qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
1,041,134 | <p>I need to show if $a$ is in $\mathbb{R}$ but not equal to $0$, and $a+\dfrac{1}{a}$ is integer, $a^t+\dfrac{1}{a^t}$ is also an integer for all $t\in\mathbb N$.
Can you provide me some hints please?</p>
| Adhvaitha | 191,728 | <p>Use induction on $$a^{t+1} + \frac1{a^{t+1}} = \left(a^t+\frac1{a^t}\right)\left(a+\frac1a\right) - \left(a^{t-1}+\frac1{a^{t-1}}\right)$$</p>
|
771,997 | <p>I'm working on a matrix extension of Fermat's Little Theorem, but I'm stuck on trying to show that if $A^p \equiv A$ mod $p$, then $A$ does not have to be diagonalizable.</p>
<p>Any help would be appreciated!</p>
<p><strong>Edit::</strong> I would like to either find a reason why $A$ does not have to be diagonalizable, or somehow be able to categorize the matrices that are / are not diagonalizable in a way that would suggest a pattern. An example would be that if $A^p \equiv A$ mod $p$, then $A$ is diagonalizable when xxx or not diagonalizable when xxx.</p>
| Pedro | 23,350 | <p>You haven't been clear about your base field. Let $k=\Bbb R$. Consider $p=2$. Then $\begin{pmatrix}1&2\\0&1\end{pmatrix}$ is congruent to its square modulo two. However, it is not diagonalizable over $\Bbb R$, say, since it is a Jordan block.</p>
|
963,125 | <p>I'm not exactly sure where to start on this one. Any help would be greatly appreciated.</p>
<p>Show that $4$ does not divide $12x+3$ for any $x$ in the integers.</p>
<p>Here's what I have so far:</p>
<p>There exist c in the integers such that 4 | 12x+3. Then 12x+3 = 4y for some y in the integers. This is a contradiction since y = 3x + 3/4 and y is in the integers.</p>
<p>Is that correct?</p>
| C.S. | 95,894 | <ul>
<li>A number which is divisible by $4$ is <strong>always even</strong> where as $12x+3$ is always <strong>Odd</strong>.</li>
</ul>
|
1,435,189 | <p>A map $T:X \rightarrow Y$ is $A, B$- measurable if the pre-image of every set in $B$ is a set in $A$, where $A,B$ are $\sigma$-algebras in $X,Y$. </p>
<p>But I was thinking that isn't any map $T: X \rightarrow Y$ measurable then? Because since pre-images behave so nicely, the pre-image of all sets from $B$ make up a $\sigma$-algebra in $X$, and thus the definition is satisfied?</p>
| Giuseppe Negro | 8,157 | <p>It is true that preimages of $B$ form a sigma-algebra in $X$. However, there is no a priori reason why this sigma-algebra should be contained in $A$.</p>
|
1,435,189 | <p>A map $T:X \rightarrow Y$ is $A, B$- measurable if the pre-image of every set in $B$ is a set in $A$, where $A,B$ are $\sigma$-algebras in $X,Y$. </p>
<p>But I was thinking that isn't any map $T: X \rightarrow Y$ measurable then? Because since pre-images behave so nicely, the pre-image of all sets from $B$ make up a $\sigma$-algebra in $X$, and thus the definition is satisfied?</p>
| Community | -1 | <p>That's not true, of course. </p>
<p>Here's an example which doesn't require us to be in any special setting: Let $X$ be any set and let $\mathcal A$ be a $\sigma$-algebra on $X$ that is a strict subset of $\mathcal P(X)$. Let $A \in \mathcal P(X) -\mathcal A$. Then: $1_A - 1_{A^C}$ is not measurable. </p>
<p>A part of what you are saying is true, though. The set $f^{-1}(\mathcal B)$ does form a $\sigma$-algebra on $X$, but it simply isn't necessarily $\subseteq \mathcal A$, the presumed $\sigma$-algebra on $X$.</p>
<p>IMO one must refer to a measurable space as $(X, \mathcal A)$ rather than just $X$ when one is still a beginner. This allows one to avoid possible misconceptions. What I mean is: had you written $T: (X, \mathcal A) \to (Y, \mathcal B)$, you wouldn't have had this confusion, I think.</p>
|
1,160,076 | <p>Consider $f:\left[a,b\right]\rightarrow \mathbb{R}$ continuous at $\left[a,b\right]$ and differentiable at $\left(a,b\right)$.<br>
$\forall x\:\ne \frac{a+b}{2}$, $f'\left(x\right)\:>\:0$, but at $x\:=\frac{a+b}{2}$, $f'\left(x\right)\:=\:0$.<br>
Show that $f$ is <strong>strictly increasing</strong> at $\left[a,b\right]$. </p>
<p>So far i don't find any approach how to handle with $x\:=\:\frac{a+b}{2}$.<br>
I know that $x\:=\:\frac{a+b}{2}$ can't be minimum or maximum.
I thought about assuming there is point $c\:\ne \frac{a+b}{2}$ such that $f\left(c\right)=f\left(\frac{a+b}{2}\right)$ and get a contradict somehow but i can't figure out how..
Any ideas? thanks in advance!</p>
| MooS | 211,913 | <p>For sure $f$ is increasing in the sense that $f(x) \geq f(y)$ whenever $x > y$. Now assume $f(x)=f(y)$ for $x > y$. Clearly, then $f$ must be constant on $[y,x]$, contradicting the fact that the derivative vanishes at only one single point.</p>
|
24,912 | <p>I have the category-theoretic background of the occasional stroll through MacLane's text, so excuse my ignorance in this regard. I was trying to learn all that I could on the subject of tensor algebras, and higher exterior forms, and I ran into the notion of cohomological determinants. Along this line of inquiry, I ran into the general use of the notion of a Picard category, and kept running into frustration in trying to find some sort of exposition of what these structures are. So where can I find out more about these structures, and about (cohomological) determinants in K-theory, which seem to be a hot topic among AG, AT, and RT researchers alike at the moment.</p>
| David Ben-Zvi | 582 | <p>Determinants are discussed (in a language relevant to this current question) in <a href="https://mathoverflow.net/questions/7124/determinant-of-a-perfect-complex/">this MO question</a>.</p>
<p>One place Picard categories naturally appear is as fundamental (aka Poincare) groupoids -- specifically those of infinite loop spaces (which are a refined homotopical version of abelian group objects in spaces, so form a natural source of abelian groups in categories.
In fact Picard categories are equivalent to spectra which have only two consecutive homotopy groups, which up to shift we may as well take to be $\pi_0$ and $\pi_1$ -- one direction is given by the fundamental groupoid.</p>
<p>The important example of the Picard category of graded lines over a field arises this way from the algebraic K-theory spectrum of the field, via the determinant line construction (see eg Beilinson's paper referred to in the answers to the above MO link). </p>
<p>Another example important in rep theory is the Picard category of sheaves of twisted differential operators. This is discussed in detail in the famous "Proof of Jantzen Conjectures" paper of Beilinson-Bernstein.</p>
|
24,912 | <p>I have the category-theoretic background of the occasional stroll through MacLane's text, so excuse my ignorance in this regard. I was trying to learn all that I could on the subject of tensor algebras, and higher exterior forms, and I ran into the notion of cohomological determinants. Along this line of inquiry, I ran into the general use of the notion of a Picard category, and kept running into frustration in trying to find some sort of exposition of what these structures are. So where can I find out more about these structures, and about (cohomological) determinants in K-theory, which seem to be a hot topic among AG, AT, and RT researchers alike at the moment.</p>
| Community | -1 | <p>Regarding the transcription of H X Sinh’s thesis <a href="https://agrothendieck.github.io/divers/GCS.pdf" rel="nofollow noreferrer">[pdf]</a> (still in progress), professor J Baez pointed out
<a href="https://categorytheory.zulipchat.com/#narrow/stream/260000-practice.3A-translation/topic/Sinh.20thesis.2E.20grothendieck" rel="nofollow noreferrer">[link]</a>:</p>
<blockquote>
<p>It may be worth mentioning that Hoàng Xuân Sính , often called "Madame Sinh" in the literature, was a Vietnamese student of Grothendieck who in her thesis studied what we now call 2-groups (monoidal categories where all objects and morphisms are invertible). She called them Gr-categories, and she classified them in terms of group cohomology.</p>
</blockquote>
<blockquote>
<p>She showed that any Gr-category gives an element of
<span class="math-container">$H^3(G,A)$</span> where G is the group of isomorphism classes of objects and A is the group of endomorphisms of the unit object. This is sometimes called the Sinh invariant. It comes from the associator, which gives a 3-cocycle thanks to the pentagon identity. Replacing the Gr-category by a monoidally equivalent one changes this 3-cocycle by a cohomologous one, so the Sinh invariant is a cohomology class.</p>
</blockquote>
<p><a href="https://pnp.mathematik.uni-stuttgart.de/lexmath/kuenzer/sinh.html" rel="nofollow noreferrer">[Gr-catégories]</a>, H X Sinh</p>
<p>The theory of Picard categories appears prominently in the thesis.</p>
|
4,629,824 | <p>Let <span class="math-container">$k$</span> be a given positive integer. I want to solve the following system of Diophantine equations: <span class="math-container">$$\begin{cases} a^2 + b^2 + c^2 = k^2 \\ b^2 = ac \end{cases}$$</span>
where <span class="math-container">$a, b, c \in \mathbb{N}$</span> are non-zero.</p>
<p>There is an OEIS sequence for the numbers which are sums of three non-zero squares, but I don't know of any general expression (some squares, such as 25, are not there).</p>
<p>I know, by substituting the second equation into the first, that <span class="math-container">$k = m^2 + n^2 + mn$</span>, <span class="math-container">$a = m^2 - n^2$</span> and <span class="math-container">$c = 2mn + n^2$</span>, where <span class="math-container">$m, n$</span> are coprime integers, but I don't really know how to use this, since plugging this back into the second equation yields a very complicated looking one (namely, when is the product a square?)</p>
<p>I'm not used to solving Diophantine equations (much less systems of them), but I have tried direct computation and seriously believe that there is no solution.</p>
<p>Are there any hints or techniques for solving this type of problem?</p>
<p>Thanks in advance!</p>
| Tomita | 717,427 | <p>According to Will Jagy's <a href="https://math.stackexchange.com/questions/2445717/how-to-solve-the-diophantine-equation-x2xyy2-r2-x-y-r-in-bbb-z?noredirect=1">answer</a>, a parametric solution of <span class="math-container">$a^2+ ac +c^2=k^2$</span> is given as follows.<br />
<span class="math-container">$(a,c)=(u^2-v^2,2uv+v^2)$</span>.<br />
Hence we get <span class="math-container">$b^2=2vu^3+v^2u^2-2v^3u-v^4$</span>.<br />
Above equation can be transformed to an elliptic curve below where <span class="math-container">$X=2u/v$</span> .<br />
<span class="math-container">$$Y^2 = X^3+X^2-4X-4$$</span>
According to <a href="https://www.lmfdb.org/EllipticCurve/Q/48/a/4" rel="nofollow noreferrer">LMFDB</a>, this elliptic curve has rank <span class="math-container">$0$</span> and has three integer solutions <span class="math-container">$(X,Y)=(-2, 0), (-1, 0), (2, 0)$</span>.<br />
Hence the problem has no positive integer solution.</p>
|
3,291,303 | <p>this very same problem appeared in a different thread but the questions was slightly different. In my case, I'm looking precisely for the answer.</p>
<p>This is how I solved it, I only need confirmation of whether this actually is correct:</p>
<p>So I assumed that 1111... (100 ones) is going to be exactly divided by 1111111 (7 ones) 14 times (100/7 = 14). Hence, if 14 times 7 equals 98 '1's, then the remainder is 2 '1's.</p>
<p>Thanks.</p>
| Robert Israel | 8,508 | <p>The first seven <span class="math-container">$1$</span>'s gives you a multiple of <span class="math-container">$1111111$</span>, same for the next seven, etc.
Since <span class="math-container">$100 = 14 \cdot 7 + 2$</span> after removing <span class="math-container">$14$</span> groups of seven <span class="math-container">$1$</span>'s you are left with two: the remainder is <span class="math-container">$11$</span>.</p>
|
156,769 | <p>Let $\{a_{n}\}$ be a sequence of real numbers, where $0<a_{n}<1$, such that $\lim_{n\to \infty} a_{n}=0$, (then every subsequence will converges to zero). Is there any way to find a subsequence of $a_{n}$ which is decreasing to 0?</p>
| ncmathsadist | 4,154 | <p>Yes. Put $n_1 = 1$. Suppose $n_1<n_2 < \cdots n_k$ are chosen. Choose $n_{k+1}> n_k$ so that $a_{n_{k+1}} < \min\{a_{n_j}, 1\le j \le k\}$. Such a sequence will fulfill your specification.</p>
|
3,276,124 | <p>Consider the equation</p>
<p><span class="math-container">$$-242.0404+0.26639x-0.043941y+(5.9313\times10^{-5})\times xy-(3.9303\times{10^{-6}})\times y^2-7000=0$$</span></p>
<p>with <span class="math-container">$x,y>0$</span>. If you plot it, it'll look like below:</p>
<p><a href="https://i.stack.imgur.com/OqfFa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OqfFa.png" alt="enter image description here"></a></p>
<p>Now, I want to find the corner point/inflection point in this equation/graph, which roughly should be somewhere here. This is my manually pinpointed approximation, using my own eyes:</p>
<p><a href="https://i.stack.imgur.com/JawkV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JawkV.png" alt="enter image description here"></a></p>
<p>Any help on how to mathematically find this point would be really helpful.</p>
<p><strong>UPDATE</strong>
Based on Adrian's answer, I've got the following <span class="math-container">$(1.1842*10^{-4},0.6456*10^{-4})$</span> (wondering what can cause this slight error?):</p>
<p><a href="https://i.stack.imgur.com/oGXyw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oGXyw.png" alt="enter image description here"></a></p>
<p>The actual corner point seems a little bit far from the one found by Adrian's approach (why?):</p>
<p><a href="https://i.stack.imgur.com/9j0tN.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9j0tN.jpg" alt="enter image description here"></a></p>
<p><strong>Update 2</strong>
The problem was the aspect ratio of my drawing, I fixed the aspect ratio and Adrian's answer looks pretty accurate:</p>
<p><a href="https://i.stack.imgur.com/vzpoI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vzpoI.png" alt="enter image description here"></a></p>
| Adrian Keister | 30,813 | <p>We first simplify the expression, and then solve for <span class="math-container">$x:$</span>
<span class="math-container">\begin{align*}
-242.0404+0.26639x-0.043941y+\left(5.9313\times10^{-5}\right)xy-\left(3.9303\times{10^{-6}}\right) y^2-7000&=0 \\
0.26639x-0.043941y+\left(5.9313\times10^{-5}\right)xy-\left(3.9303\times{10^{-6}}\right) y^2-7242.0404&=0
\end{align*}</span>
<span class="math-container">\begin{align*}
x\left[0.26639+\left(5.9313\times10^{-5}\right)\!y\right]&=7242.0404+\left(3.9303\times{10^{-6}}\right) y^2+0.043941y \\
x&=\frac{7242.0404+\left(3.9303\times{10^{-6}}\right) y^2+0.043941y}{0.26639+\left(5.9313\times10^{-5}\right)\!y}.
\end{align*}</span>
We see that <span class="math-container">$x$</span> is a function of <span class="math-container">$y$</span>, with domain all real numbers except <span class="math-container">$-0.26639/\left(5.9313\times 10^{-5}\right).$</span> Just invert the function (corresponds to reflecting about the line <span class="math-container">$y=x$</span>). We have
<span class="math-container">$$y(x)=\frac{7242.0404+\left(3.9303\times{10^{-6}}\right) x^2+0.043941x}{0.26639+\left(5.9313\times10^{-5}\right)\!x}. $$</span>
I would say that the corner you're after is a point where <span class="math-container">$y'(x)=-1$</span>. So we have
<span class="math-container">$$y'(x)=\frac{-1.18771\times 10^{8}+595.215x+0.0662637x^2}{(4491.26+x)^2}. $$</span>
Setting <span class="math-container">$y'(x)=-1$</span> and solving for <span class="math-container">$x,$</span> we find that
<span class="math-container">$$x=-15104.6,\;6122.12, $$</span>
with corresponding
<span class="math-container">$$y=-11874.4,\; 12165.6, $$</span>
respectively. So the point you're after (swapping <span class="math-container">$x$</span> and <span class="math-container">$y$</span> again) is
<span class="math-container">$$(12165.6, 6122.12). $$</span>
Incidentally, if you're "eyeballing it", you should be aware that the aspect ratio of your graph will greatly influence where you think the corner is. I recommend forcing an aspect ratio of <span class="math-container">$1,$</span> before you say where you think the corner is.</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Pietro Majer | 6,101 | <p>Countably many little dwarfs are going to their everyday work to the mine. They are marching and singing in a well-ordered line (by natural numbers), so that number 1 watches the backs of all the other ones, and, in general, number <em>n</em> watches the backs of all the others from <em>n+1</em> on. Suddenly, an evil wizard appears on the top of a small hill, and magically puts a name on the back of each dwarf. Any name may be used, even more than once: existing ones, old-fashioned ones, or just weird sounds sprang out of his sick imagination, included grunts, sneezes and any snort-like name (you may enjoy providing your listerners with examples if they ask for). Then, he claims that, at his signal, everybody has to guess his own name, and say it loudly, all together. Whoever fails, will disappear immediately. Poor dwarfs are not new to these bully spells, and do have a strategy, that allows all but finitely many of them to survive. How do they do? To formalize, we may think the evil wizard has attached a real number to each dwarf's back. </p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| BlueRaja | 2,883 | <p><em>Cryptography riddle - that's a branch of mathematics, right? :)</em></p>
<p>Jan and Maria have fallen in love (via the internet) and Jan wishes to mail her a ring. Unfortunately, they live in the country of Kleptopia, where anything sent through the mail will be stolen unless it is enclosed in a padlocked box. Jan and Maria each have plenty of padlocks, but none to which the other has a key. How can Jan get the ring safely into Maria’s hands?</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| BlueRaja | 2,883 | <p>Most of us know that, being deterministic, computers cannot generate true <a href="http://en.wikipedia.org/wiki/Pseudorandom_number_generator">random numbers</a>.</p>
<p>However, let's say you have a box which generates truly random binary numbers, but is biased: it's more likely to generate either a <code>1</code> or a <code>0</code>, but you don't know the exact probabilities, or even which is more likely <em>(both probabilities are > 0 and sum to 1, obviously)</em></p>
<p>Can you use this box to create a unbiased random generator of binary numbers?</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Richard Stanley | 2,807 | <ol>
<li><p>Alice shuffles an ordinary deck of cards and turns the cards face
up one at a time while Bob watches. At any point in this process
before the last card is turned up, Bob can guess that the next
card is red. Does Bob have a strategy that gives him a probability
of success greater that .5?</p></li>
<li><p>Let $x_1, x_2, \dots, x_n$ be $n$ points (in that order) on the
circumference of a circle. Dana starts at the point $x_1$ and
walks to one of the two neighboring points with probability $1/2$
for each. Dana continues to walk in this way, always moving
from the present point to one of the two neighboring points with
probability $1/2$ for each. Find the probability $p_i$ that the
point $x_i$ is the last of the $n$ points to be visited for the
first time. In other words, find the probability that when $x_i$ is
visited for the first time, all the other points will have already
been visited. For instance, $p_1=0$ (when $n>1$), since $x_1$ is
the <em>first</em> of the $n$ points to be visited. </p></li>
<li><p>Let $\pi$ be a random permutation of $1,2,\dots,n$ (from the
uniform distribution). What is the probability that 1 and 2 are in
the same cycle of $\pi$?</p></li>
<li><p>Choose $n$ points at random (uniformly and independently) on the
circumference of a circle. Find the probability $p_n$ that all the
points lie on a semicircle. (For instance, $p_1 = p_2 = 1$.) More
generally, fix $\theta<2\pi$ and find the probability that the $n$
points lie on an arc subtending an angle $\theta$ .</p></li>
</ol>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Ian Agol | 1,345 | <p>Suppose 100 ants are placed randomly (with random orientations) along a yard stick. Each ant walks at a pace of an inch a minute. Each time two ants meet, they instantaneously reverse direction, and if an ant meets the end of the yardstick, it instantaneously reverses direction.<br>
Do the ants ever return to their starting positions? At what time? (A yardstick is 36 inches long.)</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Andrej Bauer | 1,176 | <p>It is very important that you tell these two puzzles in the correct order, i.e., first the first puzzle and then the second one. The first puzzle is very easy but messes with people's minds in just the right way. In my experience some mathematicians are driven crazy by the second puzzle.</p>
<p><strong>Puzzle 1:</strong>
Grandma made a cake whose base was a square of size 30 by 30 cm and the height was 10 cm. She wanted to divide the cake fairly among her 9 grandchildren. How should she cut the cake?</p>
<p><strong>Puzzle 2:</strong>
Grandma made a cake whose base was a square of size 30 by 30 cm and the height was 10 cm. She put chocolate icing on top of the cake and on the sides, but not on the bottom. She wanted to divide the cake fairly among her 9 grandchildren so that each child would get an equal amount of the cake <em>and</em> the icing. How should she cut the cake?</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| villemoes | 4,747 | <p>A magician places $N = 64$ coins in a row on a table, then leaves the room. A person from the audience is then asked to flip each coin however he likes [so there are $2^{64}$ possible states]. He is also asked to mention a number between 1 and $N$. After this, the magician's assistant flips exactly one coin. The magician reenters the room, looks at the coins and "guesses" the number chosen by the audience.</p>
<p>What is their strategy? For which values of $N$ can the trick be performed?</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| doetoe | 8,216 | <p>You and your adversary have a sufficiently large bag of identical coins, and are seated on opposite sides of a rectangular table. You take turns placing coins on the table. The first one that cannot put a coin on the table without overlapping any other coin loses. What is your strategy to always win if you're allowed to start?</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Steven Landsburg | 10,503 | <p>Let $I$ be the set of irrational numbers, with the usual topology. Is $I$ homeomorphic to $I\times I$?</p>
<p>Edited to add: In fact, it's an even better puzzle if you replace $I$ with $Q$, the rational numbers.</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Jim Ward | 12,534 | <p>There was a puzzle in the Journal of Recreational Mathematics. I apologize if I am telling it incorrectly. A wire is stretched between two telephone poles. A flock of crows lands simultaneously on the wire. When they land, each crow looks at his nearest neighbor. What percentage of the flock are looking at a crow that is looking back at it?</p>
|
4,159,227 | <p>Given the problem of finding the Laplace transform of the function <span class="math-container">$$f(t)=t^ne^{at}$$</span> with <span class="math-container">$n\in\mathbb{N}$</span> and <span class="math-container">$a\in\mathbb{R}$</span>, I realize it can be shown the transform is <span class="math-container">$$\frac{n!}{(s-a)^{n+1}}$$</span> by more than one direct method. However, I'd like to show this using strong induction. I began the problem by showing that
<span class="math-container">\begin{align*}
\mathcal{L}\{te^{at}\}=\int_{0}^{\infty}te^{-(s-a)t}\,dt&=\frac{1}{(s-a)^2},\\
\mathcal{L}\{t^2e^{at}\}=\int_{0}^{\infty}t^2e^{-(s-a)t}\,dt&=\frac{2}{(s-a)^3},\text{ and}\\
\mathcal{L}\{t^3e^{at}\}=\int_{0}^{\infty}t^3e^{-(s-a)t}\,dt&=\frac{6}{(s-a)^4}.
\end{align*}</span>
I originally did this so I could personally see the pattern. Then, I began the proof as follows.</p>
<p><span class="math-container">$\textbf{Proof:}$</span> Let <span class="math-container">$P(n)$</span> be the statement
<span class="math-container">$$P(n): \mathcal{L}\{t^ne^{at}\}=\frac{n!}{(s-a)^{n+1}},$$</span>
for all <span class="math-container">$n\in\mathbb{N}.$</span> We have already shown <span class="math-container">$P(1), P(2), P(3)$</span> are true, so the base case has been proven.</p>
<p><em>Inductive Step:</em> Assume <span class="math-container">$P(k)$</span> is true for all <span class="math-container">$k$</span>. We must then show that <span class="math-container">$P(k)\implies P(k+1).$</span>
We see
<span class="math-container">\begin{align*}
\mathcal{L}\{t^{k+1}e^{at}\}&=\frac{(k+1)!}{(s-a)^{k+2}}\\
&=\frac{k+1}{s-a}\frac{k!}{(s-a)^{k+1}}\\
&=\frac{k+1}{s-a}\mathcal{L}\{t^{k}e^{at}\} \quad\text{(by the inductive step)}
\end{align*}</span>
At this point, I'm not really sure where to go. I'm pretty awful at proofs, so I assume I'm probably not going in the right direction after the inductive hypothesis.</p>
| Gaurang | 879,859 | <p>My attempt :</p>
<p><span class="math-container">$$ \mathcal{L}\{t^ne^{at}\}= \frac{n!}{(s-a)^{n+1}}$$</span></p>
<p>For <span class="math-container">$P(k):$</span></p>
<p><span class="math-container">$$ \tag{1}\mathcal{L}\{t^ke^{at}\}= \frac{k!}{(s-a)^{k+1}}$$</span></p>
<p>Need to prove for <span class="math-container">$P(k+1) :$</span></p>
<p><span class="math-container">$$\tag{2}\mathcal{L}\{t^{k+1}e^{at}\}= \frac{(k+1)!}{(s-a)^{k+2}}$$</span></p>
<p>Step 1 :</p>
<p><span class="math-container">$$ \mathcal{L}\{t^{k+1}e^{at}\}=\int^{\infty}_{0} \left (t^{k+1}e^{at} \right)e^{-st}dt $$</span></p>
<p><span class="math-container">$$\mathcal{L}\{t^{k+1}e^{at}\}=\int^{\infty}_{0} \left (t^{k+1}e^{-t(s-a)} \right)dt $$</span></p>
<p>Using integration by part :</p>
<p><span class="math-container">$$ \int udv= uv -\int v du$$</span></p>
<p><span class="math-container">$u = t^{k+1}$</span></p>
<p><span class="math-container">$ \frac{du}{dt}=(k+1)~t^k$</span></p>
<p><span class="math-container">$ \frac{dv}{dt}= e^{-t(s-a)}$</span></p>
<p><span class="math-container">$ v = - \frac{1}{s-a}~e^{-t(s-a)}$</span></p>
<p>So :</p>
<p><strong>Note :</strong></p>
<p><span class="math-container">$$\mathcal{L}\{t^ke^{at}\}=\int^{\infty}_{0} t^{k} e^{-t(s-a)}dt$$</span></p>
<p><span class="math-container">$$\mathcal{L}\{t^{k+1}e^{at}\}= \lim_{N\to\infty} \left [- \frac{t^{k+1}}{s-a}e^{-t(s-a)} \right]^{N}_{0} + \frac{k+1}{s-a} \int^{\infty}_{0}t^{k}e^{-t(s-a)}dt$$</span></p>
<p><span class="math-container">$$\mathcal{L}\{t^{k+1}e^{at}\}=0+0+\frac{k+1}{s-a}~ \mathcal{L}\{t^ke^{at}\} $$</span></p>
<p><span class="math-container">$$\mathcal{L}\{t^{k+1}e^{at}\}= \frac{k+1}{s-a} \frac{k!}{(s-a)^{k+1}}$$</span></p>
<p>Final step, proving (2):</p>
<p><span class="math-container">$$\boxed{\mathcal{L}\{t^{k+1}e^{at}\}= \frac{(k+1)!}{(s-a)^{k+2}}}$$</span></p>
|
1,194,440 | <p>How do I prove if the following formulas are consistent?</p>
<p>∀x$\neg$S(x,x)</p>
<p>∃x P(x)</p>
<p>∀x∃y S(x,y)</p>
<p>∀x(P(x)$\to$∃y S(y,x))</p>
<p>I think I proved part of it...</p>
<p>There is at least one value for x such that P(x) is <em>True</em></p>
<p>This means that for ∀x(P(x)$\to$∃y S(y,x)) ... the P(x) part is <em>False</em> because we do not know if it is <em>True</em> for all x.</p>
<p>I am not even sure if my approach is correct, can someone please explain to me how to prove if they are consistent? Unfortunately my teacher does not provide good resources for me to figure this out on my own, Thanks.</p>
| Daniel H | 224,434 | <p>The easiest way to show that it is consistent is to explicitly construct a model for this theory. (Then we're done: recall that $T$ is consistent $\leftrightarrow$ $T$ has a model)</p>
<p>Due to axioms 1 and 3, we need to have at least 2 elements: the simplest model is then $A=\{0,1\}$, and make this into an L-Structure by $P_A=\{0\}$, $S_A=\{(0,1),(1,0)\}$. It's then pretty easy to check that this defines a model, and we are done: T is consistent. $\square$</p>
|
1,582,544 | <p>$X_1,\dots,X_n$ and $Y_1,\dots,Y_n$ are random variables that take values in $\{0,1\}$. Their distribution is unknown, each variable may have a different distribution and they might be dependent. The only known facts are:</p>
<ul>
<li>$\sum_{i=1}^n X_i = \sum_{i=1}^n Y_i$</li>
<li>For all $i$, $\Pr[X_i=0] \geq \alpha$</li>
<li>For all $i$, $\Pr[Y_i=1] \geq \alpha$</li>
</ul>
<p><strong>I would like to prove (or disprove) that $\alpha \leq 1/2$.</strong></p>
<p>The claim is obviously true for $n=1$, since in this case there are only two options:</p>
<p>A. $X_1 = Y_1 = 0$</p>
<p>B. $X_1 = Y_1 = 1$</p>
<p>Each of these options must happen with probability at least $\alpha$, but the total probability of these options is at most 1, so $\alpha$ must be at most $1/2$.</p>
<p>Is the claim also true for $n>1$?</p>
| JimmyK4542 | 155,509 | <p>Thanks to linearity of expectation, this is indeed true. Taking the expectation of both sides yields:</p>
<p>$$\mathbb{E}\left(\sum_{i = 1}^{n}X_i\right) = \mathbb{E}\left(\sum_{i = 1}^{n}Y_i\right)$$</p>
<p>$$\sum_{i = 1}^{n}\mathbb{E}(X_i) = \sum_{i = 1}^{n}\mathbb{E}(Y_i)$$</p>
<p>$$\sum_{i = 1}^{n}\left(1-\Pr[X_i = 0]\right) = \sum_{i = 1}^{n}\Pr[Y_i = 1]$$</p>
<p>Then, since $\Pr[X_i = 0] \ge \alpha$ and $\Pr[Y_i = 1] \ge \alpha$, we have: $$n-n\alpha \ge \sum_{i = 1}^{n}\left(1-\Pr[X_i = 0]\right) = \sum_{i = 1}^{n}\Pr[Y_i = 1] \ge n\alpha$$</p>
<p>Since $n-n\alpha \ge n\alpha$, we get $\alpha \le \dfrac{1}{2}$, as desired.</p>
|
4,245,205 | <p>Suppose <span class="math-container">$X$</span> is a CW-complex of dimension <span class="math-container">$n$</span>. If <span class="math-container">$e_i$</span> is an <span class="math-container">$n$</span>-cell then is <span class="math-container">$H_n(X) \to H_n(X, X \setminus e_i)$</span> the zero map? If not, then is <span class="math-container">$H_n(X, X\setminus e_i) \to H_n(X, X \setminus \{x\})$</span> the zero map where <span class="math-container">$x$</span> is a point in the interior of <span class="math-container">$e_i$</span>?</p>
<p>I cannot prove either, I suspect the proof would boil down to some fact about the homology of CW complexes which I do not know.</p>
<p>For what its worth, these statements are relevant for a proof I am reading where <span class="math-container">$X$</span> is a compact manifold of dimension <span class="math-container">$n$</span> admitting a CW structure and homology with <span class="math-container">$\mathbb{Z}_2$</span> coefficients are taken.</p>
| kamills | 497,007 | <p>I have an answer that doesn't require you to know much about the homology of CW complexes but requires a bit more work. We're going to look at the long exact sequences for the pairs you mention.</p>
<p>First, we have a simple example to use for your first question that might illuminate the general case. Let <span class="math-container">$X = S^2$</span> with CW structure given by one <span class="math-container">$0$</span>-cell, one <span class="math-container">$1$</span>-cell (with constant attaching map), and two <span class="math-container">$2$</span>-cells <span class="math-container">$e_1$</span> and <span class="math-container">$e_2$</span> that give each of the hemispheres. Then <span class="math-container">$X \smallsetminus e_1$</span> is a (open) hemisphere (i.e., the interior of <span class="math-container">$e_2$</span>), and <span class="math-container">$H_2(X, X\smallsetminus e_1)$</span> by excision is isomorphic to <span class="math-container">$H_2(e_2/\partial e_2)$</span> which is free of rank <span class="math-container">$1$</span>. This means that the long exact sequence for the pair <span class="math-container">$(X, X \smallsetminus e_1)$</span> is</p>
<p><span class="math-container">$\cdots \to H_2(X \smallsetminus e_1) \to H_2(X) \to H_2(X, X \smallsetminus e_1) \to H_1(X \smallsetminus e_1) \to \cdots$</span></p>
<p>Now, <span class="math-container">$X \smallsetminus e_1$</span> is contractible, so exactness tells us that the map <span class="math-container">$H_2(X) \to H_2(X, X\smallsetminus e_1)$</span> is an isomorphism of free rank-<span class="math-container">$1$</span> abelian groups. So, it's not the zero map.</p>
<p>What about your second question? A strategy for figuring this one out for general <span class="math-container">$X$</span> can be to consider the long exact sequences for <span class="math-container">$(X, X \smallsetminus e_i)$</span> and <span class="math-container">$(X, X \smallsetminus *)$</span> (I'll be using <span class="math-container">$*$</span> to denote <span class="math-container">$\{ x\}$</span>; it's a little easier to type.), and use naturality of the sequences with respect to the map of pairs <span class="math-container">$f: (X, X\smallsetminus e_i) \to (X, X\smallsetminus *)$</span> induced by the identity, and perhaps naturality will tell us how <span class="math-container">$f_*$</span> behaves.</p>
<p>In this case, we have two long exact sequences connected by <span class="math-container">$f_*$</span>, the relevant portion of which is</p>
<p><span class="math-container">$\require{AMScd}
\begin{CD}
H_n(X \smallsetminus e_i) @>>> H_n(X) @>>> H_n(X, X \smallsetminus e_i) @>>> H_{n-1}(X \smallsetminus e_i) \\
@Vf_*VV & @V1VV @V{f_*}VV @VVf_*V & \\
H_n(X \smallsetminus *) @>>> H_n(X) @>>> H_n(X, X \smallsetminus *) @>>> H_{n-1}(X \smallsetminus *) \end{CD}$</span></p>
<p>Now, consider the situation where <span class="math-container">$X$</span> has a single <span class="math-container">$n$</span>-cell; this means that <span class="math-container">$X \smallsetminus e_i$</span> and <span class="math-container">$X \smallsetminus *$</span> deformation retract to the <span class="math-container">$(n-1)$</span>-skeleton of <span class="math-container">$X$</span>, so the two groups on the left vanish since <span class="math-container">$H_n$</span> of an <span class="math-container">$(n-1)$</span>-dimensional CW complex is always zero. Exactness tells us that the maps <span class="math-container">$H_n(X) \to H_n(X, X\smallsetminus e_i)$</span> and <span class="math-container">$H_n(X) \to H_n(X, X\smallsetminus *)$</span> are at least injective, and commutativity of the center square implies that <span class="math-container">$f_*$</span> cannot be the zero map. If we use the example <span class="math-container">$X = S^n$</span> with usual CW structure (a single zero-cell and a single <span class="math-container">$n$</span>-cell), all the groups on the left and right vanish and we end up with a commutative square where all maps are immediately seen to be isomorphisms besides <span class="math-container">$f_*$</span>, but from that it follows that <span class="math-container">$f_*$</span> is an isomorphism.</p>
<p>This is perhaps a bit more complicated of an answer than it needs to be (see the comment that explains why <span class="math-container">$f_*$</span> is always an isomorphism), but hopefully this approach is also helpful to see. In the situation where you have a compact manifold and are using <span class="math-container">$\mathbb{Z}_2$</span> coefficients, it is always possible to have a CW structure where you have a single <span class="math-container">$n$</span>-cell, so we're always in the above situation; if you want to use integer coefficients you just need to add the hypothesis that <span class="math-container">$X$</span> is oriented.</p>
|
3,530,285 | <blockquote>
<p>Let <span class="math-container">$\text{char}(\mathbb{K}) = 0$</span>. It then follows that <span class="math-container">$AB-BA \ne 1 \, (A, B \in \Bbb K^{n \times n})$</span>.</p>
</blockquote>
<p>I first showed that <span class="math-container">$\text{trace}(AB) = \text{trace}(BA)$</span> for every <span class="math-container">$A, B \in \Bbb K^{n \times n}$</span>:</p>
<p>Let <span class="math-container">$A=(a_{ij}), B=(b_{ij}) \in \Bbb K^{n \times n}$</span>, so <span class="math-container">$AB=(c_{ij}) \in \Bbb K^{n \times n}$</span> where <span class="math-container">$c_{ij}= \sum_{k=1}^na_{ik}b_{kj}$</span>. One now gets: </p>
<p><span class="math-container">$$\text{trace}(AB) = \sum_{i=1}^n \sum_{k=1}^n a_{ik}b_{ki} = \sum_{k=1}^n \sum_{i=1}^n a_{ik}b_{ki} = \sum_{k=1}^n \sum_{i=1}^n b_{ki}a_{ik} = \text{trace}(BA)\tag{*}.$$</span></p>
<p>Now, assume <span class="math-container">$AB-BA=1$</span>. Consider</p>
<p><span class="math-container">$$AB-BA=1 \Leftrightarrow \text{trace}(AB)-\text{trace}(BA)=\text{trace}(1) \xrightarrow{(*)} \\ 0 = n,$$</span></p>
<p>which is a contradiction, so the assumption <span class="math-container">$AB-BA=1$</span> was wrong.</p>
<p>My question is: Where exactly do you need <span class="math-container">$\text{char}(\Bbb K)=0$</span> in this proof?</p>
<p>Thanks in advance!</p>
| Henno Brandsma | 4,280 | <p><span class="math-container">$0 \in \Bbb Q$</span> but <span class="math-container">$0\notin S$</span> while it <strong>is</strong> a limit point of <span class="math-container">$S$</span>. So <span class="math-container">$S$</span> does not contain all its limit points so is not closed. </p>
|
78,946 | <p>I want to find min of the function
$$\frac{1}{\sqrt{2
x^2+\left(3+\sqrt{3}\right)
x+3}}+\frac{1}{\sqrt{2
x^2+\left(3-\sqrt{3}\right)
x+3}}+\sqrt{\frac{1}{3}
\left(2 x^2+2 x+1\right)}.$$
I know, the exact value minimum is $\sqrt{3}$ at $x = 0$. With <em>Mathematica</em>, I tried </p>
<pre><code>A = 1/Sqrt[2 x^2 + (3 + Sqrt[3]) x + 3] +
1/Sqrt[2 x^2 + (3 - Sqrt[3]) x + 3] + Sqrt[(2 x^2 + 2 x + 1)/3]
NMinimize[A, {x}]
</code></pre>
<p>And I got </p>
<blockquote>
<p>{1.73205, {x -> -2.57345*10^-16}}</p>
</blockquote>
<p>When I tried </p>
<pre><code>A = 1/Sqrt[2 x^2 + (3 + Sqrt[3]) x + 3] +
1/Sqrt[2 x^2 + (3 - Sqrt[3]) x + 3] + Sqrt[(2 x^2 + 2 x + 1)/3]
Minimize[A, {x}]
</code></pre>
<p>my computer ran about 20 minutes and I did not got the result. How can I get the exact value minimum of the given function?</p>
| Coolwater | 9,754 | <p>You could use <code>FindInstance</code> to find some of the exact stationary points, when mathematica can't solve the equation:</p>
<pre><code>FindInstance[Evaluate[D[A, x]] == 0, x]
(*{{x -> 0}}*)
</code></pre>
|
1,930,933 | <blockquote>
<p>Does there exist an $n \in \mathbb{N}$ greater than $1$ such that $\sqrt[n]{n!}$ is an integer?</p>
</blockquote>
<p>The expression seems to be increasing, so I was wondering if it is ever an integer. How could we prove that or what is the smallest value where it is an integer?</p>
| P Vanchinathan | 28,915 | <p>EDIT: The comment below by bof points out the flaw in my argument.
I am not deleting this answer. This proves a much weaker statement version that instead of "never happening" this shows it cannot happen for two consecutive integers. </p>
<hr>
<p>Bertrand's postulate is the shortest way to prove this as has been suggested in other answers/comments. Here is an alternative attempt. </p>
<p>Suppose $n!=x^n,$ and $(n+1)!=y^{n+1}$ for integers $x,y$. Then we get by dividing these two equations $$ n+1 =\bigg(\frac yx\bigg) ^n y $$ As LHS is an integer so is the RHS. The only way RHS can be an integer is $y/x$ be an integer (and at least 2). That means $n+1$ is bigger than $2^{n+1}$ (because y is at least 2). But this is absurd, giving the required contradiction.</p>
|
786,301 | <p>Is there a systematic way to express the sum of two complex numbers of different magnitude (given in the exponential form), i.e find its magnitude and its argument expressed in terms of those of the initial numbers?</p>
| John Hughes | 114,036 | <p>Convert to rectangular form, add. convert back. </p>
<p>$$\begin{eqnarray}
[r, \theta] + [s, \phi]& \to&
(r \cos \theta, r \sin \theta) + (s \cos \phi, s \sin \phi) \to
(r \cos \theta + s \cos \phi, r \sin \theta + s \sin \phi)\\
&\to&
\bigg[\sqrt{(r \cos \theta + s \cos \phi)^2 + ( r \sin \theta + s \sin \phi)^2},\\ &&\text{atan2}(r \sin \theta + s \sin \phi, r \cos \theta + s \cos \phi)\bigg]
\end{eqnarray}$$</p>
<p>where I'm using brackets for polar form, and parens for $x + iy$ form. "atan2" returns an arg between $-\pi$ and $\pi$, so you may want to add $2\pi$ if the returned arg is negative. </p>
<p>The expression under the square-root can be expanded and simplified somewhat, but I find it's probably simpler to see what's going on in this form. </p>
|
72,651 | <p>$G$ is a group and $H$ is a subgroup of $G$ such that $\forall a, b$ in $G, ab\in H\implies ba\in H$. Show that $H$ is normal in $G$</p>
| Sugata Adhya | 36,242 | <p>Choose $x\in G$ and $h\in H.$ Let $a=x^{-1}.$</p>
<p>Suppose $ha=b.$ </p>
<p>Then $ba^{-1}=h$</p>
<p>$\implies a^{-1}b=h_1$ for some $h_1\in H$ (by the given condition)</p>
<p>$\implies b=ah_1$</p>
<p>$\implies ha=ah_1$</p>
<p>$\implies a^{-1}ha=h_1\in H$</p>
<p>$\implies xhx^{-1}\in H.$</p>
<p>Consequently $H\triangleleft G.$</p>
|
3,831,702 | <p>One can prove that for <span class="math-container">$x\in \mathbb{R}$</span>, the sequence
<span class="math-container">$$
u_0=x\text{ and } \forall n\in \mathbb{N},\qquad u_{n+1}=\frac{e^{u_n}}{n+1}
$$</span>
converges to <span class="math-container">$0$</span> if <span class="math-container">$x \in ]-\infty,\delta[$</span> and diverges to <span class="math-container">$+\infty$</span> if <span class="math-container">$x\in ]\delta,+\infty[$</span> for a fixed <span class="math-container">$\delta$</span>. I'm trying to find more information on the value <span class="math-container">$\delta$</span> (inequalities or expression) and on the specific sequence
<span class="math-container">$$
u_0=\delta \text{ and } \forall n\in \mathbb{N},\qquad u_{n+1}=\frac{e^{u_n}}{n+1}
$$</span>
Any reference or help are welcome. The only thing I can prove at the moment is <span class="math-container">$\ln \ln 2 \le \delta \le 1$</span>.</p>
| Community | -1 | <p>My numerics agree with Simply's result. Here's a little Maple code:</p>
<p>Define <span class="math-container">$u_n(x)$</span>:</p>
<p><a href="https://i.stack.imgur.com/BeYzO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BeYzO.png" alt="enter image description here" /></a></p>
<p>At the <span class="math-container">$n$</span>-th iteration the limit (if it exists) is given by solving: <span class="math-container">$l=\exp(l)/n$</span>:</p>
<p><a href="https://i.stack.imgur.com/tSi34.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tSi34.png" alt="enter image description here" /></a></p>
<p>so the fixed point at this level is given (as a function of <span class="math-container">$n$</span>) by:</p>
<p><a href="https://i.stack.imgur.com/1YJMq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1YJMq.png" alt="enter image description here" /></a></p>
<p>That is:</p>
<p><a href="https://i.stack.imgur.com/XXGsZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XXGsZ.png" alt="enter image description here" /></a></p>
<p>the derivative <span class="math-container">$du_n(x)/dx$</span> (as a function of <span class="math-container">$n$</span>):</p>
<p><a href="https://i.stack.imgur.com/4D0Cr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4D0Cr.png" alt="enter image description here" /></a></p>
<p>And now just check the iterates <span class="math-container">$|u_n'(x_f)|$</span>:</p>
<p><a href="https://i.stack.imgur.com/ZjBMa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZjBMa.png" alt="enter image description here" /></a></p>
<pre><code> "u'@(x_f):", 1.374557011
"u'@(x_f):", 5.413650945
"u'@(x_f):", 48.78006459
"u'@(x_f):", 29.62255784
"u'@(x_f):", 14.17327019
"u'@(x_f):", 2.493811293
"u'@(x_f):", 0.2768959061
"u'@(x_f):", 0.02871591627
"u'@(x_f):", 0.002877744731
"u'@(x_f):", 0.0002733561570
</code></pre>
<p>Although <span class="math-container">$|u_n'(x_f)|$</span> oscillates some in the beginning, the above is a strong indication that the fixed points are becoming super-attractors. So now conversely simply iterate the numerical solutions to <span class="math-container">$u_n'(x)=1$</span> to pick a good bound for the initial <span class="math-container">$x$</span>:</p>
<p><a href="https://i.stack.imgur.com/WHuet.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WHuet.png" alt="enter image description here" /></a></p>
<p>With 20 iterates, I get:</p>
<pre><code> 0.
-0.1594583215
-0.1858157472
-0.1022224792
0.03068160157
0.1493286247
0.2282136125
0.2723541387
0.2946085552
0.3050959095
0.3098066970
0.3118452094
0.3127005207
0.3130500940
0.3131896678
0.3132442219
0.3132651501
0.3132730435
0.3132759718
0.3132770420
</code></pre>
<p>which agrees to 7 dec with Simply's answer.</p>
<p><strong>Note:</strong> That's the code for estimating <span class="math-container">$\delta$</span>. If you have symbolics for the derivative of functions, it's much easier, but if you don't, just code your function for the derivative of <span class="math-container">$u_n$</span> up to level <span class="math-container">$n$</span> (for accuracy to level <span class="math-container">$n$</span>) as <span class="math-container">$g(x)$</span> and numerically solve the equation <span class="math-container">$g(x)=1$</span>. For example, for accuracy to level 2, your <span class="math-container">$u_2(x)=1/2\exp(\exp(x))$</span>, so code an estimate for the derivative as: <span class="math-container">$g(x)=\frac{u_2(x+h)-u_2(x)}{h}$</span>, for small <span class="math-container">$h$</span> (say <span class="math-container">$\sim 0.01$</span> or similar). Then numerically solve the equation: <span class="math-container">$g(x)=1$</span>. The estimate of course depends on how high your <span class="math-container">$n$</span> is and how small your <span class="math-container">$h$</span> is. The higher the <span class="math-container">$n$</span> and the smaller the <span class="math-container">$h$</span>, will improve the numerical solutions found when solving <span class="math-container">$g(x)=1$</span>. The Maple code above, does all this automatically. After you get an estimate for <span class="math-container">$\delta$</span>, you can verify that <span class="math-container">$u_n(\delta)$</span> converges and <span class="math-container">$u_n(\delta+dx)$</span> diverges for most <span class="math-container">$dx\ge\epsilon\gt 0$</span>.</p>
<p><strong>Addendum to Note #2</strong>: If you don't have a symbolic calculator, the above can be a bit of a nuisance to code in low level . The reason being that the equation <span class="math-container">$u_n'(x)=1$</span> needs to be solved numerically. The usual way to solve this is to look for roots of <span class="math-container">$g(n,x)=u_n'(x)-1$</span> by implementing some sort of numerical root finder, such as <a href="https://en.wikipedia.org/wiki/Newton%27s_method" rel="nofollow noreferrer">Newton's method</a>. The problem is that Newton's method causes overshoots with this <span class="math-container">$u_n$</span> and as a result it does not converge. Halley's method will overshoot, too (the derivative <span class="math-container">$u_n'(x)$</span> may become unbounded). So, a good method to solving <span class="math-container">$u_n'(x)=1$</span>, numerically would be the bisection method, like Simply uses. Here's then some low level code which you can translate directly if you don't feel like translating his code, to obtain some estimates for <span class="math-container">$
\delta$</span>. Assuming you have defined your <span class="math-container">$u_n(x)$</span> as above in this post, code an approximation of its derivative as:</p>
<p><a href="https://i.stack.imgur.com/MfW1U.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MfW1U.png" alt="enter image description here" /></a></p>
<p>Now pick safe upper and lower bounds for your <span class="math-container">$\delta$</span>, to use for bisecting:</p>
<p>delta:=0;
delta_b:=0.32;
N:=100;
eps:=0.01;</p>
<p>And then use bisection on your range <span class="math-container">$[0,\delta_b]$</span>, to obtain some approximations. Whenever the calculated approximation for <span class="math-container">$\delta$</span> exceeds your test (meaning : <span class="math-container">$u_n'(x)\ge 1+\epsilon$</span>), you replace <span class="math-container">$\delta_b$</span> with this <span class="math-container">$\delta$</span>, and continue with bisection on the new interval <span class="math-container">$[0,\delta_b']$</span>. For example:</p>
<p><a href="https://i.stack.imgur.com/n866p.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/n866p.png" alt="enter image description here" /></a></p>
<p>After the loop runs, ask for <span class="math-container">$\delta_b$</span>:</p>
<p>delta_b;</p>
<p>With <span class="math-container">$N=1000$</span>, I get:</p>
<p>delta_b;
0.3132776394</p>
<p>which is correct to 9 decimals. Running this with higher decimal precision (Maple uses 10 by default), will bring you of course in more digits accuracy.</p>
<hr />
<p><strong>Addendum#1</strong>:</p>
<p>Adding a little code to watch the dynamics of this sequence on the complex plane. First, modify <span class="math-container">$u_n$</span> to accept a complex argument:</p>
<p><a href="https://i.stack.imgur.com/tmiaJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tmiaJ.png" alt="enter image description here" /></a></p>
<p>Some code for constructing the Julia set of the corresponding exponential sequence:</p>
<p>W:=LambertW;</p>
<p>eps:=0.01;</p>
<p><a href="https://i.stack.imgur.com/irsvs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/irsvs.png" alt="enter image description here" /></a></p>
<p>Now we can watch the dynamics of the <span class="math-container">$u_n(z)$</span> sequence dynamically on the complex plane:</p>
<p><a href="https://i.stack.imgur.com/6N9re.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6N9re.png" alt="enter image description here" /></a></p>
<p>The above gives:</p>
<p><a href="https://i.stack.imgur.com/heeW6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/heeW6.png" alt="enter image description here" /></a></p>
<p>And magnified near the origin (without escape contours):</p>
<p><a href="https://i.stack.imgur.com/nlqdD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nlqdD.png" alt="enter image description here" /></a></p>
<p>So your <span class="math-container">$\delta_{[0.313277...,0]}$</span> is the tip of the Julia Cantor Bouquet produced by <span class="math-container">$u_n(z)$</span>. There are infinately many other such <span class="math-container">$\delta_{z}$</span>, on the plane: All the yellow bouquet tip points on the sub-bouquets of the main bouquets, on the sub-sub-bouquets and so on.</p>
<p>This Julia set is similar to the Julia set for the plain exponential iteration of <span class="math-container">$E_{\lambda}(z)=\lambda e^z$</span>, with <span class="math-container">$\lambda<1/e$</span> (more <a href="https://math.stackexchange.com/questions/3825506/properties-of-the-iterated-exponential-sequences-z-n-ez-n-1/3825570#3825570">here</a>). The only difference is that your sequence iterates <span class="math-container">$E_{\lambda_n}(z)$</span> for <span class="math-container">$\lambda_n=1/n$</span>. All iterates except the first 2, have <span class="math-container">$\lambda_n<1/e$</span>, so the Julia set is a plain Cantor Bouquet plus a Fatou domain (all colored domains in lighter shades than purple). Iterating any point from these Fatou domains will eventually cause the sequence to converge to the only convergent, the point: <span class="math-container">$z_0\sim 0.02041244406...$</span>.</p>
<p>Iterating a point picked from the bouquet itself - except an endpoint, will shoot the orbit to complex infinity. (That's why if you pick <span class="math-container">$x>0.313277...$</span>, the sequence diverges: Because such a point lies on the main "hair" of the Cantor Bouquet.) Hence, your <span class="math-container">$\delta$</span> is the tip of the main Bouquet of the Julia set.</p>
<p>(If your program allows for complex arguments, you can test other such values, for example: <span class="math-container">$\delta_{[3.1234256..,-1]}$</span>, etc.).</p>
<p>The colors indicate speed of convergence: Blues are fastest, followed by greens and finally yellows and reds slower. But everything outside the bouquet eventually goes to <span class="math-container">$z_0$</span>, including the tips of the bouquet.</p>
<p>To generalize finding more such <span class="math-container">$\delta_z$</span>, just iterate for complex solutions to <span class="math-container">$|u_n'(z)|=1$</span>.</p>
|
220,618 | <p>The cyclic group of $\mathbb{C}- \{ 0\}$ of complex numbers under multiplication generated by $(1+i)/\sqrt{2}$</p>
<p>I just wrote that this is $\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i$ making a polar angle of $\pi/4$. I am not sure what to do next. My book say there are 8 elements. </p>
<p>Working backwards, maybe the angle divides the the four quadrant into 8 areas? I have no idea</p>
| Stefan Geschke | 16,330 | <p>Draw a sketch. Do you know how to multiply complex numbers in polar coordinates?
Add the angles, multiply the lengths. Now compute the powers of your number until you reach 1.</p>
|
2,046,957 | <p>The random variable $X$ is $N(5,2)$ and $y=2X+4$.
Find:</p>
<p>a) $\eta_y$</p>
<p>b) $\sigma_y$</p>
<p>c) $f_Y(y)$</p>
<p>My attempt:</p>
<p>I have solved a and b as follow:</p>
<p>a) $\eta_y = 2\eta_X+4 = 14$</p>
<p>b) $\sigma_y^2 = 4\sigma_{x}^2 = 16, \sigma_y = 4$</p>
<p>c) how can I solve $f_Y(y)$?</p>
| Gono | 384,471 | <p>By definition:
$$\begin{align*}f_Y(y) &= P'(Y \le y) \\ &= P'(2X + 4 \le y) \\ &= P'\left( X \le \frac{y - 4}{2}\right) \\ &= \frac{1}{2}f_X\left(\frac{y - 4}{2}\right)\end{align*}$$</p>
<p>And $f_X$ you should know…</p>
|
866,885 | <p>I try to write math notes as clearly as possible. In practice, this means using letters and notation similar to what the reader is already familiar with.</p>
<p>A real function is often $f(x)$, an angle is often $\theta$, a matrix has size $m\times n$, and $i$ is often an index. The full theoretical list is long and complicated. For example, $\pi$ is very often a constant, but sometimes it's a variable for a permutation. Capital sigma $\Sigma$ can indicate summing a series, but it can also denote a matrix, as in the singular value decomposition. So things like context matter, and a great list would have to include more than just variable names. Another choice to make is how to write an inner product, for example.</p>
<p>Does such a list exist?</p>
| Joel | 85,072 | <p>In my own experience the letters $a,b,c,d,e$ are reserved for coefficients. $f, g, h$ are functions. $i,j,k$ are indices. (Sometimes $i=\sqrt{-1}$) $l,m,n$ are also indicies (and in particular natural numbers or integers)</p>
<p>$o$ isn't often used, since it can be confused with zero. Though it can be "little"-o for asymptotics.</p>
<p>$p,q$ are usually polynomials or rational functions or primes.</p>
<p>$r, s,t$ are real numbers, variables, or coefficients.</p>
<p>$u,v,w$ are variables used for coordinate transformations of the real variables $x,y,z$. Also $z,w$ are variables in complex analysis.</p>
<p>Capital $A,B$ and $M$ are matrices. $N$ is a natural number. $R$ is a radius or bound. $C$ is a constant. $T$ is a linear operator.</p>
<p>Obviously this depends on the field and person to a large extent. I have always found it amusing how difficult it can be to choose just the right letter for something. I get stuck when I am trying to use a letter for a function after $f,g,h$ are already taken.</p>
|
1,003,020 | <p>Without recourse to Dirichlet's theorem, of course.
We're going to go over the problems in class but I'd prefer to know the answer today.</p>
<p>Let $S = \{3n+2 \in \mathbb P: n \in \mathbb N_{\ge 1}\}$</p>
<p>edit:</p>
<p>The original question is "the set of all primes of the form $3n + 2$, but I was only considering odd primes because of a reason I don't remember anymore.</p>
<p>How can I show $S$ is infinite? I start by assuming it's finite. </p>
<p>I've tried: </p>
<p>Assuming that the product $(3n_1+2)(3n_2+2)...(3n_m+2)$ is of the form "something" so I can show the product contains a prime factor not in the finite list of primes, which would be a contradiction, but came up with no useful "something".</p>
<p>I tried showing that the product would not be square-free, but couldn't show that.</p>
<p>I tried showing the product or sum was both even and odd, but couldn't show that.</p>
<p>What else should I try? Or was one of the above methods correct?</p>
| mookid | 131,738 | <p><strong>Hint:</strong>
Assume there are only $m$ such primes. Consider
$$P=(3n_1+2)(3n_2+2)...(3n_m+2) = (-1)^m \mod 3$$</p>
<p>Now if $2|m, P+2 = -1\mod 3$.</p>
<p>If $2\nmid m$ then $P=1\mod 3$.</p>
|
465,945 | <p>I just need some verification on finding the basis for column spaces and row spaces.</p>
<p>If I'm given a matrix A and asked to find a basis for the row space, is the following method correct?</p>
<p>-Reduce to row echelon form. The rows with leading 1's will be the basis vectors for the row space.</p>
<p>When looking for the basis of the column space (given some matrix A), is the following method correct?</p>
<p>-Reduce to row echelon form. The columns with leading 1's <strong>corresponding</strong> to the original matrix A will be the basis vectors for the column space. </p>
<p>When looking for bases of row space/column space, there's no need in taking a transpose of the original matrix, right? I just reduce to row echelon and use the reduced matrix to get my basis vectors for the row space, and use the original matrix to correspond my reduced form columns with leading 1's to get the basis for my column space. </p>
| elbeardmorez | 15,263 | <p>yes you're correct.</p>
<p>note that row echelon form doesn't necessarily result in 'leading 1s'. it's '<strong>reduced/canonical</strong> row echelon form' that requires that form.</p>
<p>having reduced your matrix to the set of the linearly independent rows/columns via the row transformations, you can choose either the new reduced vectors with leading pivots (1s or otherwise), or the corresponding vectors from the original matrix*. they are effectively 'the same'. i'd go with the reduced vectors however, as they make any further manipulation or plotting easier</p>
<p>*see caveat raised by <strong>user84413</strong></p>
|
1,103,478 | <p>$ r = 2\cos(\theta)$ has the graph<img src="https://i.stack.imgur.com/yvLb1.png" alt="enter image description here"></p>
<p>I want to know why the following integral to find area does not work $$\int_0^{2 \pi } \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta$$</p>
<p>whereas this one does:</p>
<p>$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2} } \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta$$</p>
<p>Why do the limits of integration have to go from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$? Doesn't going from $0$ to $2\pi$ also sweep out the full circle?</p>
| colormegone | 71,645 | <p>The families of polar curves $ \ r \ = \ a \ \cos \ n \theta \ $ and $ \ r \ = \ a \ \sin \ n\theta \ $ are called "rosettes", because they form symmetrical "flower-petal" arrangements. Those with $ \ n \ $ even trace their respective curves once in a period of $ \ 2 \pi \ $ and have $ \ 2n \ $ "petals", while those with $ \ n \ $ odd trace their curves, which have $ \ n \ $ "petals", in a cycle of length $ \ \pi \ $ and will then retrace the curve in subsequent periods of $ \ \pi \ $ . (For these latter rosettes, the "negative radius" portions of the curves "fall on top of" the "positive radius" portions.)</p>
<p>The curves $ \ r \ = \ a \ \cos \ \theta \ $ and $ \ r \ = \ a \ \sin \ \theta \ $ are certainly circular (as transforming them to Cartesian coordinates will confirm), but in fact belong to the families of rosettes. They are "one-petal" rosettes, since $ \ n \ = \ 1 \ $ . So these curves are among the ones <em>retraced</em> by running through an angular interval of length $ \ 2 \pi \ $ . (This is why the integrations for these curves are "stopped" after an interval of $ \ \pi \ $ radians.)</p>
|
2,000,556 | <p>Suppose $N>m$, denote the number of ways to be $W(N,m)$</p>
<h3>First method</h3>
<p>Take $m$ balls out of $N$, put one ball at each bucket. Then every ball of the left the $N-m$ balls can be freely put into $m$ bucket. Thus we have: $W(N,m)=m^{N-m}$.</p>
<h3>Second method</h3>
<p>When we are going to put $N$-th ball, we are facing two possibilities:</p>
<ol>
<li><p>the previous $N-1$ balls have already satisfies the condition we required, i.e. each of $m$ buckets has at least one ball. Therefore, we can put the $N$-th ball into any bucket.</p></li>
<li><p>the previous $N-1$ balls have made $m-1$ buckets satisfies the condition, we are left with one empty bucket, the $N$-th ball must be put into that empty bucket. However, that empty bucket may be any one of the $m$ buckets.</p></li>
</ol>
<p>Therefore, we have the recursion formula:</p>
<p>$$
W(N,m) = m W(N-1,m) + m W(N-1,m-1)
$$</p>
<p><strong>It is obvious that the two methods are not identical, which one has flaws?</strong> I would like to know which part of the reasoning is wrong and I would also want to hear about the case when the balls are distinct.</p>
| Hypergeometricx | 168,053 | <p>First put $1$ ball into each of $m$ buckets. </p>
<p>Then you are left with $N-m$ balls. </p>
<p>Use <a href="https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)" rel="nofollow noreferrer">stars and bars</a> method. Arrange the remaining balls in a row. Add $m-1$ dividers. Compute number of ways to arrange the balls and dividers.</p>
<p>Number of ways is given by =$$\binom {(N-m)+(m-1)}{m-1}=\binom{N-1}{m-1}$$</p>
|
66,719 | <p>I know some elementary proofs of this fact. I was wondering if there's some short slick proof of this fact using the structure of the $2$-adic integers? I'm looking for a proof of this fact that's easy to remember.</p>
| Plop | 2,660 | <p>There is indeed a $2$-adic proof.
$(1+u)^{1/2} = 1+\sum_{k \geq 1} \binom{1/2}{k} u^k$ where $\binom{1/2}{k}= \frac{1/2(1/2-1) \ldots (1/2-k+1)}{k!} = 2^{-k}(-1)^{k-1}\frac{(2k-3) (2k-5) \ldots 3}{k!}$.
We would like the sum to converge, and for this we only need the general term to go to $0$.</p>
<p>$v_2\left(\binom{1/2}{k} u^k\right)=k(v_2(u)-1)-v_2(k!)$.
Furthermore, $v_2(k!) = \sum_{l \geq 1} \left\lfloor\frac{k}{2^l} \right\rfloor < k$, so if $v_2(u)>2$ the series converge and the fact that as formal series (i.e. in $\mathbb{Q}[[u]]$) $\left(1+\sum_{k \geq 1} \binom{1/2}{k} u^k\right)^2=1+u$ tells you that you really find a square root of $1+u$ by taking the limit.</p>
<p>Of course if $v_2(u)>2$ all the terms are in $\mathbb{Z}_2$, so the limit is also in $\mathbb{Z}_2$.</p>
|
913,998 | <p>What can we do with this function, so the function will be continuous in $(0,0)$?</p>
<p>$f:\mathbb{R}^2\rightarrow\mathbb{R}:(x,y) \mapsto \frac{x^2+y^2-x^3y^3}{x^2+y^2}$</p>
<p>What I think we should do, is:</p>
<p>approximate $(0,0)$ via the line $y=x$, so substitute $y=x$ and take the limit of that function, i.e. $\lim_{x\rightarrow0}$:</p>
<p>$\lim_{x\rightarrow0} 1 - \frac{x^6}{2x^2} = \lim_{x\rightarrow0}1-\frac{x^4}{2} = 1$</p>
<p>So the new function, that is continuous in $(0,0)$ is defined by:</p>
<p>$F:\mathbb{R}^2\rightarrow\mathbb{R}:(x,y) \mapsto F(x) = \begin{cases} 1 & \mbox{if } (x,y) = (0,0) \\ f(x) & \mbox{if } (x,y) \neq (0,0) \end{cases}$</p>
<p>I'm sorry for my english, but if you understand my question, could you say if I'm right? </p>
| idm | 167,226 | <p>We can observe that
$$x^4+2x^2\geq 0$$
for all $x\in\mathbb R$ and so,
$$x^4+2x^2+2\geq 2$$
for all $x\in \mathbb R$. Moreover, for $x=0$, we have that
$$0^4+2\cdot 0^2+2=2$$
and so $2$ is the minimum.</p>
|
913,998 | <p>What can we do with this function, so the function will be continuous in $(0,0)$?</p>
<p>$f:\mathbb{R}^2\rightarrow\mathbb{R}:(x,y) \mapsto \frac{x^2+y^2-x^3y^3}{x^2+y^2}$</p>
<p>What I think we should do, is:</p>
<p>approximate $(0,0)$ via the line $y=x$, so substitute $y=x$ and take the limit of that function, i.e. $\lim_{x\rightarrow0}$:</p>
<p>$\lim_{x\rightarrow0} 1 - \frac{x^6}{2x^2} = \lim_{x\rightarrow0}1-\frac{x^4}{2} = 1$</p>
<p>So the new function, that is continuous in $(0,0)$ is defined by:</p>
<p>$F:\mathbb{R}^2\rightarrow\mathbb{R}:(x,y) \mapsto F(x) = \begin{cases} 1 & \mbox{if } (x,y) = (0,0) \\ f(x) & \mbox{if } (x,y) \neq (0,0) \end{cases}$</p>
<p>I'm sorry for my english, but if you understand my question, could you say if I'm right? </p>
| Khosrotash | 104,171 | <p>$$ x^4+x^2+2=(x^2+1)^2+1\\x^2\geq 0\\x^2+1 \geq 0+1 \\(x^2+1)^2 \geq (1)^2\\so\\(x^2+1)^2+1\geq 1+1 $$</p>
|
301,724 | <p>There's an example in my textbook about cancellation error that I'm not totally getting. It says that with a $5$ digit decimal arithmetic, $100001$ cannot be represented.</p>
<p>I think that's because when you try to represent it you get $1*10^5$, which is $100000$. However it goes on to say that when $100001$ is represented in this floating point system (when it's either chopped or rounded) it comes to $100000$.</p>
<p>If what I said above is correct, does $100001$ go to $100000$ because of the fact that it can only be represented like $1*10^5$? </p>
<p>If I'm completely off the mark, clarification would be great.</p>
| Tpofofn | 4,726 | <p>I think the point is about the precision of the number that can be stored, not so much the exponent. In other words you cannot store $1.00001*10^5$ because it cannot store five decimal places of precision. Equally it would not be able to store $1.00001*10^{15}$ or $1.00001*10^{-5}$.</p>
|
458,472 | <p>Just doing some revision for ODEs and came across this problem. Find the general solution to $$u''+4u=0.$$</p>
<p>So far I've applied the characteristic polynomial:
$$\begin{array}{r c l}
\lambda^2 +4 & = & 0 \\
\lambda^2 & = & -4 \\
\lambda & = & i\sqrt{4} \\
\lambda & = & 2i, -2i. \\
\end{array}$$</p>
<p>So the general solution should be:
$$\begin{array}{l c l}
u_H & = & Ae^{2ix}+Be^{-2ix} \\
& = & A(\cos{2x}+i\sin{2x})+B(\cos{(-2x)}+i\sin{(-2x)}) \\
& = & A\cos{2x}+iA\sin{2x}+B\cos{2x}-iB\sin{2x} \\
& = & (A+B)\cos{2x}+i(A-B)\sin{2x} \\
& = & C_1\cos{2x}+iC_2\sin{2x}. \\
\end{array}$$</p>
<p>The answers have $u=C_1\cos{2x}+C_2\sin{2x}$, and my question is "what happened to the $i$?" Does it drop out somewhere or is there an error in the answers? </p>
<p>Many thanks for a quick explanation/link to the appropriate website explaining this. :)</p>
| Dan | 79,007 | <p>$i$ is a constant, and so is included in $C_2$. So you're both right! :)</p>
|
6,340 | <p>Just to clarify, I want to show that:</p>
<p>If $f$ is entire and $\int_{\mathbb{C}} |f|^p dxdy <\infty$, then $f=0$.</p>
<p>I think I can show that this is the case for $p=2$, but I'm not sure about other values of $p$...</p>
| Mariano Suárez-Álvarez | 274 | <p>Use Hölder's inequality and Cauchy's integral formula to show the function and its derivatives all vanishes at zero.</p>
|
2,438,795 | <blockquote>
<p>Why are any two initial objects of a category equivalent?</p>
</blockquote>
<p>By definition:</p>
<p>If $A,B$ are initial objects of a category $C$, then for each $X \in \text{obj } C$, there exists a unique morphisms $f : A \rightarrow X, g : B \rightarrow X$.</p>
<p>How can these be equivalent? This seems to just indicate that each $A$ and $B$ have their own set of maps.</p>
| Mr. Chip | 52,718 | <p>Any two initial objects are <em>isomorphic</em>, even though they may not literally be equal.</p>
<p>Proof: Call them $A$ and $A'$. By assumption, there are unique morphisms $a: A \to A'$, $a': A' \to A$. Then $aa': A' \to A'$ has to be $1_{A'}$, because $A'$ is initial. Similarly $a'a = 1_A$. Thus $A \cong A'$.</p>
<p>For example, in the category of sets, all singleton sets are final objects (dual to initial); they are all isomorphic as sets, even though not all equal.</p>
|
1,423,728 | <p>The definition of a limit in <a href="http://rads.stackoverflow.com/amzn/click/0321888545" rel="nofollow noreferrer">this book</a> stated like this </p>
<p><a href="https://i.stack.imgur.com/tKjaa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tKjaa.png" alt="enter image description here"></a></p>
<p>1) why we require ƒ(x) be defined on an open interval about $x_0$ ?</p>
<p>2) does the definition mean it is impossible to talk about limit of function of this sort?</p>
<p>$f(x) = \begin{cases} undefined & x \not\in \mathbb{Q}\\ 1 & x \in \mathbb{Q} \end{cases}$</p>
<p><strong>Edit</strong>: Since some similar answers to my first question is that we want to talk about two-sided limit, thus require ƒ(x) be defined on an open interval about $x_0$. However, these answers doesn't clear my intended question. Now if I restrict $x>=1$, then can we talk about the limit of that function , especially for $\displaystyle \lim_{x \to 1^{+}}f(x)$?</p>
<p>As for my second question, I found a more precise definition of limit in Courant's book Introduction_to_Calculus_and_Analysis stated like this
<a href="https://i.stack.imgur.com/8J8Wh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8J8Wh.png" alt="enter image description here"></a>
which doesn't require ƒ(x) to be defined on every point of the an open interval about $x_0$, so I think it is possible to talk about limit of that function.</p>
| R.N | 253,742 | <p>Domain $f$ here is $\mathbb{Q}$, you can talk about limit of functon, and $\lim f=1$</p>
|
131,435 | <p>Wikipedia is a widely used resource for mathematics. For example, there are hundreds of mathematics articles that average over 1000 page views per day. <a href="http://en.wikipedia.org/wiki/Wikipedia:WikiProject_Mathematics/Popular_pages" rel="noreferrer">Here is a list of the 500 most popular math articles</a>. The number of regular Wikipedia readers is increasing, while the number of editors is decreasing (<a href="http://stats.wikimedia.org/EN/SummaryEN.htm" rel="noreferrer">graphs</a>), which is causing growing concern within the Wikipedia community. </p>
<p><a href="http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics" rel="noreferrer">WikiProject Mathematics</a> is relatively active (compared to other WikiProjects, but not compared to MathOverflow!), and there is always the need for more experts who really understand the material. An editor continually faces the tension between (1) providing a lot of advanced material and (2) explaining things well, which generates many productive discussions about how mathematics articles should be written, and which topics should have their own article. </p>
<p>Regardless of the long term concerns raised about whether Wikipedia is capable of being a resource for advanced mathematics (see <a href="https://mathoverflow.net/questions/19631/can-wikipedia-be-a-reliable-and-sustainable-resource-for-advanced-mathematics">this closed question</a>), the fact is, people are attempting to learn from Wikipedia's mathematics articles right now. So improvements made to articles today will benefit the readers of tomorrow. </p>
<hr>
<p>Wikipedia is a very satisfying venue for summarizing topics you know well, and explaining things to other people, due to its large readership. Based on the number of mathematicians at MathOverflow, who are willing to spend time (for free!) answering questions and clarifying subjects for other people, it seems like there is a lot of untapped volunteer potential here. So in the interests of exposing the possible obstacles to joining Wikipedia, I would like to know:</p>
<blockquote>
<p>Why don't mathematicians spend more time improving Wikipedia articles?</p>
</blockquote>
<p>Recent efforts intended to attract new participants and keep existing ones include the friendly atmosphere of the <a href="http://en.wikipedia.org/wiki/Wikipedia:Teahouse" rel="noreferrer">Teahouse</a>, as well as WikiProject Editor Retention. </p>
<p>In case anyone is interested, my Wikipedia username is User:Mark L MacDonald (which is my real name). </p>
| Neil Strickland | 10,366 | <p>In general, I find the wikipedia maths pages to be very comprehensive and useful, and I have not found any serious errors. However, they are often not very well organized or clearly explained. I agree that it would be good if more professional mathematicians tried to help with this.</p>
<p>I rewrote the article <a href="http://en.wikipedia.org/wiki/Homology_theory">http://en.wikipedia.org/wiki/Homology_theory</a> a few years ago. I thought about working on some other related ones. However, it seemed to me that it would be best to do a general reorganization, deleting some pages, renaming others and redistributing material in a more coherent way. I did not know how to go about presenting such a proposal to the relevant community, and how to ensure that there was enough consensus that a lot of work would not go to waste, and how to make the changes offline so as not to leave an inconsistent mess while the work was going on. So I did not end up doing anything.</p>
|
131,435 | <p>Wikipedia is a widely used resource for mathematics. For example, there are hundreds of mathematics articles that average over 1000 page views per day. <a href="http://en.wikipedia.org/wiki/Wikipedia:WikiProject_Mathematics/Popular_pages" rel="noreferrer">Here is a list of the 500 most popular math articles</a>. The number of regular Wikipedia readers is increasing, while the number of editors is decreasing (<a href="http://stats.wikimedia.org/EN/SummaryEN.htm" rel="noreferrer">graphs</a>), which is causing growing concern within the Wikipedia community. </p>
<p><a href="http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics" rel="noreferrer">WikiProject Mathematics</a> is relatively active (compared to other WikiProjects, but not compared to MathOverflow!), and there is always the need for more experts who really understand the material. An editor continually faces the tension between (1) providing a lot of advanced material and (2) explaining things well, which generates many productive discussions about how mathematics articles should be written, and which topics should have their own article. </p>
<p>Regardless of the long term concerns raised about whether Wikipedia is capable of being a resource for advanced mathematics (see <a href="https://mathoverflow.net/questions/19631/can-wikipedia-be-a-reliable-and-sustainable-resource-for-advanced-mathematics">this closed question</a>), the fact is, people are attempting to learn from Wikipedia's mathematics articles right now. So improvements made to articles today will benefit the readers of tomorrow. </p>
<hr>
<p>Wikipedia is a very satisfying venue for summarizing topics you know well, and explaining things to other people, due to its large readership. Based on the number of mathematicians at MathOverflow, who are willing to spend time (for free!) answering questions and clarifying subjects for other people, it seems like there is a lot of untapped volunteer potential here. So in the interests of exposing the possible obstacles to joining Wikipedia, I would like to know:</p>
<blockquote>
<p>Why don't mathematicians spend more time improving Wikipedia articles?</p>
</blockquote>
<p>Recent efforts intended to attract new participants and keep existing ones include the friendly atmosphere of the <a href="http://en.wikipedia.org/wiki/Wikipedia:Teahouse" rel="noreferrer">Teahouse</a>, as well as WikiProject Editor Retention. </p>
<p>In case anyone is interested, my Wikipedia username is User:Mark L MacDonald (which is my real name). </p>
| Aaron Meyerowitz | 8,008 | <p>Some articles have contributors who are very possessive. It is amazing and impressive that the Wikipedia model works so well. I find it a great first pass on a variety of topics. Often I feel no need to look further. Other times I do. The price of that is that contributing is not worth it unless one is willing to watch the page and decide when to engage and when to let it go.</p>
|
131,435 | <p>Wikipedia is a widely used resource for mathematics. For example, there are hundreds of mathematics articles that average over 1000 page views per day. <a href="http://en.wikipedia.org/wiki/Wikipedia:WikiProject_Mathematics/Popular_pages" rel="noreferrer">Here is a list of the 500 most popular math articles</a>. The number of regular Wikipedia readers is increasing, while the number of editors is decreasing (<a href="http://stats.wikimedia.org/EN/SummaryEN.htm" rel="noreferrer">graphs</a>), which is causing growing concern within the Wikipedia community. </p>
<p><a href="http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics" rel="noreferrer">WikiProject Mathematics</a> is relatively active (compared to other WikiProjects, but not compared to MathOverflow!), and there is always the need for more experts who really understand the material. An editor continually faces the tension between (1) providing a lot of advanced material and (2) explaining things well, which generates many productive discussions about how mathematics articles should be written, and which topics should have their own article. </p>
<p>Regardless of the long term concerns raised about whether Wikipedia is capable of being a resource for advanced mathematics (see <a href="https://mathoverflow.net/questions/19631/can-wikipedia-be-a-reliable-and-sustainable-resource-for-advanced-mathematics">this closed question</a>), the fact is, people are attempting to learn from Wikipedia's mathematics articles right now. So improvements made to articles today will benefit the readers of tomorrow. </p>
<hr>
<p>Wikipedia is a very satisfying venue for summarizing topics you know well, and explaining things to other people, due to its large readership. Based on the number of mathematicians at MathOverflow, who are willing to spend time (for free!) answering questions and clarifying subjects for other people, it seems like there is a lot of untapped volunteer potential here. So in the interests of exposing the possible obstacles to joining Wikipedia, I would like to know:</p>
<blockquote>
<p>Why don't mathematicians spend more time improving Wikipedia articles?</p>
</blockquote>
<p>Recent efforts intended to attract new participants and keep existing ones include the friendly atmosphere of the <a href="http://en.wikipedia.org/wiki/Wikipedia:Teahouse" rel="noreferrer">Teahouse</a>, as well as WikiProject Editor Retention. </p>
<p>In case anyone is interested, my Wikipedia username is User:Mark L MacDonald (which is my real name). </p>
| Gerhard Paseman | 3,528 | <p>I will expand on one of Aaron Meyerowitz's remarks.</p>
<p>I am somewhat possessive about what I have written.
Indeed one of the hurdles I had to overcome to
participate on MathOverflow was to accept that what
I submitted could be changed by many other people.
I was upset when it happened, and the major thing
that kept me from voicing that upset was the FAQ,
which said other people could do that. However,
most of the edits were sensitive to the thrust of the
post, and turned out to be more changes to style
than serious changes to content. I have grown to
be more comfortable with posting answers on this
forum, as my input has mainly been given due respect. </p>
<p>If I were a contributing author to Wikipedia, I would
have to overcome a similar hurdle, especially as there
would be no consensus as to what is "the" information
to present. The posting would not be "mine" anymore.</p>
<p>If Wikipedia had a mechanism for including outside
references in a useful fashion, I might be more inclined
to offer material for its use. My impression of the current
system is that there is a "References" section in the article
which includes hyperlinks to other material. My suggestion
would be to enhance this so that the link expands into
essentially two documents: my version of the article, which
I agree to allow Wikipedia to keep a local unedited copy and
display repeatedly for their benefit, and a meta document
which I agree can be modified in tandem by me and Wikipedia
editors whose main purpose is to explain discrepancies,
notation changes, and other elements of context to allow
the reader to transition between the Wikipedia intro and
my version. Wikipedia could then use (or not) my version of
the article, the content of which I have control, while
maintaing editorial control over their version. Even if I
decided that my version was no longer appropriate, I could
only petition for its removal, as I had granted Wikipedia the
right to use a copy of the unedited version in perpetuity, and
I have access to the meta document to say that I think a better
version is available elsewhere. </p>
<p>It seems a little more complicated then just providing a hyperlink,
but it has the advantage that it could be maintained by Wikipedia,
the situation between editor and author is clearly defined and
separated further, and the meta document has the flexibility to
handle most of the situations that arise. Also, this kind of
mechanism would accommodate my sense of possessiveness,
and allow me to write things which I could use for my own
purposes as well as allow Wikipedia to enhance their collection.</p>
<p>Gerhard "Will Write For Venti Mochas" Paseman, 2013.05.22</p>
|
131,435 | <p>Wikipedia is a widely used resource for mathematics. For example, there are hundreds of mathematics articles that average over 1000 page views per day. <a href="http://en.wikipedia.org/wiki/Wikipedia:WikiProject_Mathematics/Popular_pages" rel="noreferrer">Here is a list of the 500 most popular math articles</a>. The number of regular Wikipedia readers is increasing, while the number of editors is decreasing (<a href="http://stats.wikimedia.org/EN/SummaryEN.htm" rel="noreferrer">graphs</a>), which is causing growing concern within the Wikipedia community. </p>
<p><a href="http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics" rel="noreferrer">WikiProject Mathematics</a> is relatively active (compared to other WikiProjects, but not compared to MathOverflow!), and there is always the need for more experts who really understand the material. An editor continually faces the tension between (1) providing a lot of advanced material and (2) explaining things well, which generates many productive discussions about how mathematics articles should be written, and which topics should have their own article. </p>
<p>Regardless of the long term concerns raised about whether Wikipedia is capable of being a resource for advanced mathematics (see <a href="https://mathoverflow.net/questions/19631/can-wikipedia-be-a-reliable-and-sustainable-resource-for-advanced-mathematics">this closed question</a>), the fact is, people are attempting to learn from Wikipedia's mathematics articles right now. So improvements made to articles today will benefit the readers of tomorrow. </p>
<hr>
<p>Wikipedia is a very satisfying venue for summarizing topics you know well, and explaining things to other people, due to its large readership. Based on the number of mathematicians at MathOverflow, who are willing to spend time (for free!) answering questions and clarifying subjects for other people, it seems like there is a lot of untapped volunteer potential here. So in the interests of exposing the possible obstacles to joining Wikipedia, I would like to know:</p>
<blockquote>
<p>Why don't mathematicians spend more time improving Wikipedia articles?</p>
</blockquote>
<p>Recent efforts intended to attract new participants and keep existing ones include the friendly atmosphere of the <a href="http://en.wikipedia.org/wiki/Wikipedia:Teahouse" rel="noreferrer">Teahouse</a>, as well as WikiProject Editor Retention. </p>
<p>In case anyone is interested, my Wikipedia username is User:Mark L MacDonald (which is my real name). </p>
| Jérémy Blanc | 23,758 | <p>One difference between MO and wikipedia is that here you get very quickly answers or remarks on what you have written, and it is then very much more attractive. Going to wikipedia for finding something for oneself, you directly have the answer you were looking for (or dont find it), but really writing new stuff takes times and you don't really obtain a feedback.</p>
<p>I do not say this is a good reason, I just say that it is a reason why people go more often to MO than edit Wikipedia's pages. </p>
|
361,060 | <blockquote>
<p>Consider the ring of Gaussian integers $D=\lbrace a+bi\mid a,b \in \mathbb{Z \rbrace}$, where $i \in \mathbb{C}$ such that $i^2=-1$. Consider the map $f$ from $D$ to $\mathbb{Z}[x]/(x^2+1)$ sending $i$ to the class of $x$ modulo $x^2+1$. Show that $f$ is a ring isomorphism.</p>
</blockquote>
<p>I got a confusion in this question. I don't understand the map sending $i$ to the class of $x$ modulo $x^2+1$. Can anyone help me to clear my confusion?</p>
<p>EDIT: I am having trouble to show the map is injective and surjective. Can anyone guide me?</p>
| Ittay Weiss | 30,953 | <p>The ring $\mathbb Z[x]/(x^2+1)$ is a quotient ring. Its elements are equivalence classes modulo the ideal $(x^2+1)$. Thus, a typical element in $\mathbb Z[x]/(x^2+1)$ is of the form $f(x)+(x^2+1)$ where $f(x)\in \mathbb Z[x]$ and $(x^2+1)$ is the ideal in $\mathbb Z[x]$ generated by $x^2+1$. So, the map $D\to \mathbb Z[x]/(x^2+1)$ sends $i$ to $x+(x^2+1)$. </p>
<p>The question as stated does not actually make any sense. The reason is that the given information does not describe a function from all of $D$. What the question is really trying to say is that there is a (in effect unique) ring homomorphism $f:\mathbb Z[x]/(x^2+1)$ with the properties that $f(a)=a+(x^2+1)$ for all $a\in \mathbb Z$ and $f(i)=x+(x^2+1)$. </p>
|
361,060 | <blockquote>
<p>Consider the ring of Gaussian integers $D=\lbrace a+bi\mid a,b \in \mathbb{Z \rbrace}$, where $i \in \mathbb{C}$ such that $i^2=-1$. Consider the map $f$ from $D$ to $\mathbb{Z}[x]/(x^2+1)$ sending $i$ to the class of $x$ modulo $x^2+1$. Show that $f$ is a ring isomorphism.</p>
</blockquote>
<p>I got a confusion in this question. I don't understand the map sending $i$ to the class of $x$ modulo $x^2+1$. Can anyone help me to clear my confusion?</p>
<p>EDIT: I am having trouble to show the map is injective and surjective. Can anyone guide me?</p>
| AlexM | 72,471 | <p>Im not sure if I'm quite correctly interpreting your question, but I'll answer to the best of my ability. Start out with $\mathbb{Z}[x]$; the image of $x$ under the map $\pi: \mathbb{Z}[x] \to \mathbb{Z}[x]/(x^2 + 1)$, which let's denote $\overline{x}$, is such that $\overline{x}^2 + 1 = 0$, or that $\overline{x}^2 = -1$. So you can see why it is reasonable to identify $i$ with $\overline{x}$ (abstractly $i$ is a symbol, that follows the "grammar rule" that $i^2 = -1$). </p>
<p>To show that if is a ring homomorphism, just show that the map is well defined, and that it is a ring homomorphism. </p>
|
2,507,565 | <p>I'm failing to find any reasonable solution to this given problem.</p>
<p>Find explicit formula of $n$-th element from given sequence:
$$\begin{cases}
a_1 = 1\\
a_2 = 2\\
a_{n+2} = 4 a_{n+1} + 4 a_n + 2^n
\end{cases}$$</p>
<p>I have tried finding some clues whilst typing out few first expressions to no avail. I also tried inserting the recursion formula instead of $a_{n+1}$ or $a_n$ though it got very messy in no time and all I have gotten was something like this:
$$a_{n+2} = 4^3 a_{n-3} + \sum\limits_{j=0}^3 (2^{n+j} + 4^{j+1}a_{n-j})$$
Which I believe would get me to:
$$a_{n+2} = 4^{n-1} a_{1} + \sum\limits_{j=0}^{n-1} (2^{n+j} + 4^{j+1}a_{n-j})$$</p>
<p>Any tips on how can I solve this one?</p>
| Cye Waldman | 424,641 | <p>I propose what is possibly a more direct solution. With problems of this type I try to reduce the original recurrence to a more familiar form, usually what I call the generalized Fibonacci form, say, $f_n=af_{n-1}+bf_{n-2}$. In this instance, let us assume that</p>
<p>$$a_n=f_n+A2^n$$</p>
<p>Then,</p>
<p>$$f_{n+2}+A2^{n+2}=4(f_{n+1}+A2^{n+1})+4(f_{n}+A2^{n})+2^{n}$$</p>
<p>Now choose $A$ such that</p>
<p>$$A2^{n+2}=4A2^{n+1}+4A2^{n}+2^{n}\\
\text{or}\\
A=-\frac{1}{8}$$</p>
<p>We are then left with</p>
<p>$$f_{n+2}=4f_{n+1}+4f_{n}$$</p>
<p>with characteristic roots</p>
<p>$$\alpha,\beta=2\pm 2\sqrt{2}$$</p>
<p>The the solution is given by</p>
<p>$$a_n=a\alpha^n+b\beta^n-2^{n-3}$$</p>
<p>where $a,b$ are to be determined from the initial conditions.</p>
|
566 | <h3>We all love a good puzzle</h3>
<p>To a certain extent, any piece of mathematics is a puzzle in some sense: whether we are classifying the homological intersection forms of four manifolds or calculating the optimum dimensions of a cylinder, it is an element of investigation and inherently puzzlish intrigue that drives us. Indeed most puzzles (cryptic crosswords aside) are somewhat mathematical (the mathematics of Sudoku for example is hidden in <a href="http://en.wikipedia.org/wiki/Latin_squares" rel="nofollow noreferrer">latin squares</a>). Mathematicians and puzzles get on, it seems, rather well.</p>
<h3>But what is a good puzzle?</h3>
<p>Okay, so in order to make this question worthwhile (and not a ten-page wade-athon through 57 varieties of the men with red and blue hats puzzle), we are going to have to impose some limitations. Not every puzzle-based answer that pops into your head will qualify as a good puzzle—to do so it must</p>
<ul>
<li><strong>Not be widely known:</strong> If you have a terribly interesting puzzle that motivates something in cryptography; well done you, but chances are we've seen it. If you saw that <em>hilarious</em> scene in the film 21, where Kevin Spacey explains the Monty hall paradox badly and want to share, don't do so here. Anyone found posting the liar/truth teller riddle will be immediately disembowelled.</li>
<li><strong>Be mathematical (as much as possible):</strong> It's true that logic <em>is</em> mathematics, but puzzles beginning '<em>There is a street where everyone has a different coloured house …</em>' are tedious as hell. Note: there is a happy medium between this and trig substitutions.</li>
<li><strong>Not be too hard:</strong> Any level of difficulty is cool, but if coming up with an answer requires more than two sublemmas, you are misreading your audience.</li>
<li><strong>Actually have an answer:</strong> Crank questions will not be appreciated! You can post the answers/hints in <a href="http://en.wikipedia.org/wiki/Rot_13" rel="nofollow noreferrer">Rot-13</a> underneath as comments as on MO if you fancy.</li>
<li><strong>Have that indefinable spark that makes a puzzle awesome:</strong> Like a situation that seems familiar, requiring unfamiliar thought …</li>
</ul>
<p>Ideally include where you found the puzzle so we can find more cool stuff like it. For ease of voting, one puzzle per post is best.</p>
<hr />
<h1>Some examples to set the ball rolling</h1>
<blockquote>
<p>Simplify <span class="math-container">$\sqrt{2+\sqrt{3}}$</span></p>
</blockquote>
<p><strong>From:</strong> problem solving magazine</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Try a two term solution</p>
</blockquote>
<hr />
<blockquote>
<p>Can one make an equilateral triangle with all vertices at integer coordinates?</p>
</blockquote>
<p><strong>From:</strong> Durham distance maths challenge 2010</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> This is equivalent to the rational case</p>
</blockquote>
<hr />
<blockquote>
<p>The collection of <span class="math-container">$n \times n$</span> <a href="http://en.wikipedia.org/wiki/Magic_squares" rel="nofollow noreferrer">Magic squares</a> form a vector space over <span class="math-container">$\mathbb{R}$</span> prove this, and by way of a linear transformation, derive the dimension of this vector space.</p>
</blockquote>
<p><strong>From:</strong> Me, I made this up (you can tell, can't you!)</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Apply the rank nullity theorem</p>
</blockquote>
| mau | 89 | <p>From <em>Mathematical Puzzles</em> by Peter Winkler:</p>
<p>Divide an hexagon in equilateral triangles, like in the figure. Now fill all the hexagon with the three kinds of diamonds made from two triangles, also shown in the figure. Prove that the number of each kind of diamond is the same.</p>
<p><img src="https://i.stack.imgur.com/THuvE.png" alt="diagram of hexagon filled with diamonds"></p>
|
566 | <h3>We all love a good puzzle</h3>
<p>To a certain extent, any piece of mathematics is a puzzle in some sense: whether we are classifying the homological intersection forms of four manifolds or calculating the optimum dimensions of a cylinder, it is an element of investigation and inherently puzzlish intrigue that drives us. Indeed most puzzles (cryptic crosswords aside) are somewhat mathematical (the mathematics of Sudoku for example is hidden in <a href="http://en.wikipedia.org/wiki/Latin_squares" rel="nofollow noreferrer">latin squares</a>). Mathematicians and puzzles get on, it seems, rather well.</p>
<h3>But what is a good puzzle?</h3>
<p>Okay, so in order to make this question worthwhile (and not a ten-page wade-athon through 57 varieties of the men with red and blue hats puzzle), we are going to have to impose some limitations. Not every puzzle-based answer that pops into your head will qualify as a good puzzle—to do so it must</p>
<ul>
<li><strong>Not be widely known:</strong> If you have a terribly interesting puzzle that motivates something in cryptography; well done you, but chances are we've seen it. If you saw that <em>hilarious</em> scene in the film 21, where Kevin Spacey explains the Monty hall paradox badly and want to share, don't do so here. Anyone found posting the liar/truth teller riddle will be immediately disembowelled.</li>
<li><strong>Be mathematical (as much as possible):</strong> It's true that logic <em>is</em> mathematics, but puzzles beginning '<em>There is a street where everyone has a different coloured house …</em>' are tedious as hell. Note: there is a happy medium between this and trig substitutions.</li>
<li><strong>Not be too hard:</strong> Any level of difficulty is cool, but if coming up with an answer requires more than two sublemmas, you are misreading your audience.</li>
<li><strong>Actually have an answer:</strong> Crank questions will not be appreciated! You can post the answers/hints in <a href="http://en.wikipedia.org/wiki/Rot_13" rel="nofollow noreferrer">Rot-13</a> underneath as comments as on MO if you fancy.</li>
<li><strong>Have that indefinable spark that makes a puzzle awesome:</strong> Like a situation that seems familiar, requiring unfamiliar thought …</li>
</ul>
<p>Ideally include where you found the puzzle so we can find more cool stuff like it. For ease of voting, one puzzle per post is best.</p>
<hr />
<h1>Some examples to set the ball rolling</h1>
<blockquote>
<p>Simplify <span class="math-container">$\sqrt{2+\sqrt{3}}$</span></p>
</blockquote>
<p><strong>From:</strong> problem solving magazine</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Try a two term solution</p>
</blockquote>
<hr />
<blockquote>
<p>Can one make an equilateral triangle with all vertices at integer coordinates?</p>
</blockquote>
<p><strong>From:</strong> Durham distance maths challenge 2010</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> This is equivalent to the rational case</p>
</blockquote>
<hr />
<blockquote>
<p>The collection of <span class="math-container">$n \times n$</span> <a href="http://en.wikipedia.org/wiki/Magic_squares" rel="nofollow noreferrer">Magic squares</a> form a vector space over <span class="math-container">$\mathbb{R}$</span> prove this, and by way of a linear transformation, derive the dimension of this vector space.</p>
</blockquote>
<p><strong>From:</strong> Me, I made this up (you can tell, can't you!)</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Apply the rank nullity theorem</p>
</blockquote>
| Community | -1 | <p>If you have 32 2x1 dominoes then you can cover an 8x8 board easily enough; if you throw away a domino and cut a 1x1 square from each end of a diagonal of the board, can you cover the remaining shape with the remaining dominoes?</p>
<p>I heard this decades ago at university.</p>
|
655,591 | <p>$S=(v_1, \cdots v_n)$ of vectors in $\mathbb{F^n}$ is a basis iff the matrix obtained by forming a matrix (call it A) of the co-ordinate vectors of $v_i$ is invertible</p>
<p>My Idea: I was able to prove the reverse direction wherein we can show that $AX =0$ has trivial solutions so linearly independent and also spans the given vector space so it is a basis.</p>
<p>I am not sure about the forward direction that is to prove that A is invertible. I am just confused about the coordinate vector for $v_i$. That is,we assume $S$ is a basis then how can each element (namely $v_i'$'s) in a basis have a coordinate vector? Wouldn't it just be a single column for different vector in the vector space?</p>
| Leo Azevedo | 120,820 | <p>HINT: The determinant of a matrix is zero iff the vectors which form it are linearly independent.</p>
<p>See also:</p>
<p><a href="https://math.stackexchange.com/questions/79356/using-the-determinant-to-verify-linear-independence-span-and-basis">Using the Determinant to verify Linear Independence, Span and Basis</a></p>
|
844,887 | <p>I am trying to prove by induction that every non-zero natural number has at least one predecessor. However, I don't know what to use as a base case, since 0 is not non-zero and I haven't yet established that 1 is the number following zero.</p>
<p>My axioms are: </p>
<ol>
<li>$0$ is a natural number. </li>
<li>if $b$ is a natural number then $S(b)$ is also a natural number. </li>
<li>$0$ is not a successor of any natural number. </li>
<li>different numbers have different successors.</li>
</ol>
<p>Any advice?</p>
| Foster Antony | 337,532 | <p>Proof of Existence: Let us assume that it was already shown that $(\forall n)((n\ \in\ \omega)\ \ \rightarrow\ \ (n^+\
\neq\ 0))$ which is another way of saying that the natural number $0$ has no immediate predecessor in
$\omega$. So,</p>
<pre><code> what about the nonzero natural numbers---Do they have an immediate predecessor?
</code></pre>
<p>Hence, it suffices only to show that:</p>
<p>Each nonzero natural number has an immediate predecessor in $\omega$. </p>
<p>To this end, we conduct induction as follows. </p>
<p>We define a set $M\ \subseteq\ \omega$ consisting of only those nonzero natural numbers which has an immediate predecessor, i.e, $$x\ \in\ M\ \
\leftrightarrow\ \ (x\ \in\ \omega)\ \land\ ((\exists y)(y\ \in\ \omega)\ \ \land\ \ (x\ =\ y^+)\ \land\ (x\ \neq\ \varnothing)).$$Next, define the set $N\ \subseteq\ \omega$ by $N\ =\ M\ \cup\ \{\varnothing\}$. Clearly, by definition of $N$ we have $\varnothing\ \in\ N$. Let $n$ be any non-empty element of $N$ (i.e., $n\ \in\ M$). Then $(n\ \in\ \omega)\ \land\ ((\exists m)(m\ \in\ \omega)\ \land\ (n\ =\ m^+)\ \land\ \ (n\ \neq\ 0))$ holds. But then it follows immediately that: $$(n^+\ \in\ \omega)\ \ \land\ \ ((\exists (m^+)^+)((m^+)^+\ \in\ \omega)\ \ \land\ \ (n^+\ =\ (m^+)^+)\ \ \land\ \ (n^+\ \neq\ 0))$$also holds. This tells us precisely, that $n^+\ \in\ N$ whenever $n\ \in\ N$. Thus, we have shown that the set $N\ =\ M\ \cup\ \{0\}$ is an inductive subset of $\omega$. Therefore, by the Principle of Finite Induction, $N\ =\ \omega$. This complete the Existence part of non-zero natural numbers having an immediate predecessor (or just predecessor).</p>
<p>Predecessors are unique too! This follows from the uniqueness of successors.</p>
|
724,302 | <p>I have read the page about category theory in wikipedia carefully, but i don't really get what this theory is.</p>
<p>Is category theory a content in ZFC-set theory? (Just like measure theory, group theory etc.) If not, is it just another formal logic system independent from the standard ZFC-set theory?</p>
<p>Following wikipedia, i think it is the latter one.</p>
<p>If category theory is another formal logic system, then is there any theorem that relates category theory and ZFC-set theory? For example, there is a theorem states that "every theorem about sets in NBG is provable in ZFC, assuming consistency of ZFC" even though NBG and ZFC are completely different set theory. Moreover, is there a theorem which tells the relation between consistencies of ZFC and category theory? Also, is the usual independence proof in ZFC no more available in category theory?</p>
<p>Moreover, how does one write a <em>formula</em> in ZFC-set theory in category theory and vise versa? Since there is no <strong>undefined notion: set</strong> in category theory, i think this is not possible..</p>
| Peter Smith | 35,151 | <p>A useful resource is <a href="http://plato.stanford.edu/entries/category-theory/">http://plato.stanford.edu/entries/category-theory/</a> which aims to explain something of the role and nature of category theory at an introductory but not merely arm-waving level. </p>
|
724,302 | <p>I have read the page about category theory in wikipedia carefully, but i don't really get what this theory is.</p>
<p>Is category theory a content in ZFC-set theory? (Just like measure theory, group theory etc.) If not, is it just another formal logic system independent from the standard ZFC-set theory?</p>
<p>Following wikipedia, i think it is the latter one.</p>
<p>If category theory is another formal logic system, then is there any theorem that relates category theory and ZFC-set theory? For example, there is a theorem states that "every theorem about sets in NBG is provable in ZFC, assuming consistency of ZFC" even though NBG and ZFC are completely different set theory. Moreover, is there a theorem which tells the relation between consistencies of ZFC and category theory? Also, is the usual independence proof in ZFC no more available in category theory?</p>
<p>Moreover, how does one write a <em>formula</em> in ZFC-set theory in category theory and vise versa? Since there is no <strong>undefined notion: set</strong> in category theory, i think this is not possible..</p>
| Giorgio Mossa | 11,888 | <p>That's a question. Well for start as shown in Mac Lane <em>Categories for the working mathematician</em> there are two different way to approach categories, functor and natural transformations:</p>
<ul>
<li>you can either regard categories as some family of sets and operation between them (eventually adding some axioms to set theory since you would like to work with large collections like the class of all sets) </li>
<li>or you can define categories as those structures which satisfy the axioms of <em>the elementary theory of categories</em>, which is a theory in first order (multi sorted) logic.</li>
</ul>
<p>Of course if you use as meta-theory ZFC (actually at least NBG for the size problems I've mentioned above) then the two definition are essentially the same, and so you can see category theory as a theory developed inside set theory.
Nonetheless just because we can interpret the axioms of category theory inside a set theory doesn't mean that we have to do so. Indeed we can interpret category theory axioms in other foundational theories such as dependent type theory.</p>
<p>From another perspective if we add to the axioms of category some other axioms we can get the first order theory of a topos with some other stuff (natural number object, axiom of choice [expressed as the property that every epimorphism is split]...) we get the axioms that characterize a category with enough structure to be almost equivalent to $\mathbf{Set}$, the category of sets and functions between them. In such theory we can develop all the usual constructions of set theory and so this theory (of a special category) can be used as a foundational theory itself.</p>
<p>So we have two ways to look at categories: as structures in set theory and as <em>theories</em> of structures, some of which can be used to rebuild mathematics.</p>
<p>This second way to look at category theory is really interesting in particular in connection with the study of constructive mathematics, since the theory of particular categories allows one to build constructive foundational theories, and category theory gives also the means to compare and study these theories.</p>
<p>I could say more but I don't want to be too long. Here are some references:</p>
<ul>
<li><p>the two definitions of category (axiomatic vs set-theoretic) can be found in Mac Lane's <em>Categories for the working mathematician</em></p></li>
<li><p>for the use of <em>category theories</em> as a foundational theories I suggest you take a look to <a href="https://golem.ph.utexas.edu/category/2012/12/rethinking_set_theory.html" rel="noreferrer">this short post of Leinster</a> and his <a href="http://arxiv.org/abs/1212.6543" rel="noreferrer">related paper on arxiv</a>.</p></li>
<li><p>For more about the interaction between category theory and set theory you can also google a little bit, some keywords for the research could be <em>categorical logic</em>, <em>topos theory</em>, <em>foundation of category theory</em>, <em>category theory as foundational theory</em>.</p></li>
</ul>
<p>Hope this helps.</p>
|
2,703,323 | <p>How can one show that the limit of the following is $1$?</p>
<p>$$\lim_{x\to 0}\frac{\frac{1}{1-x}-1}{x}=1$$</p>
| Galc127 | 111,334 | <p>$$\frac{\dfrac{1}{1-x}-1}{x}=\frac{\dfrac{1-(1-x)}{1-x}}{x}=\frac{1}{1-x}\xrightarrow{x\to 0}1$$</p>
|
549,585 | <p>Let $a_1=1$ and $a_{n+1}=ma_n+10$ for $n\geq 2$. Determine for which value of m, the sequence $(a_n)$ is convergent and find the limit.</p>
<p>I know that if if we simplify we will get $0=L(M-1)+10$, which is equal to $10/(1-M)=L$ but I have no clue what to do from there</p>
| Felix Marin | 85,343 | <p>$\newcommand{\+}{^{\dagger}}%
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
\newcommand{\dd}{{\rm d}}%
\newcommand{\isdiv}{\,\left.\right\vert\,}%
\newcommand{\ds}[1]{\displaystyle{#1}}%
\newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
\newcommand{\ic}{{\rm i}}%
\newcommand{\imp}{\Longrightarrow}%
\newcommand{\ket}[1]{\left\vert #1\right\rangle}%
\newcommand{\pars}[1]{\left( #1 \right)}%
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}%
\newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
\newcommand{\sech}{\,{\rm sech}}%
\newcommand{\sgn}{\,{\rm sgn}}%
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}%
\newcommand{\verts}[1]{\left\vert #1 \right\vert}%
\newcommand{\yy}{\Longleftrightarrow}$
\begin{align}
\color{#0000ff}{\large a_{n}}
&=
ma_{n - 1} + 10 = m^{2}a_{n - 2} + 10m + 10= m^{3}a_{n - 3} + 10m^{2} + 10m + 10
= \cdots
\\[3mm]&=
m^{n - 1}a_{1} + \sum_{k = 0}^{n - 2}10\,m^{k}
=
m^{n - 1} + 10\,{m^{n - 1} - 1\over m - 1}
=
\color{#0000ff}{\large{m^{n} + 9m^{n - 1} - 10 \over m - 1}\,,\qquad n \geq 1}
\end{align}</p>
<p>It converges whenever</p>
<ol><li>$m = -9$ since the factor $m^{9} + 9m^{n - 1}$ vanishes identically.</li>
<li>Or $\verts{m} < 1$. In that case
$\lim_{n \to \infty} m^{n} = 0 = \lim_{n \to \infty} m^{n - 1}$.</li>
</ol>
<p>Then, in both cases:
$$
\mbox{When}\quad \lim_{n \to \infty}\pars{m^{n} + 9m^{n - 1}} = 0\,,
\quad
\mbox{we get}\quad
\lim_{n \to \infty}a_{n} = {10 \over 1 - m}
$$</p>
|
3,170,629 | <p>I am trying to define a sequence.
The first few terms of the sequence are:</p>
<p><span class="math-container">$2,5,13,43,61$</span></p>
<p>Not yet found other terms because I am working with paper and pen, no software.</p>
<p>Why the first term is <span class="math-container">$5$</span>?</p>
<p>Let be <span class="math-container">$\pi(x)$</span> the celebrated prime counting function.
Well 5-<span class="math-container">$\pi(5)$</span>=<span class="math-container">$5-3$</span>=2 which is a prime.
If we repeat the same thing with the new prime <span class="math-container">$2$</span>, we have 2-<span class="math-container">$\pi(2)=1$</span>, which is not a prime. So starting the sequence from prime <span class="math-container">$5$</span>, we have the cycle <span class="math-container">$5\rightarrow 2\rightarrow 1$</span>. The arrows stop when a not prime is reached. No prime below <span class="math-container">$5$</span> has a longer cycle. Infact starting for example from <span class="math-container">$3$</span> you get <span class="math-container">$3-\pi(3)=1$</span>, which is not prime so the cycle is simply <span class="math-container">$3\rightarrow 1$</span>.
The second term of the sequence is <span class="math-container">$13$</span> because below <span class="math-container">$13$</span> no other prime has a larger cycle. Infact <span class="math-container">$13-\pi(13)=7$</span>, which is prime. Then <span class="math-container">$7-\pi(7)=3$</span>, which is prime and eventually <span class="math-container">$3-\pi(3)=1$</span>, which is not prime. So the cycle is <span class="math-container">$13\rightarrow 7\rightarrow 3\rightarrow 1$</span></p>
<p>The cycle for 43 is longer so it is the third term of the above sequence.
Could you suggest to me a nice and elegant definition for this sequence: <span class="math-container">$5,13,43,61...$</span> (I don't know if it is infinite)
Could you find other terms with Pari if you want?</p>
| Ross Millikan | 1,827 | <p>I believe your sequence continues forever but grows quickly. If <span class="math-container">$n$</span> is large, the density of primes around <span class="math-container">$n$</span> is <span class="math-container">$\log n$</span>. Since <span class="math-container">$\log n$</span> is so much smaller than <span class="math-container">$n$</span>, the chance a random <span class="math-container">$n$</span> has <span class="math-container">$k$</span> arrows is about <span class="math-container">$\frac 1{(\log n)^{k+1}}$</span>. The expected number of sequences of length <span class="math-container">$k$</span> above <span class="math-container">$10^{12},$</span> say, is then <span class="math-container">$\int_{10^{12}}^\infty \frac {dn}{(\log n)^{k+1}}$</span>. This diverges because <span class="math-container">$(\log n)^k$</span> becomes less than <span class="math-container">$n$</span> for <span class="math-container">$n$</span> large enough and we know the integral of <span class="math-container">$\frac 1n$</span> diverges. Each subtraction is only of order <span class="math-container">$\frac n{\log n}$</span>, which is small compared to <span class="math-container">$n$</span> and the log will not change much. </p>
<p>If we ask what length of sequence we expect to find among the <span class="math-container">$12$</span> digit numbers, we note that the log of these numbers is about <span class="math-container">$29$</span> and that <span class="math-container">$29^{8.5} \approx 3\cdot 10^{12}$</span>. We would expect to find some sequences of <span class="math-container">$7$</span> arrows, maybe <span class="math-container">$8$</span> or <span class="math-container">$9$</span>, and be surprised at <span class="math-container">$10$</span> or more. For <span class="math-container">$100$</span> digit numbers, the log is about <span class="math-container">$231$</span> and <span class="math-container">$231^{42.5} \approx 3\cdot 10^{100}$</span>, so we would expect some sequences of length <span class="math-container">$40$</span> or <span class="math-container">$41$</span> among the <span class="math-container">$100$</span> digit numbers.</p>
|
3,372,305 | <blockquote>
<p>When it is allowed to pull a r.v. out of the expectation, i.e.</p>
</blockquote>
<p><span class="math-container">$E(XY|Y)\overset?=YE(X|Y)\tag1$</span>.</p>
<p>or even </p>
<p><span class="math-container">$E(Y^2|Y)\overset?=Y^2\tag2$</span></p>
<p>in the computation rules it is written that if, </p>
<p><span class="math-container">$XY\ge0$</span> or both r.v. are in <span class="math-container">$L^1(\star)$</span> </p>
<p>it is allowed. But what if other conditions were given instead of <span class="math-container">$(\star)$</span>, for example </p>
<p><span class="math-container">$E(X^2|Y)=Y^2$</span> and <span class="math-container">$E(X|Y)=Y$</span></p>
<p>Is then <span class="math-container">$(1)\ \&\ (2)$</span> still valid ?</p>
| grand_chat | 215,011 | <p>I don't think the textbook explanation is correct. Equation (2) doesn't apply here, since we're mixing matrices and vectors, and equation (1) doesn't get us far enough.</p>
<p>It may be easier to derive the gradient directly from scratch. For brevity write <span class="math-container">$u:=x-\mu$</span>, and <span class="math-container">$A:=P^{-1}$</span>. So <span class="math-container">$x$</span> is a vector, and <span class="math-container">$A$</span> is a constant square matrix. We seek the gradient of <span class="math-container">$u^TAu$</span> with respect to <span class="math-container">$x$</span>. Notice that <span class="math-container">$u^TAu$</span> is a scalar, so its gradient is a vector of the same size as <span class="math-container">$x$</span>, with component <span class="math-container">$k$</span> equal to:
<span class="math-container">$$
\begin{align}
\frac\partial{\partial x_k} (u^TAu)&\stackrel{(a)}=\frac\partial{\partial x_k}\sum_i\sum_j u_iA_{ij}u_j\\
&\stackrel{(b)}=\sum_i\sum_j\frac\partial{\partial x_k}(u_iA_{ij}u_j)\\
&\stackrel{(c)}=\sum_i\sum_j\left(\frac{\partial u_i}{\partial x_k}A_{ij}u_j+u_iA_{ij}\frac{\partial u_j}{\partial x_k}\right)\\
&\stackrel{(d)}=\sum_i\sum_j\left(\delta_{ik}A_{ij}u_j+u_iA_{ij}\delta_{jk}\right)\\
&\stackrel{(e)}=\sum_j A_{kj}u_j +\sum_iu_iA_{ik}\\
&\stackrel{(f)}=(Au)_k + (A^Tu)_k\\
&=\left((A+A^T)u\right)_k
\end{align}
$$</span>
Step (a) is the definition of matrix multiplication; step (b) is linearity of the (univariate) derivative. Step (c) is the (univariate) product rule for derivatives. In step (d) we recognize the derivative is zero except when the subscripts coincide; <span class="math-container">$\delta_{ij}$</span> is the Dirac delta. In step (e) we eliminate cases where the Dirac delta is zero. In step (f) we apply the definition of matrix multiplication again; the notation <span class="math-container">$(Au)_k$</span> means the <span class="math-container">$k$</span>th component of the vector <span class="math-container">$Au$</span>.</p>
<p>Conclude: the <span class="math-container">$k$</span>th component of the gradient equals the <span class="math-container">$k$</span>th component of <span class="math-container">$(A+A^T)u$</span>. Here <span class="math-container">$A$</span> is symmetric, so this simplifies to <span class="math-container">$$\frac\partial{\partial x}u^TAu=2Au$$</span></p>
|
2,885,420 | <p>Show that V is infinite dimensional vector space if and only if there is a sequence $v_1,v_2,...$ of vectors in V such that $\{v_1,v_2,...,v_n\}$ is linearly independent for every positive integer n.</p>
| Aniruddha Deshmukh | 438,367 | <p>Dimension of a vector space is defined as the number of elements in the basis, provided the basis has a finite number of elements. In such cases, the vector space is said to be finite dimensional.</p>
<p>Therefore, a vector space is said to be infinite dimensional if the basis of the vector space has infinitely many elements. Now, a basis should be linearly independent.</p>
<p>An infinite subset $S$ of a vector space $V$ is said to be linearly independent, iff every finite subset $A \subset S$ is linearly independent in $V$. This gives directly that for every $n \in \mathbb{N}$, $\left\lbrace v_1, v_2, \cdots, v_n \right\rbrace$ is linearly independent in $V$. Here, however, each $v_i$ must belong to the basis. Otherwise, the construction of the linearly independent set would be different.</p>
|
2,885,420 | <p>Show that V is infinite dimensional vector space if and only if there is a sequence $v_1,v_2,...$ of vectors in V such that $\{v_1,v_2,...,v_n\}$ is linearly independent for every positive integer n.</p>
| DanielWainfleet | 254,665 | <p>Take an arbitrary $v_1\ne 0.$ Now for $n\in \Bbb N$ suppose that $S(n)=\{v_j:j\leq n\}$ is a set of $n$ linearly independent vectors. Since $V$ is not finite dimensional we can take some $v_{n+1}$ that does not belong to the linear span of $S(n),$ so $S(n+1)=\{v_{n+1}\}\cup S(n)$ is a linearly independentset </p>
<p>Without the Axiom of Choice we obtain, by induction on $n$, that for each $n\in \Bbb N$ there exists a linearly independent set of $n$ vectors. We MUST have the Axiom of Choice (or at least a corollary of it called Dependent Choice) to justify th existence of an infinite sequence of "Takes" to get an infinite sequence $(v_n)_{n\in \Bbb N}$ with the desired property. </p>
|
173,745 | <p>I am trying to prove the following: 'If $(X_1,d_1)$ and $(X_2,d_2)$ are separable metric spaces (that is, they have a countable dense subset), then the product metric space $X_1 \times X_2$ is separable.' It seems pretty straightforward, but I would really appreciate it if someone could verify that my proof works.</p>
<p>Since $(X_1,d_1)$ and $(X_2,d_2)$ are separable, they each contain a countable dense subspace, say $D_1 \subset X_1$ and $D_2 \subset X_2$. We will show that $D_1 \times D_2 \subset X_1 \times X_2$ is dense and countable. First, $D_1 \times D_2$ is countable since both $D_1$ and $D_2$ are.</p>
<p>Now let $x=(x_1,x_2) \in X_1 \times X_2$ and let $d$ be the product metric on $X_1 \times X_2$ (given by $d(x,y)=(\displaystyle\sum_{i=1}^2 d_i(x_i,y_i)^2)^{1/2}$). We will show that every open ball $B_d(x,\varepsilon)$ containing $x=(x_1,x_2)$ also contains a distinct point of $D_1 \times D_2$. Let $a_1 \in B_{d_1}(x_1,\frac{\sqrt{2}}{2}\varepsilon)\cap D_1$ and let $a_2 \in B_{d_2}(x_2,\frac{\sqrt{2}}{2}\varepsilon)\cap D_2$ (such points exist because $D_1$ and $D_2$ are dense in $X_1$ and $X_2$, respectively.) Letting $a=(a_1,a_2)$, we then have $d(x,a)=(\displaystyle\sum_{i=1}^2 d_i(x_i,a_i)^2)^{1/2})=(d_1(x_1,a_1)^2 + d_2(x_2,a_2)^2)^{1/2} < ((\frac{\sqrt{2}}{2}\varepsilon)^2 + (\frac{\sqrt{2}}{2}\varepsilon)^2)^{1/2}=\varepsilon$, so we have that $a \in B_d(x,\varepsilon)$, so $D_1 \times D_2$ is dense in $X_1 \times X_2$. Then since $D_1 \times D_2 \subset X_1 \times X_2$ is a countable dense subspace of $X_1 \times X_2$, we have that $X_1 \times X_2$ is separable.</p>
<p>I can see how this would easily generalize to finite products, but does it also extend to countable products?</p>
| DanielWainfleet | 254,665 | <p>You can bypass a lot of details in proving that $Y=X_1\times X_2$ is separable by observing that for $ z=(p_1,p_2) \in Y$ and for $r>0$, we have $I_1\times I_2 \subset B_d(z.r)$, where $I_1=B_{d_1}(p_1,r/2)$ and $I_2=B_{d_2}(p_2,r/2)$. So if $D_1$ is dense in $X_1$ and $D_2$ is dense in $X_2$ then $$\phi\ne (D_1 \cap I_1)\times (D_2 \cap I_2)=(D_1\times D_2)\cap (I_1\times I_2)\subset (D_1\times D_2)\cap B_d(z,r).$$</p>
|
3,699,303 | <p>Let <span class="math-container">$R$</span> be a commutative ring with <span class="math-container">$1$</span>. This is a simple question, but I can't see whether it is true or not.</p>
<p>Suppose <span class="math-container">$R$</span> does not have a "unit" of (additive) order <span class="math-container">$2$</span>, i.e. , a unit <span class="math-container">$u$</span> such that <span class="math-container">$u=-u$</span>. Then it is necessarily true that <span class="math-container">$R$</span> does not have an element of order <span class="math-container">$2$</span>?</p>
| David Popović | 549,692 | <p>This is not true. For example, consider the ring <span class="math-container">$\mathbb{Z}[X]/(2X)$</span>. The units in this ring are only <span class="math-container">$\pm 1$</span>, but <span class="math-container">$X+X=0$</span>.</p>
|
175,576 | <p><a href="https://i.stack.imgur.com/wtjxT.png" rel="noreferrer"><img src="https://i.stack.imgur.com/wtjxT.png" alt="enter image description here"></a></p>
<p>This should work? But it is giving me the wrong t values</p>
| Bill | 18,890 | <p>If you didn't have a vast knowledge of all the internal functions provided by Mathematica then you might try a more brute force approach.</p>
<p>Trying a variety of starting values for t1 and t2 seems to usually come up with the same value for t1 and t2. Thus it looks like <code>FindRoot</code> might not have sufficiently good starting values to be able to find different t1 and t2 that give you your intersection. But from your graph you have a pretty good estimate of <code>r[t]</code> so lets look for points near that intersection and which have a substantially different values for t.</p>
<pre><code>rt[t_] := {6 Cos[2.9 Pi t], Cos[3.5 Pi t] + 5 t, t};
Select[Table[rt[t], {t, 0, 1, .001}], 3.2<#[[2]]<3.4 && -1<#[[1]]<0 &]
</code></pre>
<p>which gives us</p>
<pre><code>{...
{-0.067857, 3.39995, 0.516},
{-0.0508932, 3.31707, 0.863},
...
}
</code></pre>
<p>which hints that t1 ~= .516 and t2 ~= .863 so we try</p>
<pre><code>r[t_] := {6 Cos[2.9 Pi t], Cos[3.5 Pi t] + 5 t};
sol = FindRoot[r[t1] - r[t2], {{t1, .516}, {t2, .863}}]
</code></pre>
<p>and that correctly finds two distinct t1 and t2 for your problem.</p>
|
175,576 | <p><a href="https://i.stack.imgur.com/wtjxT.png" rel="noreferrer"><img src="https://i.stack.imgur.com/wtjxT.png" alt="enter image description here"></a></p>
<p>This should work? But it is giving me the wrong t values</p>
| Ulrich Neumann | 53,677 | <p>Without special knowledge NMinimize solves the problem straight forward: </p>
<pre><code>NMinimize[{1/(t2 - t1)^2 (*force t1!=t2*), r[t1] == r[t2], 0 < t1 < t2 < 1}, {t1, t2}]
(*{7.86584, {t1 -> 0.511377, t2 -> 0.867933}}*)
</code></pre>
|
164,103 | <p>I have some questions. </p>
<hr>
<p>The first one is about the product of Prikry's forcing.
Let $\kappa$ be a measurable cardinal, $U_1, U_2$ be normal measures on $\kappa$ and let $\mathbb{P}_{U_1}, \mathbb{P}_{U_2}$ be the corresponding Prikry forcings. Let $G\times H$ be $\mathbb{P}_{U_1}\times \mathbb{P}_{U_2}-$generic over $V$. It is easily seen that in the extension there are new subsets of $\omega$ (for example if $(x_n: n<\omega), (y_n: n<\omega)$ are the Prikry sequences added by $G, H$ respectively, then $\{ n<\omega: x_n < y_n\}$ is such a set). </p>
<blockquote>
<p><strong>Question 1.1</strong> Is $\kappa$ preserved in the extension $V[G\times H]$? Do $V$ and $V[G\times H]$ have the same cardinals?</p>
</blockquote>
<p><strong>Remark.</strong> Though the question remained unanswered in general, but by Yair Hayut's very nice answer, given a normal measure $U$ on $\kappa, \mathbb{P}_U^2$ does not collapse cardinals. His proof extends easily to show that for any natural number $n>1, \mathbb{P}_U^n$ is forcing isomorphism to $\mathbb{P}_U\times \mathbb{C},$ hence it preserves cardinals.</p>
<blockquote>
<p><strong>Question 1.2.</strong> What is the least cardinal $\lambda$ such that $\mathbb{P}_U^\lambda$ collapses some cardinals? What is the least cardinal $\delta$ such that $\mathbb{P}_U^\delta$ collapses $\kappa$? Are $\lambda$ and $\delta$ equal?</p>
</blockquote>
<hr>
<p>My second question is about Cohen forcing. Let $\kappa$ be a Mahlo cardinal, let $\mathbb{P}$ be the reverse Easton iteration of $Add(\alpha,1)$ for all inaccessible cardinals $\alpha\leq \kappa,$ and let $G$ be $\mathbb{P}-$generic over $V$. </p>
<blockquote>
<p><strong>Question 2.</strong> Suppose $\alpha$ is an inaccessible cardinal $\leq \kappa.$ Is there an $H\in V[G]$ which is $Add(\alpha,1)^V-$generic over $V$? (of course the answer is yest for the least inaccessible).</p>
</blockquote>
| Joel David Hamkins | 1,946 | <p>Concerning question 2, Yair Hayut's comment is exactly right, and this kind of situation is just the kind of situation that I had aimed to analyze with those results.</p>
<p>The basic fact is that if one performs forcing $\mathbb{P}_0$ of size $\delta$ followed by nontrivial strategically $\leq\delta$-closed forcing $\mathbb{Q}$, then the extension $V\subset V[g][H]$ has the <em>$\delta^+$-approximation</em> property, which means that any set $A\in V[g][H]$ with $A\subset V$ and $A\cap a\in V$ whenever $a$ has size at most $\delta$ in $V$, then $A$ is already in $V$. Thus, there can be in the extension no <em>fresh</em> subsets of a larger regular cardinal $\alpha$, where $A\subset\alpha$ is <em>fresh</em> over $V$ if $A\notin V$ but all initial segments of $A$ below $\alpha$ are in $V$. </p>
<p>Your iteration admits a closure point at the least inaccessible, and so it can add no fresh subsets above the least inaccessible. In particular, it adds no $V$-generic Cohen sets using $\text{Add}(\alpha,1)^V$, since this forcing would add a fresh subset of $\alpha$.</p>
<p>You can find details in my paper <em>Joel David Hamkins</em>, <a href="http://dx.doi.org/10.4064/fm180-3-4"><strong>Extensions with the approximation and cover properties have no new large cardinals</strong></a>, <em>Fund. Math.</em> <strong>180</strong> (2003), no. 3, 257--277. see also <a href="http://jdh.hamkins.org/approximation-and-cover-properties/">my blog</a>. </p>
<p>One can use this idea to show that if you add a Cohen subset to $\kappa$ and then to $\lambda>\kappa$, then you kill all supercompact cardinals between $\kappa$ and $\lambda$. </p>
|
1,144,540 | <p>I'm looking to determine the convergence of $\{e^{in}\}_{n=1}^\infty$ in $\mathbb{C}$, where $n \in \mathbb{N}$. Using the definition of convergence that iff $\exists \ \epsilon > 0: \forall \ n_0 \in \mathbb{N}, \exists \ n \ge n_0, |z_n - z| \le \epsilon$. </p>
<p>Using $z_n = e^{in} = \cos{n} + i\sin{n}$, and anticipating that the sequence is divergent, I am looking to show a that $|z_n - z| \ge \epsilon$ but I'm not sure how, since $n$ is in radians, not degrees. Thank you in advance for any guidance!</p>
| Chival | 118,564 | <p>A calculation of <em>Laplace transformation</em> of Poisson $\big(\lambda_i\big)$ shows us , $\forall u \in\mathbb{R}_+$,<br>
$$\mathbb{E}\Big[ \, e^{-u\cdot X_i} \, \Big] = \exp\big[\, -\lambda_i\cdot\big(1- e^{-u}\big) \,\big] \, \,\Longrightarrow\quad \mathbb{E}\Big[\, s^{X_i} \, \Big] = \exp\big[\, -\lambda_i\cdot\big(1- s\big) \,\big]\, \text{with $s = e^{-u}\in\big(0,1\big)$},$$
then it follows from dominated convergence and finite-additivity of Poisson distributions that<br>
\begin{align*}
\mathbb{E}\left\{ \,\, s^{\sum_{i\geq 1}X_i } \,\,\right\} &= \mathbb{E}\left\{ \,\, \lim_{k\to+\infty}s^{\sum_{i = 1}^k X_i } \,\,\right\} = \lim_{k\to+\infty} \mathbb{E}\left\{ \,\, s^{\sum_{i = 1}^k X_i } \,\,\right\} = \lim_{k\to+\infty}\exp\left(\,\, -\sum_{i=1}^k \lambda_i\cdot(1-s)\,\, \right) \\
&= \exp\left(\,\, -(1-s)\sum_{i\geq1} \lambda_i\,\, \right)\,\,,\quad\text{for $s\in(0,1)$.}
\end{align*}
So the sum of an independent sequence of Poisson random variables is Poisson with parameter equal to the sum of their rates. And if the sum of rates$=\infty$, then $\sum_{i=1}^\infty X_i = +\infty$ a.s.</p>
|
3,176,482 | <p>In my Econometrics class yesterday, our teacher discussed a sample dataset that measured the amount of money spent per patient on doctor's visits in a year. This excluded hospital visits and the cost of drugs. The context was a discussion of generalized linear models, and for the purposes of Stata, he found that a gamma distribution worked best. Good enough.</p>
<p>But I started thinking about the dataset itself, and the pdf curve "total dollars per patient" might follow. I'm thinking of it as two variables: </p>
<p>1) The number of visits per person (k). Let's use a Poisson distribution with λ=1, so p(0)=0.368, p(1)=0.368, p(2)=0.184, etc.</p>
<p>2) The cost of each visit (x). Let's use a normal curve with a mean of 50 dollars and a SD of 5 dollars. A person's total expenditure would be the sum of normally distributed variables, or x = N(k50, k25) for k=0 to infinity visits. So a person with no visits would spend nothing, ~.95 of people with one visit would spend between 40 and 60, ~.95 of those with two visits would spend between ~85.86 and ~114.14, three visits would spend ~132.68 and ~167.32, etc.</p>
<p>I think a pdf curve would be some combination of these two functions. My guess is that it would look like a vertical line a 0 and a series of decreasing, ever flattening humps at 50, 100, 150 dollars, etc, where the integral of x=(0, inf.) equals 0.632 (because zero visits is p=.0368)</p>
<p>Can anyone help me put the pieces together? What would that function be? I've taken through Calc III (multivariate/vector calculus), FYI. Thanks.</p>
| Henry | 6,460 | <p>There is a minor problem with using the normal distribution here as it has a positive probability of giving negative values</p>
<p>That being said, your idea of what the density in your model might look like is reasonable. Try the following R function:</p>
<pre><code>plotdensity <- function(meanvisitcost, sdvisitcost, meank){
maxkconsidered <- max(1, qpois(1 - 10^-7, meank))
kconsidered <- 0:maxkconsidered
dkconsidered <- dpois(kconsidered, meank)
dcostgivenk <- function(x, k){ dnorm(x, meanvisitcost*k, sdvisitcost*sqrt(k)) }
dcost <- function(x){ outer(x, kconsidered, dcostgivenk) %*% dkconsidered }
curve(dcost, from=0, to=6*meanvisitcost*max(1,meank), n=100001, col="red")
segments(0, 0, 0, 10^7, col="red")
}
</code></pre>
<p>Then with your parameters of <span class="math-container">$50,5,1$</span> you get the following density, where the spike at the left corresponds to a probability (not density) of <span class="math-container">$e^{-1}\approx 0.368$</span></p>
<pre><code>plotdensity(50, 5, 1)
</code></pre>
<p><a href="https://i.stack.imgur.com/2prDz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2prDz.png" alt="density graph plotdensity(50, 5, 1)"></a></p>
<p>while with BruceET's parameters of <span class="math-container">$100,20,5$</span> you get the following, where the spike at the left corresponds to a probability (not density) of <span class="math-container">$e^{-4}\approx 0.018$</span></p>
<pre><code>plotdensity(100, 20, 5)
</code></pre>
<p><a href="https://i.stack.imgur.com/sjypx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sjypx.png" alt="enter image description here"></a></p>
|
2,441,488 | <p>We are not unfamiliar to imaginary angles, so what can be $\tan(a+ib)$</p>
<blockquote>
<p>If$$\tan(a+ib)=x+iy$$then find $$x,y$$</p>
</blockquote>
<p>My attempt,</p>
<p>Let $x$ and $z$ be two complex bombers such that,
$$\tan(z)=x$$
$$z=\tan^{-1} (x)$$
$$z=\tan^{-1}(re^{i\theta}).$$
But this doesn't seem to get me anywhere, how do we proceed?</p>
| Bernard | 202,857 | <p>Use the relations between complex circular and hyperbolic functions, mainly
$$\tan iz=i\tanh z.$$
From the addition formula, we obtain
$$\tan(a+ib)=\frac{\tan a+\tan ib}{1-\tan a\tan ib}=\frac{\tan a+i\tanh b}{1-i\tan a\tanh b}$$
If you want a cartesian form of the result, multiplying by the conjugate expression f the denominator, you get
$$\frac{\tan a(1-\tanh^2b)+i\tanh b(1+\tan^2a)}{1+\tan^2a\tanh^2b}.$$</p>
|
1,728,910 | <p>Whenever I get this question, I have a hard time with it. </p>
<p>An example of a problem:</p>
<p>In the fall, the weather in the evening is <em>dry</em> on 40% of the days, <em>rainy</em>
on 58% of days and <em>snowy</em> 2% of the days. </p>
<p>At noon you notice clouds in the sky. </p>
<p>Clouds appear
at noon on 10% of the days that have <strong>dry evenings</strong>, </p>
<p>25% of the days that have <strong>rainy evenings</strong>, </p>
<p>and
35% of the days that have <strong>snowy evenings</strong>.</p>
<p>From that I get:</p>
<p>$P(S)$ = "prob. snows" = $.02$</p>
<p>$P(R)$ = "prob. rains" = $.58$</p>
<p>$P(D)$ = "prob. is dry" = $.40$</p>
<p>$P(C|S)$ = "prob. that it is cloudy given that it snows" = $.35$</p>
<p>$P(C|R)$ = "prob. that it is cloudy given that it rains" = $.25$</p>
<p>$P(C|D)$ = "prob. that it is cloudy given that it is dry" = $.10$</p>
<p>However, the question given is:</p>
<p>Find the probability of snow in the evening given that you saw clouds at noon.</p>
<p>I interpret this as:</p>
<p>$$P(S|C)$$</p>
<p>I can't get it right. Is there a formula to it without having P(C)? </p>
| Rubicon | 318,714 | <p>Using Bayes' Theorem and the law of total probability: $P(S | C) = 0.0365$</p>
|
13,432 | <p>I want to turn a sum like this</p>
<pre><code>sum =a-b+c+d
</code></pre>
<p>Into a List like this: </p>
<pre><code>sumToList[sum]={a,-b,c,d}
</code></pre>
<p>How can I achieve this?</p>
| b.gates.you.know.what | 134 | <p>Try :</p>
<pre><code>a - b + c + d /. Plus -> List
</code></pre>
<p>You can have a look at <code>a - b + c + d //FullForm</code> to see why this works.</p>
|
4,169,445 | <p>I'm getting stuck on perhaps a simple step in the Hille-Yosida theorem from 13.37 in Rudin's functional analysis.
I wonder if someone has had this same difficulty before or knows how to get around it -</p>
<p><strong>Setup:</strong> <span class="math-container">$A$</span> is a densely defined operator with domain <span class="math-container">$\mathcal{D}(A)$</span> in a Banach space <span class="math-container">$X$</span> and there are constants <span class="math-container">$C, \gamma >0$</span> such that for every <span class="math-container">$\lambda > \gamma$</span> and <span class="math-container">$m\in \mathbb{N}$</span>,
<span class="math-container">$$
\| (\lambda I - A)^{-m}\| \leq C(\lambda - \gamma)^{-m}.
$$</span>
The claim is then that <span class="math-container">$A$</span> is the <em>infinitesimal generator</em> of a semi-group of operators.
For small <span class="math-container">$\varepsilon$</span>, the bounded operator <span class="math-container">$S(\varepsilon)$</span> is defined to be <span class="math-container">$(I-\varepsilon A)^{-1}:X \to \mathcal{D}(A)$</span>.
It follows from the definitions that <span class="math-container">$AS(\varepsilon) = \varepsilon^{-1}(S(\varepsilon)-I)$</span> from which one can show that <span class="math-container">$e^{tAS(\varepsilon)}$</span> converges weakly to a bounded <span class="math-container">$Q(t)$</span>. Moreover, <span class="math-container">$\{Q(t) \}$</span> gives a semi-group.
Thus it has an infinitesimal generator <span class="math-container">$\tilde{A}$</span>.
Using the resolvent formulas for <span class="math-container">$AS(\varepsilon)$</span> and <span class="math-container">$\tilde{A}$</span>, we have for all <span class="math-container">$x$</span> and <span class="math-container">$\lambda$</span> sufficiently large,
<span class="math-container">$$
(\lambda I - \tilde{A})^{-1}x = \int_0^\infty e^{-\lambda t} Q(t)x dt
$$</span>
and
<span class="math-container">$$
(\lambda I - AS(\varepsilon))^{-1}x = \int_0^\infty e^{-\lambda t} e^{tAS(\varepsilon)}x dt.
$$</span>
One can easily justify the limit
<span class="math-container">$$
\lim_{\varepsilon \to 0} \int_0^\infty e^{-\lambda t} e^{tAS(\varepsilon)}x dt = \int_0^\infty e^{-\lambda t} Q(t)x dt.
$$</span></p>
<p><strong>Question:</strong> In order to compare <span class="math-container">$\tilde{A}$</span> and <span class="math-container">$A$</span>, how does one see that
<span class="math-container">$$
\lim_{\varepsilon \to 0} (\lambda I - AS(\varepsilon))^{-1}x = (\lambda I - A)^{-1}x?
$$</span></p>
<p>THANK YOU!</p>
| user247327 | 247,327 | <p>A product of differentials is <em>anti-symmetric</em>. That is, <span class="math-container">$dr\,d\theta= -d\theta \,dr$</span>. It is also follows that <span class="math-container">$dr\,dr= -dr\,dr$</span> so that <span class="math-container">$dr\,dr= 0$</span> and <span class="math-container">$d\theta \,d\theta= -d\theta \,d\theta$</span> so that <span class="math-container">$d\theta \,d\theta= 0$</span>.</p>
<p>If <span class="math-container">$x= r\cos(\theta)$</span> and <span class="math-container">$y= r\sin(\theta)$</span>
then <span class="math-container">$dx= \cos(\theta)\,dr- r \sin(\theta)\,d\theta$</span> and
<span class="math-container">$dy= \sin(\theta)\,dr+ r \cos(\theta)\,d\theta$</span> so that</p>
<p><span class="math-container">\begin{align}
dx\,dy = {} & (\cos(\theta)\,dr- r\sin(\theta)\,d\theta)(\sin(\theta)\,dr+ r \cos(\theta)\,d\theta) \\[8pt]
= {} & \cos(\theta)\sin(\theta)\,dr\,dr+ r\cos^2(\theta) \, dr \, d\theta \\
& {} - r \sin^2(\theta)\,d\theta \,dr- r^2 \sin(\theta) \cos(\theta) \,d\theta \,d\theta \\[8pt]
= {} & 0+ r \cos^2(\theta)\,dr \,d\theta+ r \sin^2 \,dr\,d\theta+ 0 \\[8pt]
= {} & r \,dr \,d\theta
\end{align}</span></p>
|
1,206,528 | <p>Find the matrix $A^{50}$ given</p>
<p>$$A = \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix}$$ as well as for $$A=\begin{bmatrix} 2 & 0 \\ 2 & 1\end{bmatrix}$$</p>
<p>I was practicing some questions for my exam and I found questions of this form in a previous year's paper.</p>
<p>I don't know how to do such questions.</p>
<p>Please assist over this question.</p>
<p>Thank You</p>
| DanielV | 97,045 | <p>To diagonalize a matrix, you first need to find the eigenvalues, solve:</p>
<p>$$\det(A - \lambda I ) = 0$$
$$\det \begin{bmatrix} 2 - \lambda & 0 \\ 2 & 1 - \lambda \end{bmatrix} = 0$$</p>
<p>$$(1 - L)(2 - L) = 0$$
$$L \in \{1, 2\}$$</p>
<p>So the diagonal values are $1$ and $2$.</p>
<p>Then you need to find the eigenvectors using nullspace basis:</p>
<p>$$\begin{align}
V_1 &= {\rm nullspace}(A - 1\times I) \\
& = {\rm nullspace}\begin{bmatrix} 1 & 0 \\ 2 & 0 \end{bmatrix} \\
& = \begin{bmatrix} 0 \\ 1 \end{bmatrix}
\end{align}$$<br>
and
$$\begin{align}
V_2 &= {\rm nullspace}(A - 2\times I) \\
& = {\rm nullspace}\begin{bmatrix} 0 & 0 \\ 2 & -1 \end{bmatrix} \\
& = \begin{bmatrix} 1 \\ 2 \end{bmatrix}
\end{align}$$</p>
<p>So with the eigenvalues and eigenvectors you can contruct:</p>
<p>$$\begin{align}
A^{50} &= \left([V_1 ~ V_2]A [V_1~V_2]^{-1}\right)^{50} \\
& = [V_1 ~ V_2]A^{50} [V_1~V_2]^{-1} \\
& =
\begin{bmatrix} 0 & 1 \\ 1 & 2 \end{bmatrix}
\begin{bmatrix} 1^{50} & 0 \\ 0 & 2^{50} \end{bmatrix}
\begin{bmatrix} -2 & 1 \\ 1 & 0 \end{bmatrix} \\
&= \begin{bmatrix} 2^{50} & 0 \\ 2^{51} - 2 & 1 \end{bmatrix}
\end{align}$$</p>
|
637,379 | <p>So I have a $10$ gallon aquarium slightly salty aquarium...</p>
<p>When I add water the water is at $.7$% salt. ($7$ part per thousand)</p>
<p>I let $1$ gallon evaporate, at which point I have a $\frac{7}{9}$% salinity
I then drain $1$ additional gallon. now I have $8$ gallons of $\frac{7}{9}$% salinity water.</p>
<p>I then replace the missing $2$ gallons with $.7$% salinity water again.</p>
<p>If I perform this over and over does the water continue to get ever saltier, or does it approach some limit.</p>
| John | 7,163 | <p>Let's say that your 10 gallons of (pure) water weighs 37.85 kg. Then to reach 0.7% salinity you'd add $0.26495$ kg of salt.</p>
<p>Let's call this amount of salt $x$. If you add two gallons water with this salinity, you'll be adding $x/5$. If the initial salinity is different, then it will be some non-negative fraction of $x$, $y = ax$, where $a \geq 0$.</p>
<p>One-ninth of the salt leaves the aquarium by the time it gets down to eight gallons. (One gallon evaporates, leaving nine gallons still with the original amount of salt. One gallon is drained, taking one-ninth of the water, and one-ninth of the salt.)</p>
<p>Call $S(n)$ the amount of salt after $n$ cycles. So if $S(0) = y$, then</p>
<p>$$S(1) = \frac{8}{9}y + \frac{x}{5},$$</p>
<p>or more generally,</p>
<p>$$S(n) = \frac{8}{9}S(n-1) + \frac{x}{5},$$</p>
<p>so that</p>
<p>$$S(n) = \left(\frac{8}{9}\right)^n S(0) + \sum_{k=0}^{n-1}\left(\frac{x}{5}\right)\left(\frac{8}{9}\right)^k = \left(\frac{8}{9}\right)^n S(0) + \frac{9x}{5} \left(1 - \left(\frac{8}{9}\right)^n\right).$$</p>
<p>If $S(0) = ax$, then</p>
<p>$$S(n) = \left(\frac{8}{9}\right)^n \left[\left(a - \frac{9}{5}\right)x\right] + \frac{9x}{5}.$$</p>
<p>Hence, regardless of the initial concentration, you'll end up with about 80% stronger salinity by doing what you're doing ($S(n) \to \frac{9x}{5}$ as $n \to \infty$). If $a > \frac{9}{5}$, the concentration will decrease over time. If $a < \frac{9}{5}$, it will increase over time.</p>
|
1,358,927 | <p>I have to solve this problem using integration by parts. I am new to integration by parts and was hoping someone can help me.</p>
<p>$$\int\frac{x^3}{(x^2+2)^2} dx$$</p>
<p>Here is what I have so far:</p>
<p>$$\int udv = uv-\int vdu $$</p>
<p>$$u=x^2+2$$ Therefore, $$xdx=\frac{du}{2}$$
$$dv=x^3$$
Therefor, $$v=3x^2$$</p>
| MCT | 92,774 | <p>Though it isn't perfect, you can create a "saw-tooth" function which is pretty much a periodic linear function. For example, if you want to take $\mod 3$, you have the function $f$ which has period $3$ and looks like $f(x) = x$ for $x \in [0, 3)$. </p>
<p>You can approximate this saw-tooth function using a Fourier series which involves harmonics of the form $e^{2 \pi i n x}$.</p>
|
2,431,052 | <p>Let $M$ be a an oriented riemannian manifold.
I have seen the following definition for the Hodge-star operator acting on a differential form. Starting with $\beta\in \Omega^p(M)$ we have
$$\alpha \wedge \star \beta = \left<\alpha,\beta\right>\text{vol} ~~\forall \alpha \in \Omega^p(M)$$</p>
<p>where $ \left<\alpha,\beta\right> = \left<e^1\wedge\ldots\wedge e^p, f^1\wedge \ldots \wedge f^p\right> = \det[\left<e^i,f^j\right>]$. </p>
<p>My question is if we have a Lie algebra (of a matrix group) valued form $F\in \Omega^p(M,g)$. How do we now define the Hodge star operator ?. I am asking this question because I want to understand expressions such as
$$ F \wedge \star F$$
in the context of Yang-Mills theory.
Thank you very much.</p>
<p>EDIT:
I think my confusion really comes from the fact wheter I should use the commutator or the matrix product as the first answer kindly mentioned.
In the case of Yang-Mills functional is it meant as: $$
L_{YM}= \int_{M}Tr(F\wedge\star F), ~~F = \sum\limits_{j}\omega_j \otimes g_j\in \Omega^2(M,g) \\Tr(F\wedge \star F) = \sum\limits_{j,k}\left<\omega_j, \omega_k\right>\text{vol}~\text{Tr}([g_j,g_k])$$
or is it meant as
$$F\in \Omega^2(M,g) \\Tr(F\wedge \star F) = \sum\limits_{j,k}\left<\omega_j, \omega_k\right>\text{vol}~\text{Tr}(g_jg_k).$$
Thank you for you help.</p>
| Isao Sauzedde | 483,390 | <p>If I'm not misteaking, the first expression you propose vanishes as the trace is conjugaison invariant. Moreover YM action is not suppose to measure only an error of commutation (with G=U(1), YM should be different from 0).</p>
|
1,571,083 | <p>Given the 2 terms
$$ \frac{k + a}{k + b}$$
and $$\frac{a}{b}$$
with $a, b, k \in \mathbb{R^+}$ and $a > b$</p>
<p>I want to show, that the first term is always bigger than the second one.</p>
<p><strong>My try</strong>
$$
\frac{k + a}{k + b} > \frac{a}{b} \\
\frac{k + b + (a - b)}{k + b} > \frac{b + (a - b)}{b} \\
1 + \frac{a - b}{k + b}> 1 + \frac{a - b}{b} \\
\frac{1}{k + b} > \frac{1}{b} \\
k + b > b
$$
the problem is, that I think the inequality sign has to change to $<$ in the last step because of calculating the inverse on both sides. That would mean
$$ k + b < b$$
which is nonsense.</p>
| paw88789 | 147,810 | <p>We might also consider what is the chance that someone (not a specific individual chosen in advance) wins twice in a row. This is a bit tricky since there are different numbers of entrants. But assuming that all $25$ people in the second contest were also in the first contest, the probability of someone winning both times is </p>
<p>$$1-\left(\frac{773}{775}\right)^{25}\approx .0626$$</p>
<p>So there is about a $6\%$ chance that someone would win twice in a row.</p>
|
1,429,000 | <p>Prove that for any integer $n$, the integer $n^2 + 7n + 1$ is odd.</p>
<p>I have $n=2k+1$ for some $k\in Z$</p>
<p>I really do not how to do this problem. any help in understanding would be greatly appreciated.</p>
| lhf | 589 | <p>$n^2 + 7n + 1= n(n+1)+6n+1$</p>
<p>$n(n+1)=2\binom{n}{2}$ and so is even.</p>
<p>$6n+1=2(3n)+1$ and so is odd.</p>
<p>So the sum is odd.</p>
|
2,579,137 | <p>According to the definition of harmonic number $H_n = \sum\limits_{k=1}^n\frac{1}{k}$.</p>
<p>How we can define $H_{n+1}$ and $H_{n+\frac{1}{2}}$? </p>
| Robert Israel | 8,508 | <p>$$H_{n+1} = \sum_{k=1}^{n+1} \dfrac{1}{k} = H_n + \frac{1}{n+1}$$</p>
<p>$H_{n+1/2}$ is not defined by the formula for $H_n$, unless you want to just say it's the same as $H_n$. However, one function whose restriction to the positive integers is $H_n$ is $\psi(n+1)+\gamma$, where $\psi$ is the <a href="https://en.wikipedia.org/wiki/Digamma_function" rel="nofollow noreferrer">digamma function</a> and $\gamma$ is the <a href="https://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant" rel="nofollow noreferrer">Euler-Mascheroni constant</a>. This would make</p>
<p>$$ H_{n+1/2} = \psi(n+3/2)+ \gamma$$</p>
<p>EDIT: Since $$\psi(n+3/2) = 2 \psi(2n+2) - \psi(n+1) - 2 \ln 2$$</p>
<p>you could also write
$$ H_{n+1/2} = 2 H_{2n+1} - H_n - 2 \ln 2 $$</p>
|
1,283,541 | <p>Find the least value of $f(x)=3^{-x+1} + e^{-x-1}$.</p>
<p>I tried to use the maxima/minima concept but it was of no use. Please help.</p>
| tomi | 215,986 | <p>Are you happy to start from:</p>
<p>$\sin x=x-\frac {x^3}{3!}+\frac{x^5}{5!}-...$</p>
<p>$\frac{\sin x}x=1-\frac {x^2}{3!}+\frac{x^4}{5!}-...$</p>
<p>$\cos x=1-\frac {x}{2!}+\frac{x^4}{4!}-...$</p>
<p>$\cos x-\frac{\sin x}x=-x^2\left(\frac 1{2!}-\frac 1{3!}\right)+x^4\left(\frac 1 {4!}-\frac 1{5!}\right)+...$</p>
|
1,283,541 | <p>Find the least value of $f(x)=3^{-x+1} + e^{-x-1}$.</p>
<p>I tried to use the maxima/minima concept but it was of no use. Please help.</p>
| Jack D'Aurizio | 44,121 | <p>Both:
$$\frac{\sin x}{x}\leq 1,\qquad\frac{\tan x}{x}\geq 1$$
can be seen as convexity inequalities. The first one follows from the fact that $\frac{d}{dx}\sin x = \cos x$ is decreasing on $[0,\pi]$ and the second one follows from the fact that $\frac{d}{dx}\tan x=\frac{1}{\cos^2 x}$ is increasing over the same interval.</p>
|
214,520 | <p>I'm trying to make tables like in the image to calculate max and min of the expressions.
Whenever I change the max and min of <code>a</code> and <code>b</code> in the upper table then the <code>MixEx</code>and <code>MaxEx</code> are updated automatically.</p>
<p>How can I do that in Mathematica?
<a href="https://i.stack.imgur.com/pwvP4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pwvP4.jpg" alt="enter image description here"></a></p>
| Nasser | 70 | <p>One way is to use <code>Minimize</code> and <code>Maximize</code> with constraint.</p>
<p>Need to make sure min is smaller than max, else you'll get unexpected results. </p>
<p><a href="https://i.stack.imgur.com/cI7R6.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cI7R6.gif" alt="enter image description here"></a></p>
<pre><code>Manipulate[
Module[{a, b},
Quiet@Grid[{
{"expression", "Min", "Max"},
{"a+b",
First@Minimize[{a + b, minA < a < maxA && minB < b < maxB}, {a, b}],
First@Maximize[{a + b, minA < a < maxA && minB < b < maxB}, {a, b}]},
{"-a", First@Minimize[{-a, minA < a < maxA}, a],
First@Maximize[{-a, minA < a < maxA}, a]},
{"a+2 b", First@Minimize[{a + 2 b, minA < a < maxA && minB < b < maxB}, {a, b}],
First@Maximize[{a + b, minA < a < maxA && minB < b < maxB}, {a,b}]},
{"-b", First@Minimize[{-b, minB < b < maxB}, b],
First@Maximize[{-b, minB < b < maxB}, b]}
}, Frame -> All
]
],
{{minA, 1, "Min of a"}, 0, 10, 1, Appearance -> "Labeled"},
{{maxA, 3, "Max of a"}, 0, 10, 1, Appearance -> "Labeled"},
{{minB, 2, "Min of b"}, 0, 10, 1, Appearance -> "Labeled"},
{{maxB, 6, "Max of b"}, 0, 10, 1, Appearance -> "Labeled"},
TrackedSymbols :> {minA, minB, maxA, maxB}
]
</code></pre>
|
2,245,113 | <p>I want to calculate:</p>
<p>$2^{33} \mod 25$</p>
<p>Therefore I calculate first the Euler's phi function of 25:</p>
<p>$\phi(25) = 5^{2-1} (5-1)$ = 20</p>
<p>Now I can continue with Fermat's little problem:</p>
<p>$2^{33} \mod 25$ = $(2^{20} * 2^{13}) \mod 25 \equiv 2^{13} \mod 25$</p>
<p>How can I proceed to simplify the result?
I tried the following 2 approaches and I'm not sure what kind of operations are allowed and which are not. In both ways, I'll get the same result but it is wrong:</p>
<h2>1. Approach</h2>
<p>$2^{13} \mod 25 \equiv ((2^5 \mod 25)^2 * 2^3) \mod 25 \equiv$<br>
$\equiv ((7^2 \mod 25) * (2^3 \mod 25)) \mod 25 \equiv 24 * 7 \mod 25 \equiv$<br>
$\equiv 168 \mod 25 = 18$</p>
<h2>2. Approach</h2>
<p>$2^{13} \mod 25 \equiv ((2^5 \mod 25)^2 * 2^3) \mod 25 \equiv$<br>
$\equiv 7^2 * 2^3 \mod 25 \equiv 1568 \mod 25 = 18$</p>
| Luiz Cordeiro | 58,818 | <p>Your approach is right, but the next-to-last steps on both your calculations are wrong: in the first approach, $2^3\bmod{25}=8\bmod25$, and not $7$, so $$7^2*2^3\bmod25=24*8\bmod25=192\bmod25=17\bmod25$$.</p>
<p>In the second approach you susbsituted $7^2*2^3$ by 1568, which I don't understand.</p>
|
2,245,113 | <p>I want to calculate:</p>
<p>$2^{33} \mod 25$</p>
<p>Therefore I calculate first the Euler's phi function of 25:</p>
<p>$\phi(25) = 5^{2-1} (5-1)$ = 20</p>
<p>Now I can continue with Fermat's little problem:</p>
<p>$2^{33} \mod 25$ = $(2^{20} * 2^{13}) \mod 25 \equiv 2^{13} \mod 25$</p>
<p>How can I proceed to simplify the result?
I tried the following 2 approaches and I'm not sure what kind of operations are allowed and which are not. In both ways, I'll get the same result but it is wrong:</p>
<h2>1. Approach</h2>
<p>$2^{13} \mod 25 \equiv ((2^5 \mod 25)^2 * 2^3) \mod 25 \equiv$<br>
$\equiv ((7^2 \mod 25) * (2^3 \mod 25)) \mod 25 \equiv 24 * 7 \mod 25 \equiv$<br>
$\equiv 168 \mod 25 = 18$</p>
<h2>2. Approach</h2>
<p>$2^{13} \mod 25 \equiv ((2^5 \mod 25)^2 * 2^3) \mod 25 \equiv$<br>
$\equiv 7^2 * 2^3 \mod 25 \equiv 1568 \mod 25 = 18$</p>
| Steven Alexis Gregory | 75,410 | <p>What I would do....</p>
<p>Express $13$ as a sum of powers of $2$: $13 = 8 + 4 + 1$</p>
<p>$2^4 \equiv 16 \pmod{25}$</p>
<p>$2^8 \equiv 256 \equiv 6 \pmod{25}$</p>
<p>$2^{13} \equiv 6 \times 16 \times 2 \equiv -4 \times 2 \equiv 17 \pmod{25}$</p>
|
804,963 | <p>Let $f,g:\mathbb{R}\longrightarrow \mathbb{R}$ be Lebesgue measurable. If $f$ is Borel measurable, then $f\circ g$ is Lebesgue mesuarable. In general, $f\circ g$ is not necessarily Lebesgue measurable. Is there any counterexample?</p>
| Poor Math Guy | 136,786 | <p>Recall that a function <span class="math-container">$f$</span> is Lebesgue (reps. Borel measurable) if <span class="math-container">$f^{-1}(B)$</span> is a Lebesgue (resp. Borel measurable) set whenever <span class="math-container">$B$</span> is a Borel set.</p>
<p>Observe from <span class="math-container">$(f\circ g)^{-1}(B)=g^{-1}\circ f^{-1}(B)$</span>, if <span class="math-container">$f$</span> is not Borel measurable, then we cannot ensure <span class="math-container">$f^{-1}(B)$</span> is Borel, and hence the statement is false in general.</p>
<p>From above, it remains to show that we can find a function such that <span class="math-container">$f^{-1}(B)$</span> is not Borel even <span class="math-container">$B$</span> is so. The Cantor function always do a good job in finding the counter-examples.</p>
<p>Let <span class="math-container">$f:[0,1]\longrightarrow[0,1]$</span> be the standard Cantor function, define from <span class="math-container">$f$</span> that <span class="math-container">$$g(x)=\inf\{y\in[0,1]:f(x)=y\}$$</span></p>
<p><span class="math-container">$g$</span> is increasing, one-to-one and map <span class="math-container">$[0,1]$</span> onto the Cantor set <span class="math-container">$C$</span>, the preimage of any Borel set under <span class="math-container">$g$</span> is again Borel.</p>
<p>Pick <span class="math-container">$E$</span> to be a Lebesgue non-measurable set in <span class="math-container">$R$</span>, standard example is constructed by Vitalli. <span class="math-container">$F=g(E)\subset C$</span> is measurable by completeness of Lebesgue measure, and Cantor set is of measure zero. <span class="math-container">$F$</span> is the desired non-Borel-measurable set, as if <span class="math-container">$F$</span> is Borel measurable, then <span class="math-container">$E=g^{-1}(F)$</span> is Borel measurable by above consideration, but it's abusrd from our assumption that <span class="math-container">$E$</span> is not Lebesgue measurable.</p>
|
2,249,036 | <p>I was watching a video on PDEs and when arriving at the part of Fourier Series, the professor said:</p>
<blockquote>
<p>And one of the most fascinating reads I ever had was a paper by Riemann on the history of this [Fourier Series].</p>
</blockquote>
<p>I tried looking for it but didn't succeed, and I was wondering if anyone could provide a link or a reference to this paper.</p>
<p>Thank you very much.</p>
<p>In case you need it, the link of the video is: <a href="https://www.youtube.com/watch?v=cf8rgx60IKA" rel="nofollow noreferrer">https://www.youtube.com/watch?v=cf8rgx60IKA</a> on the minute 14:21 (The audio on that specific part is messed up, so be careful if you are using headphones).</p>
| Zain Patel | 161,779 | <p>Draw two position vectors, $\mathbf{v}_1$ and $\mathbf{v}_2$ with unit magnitude and at angles $\alpha, \beta$ to the positive $x$-axis. Then the angle between the two is $\alpha - \beta$ (assuming $\alpha > \beta$ w.l.o.g). But $\mathbf{v}_1 \cdot \mathbf{v}_2$ is the cosine of the angle between them. So $\cos (\alpha - \beta) = \mathbf{v}_1 \cdot \mathbf{v}_2$. </p>
<p>But remember that the two vectors lie on the unit circle and have components $\mathbf{v}_1 = \cos \alpha \mathbf{i} + \sin \alpha \mathbf{j}$ and $\mathbf{v}_2 = \cos \beta \mathbf{i} + \sin \beta \mathbf{j}$. By the definition of the dot product $$\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta.$$</p>
<hr>
<p>To solve the given problem: note that $\beta \mapsto -\beta$ gives $$\cos (\alpha - (-\beta)) = \cos \alpha \cos (-\beta) + \sin \alpha \sin (-\beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta = \cos(\alpha + \beta)$$ using the oddity of $\sin$ and even-ness of $\cos$.</p>
|
2,786,138 | <blockquote>
<p>Find all triples of prime numbers <span class="math-container">$(p,q,r)$</span> such that <span class="math-container">$$p^q+q^r=r^p.$$</span></p>
</blockquote>
<p>I proved that when <span class="math-container">$r=2$</span>, the equation becomes <span class="math-container">$$p^q+q^2=2^p.$$</span> Then I tried to use reciprocity laws and Fermat's little theorem. I could prove that <span class="math-container">$p\equiv 7\pmod 8$</span> and that <span class="math-container">$p>q$</span>.</p>
<p>The equation appeared in some olympiad . They asked to prove that <span class="math-container">$r=2$</span>. So I am trying to find at least one triple.</p>
| Piquito | 219,998 | <p>So $$p^q+q^2=2^p \qquad(*)$$ It is easy to show that $\boxed{p\gt q}$ or just look at the graph of the function $x ^ y + y ^ 2 = 2 ^ x$ for $x,y\gt1$.</p>
<p>If $p=4n+1$ then from $(*)$,
$$q=4m\pm1\Rightarrow4N+1+16m^2\pm8m+1=2N+8m^2\pm4m+1=2^{p-1}\space\text{ absurde}.$$
Then $p$ must be of the form $\boxed{p=4n-1}$.</p>
<p>Put $p=an-1$ where $n$ is odd and $a=2^r,\space r\ge 2$.
$$(an-1)^q=(an)^q-1+\sum_{k=1}^{q-1}(\pm1)^k\binom qk(an)^{q-k}$$ $$(an-1)^q=(an)^q-1+an\left(\sum_{k=1}^{q-2}(\pm1)^k\binom qk(an)^{q-k-1}\right)+anq$$
For $\boxed{q\equiv\pm1\pmod4}$ put $q=bm\pm1$ where $m$ is odd and $b=2^s,\space s\ge 2$.</p>
<p>The equation $(*)$ becomes
$$(an)^q+an\left(\sum_{k=1}^{q-2}(\pm1)^k\binom qk(an)^{q-k-1}\right)+anq+(bm)^2\pm2bm=2^p\qquad(**)$$
This last condition, , implies</p>
<p>Note that the sum $\left(\sum\right)$ is even.</p>
<p>Now </p>
<p>►if $a\le b$ (i.e. $2^r\le2^s$) then dividing by $a$ we are done because $nq$ is odd.</p>
<p>►if $a\gt b$ (i.e. $\dfrac {a}{2b}=2^{r-s-1}\ge1$) then dividing by $2b$ we are done for all $p,q$ such that $r\gt {s+1}$ because $m$ is odd.</p>
<p>It remains the cases for which $r=s+1$, in other words and taking into account the above
$$p=2^{s+1}n-1,\space q=2^sm\pm1\space n,m \space \text{ odds},\space \text {with }p\gt q$$
This last condition, $p\gt q$ , implies
$$2^{s+1}n-1\gt2^sm\pm1\Rightarrow\begin{cases}2^sn\gt2^{s-1}m+1\\2n\gt m\end{cases}$$ Thus it remains to study the equation where $p=2^{s+1}n-1,\space q=2^sm\pm1\space n,m \space \text{ odds},\space \text {with }p\gt q$ both odd primes.
$$\boxed{\displaystyle(2^{s+1}n-1)^{2^sm\pm1}+(2^sm\pm1)^2=2^{2^{s+1}n-1}}$$</p>
<p>I will return to the possible proof of this last part. If someone wants to end this problem by proving the apparently probable impossibility of solution, let it go forward. In particular if @Servaes wants to finish his very interesting partial answer.</p>
|
519,560 | <p>I need prove that:</p>
<p>$$\int_{0}^{2\pi} \frac{R^{2}-r^{2}}{R^{2}-2Rr\cos \theta +r^{2}} d\theta= 2\pi$$</p>
<p>By deformation theorem, with $0<r<R$.</p>
<p>Professor gave us the hint to use the function $f(z)= \frac{R+z}{z(R-z)}$, and define an adequate $\gamma : [a,b]\rightarrow \mathbb{C}$ circular curve and with deformation theorem, we could find the integral. But I have been able to find the curve $\gamma$. Any advice is very helpful</p>
| Dan | 79,007 | <p>The integrand is the famous <a href="http://mathworld.wolfram.com/PoissonKernel.html">Poisson kernel</a> function! Here's a sketch of how to integrate it using the Residue Theorem:</p>
<p>Show that</p>
<p>$$
\frac{R^2-r^2}{R^2-2Rr\cos\theta +r^2} = \text{Re}\left(\frac{R+re^{i\theta}}{R-re^{i\theta}}\right)
$$</p>
<p>and hence</p>
<p>$$
\int_0^{2\pi}\frac{R^2-r^2}{R^2-2Rr\cos\theta +r^2} \,d\theta
= \int_0^{2\pi}\text{Re}\left(\frac{R+re^{i\theta}}{R-re^{i\theta}}\right)\,d\theta
= \text{Re}\int_0^{2\pi}\frac{R+re^{i\theta}}{R-re^{i\theta}}\,d\theta.
$$
View this integral as an integral along the contour $\gamma(\theta) = re^{i\theta}$ and evaluate
$$
\int_\gamma \frac{R+z}{iz(R-z)}\,dz
$$
using the Residue Theorem.</p>
|
1,267,644 | <p>Having a circle of radius <span class="math-container">$R$</span> with the center in <span class="math-container">$O(0, 0)$</span>, a starting point on the circle (e.g. <span class="math-container">$(0, R)$</span>) and an angle <span class="math-container">$\alpha$</span>, how can I move the point on the circle with <span class="math-container">$\alpha$</span> degrees? I need to get the second point where it was moved.</p>
<h2>Example</h2>
<blockquote>
<p>The red point is on <span class="math-container">$(0, R)$</span> and <span class="math-container">$\alpha$</span> is <span class="math-container">$90$</span> degrees. The violet circle is where the first point is supposed to be moved, and its coordinates are <span class="math-container">$(R, 0)$</span>. Then we consider the violet point as starting point and move it with <span class="math-container">$45$</span> degrees. The new position will be where the blue circle is.</p>
<p><img src="https://i.stack.imgur.com/Tko9F.png" alt="" /></p>
</blockquote>
| Mann | 126,204 | <p><img src="https://i.stack.imgur.com/Aa8QG.jpg" alt="enter image description here"></p>
<p>As you can see from the image, The point you require is at $-45 $ Degree from $0$ . So your required point is $R \cos (-\pi /4), R \sin (-\pi/4)$</p>
<p>Which can be equivalently written as?</p>
|
136,570 | <p>Good day everyone. I was reading the more advanced lectures on complex analysis and encountered a lot of questions, concerning the determination of complex logarithm. As far I don't even understand the concept of it, but I'll do provide you with several practical questions concerning the topic. </p>
<p>First of all, the determination of complex logarithm on open $\Omega \subset \mathbb{C}$ is continuous $f(w)$ such that:</p>
<p>$$\forall w \in \Omega \text{ }\exp(f(w))=w$$</p>
<p>So then it starts. Maybe someone could explain what presumptions does the following statement contradicts.</p>
<p>There is no continuous determination for complex logarithm in $\mathbb{C}$ \ ${0}$. </p>
<p>Second part is more practical. </p>
<p>We say that determination of complex logarithm is called principal if it given as a complement of $\mathbb{C}$ and semi axis of negative or zero reals $\Omega_{\pi}=\mathbb{C}$ \ $\{z\in \mathbb{C} : \Re(z) \leq 0\}$. Such as
$$ f(z) = \log(|z|)+ i\begin{cases} \arcsin{(y/|z|)} & x \geq 0, \\ \pi - \arcsin{(y/|z|)} & x \leq 0,\, y \geq 0, \\ -\pi - \arcsin{(y/|z|)} & x \leq 0,\, y \leq 0. \end{cases}$$</p>
<p>We can see that the argument belongs to $(-\pi,\pi]$, but I don't understand neither why a set without a negative reals define such argument neither how does it happens. After this there is an example saying that if we take $\Omega_0=\mathbb{C}$ \ $\{z \in \mathbb{C} : \Re \geq 0\}$ then the argument will be $(0,2\pi]$ but I also didn't get how this happens. What will happen if we take out non positive imaginary part? What kind of argument we will have then?</p>
<p>There is no such topic explained in wiki so maybe the deep answer will help others who encountered the same problem.</p>
| Gerry Myerson | 8,269 | <p>I think your first question is, why is there no continuous logarithm function defined in the complex plane, omitting $0$. The answer is, trace out a circle of radius $1$ centered at the origin, starting at $1$, and evaluate the logarithm function as you go around. Go ahead, do it! You will find that you either make a discontinuity somewhere along the way, or else when you get back to $z=1$ you are off by $2\pi i$ from the value you started with. </p>
<p>Now the same thing happens with the argument function. Start it with, say, the value $0$ at $z=1$, go around that circle (counterclockwise, just to be specific), and when you come back to $z=1$ the argument will be $2\pi$; discontinuity! The cure for the discontinuity is to make it impossible to go all the way around the circle. The easiest way to do that is to remove some ray that starts at the origin and goes to infinity. If the ray you remove is the non-positive reals, then the argument, starting with the value $0$ at $z=1$, will increase almost but not quite to $\pi$ as you go around counterclockwise (not reaching the excised ray), and will decrease almost but not quite to $-\pi$ as you go around the other way. </p>
<p>The non-positive imaginary part ray has argument $3\pi/2$ as you approach it counterclockwise from $z=1$, and argument $-\pi/2$ as you approach it the other way, so if you omit that ray then the argument can be taken to be between $-\pi/2$ and $3\pi/2$. </p>
<p>If I have missed the point of your question, please try again. </p>
|
3,346,543 | <p>I am aware this is a pretty big topic, but the attempts at layman's explanations I have seen either barely provide commentary on the formal proofs, or fail to provide an explanation (e.g "it gets too complex" does not really say anything)</p>
<p>Is there a good intuitive explanation as to why we fail to obtain a general solution for a 5th+ degree polynomial, and why this happens at the 5th degree and not below or above?</p>
| Dmitry Ezhov | 602,207 | <p>Formula for roots of equation <span class="math-container">$\displaystyle z^m-az^n-1=0$</span> with definite integration:</p>
<p><span class="math-container">$\displaystyle z_j=e^{2j\pi i/m}+\frac{1}{2\pi i}\left(e^{(2j+1)\pi i/m}\int_0^\infty log\left(1+a\frac{t^n}{1+t^m}e^{(2j+1)\pi in/m}\right)dt \\- e^{(2j-1)\pi i/m}\int_0^\infty log\left(1+a\frac{t^n}{1+t^m}e^{(2j-1)\pi in/m}\right)dt\right)$</span></p>
<p>where natural <span class="math-container">$m>n>0$</span>, <span class="math-container">$j=0,1,...m-1$</span> and <span class="math-container">$a$</span> is natural.</p>
<p>Based on paper <a href="http://www.mathnet.ru/links/f0548e79e3df810b5d3844c4bfdcc481/sm7494.pdf" rel="nofollow noreferrer">Лахтинъ, “Выраженiе корней трехчленнаго алгебраическаго уравненiя посредствомъ опредѣленныхъ интеграловъ” (1890)</a>.</p>
<p>Example calculation in <strong>pari/gp</strong>:</p>
<pre><code> a= 7;
m= 5; n= 2;
print("Quintic: z^5-"a"*z^2-1=0\n");
print("Galois group: "polgalois('z^5-a*'z^2-1)"\n");
print("Ordinary solution:\n"polroots('z^5-a*'z^2-1)"\n");
print("Not-ordinary solution:");
Z= [];
for(j=0, 4,
z= exp(2*j*Pi*I/m) + 1/(2*Pi*I)
*(exp((2*j+1)*Pi*I/m)*intnum(t=0, oo, log(1+a*t^n/(1+t^m)*exp((2*j+1)*Pi*I*n/m)))
- exp((2*j-1)*Pi*I/m)*intnum(t=0, oo, log(1+a*t^n/(1+t^m)*exp((2*j-1)*Pi*I*n/m))));
Z= concat(Z, [z])
);
print(Z)
</code></pre>
<p>Output:</p>
<pre><code>Quintic: z^5-7*z^2-1=0
Galois group: [120, -1, 1, "S5"]
Ordinary solution:
[1.9369100453804415363610723955778268241 + 0.E-38*I,
0.0014571193250340581699295533533713515578 - 0.37793919029580108279238966671308590927*I,
0.0014571193250340581699295533533713515578 + 0.37793919029580108279238966671308590927*I,
-0.96991214201525482635046575114228476362 - 1.6351464511815856113226711169592030156*I,
-0.96991214201525482635046575114228476362 + 1.6351464511815856113226711169592030156*I]~
Not-ordinary solution:
[1.9369100453804415363610723955778268026 + 0.E-39*I,
0.0014571193250034346617831783697118620865 + 0.37793919029582726259430943508563428674*I,
-0.96991214201522420284231937615862526337 + 1.6351464511815594315207513485866546421*I,
-0.96991214201522420284231937615862526336 - 1.6351464511815594315207513485866546421*I,
0.0014571193250034346617831783697118620703 - 0.37793919029582726259430943508563428672*I]
</code></pre>
|
3,150,645 | <p>Suppose <span class="math-container">$\dim V = \dim W$</span>. Given <span class="math-container">$\phi:V\rightarrow V$</span> and <span class="math-container">$\psi:W\rightarrow W$</span> when do we have bijective linear maps <span class="math-container">$a,b:V\rightarrow W$</span> such that
<span class="math-container">$$\require{AMScd}
\begin{CD}
V @>\phi>> V \\
@VaVV @VVbV \\
W @>\psi>>W
\end{CD}
$$</span>
commutes? </p>
<p>My intuition says that if the rank of <span class="math-container">$\phi$</span> and <span class="math-container">$\psi$</span> match there should exist <span class="math-container">$a,b$</span>. Of course with <span class="math-container">$a=b$</span>, this requires that <span class="math-container">$\phi,\psi$</span> are similar (after identifying <span class="math-container">$V$</span> and <span class="math-container">$W$</span> in any way you like), but with <span class="math-container">$a\neq b$</span> allowed, it seems this is much weaker. </p>
| user647486 | 647,486 | <p>Then that condition is just equivalence of <span class="math-container">$\phi$</span> and <span class="math-container">$\psi$</span>, which is equivalent to having the same rank. </p>
<p>To show this, take a basis <span class="math-container">$e_i^V$</span>, <span class="math-container">$i\in I_1$</span> of the kernel of <span class="math-container">$\phi$</span> and extend it to a basis <span class="math-container">$e_i^V$</span>, <span class="math-container">$i\in I_1\cup I_2$</span> of <span class="math-container">$V$</span>. Then, the elements of the basis that are not in the kernel are mapped to a basis <span class="math-container">$f_i^V$</span>, <span class="math-container">$i\in J_1$</span> of the range of <span class="math-container">$\phi$</span>. Let's assume that we index them such that <span class="math-container">$\phi(e_i^V)=f_i^V$</span> for <span class="math-container">$i\in I_1$</span>. Extend also this basis to a basis <span class="math-container">$f_i^V$</span>, <span class="math-container">$i\in J_1\cup J_2$</span> of <span class="math-container">$V$</span> (but the <span class="math-container">$V$</span> that is the codomain of <span class="math-container">$\phi$</span>). </p>
<p>Do the same business with <span class="math-container">$W$</span> and <span class="math-container">$\psi$</span> to obtain bases <span class="math-container">$e_i^W$</span>, <span class="math-container">$i\in I_1\cup I_2$</span> and <span class="math-container">$f_i^W$</span>, <span class="math-container">$i\in J_1\cup J_2$</span>. Assume that we also repeated the construction such that <span class="math-container">$\psi(e_i^W)=f_i^W$</span>, for <span class="math-container">$i\in I_1$</span>. Note that I used the same index sets, since the ranks are assumed equal. Therefore, the sets of indexes required have the same cardinality. So, we can just take them equal.</p>
<p>Finally define <span class="math-container">$a(e_i^V)=e_i^W$</span> and <span class="math-container">$b(f_i^V)=f_i^W$</span> and extend them by linearity.</p>
|
455,302 | <p>Given: $x + \sqrt x = \sqrt 3$<br>
Evaluate: $x^3 - 1 / x^3$</p>
| Matt Groff | 2,626 | <p>HINT: we can solve $x+\sqrt{x}=\sqrt{3}$, using the quadratic formula. Just use $\sqrt{x}$ as the variable.</p>
<p>From there, it's just plugging in those values for $x$ into the second equation.</p>
|
4,005,463 | <p>In the problem I am solving, table values are given for function and it says to find <span class="math-container">$f''(0.5)$</span> using second order central difference formula. I know the formula which is
<span class="math-container">$\frac{f(x+\triangle x)-2f(x)+f(x-\triangle x)}{{\triangle x}^2}$</span>. But the problem is, table values are as follows(I am listing values around <span class="math-container">$0.5$</span> only).</p>
<p><span class="math-container">$f(0.48)=1.336, f(0.5)=1.405,f(0.51)=1.481$</span></p>
<p>I am confused what value should I use as <span class="math-container">$\triangle x$</span>.</p>
| mathcounterexamples.net | 187,663 | <p><strong>Let's prove that <span class="math-container">$F$</span> is totally bounded</strong></p>
<p>Take <span class="math-container">$\epsilon \gt 0$</span>. As <span class="math-container">$\mathcal F$</span> is supposed to be equicontinuous, for all <span class="math-container">$x \in X$</span>, it exists an open neighborhood <span class="math-container">$\mathcal O_x$</span> such that <span class="math-container">$d(f(x),f(y)) \lt \epsilon/4$</span> for all <span class="math-container">$ f \in \mathcal F$</span> and all <span class="math-container">$y \in \mathcal O_x$</span>.</p>
<p>As <span class="math-container">$X$</span> is compact, we can extract from <span class="math-container">$\{\mathcal O_x \mid x \in X\}$</span> a finite cover <span class="math-container">$\mathcal O_{x_1}, \dots \mathcal O_{x_m}$</span>.</p>
<p>For all <span class="math-container">$i \in \{1, \dots, m\}$</span> <span class="math-container">$\mathcal F(x_i)$</span> is precompact and therefore so is their union <span class="math-container">$\mathcal U$</span> (a finite union of precompact subsets is precompact). Let <span class="math-container">$\{u_1, \dots, u_n\}$</span> be a finite subset of <span class="math-container">$\mathcal U$</span> such that <span class="math-container">$\mathcal U \subseteq \bigcup_{j=1}^n B(u_j, \epsilon/4)$</span> where <span class="math-container">$B(x, r)$</span> stands for the open ball centered on <span class="math-container">$x$</span> of radius equal to <span class="math-container">$r$</span>.</p>
<p>Now take <span class="math-container">$f(x) \in F$</span>. It exists <span class="math-container">$x_i$</span> such that <span class="math-container">$x \in \mathcal O_{x_i}$</span> and <span class="math-container">$u_j$</span> such that <span class="math-container">$f(x_i) \in B(u_j, \epsilon/4)$</span>. Therefore</p>
<p><span class="math-container">$$d(f(x),u_j) \le \underbrace{d(f(x),f(x_i))}_{x \in \mathcal O_{x_i}} + \underbrace{d(f(x_i),u_j)}_{f(x_i) \in B(u_j, \epsilon/4)} \le \epsilon/4 + \epsilon/4 \le \epsilon/2$$</span> which proves that <span class="math-container">$$F \subseteq \bigcup_{j=1}^n B(u_j, \epsilon/2)$$</span></p>
<p>As the diameter of each of the <span class="math-container">$B(u_j, \epsilon/2)$</span> is less or equal to <span class="math-container">$\epsilon$</span>, we can conclude that <span class="math-container">$\mathcal F$</span> is totally bounded and to the desired result.</p>
|
299,056 | <p>Let $\overline{\mathbb{Q}}$ be the algebraic closure of $\mathbb{Q}$. Let $K = \mathbb{Q}(\sqrt{d})$ and $\overline{K}$ defined to be the algebraic closure of $K$. Is it true that $\overline{K} \cong \overline{\mathbb{Q}}$?</p>
| Chris Eagle | 5,203 | <p>Algebraic closures are only defined up to isomorphism, so it doesn't make sense to ask if $\overline K$ and $\overline{\Bbb{Q}}$ are equal. It does make sense to ask if they are isomorphic, which they are: $\overline K$ is an algebraic extension of $\Bbb Q$ (since it's an algebraic extension of an algebraic extension) and is algebraically closed, and hence is an algebraic closure of $\Bbb Q$.</p>
|
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