mathbb{Z}$</span>, we have that <span class="math-container">$\phi(J)$</span> is a prime ideal of <span class="math-container">$\mathbb{Z}/n\mathbb{Z}$</span>. Is this correct? For what <span class="math-container">$p$</span> values does this hold? What if <span class="math-container">$J$</span> is not a superset of <span class="math-container">$n\mathbb{Z}$</span> as the theorem requires?</p>
| tomasz | 30,222 | <p><strong>Hint</strong>: to solve this exercise, it is enough to use the following facts:</p>
<ul>
<li>an ideal <span class="math-container">$I\unlhd R$</span> is prime if and only if <span class="math-container">$R/I$</span> is a domain,</li>
<li>the characteristic of a domain is prime.</li>
</ul>
<p>Using these, it is enough to find all quotients of <span class="math-container">${\mathbf Z}/n{\mathbf Z}$</span> of prime characteristic. It should be straightforward to check that they are, in fact, domains.</p>
|
26,986 | <p>Here are the differential equations that set's up the 11 coupled oscillators.</p>
<pre><code>new = Join[
Table[x[i]''[t] == - x[i][t] +
0.1*(x[i + 1][t] - 2*x[i][t] + x[i - 1][t]), {i, 1,
9}], {x[0]''[t] == -x[0][t], x[10]''[t] == x[9][t], x[0][0] == 1,
x[0]'[0] == 1, x[1]'[0] == 0, x[1][0] == 0},
Table[x[i][0] == 0, {i, 2, 10}], Table[x[i]'[0] == 0, {i, 2, 10}]]
</code></pre>
<p>Here are the solutions.</p>
<pre><code>Solt = NDSolve[new, Table[x[i], {i, 0, 10}], {t, 25}]
</code></pre>
<p>Here are the individual plots.</p>
<pre><code>Table[Plot[Evaluate[x[i][t] /. Solt], {t, 0, 25},
PlotRange -> All], {i, 0, 10}]
</code></pre>
<p>I am trying to figure out how to make a graph so along the x-axis are my i's from 0 to 10, and I can watch the wave move along each oscillator as time moves on. I keep getting errors in which it floods my notebook and doesn't stop unless I close the kernel.</p>
<p>This is what I have so far, and I'm not sure how to incorporate time into this.</p>
<pre><code>Plot[Evaluate[x[i][t] /. Solt], {i, 0, 10}]
</code></pre>
<p>EDIT Coupled in a circle</p>
<pre><code>Stew = Join[
Table[x[i]''[t] == - x[i][t] +
0.1*(x[i + 1][t] - 2*x[i][t] + x[i - 1][t]), {i, 1,
9}], {x[10]''[t] == - x[10][t] +
0.1*(x[0][t] - 2*x[10][t] + x[9][t]),
x[0]''[t] == - x[0][t] +
0.1*(x[1][t] - 2*x[0][t] + x[10][t])}, {x[0][0] == 0,
x[0]'[0] == 0, x[1][0] == 1, x[1]'[0] == 0.5},
Table[x[i][0] == 0, {i, 2, 10}], Table[x[i]'[0] == 0, {i, 2, 10}]];
</code></pre>
<p>The Dsolve</p>
<pre><code>Loin = NDSolve[Stew, Table[x[i], {i, 0, 10}], {t, 6.28}]
</code></pre>
<p>The individual graphs</p>
<pre><code>Table[Plot[Evaluate[x[i][t] /. Loin], {t, 0, 6.28},
PlotRange -> All], {i, 0, 10}]
</code></pre>
<p>How would I go about putting the i=0 to 10 around in a circle?</p>
| mmal | 1,210 | <p>Solution by @Kuba can be easily extended to put the oscillators on a circle.</p>
<pre><code>Loin = NDSolve[Stew, Table[x[i], {i, 0, 10}], {t, 50}];
fr[t_] = Transpose@{Most@Range[0, 2 Pi, 2 Pi/11], x[#][t]&/@Range[0, 10]} /. First@Loin;
r0 = 3;
Animate[
ListPlot[
Function[{th, r}, {(r0 + r) Cos[th], (r0 + r) Sin[th]}] @@@ fr[t],
PlotRange -> {{-5, 5}, {-5, 5}}, AspectRatio -> 1,
Prolog -> {Gray, Circle[{0, 0}, 3]}, Axes -> False,
PlotMarkers -> Automatic],
{t, 0, 50}]
</code></pre>
<p><img src="https://i.stack.imgur.com/TNJHf.gif" alt="enter image description here"></p>
|
3,547,768 | <p>Given a function <span class="math-container">$f:\mathbb{R}\rightarrow\mathbb{R}$</span> which is differentiable and given that <span class="math-container">$f(0)=2$</span> and <span class="math-container">$(e^x + 1)f'(x) = e^x f(x)$</span> for every <span class="math-container">$x \in \mathbb{R}$</span>, prove that <span class="math-container">$f(x) = e^x + 1$</span>.</p>
| Mostafa Ayaz | 518,023 | <p><strong>Hint</strong></p>
<p><span class="math-container">$${d\over dx} \ln f(x)={f'(x)\over f(x)}$$</span></p>
|
3,525,356 | <p><a href="https://i.stack.imgur.com/8OitL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8OitL.png" alt="enter image description here"></a></p>
<p>Can someone help me on this? Any help will be greatly appreciated. </p>
| Community | -1 | <p><span class="math-container">$\begin{pmatrix}1&k\\0&1\end{pmatrix}^n=\begin{pmatrix}1&nk\\0&1\end{pmatrix}$</span>, so you can replace the super diagonal <span class="math-container">$1$</span> by <span class="math-container">$k\ne0\in\Bbb Z_p$</span>.</p>
<p>Plus you can do a subdiagonal version.</p>
<p>There are others too.</p>
|
2,137,269 | <p>I just sat in on a lecture on exponential generating functions in combinatorics (I have no formal education in combinatorics myself). It was quite interesting, but I'm afraid I don't actually understand what the generating function is/does. I've tried doing some minimal research online, but everything I've seen seems to be either too complex or too general to understand well. For example, I know how to find the generating function for permutations of a finite set, $\frac{1}{1-x}$. But what role does $x $ play here, and what does the generating function tell us? I don't see how it's at all related to the species of permutations itself.</p>
| User8976 | 98,414 | <p>I recommend you to see the book <a href="https://www.math.upenn.edu/~wilf/DownldGF.html" rel="nofollow noreferrer">generating functionology</a> by Wilf. It is available online. </p>
|
3,929,893 | <p>Let <span class="math-container">$G$</span> be a finite group and <span class="math-container">$H$</span>, <span class="math-container">$K$</span> normal subgroups of <span class="math-container">$G$</span>. Prove that <span class="math-container">$G=HK$</span> if and only if <span class="math-container">$G/(H\cap K)$</span> is isomorphic to <span class="math-container">$G/H\times G/K$</span>.</p>
<p>For first part i think I have to take the function <span class="math-container">$\phi\colon G\rightarrow G/H\times G/K$</span> defined by <span class="math-container">$\phi(g)=(gH,gK)$</span> then the kernel would be <span class="math-container">$H\cap K$</span> but I am confusion how to show this function is onto and how to use the fact <span class="math-container">$G=HK$</span>?</p>
<p>For converse part since H and K be normal then <span class="math-container">$HK$</span> be a subgroup of <span class="math-container">$G$</span> so clearly <span class="math-container">$HK\subset G$</span> but now how to show that <span class="math-container">$G\subset HK$</span>?</p>
<p>It will be enough if I get a proper hint for both part.Thank you.</p>
| ArsenBerk | 505,611 | <p><strong>HINT:</strong> For the first part, we know that
<span class="math-container">$$|HK| = \frac{|H|\cdot |K|}{|H \cap K|}$$</span>
So, what can we say about <span class="math-container">$|H \cap K|$</span> if <span class="math-container">$G = HK$</span> considering <span class="math-container">$HK \le G$</span>? Also, the function <span class="math-container">$\phi$</span> you defined is not surjective unfortunately. Here, using <span class="math-container">$G = HK$</span>, we can write any element of <span class="math-container">$g \in G$</span> as <span class="math-container">$g = hk$</span> for some <span class="math-container">$h \in H$</span> and <span class="math-container">$k \in K$</span> <strong>uniquely</strong>. Now, you may try to define it as <span class="math-container">$\phi: G \to G/H \times G/K$</span> such that <span class="math-container">$\phi(g) = (kH, hK)$</span>. Then, <span class="math-container">$\phi$</span> is well-defined since <span class="math-container">$g = hk$</span> is a unique way to write <span class="math-container">$g$</span>. Now, you can either check <span class="math-container">$\phi$</span> is a surjective homomorphism or check <span class="math-container">$\phi$</span> is injective and <span class="math-container">$|G| = |G/H \times G/K|$</span> (Since <span class="math-container">$|G|$</span> is finite, these two together imply <span class="math-container">$\phi$</span> is surjective).</p>
<p>For the second part, we can write
<span class="math-container">$$\frac{|G|}{|H \cap K|} = \frac{|G|}{|H|}\cdot \frac{|G|}{|K|}$$</span>
Then, what can we say about <span class="math-container">$|HK|$</span> considering the above hint for the first part?</p>
|
4,139,003 | <p>For this I used the method of making the determinant nonzero. First I create the array</p>
<p><span class="math-container">\begin{equation}
\begin{pmatrix}
a^2 & 0 & 1\\
0 & a & 2\\
1 & 0 & 1
\end{pmatrix}
\end{equation}</span></p>
<p>Then <span class="math-container">$\det(A)=a^3-a=a(a^2-1)$</span></p>
<p>Where we can see that for the determinant to be different from zero <span class="math-container">$ a $</span> cannot take the values of
<span class="math-container">$0,1,-1$</span>.</p>
<p>Therefore the values that <span class="math-container">$ a $</span> can take so that <span class="math-container">$ A $</span> is a base of <span class="math-container">$\mathbb{R} ^ 3 $</span> are <span class="math-container">$\mathbb{R}\backslash \{0,1, -1\}$</span>.</p>
<p>This is correct?</p>
| 2'5 9'2 | 11,123 | <p>My approach would be to first put the circle at the origin. Then find where tangent lines of the given slopes interesect, and shift from there.</p>
<p>The circle <span class="math-container">$x^2+y^2=9$</span> has tangent lines at <span class="math-container">$(x,y)$</span> with slope <span class="math-container">$-\frac{x}{y}$</span>. So the slope <span class="math-container">$1$</span> tangent line is tangent at some <span class="math-container">$(a,-a)$</span>. So <span class="math-container">$2a^2=9$</span>, and <span class="math-container">$a=\frac{3}{\sqrt{2}}$</span>. The point is <span class="math-container">$\left(\frac{3}{\sqrt{2}},-\frac{3}{\sqrt{2}}\right)$</span>.</p>
<p>The slope <span class="math-container">$2$</span> tangent line is tangent at some <span class="math-container">$(-2b,b)$</span>. So <span class="math-container">$5b^2=9$</span>, and <span class="math-container">$b=\frac{3}{\sqrt{5}}$</span>. The point is <span class="math-container">$\left(\frac{-6}{\sqrt{5}},\frac{3}{\sqrt{5}}\right)$</span>.</p>
<p>So the two lines are
<span class="math-container">$$\begin{align}
y&=(x-3/\sqrt{2})-\frac{3}{\sqrt2}=x-\frac{6}{\sqrt2}\tag{EQ1}\\
y&=2(x+6/\sqrt{5})+\frac{3}{\sqrt5}=2x+\frac{15}{\sqrt{5}}\tag{EQ2}
\end{align}$$</span></p>
<p>Solve that system to find where the lines intersect:
<span class="math-container">$$\left(\text{EQ2}-\text{EQ1}\,\text{to find }x,2\text{EQ1}-\text{EQ2}\,\text{to find }y \right)$$</span>
<span class="math-container">$$\left(-\frac{15}{\sqrt{5}}-\frac{6}{\sqrt{2}},-\frac{15}{\sqrt{5}}-\frac{12}{\sqrt{2}}\right)$$</span></p>
<p>So with the original question, shift by the negative of the above to get the center:
<span class="math-container">$$\left(\frac{15}{\sqrt{5}}+\frac{6}{\sqrt{2}},\frac{15}{\sqrt{5}}+\frac{12}{\sqrt{2}}\right)$$</span></p>
|
3,714,021 | <p>I've been working with the power series <span class="math-container">$\sum_{n=0}^{\infty}n(1-2^{-n})z^n$</span> for the last few hours and I'm starting to get a little lost in trying to figure out a few things for an exercise.</p>
<p>First of all, I've managed to find the radius of convergence ofr the series with relative ease using <span class="math-container">$R=lim\frac{\vert a_n\vert}{\vert a_{n+1}\vert}$</span> as <span class="math-container">$n\rightarrow\infty$</span>. Now the radius of convergence is 1 here and I' asked about whether or not the sum converges <em>at</em> <span class="math-container">$\vert z\vert=R$</span>. I'm a little confused here as I don't really know how much I can say about it considering this is on the boundary of the convergence. I considered saying something about Abel's Theorem, but as far as I understand the theorem only says something <em>if</em> the series is indeed convergent. Besides when I set <span class="math-container">$z=1$</span> the series becomes:</p>
<p><span class="math-container">$\sum_{n=0}^{\infty}n-\frac{n}{2^n}$</span></p>
<p>The second term grows smaller and the first grows larger, suggesting the whole expression grows larger. This must mean that the series diverges here, right?</p>
<p>Secondly I'm asked to show that the sum function for the power series can be written as:</p>
<p><span class="math-container">$F(x)=\frac{x}{(1-x)^2}-\frac{2x}{(2-x)^2}$</span></p>
<p>I'm assuming this has something to do with the convergence of the series, but I'm not really able to see how to show this.</p>
<p>Any help would be deeply appreciated!</p>
| TonyK | 1,508 | <blockquote>
<p>This must mean that the series diverges</p>
</blockquote>
<p>As you have written it, it only means that the series diverges at <span class="math-container">$z=1$</span>. You want to show that the series diverges whenever <span class="math-container">$|z|=1$</span>. But you can use the same argument.</p>
<p>For the second part, rewrite <span class="math-container">$\sum n(1-2^{-n})z^n$</span> as <span class="math-container">$\sum nz^n-\sum n\left(\dfrac{z}{2}\right)^n$</span>. So now you just have to evaluate <span class="math-container">$\sum nw^n$</span> at <span class="math-container">$w=z$</span> and <span class="math-container">$w=z/2$</span>. Do you know how to do this?</p>
|
227,732 | <p>Are there any general relations between the eigenvalues of a matrix $M$ and those of $M^2$?</p>
| Emily | 31,475 | <p>If the eigenvalues are distinct, then the square matrix $A$ is diagonalizable, namely
$$A = Q^{-1}DQ.$$</p>
<p>Then, $$A^2 = (Q^{-1}DQ)^2 = Q^{-1}DQQ^{-1}DQ = Q^{-1}D^2Q.$$</p>
<p>The diagonal entries of $D^2$ are the diagonal entries of $D$, squared.</p>
|
4,477,354 | <p>Fourier series for function <span class="math-container">$f(x)=c^x$</span>, <span class="math-container">$c\in\mathbb Z$</span>, <span class="math-container">$c>1$</span> on interval <span class="math-container">$(a,b)$</span>, where <span class="math-container">$a,b\in\mathbb R$</span>, <span class="math-container">$a<b$</span>.</p>
<p>Can I use the next formulas for this case?:
<span class="math-container">$$f(x)=\frac{a_0}{2}+\sum\limits_{n=1}^{+\infty}\left[a_n\cos\left(\frac{\pi nx}{l}\right)+b_n\sin\left(\frac{\pi nx}{l}\right)\right],$$</span>
<span class="math-container">$$a_n=\frac{1}{l}\int\limits_a^bf(x)\cos\left(\frac{\pi nx}{l}\right)dx,$$</span>
<span class="math-container">$$b_n=\frac{1}{l}\int\limits_a^bf(x)\sin\left(\frac{\pi nx}{l}\right)dx,$$</span>
where <span class="math-container">$l=(b-a)/2$</span>.</p>
<p>Particularly I need a Fourier series for function <span class="math-container">$f(x)=2^x$</span> on interval <span class="math-container">$(0;1)$</span>.</p>
| V.S.e.H. | 443,030 | <p>Note that using AM-GM:
<span class="math-container">$$
x_k = x_k \prod_{i=1}^{k-1}1 \leq \frac{k-1 + x_k^k}{k}\implies kx_k\leq k-1 + x_k^k
$$</span>
for <span class="math-container">$k \geq 1$</span>. Now sum each inequality along <span class="math-container">$k$</span> and you get the desired inequality.</p>
|
6,056 | <p>Are there any good places for me to sell off my mathematics books online especially Springer and Dover books?</p>
<p>(I thought perhaps this was off topic, but then I thought everybody in math probably has the problem of having bought pricey books that they would prefer to get some money back for, at one point or another.)</p>
| Ryan Budney | 642 | <p>I find it's pretty easy to sell books on e-bay. Local used bookstores rarely go for good math books, unfortunately. </p>
<p>You might also want to try the big technical book resellers -- like Powell's Books, they're based in Portland Oregon. Most on-line searches for books (like Google Books) will search Powell's stock. So that might be one of the best options. </p>
|
2,447,549 | <p>If so, could someone give me an example?</p>
| Dirk | 379,594 | <p>$V = \mathbb{R}$ and $W = \{0\}$. Take any linear map $f : W \to V$. :)</p>
|
241,138 | <p>I think I didn't pay attention to uniform distributions because they're too easy.
So I have this problem</p>
<blockquote>
<ol>
<li>It takes a professor a random time between 20 and 27 minutes to walk from his home to school every day. If he has a class at 9.00 a.m. and he leaves home at 8.37 a.m., find the probability that he reaches his class on time.</li>
</ol>
</blockquote>
<p>I am not sure I know how to do it. I think I would use $F(x)$, and I tried to look up how to figure it out but could only find the answer $F(x)=(x-a)/(b-a)$.</p>
<p>So I input the numbers and got $(23-20)/(27-20)$ which is $3/7$ but I am not sure that is the corret answer, though it seems right to me.</p>
<p>I'm not here for homework help (I am not being graded on this problem or anything), but I do want to understand the concepts. Too often I just learn how to do math and don't "really" understand it. </p>
<p>So I would like to know how to properly do uniform distribution problems (of continuous variable) and maybe how to find the $F(x)$. I thought it was the integral but I didn't get the same answer.</p>
<p>Remember I want to understand this. Thanks so much for your time. </p>
| Aseem Dua | 48,261 | <p>In case of a double root, the method fails, we need to start over with a different assumption of the solution. This time though, we approach the problem in a more informed manner.</p>
<p>hint: when we solve a simple homogeneous system, when we have a double root, to obtain a second LI solution we simply multiply root by x.</p>
|
3,401,055 | <blockquote>
<p>If the distance of any point <span class="math-container">$(x,y)$</span> from the origin is defined as <span class="math-container">$d(x,y)=max\{|x|,|y|\}$</span> where <span class="math-container">$|.|$</span> represents the absolute value function, and <span class="math-container">$d(x,y)=a$</span> whre <span class="math-container">$a$</span> is a non-zero constant, then the locus of the resulting curve is:</p>
<p>(A) Circle</p>
<p>(B) Straight Line</p>
<p>(C) Square</p>
<p>(D) Triangle</p>
</blockquote>
<p>I know the meaning of <span class="math-container">$max\{b,c\}$</span> which gives the maximum of <span class="math-container">$b$</span> or <span class="math-container">$c$</span> as its output. How to deal with the case <span class="math-container">$max\{|x|,|y|\}$</span>? Since <span class="math-container">$d(x,y)=a$</span>, I concluded the locus must be a circle, but the answer is a square. How can this be possible?</p>
| Sangchul Lee | 9,340 | <blockquote>
<p><strong>Assumption.</strong> <span class="math-container">$V$</span> is a finite-dimensional inner product space and <span class="math-container">$A : V \to V$</span> is a linear operator such that the following condition holds:</p>
<p><span class="math-container">$$\langle v, Av\rangle \leq -\|v\|^2 \qquad \text{for all }\quad v \in V. \tag{*}$$</span></p>
</blockquote>
<p><strong>Case 1.</strong> Suppose that <span class="math-container">$V$</span> is a <span class="math-container">$\mathbb{C}$</span>-vector space. Since every eigenvalue <span class="math-container">$\lambda$</span> of <span class="math-container">$A$</span> has a unit eigenvector <span class="math-container">$v$</span>, <span class="math-container">$\text{(*)}$</span> tells that <span class="math-container">$\lambda = \langle v, Av\rangle \leq -1$</span>. Now by Jordan normal form, we know that</p>
<p><span class="math-container">$$ A = P^{-1}(D + N)P, $$</span></p>
<p>where <span class="math-container">$P$</span> is an invertible matrix, <span class="math-container">$D$</span> is the diagonal matrix consisting of eigenvalues of <span class="math-container">$A$</span>, and <span class="math-container">$N$</span> is a nilpotent matrix satisfying <span class="math-container">$DN = ND$</span>. Then</p>
<p><span class="math-container">$$ e^{tA} = P^{-1} e^{tD}e^{tN} P. $$</span></p>
<p>Write <span class="math-container">$\|\cdot\|$</span> for the operator norm. Then it is easy to check that <span class="math-container">$\| e^{tD} \| \leq e^{-t}$</span> and <span class="math-container">$\|e^{tN}\| \leq C t^{\dim V}$</span> for some constant <span class="math-container">$C > 0$</span>. So it follows that</p>
<p><span class="math-container">$$ \| e^{tA}\| \leq C' t^{\dim V} e^{-t} \to 0 $$</span></p>
<p>and the desired conclusion follows.</p>
<p><strong>Case 2.</strong> Now suppose that <span class="math-container">$V$</span> is an <span class="math-container">$\mathbb{R}$</span>-vector space. We prove the following lemma:</p>
<blockquote>
<p><strong>Lemma.</strong> Suppose that <span class="math-container">$V$</span> and <span class="math-container">$A$</span> be as in Assumption. Then <span class="math-container">$\operatorname{Re}(\lambda) \leq -1$</span> for every complex eigenvalue <span class="math-container">$\lambda$</span> of <span class="math-container">$A$</span>.</p>
<p><em>Proof.</em> Let <span class="math-container">$\lambda$</span> be an eigenvalue of <span class="math-container">$A$</span>. Regard <span class="math-container">$A$</span> as a linear operator over a complex vector space. Then there exists a a non-zero complex eigenvector <span class="math-container">$v+iw$</span> of <span class="math-container">$A$</span>, where <span class="math-container">$v, w$</span> themselves are real vectors. Then by writing <span class="math-container">$\sigma=\operatorname{Re}(\lambda)$</span> and <span class="math-container">$\xi=\operatorname{Im}(\lambda)$</span>, the identity <span class="math-container">$(\sigma+i\xi)(v+iw)=A(v+iw)$</span> yields
<span class="math-container">$$Av = \sigma v - \xi w, \qquad Aw = \xi v + \sigma w. $$</span>
From this,
<span class="math-container">$$ \begin{gathered}
(1 + \sigma)\|v\|^2 - \xi\langle v, w\rangle = \|v\|^2 + \langle v, Av\rangle \leq 0\\
(1 + \sigma)\|w\|^2 + \xi\langle v, w\rangle = \|w\|^2 + \langle w, Aw\rangle \leq 0
\end{gathered}$$</span>
Summing two inequalities tells that <span class="math-container">$\sigma + 1 \leq 0$</span>.</p>
</blockquote>
<p>Given this lemma, the proof of complex case can be easily adapted to show that <span class="math-container">$e^{tA} \to 0$</span> as <span class="math-container">$t\to\infty$</span>.</p>
|
3,401,055 | <blockquote>
<p>If the distance of any point <span class="math-container">$(x,y)$</span> from the origin is defined as <span class="math-container">$d(x,y)=max\{|x|,|y|\}$</span> where <span class="math-container">$|.|$</span> represents the absolute value function, and <span class="math-container">$d(x,y)=a$</span> whre <span class="math-container">$a$</span> is a non-zero constant, then the locus of the resulting curve is:</p>
<p>(A) Circle</p>
<p>(B) Straight Line</p>
<p>(C) Square</p>
<p>(D) Triangle</p>
</blockquote>
<p>I know the meaning of <span class="math-container">$max\{b,c\}$</span> which gives the maximum of <span class="math-container">$b$</span> or <span class="math-container">$c$</span> as its output. How to deal with the case <span class="math-container">$max\{|x|,|y|\}$</span>? Since <span class="math-container">$d(x,y)=a$</span>, I concluded the locus must be a circle, but the answer is a square. How can this be possible?</p>
| The One | 619,244 | <p>Solution without using spectral theory.</p>
<p>Choose <span class="math-container">$f:R\rightarrow R$</span> for given <span class="math-container">$v\in V $</span> with <span class="math-container">$f(t) = e^{tA}(v) \cdot e^{tA}(v) $</span></p>
<p>Then <span class="math-container">$f^{'}(t)=Ae^{tA}(v)\cdot e^{tA}(v)+e^{tA}(v)\cdot Ae^{tA}(v) = 2
(Ae^{tA}v \cdot e^{tA}(v)) \leq -2|e^{tA}(v)|^2=-2f(t)$</span></p>
<p>As <span class="math-container">$\forall t. f(t)>0 \implies \forall t. f^{'}(t)\leq-2f(t)\le0$</span>, <span class="math-container">$f(t)$</span> is non-increasing and bounded below by <span class="math-container">$0$</span>. </p>
<p>Thus, <span class="math-container">$lim_{t\rightarrow \infty }f(t)=l$</span> for some limit <span class="math-container">$l$</span>, and <span class="math-container">$lim_{t\rightarrow \infty } f^{'}(t)=0$</span>. </p>
<p>Suppose <span class="math-container">$l >0$</span>, then <span class="math-container">$lim_{t\rightarrow \infty }\frac{1}{2}f^{'}(t)=Ae^{tA}(v)\cdot e^{tA}(v) \leq - |e^{tA}(v)|^2=-l^2 < 0$</span>.</p>
<p>Yet, <span class="math-container">$lim_{t\rightarrow \infty }f^{'}(t)=0$</span>. Contradiction.</p>
<p>So, <span class="math-container">$lim_{t\rightarrow \infty }f(t)=e^{tA}(v) \cdot e^{tA}(v) =0$</span> and thus <span class="math-container">$lim_{t\rightarrow \infty }e^{tA}(v)=0$</span> for all <span class="math-container">$v \in V$</span>.</p>
|
2,722,170 | <p>I am a grade 11 student and I have to learn vectors for the IB exam. I know that to find the distance of a vector between two points in a 3 dimensional space for lets say point A and B, then you would use the formula
\begin{align*}
|AB|=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}.
\end{align*}
I am having difficulty visualizing and understanding why the coordinates of the points are being subtracted here. I really appreciate the feedback.
Sorry for the low quality of the link, but I guess I don't have enough reputation yet to insert images directly. Also, in the formula all the subscripts should be reversed. So x1 becomes x2 and vice versa.</p>
| Christian Blatter | 1,303 | <p>Assume $0<\theta<{\pi\over2}$.</p>
<p><em>Claim.</em> The resulting orbit is closed iff $\theta$ is a rational multiple of $\pi$.</p>
<p><em>Proof.</em> We draw our figure in the complex plane. Let $\ell_1$ be the real axis, and choose $\ell_2$ through the origin. We denote the points on $\ell_1$ by $p_k$ and the points on $\ell_2$ by $q_k$ and then obtain an orbit of the form
$$(p_0,q_0,p_1,q_1,p_2, \ldots)\ .$$
Define
$$z_k:=q_k-p_k\ ,\qquad{\rm resp.,}\qquad\vec z_k:=\vec{p_k q_k}\ .$$
Note that for all $k\geq0$ one has $|z_k|=L$ with a given $L>0$. Furthermore $p_k$ and $q_k$ are uniquely defined by $z_k$: Place the tail of $\vec z_k$ anywhere on $\ell_1$ and then translate $\vec z_k$ horizontally until its head is lying on $\ell_2$. The final position of the tail then is $p_k$, and the final position of the head is $q_k$. It follows that it is sufficient to analyze the sequence $(z_k)_{k\geq0}$.</p>
<p>Inspecting the orbit construction shows that
$$p_1-q_0=\overline{q_0-p_0}\ .$$
In the same way as $p_1-q_0$ results from conjugating $q_0-p_0$ with respect the real axis $\ell_1$ the next difference $q_1-p_1$ results from "conjugating" $p_1-q_0$ with respect to $\ell_2$. It follows that
$$z_1=q_1-p_1=e^{i\theta}\>\overline{e^{-i\theta}(p_1-q_0)}=e^{2i\theta}(q_0-p_0)=e^{2i\theta} z_0\ .$$
But this implies
$$z_{k+1}=e^{2i\theta} z_k\qquad(k\geq0)\ ,$$
and then, by induction,
$$z_k=e^{2ik\theta}z_0\qquad(k\geq0)\ .\qquad\square$$
If $\theta={m\over n}\pi$ in lowest terms then the period of the orbit is $n$, independently of the choice of $p_0\in \ell_1$ and $q_0\in\ell_2$, thereby fixing an $L=|q_0-p_0|>0$.</p>
|
621,544 | <p>I have multiple data-sets from a Fourier series of a function $f(t,x)$ (the data-sets where obtained by varying $x$), so $A_n=\frac{2}{T}\int_0^T{f(t,x)\sin{\frac{2\pi nt}{T}}dt}$, which seems to follow an equation of the form
$$
A=an^b
$$
I know that $b$ will be negative, since $A$ approximately equal to $0$ for large $n$. However $A$ also has noise, which means that for certain data-sets $A$ sometimes also has a few negative values.</p>
<p>This means that I will not be able to use a linear fit using
$$
\log(A)=\log(a)+b\log(n)
$$</p>
<p><a href="https://math.stackexchange.com/questions/3625/easy-to-implement-method-to-fit-a-power-function-regression">This</a> might be a related question. That has an answer which does avoid this, but secant method or Newton's method can still take some time and I wonder if there might be a faster method? </p>
<p>Maybe related to this question is how would I determine the error ($\sigma_A$) of $A_n$?</p>
<p><strong>Edit:</strong><br>
I was trying to find Fourier series of the <a href="http://en.wikipedia.org/wiki/True_anomaly" rel="nofollow noreferrer">true anomaly</a> of closed Kepler orbits (so $0\leq e<1$) as a function of eccentricity $e$, since there is no explicit solution for the position as a function of time. But there is one <a href="https://physics.stackexchange.com/questions/69380/what-are-common-methods-for-calculating-the-time-dependency-of-elliptical-orbit">the other way around</a>
$$
t(\theta)=\sqrt{\frac{a^3}{\mu}}\left(2\tan^{-1}\left(\frac{\sqrt{1-e^2}\tan{\frac{\theta}{2}}}{1+e}\right)-\frac{e\sqrt{1-e^2}\sin{\theta}}{1+e\cos{\theta}}\right)
$$
The function from which I would like to the get the Fourier series is defined as
$$
f(e,t)=\theta(t)-\frac{2\pi t}{T}
$$
It is possible to find the time derivative of the true anomaly, $\frac{\delta}{\delta t}\theta$ as a function of $\theta$ itself
$$
\dot{\theta}(\theta)=\sqrt{\frac{\mu}{a^3\left(1-e^2\right)^3}}\left(1+e\cos{\theta}\right)^2
$$
This allowed me to rewrite the integral to another integral over $\theta$, since $dt=\frac{d\theta}{\dot{\theta}}$. So
$$
A_n=\frac{2}{T}\int_{-\pi}^{\pi}{\frac{\left(\theta-\frac{2\pi t(\theta)}{T}\right)\sin\left(\frac{2\pi nt(\theta)}{T}\right)}{\dot{\theta}(\theta)}d\theta}
$$
And since $T=2\pi\sqrt{\frac{a^3}{\mu}}$, this can be further simplified to only a function of $n$ and $e$ since
$$
\frac{2\pi t(\theta)}{T}=2\tan^{-1}\left(\frac{\sqrt{1-e^2}\tan{\frac{\theta}{2}}}{1+e}\right)-\frac{e\sqrt{1-e^2}\sin{\theta}}{1+e\cos{\theta}}
$$
$$
\frac{2}{T}\frac{1}{\dot{\theta}(\theta)}=\frac{\sqrt{\left(1-e^2\right)^3}}{\pi\left(1+e\cos{\theta}\right)^2}
$$
However according to MATLAB, this integral did not have an explicit solution, so I still had to solve it numerically. But fortunately MATLAB does has functionality to solve this integral numerically within small error bounds, which is better and faster than my initial Riemann integral approach. After this I did decided to implement an iterative method to solve for $b$, since I still got quite a lot of negative values.</p>
<p>However my actual goal was actually finding expressions for $a$ and $b$ as a function of $e$. And after some trial and error I got quite good test functions which comply with the data:
<img src="https://i.stack.imgur.com/kSwSY.png" alt="enter image description here"></p>
<p>$$
a(e)=\frac{1.887 e^{0.9591}}{1-0.04663 e^{64.38}}
$$
$$
b(e)=\frac{e^4 - 2.61 e^3 + 0.7813 e^2 + 0.8935 e + 2.014 \times 10^{-3}}{1.152 e^3 - 1.08 e^2 - 0.139 e - 6.703 \times 10^{-5}}
$$
But when using this to approximate $\theta(t)$ it often seems quite off. So someone know what would be better expressions for $a$ and $b$?</p>
| Claude Leibovici | 82,404 | <p>As you know, the key problem in nonlinear regression is the starting point (initial values of the model parameters). When the problem can be linearized (such via a logarithmic transform), the problem is simpler since the solution of the linearized model wil give you good (or at least reasonable) starting values for the nonlinear model. </p>
<p>However, there are situations such as the one you describe which can be problematic. For sure, in your case, you could exclude in a first step all the data points corresponding to negative values of A(i) and proceed as Ross Millikan suggests in the above answer. </p>
<p>But let me suppose the following general situation : you have to fit a model which is nonlinear because of one single parameter ("b" in your case). If this parameter is given a fixed value, the model is linear and the problem is easy to solve. So, what I suggest you to do (I made that hundreds of time in my life) is to fix "b" at a given value and solve the problem; the result is a sum of squares. Now, change the value of "b" and repeat. This means that you will be able to plot the sum of squares of the residuals as a function of "b" and then to locate the minimum. Since you also know the values of the other parameters for the best value of "b", now you can start the nonlinear regression. </p>
<p>If you had two parameters (say "b" and "c") making the model nonlinear, I should have suggested to build a grid in order to look (contour plot for example) at the sum of squares as a function of "b" and "c". </p>
<p>I hope this is clear. If not, please post and I shall continue.</p>
<p><strong>Happy New Year</strong></p>
|
2,542,259 | <p>The question says to "find geometrically the set of points $(x,y) \in \Bbb R^2$ such that $$x+2yi=|x+yi|$$</p>
<p>I'm kinda lost on this one, I've tried solving the equation directly but I don't even know what to do with that. Here's what I have right now:</p>
<p>$$x+2yi = |x+yi|$$
$$x+2yi = \sqrt{x^2+y^2}$$
$$x^2+4yxi-4y^2=x^2+y^2$$
$$4yxi=5y^2$$
$$y=0 \lor -5y+4xi=0$$
$$y=0 \lor y=\frac{4xi}5$$</p>
<p>I don't know what to do with this... I'm reading Ahlfors' Complex Analysis but this type of exercise seems kind of weird to me.</p>
| B. Mehta | 418,148 | <p>Since the right hand side is strictly real (and non-negative), we must have $y=0$, as $(x,y) \in \mathbb{R}^2$. So, the equation becomes $x = |x|$, and the solution set is $\{(x,0) \in \mathbb{R}^2 \mid x \geq 0\}$.</p>
<p>Geometrically, this is the non-negative real axis, or the halfline $[0, \infty) \times \{0\}$.</p>
|
1,376,627 | <p>I would like to know how to rewrite the following equations:</p>
<p>$$
\frac{d (f(x))}{d(x+c)} =0\\
\frac{d^2 (f(x))}{d(x+c)^2} =0\\
$$</p>
<p>Here $x$ is a variable, $c$ is a constant and $f(x)$ is a function of x.
I would also like to know the reasoning behind the answer.</p>
| mvw | 86,776 | <p>Introducing $y=x+c$ one gets
$$
0
= \frac{d(f(x))}{d(x+c)}
= \frac{d(f(y-c))}{dy}
= f'(y-c) \cdot 1
= f'(x)
= \frac{d(f(x))}{dx}
$$
For the second derivative we get:
$$
0
= \frac{d^2(f(x))}{d(x+c)^2}
= \frac{d}{d(x+c)} \frac{d(f(x))}{d(x+c)}
= \frac{d(f'(x))}{d(x+c)} = (f')'(x) = f''(x)
= \frac{d^2(f(x))}{dx^2}
$$</p>
|
1,376,627 | <p>I would like to know how to rewrite the following equations:</p>
<p>$$
\frac{d (f(x))}{d(x+c)} =0\\
\frac{d^2 (f(x))}{d(x+c)^2} =0\\
$$</p>
<p>Here $x$ is a variable, $c$ is a constant and $f(x)$ is a function of x.
I would also like to know the reasoning behind the answer.</p>
| haqnatural | 247,767 | <p>$$\frac { d }{ dx } \left( f\left( x \right) \left( x+c \right) ^{ -1 } \right) =0\\ \frac { df\left( x \right) }{ dx } \left( x+c \right) ^{ -1 }+f\left( x \right) \frac { d\left( \left( x+c \right) ^{ -1 } \right) }{ dx } =0\\ \frac { df\left( x \right) }{ dx } \frac { 1 }{ \left( x+c \right) } -f\left( x \right) \frac { 1 }{ { \left( x+c \right) }^{ 2 } } =0$$</p>
|
286,312 | <p>I'm starting to understand how induction works (with the whole $k \to k+1$ thing), but I'm not exactly sure how summations play a role. I'm a bit confused by this question specifically:</p>
<p>$$
\sum_{i=1}^n 3i-2 = \frac{n(3n-1)}{2}
$$</p>
<p>Any hints would be greatly appreciate</p>
| amWhy | 9,003 | <p><strong>Base case:</strong> </p>
<ul>
<li><p>Let $n=1$ and test: $$\sum_{i=1}^1(3i-2)=3-2=1=\frac{1(3\cdot 1-1)}{2}$$</p></li>
<li><p>True for $n = 1$</p></li>
</ul>
<p><strong>Induction Hypothesis</strong>: </p>
<ul>
<li>Assume that it is true for $k$: assume that $$\sum_{i=1}^k(3i-2)=\frac{k(3k-1)}{2}.$$ </li>
</ul>
<p><strong>Inductive Step:</strong> </p>
<ul>
<li>Prove, using the Inductive Hypothesis as a premise, that $$\sum_{i=1}^{k+1}(3i-2)=\left(\sum_{i = 1}^k(3i - 2)\right) + (3(k+1)-2) = \frac{(k+1)(3(k+1)-1)}{2}.$$</li>
</ul>
|
2,076,984 | <p>I have already asked <a href="https://math.stackexchange.com/questions/2075949/how-to-draw-diagram-for-line-and-parabola">a similar question</a>. But the answer in that question is very difficult to understand. I am new to this concept so I am looking for an easier explanation.</p>
<blockquote>
<p>My main <strong>question</strong> is: why do we subtract things to find the area using the definite integral?</p>
</blockquote>
<p>Here are a couple of figures -</p>
<ol>
<li>Two parabolas -
<a href="https://i.stack.imgur.com/aQC8h.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aQC8h.jpg" alt="page 1"></a></li>
</ol>
<p>Area $\displaystyle = \int \left(\sqrt{x} - x^2 \right) dx$</p>
<p>Why do we subtract to find the area? Why not add?</p>
<ol start="2">
<li>Similarly in parabola and line.</li>
</ol>
<p><a href="https://i.stack.imgur.com/emac3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/emac3.png" alt="page 2"></a></p>
<p>Area $\displaystyle = \int (x + 2 - x^2)dx$</p>
| Logan Luther | 347,317 | <p>In this context, to find the area shaded, think about the area under each curve separately, and then try to find the area NOT shared by them. What will you get?</p>
<p>Since the area is a plain number, to find a difference you simply subtract two areas:<img src="https://i.stack.imgur.com/6IfNB.png" alt="The area between the two rectangles is the difference of the areas. "> </p>
<p>You can even think of it in terms of apples.
Suppose you have $15$ apples, arranged in a $3×5$ grid.</p>
<p>Now for any set of apples, the total number of apples is not about how they are arrranged, and therefore for any number of apples that we remove from the grid, we do not need to care about the way they were in the grid, but only the number removed.</p>
<p>And to see also why:</p>
<p>$$\int_a^b (f+g)(x) \mathrm dx = \int_a^b f(x) \mathrm dx + \int_a^b g(x) \mathrm dx.$$</p>
<p>Observe that the the definite integral is a linear operator, meaning it distributes over addition and this can be verified using the fundamental theorem of calculus.</p>
<p>$$h(x) =(f+g)(x)=f(x)+g(x),$$</p>
<p>$$\int_a^b h(x) \mathrm dx =H(b)-H(a)=$$ $$F(b)+G(b)-G(a)-F(a).$$</p>
<p>Rearranging terms gives:</p>
<p>$$F(b)-F(a)+G(b)-G(a)=\int_a^b f(x) \mathrm dx +\int_a^b g(x) \mathrm dx.$$</p>
|
2,076,984 | <p>I have already asked <a href="https://math.stackexchange.com/questions/2075949/how-to-draw-diagram-for-line-and-parabola">a similar question</a>. But the answer in that question is very difficult to understand. I am new to this concept so I am looking for an easier explanation.</p>
<blockquote>
<p>My main <strong>question</strong> is: why do we subtract things to find the area using the definite integral?</p>
</blockquote>
<p>Here are a couple of figures -</p>
<ol>
<li>Two parabolas -
<a href="https://i.stack.imgur.com/aQC8h.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aQC8h.jpg" alt="page 1"></a></li>
</ol>
<p>Area $\displaystyle = \int \left(\sqrt{x} - x^2 \right) dx$</p>
<p>Why do we subtract to find the area? Why not add?</p>
<ol start="2">
<li>Similarly in parabola and line.</li>
</ol>
<p><a href="https://i.stack.imgur.com/emac3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/emac3.png" alt="page 2"></a></p>
<p>Area $\displaystyle = \int (x + 2 - x^2)dx$</p>
| user39390 | 403,342 | <p>Suppose the topmost graph is 1 then 2, 3, 4 . We have to find the area between the Red-Curve and Blue line in the first quadrant ( Graph 1). When we Integrate the Red-Curve with the limit from 0 to some positive x , We get the area of red region ( Graph 2). In the same manner, When we Integrate blue line with limit from 0 to some positive x we get the area of blue region ( Graph 3). To find the area between Blue-Line and Red-Curve, We Integrate the Blue-Line with the limit x = 0 to x = 2. Similary we Integrate Red-Curve with limit x = 0 to x = 2 . Since Blue-Line is above the Red-Curve, Subtract the Blue Area from the Red Area To get Green Area ( Graph 4 )</p>
<p><a href="https://i.stack.imgur.com/yLmvUm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yLmvUm.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/FLs6Bm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FLs6Bm.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/RIRJmm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RIRJmm.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/yonsQm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yonsQm.jpg" alt="enter image description here"></a></p>
|
2,646,749 | <p>I'm solving the following problem:</p>
<blockquote>
<p>Identify all maximal ideals in the ring $\mathbb R[x]/(x^2-3x+2)$.</p>
</blockquote>
<p>I want to solve it in two ways.</p>
<p><strong>First method.</strong> Consider the surjective quotient homomorphism $\mathbb R[x]\rightarrow \mathbb R[x]/(x^2-3x+2)$ with kernel $I=(x^2-3x+2)$. The ideals of the image correspond bijectively to the ideals of $\mathbb R[x]$ containing $I$. Since $\mathbb R[x]$ is a PID, the only ideals in that ring containing $I$ are $(1),(x-1),(x-2),(x^2-3x+2)$. The image of $(1)$ is the whole ring and therefore cannot be maximal. The image of $(x^2-3x+2)$ is the zero ideal and also cannot be maximal. The images of $(x-1)$ and $(x-2)$ are $(x-1)+I$ and $(x-2)+I$ respectively. <em>How do I show rigorously that the quotient of $\mathbb R[x]/I$ by each of these ideals is a field?</em> (I think this is the most reasonable way to show that $(x-1)+I$ and $(x-2)+I$ are maximal; if not, please let me know if there is an easier way).</p>
<p><strong>Second method.</strong> I want to use the fact that the maximal ideals of $\mathbb R\times \mathbb R$ are $(0)\times \mathbb R$ and $\mathbb R\times (0)$. To this end define a ring homomorphism $\phi: \mathbb R[x]\rightarrow \mathbb R \times \mathbb R$ by $f(x)\mapsto (f(1),f(2))$. This is an empmorphism with kernel $I$. By the first isomorphism theorem it induces an isomorphism $\mathbb R[x]/I\simeq \mathbb R \times \mathbb R$. Now we need to find the inverse images of $(0)\times \mathbb R$ and $\mathbb R\times (0)$ under $\phi$ and take their images under the quotient map $\mathbb R[x]\rightarrow \mathbb R[x]/I$. We have $\phi^{-1}(1,0)=-x+2; \phi^{-1}(0,1)=x-1$. This implies $\phi^{-1}(\mathbb R\times (0))=(x-2)$ and $\phi^{-1}((0)\times \mathbb R)=(x-1)$ because for example $-x+2\in \phi^{-1}(\mathbb R\times (0)) $ and since the inverse image of any ideal is an ideal, $(x-2)$ lies in $\phi^{-1}(\mathbb R\times (0))$. <em>But why there is nothing else in $\phi^{-1}(\mathbb R\times (0))$ except $(x-2)$?</em></p>
| David Hill | 145,687 | <p>For method 1, the correspondence theorem proves this immediately: There is a 1-1 order preserving correspondence between the set of ideals in $R/I$ and the set of ideals in $R$ that contain $I$.</p>
<p>For method 2, the Chinese Remainder theorem gives an isomorphism
$$\mathbb{R}[x]/(x^2-3x+2)\cong \mathbb{R}[x]/(x-2)\oplus\mathbb{R}[x]/(x-1)\cong \mathbb{R}\oplus\mathbb{R}.$$</p>
|
1,255,629 | <p>Show that </p>
<p>$$\sin\left(\frac\pi3(x-2)\right)$$ </p>
<p>is equal to </p>
<p>$$\cos\left(\frac\pi3(x-7/2)\right)$$</p>
<p>I know that $\cos(x + \frac\pi2) = −\sin(x)$ but i'm not sure how i can apply it to this question.</p>
| ZeroCool | 109,653 | <p>Maybe you can try
\begin{align*}\sin\left(\frac{\pi}{3}(x-2)\right) &= \sin\left(\frac{\pi}{3}x-\frac{2\pi}{3}\right)\\
&=\sin\left( \pi - \frac{\pi}{3}x+\frac{2\pi}{3}\right) \\
&=\sin\left(\frac{5\pi}{3}-\frac{\pi}{3}x\right) \\
&=\sin\left(\frac{\pi}{2}+\frac{7}{2}\frac{\pi}{3} -\frac{\pi}{3}x\right)\\
&=\cos\left(\frac{\pi}{3}\left(x-\frac{7}{2}\right)\right).
\end{align*}</p>
|
1,313,056 | <p>Got this matrix: </p>
<p>\begin{bmatrix} 1 & 2 \\ -2 & 5 \end{bmatrix}</p>
<p>I should determine if the matrix is diagonalizable or not.
I found the eigenvalues ( only one) = 3.
My eigenvector is then \begin{bmatrix} 1 \\ 1 \end{bmatrix}
This matrix is not diagonizable (from my teachers notes) but i don't know why, can someone explain this? </p>
| marwalix | 441 | <p>If a $2\times 2$ matrix $A$ with one double eigenvalue say $\lambda$ is diagonalisable. It means there exists an invertible matrix $P$ such that</p>
<p>$$A=P\cdot\lambda I_2\cdot P^{-1}=\lambda P\cdot P^{-1}=\lambda I_2$$</p>
<p>This means the only $2\times 2$ with a double eigenvalue that is diagonalisable is diagonal</p>
|
446,959 | <p>I am searching two simple/efficient/generic algorithms to generate a uniform distribution of random points:</p>
<ul>
<li>in the volume of a n-dimensional hypersphere</li>
<li>on the surface of a n-dimensional hypersphere</li>
</ul>
<p>knowing the dimension $n$, the center of the hypersphere $\vec{x}$ and its radius $r$.</p>
<p>How to do that ?</p>
| nbubis | 28,743 | <p><a href="http://mathworld.wolfram.com/HyperspherePointPicking.html" rel="nofollow">Aperantly</a>, for picking points on the <em>surface</em> of a hypersphere is to generate $n$ Gaussian random variables $x_1, x_2, ...x_n$, and then use the vectors:</p>
<p>$$\frac{1}{\sqrt{x_1^2+x_2^2+...+x_n^2}} \left(
\begin{array}{c}
x_1\\ x_2\\ ..\\ x_n\\
\end{array} \right)$$
Which will be uniformly distributed on the the surface of the hypersphere.</p>
|
312,696 | <p>I have three original points $pt_1, pt_2, pt_3$ which if transformed by an unknown matrix $M$ turn into points $gd_1, gd_2, gd_3$ respectively. How can I find the matrix $M$ (all points are in 3-dimensional space)?</p>
<p>I understand that for original points holds $M\cdot pt_i = gd_i$, so combining all $pt_i$ into matrix $PT$ and all $gd_i$ into $GD$ I'd get a matrix equation $M\cdot PT=GD$ with unknown $M$.</p>
<p>However, many math packages solve matrix equations in form of $A\cdot x=B$, where $x$ is unknown.</p>
<p>Is my idea of combining points into matrices correct and if so how can I solve my matrix equation?</p>
| Wesner Moise | 168,911 | <p>I don't believe the accepted answer works for translations. If we have a three-dimensional point and a 3x3 transformation matrix, the zero point will always map to the zero point. </p>
<p>A general affine transformation requires a four-dimensional matrix.
It may be possible to use a four-dimensional point with the fourth component of the point equal to 1.</p>
|
1,680,210 | <p>Three cards are drawn at random from a full deck. what is the probability of getting a three , a seven, and an ace?</p>
<p>Can someone help me with this question please, thanks!</p>
| Barry Cipra | 86,747 | <p>Inspired by Ojas's answer (which I think is incomplete because it seems to assume <span class="math-container">$a$</span> is positive):</p>
<p><span class="math-container">$x=0\implies a\mid90$</span></p>
<p><span class="math-container">$x=1\implies a\mid92$</span></p>
<p>Thus <span class="math-container">$a\mid2$</span>, so <span class="math-container">$a\in\{\pm1,\pm2\}$</span></p>
<p><span class="math-container">$x=-1\implies(a+2)\mid88\implies a\in\{-1,2\}$</span></p>
<p>Finally, we can rule out <span class="math-container">$a=-1$</span> because <span class="math-container">$x^2-x-1$</span> has a root in <span class="math-container">$[-1,0]$</span> while <span class="math-container">$x^{13}+x+90$</span> clearly does not. Thus <span class="math-container">$a=2$</span> is the only remaining possibility.</p>
<p>Remark: This doesn't prove that <span class="math-container">$x^2-x+2$</span> necessarily <em>is</em> a factor of <span class="math-container">$x^{13}+x+90$</span>. All it proves is the following: If <span class="math-container">$a$</span> is an integer such that <span class="math-container">$x^2-x+a$</span> divides <span class="math-container">$x^{13}+x+90$</span>, then <span class="math-container">$a=2$</span>.</p>
|
4,285,189 | <p>I am aware of the following identity</p>
<p><span class="math-container">$\det\begin{bmatrix}A & B\\ C & D\end{bmatrix} = \det(A)\det(D - CA^{-1}B)$</span></p>
<p>When <span class="math-container">$A = D$</span> and <span class="math-container">$B = C$</span> and when <span class="math-container">$AB = BA$</span> the above identity becomes</p>
<p><span class="math-container">$\det\begin{bmatrix}A & B\\ B & A\end{bmatrix} = \det(A)\det(A - BA^{-1}B) = \det(A^2 - B^2) = \det(A-B)\det(A+B)$</span>.</p>
<p>However, I couldn't prove this identity for the case where <span class="math-container">$AB \neq BA$</span>.</p>
<p><strong>EDIT:</strong> Based on @Trebor 's suggestion.</p>
<p>I think I could do the following.</p>
<p><span class="math-container">$\det\begin{bmatrix}A & B\\ B & A\end{bmatrix} =
\det\begin{bmatrix}A & B\\ B-A & A-B\end{bmatrix} = \det(A^2-B^2) = \det(A-B)\det(A+B)$</span>.</p>
| Just a user | 977,740 | <p><span class="math-container">$\begin{pmatrix} A & B \\ B & A \end{pmatrix}\xrightarrow{\text{row1 -= row2}}\begin{pmatrix} A-B & B-A \\ B & A\end{pmatrix}\xrightarrow{\text{col2 += col1}}\begin{pmatrix} A-B & O \\ B & A+B\end{pmatrix}$</span></p>
|
2,680,297 | <p>I had been given in my complex analysis examination the following problem. </p>
<p><strong>Evaluate the following integral by using contour integration:</strong></p>
<p>$$
\int_0^{\pi}\dfrac{\sin(2\theta)}{5-3\cos(\theta)}d\theta
$$</p>
<p>The answer to which is:
$$
\dfrac{2}{9}(log_e(1024)-6).
$$ </p>
<p>I tried to get this answer trying different ways but I just could not find a way to do it. </p>
<p>A thing to notice is that there is no $\pi$ term occurring in the final answer which means whatever the contour is which will give the answer easily is such that by integrating along that contour the value which we will get will be proportional to $\dfrac{1}{i\pi }$. This is because the $2\pi i$ term multiplying the residue should be canceled out as the actual answer does not have $\pi$ dependence. The $pi$ in the denominator gives a hint to me that this going to be complicated as in complex analysis at least in introductory courses nowhere $\pi$ comes in the denominator unless the problem is intentionally set up that way.</p>
<p>Another thing to notice is that if we have to use a contour that covers the angle from $0$ to $\pi$. Usually, the limits given in such problems is $0$ to $2\pi$ which is easy to do just by replacing $sine$ and $cosine$ term as follows:
$$
sin(\theta) \rightarrow \dfrac{1}{2i}(z-\dfrac{1}{z})
$$
$$
cos(\theta) \rightarrow \dfrac{1}{2}(z+\dfrac{1}{z})
$$
$$
d\theta \rightarrow \dfrac{dz}{iz}
$$
and then carrying out the contour integral along $|z|=1$ contour. The limits of the integral could be converted to $0$ to $2\pi$ by proper substitution and then the contour integral by above procedure could be carried out but that would not help as it will bring in fractional order poles (as $cos\dfrac{\theta}{2}$ term will appear) which residue theory can't deal with.</p>
<p><strong>So, what I need is a hint on how to do it.</strong> </p>
| K B Dave | 534,616 | <p>Your question is basically whether
$$\int_0^{\pi}\frac{\sin 2\theta \,\mathrm{d}\theta}{5-3\cos\theta}=\int_{-1}^1\frac{2x\,\mathrm{d}x}{5-3x}= \tfrac{4}{9}(5\,\ln 2-3)$$
can be written as an integral about a <em>closed</em> curve of some algebraic function over the rationals. This statement is believed to be false; it is in fact an <em>open problem</em> to demonstrate as much (<a href="http://www.ihes.fr/~maxim/TEXTS/Periods.pdf" rel="nofollow noreferrer">Kontsevich and Zagier</a>, Problem 5).</p>
<p>The usual method (substituting $x=\cos\theta$) will have to do.</p>
|
1,237,618 | <p>I really need Your help.</p>
<p>I need to prove that Euclidean norm is strictly convex. I know that a function is strictly convex if $f ''(x)>0$. Can I use it for Euclidean norm and how? $||x||''=\frac{||x||^2-x^2}{||x||^3}$</p>
<p>Thank You!</p>
| science | 212,657 | <p>By the <a href="http://en.wikipedia.org/wiki/Convex_function" rel="nofollow">definition of convexity</a> and the triangle inequality we have</p>
<blockquote>
<p>$$ ||ax+(1-a)y|| \leq |a| ||x||+ |1-a|||y||,\quad 0\leq a\leq1. $$</p>
</blockquote>
|
4,179,720 | <p>I'm starting to study triple integrals. In general, I have been doing problems which require me to sketch the projection on the <span class="math-container">$xy$</span> plane so I can figure out the boundaries for <span class="math-container">$x$</span> and <span class="math-container">$y$</span>. For example, I had an exercise where I had to calculate the volume bound between the planes <span class="math-container">$x=0$</span>, <span class="math-container">$y=0$</span>, <span class="math-container">$z=0$</span>, <span class="math-container">$x+y+z=1$</span> which was easy. For the projection on the <span class="math-container">$xy$</span> plane, I set that <span class="math-container">$z=0$</span>, then I got <span class="math-container">$x+y=1$</span> which is a line.</p>
<p>However, now I have the following problem:</p>
<p>Calculate the volume bound between:</p>
<p><span class="math-container">$$z=xy$$</span></p>
<p><span class="math-container">$$x+y+z=1$$</span></p>
<p><span class="math-container">$$z=0$$</span></p>
<p>now I know that if I put <span class="math-container">$z=0$</span> into the second equation I get the equation <span class="math-container">$y=1-x$</span> which is a line, but I also know that <span class="math-container">$z=xy$</span> has to play a role in the projection. If I put <span class="math-container">$xy=0$</span> I don't get anything useful. Can someone help me understand how these projections work and how I can apply it here?</p>
| Ritam_Dasgupta | 925,091 | <p>For convenience, imagine the round table to be the unit circle. It is clear that we have <span class="math-container">$4$</span> seats, <span class="math-container">$2$</span> each on <span class="math-container">$x$</span>- and <span class="math-container">$y$</span>- axes, and <span class="math-container">$2$</span> each along <span class="math-container">$y=x$</span> and <span class="math-container">$x+y=0$</span>. Let us consider the <span class="math-container">$4$</span> adjacent seats, <span class="math-container">$(0,1), (\frac {1}{\sqrt 2}, \frac {1}{\sqrt 2}), (1,0), (\frac {1}{\sqrt 2}, -\frac {1}{\sqrt 2})$</span>. We can seat <span class="math-container">$1$</span> member of each couple here, either the male or the female, and then we can fix the other member to sit opposite to her spouse. There are <span class="math-container">$(\binom 21)^4$</span> ways to choose one member of each couple, and then there are <span class="math-container">$4!$</span> ways to seat them. Once these are seated, the rest of the arrangement is fixed. Note that we must divide by <span class="math-container">$8$</span> afterwards, because although we have marked coordinate axes for this solution, in reality there are none, and hence we have <span class="math-container">$8$</span> possible starting positions (for example, here we took starting point to be <span class="math-container">$(0,1)$</span>).
This gives us the total number of favorable cases, as <span class="math-container">$48$</span>.</p>
<p>Total possible cases is, of course, <span class="math-container">$\frac {8!}{8}=7!$</span>. Hence probability comes out to be:
<span class="math-container">$$P=\frac {48}{5040}=\frac {1}{105}$$</span></p>
|
267,121 | <h2>UPDATE on FINAL RESULT</h2>
<p>Thanks to @SquareOne effort I generated higher-resolution videos with smoothing transitions that can be seen here:</p>
<ul>
<li><p><a href="https://www.linkedin.com/feed/update/urn:li:activity:6926902980323512320/" rel="noreferrer">https://www.linkedin.com/feed/update/urn:li:activity:6926902980323512320/</a></p>
</li>
<li><p><a href="https://twitter.com/superflow/status/1521191832012705792" rel="noreferrer">https://twitter.com/superflow/status/1521191832012705792</a></p>
</li>
</ul>
<p>I might post my version of @SquareOne code with some bug corrections later. I am grateful to this community and @SquareOne for outstanding support.</p>
<h2>INTRO & <em>BOUNTY</em> TARGET</h2>
<p>Dear friends, as you know there is currently an ongoing war in Ukraine: <strong><a href="https://war.ukraine.ua" rel="noreferrer">https://war.ukraine.ua</a></strong></p>
<blockquote>
<p><em><strong>I need your help on some coding in image/video processing, which is very simple to formulate, but not obvious in execution. Smooth exactly-timed transitions between video frames and perfect alignment of video frames is a key challenge here. BOUNTY TARGET IS EXPLAINED AT THE END OF THE POST.</strong></em></p>
</blockquote>
<h2>DATA Description</h2>
<p><em>United States' <a href="https://www.understandingwar.org/" rel="noreferrer">Institute for the Study of War</a> (ISW, "a non-partisan, non-profit, public policy research organization") performs daily research and publishes daily maps of the battlefield. Their work is public and gives many references to data sources they use. For instance...</em></p>
<blockquote>
<p><strong>The whole-Ukraine overview map from the 1st day of invasion on FEB 24, 2022 (<a href="https://www.understandingwar.org/backgrounder/ukraine-conflict-update-7" rel="noreferrer">source</a>):</strong></p>
</blockquote>
<p><a href="https://i.stack.imgur.com/bmDVR.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/bmDVR.jpg" alt="enter image description here" /></a></p>
<blockquote>
<p><strong>A recent whole-Ukraine overview map from APR 19, 2022 (<a href="https://www.understandingwar.org/backgrounder/russian-offensive-campaign-assessment-april-19" rel="noreferrer">source</a>):</strong></p>
</blockquote>
<p><a href="https://i.stack.imgur.com/YYdnd.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/YYdnd.jpg" alt="enter image description here" /></a></p>
<h2>DATA Source</h2>
<p>These maps are published almost daily with all publications gathered here: <a href="https://www.understandingwar.org/publications" rel="noreferrer">https://www.understandingwar.org/publications</a></p>
<h2><em>BOUNTY</em> TARGET</h2>
<p><em><strong>BOUNTY WILL BE AWARDED TO the CODE GENERATING BEST .MP4 VIDEO of a SEQUENCE of MAPS.</strong></em></p>
<p>Basic part for the <strong>BOUNTY</strong>:</p>
<ul>
<li><p><strong>Programatic data access</strong>. While URLs of daily articles and images follow some pattern, it is not always regular. <em>How do we write a piece of code that accesses whole-map programmatically?</em> We do not want do do this manually. Approach 1: look at the daily image URLs (<a href="https://www.understandingwar.org/sites/default/files/DraftUkraineCoTApril19%2C2022.png" rel="noreferrer">example</a>), but they are still not regular. Approach 2: look at the daily articles URLs and get 1st image from the article (<a href="https://www.understandingwar.org/backgrounder/russian-offensive-campaign-assessment-april-19" rel="noreferrer">example</a>), but they are still not regular. Maybe there are other approaches. <strong>START DAT: FEB 24. END DATE: CURRENT DAY</strong>.</p>
</li>
<li><p><strong>Each frame must have a date stamp.</strong> For example: FEB 24, 2022, FEB 25, 2022, etc.</p>
</li>
<li><p><strong>Image alignment of Ukraine border - the GREATEST challenge</strong>. All these maps-images are slightly different. Ukraine country border should NOT jump from frame to frame.</p>
</li>
<li><p><strong>Duration of a frame and smoothness of transition</strong> Each map-image (key-frame) should be held on screen 1 second. Each of 3 transitional blended frames should last 0.15 seconds. This is a toy example of how to achieve it. Imagine you have just 3 key-frames.</p>
</li>
</ul>
<p><a href="https://i.stack.imgur.com/3VgCO.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/3VgCO.jpg" alt="enter image description here" /></a></p>
<p>Build transitions via interpolated blended frames:</p>
<pre><code>frames=Values[TimeSeriesResample[TimeSeries[imglist,{0}],1/4]]
</code></pre>
<p><a href="https://i.stack.imgur.com/oL4tF.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/oL4tF.jpg" alt="enter image description here" /></a></p>
<p>Define non-inform timings as</p>
<pre><code>In[68]:= timings=Flatten[Riffle[Table[1,3],{Table[.15,3]}]]
</code></pre>
<blockquote>
<p>Out[68]= {1,0.15<code>,0.15</code>,0.15<code>,1,0.15</code>,0.15<code>,0.15</code>,1}</p>
</blockquote>
<p>Create video as</p>
<pre><code>SlideShowVideo[frames -> timings]
</code></pre>
<p><a href="https://i.stack.imgur.com/N3XNu.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/N3XNu.jpg" alt="enter image description here" /></a></p>
<p>Export to .MP4 via</p>
<p><a href="http://reference.wolfram.com/language/ref/format/MP4.html" rel="noreferrer">http://reference.wolfram.com/language/ref/format/MP4.html</a></p>
<p><strong>Thank you very much for considering this!!!</strong> Collecting data from independent sources, and displaying it in a comprehensive animation, can help to inform society in ways that numbers and unorganized static images cannot. This is the sort of thing we as a community can do best in these difficult times.</p>
| Ben Izd | 77,079 | <blockquote>
<p>With my limited knowledge I could do <strong>Programmatic data access</strong>. I hope it would be useful for other answers.</p>
</blockquote>
<blockquote>
<p>The code provided in this answer is introduced to show the procedure and the author does not hold any responsibility for its usage.</p>
</blockquote>
<p>Although some days don't have images (like <a href="https://www.understandingwar.org/backgrounder/russia-ukraine-warning-update-russian-offensive-campaign-assessment-february-26" rel="noreferrer">February 20</a>) and there isn't a consistency in naming, below code almost find all the images to this date (except for <a href="https://www.understandingwar.org/backgrounder/russian-offensive-campaign-assessment-march-20" rel="noreferrer">March 20</a> which is named really differently).</p>
<p>The code below defines a function named <code>scrapImageURLs</code> which returns a list of image URLs based on your second approach (opening the URL and looking for the image). Then use <code>URLDownload</code> to download them on your computer and process them.</p>
<p>You can use <code>"StartDate"</code> and <code>"EndDate"</code> options to filter your requests.</p>
<pre><code>ClearAll[scrapImageURLs];
Options[scrapImageURLs]:={"StartDate"->{2022,2,25},"EndDate":>Floor[AbsoluteTime[]-86400,86400]};
scrapImageURLs[OptionsPattern[]]:=Block[{startDate=Max[Floor[AbsoluteTime@OptionValue["StartDate"],86400],AbsoluteTime@{2022,2,25}],endDate=Min[Ceiling[AbsoluteTime@OptionValue["EndDate"],86400],Ceiling[AbsoluteTime[],86400]],imageURLs,imageURL={},getError=False},
imageURLs=Table[Block[{dayString=ToLowerCase@DateString[day,{"MonthName","-","DayShort"}],justDayString=ToString@DateValue[day,"DayShort"],url,response},
url=Which[day===AbsoluteTime[{2022,2,25}],
"https://www.understandingwar.org/backgrounder/russia-ukraine-warning-update-russian-offensive-campaign-assessment-february-25-2022",
day===AbsoluteTime[{2022,2,28}],
"https://www.understandingwar.org/backgrounder/russian-offensive-campaign-assessment-february-28-2022",
day<AbsoluteTime[{2022,2,28}],
"https://www.understandingwar.org/backgrounder/russia-ukraine-warning-update-russian-offensive-campaign-assessment-"<>dayString,
True,"https://www.understandingwar.org/backgrounder/russian-offensive-campaign-assessment-"<>dayString];
response=URLRead[url];
If[response["StatusCode"]===200,
imageURL=StringCases[response["Body"],"https://www.understandingwar.org/sites/default/files/draft"~~Shortest[___]~~((justDayString~~Shortest[___]~~"ukraine")|("ukraine"~~Shortest[___]~~justDayString))~~Shortest[___]~~".png",IgnoreCase->True],Print[" ERROR on "<>dayString<>"\n "<>url<>"\n \"StatusCode\": "<>ToString[response["StatusCode"]]];getError=True;];
If[Not@getError,
If[Length@imageURL=!=0,Print[First@imageURL];First@imageURL,Print[" No image found on "<>dayString<>"\n ",url];Nothing],Nothing]
]
,{day,Range[startDate,endDate,86400]}];
imageURLs
]
</code></pre>
<p>Notes:</p>
<ul>
<li>It uses <code>URLRead</code> which was introduced in version 11.</li>
<li>Since it could fail at any moment during execution, it'll print the result/error it faced while executing and return a list of valid image URLs (for days that have an image and don't get an error).</li>
<li>It assumes the image filename starts with <code>"draft"</code> and ends with <code>".png"</code> and the day and the word <code>"ukraine"</code> is also included (for example March-19 has <code>19</code> in its name)</li>
<li><strong>Please save the URLs and only request for days that you don't have.</strong></li>
</ul>
<hr />
<p>Just to avoid putting pressure on servers, here is the list of image URLs up to <code>2022-4-20</code>:</p>
<pre><code>{"https://www.understandingwar.org/sites/default/files/DraftUkraineCoTFeb25%2C2022.png",
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|
4,070,996 | <p>I'm a bit lost in this integral: <span class="math-container">$$\int \frac{1}{1+\sin^4(x)} \, \mathrm dx$$</span>
I have tried solving with Wolfram, but I was getting a cosecant solution which doesn't seem as the correct method.</p>
<p>Do you have any ideas? :)</p>
<p>EDIT:
Do you please have step-by-step solution, because I am now somewhat lost. Using the substitution <span class="math-container">$t=\tan(x)$</span>, I got to</p>
<p><span class="math-container">$$\int \left(\frac{t^2}{2t^4+2t^2+1}+\frac{1}{2t^4+2t^2+1}\right)\mathrm dt$$</span></p>
<p>By expanding with 1:
<span class="math-container">$$\int \frac{1}{1+\sin^4x}\cdot \frac{\frac{1}{\cos^4x}}{\frac{1}{\cos^4x}}\mathrm dx$$</span>
<span class="math-container">$$\int \:\frac{1}{\frac{1}{\cos^4x}\cdot \frac{\sin^4x}{\cos^4x}}\cdot \frac{1}{\cos^4x} \mathrm dx$$</span>
<span class="math-container">$$\int \:\frac{1}{\left(\frac{1}{\cos^2x}\right)^2\cdot \tan^4x}\cdot \frac{1}{\cos^2x}\cdot \frac{1}{\cos^2x}\mathrm dx$$</span></p>
<p>And using the substitution: <span class="math-container">$t=\tan\left(x\right)$</span></p>
<p><span class="math-container">$$\mathrm dt=\frac{1}{\cos^2x}\mathrm dx$$</span></p>
<p><span class="math-container">$$t^2=\tan^2\left(x\right)$$</span></p>
<p><span class="math-container">$$t^2=\frac{\sin^2x}{\cos^2x}$$</span></p>
<p><span class="math-container">$$t^2=\frac{1-\cos^2x}{\cos^2x}$$</span></p>
<p><span class="math-container">$$t^2=\frac{1}{\cos^2x}-\frac{\cos^2x}{\cos^2x}=\frac{1}{\cos^2x}-1$$</span>
<span class="math-container">$$t^2+1=\frac{1}{\cos^2x}$$</span></p>
<p>Using it:
<span class="math-container">$$\int \:\frac{t^2+1}{2t^4+2t^2+1}\mathrm dt$$</span></p>
<p>I don't think I got to the expected result but I can't seem to be able to find why…</p>
| J.G. | 56,861 | <p>Double-check my arithmetic, but here's the strategy.</p>
<p>@Bernard's suggested substitution <span class="math-container">$t=\tan x$</span> gives
<span class="math-container">\begin{align}
& \int\left(1-\frac{t^4}{(t^2\sqrt{2}+1)^2-2(\sqrt{2}-1)t^2}\right) dt \\
= {} & t-\int\frac{t^4}{(t^2\sqrt{2}+ct+1)(t^2\sqrt{2}-ct+1)} \, dt\end{align}</span>
with <span class="math-container">$c:=\sqrt{2\sqrt{2}-1}$</span>. You can do the rest with partial fractions.</p>
|
3,194,534 | <p>Integrate the following <span class="math-container">$\int\frac{x+1}{x(1+xe^x)^2}dx$</span></p>
<p>I tried to multiply the numerator and denominator by <span class="math-container">$e^{-2x}$</span></p>
<p>Getting the following simplification
<span class="math-container">$\int\frac{e^{-2x}(x+1)}{x(e^{-x}+x)^2}dx$</span></p>
<p>My next step is as follow <span class="math-container">$\int\frac{e^{-2x}(x+1)}{x^3(\frac{e^{-x}}{x}+1)^2})dx$</span>
but could not proceed from here</p>
| Dr. Sonnhard Graubner | 175,066 | <p>Hint: Write your integrand in the form<span class="math-container">$$-\frac{e^x x+e^x}{e^x x+1}-\frac{e^x x+e^x}{\left(e^x x+1\right)^2}+\frac{1}{x}+1$$</span></p>
|
1,088,687 | <p>I have this example here:</p>
<p>$\lim_{n \to \infty} \frac{n^{2}+1}{n^{2}+n+1} = 1$</p>
<p>so I have to prove that, $\forall \epsilon > 0, \exists n_{0}\in N, \forall n(n \in N) \geq n_{0} \Rightarrow |\frac{n^{2}+1}{n^{2}+n+1} - 1| < \epsilon$</p>
<p>I have done in these steps <br/> <br/>
$|\frac{n^{2}+1}{n^{2}+n+1} - 1| < \epsilon$ <br/> <br/>
$|\frac{-n}{n^{2}+n+1}| < \epsilon$ <br/> <br/>
$\frac{n}{n^{2}+n+1} < \epsilon$ <br/> <br/>
$n^{2} + n(1-\frac{1}{\epsilon}) + 1 > 0$ <br/> <br/>
D = $(1-\frac{1}{\epsilon})^{2} - 4 = (1 - \frac{1}{\epsilon} - 2)(1 - \frac{1}{\epsilon} + 2) = -(1 + \frac{1}{\epsilon})(1-\frac{1}{\epsilon}) = \frac{1}{\epsilon^{2}} - 1$ <br/> <br/></p>
<h2>$n_{1,2} = \frac{\frac{1}{\epsilon} - 1 \pm \sqrt{D}}{2}$</h2>
<p>I guess we need that <br/> <br/></p>
<h2>$n > \frac{\frac{1}{\epsilon} - 1 + \sqrt{D}}{2} = \frac{\frac{1}{\epsilon} - 1 + \sqrt{\frac{1}{\epsilon^{2}}- 1}}{2} = \frac{\frac{1}{\epsilon} - 1 + \sqrt{1 + (\frac{1}{\epsilon^{2}}- 2)}}{2} > \frac{\frac{1}{\epsilon} - 1 + \frac{1}{2}(\frac{1}{\epsilon^{2}}- 2)}{2} = \frac{\frac{1}{\epsilon} - 1 + \frac{1}{2\epsilon^{2}}- 1}{2} = \frac{\frac{1}{2\epsilon^{2}} +\frac{1}{\epsilon} -2}{2}$</h2>
<p>But I think I have done some mistakes and this last thing could be in more simple form</p>
<p>P.S: How can I verify whether my result $n(\epsilon)$ is true in the end?</p>
| Khosrotash | 104,171 | <p>$$\left|\frac{n}{n^2+n+1}\right| <\epsilon\\\left|\frac{n}{n^2+n+1}\right| <\left|\frac{n}{n^2+n}\right| <\left|\frac{n}{n(n+1)}\right| <\left|\frac{1}{n+1}\right| <\epsilon\\n+1 >\frac{1}{\epsilon}\\n>\frac{1}{\epsilon}-1\\n\geqslant \left \lfloor \frac{1}{\epsilon}-1 \right \rfloor $$</p>
|
1,509,443 | <p>Prove that for $x\geq 1$ the following inequality is true: $$\dfrac{1}{x}\leq \ln(2x+1)-\ln(2x-1).$$</p>
<p>Can anyone help with this problem? I thought about some hours but any ideas</p>
| mathlove | 78,967 | <p>Let $f(x)=\ln(2x+1)-\ln(2x-1)-\frac 1x$. Then, we have
$$f'(x)=\frac{2}{2x+1}-\frac{2}{2x-1}+\frac 1{x^2}=\frac{-1}{x^2(2x+1)(2x-1)}\lt 0$$
for $x\ge 1$ with $\lim_{x\to +\infty}f(x)=0$.</p>
|
2,152,721 | <p>I have been messing around analyzing some self-made algorithms and then I stumbled upon this summation:</p>
<p>$$\sum_{j=0}^{n}\left\lfloor{2^{i-j}}\right\rfloor$$</p>
<p>I am thinking that my common algebraic manipulations involving series won't apply because of the floor function.</p>
<p>I also tried entering this series in wolframalpha but no luck.</p>
<p>Does anyone know how to do this?</p>
| Community | -1 | <p>Irrespective of the value of $i $, whether it is or not greater than $j $, we can write our sum as: $$S = \sum_{j=0}^{i} \lfloor 2^{i-j} \rfloor + \sum_{j=i+1}^{n} \lfloor 2^{i-j} \rfloor$$</p>
<p>Case $1$: If $i \geq n $, $$S = \sum_{j=0}^{n} \lfloor 2^{i-j} \rfloor = \sum_{j=0}^{n} 2^{i-j} (\text {why ?}) $$</p>
<p>Case $2$: If $i < n $, $$S = \sum_{j=0}^{i} \lfloor 2^{i-j} \rfloor + \sum_{j=i+1}^{n} \lfloor 2^{i-j} \rfloor$$ $$= \sum_{j=0}^{i} \lfloor 2^{i-j} \rfloor + 0 (\text {why?})$$ $$= \sum_{j=0}^{i} 2^{i-j}$$</p>
<p>The final result in both cases is an easily simplifiable geometric series. Hope it helps. </p>
|
3,659,942 | <p>I have the following problem:</p>
<p>Find all <span class="math-container">$n$</span> that are natural numbers such that <span class="math-container">$\sqrt{n}+\sqrt{n+m}\in\mathbb{N}$</span>, where <span class="math-container">$m$</span> is a positive natural number that is the product of two primes that are not equal to each other.</p>
<p>I thought that I can solve it when the function under the square roots are squares of integers but I have no idea how to prove that or how to continue.</p>
| DanielWainfleet | 254,665 | <p>Without recursion.</p>
<p>Let <span class="math-container">$<$</span> be a well-order on <span class="math-container">$A.$</span> A proper initial segment (<span class="math-container">$pis$</span>) of <span class="math-container">$A$</span> is <span class="math-container">$pred_<(a)=\{a'\in A:a'<a\}$</span> for some (any) <span class="math-container">$a\in A.$</span> ("pred" for "predecessors"). I will define <span class="math-container">$a^*=\{a\}\cup pred_<(a).$</span></p>
<p>Claim: There is no order-isomorphism of <span class="math-container">$A$</span> to any subset of any <span class="math-container">$pis$</span> of <span class="math-container">$A.$</span></p>
<p>Proof. By contradiction, suppose <span class="math-container">$a\in A$</span> and <span class="math-container">$f:A\to C$</span> is an order-isomorphism with <span class="math-container">$C\subset pred_<(a).$</span> Then <span class="math-container">$f$</span> maps <span class="math-container">$a^*$</span> order-isomorphically to a subset of <span class="math-container">$pred_<(a).$</span> </p>
<p>So consider <span class="math-container">$a_0,$</span> the <span class="math-container">$<$</span>-least <span class="math-container">$a\in A$</span> such that there is an order-isomorphism of <span class="math-container">$a^*$</span> to a subset of <span class="math-container">$pred_<(a),$</span> and let <span class="math-container">$f_0$</span> be an order-isomorphism from <span class="math-container">$a_0^*$</span> to a subset of <span class="math-container">$pred_<(a_0).$</span></p>
<p>By the minimality of <span class="math-container">$a_0,$</span> if <span class="math-container">$b<a_0$</span> then <span class="math-container">$\{f_0(c):c\in b^*\}$</span> cannot be a subset of <span class="math-container">$pred_<(b)$</span> so <span class="math-container">$\exists c\le b\,(b\le f_0(c)).$</span> But <span class="math-container">$c\le b<a_0$</span> implies <span class="math-container">$f_0(c)\le f_0(b) .$</span> </p>
<p>So for any <span class="math-container">$b<a_0$</span> we have <span class="math-container">$b\le f_0(b).$</span></p>
<p>BUT now in the case <span class="math-container">$b=f_0(a)$</span> we have <span class="math-container">$[\,a_0>b\land f(a_0)=b\le f(b)\,],$</span> so <span class="math-container">$f$</span> is not an order-isomorphism, a contradiction. </p>
<p>I think the rest of the Q has been fully addressed in other Answer(s) and Comments.</p>
|
3,969,679 | <p><strong>The letters in the word GUMTREE and KOALA are rearranged to form a 12-letter word where KOALA appears precisely in order but not necessarily together. How many ways can this happen?</strong></p>
<p>So I attmepted it via this method:</p>
<p>Firstly arrange like this since KOALA must be in order but not necessarily together: (let the fullstops (.) be spots for the letters in GUMTREE. There are <span class="math-container">$7$</span> letters in GUMTREE and <span class="math-container">$6$</span> full stops so <span class="math-container">$6^7$</span>.</p>
<p>.K.O.A.L.A.</p>
<p>Butthere are two E's so <span class="math-container">$\frac{6^7}{2!}$</span>. The letters in KOALA are fixed so they have <span class="math-container">$1$</span> way each, except for A (there are two so the first A has two choices and the second A has one choice)</p>
<p>Therefore, <span class="math-container">$$2\cdot\frac{6^7}{2!}=279936$$</span></p>
<p>But the answer is 1995840 arrangements</p>
<p>I belive my method is very close but I am forgetting to multiply by something. Can someone point out my logical flaw? Otherwise the worked solutions propose <span class="math-container">$\frac{12!}{5!2!}$</span>, but I don't get why you divide by 5! for the KOALA, since they are not identical letters... regardless it would be great to understand both ways!</p>
<p>Thanks</p>
| V.G | 746,921 | <p>As you said, we have to fill <span class="math-container">$6$</span> gaps in <span class="math-container">$\text{_K_O_A_L_A_}$</span> with <span class="math-container">$7$</span> letters, so if <span class="math-container">$a$</span> letters go in the first gap, <span class="math-container">$b$</span> letters go in the second gap and so on and <span class="math-container">$f$</span> letters go in the sixth gap, we have the equation <span class="math-container">$a+b+c+d+e+f=7$</span> whose solution by <a href="https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)" rel="nofollow noreferrer">stars and bars</a> is given by <span class="math-container">$\displaystyle \binom{7+6-1}{6-1}=\binom{12}{5}$</span>. The number of ways of arranging those <span class="math-container">$7$</span> letters as shown by other answers is <span class="math-container">$\dfrac{7!}{2!}$</span>. Hence, the number of total combinations is <span class="math-container">$\displaystyle \dfrac{7!}{2!}\binom{12}{5}$</span>.</p>
|
758,158 | <p>I am trying to use <span class="math-container">$f(x)=x^3$</span> as a counterexample to the following statement. </p>
<p>If <span class="math-container">$f(x)$</span> is strictly increasing over <span class="math-container">$[a,b]$</span> then for any <span class="math-container">$x\in (a,b), f'(x)>0$</span>. </p>
<p>But how can I show that <span class="math-container">$f(x)=x^3$</span> is strictly increasing?</p>
| Adam Rubinson | 29,156 | <p>Suppose <span class="math-container">$a<b$</span>. We want to prove that <span class="math-container">$a^3 < b^3$</span>.</p>
<p>Well, <span class="math-container">$a<b, \ \therefore \exists \delta>0$</span> such that <span class="math-container">$b = a+\delta$</span>. So:</p>
<p><span class="math-container">$b^3 = (a+ \delta)^3 = a^3 + 3a^2\delta + 3\delta^2a + \delta^3 = a^3 + (3a\delta(a+\delta)+\delta^3)$</span>.</p>
<p>Now we consider five different cases:</p>
<ol>
<li><span class="math-container">$\ a<-\delta$</span></li>
<li><span class="math-container">$\ a = -\delta$</span></li>
<li><span class="math-container">$\ -\delta<a<0$</span></li>
<li><span class="math-container">$\ a=0$</span></li>
<li><span class="math-container">$\ a > 0$</span>.</li>
</ol>
<p>In cases <span class="math-container">$1, 2, 4$</span> and <span class="math-container">$5,$</span> it is fairly easy to see that <span class="math-container">$3a\delta(a+\delta)+\delta^3 > 0,\ $</span> so all that remains is case <span class="math-container">$3$</span>:</p>
<p><span class="math-container">$-\delta<a<0 \implies \exists k$</span> with <span class="math-container">$0<k<1,\ $</span> such that <span class="math-container">$a = -k\delta.\ $</span> Then, <span class="math-container">$\ a\delta(a+\delta) = -k\delta^2(1-k)\delta = -k(1-k)\delta^3$</span>, and using the fact that <span class="math-container">$k(1-k) \leq \frac{1}{4},\ $</span> we see that <span class="math-container">$-k(1-k)\delta^3 \geq -\frac{1}{4}\delta^3,\ $</span> and so <span class="math-container">$\ 3a\delta(a+\delta)+\delta^3 \geq (-\frac{3}{4}\delta^3 + \delta^3) = \frac{1}{4}\delta^3 > 0$</span>.</p>
|
2,362,500 | <p>When you are solving an integral, and you have a square root of a square, for instance</p>
<p>$$ \int \sqrt{1+\cos x} = \int \sqrt{\frac{\sin^2x}{1-\cos x}}$$</p>
<p>Do I have to take the absolute value of $\sin x$, or can I just take the positive value? </p>
<p>$$= \int \frac{|\sin x|}{\sqrt{1-\cos x}}$$ OR
$$= \int \frac{\sin x}{\sqrt{1-\cos x}}$$</p>
<p>If I have to take the absolute value, should I look for a better method when solving integrals than that one, because two solutions seem more complicated.</p>
| Tolure | 239,222 | <p>Again another way.
Since both conditions are independent of each other I did it in two steps.
The first I calculated the number of combinations for the first condition (Either A and B are on the team or they are not).</p>
<p><span class="math-container">$\;\;\displaystyle \binom{7}{3}+\binom{7}{5}= 56$</span></p>
<p>From here I broke down each term for the next condition where 1 choice excludes 1 other choice. This is equal to the combinations of not choosing either + the combination of choosing one of the two.</p>
<p><span class="math-container">$\;\;\displaystyle (\binom{5}{3}+\binom{2}{1}\binom{5}{2}) + (\binom{5}{5}+\binom{2}{1}\binom{5}{4})= 41$</span></p>
|
1,912,660 | <p>I am trying to really understand why the gradient of a function gives the direction of steepest ascent intuitively.</p>
<p>Assuming that the function is differentiable at the point in question,<br>
a) I had a look at a few resources online and also looked at this
<a href="https://math.stackexchange.com/questions/223252/why-is-gradient-the-direction-of-steepest-ascent">Why is gradient the direction of steepest ascent?</a> , a popular question on this stackexchange site.<br>
The accepted answer basically says that we multiply the gradient with an arbitrary vector and then say that the product is maximum when the vector points in the same direction as the gradient? This to me really does not answer the question, but it has 31 upvotes so can someone please point out what I am obviously missing?</p>
<p>b) Does the gradient of a function tell us a way to reach the maxima or minima? if yes, then how and which one - maxima or minima or both?<br>
Edit: I read the gradient descent algorithm and that answers this part of my question.</p>
<p>c) Since gradient is a feature of the function at some particular point - am I right in assuming that it can only point to the local maxima or minima?</p>
| Doug M | 317,162 | <p>I had first learned it as if $f(x,y,z) = k$ is a surface $\nabla f$ is a vector perpendicular to the surface. </p>
<p>i.e. the plane tangent to the surface at $\mathbf x = (x_1,y_1,z_1)$ is$\frac {\partial f}{\partial x}(\mathbf x) (x-x_1) + \frac {\partial f}{\partial y}(\mathbf x) (y-y_1) + \frac {\partial f}{\partial z}(\mathbf x)(z - z_1) = 0$</p>
<p>And $(\frac{\partial f}{\partial x}(\mathbf x), \frac{\partial f}{\partial y}(\mathbf x),\frac {\partial f}{\partial z}(\mathbf x))$ is normal to the plane.</p>
<p>$\nabla f$ is a vector perpendicular to the surface when $k$ is fixed. Now we allow $k$ some freedom, and we want to move in the direction of greatest change. Whatever direction we go has a component perpendicular to the surface, and a component parallel to the surface. If we move parallel to the surface we are not contributing to a change in $k.$ The direction of maximal change is $100%$ perpendicular to the surface.</p>
<p>If that intuition isn't working for you. The we are back to the answer you found less than satisfying.</p>
<p>$\frac {\partial f}{\partial x}$ is the change in $f$ for a change in $x.$</p>
<p>For any unit vector $u,$ $\nabla f \cdot u$ would be the change in $f$ for a change in direction $u.$</p>
<p>And we want find $u$ that maximizes
$\nabla f \cdot u = \|\nabla f\| cos\theta$</p>
<p>Which will be maximal when $\theta = 0$, or when $u$ points in the same direction as $\nabla f$</p>
<p>Does $\nabla f$ tell us the direction of steepest decent, too? It certainly does. Straight in the opposite direction.</p>
<p>$\nabla f$ does not necessarily point directly toward the local maxima or minima. It points in the direction of greatest change. If you imagine yourself climbing a hill. Straight up the hill is not necessarily the direction the peak of the mountain. You may get up the steep part and then be making a turn. </p>
|
1,912,660 | <p>I am trying to really understand why the gradient of a function gives the direction of steepest ascent intuitively.</p>
<p>Assuming that the function is differentiable at the point in question,<br>
a) I had a look at a few resources online and also looked at this
<a href="https://math.stackexchange.com/questions/223252/why-is-gradient-the-direction-of-steepest-ascent">Why is gradient the direction of steepest ascent?</a> , a popular question on this stackexchange site.<br>
The accepted answer basically says that we multiply the gradient with an arbitrary vector and then say that the product is maximum when the vector points in the same direction as the gradient? This to me really does not answer the question, but it has 31 upvotes so can someone please point out what I am obviously missing?</p>
<p>b) Does the gradient of a function tell us a way to reach the maxima or minima? if yes, then how and which one - maxima or minima or both?<br>
Edit: I read the gradient descent algorithm and that answers this part of my question.</p>
<p>c) Since gradient is a feature of the function at some particular point - am I right in assuming that it can only point to the local maxima or minima?</p>
| Matthew Leingang | 2,785 | <p>I think something missing in the accepted answer to the linked question is the connection between gradient and directional derivative. For a function $f(x,y)$ defined at $P=(x_0,y_0)$, and unit a vector $\mathbf{u}=\left<a,b\right>$, we define the <strong>directional derivative of $f$ in the direction $\mathbf{u}$ at $P$</strong> to be
$$
D_{\mathbf{u}}f(P) = \lim_{t\to 0} \frac{f(x_0 + ta,y_0+tb)-f(x_0,y_0)}{t}
$$
I've written it in two variables but the notion carries to any number of variables.</p>
<p>So the question “What is the direction of steepest ascent of $f$ at $P$?” can be translated to: “For which $\mathbf{u}$ is $D_{\mathbf{u}}f(P)$ maximized?”</p>
<p>It turns out there is an easy way to compute $D_{\mathbf{u}}f(P)$ without taking the limit. If you let $g(t) = f(x_0 + ta,y_0 + tb)$, then $D_{\mathbf{u}}f(P)=g'(0)$. And by the Chain Rule,
\begin{align*}
g'(t) &=\frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} =\frac{\partial f}{\partial x} a + \frac{\partial f}{\partial y} b\\
\implies g'(0)&= \left<\frac{\partial f}{\partial x}(x_0,y_0),\frac{\partial f}{\partial y}(x_0,y_0)\right>\cdot \left<a,b\right>
= \nabla f(x_0,y_0) \cdot \left<a,b\right>
\end{align*}
To summarize:
$$
D_{\mathbf{u}}f(P) = \nabla f(P) \cdot \mathbf{u}
$$</p>
<p>What was the question again? That's right: for which $\mathbf{u}$ is $D_{\mathbf{u}}f(P)$ the greatest? Using the identity we just proved, and the law of cosines/dot product formula, we know:
$$
D_{\mathbf{u}}f(P) = \left\Vert \nabla f(P) \right\Vert \left\Vert \mathbf{u} \right\Vert \cos\theta = \left\Vert \nabla f(P) \right\Vert \cos\theta
$$
where $\theta$ is the measure of the angle between the vectors $\nabla f(P)$ and $\mathbf{u}$. Remember that $\mathbf{u}$ is a <em>unit</em> vector (otherwise $D_{\mathbf{u}}f(P)$ would depend on the length of $\mathbf{u}$, and we really only want it to measure $f$), so $\left\Vert \mathbf{u} \right\Vert=1$. </p>
<p>Now $\cos\theta$ has maximum value $1$, achieved when $\theta = 0$. If the angle between $\nabla f(P)$ and $\mathbf{u}$ has zero measure, the vectors are actually pointing in the same direction. That is: the maximum value of $D_{\mathbf{u}}f(P)$ is $\left\Vert \nabla f(P) \right\Vert$, achieved when $\mathbf{u}$ points in the same direction as $\nabla f(P) $.</p>
|
1,506,805 | <p>$\lim _{x\to 0}\left(\frac{\cos\left(x+\frac{\pi }{2}\right)}{x}\right)\:$</p>
| Jan Eerland | 226,665 | <p>$$\lim _{x\to 0}\left(\frac{\cos\left(x+\frac{\pi}{2}\right)}{x}\right)=$$
$$\lim _{x\to 0}\left(-\frac{\sin(x)}{x}\right)=$$
$$-\left(\lim _{x\to 0}\frac{\sin(x)}{x}\right)=$$
$$-\left(\lim _{x\to 0}\frac{\frac{\text{d}}{\text{d}x}\sin(x)}{\frac{\text{d}}{\text{d}x}x}\right)=$$
$$-\left(\lim _{x\to 0}\frac{\cos(x)}{1}\right)=$$
$$-\left(\lim _{x\to 0}\cos(x)\right)=$$
$$-\left(\cos(0)\right)=$$
$$-\left(1\right)=-1$$</p>
|
3,758 | <p>When implicitly finding the derivative of: </p>
<blockquote>
<p>$xy^3 - xy^3\sin(x) = 1$</p>
</blockquote>
<p>How do you find the implicit derivative of:</p>
<blockquote>
<p>$xy^3\sin(x)$</p>
</blockquote>
<p>Is it using a <em>triple</em> product rule of sorts?</p>
| Bill Dubuque | 242 | <p><strong>HINT</strong> $\;$ <a href="http://en.wikipedia.org/wiki/Logarithmic_differentiation" rel="nofollow"><strong>logarithmic differentiation</strong></a> makes the n-ary generalization obvious:</p>
<p>$$\rm (abc)'= abc \; log(abc)' = abc \;(log\; a + log\; b + log\; c)' = abc \; \bigg(\frac{a'}{a} + \frac{b'}{b} + \frac{c'}{c}\bigg) $$</p>
<p>Obviously the same proof works for arbitrary length products yielding</p>
<p>$$\rm (abc\cdots f)' = \: abc\cdots f\;\bigg(\frac{a'}{a} + \frac{b'}{b} + \frac{c'}{c} +\:\cdots\:+ \frac{f'}{f}\bigg) $$</p>
|
300,067 | <p>They ask me to find all invertible matrices $A$ of the form: $\begin{bmatrix}a & b\\ c&d \end{bmatrix}$ and satisfying $A=A^{-1}$ and $A^t=A^{-1}$. I find that rather complex; does it have anything to do with orthogonality? Not sure. Any help please??</p>
| Ittay Weiss | 30,953 | <p>Hint: Given a 2x2 matrix, do you know how to compute its inverse? If not then figure out how to do that. Now that you do, take an arbitrary invertible 2x2 matrix (that is call its entries $a,b,c,d$), compute its inverse and solve the equations $A=A^{-1}$ and $A^t=A^{-1}$.</p>
|
763,073 | <p>I came across the symbol $|v_1 \wedge \dots \wedge v_m|^{-1}$ in a paper - this is the norm of the wedge product of vectors $v_k \in \mathbb{R}^n$ .
I thought it's meaning was self-evident until I tried to compute it. </p>
<p>First of all $|v_1 \wedge \dots \wedge v_m|$ is a <em>number</em> since we are taking the reciprocal.</p>
<p>If there were $m = n$ vectors then $|v_1 \wedge \dots \wedge v_n| = \det (v_i \cdot v_j) $ and it is the volume of the <a href="https://en.wikipedia.org/wiki/Parallelepiped" rel="nofollow">paralleliped</a> generated by the vectors $v_1, \dots, v_n$. </p>
<p><img src="https://upload.wikimedia.org/wikipedia/commons/3/3e/Parallelepiped_volume.svg" width="100"></p>
<p>If we have $m < n$ vectors, given in coordinates, we have the volume of the $m$-dimensional parallelipiped inside of $n$-dimensional space. This volume is <strong>0</strong> unless we use the $m$-dimensional volume measure.</p>
<p>In <a href="https://en.wikipedia.org/wiki/Exterior_algebra#Alternating_multilinear_forms" rel="nofollow">exterior algebra</a>, if I have the coordinates of the vectors, $v_1, \dots, v_m$, how do I compute this volume?</p>
| Umberto P. | 67,536 | <p>Write each $v_k$ as a column vector. The $m$-volume is the square root of the sum of the squares of the $m \times m$ minor determinants of the matrix $[v_1 | v_2 | \cdots | v_m]$. </p>
|
3,714,244 | <p>In a (finite horizon) sequential game of complete information, any subgame perfect Nash equilibrium (SPNE) necessarily must specify the equilibrium of the last stage game as a Nash equilibrium.</p>
<p>I am wondering if an analogous result holds for sequential games of incomplete information, that is, in a perfect Bayesian equilibrium (PBE) does the last stage game form a Bayesian Nash equilibrium (BNE)?</p>
| Community | -1 | <p>In PBE, the information sets are not subgames, and a subgame begins at a singleton information set, where the current state is common knowledge. For example, the standard signaling game with two types, two messages for the sender, and two actions for the receiver has only one subgame, the whole game. So the "last subgame" you mention in your question is not necessarily the same thing as the terminal decision nodes or terminal simultaneous move subgame in a complete information dynamic game.</p>
<p>But if you solve for a PBE and then wipe away the beliefs, the players are still mutually best-responding, so it constitutes a Bayesian Nash Eqm of the game. The problem is that, like with complete information, there might be many patterns of best responses which cannot make sense when compared across subgames. But when you solve for a PBE, usually you sketch out a set of mutual best responses and then check what the beliefs are that support those choices (or show there aren't any such beliefs). That's already a BNE for the game if there are no profitable deviations, it just might not survive the stronger criterion of Bayesian beliefs, because there is dynamically sub-optimal behavior somewhere, just as in the comparison with NE and SPNE.</p>
|
3,714,244 | <p>In a (finite horizon) sequential game of complete information, any subgame perfect Nash equilibrium (SPNE) necessarily must specify the equilibrium of the last stage game as a Nash equilibrium.</p>
<p>I am wondering if an analogous result holds for sequential games of incomplete information, that is, in a perfect Bayesian equilibrium (PBE) does the last stage game form a Bayesian Nash equilibrium (BNE)?</p>
| brunosalcedo | 161,776 | <blockquote>
<p>In a perfect Bayesian equilibrium (PBE) does the last stage game form
a Bayesian Nash equilibrium (BNE)?</p>
</blockquote>
<p>The short answer is yes, but it can be misleading. </p>
<p>First, any proper subgame of an incomplete information game will have complete information (among the players that are still making choices in the subgame). This is because the definition of a subgame does not allow you to break information sets. Therefore, in any proper subgame, BNE and NE are the same thing. </p>
<p>Second, you cannot simply define a PBE as a strategy profile that induces a BNE in every subgame. The reason is that players make choices at each information set and many information sets will not start a new subgame (as Renar pointed out in their answer). PBE must induce rational choices at all information sets, even those that do not start a new subgame. </p>
|
3,557,398 | <p>Here are two second-order differential equations.</p>
<p><span class="math-container">$$ y''+9y=\sin(2t) \tag 1 $$</span></p>
<p><span class="math-container">$$ y'' +4y =\sin(2t) \tag 2 $$</span></p>
<p>I am told to use undetermine coefficients method to solve.</p>
<p>For 1), I use <span class="math-container">$y_p=A \cos(2t)+B \sin(2t)$</span> to get <span class="math-container">$A=0$</span> and B=<span class="math-container">$\frac{1}{5}$</span> and get <span class="math-container">$y_p=\frac{1}{5} \sin(2t)$</span></p>
<p>For 2), I realize that that method doesn't work and told to do <span class="math-container">$y_p=t(A \cos(2t)+B \sin(2t)$</span> Why does it work then?</p>
| RobPratt | 683,666 | <p>Yes, it is true, and the property is called <a href="https://en.m.wikipedia.org/wiki/Subadditivity" rel="nofollow noreferrer"><em>subadditivity</em></a>.</p>
|
1,799,559 | <blockquote>
<p>Let $f:[a,b]\to\mathbb{R}$ be a continious function. Show that if $$\int_a^b f(x)g(x)dx=0$$
for all continious functions $g:[a,b]\to\mathbb{R}$ with $g(a)=g(b)=0$, then $f(x)=0$ $\forall x\in[a,b]$</p>
</blockquote>
<p>I have difficulties proving this question. Consider for example $g(x)=0$ $\forall x\in[a,b]$, then assumptions still hold but $f(x)$ can be anything (f.i. $f(x)=1\neq0$). Could anyone tell me if this reasoning is correct? </p>
| Patrick Stevens | 259,262 | <p>Your reasoning is wrong: the hypothesis asks that the integral be zero for <em>all</em> continuous $g: [a,b] \to \mathbb{R}$ with $g(a) = g(b) = 0$, not just for the particular choice $g = 0$.</p>
<p>So $f = 1$ is invalidated because of the function $g$ which is a nonlinear quadratic which is $0$ at both $a$ and $b$.</p>
|
312,012 | <p>What is the exact difference between $\arg\max$ and $\max$ of a function?</p>
<p>Is it right to say the following?</p>
<blockquote>
<p>$\arg\max f(x)$ is nothing but the value of $x$ for which the value of the function is maximal. And $\max f(x)$ means value of $f(x)$ for which it is maximum.</p>
</blockquote>
<p>More precisely,</p>
<blockquote>
<p>$\arg \max$ returns a value from the domain of the function and $\max$ returns from the range of the function?</p>
</blockquote>
| Community | -1 | <p>Yes, you are right. $\max f(x)$ is the maximum value (if it exists) of $f(x)$ as $x$ varies through some domain, while $\arg\max f(x)$ is the value of $x$ at which this maximum is attained.</p>
<p>However, there could be more than one $x$ value which gives rise to the maximum $f(x)$, in which case $\arg\max f(x)$ would be this set of values of $x$ instead.</p>
|
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