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 |
|---|---|---|---|---|
788,671 | <p>What is the ratio of the area of a triangle $ABC$ to the area of the triangle whose sides are equal in length to the medians of triangle $ABC$?</p>
<p>I see an obvious method of brute-force wherein I can impose a coordinate system onto the figure. But is there a better solution?</p>
| Quanto | 686,284 | <p><a href="https://i.stack.imgur.com/FIXn3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FIXn3.png" alt="enter image description here"></a></p>
<p>Let <span class="math-container">$X$</span>, <span class="math-container">$Y$</span> and <span class="math-container">$Z$</span> be the side midpoints. Construct the parallelogram PYBZ and connect PC.</p>
<p>By construction, PY = BZ = AZ, PY || AZ and then PYZA is also a parallelogram, which leads to AP || ZY || XC and AP = XC, and in turn the parallelogram APCX. Thus, PC = AX and the sides of the triangle PCZ are the medians of ABC.</p>
<p>The parallelogram PYZA also yields AQ = QY = <span class="math-container">$\frac14$</span>AC and then</p>
<p><span class="math-container">$$\frac{Area_{PCZ}}{Area_{ABC}}=\frac{Area_{QCZ}}{Area_{AZC}}=\frac{QC}{AC} = \frac34
$$</span></p>
|
75,925 | <p>I hope this question is focused enough – it's not about real problem I have, but to find out if anyone knows about a similar thing.</p>
<p>You probably know the <a href="https://en.wikipedia.org/wiki/Uncertainty_principle" rel="nofollow noreferrer">Heisenberg uncertainty principle</a>: For any function <span class="math-container">$g\in L^2(\mathbb{R})$</span> for which the respective expressions exist it holds that
<span class="math-container">$$
\frac{1}{4}\|g\|_2^4 \leq \int_{\mathbb{R}} |x|^2 |g(x)|^2 \,dx \int_{\mathbb{R}} |g'(x)|^2 \,dx.
$$</span></p>
<p>This inequality is not only important in quantum mechanics, but also in signal processing for the short-time Fourier transform, see <a href="https://en.wikipedia.org/wiki/Fourier_transform#Uncertainty_principle" rel="nofollow noreferrer">here</a>.</p>
<p>One can derive this by formally using integration by parts
<span class="math-container">$$
\int_{\mathbb{R}} 1\,|g(x)|^2\, dx = -\int_{\mathbb{R}} x\tfrac{d}{dx}|g(x)|^2\,dx \leq 2\int_{\mathbb{R}} |xg(x)|\,|g'(x)|\,dx
$$</span>
and Cauchy–Schwarz.</p>
<p>Now, changing just the order of the functions, you obtain this inequality
<span class="math-container">$$
\int_{\mathbb{R}} |g(x)|^2 \, dx \leq 2\int_{\mathbb{R}} |xg'(x)|\,|g(x)| \, dx
\leq \left(\int_{\mathbb{R}} |xg'(x)|^2 \, dx\right)^{1/2} \left(\int_{\mathbb{R}} |g(x)|^2 \, dx\right)^{1/2}
$$</span>
which gives
<span class="math-container">$$
\|g\|_2\leq \|xg'\|_2.
$$</span></p>
<p>Ok, this was just playing around. However, this inequality can also be motivated by an abstract consideration about uncertainty principle associated to group-related integral transforms (see my <a href="https://regularize.wordpress.com/2011/09/16/the-uncertainty-principle-for-the-windowed-fourier-transform/" rel="nofollow noreferrer">two</a> <a href="https://regularize.wordpress.com/2011/09/20/the-uncertainty-principle-for-the-one-dimensional-wavelet-transform/" rel="nofollow noreferrer">blog
posts</a>). Interestingly, the Heisenberg uncertainty principle derives from the short time Fourier transform and the last "uncertainty principle" derives from the wavelet transform.</p>
<p>The last fact bothers me: In contrast to the fact that both inequalities can be derived from two conceptually very different integral transforms (indeed both underlying groups are very different), they have a very similar formal derivation.</p>
<p>I have the following questions: Is anyone familiar with the last inequality? Could it be useful in any context? Is there some reason why these inequalities seem so entangled?</p>
| Antoine Levitt | 19,334 | <p>I find the neatest "standard" uncertainty principle is the one with commutators, see e.g. <a href="http://galileo.phys.virginia.edu/classes/751.mf1i.fall02/GenUncertPrinciple.htm" rel="nofollow">http://galileo.phys.virginia.edu/classes/751.mf1i.fall02/GenUncertPrinciple.htm</a>. I think that readily gives both your inequalities.</p>
|
694,279 | <p>I am learning convex analysis by myself and I need help.</p>
<p>How to show that if $X=U=\mathbb{R}$
and $f\left(x\right)=\frac{|x|^{p}}{p}$
then the convex conjugate $f^{*}\left(u\right)=\frac{|u|^{q}}{q}$
when $\frac{1}{p}+\frac{1}{q}=1$?
There exists a particular technique that I have to apply in order to compute the convex conjugate? </p>
| user127096 | 127,096 | <p>For practical computation I would use the fact that $\nabla f^*$ is the inverse of $\nabla f$ (<a href="http://en.wikipedia.org/wiki/Convex_conjugate#Maximizing_argument" rel="nofollow">see here</a>). By the chain rule, $$\nabla f(x) =|x|^{p-1} \nabla |x| = |x|^{p-1} \frac{x}{|x|}$$ which means the direction of $x$ stays the same but its length is raised to power $p-1$. The inverse of this map is $$u\mapsto |u|^{1/(p-1)} \frac{u}{|u|} = |u|^{q-1} \frac{u}{|u|}$$ Observing the similarity of two formulas, we arrive at $f^*(u)=|u|^q/q$.</p>
|
4,375,994 | <blockquote>
<p>Question:</p>
<p>Show that, <span class="math-container">$$\pi =3\arccos(\frac{5}{\sqrt{28}}) +
3\arctan(\frac{\sqrt{3}}{2}) ~~~~~~ (*)$$</span></p>
</blockquote>
<p><em>My proof method for this question has received mixed responses. Some people say it's fine, others say that it is a verification, instead of a proof.</em></p>
<p>Proof: <span class="math-container">$$\pi =3\arccos(\frac{5}{\sqrt{28}}) + 3\arctan(\frac{\sqrt{3}}{2})\iff \frac{\pi}{3} = \arccos(\frac{5}{\sqrt{28}}) + \arctan(\frac{\sqrt{3}}{2})
$$</span><span class="math-container">$$\iff \frac{\pi}{3} = \arctan(\frac{\sqrt{3}}{5})+\arctan(\frac{\sqrt{3}}{2})$$</span></p>
<p>As <span class="math-container">$\arccos(\frac{5}{\sqrt{28}})=\arctan(\frac{\sqrt{3}}{5})$</span></p>
<p>The plan now is to apply the tangent function to both sides, and show that LHS=RHS using the tangent addition formula to expand it out.</p>
<p>I.e. <span class="math-container">$$\tan(\frac{\pi}{3}) = \tan\bigg(\arctan(\frac{\sqrt{3}}{5})+\arctan(\frac{\sqrt{3}}{2}\bigg)$$</span></p>
<p><span class="math-container">$$\iff \sqrt{3} = \frac{\frac{\sqrt{3}}{5}+\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{5} \frac{\sqrt{3}}{2}}$$</span></p>
<p>and the RHS will reduce down to <span class="math-container">$\sqrt{3}$</span>. Hence LHS=RHS.</p>
<p>Some things that I've noticed about this method of proof:</p>
<ul>
<li>It could be used to (incorrectly) prove that <span class="math-container">$$\frac{\pi}{3}+\pi = \arccos(\frac{5}{\sqrt{28}}) + \arctan(\frac{\sqrt{3}}{2})$$</span></li>
</ul>
<p>So because this method of proof can be used to prove things true, that are obviously false, that means it can't be used?</p>
<ul>
<li>Instead of proving (*), wouldn't this method of proof actually prove that? <span class="math-container">$$\arccos(\frac{5}{\sqrt{28}})+\arctan(\frac{\sqrt{3}}{2})=\frac{\pi}{3} + \pi k$$</span></li>
</ul>
<p>for some <span class="math-container">$k\in \mathbb{Z}$</span> which we must find. In this case being when <span class="math-container">$k=0$</span>.</p>
| user2661923 | 464,411 | <p>To show:</p>
<p><span class="math-container">$$\text{arctan}\left(\frac{\sqrt{3}}{5}\right) ~+~ \text{arctan}\left(\frac{\sqrt{3}}{2}\right) ~~<~~ \frac{\pi}{2}.$$</span></p>
<p>In fact this conclusion is immediate by the following analysis.</p>
<p>Let <span class="math-container">$~\displaystyle \theta ~~=~~ \text{arctan}\left(\frac{\sqrt{3}}{5}\right) ~+~ \text{arctan}\left(\frac{\sqrt{3}}{2}\right).$</span></p>
<p>Then,</p>
<p><span class="math-container">$$\tan(\theta) =
\frac{\frac{\sqrt{3}}{5} + \frac{\sqrt{3}}{2}}{1 - \left[\frac{\sqrt{3}}{5} \times \frac{\sqrt{3}}{2}\right]}. \tag1 $$</span></p>
<p>In (1) above, the numerator is clearly positive.</p>
<p>Further, since <span class="math-container">$~\displaystyle \left[\frac{\sqrt{3}}{5} \times \frac{\sqrt{3}}{2}\right] = \frac{9}{10} < 1$</span>,</p>
<p>the denominator in (1) above is also clearly positive.</p>
<p>Therefore, since <span class="math-container">$\tan(\theta) > 0$</span>, either <span class="math-container">$\theta$</span> is in the 1st quadrant, or <span class="math-container">$\theta$</span> is in the 3rd quadrant.</p>
<p>However, by the definition of the arctan function, <span class="math-container">$\theta$</span> is the sum of two angles, each of which are between <span class="math-container">$(0)$</span> and <span class="math-container">$(\pi/2)$</span>. Therefore, <span class="math-container">$\theta$</span> can not be in the 3rd quadrant. Therefore, <span class="math-container">$\theta$</span> must be in the first quadrant.</p>
<hr />
<p><strong>Edit</strong><br>
Thanks to ryang for pointing out a couple of analytical gaps that need to be filled:</p>
<ul>
<li><p>By definition, <span class="math-container">$\text{arccos}\left(\frac{5}{\sqrt{28}}\right)$</span> is in the 1st quadrant. Similarly <span class="math-container">$\text{arctan}\left(\frac{3}{\sqrt{2}}\right)$</span> is also in the 1st quadrant. Therefore, <span class="math-container">$\theta$</span> is in fact the sum of <span class="math-container">$2$</span> angles, each of which are in the first quadrant.</p>
</li>
<li><p>In general <span class="math-container">$\tan(\theta) = \tan(\pi/3)$</span> does not imply that <span class="math-container">$\theta = (\pi/3)$</span>. However, if you add the constraint that <span class="math-container">$\theta$</span> is in the 1st quadrant, then the implication holds. Since I deduced that <span class="math-container">$\theta$</span> is in the 1st quadrant, the analysis holds.</p>
</li>
</ul>
|
4,601,727 | <p>I'm aware that the title might be a bit off, I am unsure on how to describe this.</p>
<p>For <span class="math-container">$n\in \mathbb{N}$</span>, define <span class="math-container">$n+1$</span> independent random variables <span class="math-container">$X_0, \ldots , X_n$</span> which are uniformly distributed over the interval <span class="math-container">$[0,1]$</span>, We focus on this group:<span class="math-container">$$S=\{X_i|i\geq 1 , X_i<X_0\}$$</span>
For all <span class="math-container">$0\leq k\leq n$</span>, show that: <span class="math-container">$$P(|S|=k)=\int_0^1 {n\choose k}\cdot x^k\cdot (1-x)^{n-k}$$</span></p>
<p>I've reduced this to the following:
<span class="math-container">$$P(|S|=k)=P(exactly \ k \ elements\ are\ bigger\ than\ X_0)$$</span>
Due to independence, we can write:
<span class="math-container">$$=P(X_0 <X_i)\cdot \ \ldots \ \cdot P(X_0<X_{i+k}) , i\in\{1, \ldots ,n-k\}$$</span></p>
<p>I'm stuck here, cant find how to calculate <span class="math-container">$P(X_0<X_i)$</span>, which prevents me from proving the statement.</p>
<p>I'm aware that <span class="math-container">$n\choose k$</span> is because we're 'checking' every k-sized group out of the n available R.Vs,</p>
<p><span class="math-container">$x^k%$</span> which translates to <span class="math-container">$P(X_0<X_{i\rightarrow (i+k)})$</span> and <span class="math-container">$(1-x)^{n-k}$</span> to 'disable' the other R.Vs from being bigger than <span class="math-container">$X_0$</span></p>
| Aishgadol | 861,510 | <p>The answer I've been able to compute, using the following formula:
<span class="math-container">$$ \mathbb{P}(X\in A)=\int_{-\infty}^{\infty}\mathbb{P}(X\in A |Y=y)f_Y(y)dy$$</span></p>
<p>We know that the boundaries are <span class="math-container">$[0,1]$</span>.</p>
<p>Therefor, We're looking to find <span class="math-container">$\int_{0}^{1}\mathbb{P}(|S|=k|X_0=x)\cdot f_{X_0}(x)dx$</span></p>
<p>Since the R.Vs are independent, we're interested in <span class="math-container">$(\mathbb{P}(X_0>X_i))^k\cdot (\mathbb{P}(X_0\leq X_j))^{n-k}$</span> where <span class="math-container">$i,j$</span> dont really mean anything, since all RVs have similar PDF/CDFs.</p>
<p>Since <span class="math-container">$\mathbb{P}$</span> is calculated using the CDF, and we're looking to check every k-sized subset of <span class="math-container">$X_1, \ldots , X_n$</span>, we'll multiply the above in <span class="math-container">$n\choose k$</span> to get <span class="math-container">$${n\choose k}\cdot (\mathbb{P}(X_0>X_i))^k\cdot (\mathbb{P}(X_0\leq X_j))^{n-k}$$</span></p>
<p>We can rewrite that as:<span class="math-container">$${n\choose k}\cdot (\mathbb{P}(x_0>X_i))^k\cdot (\mathbb{P}(x_0\leq X_j))^{n-k}\cdot f_{X_0}(x_0)$$</span>
We note that <span class="math-container">$f_{X_0}(x_0)=\frac{1}{1-0}=1$</span>, now we need to integrate in order to get the CDF, which gets us to:<span class="math-container">$$\int_{0}^{1}{n\choose k}\cdot (\mathbb{P}(x_0>X_i))^k\cdot (\mathbb{P}(x_0\leq X_j))^{n-k}dx_0$$</span>
<span class="math-container">$\mathbb{P}(x_0>X_i) $</span> is denoted by <span class="math-container">$\frac{x-a}{b-a}=\frac{x}{1}=x$</span>, and naturally the complement would be <span class="math-container">$\mathbb{P}(x_0\leq X_i)=1-\mathbb{P}(x>X_i)=1-x$</span>. we plug these into the above equation, to recieve:
<span class="math-container">$$\int_{0}^{1}{n\choose k}\cdot x^k\cdot (1-x)^{n-k}$$</span>
Even though this answer was not thought of by myself, I tried my best to understand and implement this, if there are any mistakes in my work, please point them out.
Thank you for all your assistance!</p>
|
18,459 | <p>Does anyone know of any studies or have personal experience dealing with difficulties (if any) faced by students studying mathematics if they come from countries which use languages written from right-to-left or top-down? </p>
<p>I have been wondering about this recently because I have been working on supplemental study guides. It occured to me that mathematics covering everything from arithmetic to calculus is heavily viewed in terms of a left-to-right perspective (e.g. digits with the highest place value are the leftmost ones in a numeral and are thus read first; the positive x-axis extends to the right; a line with positive slope is seen as one which goes up when viewed from left-to-right)</p>
<p>Then again there are many accomplished mathematicians who come from countries which speak and write these languages so maybe it's not that difficult of a conceptual jump after all. I think I for one would have difficulty though...</p>
| IrbidMath | 5,993 | <p>Am Arabian Mathematician. In my country in all school we used to write math from right to left i scored very well in 12th grade 179/200. But in the universities we use English books so we write from left to right it was hard for me to get through that in the first course I scored 65/100 by the time I used to that and now am PhD student in the US. So my opinion it is not that hard by practice 100 or more problems normal student should be fine.</p>
|
3,378,004 | <p>If <span class="math-container">$H$</span> and <span class="math-container">$K$</span> are abelian subgroups of a group <span class="math-container">$G$</span>, then <span class="math-container">$H\cap K$</span> is a normal subgroup of <span class="math-container">$\left\langle H\cup K\right\rangle$</span>.</p>
<p>I proved <span class="math-container">$H\cap K$</span> is a subgroup and need to prove it is normal subgroup of <span class="math-container">$\left\langle H\cup K\right\rangle$</span>. But isn't it obvious that <span class="math-container">$H\cap K$</span> is normal? All the elements in <span class="math-container">$H\cap K$</span> communicates with all the elements in <span class="math-container">$H$</span> and <span class="math-container">$K$</span>, hence <span class="math-container">$H\cap K$</span> is normal. Am I missing something?</p>
| José Carlos Santos | 446,262 | <p>Yes, every element of <span class="math-container">$H\cap K$</span> commutes with every element of <span class="math-container">$H$</span> and with every element of <span class="math-container">$K$</span>. In my opinion, you should add a proof of that fact that it follows from this that every element of <span class="math-container">$H\cap K$</span> commutes with every element of <span class="math-container">$\langle H\cup K\rangle$</span>. You can say, for instance, that every element of <span class="math-container">$\langle H\cup K\rangle$</span> can be written as a product of elements of <span class="math-container">$H\cup K$</span> and then use that fact.</p>
|
153,902 | <p>Let $A_i$ be open subsets of $\Omega$. Then $A_0 \cap A_1$ and $A_0 \cup A_1$ are open sets as well.</p>
<p>Thereby follows, that also $\bigcap_{i=1}^N A_i$ and $\bigcup_{i=1}^N A_i$ are open sets.</p>
<p>My question is, does thereby follow that $\bigcap_{i \in \mathbb{N}} A_i$ and $\bigcup_{i \in \mathbb{N}} A_i$ are open sets as well?</p>
<p>And what about $\bigcap_{i \in I} A_i$ and $\bigcup_{i \in I} A_i$ for uncountabe $I$?</p>
| Davide Giraudo | 9,849 | <p>An arbitrary union (coutable or not) of open sets is open, but even for a countable intersection it's not true in general. For example, when $\Omega$ is the real line endowed with the usual topology, and $A_i:=\left(-\frac 1i,\frac 1i\right)$, $A_i$ is open but $\bigcap_{i\in \Bbb N}A_i=\{0\}$ which is not open. </p>
|
158,810 | <p>Let $ (\hat i, \hat j, \hat k) $ be unit vectors in Cartesian coordinate and $ (\hat e_\rho, \hat e_\theta, \hat e_z)$ be on spherical coordinate.
Using the relation, $$ \hat e_\rho = \frac{\frac{\partial \vec r}{\partial \rho}}{ \left | \frac{\partial \vec r}{\partial \rho} \right |}, \hat e_\theta = \frac{\frac{\partial \vec r}{\partial \theta}}{ \left | \frac{\partial \vec r}{\partial \theta} \right |}, \;\; \hat e_z = \frac{\frac{\partial \vec r}{\partial z}}{ \left | \frac{\partial \vec r}{\partial z} \right |} $$
We have the relation $$\begin{bmatrix} \hat e_{\rho}\\ \hat e_{\theta}\\ \hat e_{z} \end{bmatrix} = \begin{bmatrix} \cos \phi & \sin \phi & 0\\ -\sin \phi & \cos \phi & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \hat i\\ \hat j\\ \hat k \end{bmatrix}$$
$$\text { Let } A = \begin{bmatrix} \cos \phi & \sin \phi & 0\\ -\sin \phi & \cos \phi & 0\\ 0 & 0 & 1 \end{bmatrix}$$
To express unit vectors of Cartesian coordinate in Spherical coordinates, the author uses,
$$\hat i = \frac { \begin{vmatrix} \hat e_{\rho} & \sin \phi & 0\\ \hat e_{\theta} & \cos \phi & 0\\ 0 & 0 & 1 \end{vmatrix} }{|A|} \\ \hat j = \frac { \begin{vmatrix} \cos \phi & \hat e_{\rho} & 0\\ -\sin \phi & \hat e_{\theta} & 0\\ 0 & 0 & 1 \end{vmatrix} }{|A|} \\\hat k = \frac { \begin{vmatrix} \cos \phi & \sin \phi & 0\\ -\sin \phi & \cos \phi & 0\\ 0 & 0 & \hat e_z \end{vmatrix} }{|A|}
$$
Which I cannot understand! Can anyone help me to understand it?
$$ \begin{bmatrix} \hat i\\ \hat j\\ \hat k \end{bmatrix} = \begin{bmatrix} \cos \phi & \sin \phi & 0\\ -\sin \phi & \cos \phi & 0\\ 0 & 0 & 1 \end{bmatrix}^{-1} \begin{bmatrix} \hat e_{\rho}\\ \hat e_{\theta}\\ \hat e_{z} \end{bmatrix} $$
Looks intuitive but certainly the previous way is faster. I would like to know above relation works if it works.
Thank you!!</p>
| Chloé Seppälä Zetterberg | 514,885 | <p>A far more simple method would be to use the gradient.<br/>
Lets say we want to get the unit vector $\boldsymbol { \hat e_x } $. What we then do is to take $\boldsymbol { grad(x) } $ or $\boldsymbol { ∇x } $.<br/>
This; $\boldsymbol ∇ $, is the nabla-operator. It is a vector containing each partial derivative like this...<br/><br/>
$\boldsymbol { ∇= ( \frac {∂} {∂x}, \frac {∂} {∂y}, \frac {∂} {∂z}) } $<br/><br/>
When we take the gradient of x we get this...<br/><br/>
$\boldsymbol { ∇x= ( \frac {∂x} {∂x}, \frac {∂x} {∂y}, \frac {∂x} {∂z})=(1,0,0)=\hat e_x } $<br/><br/>
To get the unit vector of $\boldsymbol x$ in cylindrical coordinate system we have to rewrite $x$ in the form of $\boldsymbol {r_c}$ and $\boldsymbol {\phi}$. <br/><br/>
$\boldsymbol {x= r_c cos(x) } $ <br/><br/>
Now you have to use the more general definition of nabla ($\boldsymbol ∇ $).<br/>
Lets say we have a curve-linear coordinate system where the position vector is defined like this... <br/><br/>
$\boldsymbol {\vec r = u_1 \hat e_{u1} + u_2 \hat e_{u2} + u_3 \hat e_{u3}} $ <br/><br/>
... Then the nabla operator for that coordinate system is as follows...<br/><br/>
$\boldsymbol { ∇ = \frac {1}{h_1} \frac {∂}{∂u_1} \hat e_{u1} + \frac {1}{h_2} \frac {∂}{∂u_2} \hat e_{u1} + \frac {1}{h_3} \frac {∂}{∂u_3} \hat e_{u1}} $<br/><br/>
"$\boldsymbol { h_n } $" is the scale factor to the variable "$\boldsymbol { u_n } $". The scale-factor is defined as: $\boldsymbol {h_n = \frac {\partial \vec r}{\partial u_n}}$<br/>
For cylindrical coordinates the position vector is defined as: $\boldsymbol {\vec r = r_c \hat e_{rc} + z \hat e_z }$<br/>
With some simple math we can get the scale factors and they are...<br/><br/>
$\boldsymbol {h_{rc} = 1 \ \ ,\ h_{\phi} = r_c \ \ ,\ h_z = 1}$<br/><br/>
We already know that in cylindircal cooridnates $\boldsymbol x $ is defined as $\boldsymbol {x = r_c cos(x)}$, so now we can get the gradient.<br/><br/>
$\boldsymbol { ∇x = ∇(r_c cos(x))= \frac {\partial (r_c \cos(x))}{\partial r_c} \hat e_{rc} + \frac {1}{r_c} \frac {\partial (r_c \cos(x))}{\partial \phi} \hat e_{\phi} + \frac {\partial (r_c cos(x))}{\partial z} \hat e_{z}}$<br/><br/>
The result from this gradient is then...<br/><br/>
$\boldsymbol {\hat e_{x} = \cos(\phi)\hat e_{rc} - \sin(\phi) \hat e_{\phi}}$<br/><br/>
When the same method is applied to $\boldsymbol y$, where $\boldsymbol {y = r_c sin(\phi) }$, we get with ease that...<br/><br/>
$\boldsymbol {\hat e_{y} = sin(\phi)\hat e_{rc} + \cos(\phi) \hat e_{\phi}}$<br/><br/>
Hope it helped!<br/>
I also hope the use of $\boldsymbol \phi $ instead of $\boldsymbol \theta $ and $\boldsymbol {r_c} $ instead of $\boldsymbol \rho $ wasn't to confusing. As a physics student I am more used to the $\boldsymbol {(r_c,\phi,z)}$ standard for cylindrical coordinates.</p>
|
46,631 | <p>I'm writing a program to play a game of <a href="http://en.wikipedia.org/wiki/Pente" rel="noreferrer">Pente</a>, and I'm struggling with the following question:</p>
<blockquote>
<p>What's the best way to detect patterns on a two-dimensional board?</p>
</blockquote>
<p>For example, in Pente a pair of neighboring stones of the same color can be captured when they are flanked from both sides by an opponent; how can we find all the stones that can be captured with the next move for the following board?</p>
<p><img src="https://i.stack.imgur.com/cV3Gb.png" alt="sample board"></p>
<p>Below I show one possible straightforward solution, but with a defect: it's hard to extend it for other interesting patterns, i.e. three stones of the same color in a row surrounded by empty spaces, or four stones of the same color in a row which are flanked from one side but open from another, etc.</p>
<blockquote>
<p>I'm wondering whether there is a way to define a DSL for detecting 2-dimensional structures like that on a board - sort of a <em>2D pattern matching</em>.</p>
</blockquote>
<p>P.S. I would also appreciate any advice on how to simplify the code below and make it more idiomatic - for example, I don't really like the way how <code>sortStones</code> is defined.</p>
<h2>Straightforward solution</h2>
<p>Here is one way to solve this problem (see below for graphics primitives to generate and display random boards):</p>
<ul>
<li>Enumerate all subsets of 3 stones from the board above</li>
<li>Select those that form an <em>AABE</em> or <em>ABBE</em> pattern, where E denotes an unoccupied space</li>
</ul>
<p>Lets store the board as a list of black and white stones,</p>
<pre><code>a = {black[2, 1], black[4, 3], black[2, 5], black[4, 2], black[5, 3],
black[1, 2], black[1, 3], black[5, 4], black[1, 5], white[3, 1],
white[4, 1], white[4, 4], white[3, 5], white[3, 4], white[5, 1],
white[5, 2], white[3, 3], white[1, 1]}
</code></pre>
<p>First, we define <code>isTriple</code> which checks whether three stones sorted by their x and y coordinates are in the same row next to each other and follow an ABB or AAB pattern:</p>
<pre><code>isTriple[{a_, b_, c_}] := And[
(* A A B or A B B *)
Head[a] != Head[c] /. {black -> 1, white -> 0},
(* x and y coordinates are equally spaced *)
a[[1]] - b[[1]] == b[[1]] - c[[1]],
a[[2]] - b[[2]] == b[[2]] - c[[2]],
(* and are next to each other *)
Abs[a[[1]] - b[[1]]] <= 1,
Abs[a[[2]] - b[[2]]] <= 1]
</code></pre>
<p>Next, we determine the coordinates and the color of the stone that will kill the pair:</p>
<pre><code>killerStone[{a_, b_, c_}] :=
If[Head[a] == Head[b] /. {black -> 1, white -> 0},
Head[c][2 a[[1]] - b[[1]], 2 a[[2]] - b[[2]]],
Head[a][2 c[[1]] - b[[1]], 2 c[[2]] - b[[2]]]]
</code></pre>
<p>Finally, we only select those triples where killer stone's space is not already occupied:</p>
<pre><code>sortStones[l_] :=
Sort[l, OrderedQ[{#1, #2} /. {black -> List, white -> List}] &]
triplesToKill[board_] := Module[
{triples = Select[sortStones /@ Subsets[board, {3}], isTriple]},
Select[triples,
Block[
{ks = killerStone[#]},
FreeQ[board, _[ks[[1]], ks[[2]]]]] &]]
displayBoard[a, #] & /@ triplesToKill[a] //
Partition[#, 3, 3, {1, 1}, {}] & // GraphicsGrid
</code></pre>
<p><img src="https://i.stack.imgur.com/QHj2c.png" alt="straightforward solution"></p>
<h2>Graphics primitives</h2>
<pre><code>randomPoints[n_] := RandomSample[Block[{nn = Ceiling[Sqrt[n]]},
Flatten[Table[{i, j}, {i, 1, nn}, {j, 1, nn}], 1]], n];
(* n is number of moves = 2 * number of points *)
randomBoard[n_] := Module[
{points = randomPoints[2 n]},
Join[
Take[points, n] /. {x_, y_} -> black[x, y],
Take[points, -n] /. {x_, y_} -> white[x, y]
]]
grid[minX_, minY_, maxX_, maxY_] :=
Line[Join[
Table[{{minX - 1.5, y}, {maxX + 1.5, y}}, {y, minY - 1.5, maxY + 1.5,
1}],
Table[{{x, minY - 1.5}, {x, maxY + 1.5}}, {x, minX - 1.5, maxX + 1.5,
1}]]];
displayBoard[board_] := Module[
{minX = Min[First /@ board], maxX = Max[First /@ board],
minY = Min[#[[2]] & /@ board], maxY = Max[#[[2]] & /@ board], n},
Graphics[{
grid[minX, minY, maxX, maxY],
board /. {
black[n__] -> {Black, Disk[{n}, .4]},
white[n__] -> {Thick, Circle[{n}, .4], White, Disk[{n}, .4]}
}}, ImageSize -> Small, Frame -> True]];
displayBoard[board_, points_] := Show[
displayBoard[board],
Graphics[
Map[{Red, Disk[{#[[1]], #[[2]]}, .2]} &, points]]]
</code></pre>
| Pellesatansfant | 13,904 | <p><strong>EDIT</strong></p>
<p>Making this code runnable with Java reloader.</p>
<ol>
<li><p>Load the <a href="https://mathematica.stackexchange.com/questions/6144/looking-for-longest-common-substring-solution/6376#6376">Java reloader</a> (run the code from that post. For Mac OS X, see the comments below the post for a link to the Mac version)</p></li>
<li><p>Compile the class:</p></li>
</ol>
<p>-</p>
<pre><code>JCompileLoad @
"package javaapplicationsim;
/**
* @author developer
*/
public class JavaApplicationSIM {
final byte E = 0; // EDGE
final byte _ = 1; // EMPTY CELL
final byte B = 2; // BLACK
final byte W = 3; // WHITE
byte [][] board = new byte[][] {
{ E, E, E, E, E, E, E, E, E, E },
{ E, _, _, _, _, _, _, _, _, E },
{ E, _, B, B, W, _, _, _, _, E },
{ E, _, _, _, W, W, B, B, _, E },
{ E, _, B, _, W, B, B, B, _, E },
{ E, _, B, _, _, B, _, _, _, E },
{ E, _, B, _, _, B, _, _, _, E },
{ E, _, W, B, W, W, W, W, _, E },
{ E, _, _, _, _, _, _, _, _, E },
{ E, E, E, E, E, E, E, E, E, E }
};
private void drawBoard() {
for( int row=0; row<board.length; row++ ) {
String ch = \"\";
for( int col=0; col<board[row].length; col++ ) {
switch( board [row] [col] ) {
case E : ch = \"+\"; break;
case _ : ch = \" \"; break;
case B : ch = \"B\"; break;
case W : ch = \"W\"; break;
}
System.out.print( ch );
}
System.out.println();
}
}
private void count( int dx, int dy, int row, int col, int endColor ) {
boolean done = false;
boolean reachedEndColor = false;
int x = col;
int y = row;
int len = 0;
do {
x = x + dx;
y = y + dy;
if( board [y] [x] == E ) {
// reached an edge, must end the traversal!
done = true;
}
if( board [y] [x] == _ ) {
// reached an empty cell
done = true;
}
if( board [y] [x] == endColor ) {
// reached the opposite side that has the same color
reachedEndColor = true;
}
if( !done && !reachedEndColor ) {
// the color of the current cell must be the color of the other player
// keep on with the search
len = len + 1;
}
} while( !done && !reachedEndColor );
if( reachedEndColor && len > 0 ) {
System.out.println( \"Len = \" + len + \" from pos (\" + row + \" , \" + col + \"), dir (\" + dy + \" , \" + dx + \")\" );
}
}
private void solve( byte endColor ) {
for( int row=1; row<=8; row++ ) {
for( int col=1; col<=8; col++ ) {
if( board [row] [col] == _ ) {
// the cell must be empty (since the new brick is supposed to be placed there!)
count( -1, 0, row, col, endColor ); // LEFT
count( -1, -1, row, col, endColor ); // LEFT + UP
count( 0, -1, row, col, endColor ); // UP
count( 1, -1, row, col, endColor ); // RIGHT + UP
count( 1, 0, row, col, endColor ); // RIGHT
count( 1, 1, row, col, endColor ); // RIGHT + DOWN
count( 0, 1, row, col, endColor ); // DOWN
count( -1, 1, row, col, endColor ); // LEFT + DOWN
}
}
}
}
/**
* @param args the command line arguments
*/
public static void main( String [] args ) {
JavaApplicationSIM sim = new JavaApplicationSIM();
sim.drawBoard();
sim.solve( sim.B );
}
}"
</code></pre>
<ol start="3">
<li><p>Run the code as </p>
<pre><code>ShowJavaConsole[]
JavaApplicationSIM`main[{}]
</code></pre></li>
</ol>
<p>This program produces the following output (on the console):</p>
<p>(first it shows the board)</p>
<pre><code>++++++++++
+ +
+ BBW +
+ WWBB +
+ B WBBB +
+ B B +
+ B B +
+ WBWWWW +
+ +
++++++++++
</code></pre>
<p>Then the program tells all the positions that will qualify as a place to put the BLACK color.</p>
<pre class="lang-none prettyprint-override"><code>Len = 2 from pos (1 , 3), dir (1 , 1)
Len = 1 from pos (2 , 5), dir (0 , -1)
Len = 1 from pos (2 , 5), dir (1 , 0)
Len = 2 from pos (3 , 3), dir (0 , 1)
Len = 1 from pos (3 , 3), dir (1 , 1)
Len = 1 from pos (4 , 3), dir (0 , 1)
Len = 1 from pos (7 , 1), dir (0 , 1)
Len = 4 from pos (7 , 8), dir (0 , -1)
Len = 1 from pos (8 , 2), dir (-1 , 0)
Len = 1 from pos (8 , 3), dir (-1 , 1)
Len = 1 from pos (8 , 5), dir (-1 , 0)
Len = 1 from pos (8 , 7), dir (-1 , -1)
</code></pre>
<p>One can transfer the result back to Mathematica from Java with a bit more work. </p>
|
1,158,489 | <p>Is it the case that, as $N\to\infty$, $$\binom{2N}{N+j}_q\to (-1)^j,$$ where convergence of the $q$-binomial coefficient is seen as a convergence of formal power series in the variable $q$? </p>
| Community | -1 | <p>$$A_{N,j}=\binom{2N}{N+j}_q=\frac{(1-q^{2N})(1-q^{2N-1})\ldots(1-q^{2N-N-j+1})}{(1-q)(1-q^2)\ldots(1-q^{N-j})}$$</p>
<p>Take $\log$ and expand using that $$\log(1-x)=\sum_{n=1}^{\infty}\frac{x^n}{n}.$$</p>
<p>We obtain that $$\begin{align}\log(A_{N,j})&=\sum_{k=N-j+1}^{2N}\log(1-q^{k})-\sum_{k=1}^{N-j}\log(1-q^k)\\&=\sum_{k=N-j+1}^{2N}\sum_{r=1}^{\infty}\frac{q^{kr}}{r}-\sum_{k=1}^{N-j}\sum_{r=1}^{\infty}\frac{q^{kr}}{r}\end{align}$$</p>
<p>The first group with the double summation tends to $0$ as $N\to\infty$ because all exponents are above $N-j+1$. </p>
<p>For the second we have $$\sum_{k=1}^{N-j}\sum_{r=1}^{\infty}\frac{q^{kr}}{r}=\left(\frac{q^1}{1}+\frac{q^{2}}{2}+\cdots\right)+\left(\frac{q^2}{1}+\frac{q^{4}}{2}+\cdots\right)+\left(\frac{q^3}{1}+\frac{q^{6}}{2}+\cdots\right)+\cdots+\left(\frac{q^{N-j}}{1}+\frac{q^{2(N-j)}}{2}+\cdots\right)\\=a_1q+a_2q^2+\cdots$$</p>
<p>where $a_i$ is the sum of the reciprocals of the divisors of $i$. This series is not constant and its exponential is not either.</p>
|
394,085 | <p>How is it possible to establish proof for the following statement?</p>
<p>$$n = \frac{1}{2}(5x+4),\;2<x,\;\text{isPrime}(n)\;\Rightarrow\;n=10k+7$$</p>
<p>Where $n,x,k$ are $\text{integers}$.</p>
<hr>
<p>To be more verbose:</p>
<p>I conjecture that;</p>
<p>If $\frac{1}{2}(5x+4),\;2<x$ is a prime number, then $\frac{1}{2}(5x+4)=10k+7$</p>
<p>How could one prove this?</p>
| Berci | 41,488 | <p>Since $n=5\,x/2\ +2$ is integer, we have that $x$ is even, and that $n$ gives remainder $2$ modulo $5$ (written $n\equiv 2\pmod{5}$). So, modulo $10$ there are only two possibilities: $2$ and $7=5+2$. As all primes $>2$ are odd, but $10k+2$ is even, it can't be prime.</p>
<p>(So, instead of 'prime' and $x>2$ you could have also said simply 'odd'.)</p>
|
425,969 | <p>It seems striking that the cardinalities of <span class="math-container">$\aleph_0$</span> and <span class="math-container">$\mathfrak c = 2^{\aleph_0}$</span> each admit what I will call a "homogeneous cyclic order", via the examples of <span class="math-container">$ℚ/ℤ$</span> and <span class="math-container">$ℝ/ℤ$</span>. By which I mean a cyclic order (as defined in <a href="https://ncatlab.org/nlab/show/cyclic+order" rel="nofollow noreferrer">https://ncatlab.org/nlab/show/cyclic+order</a>) such that for any two elements <span class="math-container">$x, y$</span> of the cardinal, there is a bijection of the cardinal to itself taking <span class="math-container">$x$</span> to <span class="math-container">$y$</span> and preserving the cyclic order.</p>
<p>In ZFC, is there any reason to believe that either a) all infinite cardinals admit a homogeneous cyclic order, or b) there exists an infinite cardinal admitting no homogeneous cyclic order?</p>
| Joseph Van Name | 22,277 | <p>Every totally ordered abelian group is automatically homogeneous as a total order, and every totally ordered set that is homogeneous as a total order is automatically homogeneous as a cyclic order. Therefore, every totally ordered abelian group is automatically homogeneous as a cyclic order.</p>
<p>Let <span class="math-container">$\lambda$</span> be an infinite cardinal. Let <span class="math-container">$G_{\alpha}$</span> be a totally ordered abelian group for <span class="math-container">$\alpha<\lambda$</span>. Then the abelian group coproduct <span class="math-container">$\oplus_{\alpha<\lambda}G_{\alpha}$</span> is an abelian group which is totally ordered by letting
<span class="math-container">$(x_{\alpha})_{\alpha<\lambda}<(y_{\alpha})_{\alpha<\lambda}$</span> precisely when there is some ordinal <span class="math-container">$\gamma$</span> with <span class="math-container">$x_\gamma<y_\gamma$</span> but where <span class="math-container">$x_\beta=y_\beta$</span> for <span class="math-container">$\beta<\gamma$</span>.</p>
<p>Since <span class="math-container">$|\oplus_{\alpha<\lambda}G_\alpha|=\sum_{\alpha<\lambda}|G_\alpha|$</span>, we can make <span class="math-container">$\oplus_{\alpha<\lambda}G_\alpha$</span> have any infinite cardinality.</p>
|
3,624,662 | <p><a href="https://i.stack.imgur.com/kwAMn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kwAMn.png" alt=" c"></a></p>
<p>In my mind, I can think of below example which seems to work.</p>
<p>If <span class="math-container">$(X,T) = \mathbb{R}$</span>, and <span class="math-container">$A = (0,\infty)$</span>, then as far as I know it comes in the standard (order) topology of <span class="math-container">$\mathbb{R}$</span>, but what I dont know is whether it will be a proper closed subset of <span class="math-container">$\mathbb{R}$</span>. If it works, then it will be perfect b/c for all positive integer <span class="math-container">$i$</span>, if I let <span class="math-container">$P_i$</span> be the open interval <span class="math-container">$(0,i)$</span>, then clearly <span class="math-container">$A \subset \cup$</span> <span class="math-container">$O_i$</span>, where <span class="math-container">$i = 1$</span> to <span class="math-container">$\infty$</span>, </p>
<p>however there doesnt exist <span class="math-container">$i_1, i_2,......, i_n$</span>, such that <span class="math-container">$A \subset (0,i_1),(0,i_2),....,(0,i_n)$</span>, therefore by the definition of compactness we can see that we dont have a finite sub-cover, so its not compact.</p>
<p>Kindly check my proof, let me know if there is anything wrong. Also, give it better style and notation if required. </p>
| peek-a-boo | 568,204 | <p>Let <span class="math-container">$A = [0, \infty)$</span>, then its complement <span class="math-container">$A^c = (-\infty, 0)$</span> is an open interval, and hence <span class="math-container">$A^c$</span> is an open set. Thus, by definition, <span class="math-container">$A$</span> is a closed set.</p>
<p>To show it is not compact, consider the following open cover: <span class="math-container">$\{(-n,n)| \, \, n \in \Bbb{N}\}$</span>. This open cover in fact covers all of <span class="math-container">$\Bbb{R}$</span>, so in particular covers <span class="math-container">$A$</span>. Clearly, there is no finite subcover. Hence, <span class="math-container">$A$</span> is not compact.</p>
|
3,624,662 | <p><a href="https://i.stack.imgur.com/kwAMn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kwAMn.png" alt=" c"></a></p>
<p>In my mind, I can think of below example which seems to work.</p>
<p>If <span class="math-container">$(X,T) = \mathbb{R}$</span>, and <span class="math-container">$A = (0,\infty)$</span>, then as far as I know it comes in the standard (order) topology of <span class="math-container">$\mathbb{R}$</span>, but what I dont know is whether it will be a proper closed subset of <span class="math-container">$\mathbb{R}$</span>. If it works, then it will be perfect b/c for all positive integer <span class="math-container">$i$</span>, if I let <span class="math-container">$P_i$</span> be the open interval <span class="math-container">$(0,i)$</span>, then clearly <span class="math-container">$A \subset \cup$</span> <span class="math-container">$O_i$</span>, where <span class="math-container">$i = 1$</span> to <span class="math-container">$\infty$</span>, </p>
<p>however there doesnt exist <span class="math-container">$i_1, i_2,......, i_n$</span>, such that <span class="math-container">$A \subset (0,i_1),(0,i_2),....,(0,i_n)$</span>, therefore by the definition of compactness we can see that we dont have a finite sub-cover, so its not compact.</p>
<p>Kindly check my proof, let me know if there is anything wrong. Also, give it better style and notation if required. </p>
| Community | -1 | <p>By Heine-Borel, a compact subset of <span class="math-container">$\Bbb R^n$</span> is closed and bounded. </p>
<p>So take any closed, unbounded set.</p>
|
97,449 | <p>I am trying to compute $\chi(\mathbb{C}\mathrm{P}^2)$ using only elementary techniques from differential topology and this is proving to be trickier than I thought. I am aware of the usual proof for this result, which uses the cellular decomposition of $\mathbb{C}\mathrm{P}^2$ to get $\chi(\mathbb{C}\mathrm{P}^2) = 3$, but I would like to find a proof of this result that relies on concepts like indices of isolated zeros on a vector field. So for the purposes of this question, I would like to utilize the following definition of the Euler characteristic: For a closed orientable manifold $M$ we define $\chi(M) = \sum_i \mathrm{Ind}_{d_i} \mathrm{v}$ where $\mathrm{v}$ is a vector field on $M$ with isolated zeros.</p>
<p>In my first attempt at this problem I thought about finding a vector field on $\tilde{\mathrm{v}}$ on $S^5$, and then using the identification $\mathbb{C}\mathrm{P}^2 \cong S^5/\mathrm{U}(1)$, seeing if $\tilde{\mathrm{v}}$ descended to a vector field on $\mathbb{C}\mathrm{P}^2$ with isolated zeros that lent itself to computing the Euler characteristic. I had difficulty making this work out, so I am unsure if this is a good approach to tackling the problem. Any insights?</p>
| Igor Rivin | 11,142 | <p>There is a canonical way to construct <em>holomorphic</em> vector fields on $\mathbb{C}P^2,$ and that way is <a href="http://www.springer.com/us/book/9783764375355" rel="nofollow noreferrer">described in Zoladek's "Monodromy Group", page 335.</a> If you read the description, it will be pretty clear what the index is (note that if the vector field is holomorphic, it is given by at most quadratic polynomials, otherwise there are poles. There is a large literature [much of it by Iliashenko] studying such fields and the associated foliations.</p>
|
97,449 | <p>I am trying to compute $\chi(\mathbb{C}\mathrm{P}^2)$ using only elementary techniques from differential topology and this is proving to be trickier than I thought. I am aware of the usual proof for this result, which uses the cellular decomposition of $\mathbb{C}\mathrm{P}^2$ to get $\chi(\mathbb{C}\mathrm{P}^2) = 3$, but I would like to find a proof of this result that relies on concepts like indices of isolated zeros on a vector field. So for the purposes of this question, I would like to utilize the following definition of the Euler characteristic: For a closed orientable manifold $M$ we define $\chi(M) = \sum_i \mathrm{Ind}_{d_i} \mathrm{v}$ where $\mathrm{v}$ is a vector field on $M$ with isolated zeros.</p>
<p>In my first attempt at this problem I thought about finding a vector field on $\tilde{\mathrm{v}}$ on $S^5$, and then using the identification $\mathbb{C}\mathrm{P}^2 \cong S^5/\mathrm{U}(1)$, seeing if $\tilde{\mathrm{v}}$ descended to a vector field on $\mathbb{C}\mathrm{P}^2$ with isolated zeros that lent itself to computing the Euler characteristic. I had difficulty making this work out, so I am unsure if this is a good approach to tackling the problem. Any insights?</p>
| Johannes Ebert | 9,928 | <p>Take a $3 \times 3$ complex diagonal matrix $A$ with distinct nonzero diagonal entries. The 1-parameter subgroup $exp(At)$ acts on $CP^2$; the fixed points are the lines in $C^3$ containing eigenvectors of $A$. There are $3$ of them and the derivative of the action is a vector field with $3$ zeroes. As the vector field is holomorphic, the index at each zero is $+1$.</p>
|
1,572,351 | <p>Solve the differential equation;</p>
<p>$(xdx+ydy)=x(xdy-ydx)$</p>
<p>L.H.S. can be written as $\frac{d(x^2+y^2)}{2}$ but what should be done for R.H.S.?</p>
| SchrodingersCat | 278,967 | <p>As much as I can see, there is no standard solution. <br>
From <a href="http://www.wolframalpha.com/input/?i=%5B%2F%2Fmath%3Axdx%2Bydy%3Dx(xdy-ydx)%2F%2F%5D" rel="nofollow noreferrer">Wolfram Alpha</a>, I have obtained a very complicated solution. You can check the link, if you wish.</p>
<p><a href="https://i.stack.imgur.com/k2sRl.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/k2sRl.gif" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/nCQvz.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nCQvz.gif" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/YavQF.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YavQF.gif" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/kxnvr.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kxnvr.gif" alt="enter image description here"></a></p>
|
1,572,351 | <p>Solve the differential equation;</p>
<p>$(xdx+ydy)=x(xdy-ydx)$</p>
<p>L.H.S. can be written as $\frac{d(x^2+y^2)}{2}$ but what should be done for R.H.S.?</p>
| Empy2 | 81,790 | <p>Let $x^2=z+y$. The equation $(xdx+ydy)=x(xdy-ydx)$ becomes
$$(dz+dy)+2ydy=2(z+y)dy-y(dz+dy)\\
(1+y)dz=(2z-y-1)dy\\
y=w-1\\
wdz=(2z-w)dw\\
\frac{dz}{w^2}-\frac{2zdw}{w^3}=-\frac{dw}{w^2}\\
\frac{z}{w^2}-\frac1w=const$$</p>
|
3,110,508 | <p>I read that implication like a=>b can be proof using the following steps :
1) suppose a true.
2) Then deduce b from a.
3) Then you can conclude that a=>b is true.</p>
<p>Actually my real problem is to understand why step 1 and 2 are sufficient to prove that a=>b is true. I mean, how can you prove the truth table of a=>b just using 1 and 2 ?
I know that implication a=>b is actually defined as "not(a) or b". How can steps 1 and 2 can prove that a and b are related like "not(a) or b" ?</p>
| Dr. Mathva | 588,272 | <p>Just let <span class="math-container">$k=-y$</span> to obtain <span class="math-container">$$x^3-y^3=(x-y)(x^2+xy+y^2)\iff x^3+k^3=(x+k)(x^2-xk+k^2)$$</span>
Note that the same works whenever <span class="math-container">$n$</span> is odd. Given <span class="math-container">$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\dots + xy^{n-2}+y^{n-1})$$</span>
Let <span class="math-container">$k=-y$</span> and ...</p>
<p>See how it works? Try to explain why it doesn't work whenever <span class="math-container">$n$</span> is even!</p>
|
3,405,914 | <blockquote>
<p>It's known that <span class="math-container">$\lim_{n \to \infty} \left(1 + \frac{x}{n} \right)^n = e^x$</span>.</p>
<p>Using the above statement, prove <span class="math-container">$\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$</span>.</p>
</blockquote>
<h2>My attempt</h2>
<p>Obviously, we want to reach a statement such as <span class="math-container">$$\lim_{n \to \infty} \left(1 + \frac{-2}{n}\right)^n \quad \text{ or } \quad \lim_{n \to \infty} \left(1 + \frac{-1}{n}\right)^n \cdot \lim_{n \to \infty} \left(1 + \frac{-1}{n}\right)^n$$</span>
in order to be able to apply the above condition. However, I was unable to achieve this. The furthest I've got was the following:</p>
<p><span class="math-container">\begin{align}
\left(\frac{n-2}{3n+1}\right)^{2n} &= \left( \frac{9n^2 - 12n + 4}{9n^2 + 6n + 1} \right)^n\\
&= \left(1 + \frac{-18n+3}{9n^2+6n+1}\right)^n\\
f(n)&= \left(1 + \frac{-2 + \frac{3}{n}}{n+\frac{2}{3} + \frac{1}{9n}} \right)^n
\end{align}</span></p>
<p>It seems quite obvious that <span class="math-container">$\lim_{n \to \infty} \left(f(n)\right) = \left(1 + \frac{-2}{n + \frac{2}{3}}\right)^n$</span>, however, this is not exactly equal to the statement given above. Are you able to ignore the constant and apply the condition regardness? If so, why? How would you go about solving this problem?</p>
| Z Ahmed | 671,540 | <p><span class="math-container">$$L=\lim_{n \rightarrow} \left(\frac{3n-2}{3n+1} \right)^{2n}= \lim_{n \rightarrow \infty} \frac{\left([1-2/(3n)]^{3n/2}\right)^{4/3}}{\left([1+1/(3n)]^{3n}\right)^{2/3}}=\frac{ e^{-4/3}}{e^{2/3}}=e^{-2}$$</span></p>
|
1,458,960 | <p>I have been working on this problem for about(over) two months now. This is <strong>not school work</strong>.<br>
The final objective of this problem is to find the <strong>area</strong> of these <strong>11 cylinders</strong> which are enclosed in a sphere.</p>
<p>In the image below, every line represents the center of a cyllinder.</p>
<p>$a = 60^{\circ}$ and $b = 30^{\circ}$<br>
$r_{cylinder} = 0.5mm$<br>
$length_{each\,cylinder} = 1.5mm$</p>
<p>Notice that although the cylinders are <code>1.5mm</code> in length, they are completely connected, like if grabbed the image below and enlarged, meaning that the base will always be inside of the total object.</p>
<p><a href="https://i.stack.imgur.com/aKMCt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aKMCt.jpg" alt="from above"></a></p>
<p>This second picture shows the that there is also two other cylinders perpendicular to the other 9.</p>
<p>This problem came to me when I read about the Steinmetz Solid, and wondered what would happen if I aligned them in this specific formation, simple curiosity.</p>
<p><a href="https://i.stack.imgur.com/YVrn0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YVrn0.jpg" alt="from side"></a></p>
<p>Then, all of this is enclosed in a sphere of a radius of <code>1.5mm</code></p>
| Archis Welankar | 275,884 | <p>For this problem consider sun .The ball is centre and rays are the cylinders so the area of the 11 cylinders would be a shape like a flat tyre like cylinder whose radius is length of the small cylinders and height is the diameter of small cylinders .This my way of thinking. Soory if I have interpreted something wrong.!!</p>
|
1,458,960 | <p>I have been working on this problem for about(over) two months now. This is <strong>not school work</strong>.<br>
The final objective of this problem is to find the <strong>area</strong> of these <strong>11 cylinders</strong> which are enclosed in a sphere.</p>
<p>In the image below, every line represents the center of a cyllinder.</p>
<p>$a = 60^{\circ}$ and $b = 30^{\circ}$<br>
$r_{cylinder} = 0.5mm$<br>
$length_{each\,cylinder} = 1.5mm$</p>
<p>Notice that although the cylinders are <code>1.5mm</code> in length, they are completely connected, like if grabbed the image below and enlarged, meaning that the base will always be inside of the total object.</p>
<p><a href="https://i.stack.imgur.com/aKMCt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aKMCt.jpg" alt="from above"></a></p>
<p>This second picture shows the that there is also two other cylinders perpendicular to the other 9.</p>
<p>This problem came to me when I read about the Steinmetz Solid, and wondered what would happen if I aligned them in this specific formation, simple curiosity.</p>
<p><a href="https://i.stack.imgur.com/YVrn0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YVrn0.jpg" alt="from side"></a></p>
<p>Then, all of this is enclosed in a sphere of a radius of <code>1.5mm</code></p>
| Ian Miller | 278,461 | <p>Is this what you meant? (Sorry I can't put an image in a comment.)</p>
<p><a href="https://i.stack.imgur.com/evxeb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/evxeb.png" alt="Your thing"></a></p>
<p>The sphere is partially transparent so you can see the cylinders inside.</p>
|
2,704,770 | <p>I need help in calculating this strange limit.</p>
<p>$$
\lim_{n \to \infty} n^2 \int_{0}^{\infty} \frac{sin(x)}{(1 + x)^n} dx
$$</p>
| Felix Marin | 85,343 | <p>$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\lim_{n \to \infty}\bracks{n^{2}\int_{0}^{\infty}
{\sin\pars{x} \over \pars{1 + x}^{n}}\,\dd x} & =
\lim_{n \to \infty}\bracks{n^{2}\int_{0}^{\infty}
\sin\pars{x}\exp\pars{-n\ln\pars{1 + x}}\,\dd x}
\\[5mm] & =
\lim_{n \to \infty}\bracks{n^{2}\int_{0}^{\infty}
x\,\exp\pars{-nx}\,\dd x} =
\lim_{n \to \infty}\int_{0}^{\infty}
x\,\exp\pars{-x}\,\dd x
\\[5mm] & = \bbx{1}
\end{align}</p>
<blockquote>
<p>See <a href="https://en.wikipedia.org/wiki/Laplace%27s_method" rel="nofollow noreferrer">Laplace's Method</a>.</p>
</blockquote>
|
2,772,895 | <p>I've noticed one classical way of defining certain topologies is to define them as the "weakest" (or coarsest) topology such that a certain set of functions is continuous. For example,</p>
<blockquote>
<p>The <strong>product topology</strong> on <span class="math-container">$X=\prod X_i$</span> is the weakest topology such that the canonical projections <span class="math-container">$p_i : X\to X_i$</span> are continuous.</p>
<p>The <strong>weak*-topology</strong> on a Banach space <span class="math-container">$X$</span> is the weakest topology such that the <em>evaluation map</em> (the natural isomorphism from <span class="math-container">$X$</span> to <span class="math-container">$X^{**}$</span>, <span class="math-container">$J(x)(\phi) = \phi(x)$</span>) is continuous.</p>
</blockquote>
<p>I have a very poor intuition behind what these "mean". When I look at the definition of a topology, the ones that make the most sense to me are the ones in which the open sets (or at least a base of open sets) are explicitly constructed, as in the Euclidean topology (or a general metric space topology, or a norm induced topology).</p>
<p>I get a little stuck when the definition of the topology is given in some "abstract" sense, where the open sets are "chosen" to satisfy a certain other property. How am I supposed to visualize the open sets in these spaces, or work with them?</p>
<p>If <span class="math-container">$\tau$</span> is the weakest topology on <span class="math-container">$X$</span> such that <span class="math-container">$f : X\to Y$</span> is continuous, is it correct to imagine a base for the open sets to be the preimage of all open sets in <span class="math-container">$Y$</span> under <span class="math-container">$f$</span>? This follows directly from the definition of a continuous function on topological spaces. Is this always the coarsest topology?</p>
<p>Furthermore, what benefit do these topologies provide? What interesting, and potentially theoretically useful, properties do they possess? Why should I care about them?</p>
| Billy | 13,942 | <blockquote>
<p>"weakest"</p>
</blockquote>
<p>Fix a space X, and let Top(X) be the set of topologies on X. That is, an element T of Top(X) can be thought of as a subset of the power set P(X) (satisfying axioms). Now, Top(X) naturally forms a poset under $\subseteq$: that is, if S and T are elements of Top(X), you can think of S as "smaller than" T if $S\subseteq T$, i.e. all S-open sets are also T-open (but not necessarily conversely). The indiscrete topology is the "smallest" element of this poset, and the discrete topology is the "largest" element.</p>
<p>The words "coarse" and "fine" are synonyms in this context for "small" and "large" respectively. That is, the indiscrete topology is the coarsest possible topology (it has very few open sets; it smooshes all points of X together into one big open set, so that you can't tell the points or subsets apart with open sets), and the discrete topology is the finest possible topology (it has lots of open sets; you can tell lots of things apart). If X is e.g. $\mathbb{C}^n$, then you could also define the Euclidean topology, or the (weaker) Zariski topology.</p>
<p>Note the following: if S and T are both elements of Top(X), then $S\cap T$ is too. (Prove it!)</p>
<p>The "weakest" topology subject to some conditions, then, is the smallest possible topology that satisfies those conditions. Put another way, it's the intersection of all the topologies that satisfy those conditions. Put yet another way, if you are happier with algebra than topology, you might like to think of this as the topology <em>generated</em> by some open sets.</p>
<p>$f: X\to Y$ is continuous if, for every open U in Y, the set $f^{-1}(U)$ is open in X. So the <em>weakest</em> topology on $X$ such that $f:X\to Y$ is continuous is the smallest one that contains $f^{-1}(U)$ for every open U in Y. That's the intersection of all those T in Top(X) such that $f^{-1}(U)\in T$ for all open U in Y. And so on.</p>
|
834,678 | <p>An object $X$ is a <em>generator</em> of a category $\mathcal{C}$ if the functor $Hom_{\mathcal{C}}(X,\_) : \mathcal{C} \rightarrow Set$ is faithful. </p>
<p>I encountered the notion in the context of Morita-equivalence of rings, but I don't understand what its use is.
Why is $X$ called a "generator"? What does it generate? </p>
| Community | -1 | <p>While I don't know the <em>actual</em> etymology, I interpret the word "generate" as generating elements of the category.</p>
<p>There is a notion of a "generalized element" of an object: a generalized element of $A$ is simply an arrow with target $A$. This is the "right" notion of element: for example, a morphism $f : A \to B$ is monic if and only if $f(x) = f(y) \implies x=y$ for all generalized elements $x,y$ of $A$.</p>
<p>However, allowing arbitrary objects as domains of a generalized element is unwieldy. $X$ is a generator if the class of generalized elements with domain $X$ has "enough" elements.</p>
<p>In the element language, $X$ is a generator is the same thing as saying that, for $f,g : A \to B$, $\left(\forall x: f(x) = g(x)\right) \implies f = g$ when $x$ ranges over all generalized elements of $A$ that come from $X$.</p>
<p>You can even think of replacing an object with its class of generalized elements... or just the class of generalized elements with domain $X$. This is actually the functor $\hom(X, -)$.</p>
<p>$X$ is a generator if and only if the functor $\hom(X, -)$ is faithful. We can think of this as saying that $X$ generates enough elements of the category to faithfully represent its structure.</p>
<p>(note that you can consider a whole class of objects rather than just a single object)</p>
<p>As a counterexample, $\mathbf{F}_p[x]$ doesn't generate $\mathbf{CRng}$: the image of $\hom(\mathbf{F}_p[x], -)$ is faithful on the subcategory of $\mathbf{F}_p$-algebras, but it collapses all other rings into the empty set.</p>
|
3,572,842 | <p><strong>Context:</strong> 1st year BSc Mathematics, Vectors and Mechanics module, constant circular motion.</p>
<p>This may be trivial, but can someone tell me what's wrong with the following reasoning?</p>
<p><span class="math-container">$$\underline{e_r}=\underline{i}\cos\theta+\underline{j}\sin\theta=(1,\theta) \;\;(1);$$</span>
<span class="math-container">$$\underline{e_\theta}=\frac{d(\underline{e_r})}{d\theta}=-\underline{i}\sin\theta+\underline{j}\cos\theta \;\;(2);$$</span>
so
<span class="math-container">$$(1),(2):\;\; \underline{e_\theta}=\frac{d}{d\theta}((1,\theta))=(1,\frac{d\theta}{d\theta})=(1,1) \;\;(3),$$</span>
so
<span class="math-container">$$(2),(3): \;\; (1,1)=-\underline{i}\sin\theta+\underline{j}\cos\theta \;\; (4),$$</span>
an undesirable conclusion.</p>
| emacs drives me nuts | 746,312 | <blockquote>
<p>Are there some available computational results much beyond <span class="math-container">$c≤100$</span> that would confirm intuition that really <span class="math-container">$a+b$</span> is almost-always a prime number or this is just some instance of the rule that this is a small sample so it is not quite unusual to have this high percentage of primes?</p>
</blockquote>
<p>IMO, it's just because you are using small Pythagorean triples only.</p>
<p>I've conducted some tests for larger primitive triples around<span class="math-container">$^3$</span> <span class="math-container">$c_0=10^{50}$</span>. Let's call a triple <span class="math-container">$(a,b,c)$</span> <em>"prime"</em> iff <span class="math-container">$c\in\Bbb P$</span>. For example, <span class="math-container">$(3,4,5)$</span> is a prime triple whereas <span class="math-container">$(16, 63, 65)$</span> is not because <span class="math-container">$65=5\cdot13$</span>.</p>
<p>For the prime triples in <span class="math-container">$c_0-10^6\cdots c_0+10^6$</span>:</p>
<ul>
<li>From the 8770 prime<span class="math-container">$^1$</span> triples <span class="math-container">$(a,b,c)$</span> with <span class="math-container">$|c-10^{50}|\leqslant 10^6$</span>, <span class="math-container">$$\text{For 291 such triples there is }a+b\in\Bbb P.\text{ This is }291 / 8770 = 3.3\%.$$</span></li>
</ul>
<p>For generic primitive triples, such computations are less inconvenient because you have to factor <span class="math-container">$c\approx c_0$</span>, and for that reason I used the smaller interval <span class="math-container">$10^{50}\pm10^5$</span>.</p>
<ul>
<li>From the 4544 primitive triples with <span class="math-container">$|c-10^{50}|\leqslant 10^5$</span>, <span class="math-container">$$\text{For 128 such triples there is }a+b\in\Bbb P.\text{ This is }128 / 4544 = 2.8\%.$$</span></li>
</ul>
<p>I am not familiar with statistics and to tell whether this is statistically significant. Of course also non-prime triples contribute to <span class="math-container">$a+b\in\Bbb P$</span>.</p>
<h2>Example:</h2>
<p>Three triples are contributed<span class="math-container">$^2$</span> by <span class="math-container">$
c=10^{50}+3549 = \small 34457 \cdot 38411348312521 \cdot 3467788412123489 \cdot 21787650199356253$</span>:<span class="math-container">$\def\Im{\mathfrak{Im}} \def\Re{\mathfrak{Re}}$</span>
<span class="math-container">\begin{array}{r|c|c}
\hfill z \stackrel\sim= \sqrt{a+ib\,} \hfill & c & a+b \in\Bbb P\\
\hline
\small 1296976664481873740241045 + 9915535867102164494675782\,i & 10^{50}+3549 & \\
\small 2080508524790029291482210 + 9781180106627012523222107\,i & 10^{50}+3549 & \\
\small 2624897991498247705333205 + 9649347674025869875255782\,i & 10^{50}+3549 & \\
\small 3132759182644749181744955 + 9496621499436260202476718\,i & 10^{50}+3549 & \\
\small 3224036380807234324896507 + 9466022893234063478352050\,i & 10^{50}+3549 & * \\
\small 6117622915805371444867450 + 7910416541498493375147243\,i & 10^{50}+3549 & * \\
\small 6550934381660341323983205 + 7555478722568176319806718\,i & 10^{50}+3549 & * \\
\small 7011237840726321813081643 + 7130395777288053485323290\,i & 10^{50}+3549 & \\
\end{array}</span>
In order to not clutter up your screens, I am using a notation of triples as a complex number <span class="math-container">$z=x+iy$</span>. You'll get a triple <span class="math-container">$(a,b,c)$</span> by taking
<span class="math-container">\begin{align}
a &= |\Re(z^2)| = |x^2-y^2| \\
b &= |\Im(z^2)| = |2xy| \\
c &= |z|^2 = x^2+y^2 \\
\end{align}</span>
This is not exactly the same <span class="math-container">$z$</span> you get from taking a square root of <span class="math-container">$a+ib$</span> because I canonicalized <span class="math-container">$z$</span> so that <span class="math-container">$0\leqslant \Re(z)\leqslant \Im(z)$</span>, indicated by <span class="math-container">$z\stackrel\sim=\sqrt{a+ib\,}$</span>. This means you might need to swap real and imaginary part and / or to take absolute values, but for Pythagorean triples that does not matter.</p>
<hr>
<p><span class="math-container">$^1$</span>This is exactly the number of primes that are 1 mod 4 in that interval, i.e.
<span class="math-container">$$8770 = \#(\Bbb P\cap(1+4\Bbb Z)\cap[10^{50}-1\,000\,000, 10^{50}+1\,000\,000])$$</span></p>
<p><span class="math-container">$^2$</span>An integer <span class="math-container">$n$</span> yields primitive triples iff it only has prime divisors that are 1 mod 4. If <span class="math-container">$m$</span> is the number of such divisors (without multiplicity), then <span class="math-container">$n$</span> contributes <span class="math-container">$2^{m-1}$</span> primitive triples. The triple is prime iff <span class="math-container">$n$</span> is prime. Because <span class="math-container">$10^{50}+3549$</span> has 4 distinct such prime divisors, it generates 8 primitive Pythagorean triples.</p>
<p><span class="math-container">$^3$</span> There is nothing special about <span class="math-container">$10^{50}$</span> except that it is conveniently written down. Similar computations for other <span class="math-container">$c_0$</span>'s show similar results.</p>
|
791,535 | <blockquote>
<p>Let $f,g:\mathbb [a,b] \to \mathbb [a,b]$ be monotonically increasing functions
such that $f\circ g=g\circ f$</p>
<p>Prove that $f$ and $g$ have a common fixed point.</p>
</blockquote>
<p>I found this problem in a problem set, it's quite similar to this <a href="https://math.stackexchange.com/questions/791411/real-analysis-homework-fixed-point-ordered-sets-hint/791470?noredirect=1#comment1638923_791470">Every increasing function from a certain set to itself has at least one fixed point</a> but I can't solve it.</p>
<p>I think it's one of those tricky problems where you need to consider a given set and use the LUB... I think $\{x \in [a,b]/ x < f(x) \text{and} x< g(x) \}$ is a good one.</p>
<p>Any hint ?</p>
| Lutz Lehmann | 115,115 | <p>Let <span class="math-container">$z$</span> be a fixed point of <span class="math-container">$f$</span>. Then <span class="math-container">$$f(g(z))=g(f(z))=g(z),$$</span> so also <span class="math-container">$g(z)$</span> is a fixed point of <span class="math-container">$f$</span>. Now if they were unique, then <span class="math-container">$g(z)=z$</span>. In general, <span class="math-container">$g$</span> induces a <s>permutation</s> map of the fixed points of <span class="math-container">$f$</span>, now argue with monotonicity... </p>
<hr>
<p>The analogy should be the case of eigenvectors of commuting matrices, if <span class="math-container">$v$</span> is an eigenvector of <span class="math-container">$A$</span> and <span class="math-container">$AB=BA$</span>, then <span class="math-container">$B^kv$</span> are all eigenvectors of <span class="math-container">$A$</span> with the same eigenvalue, so there is a minimal polynomial with <span class="math-container">$p(B)v=0$</span>, with degree smaller than the dimension of the eigenspace of <span class="math-container">$A$</span> containing <span class="math-container">$v$</span>, and then it gets more complicated.</p>
<hr>
<p>Consider the sequence <span class="math-container">$g^k(z)$</span> and its limit points. Resp. if avoiding limits, start with the smallest fixed point of <span class="math-container">$f$</span> and consider the least upper bound of the sequence.</p>
<hr>
<p><strong>Perhaps an even better idea</strong> is to consider the fixed points of <span class="math-container">$h=f∘g=g∘f$</span>. The set
<span class="math-container">$$
B=\{x:h(x)\le x\}
$$</span>
is non-empty, bounded and contains all potential fixed points. As before,
<span class="math-container">$$
z=\inf B
$$</span>
actually is one of the fixed points. Now both <span class="math-container">$f(z)$</span> and <span class="math-container">$g(z)$</span> are also fixed points of <span class="math-container">$h$</span>, since <span class="math-container">$h(f(z))=f(g(f(z)))=f(h(z))=f(z)$</span> etc. using commutativity of the functions <span class="math-container">$f$</span> and <span class="math-container">$g$</span>, so must be contained in <span class="math-container">$B$</span>. Which means by the construction of <span class="math-container">$z$</span> as greatest lower bound of <span class="math-container">$B$</span>
<span class="math-container">$$
f(z)\ge z\text{ and }g(z)\ge z.
$$</span>
By monotonicity, <span class="math-container">$g(z)\le g(f(z))=h(z)=z$</span> and similarly <span class="math-container">$f(z)\le z$</span> follow, so <span class="math-container">$z$</span> is also a common fixed point of <span class="math-container">$f$</span> and <span class="math-container">$g$</span>.</p>
|
1,134,145 | <p>A set S is bounded if every point in S lies inside some circle |z| = R other it is unbound. Without appealing to any limit laws, theorems, or tools from calculus, prove or disprove that the set {$\frac{z}{z^2 + 1}$; z in R} is bounded.</p>
<p>I imagine that it's simple, but I have no clue where to start due to the restrictions. Thanks</p>
| AlexR | 86,940 | <p><strong>Hint</strong><br>
Estimate case-wise $\frac{|z|}{z^2+1}$ for $|z|<1$ and $|z|\ge 1$:
$$\frac{|z|}{z^2 + 1} \le \frac{|z|}{z^2} = \frac1{|z|} \stackrel{|z|\ge 1}\le 1$$
And for $|z| < 1$ use $z^2 \ge 0$.<br>
This will give you the bound $S\subset[-1,1]$.<br>
An optimal bound is $[-\frac12, \frac12]$, this can be found using $z^2 \pm 2z + 1 = (1 \pm z)^2 \ge 0$.</p>
|
393,712 | <p>I studied elementary probability theory. For that, density functions were enough. What is a practical necessity to develop measure theory? What is a problem that cannot be solved using elementary density functions?</p>
| krishnab | 38,239 | <p>I come from a probability and statistics background, so I kinda get where the OP is coming from. The basic question is about the motivation behind learning measure theory when working with probability. And sure enough, when I look at many measure theory texts, the motivation for learning measure theory--Lebesgue measure in particular--is that certain pathological sets don't work well within the framework of Jordan measure or the Riemann integral. The motivation that Stein and Shakarchi, as well as Tao give, include sets like the Cantor set, or cosets of irrational numbers, as the reasons why we need a more robust theory for integration.</p>
<p>This motivation about pathological sets makes sense if you are coming from an analysis background, because so much effort is put into defining number systems, and functions on those number systems. So the Cantor set is an important set within the context of Analysis. However, I have never seen anyone try to assign probability to the Cantor set in any practical application. I am not saying that someone has not already done this, it is more likely that this problem was solved long ago--though I have not checked.</p>
<p>I think the more sensible motivation for why to learn measure theory as a probabilist or statistician has to do with finding a way to measure the probability of infinite sets of events. In probability and statistics we need a way to identify the probability that a continuous random variable takes a value less than or equal to a specific value, for example the probability density function
<span class="math-container">$$
pdf(X) := \Pr(X \leq x_0)
$$</span>
where <span class="math-container">$x_0$</span> is some particular value within the support of <span class="math-container">$X$</span>. For example, say we want <span class="math-container">$pdf(X) = \Pr(X \leq 0.5)$</span>. Given that <span class="math-container">$X$</span> is continuous, the set of all possible values of <span class="math-container">$X$</span>--also known as the support of <span class="math-container">$X$</span>--is the set <span class="math-container">$\{x : x \in [0,1], x \in \mathbb{R} \}$</span>. The set of all possible values in the event of interest: namely <span class="math-container">$ X \leq 0.5$</span> is the set <span class="math-container">$\{x : x \in [0,0.5], x \in \mathbb{R}\}$</span>.</p>
<p>Both the support of <span class="math-container">$X$</span> and the event of interest are infinite sets. Intuitively, the size or cardinality of the support should be larger than the cardinality of the event of interest--though lacking a theory of measure we have no way to make that notion precise. But think if you were to try and figure out the frequency or probability of this event. You would have to divide <span class="math-container">$\infty \div \infty$</span>.</p>
<p>So measure gives us a way to assign probability to sets of event where each individual event has zero probability. Another way of saying this is that measure theory gives us a way to define the expectations and pdfs for continuous random variables. Of course, most of this theory is usually towards the end of a book on measure theory textbook. But when you get to the parts about Fatou's lemma and Dominated Convergence, etc., then the applications to probability become more apparent.</p>
<p>From my own experience I actually rather liked learning measure theory. It really is not that difficult once you understand the intent. It just took a second to shift my focus from a probability frame of reference to an analysis frame of reference, but the arguments are generally pretty intuitive. Good luck.</p>
|
1,366,023 | <p>Here's a problem I was just working on:</p>
<blockquote>
<p>Let $f$ have an essential singularity at $0$. Show that there is a sequence of points $z_n \to 0$ such that $z_n^n f(z_n)$ tends to infinity.</p>
</blockquote>
<p>I know already that there exists a sequence $z_n \to 0$ such that $f(z_n)$ tends to any complex number I want, hence I can get a sequence that tends to infinity. The problem is that I need this sequence to tend to infinity really fast. </p>
<p>What I did so far was look at the function $g_n(z) := z^n f(z)$, $n \geq 1$. This obviously also has an essential singularity at $0$, so I can find a sequence $z_{n1}, z_{n2}, ...$ such that $\lim\limits_k g_n(z_{nk}) = \infty$. Do you think it's possible to extract from the array $z_{nk}, (n,k) \in \mathbb{N}^2$ a subsequence $z_l$ such that $g_l(w_l) \to \infty$ as $l \to \infty$? I tried for awhile but I'm just not very good at these kinds of arguments.</p>
| Theo Bendit | 248,286 | <p>You should be able to extract such a sequence from $g_n(z_{n,k})$. To put this sequence property slightly differently, you know that for any $n \in \mathbb{N}$, any complex number $w$, and any $\varepsilon > 0$, you can find a complex number $z$ such that $|g_n(z) - w| < \varepsilon$. So, fix $n \in \mathbb{N}$, choose $\varepsilon = 1$, and $w = n$, and you get a sequence of corresponding $z_n$'s satisfying,
$$|g_n(z_n) - n| < 1.$$
Basically, $g_n(z_n)$ is a sequence that's always within distance $1$ of $n$, so it shouldn't tend to anything other than $\infty$. To prove this, use reverse triangle inequality. We have,
$$||g_n(z_n)| - n| \le |g_n(z_n) - n| < 1,$$
so
$$-1 + n < |g_n(z_n)| < 1 + n.$$
By the squeeze theorem, $|g_n(z_n)| \rightarrow \infty$.</p>
|
3,011,758 | <p>So I was going through a sum for </p>
<p>Prove <span class="math-container">$ex \leq e^x$</span> , <span class="math-container">$\forall x \in \mathbb{R} $</span></p>
<p>I took <span class="math-container">$g(x) = e^x - ex$</span></p>
<p>Then <span class="math-container">$g'(x)= e^x - e$</span></p>
<p>I understood the case when <span class="math-container">$x>1$</span> function is strictly increasing i.e <span class="math-container">$g'(x) >0$</span> then <span class="math-container">$e^x>ex$</span> but what about when <span class="math-container">$x \leq 1$</span> </p>
| gb2017 | 496,656 | <p>it is easy to see <span class="math-container">$x=1$</span> is the absolute minimum, that is, <span class="math-container">$ f(1)\leq f(x)$</span> for all <span class="math-container">$x\in R$</span> so that <span class="math-container">$0\leq ex-e^x$</span>.Thus <span class="math-container">$ex\leq e^x$</span> for all <span class="math-container">$x\in R.$</span></p>
|
1,220,923 | <p>Find the value of the integral
$$\int_0^\infty \frac{x^{\frac25}}{1+x^2}dx.$$
I tried the substitution $x=t^5$ to obtain
$$\int_0^\infty \frac{5t^6}{1+t^{10}}dt.$$
Now we can factor the denominator to polynomials of degree two (because we can easily find all roots of polynomial occured in the denominator of the former integral by using complex numbers) and then by using partial fraction decomposition method find the integral!</p>
<p>Is there any simple method to find the integral value??!!</p>
| k1.M | 132,351 | <p>With some substitutions and ordinary works on the integral we have
$$\int_0^\infty \frac{x^{\frac25}}{1+x^2}dx=\frac52\int_{-\infty}^\infty\frac{t^6}{1+t^{10}}dt$$
For every $R\gt0$ we have
$$\int_{C}\frac{z^6}{1+z^{10}}dz=\int_{-R}^R\frac{t^6}{1+t^{10}}dt+\int_{C_R}\frac{z^6}{1+z^{10}}dz$$
where $C_R$ is the contour $Re^{it}$ with $t\in [0,\pi]$ and $C=C_R\cup [-R,R]$ with positive direction.</p>
<p>now by using residue theorem on the LHS of the equation above, and letting $R\to\infty$ we obtain
$$\sum_{i=1}^5 Res\left(\frac{z^6}{1+z^{10}},\zeta^i\right)=\int_{-\infty}^\infty\frac{t^6}{1+t^{10}}dt$$
because $\int_{C_R}\frac{z^6}{1+z^{10}}dz=o(R)$.<br>
For calculating residues we can use following limits
$$Res\left(\frac{z^6}{1+z^{10}},\zeta^i\right)=\lim_{z\to\zeta^i}\frac{z^6(z-\zeta^i)}{1+z^{10}}$$</p>
|
4,246,048 | <p>As I understand it, Cantor defined two sets as having the same cardinality iff their members can be paired 1-to-1. He applied this to infinite sets, so ostensibly the integers (Z) and the even integers (E) have the same cardinality because we can pair each element of Z with exactly one element of E.</p>
<p>For infinite sets, this definition seems problematic no matter which direction we come at it from: We don't know up front that two infinite sets have the same cardinality, so we cannot conclude that their elements can be <em>exactly</em> paired. And we do not know up front that two infinite sets' elements can be paired up exactly (because we don't know with certainty what happens beyond the finite cases we can verify). So we cannot conclude that their cardinalities are the same. The definition above therefore seems useless, since we cannot start from either side of the "iff".</p>
<p>It might be argued that if we state it as follows: "For each element e of E, pair it with element e/2 of Z," then we have expressed the general case symbolically, and it works. But we can only verify that for finite values of E and Z. We can't know what happens beyond finite elements of those sets. So expressing it symbolically does not seem to help.</p>
<p>Why is Cantor's definition not circular and therefore useless for deciding the question of infinite set cardinalities?</p>
| Mark Saving | 798,694 | <p>I'm not exactly sure what the supposed problem is, but I can walk through some basic results to hopefully shed light on what's going on.</p>
<p>Two basic notions in set theory are sets and functions. I'll assume you agree that the notion of a set and the notion of a function make sense.</p>
<p>Consider a function <span class="math-container">$f : A \to B$</span>.</p>
<p><span class="math-container">$f$</span> is said to be injective iff for all <span class="math-container">$x, w \in A$</span>, if <span class="math-container">$f(x) = f(w)$</span> then <span class="math-container">$x = w$</span>.</p>
<p><span class="math-container">$f$</span> is said to be surjective iff for all <span class="math-container">$b \in B$</span>, there exists <span class="math-container">$a \in A$</span> such that <span class="math-container">$f(a) = b$</span>.</p>
<p><span class="math-container">$f$</span> is said to be bijective iff <span class="math-container">$f$</span> is both surjective and injective.</p>
<p>We say that <span class="math-container">$A \preccurlyeq B$</span> if and only if there exists an injective function <span class="math-container">$f : A \to B$</span>. We say <span class="math-container">$A \approx B$</span> if and only if there exists a bijective function <span class="math-container">$f : A \to B$</span>.</p>
<p>Cantor's insight was that it is possible to assign each set <span class="math-container">$A$</span> something called a "cardinality". The cardinality of <span class="math-container">$A$</span> is denoted <span class="math-container">$|A|$</span>. The assignment is done in such a way that for all sets <span class="math-container">$A$</span>, <span class="math-container">$B$</span>, we have <span class="math-container">$|A| = |B|$</span> if and only if <span class="math-container">$A \approx B$</span>. The technical details of this are a bit complicated, but in the case that <span class="math-container">$A$</span> has exactly <span class="math-container">$n$</span> elements, <span class="math-container">$|A| = n$</span>.</p>
<p>It turns out that once we define cardinality, we can also define the comparison operator <span class="math-container">$\leq$</span> on cardinalities. This operator is defined so that <span class="math-container">$|A| \leq |B|$</span> if and only if <span class="math-container">$A \preccurlyeq B$</span>.</p>
<p>In practice, when one is first learning about cardinality, you should always take <span class="math-container">$|A| = |B|$</span> as a shorthand for <span class="math-container">$A \approx B$</span>, and you should always take <span class="math-container">$|A| \leq |B|$</span> as a shorthand for <span class="math-container">$A \preccurlyeq B$</span>.</p>
<p>We can prove explicitly that <span class="math-container">$\mathbb{Z}$</span> and <span class="math-container">$E$</span> have the same cardinality by considering the function <span class="math-container">$f : \mathbb{Z} \to E$</span> defined by <span class="math-container">$f(n) = 2n$</span>.</p>
<p>To show that <span class="math-container">$f$</span> is a well-defined function, note that <span class="math-container">$E$</span> is defined to be <span class="math-container">$\{2n \mid n \in \mathbb{Z}\}$</span>. Therefore, <span class="math-container">$f(n) \in E$</span> for all <span class="math-container">$n \in \mathbb{Z}$</span>.</p>
<p>To show that <span class="math-container">$f$</span> is injective, consider <span class="math-container">$a, b \in \mathbb{Z}$</span> and suppose that <span class="math-container">$f(a) = f(b)$</span>. Then <span class="math-container">$2a = 2b$</span>. Dividing both sides by 2 gives <span class="math-container">$a = b$</span>.</p>
<p>To show that <span class="math-container">$f$</span> is surjective, consider some <span class="math-container">$e \in E$</span>. Then by the definition of <span class="math-container">$E$</span>, there exists <span class="math-container">$n \in \mathbb{Z}$</span> such that <span class="math-container">$2n = e$</span>. Then for this <span class="math-container">$n$</span>, we have <span class="math-container">$f(n) = e$</span>.</p>
<p>Therefore, <span class="math-container">$f : \mathbb{Z} \to E$</span> is a bijection. Thus, we have <span class="math-container">$\mathbb{Z} \approx E$</span>. Using the notation of cardinalities, we have <span class="math-container">$|\mathbb{Z}| = |E|$</span>.</p>
<p>There's nothing circular about any of it.</p>
|
2,962,203 | <p>I got stuck at : <span class="math-container">$a^2/b^2 = 12+2 \sqrt 35$</span></p>
<p>I understand that <span class="math-container">$12$</span> is rational and now I need to prove that <span class="math-container">$\sqrt{35}$</span> is irrational.</p>
<p>so I defined <span class="math-container">$∀c,d∈R$</span> while <span class="math-container">$d$</span> isn't <span class="math-container">$0$</span> that:
<span class="math-container">$c^2/d^2 = \sqrt 35$</span>
so <span class="math-container">$- c^2=(d^2)\sqrt{35}$</span>
It means that <span class="math-container">$c$</span> divide with <span class="math-container">$5$</span> and <span class="math-container">$7$</span>?
Also, how do I prove that if for example <span class="math-container">$X^2/4$</span> then <span class="math-container">$X/4$</span>? </p>
| Barry Cipra | 86,747 | <p>Here's a different approach, assuming you know that <span class="math-container">$\sqrt5$</span> and/or <span class="math-container">$\sqrt7$</span> are irrational.</p>
<p>If <span class="math-container">$\sqrt7+\sqrt5$</span> were rational, then </p>
<p><span class="math-container">$${2\over\sqrt7+\sqrt5}=\sqrt7-\sqrt5$$</span></p>
<p>would also be rational, in which case</p>
<p><span class="math-container">$$\sqrt7={\sqrt7+\sqrt5\over2}+{\sqrt7-\sqrt5\over2}$$</span></p>
<p>and</p>
<p><span class="math-container">$$\sqrt5={\sqrt7+\sqrt5\over2}-{\sqrt7-\sqrt5\over2}$$</span></p>
<p>would be rational as well. </p>
|
463,139 | <p>I have this:</p>
<p>Case 1)</p>
<p><img src="https://i.stack.imgur.com/xEEFQ.png" alt="enter image description here"></p>
<p>If <em>f</em> is a pair function $f(-x)=f(x)$ then $\int_{-a}^a f(x)dx=2\int_0^af(x)dx$</p>
<p>Case 2)</p>
<p><img src="https://i.stack.imgur.com/mnHbB.png" alt="enter image description here"></p>
<p>If $f$ is a inpair function $f(-x)=-f(x)$ then $\int_{-a}^a f(x)dx=0$</p>
<p>I understand the reasons for the case 1 to be double area and for case 2 to be zero, but, I'll be grateful if some one can tell me a little more of this <strong>symmetry aspect</strong> </p>
<p><strong>How can I realize when and where this symmetry exist in a function</strong>
Particularly will be helpful if someone explain how to interpret $f(-x)=-f(x)$ and $f(-x)=f(x)$ <strong>I'd to understand better what do they imply</strong></p>
| Jean-Sébastien | 31,493 | <p>There are many reasons why we call those even and odd functions. One is that the taylor series of an even function only includes even powers whereas the Taylor series of an odd function only includes odd power.</p>
<p>But to recognize when and what type of symmetry there is, the most useful property is that they are called even/odd because they often act similar to even/odd numbers:</p>
<ul>
<li>The sum of two even functions is even, and any constant multiple of an even function is even.</li>
<li>The sum of two odd functions is odd, and any constant multiple of an odd function is odd.</li>
<li>The difference between two odd functions is odd.</li>
<li>The difference between two even functions is even.</li>
<li>The product of two even functions is an even function.</li>
<li>The product of two odd functions is an even function.</li>
<li>The product of an even function and an odd function is an odd function.</li>
<li>The quotient of two even functions is an even function.</li>
<li>The quotient of two odd functions is an even function.</li>
<li>The quotient of an even function and an odd function is an odd function.</li>
</ul>
<p>With that in mind, you need to have a couple basic even/odd functions in your sleeve.
$x^n$ is even for $n$ even and odd for $n$ odd (even for negative integers). the function $\cos$ is even while $\sin$ is odd. What would $\tan$ be? The absolute value $|x|$ is even.</p>
<p>When looking at some functions $f(x)=g(x)/h(x)$, use the above properties to determine the "parity" of $g(x)$ and $h(x)$ and then check their quotient.</p>
<p>Note however that some functions are neither even nor odd, for example $x^3-x+1$.</p>
|
2,642,144 | <p>How would I prove or disprove the following statement?
$ \forall a \in \mathbb{Z} \forall b \in \mathbb{N}$ , if $a < b$ then $a^2 < b^2$</p>
| Community | -1 | <p>If you choose $a$ as negative number and $b$ as a positive number with module less then $a$ you find a counter example.</p>
<p>In fact your statement is false $\forall a\in\mathbb{Z},a<0\forall b\in\mathbb{N}:|a|^2\geq b^2$</p>
<p>As @kevin said if you take $a=-5$ and $b=1$ the statement is false.</p>
|
3,213,464 | <p>Does 22.449 approximate to 22 or 23?
If we see it one way
<span class="math-container">$22.449≈22$</span>
But on the other hand
<span class="math-container">$22.449≈22.45≈22.5≈23$</span>
Which one is correct?</p>
| Borelian | 290,706 | <p><a href="https://en.wikipedia.org/wiki/IEEE_754" rel="nofollow noreferrer">IEEE 754</a> is a standard that defines how to round a number to the nearest integer (at least for computers). There are many conventions, and in some of them 22.5 is rounded to either 22 or 23. However, in all the cases 22.449 is rounded to 22, because it is closer to 22 than to 23.</p>
|
1,716,656 | <p>I am having trouble solving this problem</p>
<blockquote>
<p>Julie bought a house with a 100,000 mortgage for 30 years being repaid with payments at the end of each month at an interest rate of 8% compounded monthly. If Julie pays an extra 100 each month, what is the outstanding balance at the end of 10 years immediately after the 120th payment?</p>
</blockquote>
<p>My attempt:</p>
<p>I first want to find the deposit per month.</p>
<p>I let $D$ be the deposit per month and since it increases by 100 each payment, I used an increasing annuity,</p>
<p>$D*100(Ia_{30|0.08}) = 100,000$</p>
<p>However, the $D$ I got was 8.12, which is clearly not right.</p>
<p>Can someone help?</p>
| Sava B. | 326,626 | <p>Could use the multiplication rule:</p>
<p>The probability of Die 1 landing on 1-4 is 4/6.
The probability of the Die 2 landing on the number that's Die1+2 is then 1/6. </p>
<p>(4/6) * (1/6) = 4/36</p>
<p>We multiply this by 2 to account the scenario where Die 2 is the 1-4 die, and then Die 1 is two higher than Die 2. So, 8/36. </p>
|
2,426,263 | <p>Nowadays, the most widely-taught model of computation (at least in the English-speaking world) is that of Turing Machines, however, it wasn't the first Turing-Complete model out there: μ-recursive functions came a few years earlier, and λ-calculus came a year earlier. Why is it that Turing machines are so popular today. Intuitive appeal? The depth or notoriety of Turing's work? Or is the model perceived to be "natural" in some sense, and if so, why?</p>
| J.-E. Pin | 89,374 | <p>There are several reasons of the popularity of Turing machines: </p>
<p>(1) Turing machines fit well in a course of automata theory: you can start by introducing simpler models, like finite automata or pushdown automata and then move to TMs.</p>
<p>(2) The statement of the most famous open problem in theoretical computer science, namely P = NP, and more generally <a href="https://en.wikipedia.org/wiki/Computational_complexity_theory#Machine_models_and_complexity_measures" rel="nofollow noreferrer">Computational complexity theory</a> heavily relies on TMs.</p>
<p>(3) The notion is quite flexible and admits several extensions like <a href="https://en.wikipedia.org/wiki/Alternating_Turing_machine" rel="nofollow noreferrer">Alternating TMs</a>, <a href="https://en.wikipedia.org/wiki/Probabilistic_Turing_machine" rel="nofollow noreferrer">Probabilistic TMs</a>, <a href="https://en.wikipedia.org/wiki/Quantum_Turing_machine" rel="nofollow noreferrer">Quantum TMs</a>, just to name a few.</p>
<p>Some people might also be perhaps fascinated by the <a href="https://en.wikipedia.org/wiki/Busy_beaver" rel="nofollow noreferrer">Busy Beaver problem</a> (see
<a href="https://arxiv.org/pdf/1704.08752.pdf" rel="nofollow noreferrer">Busy Beaver Scores and Alphabet Size</a> for recent results) and its applications to <a href="https://cs.stackexchange.com/questions/66486/goldbach-conjecture-and-busy-beaver-numbers">Goldbach's conjecture</a>: there is a <a href="https://gist.github.com/jms137/cbb66fb58dde067b0bece12873fadc76" rel="nofollow noreferrer">47-state, 2-symbol, 1-tape Turing machine that halts iff Goldbach's conjecture is false</a>.</p>
|
1,411,305 | <p>I have been trying to solve the following problem:</p>
<blockquote>
<p>What is the probability that among 3 random digits, there appear
exactly 2 different ones?</p>
</blockquote>
<p>The formula for no repititions is:</p>
<pre><code>(n*(n-1)...(n-r+1))/n^r
</code></pre>
<p>So, for the first digit there are 10 possibilities, for the second 9.
Next, the third digit should be the same as previous two: it is binomial(2,1)
And there are total 10^3 possibilities of ordering with replacement:
My formula is:</p>
<pre><code>(10*9*binomial(2,1))/10^3
</code></pre>
<p>But it leads to the wrong answer.
Could you please help me understand where is my mistake and the logic of solving such problems?
Thank you </p>
| vonbrand | 43,946 | <p>Your mistake is that you don't consider the case where the first digit repeats. You have the following exhaustive and exclusive cases:</p>
<ul>
<li><strong>The first digit repeats:</strong> There are $10$ options for the first digit, which can appear in any of the $2$ remaining positions, and the other digit is one in $9$. There are $10 \cdot 2 \cdot 9 = 180$ cases here</li>
<li><strong>The first digit doesn't repeat:</strong> You can select the first digit in $10$ ways, and the other two (which repeat) in $9$ ways, for a total of $10 \cdot 9 = 90$</li>
</ul>
<p>Adding both gives $180 + 90 = 270$ cases. </p>
|
2,243,542 | <p>In a previous question here <a href="https://math.stackexchange.com/q/2240195/369757">Can we define the Cantor Set in this way?</a></p>
<p>we defined a family of sets $ \left\{ C_0,C_1,C_2,C_3,\dots \right\}$</p>
<p>We can call this set $S_1$ , where the values of these elements is</p>
<p>$C_0 = \left\{ 0.0 \right\}$</p>
<p>$C_1 = \left\{ 0.0 , 0.2 \right\}$</p>
<p>$C_2 = \left\{ 0.0 , 0.2 ,0.02 , 0.22 \right\}$</p>
<p>$C_3 = \left\{ 0.0 , 0.2 ,0.02 , 0.22 ,0.002 , 0.202 ,0.022 , 0.222 \right\}$</p>
<p>and so on...</p>
<p>We can then take the union of $\bigcup S_1$ and get some countable set $X$</p>
<p>We will define a set $S_2$ to be all the members of $S_1$ that are not redundant.</p>
<p>$C_0$ is redundant because it can be removed from $S_1$ and the union will still be the same.</p>
<p>In fact all the elements of $S_1$ are redundant so our set</p>
<p>$\bigcup S_2 = \left\{ \not C_0,\not C_1,\not C_2,\not C_3,\dots \right\} = \varnothing \neq X \to$ Contradiction</p>
<p>We removed all the sets from the union that did not contribute any information, and all the information disappeared.</p>
<p>What went wrong here?</p>
| Ivan Hieno | 369,757 | <p>Casting this in terms of the set of natural numbers, there seems to be no contradiction if $ \mathbb{N} \in S_1 $.</p>
<p>Let:$$ S_1 = \left\{ C_m = \left\{ n \in \mathbb{N} : n \le m \right\} : m \in \mathbb{N} \right\} $$
Then:$$C_0= \left\{ 0 \right\}$$</p>
<p>$$C_1= \left\{ 0,1 \right\}$$</p>
<p>$$C_2= \left\{ 0,1,2 \right\}$$</p>
<p>$$\dots$$</p>
<p>$$ S_1 = \left\{ C_0,C_1,C_2,C_3,\dots , \mathbb{N} \right\} $$</p>
<p>$$ \mathbb{N} \in S_1 $$</p>
<p>$$\bigcup S_1 = \mathbb{N} $$</p>
<p>Let:$$ S_2 = \left\{ C_n\not\subset (X \in S_1) : n \in \mathbb{N} \right\} $$</p>
<p>Then:$$ S_2 = \left\{ \not C_0,\not C_1,\not C_2,\not C_3,\dots , \mathbb{N} \right\} $$</p>
<p>$$S_2 = \left\{\mathbb{N} \right\} $$</p>
<p>$$\bigcup S_2 = \mathbb{N} $$</p>
<p>It only gets ugly if $ \mathbb{N} \notin S_1 $</p>
|
12,717 | <p>In the familiar case of (smooth projective) curves over an algebraically closed fields, (closed) points correspond to DVR's.</p>
<p>What if we have a non-singular projective curve over a non-algebraically closed field? The closed points will certainly induce DVR's, but would all DVR's come from closed points? Is there a characterization of the DVR's that aren't induced by closed points?</p>
<p>And how about for a general projective variety that is regular in codimension 1 (both for algebraically closed and non-algebraically closed)? Point of codimension 1 induce DVR's. Do they induce all of them? What is the characterization of the ones they do induce?</p>
<p>How about complete integral schemes that are regular in codimension 1?</p>
| Qing Liu | 3,485 | <p>To complete partly the answer of Emerton, the picture for DVR is relatively clear. Let $X$ be an integral noetherian scheme and let $R$ be a DVR with field of fractions equal to the field of rational functions $k(X)$ on $X$. Suppose that $R$ has a center $x\in X$ (e.g. if $X$ is proper over a subring of $R$). Let $k_R$ be the residue field of $R$. Then $k_R$ has transcendental degree over $k(x)$ bounded by $\dim O_{X,x} -1$. Suppose further that $X$ is universally catenary and Nagata (e.g. $X$ is excellent), then the equality holds if and only if the center of $R$ in some $X'$ proper and birational over $X$ is a regular point of codimension 1. This is a theorem of Zariski. See M. Artin: ''Néron Models'', § 5, in Cornell & Silverman: "Arithmetic Geometry". </p>
|
1,442,240 | <p>I have a little question, that run threw my thoughts, when i saw this exercise: $$\lim _{x\to \infty }\left(\frac{\int _{sin\left(x\right)}^xe^{\sqrt{t^2+1}}dt}{e^{\sqrt{x^2-1}}}\:\right)$$</p>
<p>Of course I want to implement here Lophital's rule, but without showing a calculation, is there intuitive and logical explanation why is the nominator (the integral) is $\infty$ ?</p>
<p>I asked myself this question, because I remember that double integral on constant 1 for exmaple is the the area, and triple on 1 is the velocity, so here I tried to think what is an integral with the same domain but with constant 1? </p>
<p>$$\int _{sin\left(x\right)}^x1dt\:$$</p>
<p>That is defintly diverges because $x-sin(x)$ when x "aproaches" infnity.
And here, as I conclude, there isn't too much geometric meaning to that kind of integral, where the domain is between functions (like here x and sin x), I mean I can't say that it is the area that bounded by x and sin(x), the integral on 1 is just the result of substraction, right ? </p>
<p>So, I dont have any geometrical intuition on that integral, except one thing I see clearly: $$$$ As I said, the integral on 1 diverges and the function $e^{\sqrt{x^2+1}}$ is defintly increasing function, but it is enough to say that it is bigger then 1 for each $x$ and that's why (one of the reasons the limit is $\infty$.</p>
| Jack D'Aurizio | 44,121 | <p><strong>Hint:</strong> By De l'Hopital theorem,
$$ \lim_{x\to +\infty} e^{-x}\int_{0}^{x}e^{\sqrt{t^2+1}}\,dt = \lim_{x\to +\infty}\exp\left(\sqrt{x^2+1}-x\right)=\exp(0)=\color{red}{1}.$$</p>
|
954,419 | <p>I am teaching myself mathematics, my objective being a thorough understanding of game theory and probability. In particular, I want to be able to go through A Course in Game Theory by Osborne and Probability Theory by Jaynes.</p>
<p>I understand I want to cover a lot of ground so I'm not expecting to learn it in less than a year or maybe even two. Still I'm fairly certain it's not impossible.</p>
<p>However I would like to have a study plan more or less fleshed out just to know I'm on the right track. There were some other questions related to self learning math here but I couldn't find one like mine.</p>
<p>I'd appreciate some feedback.</p>
<p>Calc I + II: no book, I already know basic calculus</p>
<ul>
<li>Differential equations: MIT's OCW lectures </li>
<li>Calc III: Stewart's Multivariable calculus</li>
<li>Linear Algebra: Strang, Gilbert, Linear Algebra and Its Applications complemented with MIT's OCW lectures <strong>OR</strong> Linear Algebra Done Right</li>
</ul>
<p>Until here I am more or less certain on what I want to study, but I'm totally confused on what to learn next. Jayne's book states that you need to be familiar with applied mathematics.</p>
<p>After reading about applied mathematics, I came up with this plan to be done after finishing what I mentioned earlier (in order of course, not all at the same time):</p>
<ol>
<li>Topology A: Munkres, part I.</li>
<li>Real analysis: Still not sure about the material, probably Abbott or Rudin.</li>
<li>Complex Analysis: No idea about the material</li>
<li>Group Theory: Rotman, An Introduction to the Theory of Groups</li>
<li>Topology B: Munkres, part II.</li>
</ol>
<p>And then finally,
Jayne's Probability Theory and game theory.</p>
<p>Am I missing something here? Some of these books such as Rotman's are aimed at a graduate level, is it foolish to think I will understand them?</p>
| Shane | 61,746 | <p>Perhaps this should be a comment, but I don't have the reputation. I TA a class that uses the Osborne and Rubinstein text you mention. I can't speak in depth regarding your goals in probability, but here's my advice on Game Theory:</p>
<p>Unless you're looking to publish research in game theory, the mathematics required is not that substantial. You already have a background in Calculus. On top of that, I would do a little work in analysis (I recommend Rudin) and maybe some optimization - <em>Mathematics for Economists</em> by Simon and Blume is a good reference for that. If you're doing more advanced problems, then you'll want to get good at proofs, particularly direct, by contradiction and by induction. Again, Rudin's good for that. Analysis is hard work - if this is your first time doing it, don't be intimidated. Just slog through it! You could take some time on fixed point and separating hyperplane theorems, but you'll rarely use them.</p>
<p>Once you have that under your belt, you're more than ready to start game theory. I might recommend you start with a slightly easier game theory book and then move up to <em>A Course in Game Theory</em>. My favorite undergrad GT textbook is <em>Strategy</em> by Watson. But if you like Osborne, then he also has an introductory text called <em>An Introduction to Game Theory</em>. I'd start with skimming through one those texts before tackling <em>A Course in Game Theory</em>, which goes a little deeper and provides a little less explanation.</p>
<p>Sometimes students come to me and say that they're struggling because they can't do the math required in game theory. Very rarely is this actually the case. The real issue is that game theory is notationally very dense and the professor I TA for values formalism (as do I). For students who have not done a lot of math, this notational density really intimidates them. Anyway - don't let anybody tell you that Game Theory is all about difficult math - it isn't. It just look like it is!</p>
<p>By the way, I think it's admirable what you're doing and I wish you all the best in your self-learning.</p>
|
300,867 | <p>I am having a difficult time understanding where I went wrong with the following:
$$\begin{matrix}4x-y = 1 \\ 2x+3y = 3 \end{matrix} $$
$$\begin{matrix}4x-y = -3 \\ 2x+3y = 3 \end{matrix} $$</p>
<p>I found the inverse of the common coefficient matrix of the systems:
$$A^{-1} \begin{cases} \frac3{14}, \frac1{14} \\ \\ -\frac17, \frac27 \end{cases} $$</p>
<p>The issue is this question:
Find the solutions to the two systems by using the inverse, i.e. by evaluating $A^{-1}B$ where $B$ represents the right hand side (i.e. $B = \begin{bmatrix}1 \\ 3 \end{bmatrix}$ for system (a) and $B = \begin{bmatrix}-3 \\ 3 \end{bmatrix}$ for system (b)). </p>
<p>Now whatever I find for x and y for both solutions keep coming as wrong. I am thinking, I might of read the question wrongly....I am using the negative values to find the x and y. Not too sure how to go at this now...</p>
| Mariano Suárez-Álvarez | 274 | <p>If you want to do this, you can always find an arc-connected compact subset $K\subseteq U$ such that $C\subseteq K$. Then do your argument for $K$ and, since you can restrict the conclusion to $C$, you are happy.</p>
|
3,165,781 | <p>Ok, I am a bit confused.</p>
<p>Does first countability mean that for every element <span class="math-container">$x$</span> in the space, there is a collection of open sets and each of those open sets are countable containing <span class="math-container">$x$</span> and for every open neighborhood of <span class="math-container">$x$</span>, one of those countable open set is contained in. </p>
<p>OR</p>
<p>For every element <span class="math-container">$x$</span> in the space, there is a collection of open sets, where the number of sets in this collection is countably many, that contains <span class="math-container">$x$</span> and for every neighborhood of <span class="math-container">$x$</span>, one of those open sets (not necessarily countable) is contained in?</p>
<p>Thank you in advanced. </p>
| Kavi Rama Murthy | 142,385 | <p>The first one is wrong. Even in the real line open sets are all uncountable so the condition there is not satisfied. The second definition is correct. </p>
|
1,395,619 | <p>One of my friend asked this doubt.Even in lower class we use both as synonyms,he says that these two concepts have difference.Empty set $\{ \}$ is a set which does not contain any elements,while null set ,$\emptyset$ says about a set which does not contain any elements.</p>
<p>I could not make out that...is his argument correct ? if so how ?</p>
| Khallil | 99,916 | <p>They aren't the same although they were used interchangeable way back when.</p>
<blockquote>
<p>In mathematics, a <a href="https://en.wikipedia.org/wiki/Null_set" rel="noreferrer">null set</a> is a set that is negligible in some sense. For different applications, the meaning of "negligible" varies. In measure theory, any set of measure 0 is called a null set (or simply a measure-zero set). More generally, whenever an ideal is taken as understood, then a null set is any element of that ideal.</p>
</blockquote>
<p>Whereas an empty set is defined as:</p>
<blockquote>
<p>In mathematics, and more specifically set theory, the empty set is the unique set having no elements; its size or cardinality (count of elements in a set) is zero. Some axiomatic set theories ensure that the empty set exists by including an axiom of empty set; in other theories, its existence can be deduced. Many possible properties of sets are trivially true for the empty set.</p>
</blockquote>
|
320,452 | <p>For any positive integer <span class="math-container">$n\in\mathbb{N}$</span> let <span class="math-container">$S_n$</span> denote the set of all bijective maps <span class="math-container">$\pi:\{1,\ldots,n\}\to\{1,\ldots,n\}$</span>. For <span class="math-container">$n>1$</span> and <span class="math-container">$\pi\in S_n$</span> define the <em>neighboring number</em> <span class="math-container">$N_n(\pi)$</span> as the minimum distance of <span class="math-container">$\pi$</span>-neighbors, or more formally: <span class="math-container">$$N_n(\pi) = \min \big(\big\{|\pi(k)-\pi(k+1)|:k\in\{1,\ldots,n-1\}\big\}\cup \big\{|\pi(1) - \pi(n)|\big\}\big).$$</span></p>
<p>For <span class="math-container">$n>1$</span> let <span class="math-container">$E_n$</span> be the expected value of the neighboring number of a member of <span class="math-container">$S_n$</span>.</p>
<p><strong>Question.</strong> Do we have <span class="math-container">$\lim\sup_{n\to\infty}\frac{E_n}{n} > 0$</span>?</p>
| Michael Lugo | 143 | <p>Some quick simulation leads me to conjecture that <span class="math-container">$\lim_{n \to \infty} E_n$</span> exists and is around <span class="math-container">$1.15$</span>. </p>
<p>The number of permutations in <span class="math-container">$S_n$</span> with <span class="math-container">$N_n(\pi) = 1$</span> is <a href="https://oeis.org/A129535" rel="nofollow noreferrer">https://oeis.org/A129535</a>. It appears again from numerical evidence that <span class="math-container">$\lim_{n \to \infty} P(N_n > 1) = e^{-2}$</span>, where <span class="math-container">$N_n$</span> is chosen uniformly at random from <span class="math-container">$S_n$</span>. This is consistent with heuristics. Namely, for <span class="math-container">$k = 1, 2, \ldots, n-1$</span>, let <span class="math-container">$X_k = |\pi(k) - \pi(k+1)|$</span> be a random variable, where <span class="math-container">$\pi$</span> is a permutation on <span class="math-container">$[n]$</span> chosen uniformly at random. Then <span class="math-container">$P(X_k = 1) \approx 2/n$</span> and the <span class="math-container">$X_k$</span> are approximately independent, so <span class="math-container">$P(N_n > 1) = P(X_1 > 1, X_2 > 1, \ldots, X_{n-1} > 1) \approx (1-2/n)^{n-1} \approx e^{-2}$</span>.</p>
<p>Furthermore <span class="math-container">$P(N_n > 2) = P(X_1 > 2, X_2> 2, \ldots, X_{n-2} > 2) \approx (1-4/n)^{n-1} \approx e^{-4}$</span> and similarly <span class="math-container">$P(N_n > k) \approx e^{-2k}$</span>. So <span class="math-container">$E_n$</span> should be a geometric random variable with <span class="math-container">$p = 1 - e^{-2}$</span> and we should have <span class="math-container">$\lim_{n \to \infty} E_n = 1/(1-e^{-2}) \approx 1.1565$</span>.</p>
|
909,741 | <blockquote>
<p><strong>ALREADY ANSWERED</strong></p>
</blockquote>
<p>I was trying to prove the result that the OP of <a href="https://math.stackexchange.com/questions/909712/evaluate-int-0-frac-pi2-ln1-cos-x-dx"><strong><em>this</em></strong></a> question is given as a hint.</p>
<p>That is to say: <em>imagine that you are not given the hint and you need to evaluate</em>:</p>
<blockquote>
<p>$$I = \int^{\pi/2}_0 \log{\cos{x}} \, \mathrm{d}x \color{red}{\overset{?}{=} }\frac{\pi}{2} \log{\frac{1}{2}} \tag{1}$$ </p>
</blockquote>
<p><em>How would you proceed?</em></p>
<hr>
<p>Well, I tried the following steps and, despite it seems that I am almost there, I have found some troubles:</p>
<ul>
<li>Taking advantage of the fact: $$\cos{x} = \frac{e^{ix}+e^{-ix}}{2}, \quad \forall x \in \mathbb{R}$$</li>
<li>Plugging this into the integral and performing the change of variable $z = e^{ix}$, so the line integral becomes a contour integral over <em>a quarter of circumference of unity radius centered at $z=0$</em>, i.e.:
$$ I = \frac{1}{4i} \oint_{|z|=1}\left[ \log{ \left(z+\frac{1}{z}\right)} - \log{2} \right] \, \frac{\mathrm{d}z }{z}$$</li>
</ul>
<blockquote>
<p>$\color{red}{\text{We cannot do this because the integrand is not holomorphic on } |z| = 1 }$</p>
</blockquote>
<ul>
<li>Note that the integrand has only one pole lying in the region enclosed by the curve $\gamma : |z|=1$ and it is holomorphic (is it?) almost everywhere (except in $z =0$), so the residue theorem tells us that:</li>
</ul>
<p>$$I = \frac{1}{4i} \times 2\pi i \times \lim_{z\to0} \color{red}{z} \frac{1}{\color{red}{z}} \left[ \underbrace{ \log{ \left(z+\frac{1}{z}\right)} }_{L} - \log{2} \right] $$</p>
<ul>
<li>As I said before, it seems that I am almost there, since the result given by eq. (1) follows iff $L = 0$, which is not true (I have tried L'Hôpital and some algebraic manipulations).</li>
</ul>
<p>Where did my reasoning fail? Any helping hand?</p>
<p>Thank you in advance, cheers!</p>
<hr>
<p>Please note that I'm not much of an expert in either complex analysis or complex integration so please forgive me if this is trivial.</p>
<hr>
<p>Notation: $\log{x}$ means $\ln{x}$.</p>
<hr>
<p>A graph of the function $f(z) = \log{(z+1/z)}$ helps to understand the difficulties:</p>
<p><img src="https://i.stack.imgur.com/jAjTP.png" alt="enter image description here"></p>
<p>where $|f(z)|$, $z = x+i y$ is plotted and the white path shows where $f$ is not holomorphic.</p>
| dustin | 78,317 | <p>As shown by @idm, we have that
$$
\int_0^{\pi/2}\ln(\cos(x))dx = \int_0^{\pi/2}\ln(\sin(x))dx.
$$
We can exploit this identity for another one and use Feynman's method (differentiating under the integral).
Consider
$$
\int_0^{\pi/2}x\cot(x)dx.\tag{1}
$$
By integration by parts, we have
$$
\int_0^{\pi/2}x\cot(x)dx = x\ln(\sin(x))\Bigr|_0^{\pi/2} - \int_0^{\pi/2}\ln(\sin(x))dx = - \int_0^{\pi/2}\ln(\sin(x))dx
$$
since $\lim_{x\to 0}x\ln(\sin(x)) = 0$. Therefore, we can evaluate the negative of equation $(1)$.
\begin{align}
I(\alpha) &= \int_0^{\pi/2}\arctan(\alpha\tan(x))\cot(x)dx\tag{2}\\
I'(\alpha) &= \int_0^{\pi/2}\frac{\partial}{\partial\alpha}\Bigl[\arctan(\alpha\tan(x))\cot(x)\Bigr]dx\\
&= \int_0^{\pi/2}\frac{dx}{\alpha^2\tan^2(x) + 1}\\
&= \frac{\pi}{2(\alpha +1)}\\
I(\alpha) &= \frac{\pi}{2}\ln(\alpha + 1) + C
\end{align}
Thus, $I(0)\Rightarrow C=0$ so
$$
I(\alpha) = \frac{\pi}{2}\ln(\alpha + 1)
$$
We recover equation $(1)$ from equation $(2)$ when $\alpha = 1$ so
$I(1) = \frac{\pi}{2}\ln(2)$ and since
$$
\int_0^{\pi/2}\ln(\sin(x))dx = -\int_0^{\pi/2}x\cot(x)dx = -\frac{\pi}{2}\ln(2),
$$
we have
$$
\int_0^{\pi/2}\ln(\sin(x))dx = -\frac{\pi}{2}\ln(2),
$$</p>
|
792,924 | <p>If a quantity can be either a scalar or a vector, how would one call that property? I could think of scalarity but I don't think such a term exists.</p>
| Mikhail Katz | 72,694 | <p>No because by the Cauchy-Schwarz inequality, $\int_0^1 x^n f(x)dx\leq \sqrt{\int_0^1 x^{2n}dx}\sqrt{\int_0^1 f^2(x)dx}$,where the second factor is constant whereas the first factor tends to zero. This shows there is not even an $L^2$ function with such a property.</p>
|
347,385 | <p>Assume $f(x) \in C^1([0,1])$,and $\int_0^{\frac{1}{2}}f(x)\text{d}x=0$,show that:
$$\left(\int_0^1f(x)\text{d}x\right)^2 \leq \frac{1}{12}\int_0^1[f'(x)]^2\text{d}x$$</p>
<p>and how to find the smallest constant $C$ which satisfies
$$\left(\int_0^1f(x)\text{d}x\right)^2 \leq C\int_0^1[f'(x)]^2\text{d}x$$</p>
| math110 | 58,742 | <p>solutin 2:</p>
<p>by Schwarz,we have
$$\int_{0}^{\frac{1}{2}}[f'(x)]^2dx\int_{0}^{\frac{1}{2}}x^2dx\ge\left(\int_{0}^{\frac{1}{2}}xf'(x)dx\right)^2=\left[\dfrac{1}{2}f(\dfrac{1}{2})-\int_{0}^{\frac{1}{2}}f(x)dx\right]^2$$
so
$$\int_{0}^{\frac{1}{2}}[f'(x)]^2dx\ge 24\left[\dfrac{1}{2}f(\dfrac{1}{2})-\int_{0}^{\frac{1}{2}}f(x)dx\right]^2$$
the same methods,we have
$$\int_{\frac{1}{2}}^{1}[f'(x)]^2dx\ge 24\left[\dfrac{1}{2}f(\dfrac{1}{2})-\int_{0}^{\frac{1}{2}}f(x)dx\right]^2$$</p>
<p>and use $2(a^2+b^2)\ge (a+b)^2$</p>
<p>then we have </p>
<p>$$\int_{0}^{1}[f'(x)]^2dx\ge 12\left(\int_{0}^{1}f(x)dx-2\int_{0}^{\frac{1}{2}}f(x)dx\right)^2$$</p>
|
706,980 | <p>If I know that $f(z)$ is differentiable at $z_0$, $z_0 = x_0 + iy_0$.
How do I prove that $g(z) = \overline{f(\overline{z})}$ is differentiable at $\overline z_0$?</p>
| BigM | 90,395 | <p>Let $f(z)=f(x+iy)=u(x,y)+iv(x,y)$, then $f(\bar{z})=f(x-iy)=u(x,-y)+iv(x,-y)\Rightarrow \overline{f(\bar{z})}=u(x,-y)-iv(x,-y)$.Put $u_{1}(x,y)=u(x,-y)$ and $v_1(x,y)=-v(x,-y)$.</p>
<p>We have $\frac{\partial}{\partial x}u_1=\frac{\partial }{\partial x}u(x,-y)=\frac{\partial }{\partial x}u(x,-y)$ and $\frac{\partial}{\partial y}v_1=-\frac{\partial }{\partial y}v(x,-y)$. If $f$ satisfies Cauchy-Riemann equations then $\frac{\partial }{\partial x}u(x,-y)=\frac{\partial }{\partial y}v(x,-y)=-\frac{\partial}{\partial y}v(x,-y)$. So $\frac{\partial u_1}{\partial x}=\frac{\partial v_1}{\partial y}$. Simlarly you can show $\frac{\partial u_1}{\partial y}=-\frac{\partial v_1}{\partial x}$. Actually this is biconditional since arrows are reversible. $f(z)$ is holomorphic iff $\overline{f(\overline{z})}$ is holomorphic.</p>
|
3,288,815 | <p>I'm reading theorem 3.11 in Rudin's RCA what says <span class="math-container">$L^p$</span> space is a complete metric space.
At the end of the proof, Rudin says that "Then <span class="math-container">$\mu(E)=0$</span>, and on the complement of <span class="math-container">$E$</span> the sequence <span class="math-container">${f_n}$</span> converges uniformly to a bounded function <span class="math-container">$f$</span>".
Why <span class="math-container">${f_n}$</span> converges uniformly?
I see <span class="math-container">$$E^c=\{f_n : |f_n (x)|\leq \|f_n\|_\infty \text{ and } |f_n-f_m|\leq \|f_n-f_m\|_\infty \}$$</span></p>
<p><img src="https://i.stack.imgur.com/EqFah.jpg" alt=""></p>
| Kavi Rama Murthy | 142,385 | <p>Since <span class="math-container">$\|f_n(x)-f_m(x)| \leq \|f_n-f_m\|_{\infty}$</span> for all <span class="math-container">$n,m$</span>, for <span class="math-container">$x$</span> not in <span class="math-container">$E$</span> and since <span class="math-container">$\|f_n-f_m\|_{\infty} \to 0$</span> the convergence is uniform. Remember that this proof is for the case <span class="math-container">$p=\infty$</span> where we are starting with a Cauchy sequence in <span class="math-container">$L^{\infty}$</span>.</p>
|
490,802 | <p>Is $(x,3,5)$ a plane, for $x\in\mathbb{R}$?</p>
<p>I know that if two of the coordinates are "arbitrary", like $(x,y,4)$or $(3,y,z)$, then it creates a plane (for $x,y,z\in \mathbb{R}).$</p>
<p>Is there a way to tell if it would create a plane in $\mathbb{R}^3?$</p>
| imranfat | 64,546 | <p>It doesn't. It is a line in space parallel to the x-axis. Graph the points (-1,3,5) , (0,3,5) , (1,3,5) , (2,3,5) etc. What do you notice?</p>
|
4,190,301 | <p>I found <span class="math-container">$\tilde{R}$</span> in a mathematical text, and I would like to know how this is pronounced. I tried to search on the internet but was not able to find anything related.</p>
| Lee Mosher | 26,501 | <p>That symbol on top is a <a href="https://en.wikipedia.org/wiki/Tilde" rel="noreferrer">tilde</a>.</p>
<p>So you can pronounce that "arr tilde" or sometimes "tilde arr", depending on your preference. I've gotten used to "tilde arr" because that's how it's typed in LaTeX, namely \tilde R which when you put dollar signs around it, becomes <span class="math-container">$\tilde R$</span> (which, I see, is pretty much what you typed in your post).</p>
|
2,155,589 | <p>I'm reading a computer science book that gives several functions, in the mathematical sense. There are two that are the basis of this question.</p>
<p>These are equations used to convert a number represented in base ten to a bit representation using two's complement and back.</p>
<p>One function makes the conversion from binary two's complement to decimal, $B2T_w$, defined as $$B2T_w(x) = -x_{w-1}2^{w-1} + \sum_{i=0}^{w-2} x_i2^i$$ where x is a vector of length w.</p>
<p>It then goes to define another function, $T2B_w$ as the inverse of $B2T_w$; which it does not define using mathematical notation.</p>
<p>I understand how to convert between a two's complement bit representation of a number to it's decimal representation, but for the sake of understanding I'd like to know how to derive the inverse of $B2T_w$</p>
<p>How do I find $B2T_w^{-1}$?</p>
| Veridian Dynamics | 408,632 | <p>Let $\epsilon >0$. Take $\delta = \min\{1,\frac{\epsilon}{3(1+|z_{0}|)^{2}}\}$. If $|z-z_{0}|<\delta$, then $|z|\leq|z-z_{0}|+|z_{0}|< \delta+|z_{0}|$. So,
$$|z^{3}-z_{0}^{3}|=|z-z_{0}||z^{2}+zz_{0}+z_{0}^{2}|< \delta((\delta+|z_{0}|)^{2}+(\delta+|z_{0}|)||z_{0}|+|z_{0}|^{2})\leq3\delta (1+|z_{0}|)^{2}\leq\epsilon$$</p>
|
3,819,202 | <p>Can anyone explain to solve the identity posted by my friend <span class="math-container">$$2\cos12°= \sqrt{2+{\sqrt{2+\sqrt{2-\sqrt{2-...}}} }}$$</span> which is an infinite nested square roots of 2. <strong>(Pattern <span class="math-container">$++--$</span> repeating infinitely)</strong></p>
<p>Converging to finite nested radical of <span class="math-container">$2\cos12° = \frac{1}{2}\times\sqrt{9+\sqrt5+\sqrt{(30-6\sqrt5)}}$</span></p>
<p>The finite nested radical, I was able to derive <span class="math-container">$\cos12° = \cos(30-18)°$</span> as follows</p>
<p><span class="math-container">$$\cos30°\cdot\cos18° + \sin30°\cdot\sin18°$$</span>
<span class="math-container">$$= \frac{√3}{2}\cdot\frac{\sqrt{2+2\cos36°}}{2}+\frac{1}{2}\cdot\frac{\sqrt{2-2\cos36°}}{2}$$</span>
Where <span class="math-container">$\cos18° = \frac{\sqrt{2+2\cos36°}}{2}$</span> (by Half angle cosine formula) and
<span class="math-container">$\sin18° = \frac{\sqrt{2-2\cos36°}}{2}$</span> (solving again by half angle cosine formula)
<span class="math-container">$2\cos36° =\frac{ \sqrt5 +1}{2}$</span> which is golden ratio</p>
<p><span class="math-container">$\frac{\sqrt3}{2}\cdot\frac{\sqrt{10+2\sqrt5}}{4}+ \frac{1}{2}\cdot\frac{\sqrt{5}-1}{4} = \frac{\sqrt{30+6\sqrt5}}{8}+ \frac{\sqrt5-1}{8}$</span></p>
<p>Further steps finally lead to the finite nested radical</p>
<p>Method actually I tried to solve infinite nested square roots of 2 is as follows.</p>
<p><span class="math-container">$2\cos\theta = \sqrt{2+2\cos2\theta}$</span> and
<span class="math-container">$2\sin\theta = \sqrt{2-2\cos2\theta}$</span></p>
<p>Now simplifying infinite nested square roots of 2, we will get the following as simplified nested radical
<span class="math-container">$$2\cos12° = \sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-2\cos12°}}}}$$</span></p>
<p>Simplifying step by step as follows</p>
<p><span class="math-container">$2\cos12° = \sqrt{2+\sqrt{2+\sqrt{2-2\sin6°}}}$</span> then</p>
<p><span class="math-container">$2\cos12° = \sqrt{2+\sqrt{2+\sqrt{2-2\cos84°}}}$</span> (by <span class="math-container">$\sin\theta = \cos(90-\theta)$</span></p>
<p><span class="math-container">$2\cos12° = \sqrt{2+\sqrt{2+2\sin42°}}$</span></p>
<p><span class="math-container">$2\cos12° = \sqrt{2+\sqrt{2+2\cos48°}}$</span></p>
<p><span class="math-container">$2\cos12° = \sqrt{2+2\cos24°}$</span></p>
<p><span class="math-container">$2\cos12° = 2\cos12°$</span></p>
<p>We are back to <span class="math-container">$\sqrt1$</span></p>
<p>Actually this is how I got stuck!</p>
<p>But for infinite nested square roots of 2(as depicted), if I run program in python I am able to get good approximation ( Perhaps if we run large number of nested square roots in python we get more number of digits matching the finite nested radical), because I'm not able get anywhere solving such a kind of infinite cyclic nested square roots of 2.</p>
<p>Dear friends, is there anyway to find the solution by any other means like solving infinite nested square roots</p>
<p>Thanks in advance.</p>
| Sivakumar Krishnamoorthi | 686,991 | <p>Somehow I got the answer from my subsequent post (after a long homework for cyclic infinite nested square roots of 2)</p>
<p>Sivakumar Krishnamoorthi (<a href="https://math.stackexchange.com/users/686991/sivakumar-krishnamoorthi">https://math.stackexchange.com/users/686991/sivakumar-krishnamoorthi</a>), Solving cyclic infinite nested square roots of 2 as cosine functions, URL (version: 2020-09-26): <a href="https://math.stackexchange.com/q/3841605">https://math.stackexchange.com/q/3841605</a></p>
<p><span class="math-container">$2\cos48°$</span> or <span class="math-container">$2\cos\frac{4}{15}π$</span> is cyclic infinite nested square roots of 2 of form <span class="math-container">$cin\sqrt2[2-2+]$</span> i.e. <span class="math-container">$\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+...}}}}$</span></p>
<p>According to half angle cosine formula within next 2 steps we get <span class="math-container">$2\cos12°$</span> or <span class="math-container">$2\cos\frac{π}{9}$</span> as <span class="math-container">$cin\sqrt2[2+2-]$</span> i.e. cyclic infinite nested square roots of 2 as <span class="math-container">$\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-...}}}}$</span></p>
<blockquote>
<p><strong>Therefore <span class="math-container">$2\cos12°$</span> as a multiple of 3 it can be solved as nested radicals as in LHS of my question and on RHS it is cyclic infinite nested square roots of 2 which is also equivalent. As both the result belong to same <span class="math-container">$\cos \frac{π}{15}$</span> cyclic infinite nested square roots of 2 on RHS and finite nested radical are the same in terms of results.</strong></p>
</blockquote>
|
1,436,867 | <p>I don´t know an example wich $ \rho (Ax,Ay)< \rho (x,y) $ $ \forall x\neq y $ is not sufficient for the existence of a fixed point .
can anybody help me? please</p>
| eudes | 265,976 | <p>Even simpler example, although not on whole $\mathbb R$:<br>
$$f\!: \mathbb R\setminus\{0\}\to\mathbb R,\quad f(x) = \frac 12 x.$$
(This is what ThePortakal said in commment.)</p>
|
71,636 | <p>For a self-map $\varphi:X\longrightarrow X$ of a space $X$, many important notions of entropy are defined through a limit of the form $$\lim_{n\rightarrow\infty}\frac{1}{n}\log a_n,$$ where in each case $a_n$ represents some appropriate quantity (see, for example, <a href="https://mathoverflow.net/questions/69218/if-you-were-to-axiomatize-the-notion-of-entropy/69231#69231">this answer</a> to one of my previous questions.) Let $h(\varphi)$ denote a typical entropy that is defined by a limit as above and <strong>after</strong> Ian's example, assume that $h(\varphi)>0$. Does anyone know if limits of the form
\begin{equation}
\lim_{n\rightarrow\infty}\ \ \frac{a_n}{\exp(n\cdot h(\varphi))}
\end{equation} have been studied anywhere? I will appreciate any possible information about such limits. For example, is there a known case where the limit exists? If so, what is the limit called? etc.</p>
<p><strong>EDIT</strong>: As pointed out later by Ian, even if we assume $h(\varphi)>0$ this limit may not exist. I was curios to know if there were instances where the limit is known to exist. Or even better, can one characterize self-maps $\varphi$ for which the limit exists? </p>
| Vaughn Climenhaga | 5,701 | <p>The limit exists for the first two examples that come to mind, namely topological entropy on the full shift and on certain simple Markov shifts.</p>
<p>If $X \subset \Sigma_d^+ = \{1,2, \dots, d\}^{\mathbb{N}}$ and $\sigma$ is the shift map, then for the topological entropy the quantity $a_n$ denotes the number of words of length $n$ that appear in some sequence $x\in X$. If $X$ is the full shift, then $a_n = d^n$, the entropy is $h = \log d$, and we quickly see that $a_n / e^{nh} = 1$ for all $n$.</p>
<p>Slightly more interesting is when you have a Markov shift, say $X\subset \Sigma_2^+$ determined by the transition matrix $\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$.</p>
<p>In this case it's not hard to show that the sequence $a_n$ is actually the Fibonacci sequence, and thus writing $\phi = \frac {1+\sqrt 5}2$ and $\psi = \frac { 1-\sqrt 5}2$, we have
$$
a_n = \frac 1{\sqrt 5} (\phi^{n+2} - \psi^{n+2}).
$$
Since $|\psi|<1$, this shows that the limit of $a_n / e^{nh}$ exists.</p>
<p>My guess is that a similar argument works for other Markov shifts and shows that the limit exists in those cases, based on obtaining a recurrence relation for $a_n$ and then an exact formula using standard tools for solving such sequences.</p>
<p>All that said, it's not immediately clear what the significance of the limit is, and I don't know of any name for it. For other interesting shifts, such as sofic shifts or shifts with specification, I'd be surprised if the limit always exists. </p>
<p>What <strong>is</strong> certainly quite important is to have conditions under which the ratio $a_n / e^{nh}$ is bounded away from $0$ and $\infty$. Such estimates are a significant part of arguments on the uniqueness of a measure of maximal entropy (and more generally uniqueness of equilibrium states), in particular the proof that such a measure satisfies a Gibbs property. For example, see Bowen's 1975 paper "<a href="http://www.springerlink.com/content/uu2n4xh52463rn14/" rel="nofollow">Some systems with unique equilibrium states</a>". (Dan Thompson and I also struggled with this not too long ago in Section 5.1 of <a href="http://arxiv.org/abs/1011.2780" rel="nofollow">this paper</a>.)</p>
<p>It turns out that in the general setting, one of those bounds is immediate -- the sequence $a_n$ is submultiplicative, and so it's not hard to show that $a_n \geq e^{nh}$ for all $n$, whatever other properties the shift space has. Getting an upper bound on $a_n / e^{nh}$ is harder and requires some sort of specification property.</p>
|
71,636 | <p>For a self-map $\varphi:X\longrightarrow X$ of a space $X$, many important notions of entropy are defined through a limit of the form $$\lim_{n\rightarrow\infty}\frac{1}{n}\log a_n,$$ where in each case $a_n$ represents some appropriate quantity (see, for example, <a href="https://mathoverflow.net/questions/69218/if-you-were-to-axiomatize-the-notion-of-entropy/69231#69231">this answer</a> to one of my previous questions.) Let $h(\varphi)$ denote a typical entropy that is defined by a limit as above and <strong>after</strong> Ian's example, assume that $h(\varphi)>0$. Does anyone know if limits of the form
\begin{equation}
\lim_{n\rightarrow\infty}\ \ \frac{a_n}{\exp(n\cdot h(\varphi))}
\end{equation} have been studied anywhere? I will appreciate any possible information about such limits. For example, is there a known case where the limit exists? If so, what is the limit called? etc.</p>
<p><strong>EDIT</strong>: As pointed out later by Ian, even if we assume $h(\varphi)>0$ this limit may not exist. I was curios to know if there were instances where the limit is known to exist. Or even better, can one characterize self-maps $\varphi$ for which the limit exists? </p>
| Barbara Schapira | 30,691 | <p>In the case of the geodesic flow acting on the unit tangent bundle of a compact negatively curved manifold, if $a_n$ is the number of closed geodesics of length at most $n$, and $h$ the topological entropy of the geodesic flow, Margulis proved that $a_n$ is equivalent to $\frac{e^{hn}}{hn}$. </p>
<p>The original article has only 2 pages, but his phd thesis was relatively recently published as a book. </p>
|
197,603 | <p>I'm a newcomer in topology, so I have many things chaotic in my minds, so I hope you could help me.
In order topology, an basis has structure $(a,b)$, right. This is no problem when considering a topology like R, but, what if the number of elements between a and b is finite, so we can write $$(a,b) = [a_1, b_1]$$ which is not open, right?
Can any one explain to me. Thanks</p>
| G. R. | 41,674 | <p>In some ordered sets like $[0,1]$, in order to get a base for the order topology, you need consider too the intervals of the form $(\leftarrow,a)$ and $(b,\rightarrow)$.</p>
|
2,038,189 | <p>(Note: I didn't learn how to solve equations the conventional way; instead I was just taught to "move numbers from side to side", inverting the sign or the operation accordingly. I am learning the conventional way though because I think it makes the process of solving equations clearer. That being said, I apologize if this question is too "basic".)</p>
<p>I know that when I have an equality such as $5 = \frac{x}{2}$ I have to multiply both sides by 2 to get the answer.</p>
<p>However, what is the process behind $5 = \frac{2}{x} \Leftrightarrow \frac{2}{5} = x$ ?</p>
<p>I know that when I have an equation which the variable is in the denominator I have to move the numerator to the other side and make it the numerator and the number that's already there the denominator, but I don't really know why that is or how that's done "mathematically".</p>
<p>I have a theory:</p>
<ul>
<li>Invert both sides and then multiply both sides by 2;</li>
</ul>
<p>Is this correct?</p>
| MattG88 | 159,928 | <p>You can do it by means the second equivalence principle: "multiplying or dividing both sides of an equation by a <strong>non-zero</strong> constant" we obtain an equivalent equation. This is the basis of your calculations in the example $5=\frac{2}{x}$.</p>
|
229,966 | <p>I want to put a title to the plotlegends I am using. I get a solution <a href="https://mathematica.stackexchange.com/questions/201353/title-for-plotlegends">here</a> which says to use <code>PlotLegends -> SwatchLegend[{0, 3.3, 6.7, 10, 13, 17, 20}, LegendLabel -> "mu"]</code>. But I also want to place the legends where I want like using <code>PlotLegends -> Placed[Range[1, 6, 1], {0.2, 0.3}]</code>.</p>
<p>How can I do both?</p>
<p>Edit:
As the answer was given, the wrapping works, but there is still a problem, i.e my plot looks like this: <a href="https://i.stack.imgur.com/Mad6w.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Mad6w.png" alt="enter image description here" /></a></p>
<p>Where I want to name the legends as "H".
But when I do that Swatchlegend thing it becomes like this :
<a href="https://i.stack.imgur.com/In3YR.png" rel="noreferrer"><img src="https://i.stack.imgur.com/In3YR.png" alt="enter image description here" /></a></p>
<p>I want to keep the markers and colors same.
What should I do?</p>
| kglr | 125 | <p>You can use <code>PlotLegends - > Placed[labels, Top, Labeled[#, legendlabel, Top] &]</code> .</p>
<p>Using the example input from Tim Laska's answer:</p>
<pre><code>mu = {0, 3.3, 6.7, 10, 13, 17, 20};
fns = Table[n^(1/p), {p, 7}, {n, 10}];
ListLinePlot[fns, PlotMarkers -> Automatic,
PlotLegends -> Placed[mu, Top, Labeled[#, Style["mu", 16], Top] &]]
</code></pre>
<p><a href="https://i.stack.imgur.com/dtlcE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dtlcE.png" alt="enter image description here" /></a></p>
<p>You can further embellish the appearance using the third argument of <code>Placed</code>:</p>
<pre><code>ListLinePlot[fns, PlotMarkers -> Automatic,
PlotLegends -> Placed[mu, Top, Panel @ Labeled[#, Style["mu", 16], Top] &]]
</code></pre>
<p><a href="https://i.stack.imgur.com/UlQu6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UlQu6.png" alt="enter image description here" /></a></p>
<pre><code>ListLinePlot[fns, PlotMarkers -> Automatic,
PlotLegends -> Placed[mu, Top,
Framed[#, FrameStyle -> Directive[Thin, Gray],
RoundingRadius -> 5, ImageMargins -> 5] & @
Labeled[#, Style[Row[{Spacer[5], "mu"}], 24, "Section"], {{Top, Left}}] &]]
</code></pre>
<p><a href="https://i.stack.imgur.com/1iFo8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1iFo8.png" alt="enter image description here" /></a></p>
|
275,775 | <p>For the FrameLabel, I have:</p>
<pre><code>Style["\[NumberSign] Humans per city", FontFamily -> "Latin Modern Math"]
</code></pre>
<p>How can I make only "Humans" in the label to be italic?</p>
| lericr | 84,894 | <p>Something like this might work:</p>
<pre><code>Row[
{"\[NumberSign] ", Style["Humans", FontSlant -> Italic], " per city"},
BaseStyle -> {FontFamily -> "Latin Modern Math"}]
</code></pre>
<p>If you don't want to insert your own spaces, you can add a spacer argument to Row:</p>
<pre><code>Row[
{"\[NumberSign]", Style["Humans", FontSlant -> Italic], "per city"},
" ",
BaseStyle -> {FontFamily -> "Latin Modern Math"}]
</code></pre>
|
275,775 | <p>For the FrameLabel, I have:</p>
<pre><code>Style["\[NumberSign] Humans per city", FontFamily -> "Latin Modern Math"]
</code></pre>
<p>How can I make only "Humans" in the label to be italic?</p>
| Chris Degnen | 363 | <p>As per <a href="https://mathematica.stackexchange.com/a/11002/363">WReach</a></p>
<pre><code>Style[
StringJoin["\[NumberSign] ",
ToString[Style["Humans", Italic], StandardForm], " per city"],
FontFamily -> "Latin Modern Math"]
</code></pre>
<blockquote>
<p># <em>Humans</em> per city</p>
</blockquote>
|
1,363,860 | <p>This problem is for my own exploration, not for class. The problem goes as follows:</p>
<blockquote>
<p>There are <span class="math-container">$n$</span> pairs of people with restraining orders against one another. However, all <span class="math-container">$2n$</span> people are friends with the other <span class="math-container">$2n-2$</span> people (does not include the person they have the restraining order against). So, all <span class="math-container">$2n$</span> people are at a party. Each person wants to stand closer than <span class="math-container">$1$</span> to every person he is friends with. However, he must stand further than <span class="math-container">$1$</span> from the person who has a restraining order against him. We assume that the party is held in <span class="math-container">$\mathbb{R}^2$</span>.</p>
</blockquote>
<blockquote>
<p>How large can <span class="math-container">$n$</span> be before this becomes impossible?</p>
</blockquote>
<p>I conjecture than <span class="math-container">$n$</span> must be finite. However, it would also be useful if <span class="math-container">$n$</span> could be proven to be strictly countable.</p>
<p><strong>EDIT:</strong> This can be done for all positive integers <span class="math-container">$n$</span>. Can this be done for uncountably many pairs?</p>
| Marcus M | 215,322 | <p>This can be done for all $n$. Order the $2n$ people so that person $k$ has a restraining order against person $k + n$ (and vice versa) for every $k$. Then put person $k$ on a circle centered at the origin; more specifically, position them at $r e^{2\pi k i/ 2n}$, with $r$ to be decided later. Then on the opposite side of the circle from the person whom they have a restraining order against, and is thus a distance of $2r$ away. The person they are second furthest from they are a distance of $$ |r e^{2\pi ki /2n} - r e^{2 \pi i (k + n + 1)/2n} | = r| - 1 - e^{i \pi /n} |.$$</p>
<p>We thus must choose $r$ so that $2r > 1$ and $r| - 1- e^{i\pi/n}| < 1$. We thus can pick $$ r = \frac{1}{2}\left(\frac{1}{2} + \frac{1}{| - 1- e^{i\pi/n}|}\right).$$ Then since $| - 1- e^{i\pi/n}| < 2$, we have </p>
<p>$$ r = \frac{1}{2}\left(\frac{1}{2} + \frac{1}{| - 1- e^{i\pi/n}|}\right) > \frac{1}{2}\left(\frac{1}{2} + \frac{1}{2}\right) = \frac{1}{2}$$ and
$$ r = \frac{1}{2}\left(\frac{1}{2} + \frac{1}{| - 1- e^{i\pi/n}|}\right) < \frac{1}{2}\left(\frac{1}{| - 1- e^{i\pi/n}|} + \frac{1}{| - 1- e^{i\pi/n}|}\right) = \frac{1}{| - 1- e^{i\pi/n}|}$$
as desired.</p>
|
3,570,688 | <p>For example, if a ball can be any of 3 colors, then the number of configurations (with repetition of colors) of 2 balls is <span class="math-container">$(3+2-1)C_{2} = 4C_{2} = 6$</span> Why?</p>
| LHF | 744,207 | <p>A similar idea as the accepted answer after rewriting the condition as:</p>
<p><span class="math-container">$$\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1$$</span></p>
<p>and using Cauchy-Schwarz:</p>
<p><span class="math-container">$$
\begin{aligned}
\frac{a+1}{a+2}&=\frac{1}{b+2}+\frac{1}{c+2}\\
&=5\cdot\frac{1}{5(b+2)}+7\cdot\frac{1}{7(c+2)}\\
&\geq \frac{144}{25b+49c+148}
\end{aligned}
$$</span></p>
<p>Therefore</p>
<p><span class="math-container">$$25b+49c \geq \frac{144}{a+1}-4$$</span></p>
<p>and using AM-GM:</p>
<p><span class="math-container">$$
\begin{aligned}
9a+25b+49c &\geq 9a+\frac{144}{a+1}-4\\
&=9(a+1)+\frac{144}{a+1}-13\\
& \geq 2\sqrt{9(a+1)\cdot \frac{144}{a+1}}-13\\
& = 59
\end{aligned}
$$</span></p>
|
1,702,616 | <p>I was working on a programming problem to find all 10-digit perfect squares when I started wondering if I could figure out how many perfects squares have exactly N-digits. I believe that I am close to finding a formula, but I am still off by one in some cases.</p>
<p>Current formula where $n$ is the number of digits:</p>
<p>$\lfloor\sqrt{10^n-1}\rfloor - \lfloor\sqrt{10^{n-1}}\rfloor$ </p>
<p>How I got here: </p>
<ol>
<li>Range of possible 10-digit numbers is from $10^9$ to $10^{10}-1$</li>
<li>10-digit perfect squares should fall into the range of $\sqrt{10^9}$ to $\sqrt{10^{10}-1}$</li>
</ol>
<p>Results of program for $n = 1,2,3,4,5$:</p>
<pre><code>DIGITS=1, ACTUAL_COUNT=3, COMPUTED_COUNT=2
DIGITS=2, ACTUAL_COUNT=6, COMPUTED_COUNT=6
DIGITS=3, ACTUAL_COUNT=22, COMPUTED_COUNT=21
DIGITS=4, ACTUAL_COUNT=68, COMPUTED_COUNT=68
DIGITS=5, ACTUAL_COUNT=217, COMPUTED_COUNT=216
</code></pre>
<p>Program: </p>
<pre><code>#!/usr/bin/perl
use strict;
use warnings;
sub all_n_digit_perfect_squares {
my ($n) = @_;
my $count = 0;
my $MIN = int( sqrt( 10**($n-1) ) );
my $MAX = int( sqrt( (10**$n)-1 ) );
foreach my $i ( $MIN .. $MAX ) {
if ( ($i * $i) >= 10**($n-1) ) {
$count++;
}
}
print "DIGITS=$n"
. ", ACTUAL_COUNT=$count"
. ", COMPUTED_COUNT="
. ($MAX-$MIN),
"\n";
return;
}
all_n_digit_perfect_squares($_) for (1..5);
</code></pre>
<p>Any advice on where I went wrong would be helpful.</p>
| gnasher729 | 137,175 | <p>If you think about it, you should have a formula that says the number of squares is f (n) - f (n-1) for some function f, so that every perfect square is counted exactly once if you calculate the squares from 10^1 to 10^2, from 10^2 to 10^3 and so on. </p>
<p>In your formula, the squares 100, 10,000, 1,000,000 and so on are not counted at all. For example, for 3 digit numbers the squares are from 10^2 to 31^2, that's 22 numbers. You calculate 31 - 10 = 21. Change your formula to </p>
<p>$\lfloor\sqrt{10^n-1}\rfloor - \lfloor\sqrt{10^{n-1}-1}\rfloor$ </p>
|
115,483 | <p>Edited:</p>
<p>I guess </p>
<p>$$H^2_{(x,y)}\left(\frac{\Bbb Z[x,y]}{(5x+4y)}\right)=0$$</p>
<p>We know that if $\operatorname{Supp} H^i_I(M)\subseteq V(I)\cap \operatorname{Supp}(M)$, then
$$\operatorname{Supp} H^2_{(x,y)}\frac{\Bbb Z[x,y]}{(5x+4y)})\subseteq V((x,y))\cap V((5x+4y))=V((x,y))=\lbrace(x,y) \rbrace\cup \lbrace(x,y,p) \rbrace$$ where $p$ is prime number.</p>
<p>If we could show that for every $P\in V((x,y))$, $$\left(H^2_{(x,y)}\left(\frac{\Bbb Z[x,y]}{(5x+4y)}\right)\right)_P=0$$</p>
<p>then it is done.</p>
<p><strong>background:</strong>
$H^i_I(M)$ means $i$-th local cohomology module of $M$ with respect to ideal $I$.
$V(I)=\lbrace P\in Spec(R); I\subseteq P\rbrace $ and $\operatorname{Supp}(M)=\lbrace P\in \operatorname{Spec}(R) ; M_p\neq 0\rbrace$and $M_P$ means $M$ localized at prime ideal $P$. Furthermore $\operatorname{Supp}(R/I)=V(I)$.</p>
| Damian Rössler | 17,308 | <p>I wasn't able to use the method you suggest. Here is a different proof.
Let $M$ be any ${\bf Z}[x,y]$-module. We have (see Brodmann and Sharp, Th. 1.3.8),
$$
H^2_{(x,y)}(M)=\varinjlim_n \operatorname{Ext}^2({\bf Z}[x,y]/(x,y)^n,M)
$$
where the limit is an inductive limit. Now we have
$$
\varinjlim_n \operatorname{Ext}^2({\bf Z}[x,y]/(x,y)^n,M)=\varinjlim_n \operatorname{Ext}^2({\bf Z}[x,y]/(x^n,y^n),M)
$$
because for any $n$, we have $(x^n,y^n)\subseteq (x,y)^n$ and
$(x,y)^{2n}\subseteq (x^n,y^n)$. Now the ideal $(x^n,y^n)$ defines a regular closed immersion
into $\operatorname{Spec} {\bf Z}[x,y]$, so that we have the "fundamental local isomorphism" (see Hartshorne, Residues and duality, Prop. 7.2): ...
EDIT: the end of the original argument was wrong. I don't know how to make it work. </p>
|
4,487,654 | <blockquote>
<p>Demonstrate recursively that</p>
<p><span class="math-container">$$\prod_{k = 0}^\infty (1 + x^{2^k}) = \frac{1}{1-x}$$</span></p>
</blockquote>
<p><strong>My work:</strong></p>
<p>Define</p>
<p><span class="math-container">$$a_n = \prod_{k = 0}^n (1 + x^{2^k}) = (1 + x^{2^n})a_{n - 1} \iff a_n - (1 + x^{2^n})a_{n - 1} = 0$$</span>
<span class="math-container">$$A(x) = \sum_{n = 0}^\infty a_nx^n = a_0 + \sum_{n = 1}^\infty a_{n}x^n = 1 + x + \sum_{n = 1}^\infty a_{n}x^n$$</span>
<span class="math-container">$$\implies xA(x) = \sum_{n = 1}^\infty a_{n-1}x^n$$</span></p>
<p>Therefore,</p>
<p><span class="math-container">\begin{align}
A(x) - (1 + x^{2^n})xA(x) &= 1 + x + \sum_{n = 1}^\infty a_{n}x^n - \sum_{n = 1}^\infty (1 + x^{2^n})a_{n-1}x^n\\
A(x) \left(1 - x(1 + x^{2^n})\right) &=1 + x + \sum_{n = 1}^{\infty} (a_n - (1 + x^{2^k})a_{n - 1})x^n\\
A(x) \left(1 - x(1 + x^{2^n})\right) &=1 + x\\
A(x) &= \frac{1 + x}{1 - x(1 + x^{2^n})}\\
&= \frac{1 + x}{1 - x - x^{2^n + 1}}
\end{align}</span></p>
<p>To find <span class="math-container">$a_n$</span>, I now want to transform <span class="math-container">$A(x)$</span> in the form <span class="math-container">$\sum_{k = 0}^\infty a_nx^n$</span> and then take the limit as <span class="math-container">$n \to \infty$</span>. However, I’m unsure how to do this since I don’t think the expression can be decomposed into partial fractions.</p>
<p><strong>My question:</strong></p>
<ul>
<li>Is my approach correct?</li>
<li>How do I continue?</li>
</ul>
<p><strong>Note:</strong> I know that there are <a href="https://math.stackexchange.com/a/4487579/1072645">other solutions</a>, but I specifically want to see if defining a recurrence can yield a solution.</p>
| Igor Rivin | 109,865 | <p>Write your equation as <span class="math-container">$\frac{\sin(\pi x)}{x} = \frac{1}{n}.$</span> It is clear that any solution is at most <span class="math-container">$n$</span>, since if <span class="math-container">$x > n,$</span> the LHS is smaller than <span class="math-container">$1/n.$</span> Thus, the limit has to be zero.</p>
|
576,379 | <p>I know how to show that $f(x)=x^2$ is uniformly continuous, but I am confused when it is $x^2 +x$</p>
| user1337 | 62,839 | <p><strong>Hint:</strong></p>
<p>The collection of all uniformly continuous functions in a given interval form a vector space.</p>
|
1,506,741 | <p>Here's an exercise: </p>
<p>Let $(X,M,\mu)$ be a measure space with $\mu(X)<\infty$. Let $N\subseteq M$ be a $\sigma$-algebra. If $f\geq 0$ is $M$-measurable and $\mu$-integrable, then there exists some $N$-measurable and $\mu$-integrable function $g\geq 0$ such that
$$
\int_E g \, d\mu=\int_E f \, d\mu,~~~~E\in N.
$$</p>
<p>My proof does not involve the finite measure property: </p>
<p>Let $E\in N$, define
$$
{\frak{A}}_E = \left\{\sum_{k=1}^n a_k \mu(E_k): a_k \geq 0, E_k \in N, \biguplus_{k=1}^{n_m}E_k^{(m)} = E, \sum_{k=1}^n a_k \chi_{E_k} \leq f, k=1,\ldots,n, n\in {\bf N}\right\}.
$$
In particular, for $E=X$, find some sequence $(s_{m})$ that of the form
\begin{align*}
s_m=\sum_{k=1}^{n_m} a_k^{(m)} \chi_{E_k^{(m)}}\leq f,~~~~a_{k}^{(m)}\geq 0,~~~~E_k^{(m)}\in N,~~~~\biguplus_{k=1}^{n_m}E_k^{(m)}=X,~~~~k=1,\dots,n_m,
\end{align*}
where $n_m \in{\bf{N}}$, $m=1,2,\ldots$ such that
\begin{align*}
(N)\int_X s_m \, d\mu=\sum_{k=1}^{n_m} a_k^{(m)} \mu\left(E_k^{(m)}\right)\rightarrow \sup{\frak{A}}_X
\end{align*}
as $m\rightarrow\infty$, where the notation $(N)$ indicates that the integration is taken under the measure space $(X,N,\mu)$, $\displaystyle\biguplus$ means that the union is disjoint. We note that as $f$ is $\mu$-integrable, we have $\sup{\frak{A}}_{E}<\infty$ for each $E\in N$, and hence $\sup{\frak{A}}_E + \sup{\frak{A}}_{E^c} \leq \sup{\frak{A}}_X$, so
$$
(N)\int_E s_m \, d\mu=(N)\int_X s_m \, d\mu-(N)\int_{E^c} s_m \, d\mu\geq(N)\int_X s_m \, d\mu-\sup{\frak{A}}_{E^c},
$$
and hence
$$
\liminf_{m\rightarrow\infty}\left[(N)\int_E s_m \, d\mu\right] \geq \sup{\frak{A}}_X -\sup{\frak{A}}_{E^c}\geq\sup{\frak{A}}_E.
$$
Since
$$
\int_E s_m \, d\mu\leq\sup{\frak{A}}_E,
$$
we deduce that
$$
\lim_{m\rightarrow\infty} \left[(N)\int_E s_m \, d\mu\right]=\sup{\frak{A}}_E.
$$
By construction $s_{m}$ is $N$-measurable, if we let $g(x)=\limsup_{m\rightarrow\infty}s_{m}(x)$, we have $g\geq 0$ is $N-$measurable. Since $s_m \leq f$ for each $m=1,2,\ldots$, if we view $s_{m}$ as $M$-measurable, a varied Fatou's Lemma gives
\begin{align*}
\limsup_{m\rightarrow\infty}\left[(M)\int_E s_m \, d\mu\right] \leq (M) \int_E \limsup_{m\rightarrow\infty} s_m \, d\mu=(M)\int_E g \, d\mu,
\end{align*}
since
\begin{align*}
(M)\int_E s_m \, d\mu=(N)\int_E s_m \, d\mu
\end{align*}
and that
\begin{align*}
(M)\int_E g \, d\mu \leq (M)\int_E f \, d\mu,
\end{align*}
we conclude that
\begin{align*}
\sup{\frak{A}}_{E}\leq(M)\int_{E}fd\mu.
\end{align*}
On the other hand, given $b_k\geq 0$, $B_k\in M$, $k=1,\ldots,l$, $\displaystyle\biguplus_{k=1}^l B_k=E$ and that
$$
b=\sum_{k=1}^l b_k \chi_{B_k}\leq f,
$$
we have
\begin{align*}
b&=\sum_{k=1}^l b_k \chi_{B_k}\\
&=\sum_{k=1}^l b_k \chi_{B_k\cap E}\\
&=\left(\sum_{1\leq k\leq l: B_k \cap E\ne\emptyset} b_k\right)\chi_E,
\end{align*}
if $\{1\leq k\leq l: B_k\cap E\ne\emptyset\}$ is empty, we set
\begin{align*}
\sum_{1\leq k\leq l: B_k\cap E\ne\emptyset}b_k=0,
\end{align*}
so
\begin{align*}
(M)\int_E b \, d\mu=\left(\sum_{1\leq k\leq l: B_k\cap E \ne \emptyset} b_k \right)\mu(E) \leq \sup{\frak{A}}_E,
\end{align*}
since $(M)\displaystyle\int_E f \, d\mu$ is defined by the supremum of the set of all such integrals of $b$ taken in $(X,M,\mu)$, we find that
\begin{align*}
(M)\int_E f \, d\mu \leq \sup{\frak{A}}_E,
\end{align*}
the result follows.</p>
<p>Can anyone point out the mistakes in this proof, if they exist?</p>
| Michael Hardy | 11,667 | <p>For $E \in N$, let
$$
\lambda(E) = \int_E f\, d\mu.
$$
Then the measure $\lambda$ on $N$ is absolutely continuous to the measure $\mu$ on $N$ (i.e. the restriction to $N$ of the measure $\mu$ on $M$). Therefore by the Radon–Nikodym theorem, there is an $N$-measurable function $g=\dfrac{d\lambda}{d\mu}$ such that for $E\in N$ we have
$$
\lambda(E) = \int_E g\,d\mu = \int_E f\,d\mu.
$$</p>
|
1,506,741 | <p>Here's an exercise: </p>
<p>Let $(X,M,\mu)$ be a measure space with $\mu(X)<\infty$. Let $N\subseteq M$ be a $\sigma$-algebra. If $f\geq 0$ is $M$-measurable and $\mu$-integrable, then there exists some $N$-measurable and $\mu$-integrable function $g\geq 0$ such that
$$
\int_E g \, d\mu=\int_E f \, d\mu,~~~~E\in N.
$$</p>
<p>My proof does not involve the finite measure property: </p>
<p>Let $E\in N$, define
$$
{\frak{A}}_E = \left\{\sum_{k=1}^n a_k \mu(E_k): a_k \geq 0, E_k \in N, \biguplus_{k=1}^{n_m}E_k^{(m)} = E, \sum_{k=1}^n a_k \chi_{E_k} \leq f, k=1,\ldots,n, n\in {\bf N}\right\}.
$$
In particular, for $E=X$, find some sequence $(s_{m})$ that of the form
\begin{align*}
s_m=\sum_{k=1}^{n_m} a_k^{(m)} \chi_{E_k^{(m)}}\leq f,~~~~a_{k}^{(m)}\geq 0,~~~~E_k^{(m)}\in N,~~~~\biguplus_{k=1}^{n_m}E_k^{(m)}=X,~~~~k=1,\dots,n_m,
\end{align*}
where $n_m \in{\bf{N}}$, $m=1,2,\ldots$ such that
\begin{align*}
(N)\int_X s_m \, d\mu=\sum_{k=1}^{n_m} a_k^{(m)} \mu\left(E_k^{(m)}\right)\rightarrow \sup{\frak{A}}_X
\end{align*}
as $m\rightarrow\infty$, where the notation $(N)$ indicates that the integration is taken under the measure space $(X,N,\mu)$, $\displaystyle\biguplus$ means that the union is disjoint. We note that as $f$ is $\mu$-integrable, we have $\sup{\frak{A}}_{E}<\infty$ for each $E\in N$, and hence $\sup{\frak{A}}_E + \sup{\frak{A}}_{E^c} \leq \sup{\frak{A}}_X$, so
$$
(N)\int_E s_m \, d\mu=(N)\int_X s_m \, d\mu-(N)\int_{E^c} s_m \, d\mu\geq(N)\int_X s_m \, d\mu-\sup{\frak{A}}_{E^c},
$$
and hence
$$
\liminf_{m\rightarrow\infty}\left[(N)\int_E s_m \, d\mu\right] \geq \sup{\frak{A}}_X -\sup{\frak{A}}_{E^c}\geq\sup{\frak{A}}_E.
$$
Since
$$
\int_E s_m \, d\mu\leq\sup{\frak{A}}_E,
$$
we deduce that
$$
\lim_{m\rightarrow\infty} \left[(N)\int_E s_m \, d\mu\right]=\sup{\frak{A}}_E.
$$
By construction $s_{m}$ is $N$-measurable, if we let $g(x)=\limsup_{m\rightarrow\infty}s_{m}(x)$, we have $g\geq 0$ is $N-$measurable. Since $s_m \leq f$ for each $m=1,2,\ldots$, if we view $s_{m}$ as $M$-measurable, a varied Fatou's Lemma gives
\begin{align*}
\limsup_{m\rightarrow\infty}\left[(M)\int_E s_m \, d\mu\right] \leq (M) \int_E \limsup_{m\rightarrow\infty} s_m \, d\mu=(M)\int_E g \, d\mu,
\end{align*}
since
\begin{align*}
(M)\int_E s_m \, d\mu=(N)\int_E s_m \, d\mu
\end{align*}
and that
\begin{align*}
(M)\int_E g \, d\mu \leq (M)\int_E f \, d\mu,
\end{align*}
we conclude that
\begin{align*}
\sup{\frak{A}}_{E}\leq(M)\int_{E}fd\mu.
\end{align*}
On the other hand, given $b_k\geq 0$, $B_k\in M$, $k=1,\ldots,l$, $\displaystyle\biguplus_{k=1}^l B_k=E$ and that
$$
b=\sum_{k=1}^l b_k \chi_{B_k}\leq f,
$$
we have
\begin{align*}
b&=\sum_{k=1}^l b_k \chi_{B_k}\\
&=\sum_{k=1}^l b_k \chi_{B_k\cap E}\\
&=\left(\sum_{1\leq k\leq l: B_k \cap E\ne\emptyset} b_k\right)\chi_E,
\end{align*}
if $\{1\leq k\leq l: B_k\cap E\ne\emptyset\}$ is empty, we set
\begin{align*}
\sum_{1\leq k\leq l: B_k\cap E\ne\emptyset}b_k=0,
\end{align*}
so
\begin{align*}
(M)\int_E b \, d\mu=\left(\sum_{1\leq k\leq l: B_k\cap E \ne \emptyset} b_k \right)\mu(E) \leq \sup{\frak{A}}_E,
\end{align*}
since $(M)\displaystyle\int_E f \, d\mu$ is defined by the supremum of the set of all such integrals of $b$ taken in $(X,M,\mu)$, we find that
\begin{align*}
(M)\int_E f \, d\mu \leq \sup{\frak{A}}_E,
\end{align*}
the result follows.</p>
<p>Can anyone point out the mistakes in this proof, if they exist?</p>
| David C. Ullrich | 248,223 | <p><strong>NEW:</strong> In fact the result <em>always</em> fails unless the restriction $\mu|_N$ is $\sigma$-finite. See below.</p>
<p>First the simple counterexample showing that the result <em>may</em> fail without some finiteness hypothesis:</p>
<p>Without the assumption that $\mu(X)<\infty$ the result is false. Simple counterexample:</p>
<p>Say $(X,M,\mu)=(\Bbb R,M,m)$, where $M$ is the algebra of Lebesgue measurable sets and $m$ is Lebesgue measure. Say $f=\chi_{[0,1]}$, and $N=\{\Bbb R,\emptyset\}$. An $N$-measurable function must be constant.</p>
<p>So the result you say you've proved would show that there is a <em>constant</em> $c$ with $$\int_{\Bbb R}c\,dx=1,$$clearly impossible.</p>
<p><strong>Comment.</strong> If we assume that $\mu(X)<\infty$ then the result is a simple standard application of the Radon-Nikodym theorem. We need $\mu(X)<\infty$ because of the $\sigma$-finiteness hypothesis in R-N. So one might think that assuming that $\mu$ is $\sigma$-finite would be enough. Not so. Curiously, although assuming $\mu$ finite certainly implies that the restriction of $\mu$ to $N$ is finite, assuming $\mu$ is $\sigma$-finite does <em>not</em> imply that the restriction of $\mu$ to $N$ is $\sigma$-finite (as the example shows).</p>
<hr>
<p><strong>NEW</strong></p>
<p>First two simple general measure-theory exercises:</p>
<p><strong>Exercise</strong> The measure $\mu$ is $\sigma$-finite if and only if there exists $f\in L^1(\mu)$ with $f>0$ almost everywhere.</p>
<p><strong>Exercise</strong> A measurable function $f$ is strictly positive almost everywhere if and only if $\int_E f\,d\mu>0$ for every measurable set $E$ with $\mu(E)>0$.</p>
<p><strong>Theorem</strong> Suppose that $(X,M,\mu)$ is a $\sigma$-finite measure space and $N$ is a sub-$\sigma$-algebra of $M$. The following are equivalent:</p>
<p>(i) There exist sets $E_n\in N$ with $\mu(E_n)<\infty$ and $\bigcup E_n=X$.</p>
<p>(ii) For every $f\in L^1(\mu)$ there exists an $N$-measurable function $g$ such that $\int_Eg\,d\mu=\int_Ef\,d\mu$ for every $E\in N$.</p>
<p><strong>Proof</strong> One direction is perfectly standard. Assume (i). That says that $\mu|_N$ is $\sigma$-finite. Define $\nu:N\to\Bbb C$ by $\nu(E)=\int_E f\,d\mu$. Then $\nu$ is a complex measure on $N$ and $\nu<<\mu|_N$, so Radon-Nikodym says that there exists $g\in L^1(\mu|_N)$ with $\nu(E)=\int_E g\,d\mu$ for all $E\in N$.</p>
<p>Now assume (ii). Since $\mu$ is $\sigma$-finite there exists $f\in L^1(\mu)$ with $f>0$ almost everywhere. Choose $g\in L^1(\mu|_N)$ as in (ii). Now if $E\in N$ and $\mu(E)>0$ then $\int_E g\,d\mu=\int_Ef\,d\mu>0$, since $f>0$ almost everywhere. Hence $g>0$ almost everywhere. Hence $\mu|_N$ is $\sigma$-finite. QED.</p>
|
134,937 | <p>Let $p \equiv q \equiv 3 \pmod 4$ for distinct odd primes $p$ and $q$. Show that $x^2 - qy^2 = p$ has no integer solutions $x,y$.</p>
<p>My solution is as follows.</p>
<p>Firstly we know that as $p \equiv q \pmod 4$ then $\big(\frac{p}{q}\big) = -\big(\frac{q}{p}\big)$</p>
<p>Assume that a solution $(x,y)$ does exist and reduce the original equation modulo $q$ to get $x^2 \equiv p \pmod q.$ This implies $\big(\frac{p}{q} \big) = 1 $. </p>
<p>Now if we reduce the original equation modulo $p$ then $x^2 \equiv qy^2 \pmod p \ (*)$. To get a contradiction I want to show that $\big(\frac{q}{p}\big) =1$ and to do this I need to prove that $gcd(p.y) = 1$ as this means we can divide both sides of ($*$) by $y^2$. But how do you prove this?</p>
| Kerry | 7,887 | <p>You probably tried too hard on this. We know $x^{2},y^{2}\equiv 1/0\pmod{4}$. So $x^{2}$ choices are $0,1$. We have $0-1=3$, $1-2=3$. $qy^{2}$ cannot be $1\pmod{4}$. Thus $x$ must be odd and $y$ be even. But if that is so $qy^{2}$ must be $0\pmod{4}$, which contradicts with our assumption $qy^{2}\equiv 2\pmod{4}$. This showed the original equation has no solution. </p>
|
70,946 | <p>I'm an REU student who has just recently been thrown into a dynamical system problem without basically any background in the subject. My project advisor has told me that I should represent regions of my dynamical system by letters and look at the sequence of letters formed by the trajectory of a point under the iteration of my map.</p>
<p>He claims that it's a common result that if two points share the same sequence, then this sequence of letters is periodic. I've asked around among some of the other students, and they said that this is sometimes called symbolic dynamics, but none of them remembers this sort of result. I've also searched the internet, but it's possible that my google-fu is weak, since I didn't find any answers that way.</p>
<p>To go one step further, there are obvious cases where it is false- take $S^1\times I$, and encode the regions as $A$ corresponds to $[0,\pi)\times I$ and $B$ corresponds to $[\pi,2\pi)\times I$ with map $f(x,y)=(x+1\mod{2\pi},y)$. Obviously any two points $(x,y)$ and $(x,z)$ with $y\neq z$ will have the same sequence, but since 1 is an irrational multiple of $2\pi$, the trajectory will never be periodic.</p>
<p>I'm interested in the general theory and common techniques applied to the question:</p>
<blockquote>
<p>Represent a dynamical system by associating symbols with regions of the space. When is it true that if two distinct points's trajectories have the same sequence of symbols, then the sequence of symbols is periodic?</p>
</blockquote>
<p>Any answers, examples, or specific references would be greatly appreciated.</p>
| Anthony Quas | 11,054 | <p>I think symbolic dynamics is the study of what you get <em>after</em> you've introduced your partition and coded points by their itinerary. Your question is essentially "when is the symbolic dynamics a faithful representation of the original dynamical system?".</p>
<p>As Vaughn says, the thing you're looking for is expansiveness - effectively a guarantee that if you start with two different points then sooner or later they end up in a different element of the partition. It sounds as though the paper mentioned by Gjergji is very close to what you're looking for. </p>
<p>Other explicit examples where the symbolic dynamics represents things faithfully:
Axiom A maps (Smale); Geodesic flows (Bowen and Ratner); Expanding maps. More generally there is a beautiful paper by Boyle and Lind "expansive subdynamics" on expansiveness for higher-dimensional actions (when you don't just have a single transformation acting but a bunch of commuting transformations). </p>
|
2,353,190 | <p>Let $f(x)=\dfrac{1+x}{1-x}$ The nth derivative of f is equal to:</p>
<ol>
<li>$\dfrac{2n}{(1-x)^{n+1}} $</li>
<li>$\dfrac{2(n!)}{(1-x)^{2n}} $</li>
<li>$\dfrac{2(n!)}{(1-x)^{n+1}} $</li>
</ol>
<p>by Leibniz formula </p>
<p>$$ {\displaystyle \left( \dfrac{1+x}{1-x}\right)^{(n)}=\sum _{k=0}^{n}{\binom {n}{k}}\ (1+x)^{(k)}\ \left(\dfrac{1}{1-x}\right)^{(n-k)}}$$</p>
<p>using the hint </p>
<ul>
<li>$\dfrac{1+x}{1-x}=\dfrac{2-(1-x)}{1-x}=\dfrac2{1-x}-1$ and </li>
<li>$\left(\dfrac{1}{x}\right)^{n}=\dfrac{(-1)^{n}n!}{x^{n+1}}$</li>
</ul>
<p>so </p>
<p>$${\displaystyle \left( \dfrac{1+x}{1-x} \right)^{(n)} = \left( \dfrac{2}{1-x}-1 \right)^{(n)}=2\dfrac{ (-1)^{n}n! }{ (1-x)^{n+1} } } $$ but this result isn't apear in any proposed answers</p>
<p>what about the method of <strong>Lord Shark the Unknown</strong></p>
<p>tell me please this way holds for any mqc question contain find the n th derivative so it's suffice to check each answer in y case i will start with first </p>
<ul>
<li>let $f_n(x)=\dfrac{2n}{(1-x)^{n+1}}$ then $f_{n+1}(x)=\dfrac{2(n+1)}{(1-x)^{n+2}}$ do i have $f'_{n}=f_{n+1}$ let calculate $$ f'_n=\dfrac{-2n(n+1)}{(1-x)^{n+2}}\neq f_{n+1}$$</li>
<li>let $f_n(x)=\dfrac{2(n!)}{(1-x)^{2n}}$ then $f_{n+1}(x)=\dfrac{2((n+1)!)}{(1-x)^{2(n+1)}}$ do i have $f'_{n}=f_{n+1}$ let calculate $$ f'_n=\dfrac{-2(n!)(2n)}{(1-x)^{4n}}\neq f_{n+1}$$</li>
<li>let $f_n(x)=\dfrac{2(n!)}{(1-x)^{n+1}}$ then $f_{n+1}(x)=\dfrac{2((n+1)!)}{(1-x)^{n+2}}$ do i have $f'_{n}=f_{n+1}$ let calculate $$ f'_n=\dfrac{2(n!)(n+1)}{((1-x)^{n+1})^{2}}=\dfrac{2((n+1)!)}{(1-x)^{2n+2}}\neq f_{n+1}$$</li>
</ul>
| egreg | 62,967 | <p>You can do it with Leibniz's formula (not that it's easier than without it); just consider that
$$
(1+x)^{(k)}=
\begin{cases}
1+x & k=0 \\
1 & k=1 \\
0 & k>1
\end{cases}
$$
so the formula gives
$$
\left( \frac{1+x}{1-x}\right)^{\!(n)}=
(1+x)\left(\dfrac{1}{1-x}\right)^{\!(n)}+
n\left(\frac{1}{1-x}\right)^{\!(n-1)}
$$
Now make a conjecture about
$$
\left(\frac{1}{1-x}\right)^{\!(n)}
$$</p>
|
1,392,661 | <p>For a National Board Exam Review: </p>
<blockquote>
<p>Find the equation of the perpendicular bisector of the line joining
(4,0) and (-6, -3)</p>
</blockquote>
<p>Answer is 20x + 6y + 29 = 0</p>
<p>I dont know where I went wrong. This is supposed to be very easy:</p>
<p>Find slope between two points:</p>
<p>$${ m=\frac{y^2 - y^1}{ x^2 - x^1 } = \frac{-3-0}{-6-4} = \frac{3}{10}}$$</p>
<p>Obtain Negative Reciprocal:</p>
<p>$${ m'=\frac{-10}{3}}$$</p>
<p>Get Midpoint fox X</p>
<p>$${ \frac{-6-4}{2} = -5 }$$</p>
<p>Get Midpoint for Y</p>
<p>$${ \frac{-0--3}{2} = \frac{3}{2} }$$</p>
<p>Make Point Slope Form: </p>
<p>$${ y = m'x +b = \frac{-10}{3}x + b}$$</p>
<p>Plugin Midpoints in Point Slope Form</p>
<p>$${ \frac{3}{2} = \frac{-10}{3}(-5) + b}$$</p>
<p>Evaluate b</p>
<p>$${ b = \frac{109}{6}}$$</p>
<p>Get Equation and Simplify</p>
<p>$${ y = \frac{-10}{3}x + \frac{109}{6}}$$
$${ 6y + 20x - 109 = 0 }$$</p>
<p>Is the problem set wrong? What am I doing wrong?</p>
| pancini | 252,495 | <p>lol. a lot of confusion on this thread. When you are computing the midpoints, what you actually do is find the distance between them and divide by two. This isn't the coordinate of the midpoint; this is just the distance form one end to the midpoint.</p>
<p>What you meant to do is take the average of the x and y <em>values</em>, not the average of their difference.</p>
<p>You should get a midpoint of $(-1,-1.5)$.</p>
|
1,392,661 | <p>For a National Board Exam Review: </p>
<blockquote>
<p>Find the equation of the perpendicular bisector of the line joining
(4,0) and (-6, -3)</p>
</blockquote>
<p>Answer is 20x + 6y + 29 = 0</p>
<p>I dont know where I went wrong. This is supposed to be very easy:</p>
<p>Find slope between two points:</p>
<p>$${ m=\frac{y^2 - y^1}{ x^2 - x^1 } = \frac{-3-0}{-6-4} = \frac{3}{10}}$$</p>
<p>Obtain Negative Reciprocal:</p>
<p>$${ m'=\frac{-10}{3}}$$</p>
<p>Get Midpoint fox X</p>
<p>$${ \frac{-6-4}{2} = -5 }$$</p>
<p>Get Midpoint for Y</p>
<p>$${ \frac{-0--3}{2} = \frac{3}{2} }$$</p>
<p>Make Point Slope Form: </p>
<p>$${ y = m'x +b = \frac{-10}{3}x + b}$$</p>
<p>Plugin Midpoints in Point Slope Form</p>
<p>$${ \frac{3}{2} = \frac{-10}{3}(-5) + b}$$</p>
<p>Evaluate b</p>
<p>$${ b = \frac{109}{6}}$$</p>
<p>Get Equation and Simplify</p>
<p>$${ y = \frac{-10}{3}x + \frac{109}{6}}$$
$${ 6y + 20x - 109 = 0 }$$</p>
<p>Is the problem set wrong? What am I doing wrong?</p>
| bubba | 31,744 | <p>Just for variety, a different approach:</p>
<p>If a point $(x,y)$ is on the perpendicular bisector, then it is equidistant from $(4,0)$ and $(-6,-3)$, so
$$
(x-4)^2 + (y-0)^2 = (x+6)^2 + (y+3)^2
$$
Multiplying out, we get
$$
x^2 -8x +16 + y^2 = x^2 +12x +36 \; + \; y^2 +6y + 9
$$
So
$$
20x + 6y + 29 = 0
$$</p>
|
3,960,404 | <p>Let <span class="math-container">$T$</span> be a tree.</p>
<p>Suppose that <span class="math-container">$T$</span> doesn’t have a perfect matching and let <span class="math-container">$M$</span> be a matching of <span class="math-container">$T$</span>, <span class="math-container">$|M| = k$</span>. Prove
that there exists a matching of the same cardinality <span class="math-container">$k$</span> which exposes at least a pendant vertex
of <span class="math-container">$T$</span>.</p>
<p>My idea:</p>
<p>If <span class="math-container">$T$</span> has not a perfect matching it means that <span class="math-container">$\exists x$</span> in <span class="math-container">$V(T)$</span> and <span class="math-container">$x$</span> is exposed. I don't know how to use this in symmetric difference to find another matching <span class="math-container">$M_1$</span>, <span class="math-container">$|M_1|=k$</span> which exposes at least a pendant vertex of <span class="math-container">$T$</span>.</p>
| Brian M. Scott | 12,042 | <p>HINT: If <span class="math-container">$M$</span> exposes some pendant vertex, we’re done, so suppose that <span class="math-container">$M$</span> covers every pendant vertex. Root <span class="math-container">$T$</span> at a non-pendant vertex <span class="math-container">$v$</span>, so that the pendant vertices are the leaves of <span class="math-container">$T$</span>. From each leaf there is a unique path to <span class="math-container">$v$</span>, and every vertex of <span class="math-container">$T$</span> is on at least one of those paths, so there is a leaf <span class="math-container">$u$</span> such that the path <span class="math-container">$P$</span> from <span class="math-container">$u$</span> to <span class="math-container">$v$</span> contains an exposed vertex. Let <span class="math-container">$w$</span> be the first exposed vertex that we encounter when travelling <span class="math-container">$P$</span> from <span class="math-container">$u$</span> to <span class="math-container">$v$</span>.</p>
<p>If the edges of <span class="math-container">$P$</span>, listed in order starting at <span class="math-container">$u$</span>, are <span class="math-container">$e_1,e_2,\ldots,e_n$</span>, show that the odd-numbered edges of <span class="math-container">$P$</span> are in <span class="math-container">$M$</span>, and the even-numbered edges are not in <span class="math-container">$M$</span>. Conclude that <span class="math-container">$n$</span> is even. Show that if you form <span class="math-container">$M_1$</span> by deleting from <span class="math-container">$M$</span> the odd-numbered edges of <span class="math-container">$P$</span> and adding the even-numbered edges, <span class="math-container">$M_1$</span> will be a matching that exposes <span class="math-container">$u$</span> and has the same cardinality as <span class="math-container">$M$</span>.</p>
|
661,026 | <p>prove or disprove this
$$\sum_{k=0}^{n}\binom{n}{k}^3\approx\dfrac{2}{\pi\sqrt{3}n}\cdot 8^n,n\to\infty?$$</p>
<p>this problem is from when Find this limit
$$\lim_{n\to\infty}\dfrac{\displaystyle\sum_{k=0}^{n}\binom{n}{k}^3}{\displaystyle\sum_{k=0}^{n+1}\binom{n+1}{k}^3}=\dfrac{1}{8}?$$</p>
<p>first,follow I can't
$$\sum_{k=0}^{n}\binom{n}{k}^3\approx\dfrac{2}{\pi\sqrt{3}n}\cdot 8^n,n\to\infty?$$
.Thank you for you help</p>
| vonbrand | 43,946 | <p>This is a <a href="http://en.wikipedia.org/wiki/Quartic_function">quartic</a>, its discriminant tells all you need.</p>
|
661,026 | <p>prove or disprove this
$$\sum_{k=0}^{n}\binom{n}{k}^3\approx\dfrac{2}{\pi\sqrt{3}n}\cdot 8^n,n\to\infty?$$</p>
<p>this problem is from when Find this limit
$$\lim_{n\to\infty}\dfrac{\displaystyle\sum_{k=0}^{n}\binom{n}{k}^3}{\displaystyle\sum_{k=0}^{n+1}\binom{n+1}{k}^3}=\dfrac{1}{8}?$$</p>
<p>first,follow I can't
$$\sum_{k=0}^{n}\binom{n}{k}^3\approx\dfrac{2}{\pi\sqrt{3}n}\cdot 8^n,n\to\infty?$$
.Thank you for you help</p>
| Steve Kass | 60,500 | <p>Sketch $\frac{1}{2}x^4 - x^3 -x + 100$. There are two changes in concavity between $x=-2$ and $x=3$, so the graph can't "change direction" outside those values, and the minimum must be between them. Direct calculation shows that within that interval, the function exceeds 50 at each integer-valued $x$.</p>
<p>For there to be a zero within $0.5$ units of any of these integer points, the function would have to have an average rate of change of more than $50$ on an interval of length less than $0.5$, and by the mean value theorem, the absolute value of the derivative would have to exceed $\frac{50}{0.5}=100$ somewhere. It doesn't. The absolute value of the derivative in that interval is easily bounded above by $2|3|^3+3|3|^2+1=82$ (the sum of the absolute values of the terms of the derivative at the largest $|x|$-value.</p>
|
794,736 | <blockquote>
<p>Let $60$ students and $10$ teachers. How many arrangements are there, such that, between two teachers must be exactly $6$ students? </p>
</blockquote>
<p>I know that there are $10!$ permutations for the teachers, and there are $54$ places between them for the students. Nothing said about the edges. Hence, at the edges might be $0-6$ students. </p>
<p>I've been told the answer is $10!\cdot 60!\cdot 7$ but I'm not sure why..I'll be glad for an explanation. </p>
<p>Thanks!</p>
| Community | -1 | <p>The proof is essentially correct, but a bit too brief. The part "$g(x)=g(1)$ for all $x=0$" is written nicely. For $x<0$, I would write instead: $g(x)=g(x^2)=g(1)$, where the second equality holds since $x^2>0$. And as Jean Claude Arbaut added, the proof should conclude with $g(0)=\lim_{x\to 0}g(x) = g(1)$. </p>
|
1,115,222 | <blockquote>
<p>Suppose <span class="math-container">$f$</span> is a continuous, strictly increasing function defined on a closed interval <span class="math-container">$[a,b]$</span> such that <span class="math-container">$f^{-1}$</span> is the inverse function of <span class="math-container">$f$</span>. Prove that,
<span class="math-container">$$\int_{a}^bf(x)dx+\int_{f(a)}^{f(b)}f^{-1}(x)dx=bf(b)-af(a)$$</span></p>
</blockquote>
<p>A high school student or a Calculus first year student will simply, possibly, apply change of variable technique, then integration by parts and he/she will arrive at the answer without giving much thought into the process. A smarter student would probably compare the integrals with areas and conclude that the equality is immediate.</p>
<p>However, I am an undergrad student of Analysis and I would want to solve the problem "carefully". That is, I wouldn't want to forget my definitions, and the conditions of each technique. For example, while applying change of variables technique, I cannot apply it blindly; I must be prudent enough to realize that the criterion to apply it includes continuous differentiability of a function. Simply with <span class="math-container">$f$</span> continuous, I cannot apply change of variables technique.</p>
<p>Is there any method to solve this problem rigorously? One may apply the techniques of integration (by parts, change of variables, etc.) only after proper justification.</p>
<p>The reason I am not adding any work of mine is simply that I could not proceed even one line since I am not given <span class="math-container">$f$</span> is differentiable. However, this seems to hold for non-differentiable functions also.</p>
<p>I would really want some help. Pictorial proofs and/or area arguments are invalid.</p>
| Tom-Tom | 116,182 | <p>For completeness, the case where <span class="math-container">$f$</span> is differentiable is handled this way
<span class="math-container">$$\begin{split} I&=\int_a^bf(x)\mathrm dx+\int_{f(a)}^{f(b)}f^{-1}(x)\mathrm dx\\
& = \int_a^bf(x)\mathrm dx+\int_a^bf^{-1}\big(f(u)\big)f'(u)\mathrm du \quad
\text{(by substitution $x=f(u)$)}\\&=
\int_a^bf(x)\mathrm dx+\int_a^bxf'(x)\mathrm dx \qquad\text{(simply renaming $u$ as $x$)}\\&=
\int_a^b\left[f(x)+xf'(x)\right]\mathrm dx\qquad\text{(merging of the integrals)}\\&=
\big[xf(x)\big]_a^b=bf(b)-af(a)\qquad\text{(recognizing the derivative of $x\mapsto xf(x)$)}
\end{split}$$</span></p>
|
75,875 | <p>I am asking in the sense of isometry groups of a manifold. SU(3) is the group of isometries of CP2, and SO(5) is the group of isometries of the 4-sphere. Now, it happens that both manifolds are related by Arnold-Kuiper-Massey theorem: $\mathbb{CP}^2/conj \approx S^4$; one is a branched covering of the other, the quotient being via complex conjugation.</p>
<p>Now, for the case of a manifold and a lower dimensional submanifold, it is not rare to find that the corresponding isometry groups are subgroups one of the other. But here, which is the equivalent result? is SO(5) an "enhanced SU(3)" in some way?</p>
<p>The context of the question comes from 11D Kaluza Klein, more particularly from the classification of Einstein metrics in compact 7-manifolds. It is easy to produce from the 7-sphere metric a "squeezed sphere" whose isometry group is, instead of SO(8), just the one of $S^4 \times S^3$. But it is not known if there is some relationship between the 7-sphere and the "Witten manifolds" of the kind $CP^2 \times S^3$. </p>
<p>EDIT: to add more context, some dynkin diagrams.</p>
<pre><code>o====o SO(5), isometries of the sphere S4
o----o SU(3) are the isometries of CP2
o o SU(2)xSU(2), isometries of S2xS2. Also SO(4), so isometries of S3
</code></pre>
<p>So it seems that the quotient under conjugation has implied, or is compensated by, some change in the angles between roots, but not in the number of roots.</p>
<p>For isometries of 7-manifolds we have also some similarities.</p>
<pre><code> o o o
/
/
o----o SO(8) o----o SU(3)xSO(4) o====o SO(5)xSO(4)
\
\
o o o
</code></pre>
<p>where the first diagram is the [isometry group of] the seven sphere, the last is the squashed sphere, and the intermediate is the one I am intrigued about, as it contains the physicists standard model gauge group.</p>
<p>By the way, the last drawing makes one to ask about how triality survives in the representations of these product groups, but that is other question :-)</p>
| José Figueroa-O'Farrill | 394 | <p>Why would you think that they are related?</p>
<p>The map $\mathbb{CP}^2/\text{conjugation} \to S^4$ is only $SO(3)$-equivariant, where $SO(3) \subset SU(3)$ consists of the real matrices and $SO(3) \subset SO(5)$ is the maximal subgroup acting irreducibly on the 5-dimensional vector representation of $SO(5)$.</p>
<p>Not sure if this answers your question, though.</p>
|
2,324,850 | <p>How to find the shortest distance from line to parabola?</p>
<p>parabola: $$2x^2-4xy+2y^2-x-y=0$$and the line is: $$9x-7y+16=0$$
Already tried use this formula for distance:
$$\frac{|ax_{0}+by_{0}+c|}{\sqrt{a^2+b^2}}$$</p>
| Steven Alexis Gregory | 75,410 | <p>$$2x^2-4xy+2y^2-x-y=0$$</p>
<p>At a point $(x_0, y_0)$ on the parabola, the gradient is
$\nabla f(x_0,y_0) = (4x_0-4y_0-1, -4x_0+4y_0-1)$</p>
<p>The direction of a normal to the line $9x-7y+16=0$ is $(7,9)$. So we must have</p>
<p>\begin{align}
(4x_0-4y_0-1, -4x_0+4y_0-1) \circ (7,9) &= 0 \\
-8 x_0 + 8 y_0 - 16 &= 0 \\
y_0 = 2+x_0 \\
\hline
2 x_0^2 - 4 x_0 (2+x_0) + 2 (2+x_0)^2 - x_0 - (2+x_0) &= 0 \\
6 - 2x_0 &= 0 \\
(x_0, y_0) &= (3, 5)
\end{align}</p>
<p>The distance from the line $9x-7y+16=0$ to the point $(3,5)$ is</p>
<p>$$\frac{|9x_0-7y_0+16|}{\sqrt{9^2+7^2}}
=\frac{|27-35+16|}{\sqrt{130}} =\frac{8}{\sqrt{130}}$$</p>
|
3,979,674 | <p>Let
<span class="math-container">$$I=\int\frac{dx}{\sqrt{ax^2+bx+c}}$$</span>
I know this can be either
<span class="math-container">$$\displaystyle I=\frac{1}{\sqrt{a}}\ln\left({2\sqrt{a}\sqrt{ax^2+bx+c}+2ax+b}\right)+C$$</span>
<span class="math-container">$$\displaystyle I=-\frac{1}{\sqrt{-a}}\arcsin{\left(\frac{2ax+b}{\sqrt{b^2-4ac}}\right)}+C$$</span>
Or
<span class="math-container">$$I=\frac1{\sqrt{a}}\mathrm{arcsinh}\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)+C$$</span></p>
<p>Can you tell me the difference between these three solutions?</p>
<p>Thanks a lot</p>
| Leonard Neon | 818,617 | <p><span class="math-container">$$
\displaystyle I=\frac{1}{\sqrt{a}}\ln\left({2\sqrt{a}\sqrt{ax^2+bx+c}+2ax+b}\right)+C
\label{eq1} \tag{eq1}\\
$$</span></p>
<p><span class="math-container">$$
\displaystyle I=-\frac{1}{\sqrt{-a}}\arcsin{\left(\frac{2ax+b}{\sqrt{b^2-4ac}}\right)}+C
\label{eq2} \tag{eq2}\\
$$</span></p>
<p><span class="math-container">$$
I=\frac1{\sqrt{a}}\mathrm{arcsinh}\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)+C
\label{eq3} \tag{eq3}\\
$$</span></p>
<p><strong>First, \eqref{eq1} and \eqref{eq3} are the same.</strong>
<span class="math-container">$$\text{arcsinh} (x/a) = \ln(x + \sqrt{x^2 + a^2})-\ln a.$$</span></p>
<p><strong>Second, \eqref{eq2} and \eqref{eq3} are different.</strong></p>
<p>\eqref{eq2} holds, when <span class="math-container">$a<0$</span> and <span class="math-container">$b^2-4ac> 0$</span>.</p>
<p>\eqref{eq2} holds, when <span class="math-container">$a>0$</span> and <span class="math-container">$b^2-4ac< 0$</span>.</p>
|
217,291 | <p>I am trying to recreate the following image in latex (pgfplots), but in order to do so I need to figure out the mathematical expressions for the functions</p>
<p><img src="https://i.stack.imgur.com/jYGNP.png" alt="wavepacket"></p>
<p>So far I am sure that the gray line is $\sin x$, and that
the redline is some version of $\sin x / x$.
Whereas the green line is some linear combination of sine and cosine functions.</p>
<p>Anyone know a good way to find these functions? </p>
| Neal | 20,569 | <p>Here's a nice theorem illustrating how some basic algebraic concepts are useful in seemingly unrelated areas of mathematics. The <a href="http://en.wikipedia.org/wiki/Uniformization_theorem" rel="nofollow">uniformization theorem for surfaces</a> states that</p>
<blockquote>
<p>Every orientable topological surface $S$ admits a Riemannian metric of constant curvature $1$, $0$, or $-1$.</p>
</blockquote>
<p>What does algebra have to do with this? It's because we can rephrase this theorem using the <a href="http://en.wikipedia.org/wiki/Fundamental_group" rel="nofollow">fundamental group</a> of $S$, $\pi_1 S$:</p>
<blockquote>
<p>For every orientable topological surface, there exists a homomorphism from $\pi_1 S$ into one of $\operatorname{Isom}^+\mathbb{S}^2 = SO(3)$, $\operatorname{Isom}^+\mathbb{E}^2 = \mathbb{R}^2 \rtimes SO(2)$, or $\operatorname{Isom}^+\mathbb{H}^2 = SL(2;\mathbb{R})/\pm I$. The isometric action induced by this homomorphism is free and properly discontinuous and the quotient map is a universal covering map.</p>
</blockquote>
<p>There's a lot of background material here, so don't be intimidated if you don't fully understand the statement. Instead, look at how casually these algebraic concepts -- group, homomorphism, action, semidirect product, action -- are invoked in what seemed at first to be a purely geometric statement. </p>
<p>Moreover, in rephrasing the uniformization theorem in terms of algebra, one transforms the problem from differential geometry (computing curvature tensors on surfaces) to algebra (examining homomorphisms from a surface group into these special groups).</p>
<p>Here's another famous example: Mostow-Prasad rigidity:</p>
<blockquote>
<p>If two three-manifolds admit finite volume metrics of constant sectional curvature $-1$, then if they are homotopy equivalent, there is a homotopy from the homotopy equivalence to an isometry.</p>
</blockquote>
<p>Rephrased algebraically, </p>
<blockquote>
<p>If $\Gamma$ and $\Lambda$ are discrete cofinite subgroups of $PSL(2;\mathbb{C})$, and if $\Gamma$ and $\Lambda$ are isomorphic as abstract groups, then $\Gamma$ and $\Lambda$ are conjugate in $PSL(2;\mathbb{C})$.</p>
</blockquote>
<p>Again, without getting lost in details, note how this statement, which seems to be purely topological, can be rephrased in a statement that seems to be purely algebraic!</p>
<p>This is often a theme: take a difficult problem in geometry or topology, use a geometric construction to create an algebraic object, use algebra to study the algebraic object, and then deduce geometric or topological conclusions. </p>
<p>The take-home message here is: don't lose hope! To build up your understanding of algebra, and because there's only so much time in a semester, your class has to skimp on all the ways algebra is useful in mathematics and instead focus on building up the abstract edifice of algebra so you can work through the theorems and understand the ideas. But if you continue on in mathematics, you will begin to see these ideas casually invoked in areas that are, at first glance, completely unrelated to the pure structure of number systems.</p>
|
1,077,504 | <p>Evaluate:</p>
<p>$$\int_{0}^{\infty} \frac{1}{x^6 + 1} \,\mathrm dx$$</p>
<p>Without <strong>the use of complex-analysis.</strong></p>
<p>With complex analysis it is a very simple problem, how can this be done WITHOUT complex analysis?</p>
| Aditya Hase | 190,645 | <p>Let $\displaystyle \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathcal I=\int_0^\infty\frac{1}{1+x^6} \,\mathrm dx
$</p>
<p>$$\begin{align}
I&=\frac{1}{2}\left[ \int_0^\infty \frac{(1-x^2+x^4)+x^2+(1-x^4)}{(1+x^2)(1-x^2+x^4)} \,\mathrm dx \right]\tag{1}\\
&=
\frac{1}{2}\left[\int_0^\infty \frac{1}{1+x^2} \,\mathrm dx
+ \int_0^\infty \frac{x^2}{1+x^6} \,\mathrm dx
+ \color{grey}{\int_0^\infty \frac{1-x^2}{1-x^2+x^4} \,\mathrm dx}\right] \tag{2}\\
&=\frac{1}{2}\left[\frac\pi2+ \frac\pi{6} +\color\grey{0} \right] \tag{3}\\
&\mathcal I=\frac{\pi}{3} \tag{4}\\
\end{align}$$</p>
<blockquote>
<p>$$\int_0^\infty\frac{1}{1+x^6} \,\mathrm dx=\frac\pi3$$</p>
</blockquote>
<hr>
<p>$\text{ Explanation : }(3)$</p>
<p>$$
\small\color\grey{J=\int_0^\infty \frac{1-x^2}{1-x^2+x^4} \,\mathrm dx}
=\int_0^1 \frac{1-x^2}{1-x^2+x^4} \,\mathrm dx
+\int_1^\infty \frac{1-x^2}{1-x^2+x^4} \,\mathrm dx$$
Now substitute
$\small\displaystyle x=\frac1t$ in second integral, To get</p>
<p>$$
\small\color\grey{J=\int_0^\infty \frac{1-x^2}{1-x^2+x^4} \,\mathrm dx}
=\int_0^1 \frac{1-x^2}{1-x^2+x^4} \,\mathrm dx
-\int_0^1 \frac{1-t^2}{1-t^2+t^4} \,\mathrm dt=\color\grey0$$</p>
|
1,570,044 | <p>How many arrangements of banana such that the "b" occurs before any of the "a's"?</p>
<p>This is more an inquiry into what I did wrong in my counting. I came up with a solution of: $$\binom{3}{1} \binom{5}{3}$$ where i did C(3,1) to account for the 3 possible places the "b" could go and C(5,3) to choose the positions of the 3 "a's" Instead they rationalized it as $$ \binom{6}{4} (1)(1)$$</p>
<p>where and how did I double count? </p>
| joriki | 6,622 | <p>There are only $5$ slots for the $3$ a's if the 'b' is in the beginning. If it's in second place, there are $4$, and if it's in third place, there are $3$, so the total is</p>
<p>$$
\binom53+\binom43+\binom33=10+4+1=15=\binom64\;,
$$</p>
<p>which is the number of ways to choose $4$ out of $6$ slots for the letters "baaa" in that order.</p>
|
762,651 | <p>I have to prove that "any straight line $\alpha$ contained on a surface $S$ is an asymptotic curve and geodesic (modulo parametrization) of that surface $S$". Can I have hints at tackling this problem? It seems so general that I am not sure even how to formulate it well, let alone prove it. Intuitively, I imagine that the normal $n_{\alpha}$ to the line/curve is perpendicular to the normal vector $N_{S}$ to the surface $S$, thus resulting in the asymptoticity; alternatively, a straight line has curvature $k = 0$ everywhere, and so the result follows. Is this reasoning adequate for a proof of the first part? I also realize that both geodesics and straight lines are the paths of shortest distance between two points on given surfaces (here, both $S$), thus the straight line must be a geodesic of any surface which contains it; should I quantify this statement, though?</p>
<p>Let $\mathbb{H}^2 = \{(x, y) \in \mathbb{R}^2: y>0 \}$ be the hyperbolic plane with the Riemannian metric $(\mathbb{d}s)^2 =\frac{(\mathbb{d}x)^2+(\mathbb{d}y)^2}{y^2}$. Consider a "square" $P = \{ (x, y) \in \mathbb{H}^2: 1 \leq x,y \leq 2 \}$. I need to calculate the geodesic curvature of the sides of $P$ and, for Gaussian curvature $K$ of $P$, I have to calculate $\int_{P} (K) \mathbb{d}\sigma$, where $\mathbb{d}\sigma$ is the area measure element of $\mathbb{H}^2$. Just hints as to how to start would be helpful. (I see that I have the first fundamental form, from which I can derive the coefficients $E$, $F$, and $G$ and thereby (hopefully easily) Christoffel symbols and an expression for area, but I do not see how any of this takes the actual square into account. Only the coordinates at which I evaluate these quantities seem to come come from the square! But I would still like detailed examples of even these things, please.)</p>
| Community | -1 | <p>Short answer: We can't simply square both sides because that's exactly what we're trying to prove: $$0 < a < b \implies a^2 < b^2$$</p>
<p>More somewhat related details: I think it may be a common misconception that simply squaring both sides of an inequality is ok because we can do it indiscriminately with equalities. Let's take an example: $$x = 2$$ We know we can square both sides to get: $$x^2 = 4$$ But <em>why</em> can we square both sides? <strong>It's because squaring both sides is really just a special case of "multiplying both sides of the equality by the same thing."</strong> I think most of us are comfortable with the fact that if we have an equation such as
$$\dfrac{2y}3 = \dfrac83$$
then to solve it we multiply both sides by $3/2$:
\begin{align*}
\frac32 \cdot \dfrac{2y}3 &= \dfrac83 \cdot \frac32\\[0.3cm]
y &= 4
\end{align*}
Well, that's exactly the same concept we used when we squared both sides of $x=2$. Let's take $x=2$ and multiply both sides by $2$:
\begin{align*}
x &= 2\\
2 \cdot x &= 2 \cdot 2
\end{align*}
On the left-hand side of the last line above, we can replace the $2$ with a $x$ because <em>we already know that $x$ and $2$ are the same thing</em>. So then:
\begin{align*}
x &= 2\\
2 \cdot x &= 2 \cdot 2\\
x \cdot x &= 2 \cdot 2\\
x^2 &= 4
\end{align*}
I'll reiterate the main point for emphasis: Squaring both sides here worked because <em>we already know that $x$ and $2$ are equal</em>. This is exactly why we can't do the same thing with inequalities in general. If we have $x < 2$ then we can't square both sides, because squaring both sides is a special case of multiplying both sides by the same thing, and if $x < 2$ then $x$ and $2$ are <strong>not</strong> equal.</p>
<p>If $x < 2$ then $x^2 < 4$ is true if $x$ is, for example, $1$, and $x^2 < 4$ is false if $x$ is, for example, $-3$. So if we want to square both sides of $x < 2$ and still have a true inequality, then we need the additional restriction that $x > 0$. (Actually $x \ge 0$ is sufficient.)</p>
<p>But why is this restriction enough to make squaring both sides of the inequality ok? That's exactly what the original question wants you to answer. Yes, by the time you get to a math course involving proofs you should be very familiar with the fact that $0 < a < b \implies a^2 < b^2$, but now you actually have to explain <em>why</em> it's true. And the proof you outlined in your question (first multiply both sides by $a$, then by $b$) is exactly how we do that.</p>
|
410,105 | <p>I have this recurrence relation:</p>
<p>$$
R(1)=1, RE(1)=0, EE(1)=0$$</p>
<p>$$a(n)=R(n) + RE(n)$$</p>
<p>$$R(n)=EE(n-1)+RE(n-1),$$$$ RE(n)=R(n-1),$$$$ EE(n)=RE(n-1)
$$</p>
<p>How do I get $a(15)$?
What kind of method do I use?</p>
| N. S. | 9,176 | <p>$$R(n)=EE(n-1)+RE(n-1)=RE(n-2)+RE(n-1)=R(n-3)+R(n-2) \,.$$
$$RE(n)=R(n-1)=EE(n-2)+RE(n-2)=RE(n-3)+RE(n-2) \,.$$</p>
<p>Adding them you get</p>
<p>$$a(n)=a(n-3)+a(n-2) \,.$$</p>
<p>Solve it!</p>
|
4,402,262 | <p>A class of 24 pupils consists of 11 girls and 13 boys. To form the class committee, four of the pupils are chosen at random as "Chairperson", "Vice-Chairperson", "Treasurer", and "Secretary". Find the number of ways the committee can be formed if<br>
(i) the committee consists of at least one girl and at least one boy,<br>
(ii) the "Treasurer" and "Secretary" are both girls,<br>
(iii) The teacher requires a group of four students to represent the class in a student survey. Find the number of ways this group of students can be selected if there must be at least 1 girl and at most 2 boys.</p>
<p>My answers:</p>
<p>(i) GBBG, GBBB, GBGG</p>
<p>(11C2 x 13C2) + (11C1 x 13C3) + (11C3 x 13C1) = 9581</p>
<p>(ii) GGGG, BBGG, BGGG</p>
<p>11C4 + (11C2 x 13C2) + (11C3 x 13C1) = 6765</p>
<p>(iii) GGGG, BBGG, BGGG</p>
<p>11C4 + (11C2 x 13C2) + (11C3 x 13C1) = 6765</p>
<p>I have checked the correct answers, it shows that (i) 229944 (ii) 50820 (iii) 6765</p>
<p>I don't see my (i) is wrong, is it the correct answer for (i) of 229944 as incorrect?</p>
<p>For (ii) why it is using 11P2 x 22P2 = 50820 for the answer? Why this is a permutation question?</p>
| Acccumulation | 476,070 | <p>It wasn't clear at first that these are three different questions, rather than three conditions for the answer. Your teacher should write questions more clearly.</p>
<p>(i) There are <span class="math-container">$24P4$</span> ways to choose <span class="math-container">$4$</span> people from a set of <span class="math-container">$24$</span> where order matters (combinations are for when there aren't distinguished positions). There are <span class="math-container">$11P4$</span> ways to choose an all-girl committee, and <span class="math-container">$13P4$</span> ways to choose an all-boy committee. So there are <span class="math-container">$24P4-(11P4+13P4)$</span> ways to choose a committee that is neither all-girl nor all-boy.</p>
<p>(ii) There are <span class="math-container">$11P2$</span> ways to choose the Treasurer and Secretary. There are then <span class="math-container">$22$</span> people left to choose for <span class="math-container">$2$</span> positions, so <span class="math-container">$22P2$</span> ways to choose the remaining positions, giving <span class="math-container">$(11P2)*(22P2)$</span> in total.</p>
|
292,831 | <p>Usually the question whether the <a href="https://en.wikipedia.org/wiki/Diamond_principle" rel="noreferrer">diamond principle</a> $\diamondsuit(\kappa)$ holds for some large cardinal $\kappa$ only concerns large cardinal notions of very low consistency (among the weakly compacts). Partly since it <em>does</em> hold for all <a href="http://cantorsattic.info/Ineffable#Subtle_cardinal" rel="noreferrer">subtle cardinals</a>, which are only barely stronger than the weakly compacts, and pretty much every large cardinal notion below a weakly compact has been shown to consistently <em>not</em> satisfy it (see <a href="https://mathoverflow.net/questions/137036/failure-of-diamond-at-large-cardinals">Failure of diamond at large cardinals</a> and <a href="https://arxiv.org/abs/1705.01611" rel="noreferrer">Ben Neria ('17)</a>).</p>
<p>That subtle cardinals satisfy diamond of course means that almost all large cardinals <em>do</em> satisfy it as well, but there are some strange ones lying around though, including <a href="https://en.wikipedia.org/wiki/Woodin_cardinal" rel="noreferrer">Woodin cardinals</a> and inaccessible <a href="http://cantorsattic.info/Jonsson" rel="noreferrer">Jónsson cardinals</a>. Is anything known about diamond holding for any of these two?</p>
| Yair Hayut | 41,953 | <p>This is a partial answer. I will show that if $\delta$ is Woodin then $\diamondsuit_\delta$ holds. </p>
<p><strong>Claim:</strong> Any Woodin cardinal is subtle.</p>
<p><strong>Proof:</strong> Let $\delta$ be a Woodin cardinal. Let $\vec{A} = \langle A_\alpha \mid \alpha < \delta\rangle$ be a sequence a sets, $A_\alpha \subseteq \alpha$ and let $C$ be a club in $\delta$. We want to find $\alpha < \beta$ in $C$ such that $A_\alpha = A_\beta \cap \alpha$. </p>
<p>Since $\delta$ is Woodin, there is a cardinal $\kappa < \delta$ which is $\vec{A} \times C$-strong up to $\delta$. Thus, $\kappa \in C$ and there is an elementary emebedding $j\colon V\to M$, such that :</p>
<ul>
<li>$\mathrm{crit}\ j = \kappa$, </li>
<li>$j(\vec{A}) \restriction \kappa + 1 = \vec{A} \restriction \kappa + 1$, </li>
</ul>
<p>In $M$, $j(\vec{A})(j(\kappa)) \cap \kappa = j(\vec{A})(\kappa) = A_\kappa$ and $\kappa, j(\kappa) \in j(C)$. By elementarity, there is $\alpha < \kappa$ in $C$ such that $A_\alpha = A_\kappa \cap \alpha$. </p>
|
241,612 | <blockquote>
<p>Find all eigenvalues and eigenvectors:</p>
<p>a.) $\pmatrix{i&1\\0&-1+i}$</p>
<p>b.) $\pmatrix{\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta}$</p>
</blockquote>
<p>For a I got:
$$\operatorname{det} \pmatrix{i-\lambda&1\\0&-1+i-\lambda}= \lambda^{2} - 2\lambda i + \lambda - i - 1
$$</p>
<p>For b I got:
$$\operatorname{det} \pmatrix{\cos\theta - \lambda & -\sin\theta \\ \sin\theta & \cos\theta - \lambda}= \cos^2\theta + \sin^2\theta + \lambda^2 -2\lambda \cos\theta = \lambda^2 -2\lambda \cos\theta +1$$</p>
<p>But how can I find the corresponding eigenvalues for a and b? </p>
| dantopa | 206,581 | <p><strong>Basic tools</strong></p>
<p>For $2 \times 2$ matrices the characteristic polynomial is:
$$
p(\lambda) = \lambda^{2} - \lambda\, \text{tr }\mathbf{A} + \det \mathbf{A}
$$</p>
<p>The roots of this function are the eigenvalues, $\lambda_{k}$, k=1,2$.</p>
<p>The eigenvectors solve the eigenvalue equation
$$
\mathbf{A} u_{k} = \lambda_{k} u_{k}
$$
<hr>
<strong>Case 1</strong></p>
<p>$$
\mathbf{A} =
%
\left(
\begin{array}{cc}
i & 1 \\
0 & -1+i \\
\end{array}
\right)
$$
The trace and determinant are
$$
\text{tr } \mathbf{A} = -1 + 2 i, \qquad \det \mathbf{A} = -1 - i
$$
The characteristic polynomial is
$$
p(\lambda) = \lambda ^2+(1-2 i) \lambda +(-1-i)
$$
The roots of this polynomial are the eigenvalues:
$$ \lambda \left( \mathbf{A} \right) = \left\{
i, -1 + i
\right\}
$$</p>
<p><em>First eigenvector:</em>
$$
\begin{align}
\left( \mathbf{A} - \lambda_{1} \mathbf{I}_{2} \right) u_{1} &= \mathbf{0} \\[3pt]
%
\left(
\begin{array}{cc}
i & 1 \\
0 & -1+i \\
\end{array}
\right) -
\left( -1 + i \right)
%
\left(
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{cc}
u_{x} \\
u_{y} \\
\end{array}
\right)
&=
\left(
\begin{array}{c}
0 \\
0 \\
\end{array}
\right) \\
%
\left(
\begin{array}{cc}
u_{x} + u_{y}\\
0 \\
\end{array}
\right)
&=
\left(
\begin{array}{c}
0 \\
0 \\
\end{array}
\right)
%
\end{align}
$$
The result is
$$
u_{1} =
\left(
\begin{array}{cc}
u_{x} \\
u_{y} \\
\end{array}
\right)
=
\alpha
\left(
\begin{array}{r}
1 \\
-1 \\
\end{array}
\right), \quad \alpha \in \mathbb{C}
$$</p>
<p><em>Second eigenvector:</em>
$$
\begin{align}
\left( \mathbf{A} - \lambda_{2} \mathbf{I}_{2} \right) u_{2} &= \mathbf{0} \\[3pt]
%
\left(
\begin{array}{cr}
0 & 1 \\
0 & -1 \\
\end{array}
\right)
%
\left(
\begin{array}{cc}
u_{x} \\
u_{y} \\
\end{array}
\right)
&=
\left(
\begin{array}{c}
0 \\
0 \\
\end{array}
\right) \\
%
\end{align}
$$
The result is
$$
u_{2} =
\alpha
\left(
\begin{array}{c}
1 \\
0 \\
\end{array}
\right), \quad \alpha \in \mathbb{C}
$$
<hr>
<strong>Case 2</strong></p>
<p>$$
\mathbf{A} =
%
\left(
\begin{array}{cr}
\cos (\theta ) & -\sin (\theta ) \\
\sin (\theta ) & \cos (\theta ) \\
\end{array}
\right)
$$
The trace and determinant are
$$
\text{tr } \mathbf{A} = 2 \cos \theta, \qquad \det \mathbf{A} = \cos^{2} \theta + \sin^{\theta} = 1
$$
The characteristic polynomial is
$$
p(\lambda) = \lambda ^2+(1-2 i) \lambda +(-1-i)
$$
The roots of this polynomial are the eigenvalues:
$$ \lambda \left( \mathbf{A} \right) = \left\{
\cos \theta -i \sin \theta ,\cos \theta +i \sin \theta
\right\}
$$</p>
<p><em>First eigenvector:</em>
$$
\begin{align}
\left( \mathbf{A} - \lambda_{1} \mathbf{I}_{2} \right) u_{1} &= \mathbf{0} \\[3pt]
%
\left(
\begin{array}{cr}
i \sin \theta & -\sin \theta \\
\sin \theta & i \sin \theta \\
\end{array}
\right)
%
\left(
\begin{array}{cc}
u_{x} \\
u_{y} \\
\end{array}
\right)
&=
\left(
\begin{array}{c}
0 \\
0 \\
\end{array}
\right) \\
%
\left(
\begin{array}{cc}
-u_{y} \sin \theta +i u_{x} \sin \theta \\
u_{x} \sin \theta +i u_{y} \sin \theta
\end{array}
\right)
&=
\left(
\begin{array}{c}
0 \\
0 \\
\end{array}
\right)
%
\end{align}
$$
The result is
$$
u_{1} =
\left(
\begin{array}{cc}
u_{x} \\
u_{y} \\
\end{array}
\right)
=
\alpha
\left(
\begin{array}{r}
i \\
-1 \\
\end{array}
\right), \quad \alpha \in \mathbb{C}
$$</p>
<p><em>Second eigenvector:</em>
$$
\begin{align}
\left( \mathbf{A} - \lambda_{2} \mathbf{I}_{2} \right) u_{2} &= \mathbf{0} \\[3pt]
%
\left(
\begin{array}{rr}
-i \sin \theta & -\sin \theta \\
\sin \theta & -i \sin \theta \\
\end{array}
\right)
%
\left(
\begin{array}{c}
u_{x} \\
u_{y} \\
\end{array}
\right)
&=
\left(
\begin{array}{c}
0 \\
0 \\
\end{array}
\right) \\
%
\end{align}
$$
The result is
$$
u_{2} =
\alpha
\left(
\begin{array}{c}
i \\
1 \\
\end{array}
\right), \quad \alpha \in \mathbb{C}
$$</p>
|
1,218,582 | <p>I was presented with the function $max (|x|,|y|)$ which should output a maximum value of given two.... I can only suppose this one creates some body in $\mathbb{R}^3$ but how do you sketch it and what does it mean in $\mathbb{R}^3$? for that matter in $\mathbb{R}^2$ I cant really imagine it also.. </p>
| Hagen von Eitzen | 39,174 | <p>It's a top-down pyramid, the faces lying in the four planes $z=x$, $z=-x$, $z=y$, $z=-y$.</p>
|
1,215,542 | <p>I am not sure why in my textbook there is a long proof for this because on the pages before this was prooven:</p>
<p>Let $S = \{ x \mid a \leq x \leq b \hbox{ and } x \in \mathbb{R} \}$. Then $\sup S = b$ must hold.</p>
<p>EDIT: also $\inf S=a$ follows. But the proof for $\sup S$ does not depend on there being a lower bound to $S$ this can be modified to be:</p>
<p>Let $S = \{ x \mid x \leq b \hbox{ and } x ∈ \mathbb{R} \}$. Then $\sup S = b$ must hold.</p>
<p>With this you can instantly proove $\sup (S + T) = \sup (S) + \sup (T)$
since from $s \leq \sup S$ and $t\leq \sup T$ we get $s+t \leq \sup S + \sup T$ and so we get $\sup (S+T)=\sup S + \sup T$.</p>
<p>I just don't see why the book goes the long way when we already just did the work. Or am I just getting some logic completely wrong here?</p>
| Regret | 184,794 | <p>The problem is that $S+T$ is not defined as</p>
<p>$$S+T=\{x\in\Bbb R\mid\inf S + \inf T\le x\le\sup S+\sup T \}$$</p>
<p>It is defined as</p>
<p>$$S+T=\{s+t\mid s\in S,t\in T\}$$</p>
<p>So you can not use your previous result that if $X=\{x\in\Bbb R\mid a\le x \le b\}$ then $\sup X=b$. While it is true that $s+t\le \sup S+\sup T$, this is not how the set is defined.</p>
|
1,215,542 | <p>I am not sure why in my textbook there is a long proof for this because on the pages before this was prooven:</p>
<p>Let $S = \{ x \mid a \leq x \leq b \hbox{ and } x \in \mathbb{R} \}$. Then $\sup S = b$ must hold.</p>
<p>EDIT: also $\inf S=a$ follows. But the proof for $\sup S$ does not depend on there being a lower bound to $S$ this can be modified to be:</p>
<p>Let $S = \{ x \mid x \leq b \hbox{ and } x ∈ \mathbb{R} \}$. Then $\sup S = b$ must hold.</p>
<p>With this you can instantly proove $\sup (S + T) = \sup (S) + \sup (T)$
since from $s \leq \sup S$ and $t\leq \sup T$ we get $s+t \leq \sup S + \sup T$ and so we get $\sup (S+T)=\sup S + \sup T$.</p>
<p>I just don't see why the book goes the long way when we already just did the work. Or am I just getting some logic completely wrong here?</p>
| egreg | 62,967 | <p>If $S=\{x\in\mathbb{R}:a\le x\le b\}$, then it's of course true that $b=\sup S$: indeed, $b\le x$, for all $x\in S$ and $b\in S$, so $b$ is actually the maximum of $S$.</p>
<p>When you have two sets $S$ and $T$ (subsets of $\mathbb{R}$), then
$$
S+T=\{x+y:x\in S,y\in T\}
$$
is the set of all numbers that can be expressed as the sum of an element in $S$ with an element in $T$.</p>
<p>Assume both sets are bounded (and non empty). We wish to prove that
$$
\sup(S+T)=\sup S+\sup T
$$</p>
<p>If $x\in S$ and $y\in T$, then $x\le\sup S$ and $y\le\sup T$, so $x+y\le\sup S+\sup T$. Therefore $\sup S+\sup T$ is an upper bound for $S+T$ and, by definition of supremum, $\sup(S+T)\le\sup S+\sup T$.</p>
<p>Conversely, we want to see that, for every $\varepsilon>0$, $\sup S+\sup T-\varepsilon$ is <em>not</em> an upper bound for $S+T$, so it will follow that $\sup S+\sup T$ is the <em>least</em> upper bound of $S+T$.</p>
<p>Since $\sup S-\frac{\varepsilon}{2}$ is not an upper bound of $S$, there exists $x\in S$ such that $\sup S-\frac{\varepsilon}{2} < x$. Similarly, there exists $y\in T$ such that $\sup T-\frac{\varepsilon}{2}<y$.</p>
<p>It follows that
$$
\sup S+\sup T-\varepsilon<x+y
$$
Since $x+y\in S+T$, we have what we wanted.</p>
<hr>
<p>Note that the set $(S+T)'=\{x\in\mathbb{R}:x\le\sup S+\sup T\}$ can be <strong><em>very</em></strong> different from $S+T$. Just take $S=\{1\}$ and $T=\{2\}$; then $S+T=\{3\}$ (one element set), while $(S+T)'$ contains infinitely many elements.</p>
|
4,195,399 | <p>Given that <span class="math-container">$a,b,c > 0$</span> are real numbers such that <span class="math-container">$$\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}\le 1,$$</span> prove that <span class="math-container">$$\frac{1}{b+c+1}+\frac{1}{c+a+1}+\frac{1}{a+b+1}\ge 1.$$</span></p>
<hr />
<p>I first rewrote <span class="math-container">$$\frac{1}{a+b+1} = 1 - \frac{a+b}{a+b+1},$$</span> so the second inequality can be rewritten as <span class="math-container">$$\frac{b+c}{b+c+1} + \frac{c+a}{c+a+1} + \frac{a+b}{a+b+1} \le 2.$$</span> Cauchy-Schwarz gives us <span class="math-container">$$\sum \frac{a+b}{a+b+1} \geq \frac{(\sum \sqrt{a+b})^2}{\sum a+ b+ 1}.$$</span> That can be rewritten as <span class="math-container">$$\frac{2(a+b+c) + 2\sum \sqrt{(a+b)(a+c)}}{2(a+b+c) + 3},$$</span> which is greater than or equal to <span class="math-container">$$\frac{2(a+b+c) + 2 \sum(a + \sqrt{bc})}{2(a+b+c) + 3} = \frac{4(a+b+c) + 2 \sum \sqrt{bc}}{2(a+b+c) + 3} \geq 2,$$</span> which is the opposite of what I want. Additionally, I'm unsure of how to proceed from here.</p>
| achille hui | 59,379 | <p>Since the two inequalities are completely symmetric in <span class="math-container">$a,b,c$</span>. WLOG, we only need to study the case <span class="math-container">$a \ge b \ge c$</span>.</p>
<p>Let <span class="math-container">$\Lambda = a + b + c + 1$</span>. The two inequalities can be rewritten
as</p>
<p><span class="math-container">$$\sum_{cyc}\frac{a}{\Lambda -a } \stackrel{def}{=}\frac{a}{b+c+1} + \frac{b}{c+a+1} + \frac{c}{a+b+1} \le 1\\
\sum_{cyc}\frac{1}{\Lambda-a}\stackrel{def}{=}\frac{1}{b+c+1}
+ \frac{1}{c+a+1} + \frac{1}{a+b+1} \stackrel{?}{\ge} 1
$$</span></p>
<p>Notice for <span class="math-container">$x \in (0,\Lambda)$</span>, the map <span class="math-container">$x \mapsto \frac{1}{\Lambda - x}$</span> is increasing. We have</p>
<p><span class="math-container">$$a \ge b \ge c \quad\implies\quad \frac{1}{\Lambda - a} \ge \frac{1}{\Lambda - b} \ge \frac{1}{\Lambda - c}$$</span></p>
<p>By <a href="https://en.wikipedia.org/wiki/Rearrangement_inequality" rel="nofollow noreferrer">Rearrangement inequality</a>, we have</p>
<p><span class="math-container">$$\begin{align}
1 \ge \sum_{cyc}\frac{a}{\Lambda-a} \ge \sum_{cyc}\frac{b}{\Lambda-a}
= \frac{b}{\Lambda-a} +\frac{c}{\Lambda-b} + \frac{a}{\Lambda-c}
\\
1 \ge \sum_{cyc}\frac{a}{\Lambda-a} \ge \sum_{cyc}\frac{c}{\Lambda-a}
= \frac{c}{\Lambda-a} +\frac{a}{\Lambda-b} + \frac{b}{\Lambda-c}
\end{align}
$$</span>
From these, using the decomposition you have, one obtain:</p>
<p><span class="math-container">$$\sum_{cyc}\frac{1}{\Lambda-a}
= \sum_{cyc}\left(1 - \frac{b+c}{\Lambda-a}\right) = 3 -
\sum_{cyc}\frac{b+c}{\Lambda - a}
\ge 3 - (1+1) = 1
$$</span></p>
|
21,262 | <p><strong>Bug introduced in 9.0 and fixed in 11.1</strong></p>
<hr>
<p><code>NDSolve</code> in Mathematica 9.0.0 (MacOS) is behaving strangely with a piecewise right hand side. The following code (a simplified version of my real problem):</p>
<pre><code>sol = NDSolve[{x'[t] ==
Piecewise[{{2, 0 <= Mod[t, 1] < 0.5},
{-1, 0.5 <= Mod[t, 1] < 1}}
], x[0] == 0}, x, {t, 0, 1}];
Print[x[1] /. sol[[1]]];
</code></pre>
<p>gives the correct answer of 0.5 about 50% of the time, but often returns -0.5 and -1 instead. Rerunning it gives apparently random results. It always gives the correct result in Mathematica 8.</p>
<p>Here's what I've figured out so far:</p>
<ol>
<li>It apparently has something to do with the <code>Mod[t,1]</code>, because it works fine with just "t" in the <code>Piecewise</code>. Unfortunately I'm looking at a piecewise periodic system (not just from t=0 to 1).</li>
<li>It's only the first segment of the solution from t=0 to t=0.5 that varies from run to run.</li>
<li>Using initial condition <code>x[10^-100]==0</code> fixes the problem, but this is an ugly hack.</li>
</ol>
<p>Can anyone replicate this strange behavior, know what's behind it, or have a better suggested fix?</p>
| Chris K | 6,358 | <p>Following a lead from Albert Retey, I found an <code>NDSolve</code> option that fixes this problem:</p>
<pre><code>sol = NDSolve[{x'[t] ==
Piecewise[{{2, 0 <= Mod[t, 1] < 0.5},
{-1, 0.5 <= Mod[t, 1] < 1}}
], x[0] == 0}, x, {t, 0, 1}, Method -> {"DiscontinuityProcessing" -> False}];
Print[x[1] /. sol[[1]]];
</code></pre>
|
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