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 |
|---|---|---|---|---|
129,261 | <p>I need to prove several inequalities trivially. (i.e. without using AM-GM, re-arrangement etc). I just keep hitting a blank. Could anyone help?</p>
<p>$$x^{4}+y^{4}+z^{4}\geq x^{2}yz+xy^{2}z+xyz^{2}$$</p>
| Rijul Saini | 27,729 | <p>I'm guessing Sum-of-Squares representation would amount to a 'trivial' proof, right? How about expanding the following (this further proves that your inequality holds for all reals, not just positive ones..): $$\sum_{cyclic}(x^2 - y^2)^2 + \sum_{cyclic}x^2(y-z)^2 \ge 0$$</p>
|
1,978,935 | <p>Let $f,g : [a,b] \rightarrow \mathbb{R}$ be continuous. We know that $f$ and $g$ have maximal values, as they are continuous on a closed interval. Let $M_f$ be the maximal value of $f$, and $M_g$ the maximal value of $g$. Show that if $M_f$ = $M_g$, then there exists $\psi \in [a,b]$ with $f(\psi) = g(\psi)$</p>
<p>Would it suffice to show that $\psi$ = maximal values, and show that this is an example which shows the exist of such a $\psi$?</p>
| ec92 | 34,552 | <p>Suppose $f(x_1) = M_f$ and $g(x_2) = M_f = M_g$. </p>
<p>If $x_1 = x_2$, you're done. </p>
<p>Otherwise, consider the interval $[x_1, x_2]$ (or $[x_2, x_1]$ if $x_2 < x_1$).
The function $f-g$ is continuous on this interval, it's nonnegative at $x_1$, and nonpositive at $x_2$. Thus there must be a point in the interval where $f-g$ is zero. This point is your $\psi$. </p>
|
136,264 | <p>I have a question concerning the stability analysis for a kind of differential equation taking the form
$$\dot x=Ax+Bw,$$
where $A\in \mathbb{R}^{n \times n}$, $B\in \mathbb{R}^{n \times m}$ are constant matrices and
$w \in \mathbb{R}^m$ is a normal random variable, i.e., $w\sim \mathcal{N}(0,W)$ with $W$ be a symmetric and positive definite matrix.</p>
<p>Since the zero is not an equilibrium of the system, the Lyapunov analysis does not make sense. When the input-to-state stability analysis is considered, the robust control theory does not apply due to the unboundness of $w$. By resorting to stochastic stability in the sense of mean square or almost surely, the Ito formula seems to be invalid.</p>
<p>HOW to carry out the stability analysis of this kind of systems? Any pointer will be helpful. Thanks!</p>
| Nick Gill | 801 | <p>The Australian Mathematical Society have produced a ranking:</p>
<p><a href="http://www.austms.org.au/Rankings/0101_AustMS_final_ranked.html">http://www.austms.org.au/Rankings/0101_AustMS_final_ranked.html</a></p>
<p>It is widely used (for instance, by my own institution in the UK).</p>
<p>When choosing where to submit I also make use of the following ranking of journal prices/ value:</p>
<p><a href="http://www.mathematik.uni-bielefeld.de/~rehmann/BIB/AMS/Price_per_Volume.html">http://www.mathematik.uni-bielefeld.de/~rehmann/BIB/AMS/Price_per_Volume.html</a></p>
<p>Of course it's a different type of ranking, but you might argue that it's a lot less subjective!</p>
|
164,629 | <p>Probably this is well known to those who know it.</p>
<p>Got an argument and numerical support that over
number fields elliptic curves in minimal models
might have unbounded number of integral points,
the number depending on the degree of the field.</p>
<p>Set $f(x)=x^3+ax+b$ and consider the curve
$E: y^2=f(x)$.</p>
<p>Chose $x_1 \ldots x_n$ such that $f(x_n)$ is
prime and work in $K=\mathbb{Q}[\sqrt{f(x_1)},\ldots\,\sqrt{f(x_n)}]$.</p>
<p>$E$ has the obvious $n$ points $(x_n,\sqrt{f(x_n)})$.</p>
<p>Experimentally for $f(x)=x^3-x+1$ over
$\mathbb{Q}[\sqrt{7},\sqrt{61},\sqrt{211},\sqrt{337},\sqrt{991}]$ the five points are linearly independent according to sage so the
rank is at least $5$.</p>
<p>Computing the absolute field is not efficient for me.</p>
<p>Over the rationals there is a conjecture relating
the number of integral points to the rank,
is there a similar conjecture for number fields?</p>
<p>Is there an example (with few primes) when in
this construction the points are linearly dependent?</p>
<p>The same argument works for higher genus.</p>
| JSE | 431 | <p>If the 5 points had a linear dependence, their coordinates could not generate a (Z/2Z)^5 - extension of Q. But they visibly do.</p>
|
93,063 | <p>There is a story I read about tiling the plane with convex pentagons.</p>
<p>You can read about it in this <a href="http://www.ivanrival.com/docs/picturepuzzling_2.pdf">article</a> on pages 1 and 2.</p>
<p>Summary of the story:
A guy showed in his doctorate work all classes of tiling the plane with convex pentagons, and proved that they are indeed all possible cases.
Later, riddle was published in popular science magazine to find all these classes. One of the readers found a tiling that did not belong to any of the classes, and so the claim and the proof turned out to be wrong.</p>
<hr>
<p>Reading this made me think about some questions.</p>
<p>Is it rare when a theorem was proved and the proof was published, and later it turned out that the theorem is wrong?</p>
<p>Can we somehow guess how many theorems out there that we think are right, but actually are wrong? I bet that if in our case, the theorem was about tiling in $R^3$ nobody would ever notice.</p>
<p>What can be the effect of such theorems on mathematics in general? Can it be a serious issue for mathematics even if the wrong theorems were not very important, but still, some stuff was based on them?</p>
| Zarrax | 3,035 | <p>I think Carl Mummert's answer is spot-on. One other thing worth mentioning is that more often than being wrong, or even having a recognizably flawed proof, are papers which are just really hard to understand. (This is sometimes compounded by language barriers... more than once I've had to deal with papers that were written in Russian, then either never translated or translated incoherently.) Some of these proofs are not complete, others are complete but hard to understand, some are mostly correct but have a few errors, and so on...</p>
<p>So people generally take a cautious approach. One good piece of advice my PhD adviser in grad school told me, never use a result whose proof you don't understand. Some people rely on proofs that they don't understand but that are accepted and many people understand, so they figure it's an acceptable risk. Others (I'm one of them) won't use any proof they don't understand. This may be more doable in some fields than others. So while there are wrong theorems out there, the effect is minimized by people recognizing this fact and proceeding with caution.</p>
|
735,101 | <p>Let $X_1$ and $X_2$ be independent random variables having the uniform density with $\alpha = 0$ and $\beta = 1$. Find expressions for the function $Y =X_1 + X_2$.</p>
<p>(a)$y \le 0$</p>
<p>(b)$0<y<1$</p>
<p>(c)$1<y<2$</p>
<p>(d)$y\ge2$</p>
<p>I'm thinking $f(x_1)=f(x_2) = 1$ for $0 \le x_1\le 1$ and $0 \le x_2\le 1$. I suppose this implies that $f(x_1,x_2) = 1$ so $F(x_1,x_2) = \int_0^y\int_0^{y-x_2} f(x_1,x_2) dx_1dx_2 $ Then I would need $\frac{d}{dy} F(x_1,x_2)$ to find an appropriate $f(y)$</p>
<p>I already have the answers to these in my book but I don't know how to come upon them as they don't correspond at all to what I have here. Where is my fault?</p>
| user111013 | 111,013 | <p>If you find $x'$ and $y'$ such that $47x' + 64y' = gcd(47,64)$ then solve $gcd(47,64)\cdot t + 70z = 1$. Then $x = x't$ and $y = y't$. This works since $gcd(47,64,70) = gcd(gcd(47,64),70)$.</p>
|
735,101 | <p>Let $X_1$ and $X_2$ be independent random variables having the uniform density with $\alpha = 0$ and $\beta = 1$. Find expressions for the function $Y =X_1 + X_2$.</p>
<p>(a)$y \le 0$</p>
<p>(b)$0<y<1$</p>
<p>(c)$1<y<2$</p>
<p>(d)$y\ge2$</p>
<p>I'm thinking $f(x_1)=f(x_2) = 1$ for $0 \le x_1\le 1$ and $0 \le x_2\le 1$. I suppose this implies that $f(x_1,x_2) = 1$ so $F(x_1,x_2) = \int_0^y\int_0^{y-x_2} f(x_1,x_2) dx_1dx_2 $ Then I would need $\frac{d}{dy} F(x_1,x_2)$ to find an appropriate $f(y)$</p>
<p>I already have the answers to these in my book but I don't know how to come upon them as they don't correspond at all to what I have here. Where is my fault?</p>
| Rodney Coleman | 73,128 | <p>Notice that $gcd(x,y,z)=gcd(x,gcd(y,z))$. First we find $a$, $b$ such that $gcd(x,gcd(y,z))=ax+bgcd(y,z)$, then $c$, $d$ such that $gcd(y,z)=cy+dz$. Finally we obtain $gcd(x,y,z)=ax+bcy+bdz$. </p>
|
2,180,398 | <p>I am not sure if I got the correct answers to these basic probability questions.</p>
<blockquote>
<p>You and I play a die rolling game, with a fair die. The die is equally likely to land on any of its $6$ faces. We take turns rolling the die, as follows. </p>
<p>At each round, the player rolling the die wins (and the game stops) if (s)he rolls either "$3$", "$4$","$5$", or "$6$". Otherwise, the next player has a chance to roll. </p>
<ol>
<li>Suppose I roll 1st. What is the probability that I win in the 1st round? </li>
<li>Suppose I roll in the 1st round, and we play the game for at most three rounds. What is the probability that<br>
a. I win?<br>
b. You win?<br>
c. No one wins? (i.e., I roll "$1$" or "$2$" in the 3rd round?)</li>
</ol>
</blockquote>
<p>My answers:</p>
<ol>
<li>$4/6$</li>
<li>a. I think it is $(4/6)^2$ . Can someone explain why it isn't $4/6 + 4/6$ ?<br>
b. I think this is $4/6$.<br>
c. I think it is $(2/6)^3$</li>
</ol>
| N. F. Taussig | 173,070 | <blockquote>
<p>What is the probability that the player who rolls first wins in the first round?</p>
</blockquote>
<p>Your answer of $4/6 = 2/3$ is correct.</p>
<blockquote>
<p>What is the probability that the player who rolls first wins the game if the game lasts at most three rounds?</p>
</blockquote>
<p>The player who rolls first if he or she wins during the first round, second round, or third round. We know that the probability that the player who rolls first has probability $2/3$ of winning in the first round. </p>
<p>For the player who rolls first to win in the second round, both players must roll a 1 or 2 in the first round, then the first player must roll a 3, 4, 5, or 6 in the second round. The probability that this event occurs is
$$\frac{2}{6} \cdot \frac{2}{6} \cdot \frac{4}{6} = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{27}$$</p>
<p>For the first player to win in the third round, both players must roll a 1 or 2 for the first two rounds, then the first player must roll a 3, 4, 5, or 6 in the third round. The probability that this occurs is
$$\frac{2}{6} \cdot \frac{2}{6} \cdot \frac{2}{6} \cdot \frac{2}{6} \cdot \frac{4}{6} = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{243}$$</p>
<p>Since these three events are mutually exclusive, the probability that the player who rolls first wins the game is
$$\frac{2}{3} + \frac{2}{27} + \frac{2}{243} = \frac{162}{243} + \frac{18}{243} + \frac{2}{243} = \frac{182}{243}$$</p>
<p>The probability that the first player wins cannot be
$$\frac{4}{6} + \frac{4}{6} = \frac{8}{6} = \frac{4}{3} > 1$$
since it is impossible for a probability to exceed $1$.</p>
<blockquote>
<p>What is the probability that the player who rolls second wins the game if the game lasts at most three rounds.</p>
</blockquote>
<p>The player who rolls second can win in the first round if the first player rolls a 1 or 2, then the second player rolls a 3, 4, 5, or 6. The probability that this event occurs is
$$\frac{2}{6} \cdot \frac{4}{6} = \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{9}$$
For the second player to win in the second round, the first and second players must both roll a 1 or 2 in the first round, the first player must roll a 1 or 2 in the second round, then the second player must roll a 3, 4, 5, or 6 in the second round. The probability that this event occurs is
$$\frac{2}{6} \cdot \frac{2}{6} \cdot \frac{2}{6} \cdot \frac{4}{6} = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{81}$$
For the second player to win in the third round, both players must roll a 1 or 2 in the first two rounds, the first player must roll a 1 or 2 in the third round, then the second player must roll a 3, 4, 5, or 6 in the third round. The probability that this event occurs is
$$\frac{2}{6} \cdot \frac{2}{6} \cdot \frac{2}{6} \cdot \cdot \frac{2}{6} \cdot \frac{2}{6} \cdot \frac{4}{6} = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{729}$$
Since these three events are mutually exclusive, the probability that the second player wins is
$$\frac{2}{9} + \frac{2}{81} + \frac{2}{729} = \frac{162}{729} + \frac{18}{729} + \frac{2}{729} = \frac{182}{729}$$</p>
<blockquote>
<p>What is the probability that neither player wins a game that lasts at most three rounds?</p>
</blockquote>
<p>For this event to occur, both players must roll a 1 or 2 in all three rounds. The probability that this event occurs is
$$\left(\frac{2}{6}\right)^6 = \left(\frac{1}{3}\right)^6 = \frac{1}{729}$$</p>
<p><strong>Check:</strong> There are three possibilities: The first player wins, the second player wins, or neither player wins. Therefore, the probabilities of these three events should add up to $1$. Observe that
$$\frac{182}{243} + \frac{182}{729} + \frac{1}{729} = \frac{546}{729} + \frac{182}{729} + \frac{1}{729} = 1$$</p>
|
1,921,879 | <blockquote>
<p>Find all positive integers $n$ for which $\dfrac{x^n + y^n + z^n}2$ is a perfect square, whenever $x$, $y$, and $z$ are integers such that $x + y + z = 0$.</p>
</blockquote>
<p>I don't even know where to start.</p>
| ShakesBeer | 168,631 | <p>Leading on from Dietrich's answer, let's take $(x,y,z)=(1,1,-2)$ and consider even $n$ of the form $n=2k$, $k \geq 2$.</p>
<p>$$\frac{x^n+y^n+z^n}{2}=\frac{2+(-2)^n}{2}=1+2^{2k-1}$$</p>
<p>Suppose $1+2^{2k-1}$ is a perfect square. It is odd, so </p>
<p>$$1+2^{2k-1}=(2l+1)^2$$</p>
<p>$$\iff 2^{2k-1}=4l^2+4l$$</p>
<p>$$\iff 2^{2k-3}=l(l+1)$$</p>
<p>Since either $l$ or $l+1$ is odd, and $2^{2k-1}$ contains only even factors, it must be that $l=1$, $l+1=2$ and $k=2$.</p>
<p>i.e. the only cases are $n=0,4$ found by Dietrich above, who did all the hard work.</p>
|
541,890 | <p>Prove or disprove:
For all integers $m$ and $n$, if $m+n$ is even then so is $m-n$.
Would you just set them even to each other because you are given $m+n$ is even?</p>
| user103074 | 103,074 | <p>The problem is when you go from 1 to 2 persons.</p>
<p>Assuming S(K) valid for k=1: that in any group of 1 human being, everybody has the same hair colour</p>
<p>Now you should be able to prove the validity of S(K+1): the equivalent to any group of 2 human beings.</p>
<p>Ok. Let's try. If you have any group of 2 human beings in a room, then you ask one of them to go out. The one who stayed is blond, for example. When the first person returns, then you ask to the blond one (who stayed) to go out now, and apply the hipothesis: being a group of 1 human being, "all" should have the same hair colour, ANY COLOUR, for example, brown hair.</p>
|
656,458 | <p>If $A$ is an $n\times n$-matrix, $A^H$ is a Hermitian Matrix and $A^S$ is a Skew Hermitian, show $A=A^H+A^S$.</p>
<p>I am having trouble working with these so far and really cannot find many characteristics except the definitions. A Hermitian is made up of reals on the diagonal and is $A^*=A$. It is skew hermitian iff the diagonal is made up of imaginary numbers and zero and $A^H =\bar A^T$</p>
| Kathleen | 89,461 | <p>If $A$ is an $n\times n$-matrix, $A^H$ is a Hermitian Matrix and $A^S$ is a Skew Hermitian, show $A=A^H+A^S$.</p>
<p>$A^H=1/2(A+A*)$</p>
<p>$A^S=1/2(A-A*)$</p>
<p>$A^H+A^S=1/2(A+A*)+1/2(A-A*)$</p>
<p>$=1/2A+1/2A*+1/2A-1/2A*$</p>
<p>$=A$ </p>
<p>QED</p>
|
924,555 | <p>My homework question:</p>
<blockquote>
<p>From the order axioms for $\mathbb{R}$, show that $0 < 1$. [<em>Hint:</em> From the field axioms, $0 \not=1$. By the trichotomy property, either $0<1$ or $4<0$. Assuming $1 < 0$, get $0 < -1$. Now use Exercise 4.]</p>
</blockquote>
<p>Exercise 4 from my textbook problems states:</p>
<p>"From the order axioms for $\mathbb{R}$, show that the set of positive real numbers, {$x \in \mathbb{R} : x > 0$}, is closed under addition and multiplication."</p>
<p>How am I expected to use Exercise 4 as directed?</p>
| Avrham Aton | 171,799 | <p>By contradiction assume $1<0$ then $0<-1 $ ,so $ -1 $ is a positive number so(by the fact that positive reals is closed under multiplication) $(-1)(-1) $must also be positive , but by assumption 1 is not positive . contradiction</p>
|
31,100 | <p>I have just begun to learn about the fundamental group.
An exercise asks me to prove that $$X=\{(x,y,z): z \ge 0\}-\{(x,y,z): y=0,0\leq z \leq 1\}$$ has trivial fundamental group.</p>
<p>What I know is:</p>
<p>1) the definition of the fundamental group.</p>
<p>2) X has trivial fundamental group iff any loop in X can be shrunk into a constant loop at the base point.</p>
<p>3) Homeomorphic (path-connected) spaces have isomorphic fundamental groups.</p>
<p>4) Any convex subset of $\mathbb{E}^n$ and $S^m,m\ge 2$ has trivial fundamental group.</p>
<p>I tried to construct a homeomorphism from X to a convex subset of $\mathbb{E}^3$ such as an area like this:
$$\{(x,y,z): -1\leq y \leq 1,z>0\}$$
But I failed.</p>
<p>Can you please help me? Thank you!</p>
| Sam Nead | 1,307 | <p>Suggestion - break $H$ (your set) into two pieces using the half plane $P = \{ (x,y,z) : y = 0, z > 1 \}$. Show that the two pieces $K^\pm$ of $H - P$ each have trivial fundamental group, as does $P$. Now assemble. </p>
<p>Two more suggestions: 1. If you want to talk about stuff, it helps to give the stuff names. 2. Look up the Seifert-van Kampen theorem. </p>
|
1,657,664 | <p>Struggling with a homework problem here and can't understand logically which one would be correct (each has different truth tables). I need to express the following statement using quantifiers, variables, and the predicates M(s), C(s), and E(s) </p>
<blockquote>
<p>"No computer science students are engineering students" </p>
</blockquote>
<p>D = set of all students</p>
<p>C(s) = "s is a computer science major"</p>
<p>E(s) = "s is an engineering student" </p>
<p>So I'm stuck between,</p>
<p>$\forall s \in D, C(s) \implies \lnot E(s)$</p>
<p>-OR-</p>
<p>$\forall s \in D, \lnot C(s) \land E(s)$ </p>
| Graham Kemp | 135,106 | <blockquote>
<p>"No computer science students are engineering students" </p>
</blockquote>
<ul>
<li>$\neg \exists s {\in} D ~(C(s)\wedge E(s))$ "there is not a student who is both".</li>
<li>$\forall s {\in} D ~(\neg C(s)\vee \neg E(s))$ "all students are either not comsci. or not eng. student"</li>
<li>$\forall s {\in} D ~(C(s)\to \neg E(s))$ "all comsci. students are not eng. students"</li>
<li>$\forall s {\in} D ~(E(s)\to \neg C(s))$ "all eng. students are not comsci. students"</li>
</ul>
|
1,515,775 | <p>So, everyone knows the famous <a href="https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem" rel="nofollow">Lagrange's four-square theorem</a>, which states, that every positive integer can be written down as the sum of $4$ square numbers. Since $4=2^2$, and $2$ represents the square numbers, could this be stated for bigger numbers too? For example, $8=2^3$, so we could state, that every positive integer can be written down as the sum of $8$ cube numbers? I tried to find a counter-example for this statement, but didn't have any success.</p>
<p>What do you think about this idea? Can you tell me a counter example or a problem with the thinking? Thanks!</p>
| Kieren MacMillan | 93,271 | <p>Because a square is both a <a href="https://en.wikipedia.org/wiki/Polygonal_number" rel="nofollow noreferrer">polygonal number</a> and a <a href="https://en.wikipedia.org/wiki/Perfect_power" rel="nofollow noreferrer">perfect power</a>, <a href="https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem" rel="nofollow noreferrer">Lagrange’s four-square theorem</a> can be generalized/extended in several ways.</p>
<p>First, it is the case $n=4$ of the <a href="https://en.wikipedia.org/wiki/Fermat_polygonal_number_theorem" rel="nofollow noreferrer">Fermat polygonal number theorem</a>, first stated without proof by Fermat (in 1638), and first proven by Cauchy (in 1813):</p>
<blockquote>
<p><strong>Theorem</strong>: Every positive integer is a sum of at most $n$ $n$-gonal numbers — <em>i.e.</em>, every positive integer can be written as the sum of three or fewer triangular numbers, and as the sum of four or fewer square numbers, and as the sum of five or fewer pentagonal numbers, and so on.</p>
</blockquote>
<p>That theorem in turn can likely be generalized/extended to <a href="https://en.wikipedia.org/wiki/Figurate_number" rel="nofollow noreferrer">figurate numbers</a>, as conjectured by Pollock (in 1850).</p>
<p>The second obvious way to generalize/extend Lagrange’s theorem is the one you’ve described, known as <a href="https://en.wikipedia.org/wiki/Waring%27s_problem" rel="nofollow noreferrer">Waring’s Problem</a>, first proposed by Waring (in 1770).</p>
<blockquote>
<p><strong>Question</strong>: For each natural number $k$, is there an associated positive integer $s$ such that every natural number is the sum of the $k$th powers of at most $s$ natural numbers?</p>
</blockquote>
<p>It is currently an open area of research in additive number theory.</p>
|
461 | <p>There is a function on $\mathbb{Z}/2\mathbb{Z}$-cohomology called <em>Steenrod squaring</em>: $Sq^i:H^k(X,\mathbb{Z}/2\mathbb{Z}) \to H^{k+i}(X,\mathbb{Z}/2\mathbb{Z})$. (Coefficient group suppressed from here on out.) Its notable axiom (besides things like naturality), and the reason for its name, is that if $a\in H^k(X)$, then $Sq^k(a)=a \cup a \in H^{2k}(X)$ (this is the cup product). A particularly interesting application which I've come across is that, for a vector bundle $E$, the $i^{th}$ Stiefel-Whitney class is given by $w_i(E)=\phi^{-1} \circ Sq^i \circ \phi(1)$, where $\phi$ is the Thom isomorphism.</p>
<p>I haven't found much more than an axiomatic characterization for these squaring maps, and I'm having trouble getting a real grip on what they're doing. I've been told that $Sq^1$ corresponds to the "Bockstein homomorphism" of the exact sequence $0 \to \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$. Explicitly, if we denote by $C$ the chain group of the space $X$, we apply the exact covariant functor $Hom(C,-)$ to this short exact sequence, take cohomology, then the connecting homomorphisms $H^i(X)\to H^i(X)$ are exactly $Sq^1$. This is nice, but still somewhat mysterious to me. Does anyone have any good ideas or references for how to think about these maps?</p>
| Elizabeth S. Q. Goodman | 1,198 | <p>I second the references to Hatcher and to Mosher & Tangora, though you can also find Steenrod's original paper. At least the first two of those start out by listing the various axioms of Steenrod squares and then construct them.</p>
<p>The reason to axiomatize the properties of Steenrod squares is that it is hard to understand the relation of different constructions to each other (I spent a semester struggling with that), and the construction is not the point. Steenrod's original construction, calculating on simplices, has the same properties as Hatcher's CW-complex construction (note above), but the former is hard to prove things with and the latter is hard to visualize. It's nice to get a grip on one construction or other but ultimately the axioms themselves and the proof that they uniquely determine the maps becomes more practical.</p>
<p>All that said, I like Hatcher's construction because it uses the fact that elements of $H^n(X; G)$ are the same as homotopy classes of maps from X to the Eilenberg-Maclane space $K(G, n)$, a connected CW-complex whose nth homotopy group is G and others are trivial. Then you can understand any cohomology operation $H^n(-; G)\to H^m(-;H)$ as a map on two spaces, $K(G,n)\to K(H,m)$. [You can make a $K(G,n)$ starting with $n$-cells of trivial boundary, attaching $n+1$-cells to make the right relations, and then continually attaching cells to kill the higher homotopy groups. $K(\mathbb{Z},1)=S^1$ doesn't need any higher cells, but $\mathbb{RP}^\infty$ is the usual $K(\mathbb{Z}/2, 1$). Sorry if you already know this.] This makes it easy to see why cohomology operations on the same group have to increase dimension, and it gives you an actual space to do Steenrod squares in. </p>
<p>I believe the following has been said above, but in brief: One reason we like Steenrod squares is that they commute with suspension, but suspension kills cup products (even though $Sq^n$ is the cup product square on $H^n(X)$, after suspension the cup product square would be $Sq^{n+1}$ but $Sq^n$ is still defined). It's also awesome that the sum $Sq^*$ of the various maps forms a ring operation of the cohomology ring. It's a good idea to look at $\mathbb{RP}^\infty$ and do Steenrod squares on that, and to think about the Steenrod algebra.</p>
|
461 | <p>There is a function on $\mathbb{Z}/2\mathbb{Z}$-cohomology called <em>Steenrod squaring</em>: $Sq^i:H^k(X,\mathbb{Z}/2\mathbb{Z}) \to H^{k+i}(X,\mathbb{Z}/2\mathbb{Z})$. (Coefficient group suppressed from here on out.) Its notable axiom (besides things like naturality), and the reason for its name, is that if $a\in H^k(X)$, then $Sq^k(a)=a \cup a \in H^{2k}(X)$ (this is the cup product). A particularly interesting application which I've come across is that, for a vector bundle $E$, the $i^{th}$ Stiefel-Whitney class is given by $w_i(E)=\phi^{-1} \circ Sq^i \circ \phi(1)$, where $\phi$ is the Thom isomorphism.</p>
<p>I haven't found much more than an axiomatic characterization for these squaring maps, and I'm having trouble getting a real grip on what they're doing. I've been told that $Sq^1$ corresponds to the "Bockstein homomorphism" of the exact sequence $0 \to \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$. Explicitly, if we denote by $C$ the chain group of the space $X$, we apply the exact covariant functor $Hom(C,-)$ to this short exact sequence, take cohomology, then the connecting homomorphisms $H^i(X)\to H^i(X)$ are exactly $Sq^1$. This is nice, but still somewhat mysterious to me. Does anyone have any good ideas or references for how to think about these maps?</p>
| Robert Bruner | 6,872 | <p>If I interpret the request a bit differently, I would say that the Steenrod operations in the cohomology of a spectrum tell you about the attachments of the cells. If $Sq^1 x = y$, then a cell dual to $y$ is attached by a map of degree 2 mod 4 to a cell dual to $x$. Similarly, $Sq^2 x = y$ tells us the attaching map is $\eta$, $Sq^4$ detects $\nu$ and $Sq^8$ detects $\sigma$. This doesn't go very far, but may help with the need to 'get a real grip on what they're doing'.</p>
<p>Next, let's assume you're really interested in homotopy, not just (co)homology. A class dual to a homology class in the image of the Hurewicz homomorphism must be indecomposable under the action of the Steenrod algebra, by naturality w.r.t. the map $S^n \longrightarrow X$. This limits the homotopy of $X$ which can be detected by the homomorpism $\pi_* X \longrightarrow H_* X$: the homomorphism $H^* X \longrightarrow H^* S^n$ can only map indecomposables non-trivially, since all classes in degrees below $n$ must go to $0$.</p>
<p>Then there are the relations. The fact that $Sq^n$ is decomposable when n is not a power of two tells us that if $y = Sq^n x$, there must be other classes between $x$ and $y$. EG, $Sq^3 x = y \neq 0$ tells us that $Sq^2 x \neq 0$ also, since $Sq^3 = Sq^1 Sq^2$. So our spectrum can't have just two cells, dual to $x$ and $y$, but must have a three cell subquotient with top cell attached by 2 (mod 4) to a cell attached by $\eta$ to the bottom cell.</p>
<p>Or, if $Sq^2 Sq^2 x = y \neq 0$ then we must also have nonzero classes $Sq^1 x$ and $Sq^2 Sq^1 x$, since $Sq^2 Sq^2 = Sq^1 Sq^2 Sq^1$, and vice versa, if $Sq^1 Sq^2 Sq^1 x = y \neq 0$ then $Sq^2 x \neq 0$ as well. This leads to an easy proof that the mod 2 Moore spectrum $M$ isn't a ring spectrum, since $2 \pi_0M = 0$ but $\pi_2 M = Z/4$, by looking at the obstruction to attaching the top cell of a putative spectrum with nonzero cohomology spanned by $x$, $Sq^1 x$, $Sq^2 Sq^1 x$, and $Sq^1 Sq^2 Sq^1 x$. More, the fact that you can only add such a top cell if you also have a class $Sq^2 x$ so that the top cell can be attached by the sum of $Sq^1$ on $Sq^2 Sq^1 x$ and $Sq^2$ on $Sq^2 x$ shows that $\eta^2$ (corresponding to the path $Sq^2 Sq^2$ from bottom to top, must lie in the Toda bracket $\langle 2, \eta, 2\rangle$, corresponding to the path $Sq^1$, $Sq^2$, $Sq^1$ from bottom to top. </p>
<p>Similarly, $y = Sq^1 Sq^2 x$ tells us that homotopy supported on a cell dual to $x$ can be acted on by $\text{v}_1$ to get $y$, literally if we have a $ku$-module and multiply by $\text{v}_1 \in ku_2$, or as the Toda bracket $\langle 2, \eta, -\rangle$ more generally. The key fact here is that $\text{v}_1 \in ku_2$ is in $\langle 2, \eta, 1_{ku} \rangle$, where $1_{ku} : S \longrightarrow ku$ is the unit. </p>
<p>Likewise, $Sq^2 Sq^1 Sq^2 x = y$ corresponds to multiplication by the generator of $ko_4$, literally for $ko$-modules, or as a bracket $\langle \eta, 2, \eta, - \rangle$ more generally. Here you have to be in a situation where $2 \nu = 0$ to form the bracket, since $\langle \eta, 2, \eta \rangle = \{ 2\nu, 6 \nu\}$. This hints that the role of $\nu$ is non-trivial in real K-theory, despite going to $0$ under the homomorphism $\pi_* S \longrightarrow \pi_* ko$ and despite the cohomology of $ko$ being induced up from the subalgebra $A(1)$ generated by $Sq^1$ and $Sq^2$. The Adem relation $Sq^2 Sq^1 Sq^2 = Sq^1 Sq^4 + Sq^4 Sq^1$ shows that $Sq^4$ must act nontrivially if $Sq^2 Sq^1 Sq^2$ does. Also, the fact that $A(1)//A(0)$ is spanned by $1$, $Sq^2$, $Sq^1 Sq^2$, and $Sq^2Sq^1Sq^2$ tells us (with a bit more work) that we can build $HZ$ as a four cell $ko$-module. </p>
<p>A good way to organize all this information is the Adams spectral sequence, which tells you that the mod $p$ cohomology of $X$ gives a decent first approximation, $\text{Ext}_{A}(H^*X,F_p)$, to the homotopy of the $p$-completion of $X$.</p>
|
388,766 | <p>I need to show that every element in $\Bbb Z/p\Bbb Z$ can be written as a sum of two squares. The case $p=2$ is trivial and $0$ is always $0^2 + 0^2$. So all I have to do is show that every element of $(\Bbb Z/p\Bbb Z)^\times$ (the group of units) can be expressed as a sum of two squares. The question hints that I should consider the set of elements of $(\Bbb Z/p\Bbb Z)^\times$ expressible as a sum of two squares.</p>
<p>Associativity is trivial, the identity element $1 = 1^2 + 0^2$ exists.
Closure holds by the identity $(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2$.
I don't know how to show that inverses exist.</p>
| Community | -1 | <p>For finite groups, to show that a subset is a subgroup if you check closure and non-emptyness you don't need to check inverses because each element has finite order: $g^{-1} = g^{o(g) - 1}$ is in the subset by closure assumption.</p>
|
3,974,394 | <p>I am relatively new to isomorphisms and I don't understand how <span class="math-container">$\varphi$</span> is surjective in this proof. I have searched online, but I still don't understand. If anyone could straight up tell me because I feel like I'm being a bit dumb.</p>
<p><a href="https://i.stack.imgur.com/oIB4A.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oIB4A.png" alt="enter image description here" /></a></p>
| azif00 | 680,927 | <p>A function <span class="math-container">$f : A \to B$</span> is <em>surjective</em> if every element of <span class="math-container">$B$</span> can be written as <span class="math-container">$f(\cdot)$</span>, that is, for any <span class="math-container">$b \in B$</span> there is <span class="math-container">$a \in A$</span> such that <span class="math-container">$b = f(a)$</span>. In your case, the function <span class="math-container">$\varphi : G \to G_1 \times \cdots \times G_n$</span> is surjective since for any <span class="math-container">$(g_1,\dots,g_n) \in G_1 \times \cdots \times G_n$</span> (this is how looks a generic element of <span class="math-container">$G_1 \times \cdots \times G_n$</span>) we have that <span class="math-container">$$(g_1,\dots,g_n) = \varphi(g_1 \cdots g_n).$$</span></p>
|
544,779 | <p>So here are the contextual statements:
1) Maya either listens to music or does her homework. If she listens to music she feels happy.If she does her homework she feels unhappy. Therefore she will not do her homework while listening to music.</p>
<p>Let P be the statement "Maya listens to Music".
Q "Maya does homework".
R "Maya feels happy".
So am I right to write it as ((P=>Q)∨(Q=>¬R) ) => ( ¬Q∧¬P ) </p>
<p>2) If I drink coffee, then I will get my assignment done on time. If I do not drink coffee, then I will feel sleepy. If I feel sleepy, then I will make mistakes. Therefore, if I will not get the assignment done on time, then I will make mistakes.</p>
<p>Let P be the statement "I drink coffee".
Q "I get the assignment done on time".
R "I will feel sleepy".
S "I will make mistakes".</p>
<p>( (P=>Q)∧(¬P=>R)∧(R=>S) ) =>( ¬Q=>S ) . Is it right?</p>
<p>In order to examine whether the arguments are right do I really need to do the truth tables..? It will be a huge one for the second statement.</p>
| Henno Brandsma | 4,280 | <p>If you start with a Hausdorff space $X$ and want to consider a compactification, which by definition is a pair $(Y, f)$ where $Y$ is a compact space (let's be the most general and not require Hausdorff) and $f: X \to Y$ is an embedding and $f[X]$ is dense in $Y$.</p>
<p>Trivial example: if $X$ is already compact (and not even Hausdorff) then $(X, 1_X)$, where $1_X$ is the identity on $X$, is a compactification. </p>
<p>Also, for any space $X$, define $Y = X \cup \{p\}$, with $p \notin X$, with the topology defined by $\mathcal{T}_X \cup\{{Y}\}$ (the original topology on $X$ with just the whole set $Y$ added, which indeed is a topology on $Y$, as is easily checked), and let $f(x) = x$ be the obvious map from $X$ into $Y$. Then $f$ is an embedding, the only neighborhood of $p$ is $Y$ and this implies both that $f[X]$ is dense in $Y$ and that $Y$ is compact as well. In fact we can add more than just one point, we can add as many as we like... The resulting spaces are all compact for trivial reasons, and even if we start with a nice metrisable space, the compactifications are at most $T_1$ and never $T_2$. So allowing <em>all</em> spaces $Y$ allows for all sorts of trivialities. </p>
<p>This is the reason that most texts assume that $Y$ is Hausdorff as well, and so even $T_4$ by standard results. Note then that $X$, being homeomorphic to $f[X] \subset Y$ is at least required to be Tychonoff (a.k.a. $T_{3\frac{1}{2}}$) itself in that case, and because every Tychonoff space embeds into some product $[0,1]^I$, it always has a compactification (take $Y$ the closure of the embedded copy of $X$, essentially). So allowing only Hausdorff only compactifications, we get that we restrict to Tychonoff spaces, all of which actually have Tychonoff (even normal) compactifications, so we stay inside the same class of spaces, as opposed to the trivial example spaces above.</p>
<p>The one-point compactification case then becomes a bit clearer too: if we just take any space $X$ and call $(Y,f)$ a one-point compactification when $Y \setminus f[X]$ has one point, then a (even Hausdorff locally compact) space $X$ can have more then one non-homeomorphic one-point compactification: $\mathbb{R}$ has the circle, as usual, and its trivial extension from above, and more as well. </p>
<p>If $X$ is Hausdorff and we assume that all compactifications are Hausdorff, we get as a theorem that a one-point Hausdorff compactification $Y$ exists iff $X$ is locally compact and then it is essentially unique as well (as all such $Y$ must be homeomorphic).</p>
<p>If $X$ is just Hausdorff, we can define a space $Y = X \cup \{p\}$ (where $p \notin X$) with
the topology $$\mathcal{T}_X \cup \left\{ \{p\} \cup (X \setminus K): K \mbox{ compact and closed in } X \right\}$$ and this space is such that the $Y$ is indeed compact but it needs not be Hausdorff: in fact this space is Hausdorff only if $X$ was locally compact to begin with. If however $X$ is locally compact, it is the same unique Hausdorff compactification as before, so this space is the natural generalisation of the Hausdorff compactification. But as said, it's not unique outside of the Hausdorff setting, though it is IMHO the most natural one. </p>
|
184,219 | <p>I know that it is the standard functionality of <code>Merge</code> to combine the values of the same keys among associations.</p>
<p>Now I would like to deal with a situation in which, in my associations, the keys are strings (English words). And I want to define the sameness as two words having the same result from <code>WordStem</code> so that "effects" and "effect" are the same key.</p>
<p>So is it possible for <code>Merge</code> to accept such a same test function for the keys, e.g., <code>Equal@@WordStem[{##}]&</code>, to determine what keys should be considered the same and be merged?</p>
<hr>
<p>A MWE goes as below:</p>
<pre><code>mergeFunc = x \[Function] {Total[#], Union[Flatten@#2]} & @@ (x\[Transpose]);
Merge[{<|"effect" -> {5, {2, 3}}|>, <|"effects" -> {4, {1, 3, 5}}|>}, mergeFunc]
Merge[{<|"effect" -> {5, {2, 3}}|>, <|"effect" -> {4, {1, 3, 5}}|>}, mergeFunc]
</code></pre>
<blockquote>
<pre><code><|"effect" -> {5, {2, 3}}, "effects" -> {4, {1, 3, 5}}|>
<|"effect" -> {9, {1, 2, 3, 5}}|>
</code></pre>
</blockquote>
<p>The second result is desired.</p>
<hr>
<p><strong>Additional requirement</strong>: no new keys should be introduced.</p>
<p>For example, what if I now have "apple" and "apples" as the keys instead of "effect" and "effects"? </p>
<p>That means I need the word stems as the criterion but not as the keys in the merged result.</p>
| Henrik Schumacher | 38,178 | <p>What about this?</p>
<pre><code>data = <|"effect" -> {5, {2, 3}}, "effects" -> {4, {1, 3, 5}}|>;
Merge[
KeyValueMap[{key, value} \[Function] WordStem[key] -> value, data],
mergeFunc
]
</code></pre>
<blockquote>
<p><|"effect" -> {9, {1, 2, 3, 5}}|></p>
</blockquote>
|
1,449,776 | <p>I have always known that $a^n=a*a*a*.....$(n times)</p>
<p>Then what exactly is the meaning if $a^0$ and why will it be equal to $1$?</p>
<p>I have checked it in the internet but everywhere the solution is based on the principle that $a^m*a^n=a^{m+n}$ and when $n=0$ it will be $a^m$ and clearly $a^0$ is equal to $1$. </p>
<p>But what exactly does $a^0$ mean does it mean $a*a*a*...$(zero times)?</p>
<p>Any help is highly appreciated.</p>
| Paul G | 273,789 | <p>The definition you give only applies when <em>n</em> is an integer greater than zero. It does not apply to <em>n</em> = 0 or negative <em>n</em> because it doesn't make sense to talk about multiplying together zero <em>a</em>'s or a negative number of <em>a</em>'s.</p>
<p>However, using the principle that you mention, we can extend this definition to zero and negative numbers. This allows us to give a meaning to <em>a</em> to the power of zero or negative exponents. That meaning is no longer "a certain number of <em>a</em>'s multiplied together" but something that works consistently with the definition we already have for positive <em>n</em>.</p>
|
1,217,557 | <p>I was tasked with drawing the contour lines of $ z = \sqrt{xy} $, which I find a bit problematic since I can see no way in which one can plot (by hand, and not with wolfram and others....) the $ z = \sqrt{xy} $ graph in $R^2( x-, y- $ projection} to begin with for this surface...</p>
<p>How can one draw this contour graph manually? </p>
| Community | -1 | <p>That function is what's called AFFINE (<a href="http://en.wikipedia.org/wiki/Affine_transformation" rel="nofollow">http://en.wikipedia.org/wiki/Affine_transformation</a>), it's not linear.</p>
|
707,193 | <p>The inequality to solve:
$$\left[\frac{-K^2+13K+44}{14-K}\right] > 0$$</p>
<p>How do I solve this?
I tried this:
$$
-K^2+13K+44 > 0 \quad \text{(multiply both sides by $14-K$)}\\
K^2-13K < 44\\
K(K-13) < 44
$$
Is this correct? Any way to get a more precise $K$ value? Thanks. </p>
| Alijah Ahmed | 124,032 | <p><strong>Hint:</strong></p>
<p>In order to maintain the same direction of the inequality you can multiply both sides by the positive term $(14-K)^2$.</p>
<p>Thus you will obtain the following inequality</p>
<p>$(14-K)(-K^2+13K+44)>0$</p>
<p>If you solve the quadratic $-K^2+13K+44=0$, you will obtain two roots $a$ and $b$ - I'll leave this for you to work it out. </p>
<p>Once you have found $a$ and $b$, you will be able to factorise $-K^2+13K+44=(a+K)(b-K)$, resulting in the inequality </p>
<p>$(14-K)(a+K)(b-K)>0$</p>
<p>Now, in order to satisfy the above inequality, either all three terms need to be positive, or two of the terms need to be negative and one term positive. </p>
<p>The task is then to find which values of $K$ will satisfy these conditions.</p>
|
501,678 | <p>Let $f(x)=x-\cos(x)$. Find all points on the graph of $y=f(x)$ where the tangent line has slope 1. (In each answer $n$ varies among all integers).</p>
<p>So far I've used the Sum derivative rule for which I have $1+\sin(x)$. So do I put in 1 in for $x$ for sin$(x)$.</p>
<p>Please Help!!</p>
| njguliyev | 90,209 | <p>Hint: $$\sqrt{\frac {1}{x}+2}-\sqrt{\frac {1}{x}} = \dfrac{2}{\sqrt{\dfrac {1}{x}+2}+\sqrt{\dfrac {1}{x}} }.$$</p>
|
501,678 | <p>Let $f(x)=x-\cos(x)$. Find all points on the graph of $y=f(x)$ where the tangent line has slope 1. (In each answer $n$ varies among all integers).</p>
<p>So far I've used the Sum derivative rule for which I have $1+\sin(x)$. So do I put in 1 in for $x$ for sin$(x)$.</p>
<p>Please Help!!</p>
| DonAntonio | 31,254 | <p>$$\sqrt{\frac1x+2}-\sqrt\frac1x=\frac2{\sqrt{\frac1x+2}+\sqrt\frac1x}=\frac{2\sqrt x}{\sqrt{2x+1}+1}\xrightarrow[x\to 0]{}\frac 0{1+1}=0$$</p>
|
374,881 | <p>I'd like to know how I can recursively (iteratively) compute variance, so that I may calculate the standard deviation of a very large dataset in javascript. The input is a sorted array of positive integers.</p>
| Schatzi | 182,154 | <p>There are two problems in the preceding answer, the first being the formula for the variance is incorrect(see the formula below for the correct version) and the second is that the formula for the recursion ends up subtracting large, nearly equal, numbers.</p>
<p>The definition for unbiased estimates of mean($\bar x$) and variance($\sigma^2$) for a sample of size n are:
$$
\bar x_n=\frac1n\sum_{k=1}^nx_k,
$$
and
$$
\bar\sigma^2_n=\frac1{n-1}\sum_{k=1}^n(x_k-\bar x_n)^2
$$</p>
<p>Define the recursion variables </p>
<p>$$
M_n = n \bar x_n=\sum_{k=1}^nx_k,
$$
and
$$
S_n = (n-1)\bar\sigma^2_n=\sum_{k=1}^n(x_k-\bar x_n)^2
$$</p>
<p>The recursion relation for $M_{n+1}$ is obvious
$$ M_{n+1} = M_n + x_{n+1} $$
and the recursion relation for $S_n$ is obtained via
$$ S_{n+1} = (x_{n+1}-\bar x_{n+1})^2+\sum_{k=1}^n(x_k-\bar x_{n+1})^2\phantom{XXXXXX}\\
\phantom{S_{n+1}} = (x_{n+1}-\bar x_{n+1})^2+\sum_{k=1}^n(x_k-\bar x_n+\bar x_n-\bar x_{n+1})^2\\
\phantom{S_{n+1}XXXXXXXXXX} = (x_{n+1}-\bar x_{n+1})^2+\sum_{k=1}^n(x_k-\bar x_n)^2+2(\bar x_n-\bar x_{n+1})\sum_{k=1}^n(x_n-\bar x_n) + \sum_{k=1}^n(\bar x_n -\bar x_{n+1})^2\\
$$
And since
$$S_n = \sum_{k=1}^n(x_k-\bar x_n)^2$$
$$\sum_{k=1}^n(\bar x_n-\bar x_{n+1})^2 = n(\bar x_n-\bar x_{n+1})^2$$ and
$$\sum_{k=1}^n(x_k-\bar x_n) = 0$$
this simplifies to
$$
S_{n+1} = S_n+(x_{n+1}-\bar x_{n+1})^2 +n(\bar x_n -\bar x_{n+1})^2
$$</p>
<p>Now, this recursion relation has the nice property that it $S_n$ a sum of squared terms, and thus cannot be negative. Written in terms of $M_n$ and $S_n$, the recursion relations are:
$$ M_{n+1} = M_n + x_{n+1} $$
$$ S_{n+1} = S_n+\left(x_{n+1}-\frac{M_n+x_{n+1}}{n+1}\right)^2 +n\left(\frac{M_n}{n} -\frac{M_n+x_{n+1}}{n+1}\right)^2
$$
and we can further simplify the recursion relation for $S_n$ to
\begin{eqnarray}
S_{n+1} &= S_n + \left(\frac{n x_{n+1}-M_n}{n+1}\right)^2+n\left(\frac{M_n-n x_{n+1}}{n(n+1)}\right)^2\\
&=S_n+ \left(1+\frac1n\right)\left(\frac{n x_{n+1}-M_n}{n+1}\right)^2\\
&=S_n+ \frac{(n x_{n+1}-M_n)^2}{n(n+1)}
\end{eqnarray}</p>
<p>So we have the simple recursion relations:
$$ M_{n+1} = M_n + x_{n+1} $$
$$ S_{n+1} = S_n + \frac{(n x_{n+1}-M_n)^2}{n(n+1)}$$</p>
<p>with the mean given by $$\bar x_n = \frac1n M_n$$ and the unbiased estimate of the variance is given by $$\sigma_n^2 = \frac1{n+1}S_n$$.</p>
|
2,679,821 | <p>$K=\{(x,y,z)\in \mathbb{R}^{3}|1\leq x^{2}+y^{2}+z^{2}\leq 2,x+y\geq0,\sqrt3x-y\leq0,z\geq0\} \rightarrow \\
\rightarrow\{spherical\quad coordinates\}\rightarrow K:1\leq r\leq \sqrt2,\quad0\leq \theta\leq \pi/2,\quad\pi/3\leq\phi\leq3\pi/4.$</p>
<p>My question is how we arrive at these inequalities/intervals for the new variables? I can understand the $r$ part, just plug all new expressions into the first inequality and we get the answer. As for the others, I don't see it. Some kind combinations of the inequalities?</p>
<p>Edit:</p>
<p><strong>Spherical coordinates</strong> $ :\\ x=r\sin{\theta}\cos{\phi}\\
y=r\sin{\theta}\sin{\phi}\\
z=r\cos{\theta}$</p>
| Bram28 | 256,001 | <p>If you are allowed to use biconditionals, you can do:</p>
<p>$$\forall x (Person(x) \rightarrow (LikesPizza(x) \leftrightarrow \neg Irrational(x)))$$</p>
|
3,333,924 | <p>I am wondering about solutions to the following differential equation:
<span class="math-container">$f(x)=C_1 \cdot f'(x+C_2) \; \forall x \in \mathbb{R} \; \exists \; C_1, C_2 \in \mathbb{R}$</span>. With <span class="math-container">$C_1, C_2$</span> being constant. Are the solutions uniquely in the family of sin/cos functions? It bugs me that I was not able to come up with a counterexample except for the trivial solution <span class="math-container">$f(x)=0$</span>. </p>
| trula | 697,983 | <p>a transcendental function f(x) gives transcendental results for most rational x
example: e^x, sin(x) etc.
the simple seaming equation e^x=x or cos(x)=x have no formula for x as result, but must be calculated numerically.
also you cn not rewrite e^x as a polynomial or a fraction of polynoms
trula</p>
|
19,996 | <p>In 1556, Tartaglia claimed that the sums<br>
1 + 2 + 4<br>
1 + 2 + 4 + 8<br>
1 + 2 + 4 + 8 + 16<br>
are alternative prime and composite. Show that his conjecture is false. </p>
<p>With a simple counter example, $1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255$, apparently it's false. However, I want to prove it in general case instead of using a specific counter example, but I got stuck :( !
I tried:<br>
The sum $\sum_{i=0}^n 2^i$ is equal to $2^{n+1} - 1$. I assumed that $2^{n+1} - 1$ is prime, then we must show that $2^{n+1} - 1 + 2^{n+1} = 2^{n+2} - 1$ is not composite. Or we assume $2^{n+1}$ is composite and we must show that $2^{n+2} - 1$ is not prime.
But I have no clue how $2^{n+2} - 1$ relates to its previous prime/composite. Any hint?</p>
| Ross Millikan | 1,827 | <p>It is known that $2^n-1$ can only be prime if $n$ is prime. This is because if $jk=n$, $2^n-1=\sum_{i=0}^{n-1}2^i=\sum_{i=0}^{j-1} 2^i \sum_{i=0}^{k-1} 2^{ij}$ So they will only continue to alternate at twin primes. In particular, $2^{6k+2}-1, 2^{6k+3}-1$ and $2^{6k+4}-1$ will all be composite</p>
|
1,921,562 | <p>Couldn't solve this indefinite integral, can someone help me? $$\int \frac {x^3+4x^2+6x+1}{x^3+x^2+x-3} dx$$</p>
| GoodDeeds | 307,825 | <p>$$x^3+4x^2+6x+1=(x^3+x^2+x-3)+3x^2+5x+4$$
Thus,
$$\int\frac{x^3+4x^2+6x+1}{x^3+x^2+x-3}dx=\int\left(1+\frac{3x^2+5x+4}{x^3+x^2+x-3}\right)dx=\int\left(1+\frac{3x^2+5x+4}{(x-1)(x^2+2x+3)}\right)dx$$</p>
<p>Let
$$\frac{3x^2+5x+4}{(x-1)(x^2+2x+3)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+2x+3}$$</p>
<p>Find A,B, and C and solve.</p>
|
103,960 | <pre><code>tt = Flatten[Table[{x, y, z, btot[x, y, z]}, {x, -1, 1, 0.1}, {y, -1, 1,0.1}, {z, -1, 1, 0.1}], 2];
ff = Interpolation[tt]
</code></pre>
<p>Till here it is working fine as it is returning the values of the interpolated function at various <code>{x,y,z}</code> points.</p>
<p>Then I want to find the gradient of this interpolated function. But when I am using</p>
<pre><code>ffd[x_,y_,z_]:= D[ff[x,y,z],{{x,y,z}}]
</code></pre>
<p>I am not getting the gradient.</p>
| Coolwater | 9,754 | <p>With</p>
<pre><code>ffd[x_,y_,z_]:= D[ff[x,y,z],{{x,y,z}}]
</code></pre>
<p>the values of x, y, and z are substituted as arguments causing differentation wrt. numbers, i.e. nonsense. Moreover, you are using SetDelayed, which differentiate once for every call, which rather should be once for all time.</p>
<p>The solution to both problem is replacing SetDelayed with Set:</p>
<pre><code>ffd[x_,y_,z_]= D[ff[x,y,z],{{x,y,z}}]
</code></pre>
|
367,116 | <p>I have a question about vacuous true and it always make me confused. If I want to prove that the empty set is the subset of all the set A, the proof is as following:
if x is in empty set, then x is in A. since x is in empty set is always false,, so the conditional statement is always true~
my question is why x is in empty set is always false, what if x=unicorn, since unicorn is in empty set so x is in empty set is true, right?
the question goes to why I can not plug a non-exist thing( like unicorn)into the variable x?</p>
| Peter Smith | 35,151 | <p>@AustinMorh makes the key point. But a footnote about unicorns. We must distinguish two different claims.</p>
<p>Suppose $U$ is the predicate satisfied by all and only unicorns, then</p>
<blockquote>
<p>$\forall x(Ux \to x \in \emptyset)$</p>
</blockquote>
<p>indeed comes out true for sensible domains (since nothing satisfies the antecedent of the conditional). 'All unicorns are members of the empty set' is another vacuous truth! [But note, we are not "plugging a non-existent thing into a variable": the variables still run over what there is in the domain.]</p>
<p>Suppose however $u$ is now a constant, purportedly naming a unicorn. In standard logic where empty names are not allowed, then $u$ is illegitimate. But take a free logic which does allow empty names. Then atomic wffs involving empty names are not true [on some accounts, they are false, on others they are neither true nor false]. So, since there are no unicorns to be named and $u$ is an empty name, no atomic wff $Fu$ is true, and in particular </p>
<blockquote>
<p>$u \in \emptyset$ </p>
</blockquote>
<p>is <em>not</em> true.</p>
<p>So what are we to make of</p>
<blockquote>
<p>if x=unicorn, since unicorn is in empty set so x is in empty set is true?</p>
</blockquote>
<p>If x is intended to be name-like, then "x is in empty set" is <em>not</em> true. The truth in the vicinity is something else, i.e. $\forall x(Ux \to x \in \emptyset)$ </p>
|
2,774,792 | <p>How would you go about proving the recursion
$$T(n) = T\left(\frac n4\right) + T\left(\frac{3n}4\right) + n$$is $\mathcal O(n\log n)$ using induction?</p>
<p>Thanks!</p>
| Henry | 6,460 | <p>Let's count all desired possibilities out the $6^3$ total potential results:</p>
<ul>
<li>${3 \choose 3} \times 3!=6$ ways of getting all three original values in some order</li>
<li>${3 \choose 2} \times {6-3 \choose 1} \times 3!= 54$ ways of getting two different original values and one non-original value in some order</li>
<li>${3 \choose 1} \times {2 \choose 1} \times \frac{3!}{2!} = 18$ ways of getting a double original value and a single original value in some order</li>
</ul>
<p>making the probability $\dfrac{6+54+18}{6^3} = \dfrac{78}{216}=\dfrac{13}{36}\approx 0.361$, the same as drhab found another way </p>
|
510,732 | <p>I am trying to think of a case where this is not true:</p>
<p>$f(n) = O(g(n))$ and $f(n) \neq \Omega(g(n))$, does $f(n) = o(g(n))$?</p>
<p>I suspect that it has to do with the varying $c$ and $n_{0}$ constants but am not completely sure. </p>
<p>Thanks!</p>
| mjqxxxx | 5,546 | <p>You can read the relations this way.</p>
<ul>
<li>$f \in O(g)$ means that $f$ eventually stays below some multiple of $g$.</li>
<li>$f \in \Omega(g)$ means that $f$ eventually stays above some multiple of $g$.</li>
<li>$f \in o(g)$ means that $f$ eventually stays below <em>any</em> multiple of $g$.</li>
</ul>
<p>So if $f\not\in \Omega(g)$, $f$ must dip below any particular multiple of $g$ infinitely often. But that doesn't mean it <em>stays</em> below that multiple. And knowing that $f\in O(g)$ doesn't help much; that just means $f$ doesn't dominate $g$. So a counterexample to your question, for instance, is obtained by taking $g(n)=1$ everywhere, $f(n)=1$ for even $n$, and $f(n)=1/n$ for odd $n$. Then $f$ is $O(g)$ (it's $\le g$), it's not $\Omega(g)$ (it dips below any constant infinitely often), and it's not $o(g)$ (it doesn't stay below $g/2$, say).</p>
|
3,684,048 | <p>I understand how to prove that algebraically using implicit differentiation:
<a href="https://i.stack.imgur.com/wvANe.png" rel="nofollow noreferrer">1</a></p>
<p>However, when I hope to gain an understanding through the graphs, I had a hard time wrapping my head around why the transformed lnx function still has the same slope at every point of x as the parent function.</p>
<p><a href="https://i.stack.imgur.com/2Ob5v.png" rel="nofollow noreferrer">Graphs of the 3 functions that I talked about</a></p>
| Community | -1 | <p>It's because of the identity <span class="math-container">$\ln(ab)=\ln a+\ln b$</span>, which holds for all <span class="math-container">$a,b>0$</span>.</p>
|
1,406,535 | <p>Let $ f$ be a function such that $|f(u)-f(v)|\leq|u-v|$ for all real $u$ and $v$ in an interval $[a,b]$.Then:<br>
$(i)$Prove that $f$ is continuous at each point of $[a,b]$.<br></p>
<p>$(ii)$Assume that $f$ is integrable on $[a,b]$.Prove that,$|\int_{a}^{b}f(x)dx-(b-a)f(c)|\leq\frac{(b-a)^2}{2}$,where $a\leq c \leq b$<br></p>
<p>I tried to solve second part,First part i could not get idea.<br>
$|\int_{a}^{b}f(x)dx-(b-a)f(c)|=|\int_{a}^{b}f(x)-f(c)dx|=\int_{a}^{b}|f(x)-f(c)|dx\leq\int_{a}^{b}|x-c|dx\leq\int_{a}^{c}(c-x)dx+\int_{c}^{b}(x-c)dx$<br></p>
<p>But i am not getting desired result,what have i done wrong in this?Or is there another method to prove it.Please help.</p>
| Sempliner | 122,727 | <p>This occurs because for Carmichael numbers like $N$ we have that, by the Chinese remainder theorem, $\mathbb{Z}/N\mathbb{Z} \cong \mathbb{Z}/p_1\mathbb{Z} \times \mathbb{Z}/p_2\mathbb{Z} \times \dots \times \mathbb{Z}/p_n\mathbb{Z}$ where the $p_i$ are its prime factors, and because for all these primes we have that $p-1|N-1$ we know that any element of this product ring raised to the $N-1$ is going to be itself again just by the fact that this obviously works factor by factor. So actually the percentage of numbers that give false positives are a rather large percentage of these numbers, in fact all of them coprime to $N$...</p>
|
181,855 | <p>In the latest <a href="http://what-if.xkcd.com/113/" rel="noreferrer">what-if</a> Randall Munroe ask for the smallest number of geodesics that intersect all regions of a map. The following shows that five paths of satellites suffice to cover the 50 states of the USA:
<img src="https://i.stack.imgur.com/gyfYt.png" alt="from what-if.xkcd.com"></p>
<p>A similar configuration where the lines are actually great circles is claimed by the author:</p>
<blockquote>
<p>They're all slightly curved, since the Earth is turning under the satellites, but it turns out that this arrangement of lines also works for the much simpler version of the question that ignores orbital motion: "How many straight (great-circle) lines does it take to intersect every state?" For both versions of the question, my best answer is a version of the arrangement above.</p>
</blockquote>
<p>There has been quite some work on similar sounding problems. For stabbing (or finding transversals of) line segments see as an example <a href="http://link.springer.com/article/10.1007%2FBF01934440" rel="noreferrer">Stabbing line segments</a> by H. Edelsbrunner, H. A. Maurer,
F. P. Preparata, A. L. Rosenberg, E. Welzl and D. Wood (and papers which reference it.)
or L.M. Schlipf's <a href="http://www.diss.fu-berlin.de/diss/receive/FUDISS_thesis_000000096077" rel="noreferrer">dissertation</a> with examples of different kinds. </p>
<p><strong>Is there an algorithmic approach known to tackle this problem (or for the
simpler problem when all regions of the map are convex)?</strong></p>
<p>In the case of the 50 states of the USA, it is of course easy to see that one great circle does not suffice: take two states (e.g. New York and Louisiana) such that all great circles that intersect those do not pass through a third state (e.g. Alaska). Similarly one can show that we need at least 3 great circles. </p>
<p>Maybe it would be helpful to consider all triples of regions that do not lie on a great circle and use this hypergraph information to deduce lower bounds.</p>
<p><strong>What are good methods to find lowers bounds?</strong></p>
<p>Randall Munroe's conjectures that 5 is optimal:</p>
<blockquote>
<p>I don't know for sure that 5 is the absolute minimum; it's possible
there's a way to do it with four, but my guess is that there isn't.
[...] If
anyone finds a way (or proof that it's impossible) I'd love to see it!</p>
</blockquote>
| The Masked Avenger | 35,626 | <p>I think the easiest route to a lower bound is to pick four states such as Hawaii, Alaska, Rhode Island, and Florida, and show that any geodesics cutting them leave too many states uncovered, or are five or more in number.
It should be possible to enumerate the maximal cutting geodesics for each pair of states, and then
argue using such numbers. Even trying all combinations of four cutting geodesics involving more than
six states should be tractable, say on order of 10^15 combinations or so.</p>
|
1,580,270 | <p>Consider the groups $G = \{0,1,2\} = \mathbb Z_3$ and $H = \{a,b,c\}$
given by the following multiplication tables:</p>
<p><a href="https://i.stack.imgur.com/hXgBb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hXgBb.jpg" alt="enter image description here"></a></p>
<p>The first one isn't really multiplication but in my notes it said it doesn't really matter.</p>
<p>So how do I show an isomorphism? The groups have the same size so they can be bijective right? But it just seems so abstract to show if there's an isomorphism... Exactly what do we have to check.</p>
| Bernard | 202,857 | <p>First check the second table indeed defines a group law. Looking at the second row and second column in the right-hand side table, you can see <span class="math-container">$b$</span> is the neutral element. Then you can check <span class="math-container">$a^{-1}=b$</span> and <span class="math-container">$b^{-1}=a$</span>. You can at once see that the law is commutative (symmetry w.r.t. the principal diagonal). Also, which is longer, but shortened by commutativity, you should check the law is associative, examining all possibilities for the triples <span class="math-container">$(a,b,c)$</span>.</p>
<p>Now you can define a bijective mapping from the first set to the second by
<span class="math-container">\begin{align*}
0&\mapsto b,\\
1&\mapsto a\quad (\text{or}\enspace b),\\
2&\mapsto c\quad (\text{or}\enspace a).
\end{align*}</span>
There remains to check it is compatible with both laws, considering all possible cases.</p>
|
2,955,780 | <p>The midpoint of a chord of length <span class="math-container">$2a$</span> is at a distance <span class="math-container">$d$</span> from the midpoint of the minor arc it cuts out from the circle. Show that the diameter of the circle is <span class="math-container">$\frac{a^2+d^2}{d}$</span> .</p>
<p>I know I have to find similar triangles, I cannot see them...</p>
| Phil H | 554,494 | <p><a href="https://i.stack.imgur.com/Gcnxa.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Gcnxa.jpg" alt="enter image description here"></a></p>
<p><span class="math-container">$r = \sqrt{r^2 - a^2} + d$</span></p>
<p><span class="math-container">$r - d = \sqrt{r^2 - a^2}$</span></p>
<p><span class="math-container">$(r - d)^2 = r^2 - a^2$</span></p>
<p><span class="math-container">$r^2 - 2rd + d^2 = r^2 - a^2$</span></p>
<p><span class="math-container">$-2rd = -a^2 - d^2$</span></p>
<p><span class="math-container">$r = \frac{a^2 + d^2}{2d}$</span></p>
<p>So <span class="math-container">$D = \frac{a^2 + d^2}{d}$</span></p>
|
3,184,802 | <p>Are there any <span class="math-container">$C^\infty$</span> real functions except the exponential family and gamma function family which has all the derivatives of same sign on an interval [a,<span class="math-container">$\infty$</span>) with a<span class="math-container">$\gt$</span>0 ?
I speculate the function is always uses exponential as building blocks and it is unique defining property of exponential functions.</p>
<p>Please provide some instances otherwise.</p>
<p>I have not been able to find any so far.</p>
| Martin R | 42,969 | <p>Choose an arbitrary sequence <span class="math-container">$(a_n)$</span> with <span class="math-container">$a_n \ge 0$</span> and <span class="math-container">$\lim_{n\to \infty} \sqrt[n]{a_n} = 0$</span>. Then the power series
<span class="math-container">$$
f(x) = \sum_{n=0}^\infty a_n x^n
$$</span>
converges on <span class="math-container">$\Bbb R$</span>, is a <span class="math-container">$C^\infty$</span> function, and <span class="math-container">$f$</span> and all its derivatives are positive on <span class="math-container">$[0, \infty)$</span>.</p>
<p>Examples are
<span class="math-container">$$
\sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \ldots \\
\cosh(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \ldots
$$</span></p>
|
2,698,357 | <p><strong>Draw a graph on $10$ vertices with no more than $20$ edges that contains no independent set of size $3$.</strong></p>
<p>So I was trying to draw the above graph but I was kind of stuck. What I basically did was draw a bipartite graph with $5$ vertices on each side and then $20$ edges randomly connecting between them. I suppose it was similar to $K5, 5$, though I don't believe it is quite the same. I was also a little confused on how to identify an independent set in a graph so it was hard for me to verify my graph.</p>
<p>If anyone knows how to approach this problem, some assistance will be highly appreciated!</p>
| Mk Utkarsh | 532,893 | <p>Take <a href="https://commons.wikimedia.org/wiki/File:Complete_bipartite_graph_K4,4.svg" rel="nofollow noreferrer">K4,4</a> and 2 vertices connected with an edge. That will result in 2 independent sets of size 4 and 2 independent sets of size 1</p>
|
43,743 | <p>An alternative title is: When can I homotope a continuous map to a smooth immersion?</p>
<p>I have a simple topology problem but it's outside my area of expertise and I worry may be rather subtle. Any help would be appreciated.</p>
<p>The set-up is the following:
Let $M$ be some (closed say) $n$ dimensional manifold and suppose that $\Sigma_1$ and $\Sigma_2$ are two closed submanifolds of $M$ of dimension $k$. Note that $\Sigma_1$ and $\Sigma_2$ are allowed to intersect (in my situation they are also embedded but I don't believe this is effects anything). Suppose in addition that $\Sigma_1$ and $\Sigma_2$ are homologous. If $k\leq n-2$ I would like a compact manifold with boundary $\Gamma$ with $\partial \Gamma=\gamma_1\cup \gamma_2$ and a smooth immersion $F:\Gamma\to M$ so that $F(\gamma_i)=\Sigma_i$. In other, words the homology between $\Sigma_1$ and $\Sigma_2$ can be realized by a smooth immersion. </p>
<p>I believe by approximation arguments one can always get a smooth such $F$ without restriction on $k$ but it need not be an immersion (especially if $k=n-1$). My gut is that when you have $k\leq n-2$ since the dimension of the image of $F$ is codimension one you have enough room to perturb it to be an immersion. That is that $F$ is homotopic rel boundary to our desired immersion.</p>
<p>Unfortunately, I don't know enough to formalize this and all my intuition comes from considering curves and domains in $\mathbb{R}^3$ so I'm afraid there may be obstructions in general.</p>
<p>References would be greatly appreciated.</p>
<p>Thanks!</p>
<p>Edit:</p>
<p>As suspected, the question is somewhat subtle . To make it tractable lets assume that $M$ is a $C^\infty$ domain in $\mathbb{R}^3$ (so is a fairly simple three-manifold with boundary) and that the $\Sigma_i$ are curves. This is where my intuition says that there should be such a smooth immersion.</p>
| András Szűcs | 36,950 | <p>The answer is known in the special case you mentioned (curves in a 3-manifold)
Actually it is true more generally: Namely
if k = n-2 and your k-dimensional oriented submanifolds $\Sigma_1$ and $\Sigma_2$ represent the same INTEGER homology class in the ambient ORIENTED n-manifold M, and they are disjoint and embedded,
then there is an embedded k+1 dimensional submanifold in the cylinder $M \times I$ with boundary $\Sigma_1$ in $M\times {0}$ and $\Sigma_2$ in $M \times {1}.$</p>
<p>This a trivial consequence of the fact, that
$K(Z,2) = MSO(2).$
The homotopy classes [M, K(Z,2)] give the 2-dimensional cohomology group, hence also the k-th homology group of M.
The homotopy classes [M, MSO(2)] give the codimension 2 embedded submanifolds up to an embedded cobordism (by the Thom construction). QED.</p>
<p>In general the answer is negative, as it was mentioned by others.</p>
|
1,134,510 | <p>Regarding My Background I have covered stuff like </p>
<p>1.Single Variable Calculus</p>
<p>2.Multivariable Calculus (Multiple Integration,Vector Calculus etc) (Thomas Finney)</p>
<p>3.Basic Linear Algebra Course (Containing Vector spaces,Linear Transformation)</p>
<p>4.Ordinary Differential Equation</p>
<p>5.Real Analysis (Sequences And series)</p>
<p>I am interested In Number theory and i am big fan of Ramanujan .I have not been through rigorous proofs before . But i want to dig deep into number theory especially area where Ramanujan was working .Anyone researcher,Professor can advice me about which preparations are needed for going into number theory and which books should i need .I will be highly obliged .I am asking this for self study or you can say pursuing research at home .</p>
<p>Note - All stuff i have covered is with help of youtube videos and self study .Thanks </p>
| ganbustein | 198,393 | <p>$26^7\cdot 7!$ is clearly way too big. The $26^7$ factor all by itself counts <em>all</em> 7-letter passwords made up from a 26-letter alphabet. Multiplying by $7!$ can only make the number bigger (a lot bigger), when the correct number should be a lot smaller because of all the incorrect combinations.</p>
<p>There are 26 ways to choose the letter that occurs 4 times, and having chosen it there are 25 ways to choose the letter that occurs 3 times. There are $7 \choose 4$ ways to pick the places where the first letter occurs. The total number of passwords is then $26 \cdot 25 \cdot {7 \choose 4}$</p>
|
2,952,028 | <p>The question asks: Find the values of k for which the line</p>
<p><span class="math-container">$y=2x-k$</span> is tangent to the circle with equation <span class="math-container">$x^2+y^2=5$</span></p>
<p>So I started by substituting,</p>
<p><span class="math-container">$x^2+(2x-k)^2=5$</span></p>
<p><span class="math-container">$x^2+4x^2-4xk+k^2=5$</span></p>
<p><span class="math-container">$5x^2+k^2-4xk-5=0$</span></p>
<p>But after this I couldn't see a way to factor it that would make able to find the discriminant and set that equal to <span class="math-container">$0$</span>.</p>
<p>Help would be appreciated.</p>
| TonyK | 1,508 | <p>Here is how I did it:</p>
<p>Differentiating <span class="math-container">$x^2+y^2=5$</span> gives <span class="math-container">$2x+2y\;dy/dx=0$</span>, or <span class="math-container">$dy/dx=-x/y$</span>. </p>
<p>The slope of the line <span class="math-container">$y=2x-k$</span> is <span class="math-container">$2$</span>. </p>
<p>So at the tangent points, <span class="math-container">$dy/dx=-x/y=2$</span>, i.e. <span class="math-container">$x=-2y$</span>.</p>
<p>Plugging this back into the equation for the circle, we get
<span class="math-container">$$(-2y)^2+y^2=5$$</span>
<span class="math-container">$$y=\pm1$$</span>
So the tangent points are <span class="math-container">$(2,-1)$</span> and <span class="math-container">$(-2,1)$</span>. Substituting these into the equation for the line gives <span class="math-container">$k=\pm5$</span>.</p>
|
67,630 | <p>I know of a theorem from Axler's <em>Linear Algebra Done Right</em> which says that if $T$ is a linear operator on a complex finite dimensional vector space $V$, then there exists a basis $B$ for $V$ such that the matrix of $T$ with respect to the basis $B$ is upper triangular.</p>
<p>The proof of this theorem is by induction on the dimension of $V$. For dim $V = 1$ the result clearly holds, so suppose that the result holds for vector spaces of dimension less than $V$. Let $\lambda$ be an eigenvalue of $T$, which we know exists for $V$ is a complex vector space.</p>
<p>Consider $U = $ Im $(T - \lambda I)$. It is not hard to show that Im $(T - \lambda I)$ is an invariant subspace under $T$ of dimension less than $V$.</p>
<p>So by the induction hypothesis, $T|_U$ is an upper triangular matrix. So let $u_1 \ldots u_n$ be a basis for $U$. Extending this to a basis $u_1 \ldots u_n, v_1 \ldots v_m$ of $V$, the proof is complete by noting that for each $k$ such that $1 \leq k \leq m$, $T(v_k) \in $ span $\{u_1 \ldots u_n, v_1 \ldots v_k\}$.</p>
<p>The proof of this theorem seems to be only using the hypothesis that $T$ has at least one eigenvalue (in a complex vector space). So I turned to the following example of a linear operator $T$ on a <em>real</em> vector space $(\mathbb{R}^3)$ instead that has one real eigenvalue:</p>
<p>$T(x,y,z) = (x, -z ,y)$. </p>
<p>In the standard basis of $\mathbb{R}^3$ the matrix of $T$ is </p>
<p>$\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{array}\right]$</p>
<p>The only real eigenvalue of this matrix is $1$ with corresponding eigenvector </p>
<p>$\left[\begin{array}{c}1 \\ 0 \\ 0 \end{array}\right]$.</p>
<p>This is where I run into trouble: If I just extend this to any basis of $\mathbb{R}^3$, then using the standard basis again will not put the matrix of $T$ in upper triangular form. Why can't the method used in the proof above be used to put the matrix of $T$ in upper triangular form?</p>
<p>I know this is not possible for if $T$ were to be put in upper triangular form, then this would mean all its eigenvalues are real which contradicts it having eigenvalues $\pm i$ as well.</p>
| Robert Israel | 8,508 | <p>The proof breaks down because $T|_U$ can't be made upper triangular, where in this case
$U = \{(0, x, y): x, y \in {\mathbb R}\}$.</p>
|
367,669 | <p><img src="https://i.stack.imgur.com/zQFyC.jpg" alt="enter image description here"></p>
<p>This is probably a very simple questions but I am not clear on Möbius transformations and how to solve this problem. I'd appreciate if somebody can point me towards a method to do these sort of questions or a webpage that explains what I need to solve this problem.</p>
| xyzzyz | 23,439 | <p>Let $f(n) = k$ if $n = 2^k \cdot (2l -1)$ for natural numbers $k, l$ -- in other words, if $2^k$ is the highest power of $2$ that divides $n$. Then $f^{-1}(k) = \{2^k \cdot 1, 2^k \cdot 3, 2^k \cdot 5, 2^k \cdot 7\ldots \}$.</p>
|
84,249 | <p>I have a data set with evenly spaced data points. The plot is frequency vs. intensity. The overall shape of the plot is an upwards curve into a plateau, this cannot be seen in the data as this is an unimportant feature. There is also an oscillation in this curve. This can be seen in the plot.</p>
<p><img src="https://i.stack.imgur.com/1d2a7.png" alt="enter image description here"></p>
<p>There are a few things I would like to do here…</p>
<ol>
<li><p>Find out the frequency of this oscillation via a Fourier transform. I have done this already but I have encountered some problems. There are large peaks at the very end of the Fourier transformed data, I believe this can be sorted by zero filling using PadLeft/Right. I have also had quite a few attempts at this but have so far been unsuccessful.</p></li>
<li><p>I would like to filter this oscillation (frequency) from the data and then re-plot the data with this frequency removed. Again, I have had a few attempts at this but to no avail. I also believe that the data can be back-converted using the inverse Fourier transform.</p></li>
</ol>
<p>The data that I am using has been attached below:</p>
<pre><code>{{2.96536, 0.104234}, {2.98246, 0.0969915}, {2.99966,
0.102057}, {3.01696, 0.0921243}, {3.03436, 0.119644}, {3.05186,
0.111209}, {3.06945, 0.114199}, {3.08716, 0.109548}, {3.10496,
0.131311}, {3.12286, 0.11789}, {3.14087, 0.136387}, {3.15898,
0.156646}, {3.1772, 0.14701}, {3.19552, 0.170584}, {3.21395,
0.135949}, {3.23248, 0.155617}, {3.25112, 0.169365}, {3.26987,
0.177859}, {3.28873, 0.166621}, {3.30769, 0.16418}, {3.32676,
0.176456}, {3.34595, 0.194153}, {3.36524, 0.191821}, {3.38465,
0.16664}, {3.40417, 0.19331}, {3.4238, 0.197461}, {3.44354,
0.190734}, {3.4634, 0.218241}, {3.48337, 0.22238}, {3.50346,
0.218784}, {3.52366, 0.215578}, {3.54398, 0.23948}, {3.56442,
0.245256}, {3.58497, 0.226847}, {3.60564, 0.219071}, {3.62643,
0.223373}, {3.64735, 0.225021}, {3.66838, 0.225226}, {3.68953,
0.233922}, {3.71081, 0.232157}, {3.73221, 0.228506}, {3.75373,
0.203525}, {3.77538, 0.244966}, {3.79715, 0.241911}, {3.81904,
0.21011}, {3.84107, 0.208428}, {3.86322, 0.227731}, {3.88549,
0.222106}, {3.9079, 0.237606}, {3.93043, 0.221255}, {3.9531,
0.19241}, {3.9759, 0.221645}, {3.99882, 0.243768}, {4.02188,
0.217034}, {4.04507, 0.203556}, {4.0684, 0.205594}, {4.09186,
0.224882}, {4.11546, 0.213087}, {4.13919, 0.205046}, {4.16306,
0.216099}, {4.18706, 0.225207}, {4.21121, 0.222689}, {4.23549,
0.214728}, {4.25992, 0.23614}, {4.28448, 0.240632}, {4.30919,
0.224024}, {4.33404, 0.239854}, {4.35903, 0.242658}, {4.38417,
0.27057}, {4.40945, 0.258658}, {4.43488, 0.265637}, {4.46045,
0.259903}, {4.48617, 0.269462}, {4.51204, 0.283}, {4.53806,
0.289011}, {4.56423, 0.30783}, {4.59055, 0.297366}, {4.61702,
0.299034}, {4.64365, 0.311518}, {4.67042, 0.305525}, {4.69736,
0.313848}, {4.72444, 0.330391}, {4.75169, 0.329848}, {4.77909,
0.322798}, {4.80665, 0.351296}, {4.83437, 0.347589}, {4.86224,
0.370153}, {4.89028, 0.35712}, {4.91848, 0.357503}, {4.94685,
0.369257}, {4.97537, 0.342726}, {5.00406, 0.361666}, {5.03292,
0.353577}, {5.06194, 0.361541}, {5.09113, 0.355642}, {5.12049,
0.356704}, {5.15002, 0.333525}, {5.17971, 0.373981}, {5.20958,
0.365135}, {5.23963, 0.354161}, {5.26984, 0.338223}, {5.30023,
0.332814}, {5.33079, 0.34522}, {5.36153, 0.339017}, {5.39245,
0.335138}, {5.42355, 0.304693}, {5.45482, 0.3446}, {5.48628,
0.302124}, {5.51791, 0.319763}, {5.54973, 0.322771}, {5.58174,
0.330913}, {5.61392, 0.32405}, {5.6463, 0.356941}, {5.67886,
0.334621}, {5.7116, 0.342564}, {5.74454, 0.371111}, {5.77767,
0.34261}, {5.81098, 0.388414}, {5.84449, 0.384681}, {5.8782,
0.399269}, {5.91209, 0.387332}, {5.94619, 0.384739}, {5.98048,
0.379754}, {6.01496, 0.400486}, {6.04965, 0.45052}, {6.08453,
0.421576}, {6.11962, 0.426608}, {6.15491, 0.426233}, {6.1904,
0.436149}, {6.2261, 0.455608}, {6.262, 0.45478}, {6.29811,
0.458522}, {6.33443, 0.471644}, {6.37096, 0.469424}, {6.4077,
0.450665}, {6.44465, 0.441694}, {6.48181, 0.4626}, {6.51919,
0.477626}, {6.55678, 0.435283}, {6.59459, 0.429262}, {6.63262,
0.44671}, {6.67087, 0.403679}, {6.70934, 0.444082}, {6.74803,
0.450073}, {6.78694, 0.442818}, {6.82608, 0.431519}, {6.86544,
0.429014}, {6.90503, 0.446844}, {6.94485, 0.439155}, {6.9849,
0.441625}, {7.02517, 0.448944}, {7.06569, 0.469756}, {7.10643,
0.467907}, {7.14741, 0.48434}, {7.18863, 0.505442}, {7.23008,
0.493814}, {7.27177, 0.480819}, {7.31371, 0.509904}, {7.35588,
0.510856}, {7.3983, 0.520773}, {7.44096, 0.539145}, {7.48387,
0.541484}, {7.52703, 0.552256}, {7.57043, 0.566155}, {7.61409,
0.579046}, {7.65799, 0.55074}, {7.70215, 0.560432}, {7.74657,
0.546139}, {7.79124, 0.568914}, {7.83617, 0.549678}, {7.88136,
0.520845}, {7.9268, 0.531402}, {7.97251, 0.529267}, {8.01849,
0.545642}, {8.06473, 0.540622}, {8.11123, 0.534639}, {8.15801,
0.527247}, {8.20505, 0.517973}, {8.25237, 0.5166}, {8.29995,
0.51883}, {8.34782, 0.534269}, {8.39595, 0.535702}, {8.44437,
0.537064}, {8.49307, 0.550537}, {8.54204, 0.552024}, {8.5913,
0.53197}, {8.64084, 0.576196}, {8.69067, 0.590832}, {8.74078,
0.602607}, {8.79119, 0.585699}, {8.84188, 0.578139}, {8.89287,
0.57396}, {8.94415, 0.588638}, {8.99573, 0.55418}, {9.0476,
0.578538}, {9.09978, 0.588798}, {9.15225, 0.618243}, {9.20503,
0.61227}, {9.25811, 0.623181}, {9.3115, 0.614365}, {9.36519,
0.582252}, {9.4192, 0.591002}, {9.47351, 0.582036}, {9.52814,
0.57551}, {9.58309, 0.579221}, {9.63835, 0.601598}, {9.69393,
0.583821}, {9.74983, 0.601753}, {9.80605, 0.616571}, {9.8626,
0.623343}, {9.91948, 0.625228}, {9.97668, 0.646208}, {10.0342,
0.640938}, {10.0921, 0.652611}, {10.1503, 0.662041}, {10.2088,
0.667227}, {10.2677, 0.676957}, {10.3269, 0.668459}, {10.3864,
0.676449}, {10.4463, 0.663062}, {10.5066, 0.671728}, {10.5671,
0.658277}, {10.6281, 0.645957}, {10.6894, 0.672471}, {10.751,
0.628736}, {10.813, 0.632566}, {10.8754, 0.636309}, {10.9381,
0.65611}, {11.0012, 0.615158}, {11.0646, 0.649256}, {11.1284,
0.632535}, {11.1926, 0.640524}, {11.2571, 0.649521}, {11.322,
0.678226}, {11.3873, 0.692686}, {11.453, 0.69979}, {11.519,
0.707412}, {11.5854, 0.702047}, {11.6523, 0.691649}, {11.7195,
0.70039}, {11.787, 0.708576}, {11.855, 0.688944}, {11.9234,
0.701633}, {11.9921, 0.679641}, {12.0613, 0.699239}, {12.1308,
0.673699}, {12.2008, 0.677758}, {12.2711, 0.671892}, {12.3419,
0.682907}, {12.4131, 0.689255}, {12.4847, 0.695831}, {12.5566,
0.708726}, {12.6291, 0.706843}, {12.7019, 0.700961}, {12.7751,
0.708156}, {12.8488, 0.735001}, {12.9229, 0.721247}, {12.9974,
0.727045}, {13.0724, 0.711983}, {13.1477, 0.747442}, {13.2236,
0.743774}, {13.2998, 0.703271}, {13.3765, 0.74117}, {13.4536,
0.739813}, {13.5312, 0.6996}, {13.6093, 0.698404}, {13.6877,
0.723868}, {13.7667, 0.709428}, {13.8461, 0.752924}, {13.9259,
0.747375}, {14.0062, 0.723421}, {14.087, 0.768379}, {14.1682,
0.756985}, {14.2499, 0.7739}, {14.3321, 0.780965}, {14.4147,
0.751474}, {14.4978, 0.797507}, {14.5814, 0.724096}, {14.6655,
0.744758}, {14.7501, 0.712119}, {14.8352, 0.740484}, {14.9207,
0.729569}, {15.0067, 0.696926}, {15.0933, 0.715435}, {15.1803,
0.716336}, {15.2679, 0.755648}, {15.3559, 0.785479}, {15.4445,
0.780429}, {15.5335, 0.810293}, {15.6231, 0.776618}, {15.7132,
0.806514}, {15.8038, 0.794361}, {15.8949, 0.768466}, {15.9866,
0.758793}, {16.0788, 0.799427}, {16.1715, 0.713711}, {16.2647,
0.779966}, {16.3585, 0.724277}, {16.4529, 0.743149}, {16.5477,
0.759417}, {16.6432, 0.783702}, {16.7391, 0.771348}, {16.8357,
0.833143}, {16.9328, 0.865346}, {17.0304, 0.838565}, {17.1286,
0.849294}, {17.2274, 0.772466}, {17.3267, 0.791529}, {17.4266,
0.768608}, {17.5271, 0.763986}, {17.6282, 0.741564}, {17.7299,
0.7283}, {17.8321, 0.748744}, {17.9349, 0.768283}, {18.0383,
0.803771}, {18.1424, 0.793137}, {18.247, 0.851768}, {18.3522,
0.887429}, {18.458, 0.864324}, {18.5645, 0.862401}, {18.6715,
0.789341}, {18.7792, 0.773965}, {18.8875, 0.827819}, {18.9964,
0.815678}, {19.106, 0.773111}, {19.2161, 0.849946}, {19.3269,
0.850296}, {19.4384, 0.834852}, {19.5505, 0.883059}, {19.6632,
0.881796}, {19.7766, 0.888633}, {19.8907, 0.862076}, {20.0054,
0.859558}, {20.1207, 0.852859}, {20.2367, 0.844693}, {20.3534,
0.792461}, {20.4708, 0.836719}, {20.5889, 0.834452}, {20.7076,
0.844025}, {20.827, 0.855403}, {20.9471, 0.864835}, {21.0679,
0.852573}, {21.1894, 0.88278}, {21.3116, 0.854075}, {21.4345,
0.823129}, {21.5581, 0.863766}, {21.6824, 0.787823}, {21.8074,
0.837579}, {21.9332, 0.832601}, {22.0596, 0.843718}, {22.1869,
0.8772}, {22.3148, 0.872614}, {22.4435, 0.884425}, {22.5729,
0.860168}, {22.7031, 0.843206}, {22.834, 0.861999}, {22.9657,
0.827936}, {23.0981, 0.842105}, {23.2313, 0.83025}, {23.3653,
0.843434}, {23.5, 0.866308}, {23.6355, 0.887162}, {23.7718,
0.879083}, {23.9089, 0.885303}, {24.0468, 0.872957}, {24.1854,
0.875103}, {24.3249, 0.851151}, {24.4652, 0.872132}, {24.6062,
0.893851}, {24.7481, 0.876361}, {24.8908, 0.891544}, {25.0344,
0.881993}, {25.1787, 0.867608}, {25.3239, 0.902699}, {25.47,
0.880534}, {25.6168, 0.883281}, {25.7646, 0.894037}, {25.9131,
0.865991}, {26.0626, 0.894638}, {26.2129, 0.884609}, {26.364,
0.902747}, {26.516, 0.894505}, {26.669, 0.891673}, {26.8227,
0.894257}, {26.9774, 0.868452}, {27.133, 0.894673}, {27.2894,
0.880052}, {27.4468, 0.89386}, {27.6051, 0.926104}, {27.7643,
0.89179}, {27.9244, 0.910141}, {28.0854, 0.889865}, {28.2474,
0.891172}, {28.4103, 0.873922}, {28.5741, 0.878855}, {28.7389,
0.890755}, {28.9046, 0.912344}, {29.0713, 0.904055}, {29.2389,
0.903534}, {29.4075, 0.892855}, {29.5771, 0.881583}}
</code></pre>
<p>I apologise for importing my data like this, if there is a more convenient way of doing so please let me know and I can import it that way.</p>
<p>Thank you very much for your help.
Stuart.</p>
<p>Addition:</p>
<p><img src="https://i.stack.imgur.com/JRUWd.png" alt="enter image description here"></p>
<p><img src="https://i.stack.imgur.com/2oU67.png" alt="enter image description here"></p>
| Histograms | 29,371 | <p>For problem #2, I recommend a combination of <code>TimeSeries</code> and <code>TimeSeriesResample</code> to obtain a regular sampling of your data, followed by an eyeball inspection of the <code>Periodogram</code>. You'll see that there is a spike around 0.015-ish in this power spectrum. </p>
<p>You can then use a <code>LowpassFilter</code> or most filters from <a href="https://reference.wolfram.com/language/guide/LinearAndNonlinearFilters.html" rel="nofollow noreferrer">Linear and Nonlinear Filtering Reference</a> </p>
<pre><code>resampledData = TimeSeriesResample[data,
ResamplingMethod -> {"Interpolation", InterpolationOrder -> 3}];
Periodogram[resampledData[[All, 2]], PlotLabel -> "Periodogram Power Spectrum dB"]
lowpassed = LowpassFilter[TimeSeries[resampledData], 0.015]["Path"];
ListLinePlot[ {data, lowpassed }, PlotStyle -> {Gray, Red}]
</code></pre>
<p><img src="https://i.stack.imgur.com/cHHRB.png" alt="enter image description here"></p>
<p>EDIT: alternative method using the first 15 coefficients of a <code>FourierDCT</code> since we ran into unusual problems...</p>
<pre><code>intp = Interpolation[data];
dt = Min[Differences[data][[2]]];
mint = Min[data[[All, 1]]];
maxt = Max[data[[All, 1]]];
resampled = Table[{t, intp[t]}, {t, mint, maxt, dt}];
ListLinePlot[{data,Transpose[{resampled[[All, 1]],
FourierDCT[
PadRight[Take[
FourierDCT[resampled[[All, 2]]], 15],
Length[resampled]], "III"]
}]}, PlotStyle -> {Gray, Red}]
</code></pre>
|
1,765,946 | <p>$\newcommand{\Sig}{\Sigma}$
Let $\Sig$ be a diagonal matrix with strictly positive entries on the diagonal. Define $V=\{B \in M_n\mid B\Sig +\Sig B^T=\Sig B +B^T \Sig \}$ (where $M_n$ is the vector space of $n \times n$ real matrices).</p>
<p><strong>What is the dimension of $V$? Is there a "nice" basis for it?</strong></p>
<p>Clearly $V$ contains the subspace of the symmetric matrices.
So $\dim V \ge \frac{n(1+n)}{2}$.</p>
<p>Of course the anser depends on $\Sig$. If $\Sig =Id$ then $V=M_n$. </p>
<p>It seems that the more identical entries, the larger is $V$.</p>
<p><strong>Is $\dim V$ only dependent on the number of different entries of $\Sig$?</strong> If so, what is the functional dependence?</p>
| b2coutts | 335,797 | <p>Let $\lambda_1, \dots, \lambda_n$ be the diagonal entries of $\Sigma$; note that $\Sigma$ is invertible, and the diagonal entries of $\Sigma$ are $\lambda_1^{-1}, \dots, \lambda_n^{-1}$. Now, we can rewrite your condition on $B$ as $B-B^T = \Sigma^{-1}(B-B^T)\Sigma$.</p>
<p>Let $B \in M_n$. For $1 \leq i,j \leq n$, let $c_{i,j}$ be the $(i,j)$ entry of $B-B^T$. Then, the $(i,j)$ entry of $\Sigma^{-1}(B-B^T)\Sigma$ is $\lambda_i^{-1}\lambda_j c_{i,j}$. So $B$ satisfies the above equation (i.e., $B \in V$) if and only if either $\lambda_i = \lambda_j$, or $c_{i,j} = 0$.</p>
<p>Since $\Sigma$ is fixed, $B \in V$ is equivalent to the constraint that $c_{i,j} = 0$ for all $i,j$ such that $\lambda_i \neq \lambda_j$. In other words, for all such $i,j$, $b_{i,j} = b_{j,i}$, where $b_{i,j}$ represents the $(i,j)$ entry of $B$.</p>
<p>So if we let $S := \{(i,j) : \lambda_i \neq \lambda_j\}$, then a basis for $V$ would be $\{E_{i,j} + E_{j,i} : (i,j) \in S\} \cup \{E_{i,j} : (i,j) \notin S\}$, where $E_{i,j}$ denotes the matrix whose $(i,j)$ entry is 1, and whose other entries are 0. Note that the left part of this union contains $|S|/2$ elements, since each $(i,j)$ and $(j,i)$ in $S$ contributes the same matrix. So $\dim V = |S|/2 + (n^2-|S|)$.</p>
<p>Note that $|S|$ (and hence $\dim V$) depends not only on the number of different entries of $\Sigma$, but how many of each there are; for instance, if its entries are $1,2,2,2$, then $|S| = 6$, but if its entries are $1,1,2,2$, then $|S| = 8$.</p>
|
945,334 | <p>Here is a lemma whose proof is as under:</p>
<blockquote>
<p>If $S \in L(X,Y)$ and lim$_{r \to 0}\frac{\|Sr\|}{\|r\|}=0$,then $S=0$.</p>
</blockquote>
<p>Proof:</p>
<p>The condition lim$_{r \to 0}\Big(\frac{\|Sr\|}{\|r\|}\Big)=0$means that for each $\epsilon \gt 0$ there is a $\delta \gt 0$ such that
$\Big(\frac{\|Sr\|}{\|r\|}\Big)\leq \epsilon $ whenever $0\lt \|r\| \lt \delta$</p>
<p>Let $\text{u}\in X$ be a non-zero vector.Choose a non-zero $t\in \mathbb R$ so that $\|t\text{u}\|\lt \delta $ .Then
$\Big(\frac{\|S(t\text{u})\|}{\|t\text{u}\|}\Big)= \Big(\frac{\|S\text{u}\|}{\|\text{u}\|}\Big)\leq \epsilon $
and therefore $\|S\text{u}\|\leq\epsilon \|\text{u}\|.$</p>
<p>This is true for any $\epsilon \gt 0.$Hence $S\text{u}=0$ for all $u \in X$ This means that $S=0$</p>
<p>$---------------------------------------$</p>
<p>I can't understand the step why did we take a vector $\text{u}\in X$ and then introduce $t$ in proof?
Please help....</p>
| Petite Etincelle | 100,564 | <p>Suppose $\{x_n\}$ is increasing and has a subsequence $\{x_{n_k}\}$ which converges to $L$. We will prove that $\{x_n\}$ itself converges to $L$.</p>
<p>For any $\epsilon > 0$, we want to find an integer $N_\epsilon$ such that $|x_n - L| \leq \epsilon$ for any $n \geq N_\epsilon$. </p>
<p>Since $\{x_{n_k}\}$ is increasing and converges to $L$, we can find $k_\epsilon$ such that for any $k \geq k_\epsilon$, $ -\epsilon < x_{n_k} - L <0$. </p>
<p>Take $N_\epsilon = n_{k_\epsilon}$, then for any $n \geq N_\epsilon,$ $ x_{n_{k_\epsilon}}\leq x_n \leq L$, so $-\epsilon \leq x_{n_{k_\epsilon}} - L\leq x_n - L \leq 0$.</p>
<p>Similarly, we can prove when $\{x_n\}$ is decreasing</p>
|
945,334 | <p>Here is a lemma whose proof is as under:</p>
<blockquote>
<p>If $S \in L(X,Y)$ and lim$_{r \to 0}\frac{\|Sr\|}{\|r\|}=0$,then $S=0$.</p>
</blockquote>
<p>Proof:</p>
<p>The condition lim$_{r \to 0}\Big(\frac{\|Sr\|}{\|r\|}\Big)=0$means that for each $\epsilon \gt 0$ there is a $\delta \gt 0$ such that
$\Big(\frac{\|Sr\|}{\|r\|}\Big)\leq \epsilon $ whenever $0\lt \|r\| \lt \delta$</p>
<p>Let $\text{u}\in X$ be a non-zero vector.Choose a non-zero $t\in \mathbb R$ so that $\|t\text{u}\|\lt \delta $ .Then
$\Big(\frac{\|S(t\text{u})\|}{\|t\text{u}\|}\Big)= \Big(\frac{\|S\text{u}\|}{\|\text{u}\|}\Big)\leq \epsilon $
and therefore $\|S\text{u}\|\leq\epsilon \|\text{u}\|.$</p>
<p>This is true for any $\epsilon \gt 0.$Hence $S\text{u}=0$ for all $u \in X$ This means that $S=0$</p>
<p>$---------------------------------------$</p>
<p>I can't understand the step why did we take a vector $\text{u}\in X$ and then introduce $t$ in proof?
Please help....</p>
| Rafael Vergnaud | 301,493 | <p>Suppose <span class="math-container">$(x_n)$</span> is monotone and contains a convergent subsequence <span class="math-container">$(x_{n_i}).$</span> </p>
<p>Given that <span class="math-container">$(x_{n_i})$</span> is convergent, it is bounded above by some upper bound <span class="math-container">$b \in \mathbb{N},$</span> that is, <span class="math-container">$x_{n_i} \leq b,$</span> <span class="math-container">$\forall i \in \mathbb{N}.$</span> </p>
<p>Suppose <span class="math-container">$(x_n)$</span> is divergent. Given its monotonicity, it follows that <span class="math-container">$(x_n)$</span> is unbounded, that is, <span class="math-container">$\forall M \in \mathbb{N},$</span> <span class="math-container">$\exists N \in \mathbb{N}$</span> such that <span class="math-container">$\forall n \geq N$</span> it follows that <span class="math-container">$x_n > M$</span> (at some point "<span class="math-container">$N$</span>," the sequence passes the boundary <span class="math-container">$M$</span> for any boundary <span class="math-container">$M \in \mathbb{N}).$</span> If <span class="math-container">$(x_n)$</span> is bounded, it is necessarily the case that <span class="math-container">$(x_n),$</span> given that it is monotone increasing, IS convergent (you can prove this).</p>
<p>So, <span class="math-container">$\exists N \in \mathbb{N}$</span> such that <span class="math-container">$\forall n \geq N$</span> it follows that <span class="math-container">$x_n > b.$</span></p>
<p>Given that <span class="math-container">$(x_{n_i})$</span> is bounded above by <span class="math-container">$b,$</span> this means that <span class="math-container">$\forall i \in \mathbb{N},$</span> <span class="math-container">$n_i < N.$</span> </p>
<p>Hence, as by definition of a subsequence it is the case that <span class="math-container">$n_1 < n_2 < \cdots n_i,$</span> the subsequence <span class="math-container">$(x_{n_i})$</span> contains fewer than <span class="math-container">$N$</span> elements (at most <span class="math-container">$N - 1$</span> elements). </p>
<p>However, a subsequence is defined as a function whose domain is the natural numbers. Given that <span class="math-container">$(x_{n_i})$</span> contains fewer than <span class="math-container">$N$</span> elements, its domain is a finite, proper subset of the natural numbers, that is, its domain is not the natural numbers. Contradiction. </p>
<p>Therefore, <span class="math-container">$(x_n)$</span> is convergent.</p>
|
1,098,253 | <p>I have got some trouble with proving that for $x\neq 0$:
$$
\frac{\arctan x}{x }< 1
$$
I tried doing something like $x = \tan t$ and playing with this with no success.</p>
| idm | 167,226 | <p>$$\frac{\arctan x}{x}<1\iff \begin{cases}\arctan x<x&if\ x>0\\ \arctan x> x&if\ x<0\end{cases}$$</p>
<p>$$(\arctan x-x)'=\underbrace{\frac{1}{x^2+1}}_{< 1\ if\ x\neq 0}-1< 0$$ therefore $$f:x\longmapsto\arctan x -x$$ is strictly decrasing. Moreover $$\arctan 0-0=0,$$ therefore $(\arctan x-x)<0$ if $x>0$ and $(\arctan x-x)>0$ if $x<0$ because $f$ is continuous on $\mathbb R$. I let you conclude.</p>
|
298,912 | <p>I was reading some basic information from Wiki about category theory and honestly speaking I have a very weak knowledge about it. As it sounds interesting, I will go into the theory to learn more if it is actually useful in practice.</p>
<p>My question is to know if category theory has some applications in practice, namely in engineering problems.</p>
<p>I have already read this <a href="https://math.stackexchange.com/questions/280166/applications-of-category-theory-and-topoi-topos-theory-in-reality">Applications of category theory and topoi/topos theory in reality</a> </p>
<p>and the answers are only about programming which are not very interesting from my point of view.</p>
<p>Any comments are welcomed, thanks in advance.</p>
| Manos | 11,921 | <p>Category theory is far from the engineering textbook level, for now. On the research level, there are a lot of instances where category theory is applied in engineering context, from electrical to biomedical engineering. Beware though: these usually come from people who try to apply category theory, rather than from people who try to solve an engineering problem and find category theory useful in doing so.</p>
|
298,912 | <p>I was reading some basic information from Wiki about category theory and honestly speaking I have a very weak knowledge about it. As it sounds interesting, I will go into the theory to learn more if it is actually useful in practice.</p>
<p>My question is to know if category theory has some applications in practice, namely in engineering problems.</p>
<p>I have already read this <a href="https://math.stackexchange.com/questions/280166/applications-of-category-theory-and-topoi-topos-theory-in-reality">Applications of category theory and topoi/topos theory in reality</a> </p>
<p>and the answers are only about programming which are not very interesting from my point of view.</p>
<p>Any comments are welcomed, thanks in advance.</p>
| Idempotent | 159,957 | <p>This too is a late answer, but in case anyone is still interested, <a href="https://golem.ph.utexas.edu/category/2007/11/category_theory_and_biology.html" rel="noreferrer">here</a> is a discussion with links about the use of category theory in biology/bioinformatics and genetics. Also, while not specifically a book on applications of category theory, the book Conceptual Mathematics by William Lawvere (an undergrad book, so not super advanced, but still a very nice read) takes a practical-minded approach to categories.</p>
|
1,765 | <p>Can anyone explain to me this behaviour? I've been having more than a couple of similar doubts these last weeks. </p>
<p>For example</p>
<pre><code>f[_?NumericQ] := 8;
</code></pre>
<p>Now, if I do</p>
<pre><code>With[{a = f[a]}, HoldForm@Block[{NumericQ = True &}, a]]
</code></pre>
<p>I get</p>
<pre><code>Block[{NumericQ = True &}, f[a]]
</code></pre>
<p>And if I do</p>
<pre><code>Block[{NumericQ = True &}, f[a]]
</code></pre>
<p>I get</p>
<pre><code>8
</code></pre>
<p>So far so good... Another so-far-so-goodie is (notice the <code>:=</code>)</p>
<pre><code>With[{a := f[a]}, Block[{NumericQ = True &}, a]]
8
</code></pre>
<p>Question: Can anyone help me understand why this output?</p>
<pre><code>With[{a = f[a]}, Block[{NumericQ = True &}, a]]
f[a]
</code></pre>
<p>Could it be that <code>With</code> (<code>=</code> version) not only evaluates and replaces, but also guarantees that the replaced expression won't be reevaluated no matter what until the <code>With</code> is exited? If that's the case I wasn't expecting that. What's happening here?</p>
<p><strong>EDIT</strong></p>
<p>Question also applies to Function</p>
<pre><code>Block[{NumericQ = True &}, #] &[f[a]]
f[a]
</code></pre>
<p>And not just those too. Everything I try behaves the same way... A couple of other examples</p>
<pre><code>Block @@ (Hold[{NumericQ = True &}, exp] /. exp -> f[a])
With[{a := Evaluate@f[a]}, Block[{NumericQ = True &}, a]]
</code></pre>
<p>both give <code>f[a]</code></p>
<p><strong>EDIT</strong></p>
<pre><code>With[{g = h}, h = 8; Print[g] ]
</code></pre>
<p>prints 8, not h, so clearly h is reevaluated inside the <code>With</code> in this case, so <code>g</code> is not so constant. </p>
<p><strong>EDIT</strong></p>
<p>Ok, another couple of examples</p>
<pre><code>In[10]:= ClearAll[h, f];
h := 8 /; NumericQ["a"];
f[_?NumericQ] := 8;
</code></pre>
<p>Now, both <code>h</code> and <code>f[a]</code> remain unevaluated</p>
<pre><code>In[21]:= {h, f[a]}
Out[21]= {h, f[a]}
</code></pre>
<p>Now, with the <code>OwnValues</code> everything works as expected</p>
<pre><code>In[17]:= With[{g = h},
Block[{NumericQ = True &}, g]]
Out[17]= 8
</code></pre>
<p>But with the <code>DownValues</code>, it doesn't </p>
<pre><code>In[19]:= With[{g = f[a]},
Block[{NumericQ = True &}, g]]
Out[19]= f[a]
</code></pre>
<p>Similarly</p>
<pre><code>f[a_?NumericQ] := 8;
g[e_] := Block[{NumericQ = True &}, e]
</code></pre>
<p><code>f[a]</code> evaluates to <code>f[a]</code>, but weirdly</p>
<pre><code>In[17]:= g[Unevaluated@f[a]]
Out[17]= 8
In[18]:= g[f[a]]
Out[18]= f[a]
</code></pre>
| FJRA | 495 | <p>You can use <code>TracePrint</code> to try to understand how evolves the evaluation step by step:</p>
<pre><code>In[23]:= TracePrint[With[{b=f[a]},Block[{NumericQ=True&},b]]]
During evaluation of In[23]:= With[{b=f[a]},Block[{NumericQ=True&},b]]
During evaluation of In[23]:= With
During evaluation of In[23]:= f[a]
During evaluation of In[23]:= f
During evaluation of In[23]:= a
During evaluation of In[23]:= NumericQ[a]
During evaluation of In[23]:= NumericQ
During evaluation of In[23]:= a
During evaluation of In[23]:= False
During evaluation of In[23]:= Block[{NumericQ=True&},f[a]]
During evaluation of In[23]:= Block
During evaluation of In[23]:= True&
During evaluation of In[23]:= Function
During evaluation of In[23]:= NumericQ=Unevaluated[True&]
During evaluation of In[23]:= Set
During evaluation of In[23]:= NumericQ=True&
During evaluation of In[23]:= True&
During evaluation of In[23]:= f[a]
During evaluation of In[23]:= f[a]
Out[23]= f[a]
</code></pre>
<p>As Verbeia said before, you can see that <code>NumericQ[a]</code> is evaluated before it goes to the <code>Block</code>. In the other hand, if you use the delayed set:</p>
<pre><code>In[24]:= TracePrint[With[{b:=f[a]},Block[{NumericQ=True&},b]]]
During evaluation of In[24]:= With[{b:=f[a]},Block[{NumericQ=True&},b]]
During evaluation of In[24]:= With
During evaluation of In[24]:= Block[{NumericQ=True&},f[a]]
During evaluation of In[24]:= Block
During evaluation of In[24]:= True&
During evaluation of In[24]:= Function
During evaluation of In[24]:= NumericQ=Unevaluated[True&]
During evaluation of In[24]:= Set
During evaluation of In[24]:= NumericQ=True&
During evaluation of In[24]:= True&
During evaluation of In[24]:= f[a]
During evaluation of In[24]:= f
During evaluation of In[24]:= a
During evaluation of In[24]:= NumericQ[a]
During evaluation of In[24]:= NumericQ
During evaluation of In[24]:= True&
During evaluation of In[24]:= a
During evaluation of In[24]:= (True&)[a]
During evaluation of In[24]:= True
During evaluation of In[24]:= 8
During evaluation of In[24]:= 8
Out[24]= 8
</code></pre>
<p>Evaluation of <code>f[a]</code> happens inside the <code>Block</code>. I changed the <code>With</code> variable to <code>b</code>, so you can see the difference, but the output is the same using <code>a</code> as variable, I don't get any <code>$RecursionLimit</code> error (using <em>Mathematica</em> 8).</p>
|
1,765 | <p>Can anyone explain to me this behaviour? I've been having more than a couple of similar doubts these last weeks. </p>
<p>For example</p>
<pre><code>f[_?NumericQ] := 8;
</code></pre>
<p>Now, if I do</p>
<pre><code>With[{a = f[a]}, HoldForm@Block[{NumericQ = True &}, a]]
</code></pre>
<p>I get</p>
<pre><code>Block[{NumericQ = True &}, f[a]]
</code></pre>
<p>And if I do</p>
<pre><code>Block[{NumericQ = True &}, f[a]]
</code></pre>
<p>I get</p>
<pre><code>8
</code></pre>
<p>So far so good... Another so-far-so-goodie is (notice the <code>:=</code>)</p>
<pre><code>With[{a := f[a]}, Block[{NumericQ = True &}, a]]
8
</code></pre>
<p>Question: Can anyone help me understand why this output?</p>
<pre><code>With[{a = f[a]}, Block[{NumericQ = True &}, a]]
f[a]
</code></pre>
<p>Could it be that <code>With</code> (<code>=</code> version) not only evaluates and replaces, but also guarantees that the replaced expression won't be reevaluated no matter what until the <code>With</code> is exited? If that's the case I wasn't expecting that. What's happening here?</p>
<p><strong>EDIT</strong></p>
<p>Question also applies to Function</p>
<pre><code>Block[{NumericQ = True &}, #] &[f[a]]
f[a]
</code></pre>
<p>And not just those too. Everything I try behaves the same way... A couple of other examples</p>
<pre><code>Block @@ (Hold[{NumericQ = True &}, exp] /. exp -> f[a])
With[{a := Evaluate@f[a]}, Block[{NumericQ = True &}, a]]
</code></pre>
<p>both give <code>f[a]</code></p>
<p><strong>EDIT</strong></p>
<pre><code>With[{g = h}, h = 8; Print[g] ]
</code></pre>
<p>prints 8, not h, so clearly h is reevaluated inside the <code>With</code> in this case, so <code>g</code> is not so constant. </p>
<p><strong>EDIT</strong></p>
<p>Ok, another couple of examples</p>
<pre><code>In[10]:= ClearAll[h, f];
h := 8 /; NumericQ["a"];
f[_?NumericQ] := 8;
</code></pre>
<p>Now, both <code>h</code> and <code>f[a]</code> remain unevaluated</p>
<pre><code>In[21]:= {h, f[a]}
Out[21]= {h, f[a]}
</code></pre>
<p>Now, with the <code>OwnValues</code> everything works as expected</p>
<pre><code>In[17]:= With[{g = h},
Block[{NumericQ = True &}, g]]
Out[17]= 8
</code></pre>
<p>But with the <code>DownValues</code>, it doesn't </p>
<pre><code>In[19]:= With[{g = f[a]},
Block[{NumericQ = True &}, g]]
Out[19]= f[a]
</code></pre>
<p>Similarly</p>
<pre><code>f[a_?NumericQ] := 8;
g[e_] := Block[{NumericQ = True &}, e]
</code></pre>
<p><code>f[a]</code> evaluates to <code>f[a]</code>, but weirdly</p>
<pre><code>In[17]:= g[Unevaluated@f[a]]
Out[17]= 8
In[18]:= g[f[a]]
Out[18]= f[a]
</code></pre>
| Szabolcs | 12 | <p>You were asking why</p>
<pre><code>f[_?NumericQ] := 8
With[{a = f[a]}, Block[{NumericQ = True &}, a]]
</code></pre>
<p>outputs <code>f[a]</code>.</p>
<p>This is because of <a href="http://reference.wolfram.com/mathematica/ref/Update.html">caching</a> of the result of <code>Condition</code>s and <code>PatternTest</code>s. Compare with this:</p>
<pre><code>With[{a = f[a]}, Block[{NumericQ = True &}, Update[]; a]]
(* ==> 8 *)
</code></pre>
<p>Generally, making global changes that might affect the outcome of a <code>Condition</code> will have unpredictable results due to caching---unless you use <code>Update[]</code> after each change.</p>
<p>Please see also the last paragraph of <a href="http://reference.wolfram.com/mathematica/tutorial/ControllingInfiniteEvaluation.html">Controlling Infinite Evaluation</a>.</p>
|
4,575,165 | <p>Let <span class="math-container">$\alpha$</span> be an arbitrary positive real number in:
<span class="math-container">$$
F_1 = \int_0^1 x^2 \left[ \int_{-1}^{+1} \frac{e^{-\alpha\sqrt{1+x^2+2xy}}(xy+1)}{(1+x^2+2xy)^{3/2}}dy\right]dx $$</span> <span class="math-container">$$
F_2 = \int_1^\infty x^2 \left[ \int_{-1}^{+1} \frac{e^{-\alpha\sqrt{1+x^2-2xy}}(xy-1)}{(1+x^2-2xy)^{3/2}}dy\right]dx
$$</span>
<b>Prove or disprove that <span class="math-container">$F_1 = F_2$</span></b>.
<br>Source of the problem are the equations (3) and (4) in
<a href="https://www.ias.ac.in/article/fulltext/reso/021/05/0447-0452" rel="nofollow noreferrer"><i>A Paradox of Newtonian Gravitation and Laplace’s Solution</i></a>
by Amitabha Ghosh and Ujjal Dey.
They have done already numerical experiments that seem to confirm equality.
<br>Quote: <i>an analytical proof showing that F1 and F2 are exactly equal will be an interesting mathematical exercise.</i>
And that's it. I have no idea how to proceed.</p>
<h3>Progress so far</h3>
I promised myself not to do numerical experiments.
But the outcome of the inner integral - the one between square brackets - is indeed terrible.
So what else would be an option?
With <a href="https://en.wikipedia.org/wiki/Maple_(software)" rel="nofollow noreferrer">MAPLE</a> 8 some values in the publication can be reproduced.
<pre>
for k from 0 to 10 do
alpha := k*0.1;
g(x,alpha) := int(exp(-alpha*sqrt(1+x^2+2*x*y))*(x*y+1)/(1+x^2+2*x*y)^(3/2),y=-1..1);
F1 := evalf(int(g(x,alpha)*x^2,x=0..1));
F2 := evalf(int(-g(x,alpha)*x^2,x=1..10^3));
end do;
</pre>
The special case <span class="math-container">$\alpha = 0$</span> gives <span class="math-container">$\,F_1=\frac{2}{3}\,$</span> and <span class="math-container">$\,F_2=0\,$</span> exactly.
So it appears that nearby <span class="math-container">$\alpha = 0$</span> the integrals must be unequal.
But MAPLE keeps calculating endlessly for those low values and I had to manually stop the program.
<p><p>
<b><a href="https://www.cantorsparadise.com/richard-feynmans-integral-trick-e7afae85e25c" rel="nofollow noreferrer">Feynman trick</a>.</b>
<span class="math-container">$F_1$</span> and <span class="math-container">$F_2$</span> are both a function of <span class="math-container">$\alpha$</span> . We can take derivatives under the integral sign and see what happens near <span class="math-container">$\alpha=0$</span>.
<span class="math-container">$$
\left.\frac{dF_1}{d\alpha}\right|_{\alpha=0} =
- \int_0^1 x^2 \left[ \int_{-1}^{+1} \frac{xy+1}{1+x^2+2xy}dy\right]dx = -\frac{1}{2} \\
\Longrightarrow \quad dF_1 = -\frac{1}{2}d\alpha
$$</span> <span class="math-container">$$
\left.\frac{dF_2}{d\alpha}\right|_{\alpha=0} =
- \int_1^\infty x^2 \left[ \int_{-1}^{+1} \frac{xy-1}{1+x^2-2xy}dy\right]dx = \infty \\
\Longrightarrow \quad dF_2 = \infty\,d\alpha
$$</span>
Leading to the following heuristics.
<span class="math-container">$F_1(\alpha)$</span> is somewhat decreasing from <span class="math-container">$F_1(0)=2/3$</span> to lower values, but the increase in <span class="math-container">$F_2(0)=0$</span> is infinitely large at that place.
Based upon this, it's impossible to <a href="https://en.wikipedia.org/wiki/Keeping_Up_Appearances" rel="nofollow noreferrer">keep up appearances</a> )-:
we can actually say nothing yet whether the outcome is <span class="math-container">$\,F_1(\alpha) = F_2(\alpha)\,$</span> for all <span class="math-container">$\,\alpha \gt 0\,$</span>.
| Simeon Albert L.M. | 1,121,514 | <p>Define
<span class="math-container">$$ F(\alpha) = \int_0^{\infty} \int_0^{\pi} {\rm e}^{-\alpha r} \sin \theta \, \cos \theta \, {\rm d}\theta \, {\rm d}r $$</span></p>
<p>Integrating yields
<span class="math-container">$$ F(\alpha) = -\frac{1}{\alpha} \left. {\rm e}^{-\alpha r} \right|_{r=0}^\infty \; \cdot \; \frac{1}{2} \left. \sin^2 \theta \right|_{\theta=0}^\pi = 0 \quad \text{for all } \alpha>0, $$</span>
but does not exist for <span class="math-container">$\alpha = 0$</span>.</p>
<p>Map the function using <span class="math-container">$u = -r \cos \theta$</span>, <span class="math-container">$v = r \sin \theta$</span> and the determinant of the Jacobian
<span class="math-container">$$ \left| J \right| = \left|
\begin{matrix}
\frac{\partial u}{\partial \theta} = r \sin \theta & \frac{\partial u}{\partial r} = -\cos \theta \\
\frac{\partial v}{\partial \theta} = r \cos \theta & \frac{\partial v}{\partial r} = \sin \theta
\end{matrix} \right| = r$$</span>
This gives
<span class="math-container">$$ F(\alpha) =
\int_0^\infty \int_0^\pi \frac{{\rm e}^{-\alpha r} \sin\theta \cos\theta}{r} \, r \, {\rm d}\theta \, {\rm d}r =
\int_0^\infty \int_{-\infty}^\infty \frac{{\rm e}^{-\alpha \sqrt{u^2 +v^2}} uv}{(u^2 +v^2)^{3/2}} \, {\rm d}u \, {\rm d}v
$$</span>
The integration limits are determined from this map:</p>
<p><a href="https://i.stack.imgur.com/YQD05.jpg" rel="nofollow noreferrer">map from r, <span class="math-container">$\theta$</span> to u, v</a></p>
<p>The origin is shifted by one, <span class="math-container">$u \rightarrow u +1$</span>, to give
<span class="math-container">$$ F(\alpha) =
\int_0^\infty \int_{-\infty}^\infty \frac{{\rm e}^{-\alpha \sqrt{u^2 +2u +1 +v^2}} (u +1)v}{(u^2 +2u +1 +v^2)^{3/2}} \, {\rm d}u \, {\rm d}v
$$</span></p>
<p>Map again using <span class="math-container">$y = -\cos\theta' = u/\sqrt{u^2 +v^2}$</span>, <span class="math-container">$x = r' = \sqrt{u^2 +v^2}$</span>, <span class="math-container">$u = xy$</span> and the determinant of the Jacobian
<span class="math-container">$$ \left| J \right| = \left|
\begin{matrix}
\frac{\partial y}{\partial u} = 1/x -u^2/x^3 & \frac{\partial y}{\partial v} = -uv/x^3 \\
\frac{\partial x}{\partial u} = u/x & \frac{\partial x}{\partial v} = v/x
\end{matrix} \right| = v/x^2$$</span></p>
<p><span class="math-container">$$ F(\alpha) =
\int_0^\infty \int_{-\infty}^\infty \frac{x^2}{v} \frac{{\rm e}^{-\alpha \sqrt{u^2 +2u +1 +v^2}} (u +1)v}{(u^2 +2u +1 +v^2)^{3/2}} \, \frac{v}{x^2} \, {\rm d}u \, {\rm d}v $$</span>
<span class="math-container">$$ = \int_0^\infty x^2 \int_{-1}^1 \frac{{\rm e}^{-\alpha \sqrt{x^2 +2xy +1}} (xy +1)}{(x^2 +2xy +1)^{3/2}} \, {\rm d}y \, {\rm d}x
$$</span>
<span class="math-container">$x$</span> acts as a radius, <span class="math-container">$y$</span> is the fraction of <span class="math-container">$x$</span> along <span class="math-container">$u$</span>. The integration limits are determined from this map:</p>
<p><a href="https://i.stack.imgur.com/BTQtl.jpg" rel="nofollow noreferrer">map from u, v to x, y</a></p>
<p>This is the same function given above by @Gary, so we have
<span class="math-container">$$ F(\alpha) = F_1 - F_2 = \int_0^{\infty} x^2 \int_{-1}^1 \frac{{\rm e}^{-\alpha \sqrt{1 +x^2 +2xy}} (xy +1)} {(1 +x^2 +2xy)^{3/2}} \, {\rm d}y \, {\rm d}x = 0 \quad \text{for all } \alpha>0, $$</span>
but does not exist for <span class="math-container">$\alpha = 0$</span>. Therefore <span class="math-container">$F_1 = F_2$</span> for all <span class="math-container">$\alpha > 0$</span>.</p>
|
4,942 | <p>I'd like to control more aspects of a <code>DateListPlot</code>, for example: shading for weekend days, and/or indicators for daytime/nighttime areas. </p>
<p>By way of illustration, here's a simple example of a set of time data points (recent questions on mathematica.stackexchange):</p>
<pre><code>questions =
First[Rest[
Import["http://api.stackexchange.com/2.0/questions?page=1&\
pagesize=100&order=desc&sort=creation&site=mathematica", "JSON"]]];
questionTimes =
Cases[questions, HoldPattern["creation_date" -> value_] :> value,
Infinity];
hours = (Mod[#, 86400] /86400) 24 & /@ questionTimes ;
DateListPlot[Transpose[{questionTimes, hours}], Filling -> Bottom,
GridLines -> False, Frame -> {True, True, False, False}]
</code></pre>
<p>and the plot is like this:</p>
<p><img src="https://i.stack.imgur.com/xwQlR.png" alt="stack exchange"></p>
<p>I can't see how to show the necessary information along the x-axis, nor how to shade different areas of the graph to show day/night.</p>
<p><em>Edit</em>: I now realise that the date/times returned by the SE API are in Unix Epoch (1970), and I hadn't noticed because I wasn't able to see the day numbers or years on my first attempt at a plot...</p>
| image_doctor | 776 | <p>Here is an alternative approach which uses colour coding of date labels to indicate Weekday/Weekend ( Green/Red ) and colour shade to indicate daytime/nightime ( Lighter/Darker ).</p>
<p>I've dropped some of the points and applied a minimum separation to help make the plot more legible in the limited display space here.</p>
<p><strong>Functions</strong> </p>
<pre><code>(* Check for the type of day of the week *)
WeekdayQ[dateString_String] := With[{weekend = {"Sat", "Sun"}},
And @@ (StringFreeQ[dateString, #] & /@ weekend)]
(* Check for time of day, adjust for time of year and location *)
DaytimeQ[dateString_String] := With[{daylight = Range[8, 18]},
MemberQ[daylight, (DateList@dateString)[[4]]]]
(* Apply colour adjustments to date strings to reflect weekday/time *)
ColorizeDate[dateString_String] := Style[dateString,
If[DaytimeQ@dateString, Lighter, Darker]@
If[WeekdayQ@dateString, Darker@Green, Red]]
(* Ease points apart enough to make the text legible *)
Options[Relax]={MinSeperation->0.03};
Relax[data_,OptionsPattern[]]:=With[{sep=OptionValue@MinSeperation(Max@data-Min@data)},
Flatten[{First@data,First@data +Accumulate[If[#<sep, sep,#]&@Differences@data]}]]
</code></pre>
<p><strong>Prep data for plotting</strong></p>
<pre><code>(* points thinned to allow plot to be readable within the online space constraints *)
questionTimes =
Reverse@Cases[questions,
HoldPattern["creation_date" -> value_] :> value,
Infinity][[Select[Range@Length@questions[[2]], OddQ]]];
shift = DateDifference[{1900, 1, 1}, {1970, 1, 1}]*24*60*60;
hours = (Mod[#, 86400]/86400) 24 & /@ questionTimes;
questionData = {ToDate /@ (questionTimes+shift), hours}\[Transpose];
</code></pre>
<p><strong>Plot the data</strong></p>
<pre><code>With[{labels =
Rotate[ColorizeDate@DateString@#, (3 \[Pi])/2] & /@ questionData[[All, 1]],
softCoords = Relax[AbsoluteTime /@ questionData[[All, 1]], MinSeperation -> 0.01]},
ListPlot[{softCoords, questionData[[All, 2]]}\[Transpose],
Ticks -> {{softCoords, labels}\[Transpose], Automatic},
Filling -> Axis, AxesOrigin -> {First@softCoords, 0}]]
</code></pre>
<p><img src="https://i.stack.imgur.com/P5d5v.png" alt="Mathematica graphics"></p>
|
28,892 | <p>I was searching on MathSciNet recently for a certain paper by two mathematicians. As I often do, I just typed in the names of the two authors, figuring that would give me a short enough list. My strategy was rather dramatically unsuccessful in this case: the two mathematicians I listed have written 80 papers together!</p>
<p>So this motivates my (rather frivolous, to be sure) question: which pair of mathematicians has the most joint papers? </p>
<p>A good meta-question would be: can MathSciNet search for this sort of thing automatically? The best technique I could come up with was to think of one mathematician that was both prolific and collaborative, go to their "profile" page on MathSciNet (a relatively new feature), where their most frequent collaborators are listed, alphabetically, but with the wordle-esque feature that the font size is proportional to the number of joint papers. </p>
<p>Trying this, to my surpise I couldn't beat the 80 joint papers I've already found. Erdos' most frequent collaborator was Sarkozy: 62 papers (and conversely Sarkozy's most frequent collaborator was Erdos). Ronald Graham's most frequent collaborator is Fan Chung: 76 papers (and conversely).</p>
<p>I would also be interested to hear about triples, quadruples and so forth, down to the point where there is no small set of winners.</p>
<hr>
<p><b>Addendum</b>: All right, multiple people seem to want to know. The 80 collaboration pair I stumbled upon is Blair Spearman and Kenneth Williams. </p>
| Victor Protsak | 5,740 | <blockquote>
I.M.Gelfand and M.I.Graev: 119
</blockquote>
<p><b>Disclaimer:</b> For the purposes of this answer, the paper count is the number given by MathSciNet, which includes book translations.</p>
|
4,272,755 | <p>Calculate the triple integral using spherical coordinates: <span class="math-container">$\int_C z^2dxdydz$</span> where C is the region in <span class="math-container">$R^3$</span> described by <span class="math-container">$1 \le x^2+y^2+z^2 \le 4$</span></p>
<p>Here's what I have tried:</p>
<p>My computation for <span class="math-container">$z$</span> is: <span class="math-container">$\sqrt{1-x^2-y^2} \le z \le \sqrt{4-x^2-y^2}$</span>, as for y I get: <span class="math-container">$-\sqrt{1-x^2}\le y \le \sqrt{4-x^2}$</span> and for x I get: <span class="math-container">$1 \le x \le 2$</span></p>
<p>The triple integral becomes:</p>
<p><span class="math-container">$$\int_{1}^2 dx \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dy \int_{-\sqrt{1-x^2-y^2}}^{\sqrt{1-x^2-y^2}}z^2dz$$</span></p>
<p>The way I have pictured theta is as so:</p>
<p><a href="https://i.stack.imgur.com/Y1fvn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Y1fvn.png" alt="enter image description here" /></a></p>
<p>Where The Red + Green is equal to <span class="math-container">$\frac{\pi}{2}$</span> But because we're only interested in the region from <span class="math-container">$1 \le x \le 2$</span> this covers only <span class="math-container">$\frac{\pi}{4}$</span></p>
<p>The integral becomes:</p>
<p>I presumed <span class="math-container">$\phi$</span> only goes from <span class="math-container">$\frac{\pi}{4}$</span> also, and we know that <span class="math-container">$r$</span> goes from <span class="math-container">$1\le r \le 2$</span></p>
<p>So the final integral becomes:</p>
<p><span class="math-container">$$\int_0^{\frac{\pi}{4}} d\theta \int_0^{\frac{\pi}{4}} \cos^2(\phi)\sin(\phi) \space d\phi \int_1^2 r^4 dr$$</span></p>
<p>Because <span class="math-container">$z^2 = r^2 \cos^2(\phi)$</span></p>
<p>However my answer that I get is <span class="math-container">$2\pi(\sqrt{2}-4)$</span> but the answer should be <span class="math-container">$\frac{124\pi}{15}$</span>. I would greatly appreciate the communities assistance</p>
| Joshua Pasa | 689,689 | <p>From the conditions you have stated above <span class="math-container">$C: 1 \leq x^2 + y^2 + z^2 \leq4$</span> by using a substitution we can see that <span class="math-container">$1 \leq r \leq 2$</span>. However there are no conditions on the other spherical components so we can see that
<span class="math-container">$$\int_C z^2dxdydz=\int_0^\pi\int_0^{2\pi}\int_1^2 z^2|J|dr d\theta d\phi$$</span>
We still need to find <span class="math-container">$z$</span> in terms of spherical coordinates which is <span class="math-container">$z=r\cos\phi$</span>. The jacobian for spherical coordinates is <span class="math-container">$r^2\sin\phi$</span> substituting this in we find that
<span class="math-container">$$\int_0^\pi\int_0^{2\pi}\int_1^2 (r\cos\phi)^2 r^2\sin\phi dr d\theta d\phi$$</span>
<span class="math-container">$$\int_0^\pi\int_0^{2\pi}\int_1^2 r^2\cos^2\phi \, r^2\sin\phi dr d\theta d\phi$$</span>
<span class="math-container">$$\int_0^\pi\int_0^{2\pi}\int_1^2 r^4\cos^2\phi \,\sin\phi dr d\theta d\phi$$</span>
<span class="math-container">$$\int_0^\pi\int_0^{2\pi} \frac{31}{5}\cos^2\phi \,\sin\phi d\theta d\phi$$</span>
<span class="math-container">$$\int_0^\pi \frac{62\pi}{5}\cos^2\phi \,\sin\phi \,d\phi$$</span>
<span class="math-container">$$ \frac{62\pi}{5}\int_0^\pi\cos^2\phi \,\sin\phi \,d\phi$$</span>
Use <span class="math-container">$u$</span> substitution. Let <span class="math-container">$u = \cos\phi$</span> which means <span class="math-container">$du = -\sin\phi d\phi$</span>. This means that the integral ranges from <span class="math-container">$u=1$</span> to <span class="math-container">$u=-1$</span>
<span class="math-container">$$ -\frac{62\pi}{5}\int_1^{-1} u^2 du$$</span>
<span class="math-container">$$ \frac{62\pi}{5}\int_{-1}^{1} u^2 du$$</span>
<span class="math-container">$$ \frac{62\pi}{5} \left(\frac{1^3}{3} - \frac{-1^3}{3}\right) $$</span>
<span class="math-container">$$ \frac{62\pi}{5} \left(\frac{2}{3}\right) $$</span>
<span class="math-container">$$ \frac{124\pi}{15} $$</span></p>
|
3,760,450 | <p>I have a class in numerical mathematics, and I received several tasks I should answer. I am not a mathematician, and this is a bit out of my mind range, and I would be grateful for answers.
Question is as follows:</p>
<p>Let <span class="math-container">$x, y \in \Bbb{C}^n$</span> be arbitrary vectors. Show what can be the rank of matrix <span class="math-container">$xy^* \in \Bbb{C}^{n×n}$</span>. If rank of matrix <span class="math-container">$xy^*$</span> is equal to zero, describe (in most general terms) what should vectors <span class="math-container">$x$</span> and <span class="math-container">$y$</span> look like.</p>
<p>Edit:</p>
<p>From what I know, multiplying vector <span class="math-container">$x$</span> with conjugate transpose of vector <span class="math-container">$y$</span>, both of which are dimension <span class="math-container">$n$</span>, will result in square matrix of dimension <span class="math-container">$n\times n$</span>. Maximum number of diagonal elements that can be independent are <span class="math-container">$n$</span>, so rank of that matrix can vary from zero to n. (Please correct me if I am completely wrong in this)
And only way the rank can be zero is if all diagonal elements are zero, so <span class="math-container">$x$</span> and <span class="math-container">$y$</span> should be null vectors. These are my thoughts, but I am not sufficiently familiar with linear algebra to be sure. Please help</p>
| Community | -1 | <p><span class="math-container">$f(-x)=x^{2}-|x|=f(x)$</span></p>
<p>Hence <span class="math-container">$x^{2}-|x| \quad$</span> is an even function.</p>
<p>For <span class="math-container">$x<0$</span>
<span class="math-container">$f(x)=x^{2}+x$</span> (By symmetry we can plot the the total graph)</p>
<p>f(x) is a quadratic equation. whose vertex point will be (<span class="math-container">$\frac{-b}{2a},\frac{-d}{4a}$</span>)</p>
<p><a href="https://i.stack.imgur.com/oTsbN.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oTsbN.jpg" alt="hope it helps" /></a></p>
|
85,343 | <p>I am looking for a reference to study classical (i.e., not quantized) Yang-Mills theory. </p>
<p>Most of the sources I find focus on mathematical aspects of the theory, like Bleecker's book <em>Gauge theory and variational principles</em>, or Baez & Muniain's <em>Gauge fields, knots and gravity</em>.</p>
<p>But I am more interested something similar to the standard development of electromagnetism, as can be found, for example, in Landau & Lifchitz's course on theoretical physics. To be more exact, I would like to learn about the field equations (Yang-Mills and Einstein equations), but also about the corresponding Lorentz Law, the energy of the Yang-Mills field,... and the analogous concepts of what is made for the electromagnetism.</p>
<p>That is, I look for a rigorous exposition where, at the same time, I could learn whether it is possible to prove, at the classical level, the quick decrease of the strong interaction, and things like that.</p>
| Jon | 19,520 | <p>I would add: Atiyah, Michael F. (1979), <em>Geometry of Yang–Mills fields</em> and then the book of the same author about gauge theories: Atiyah, Michael F. (1988e), <em>Collected works. Vol. 5 Gauge theories</em></p>
|
1,064,091 | <p>I am asked to generate 200 and 1000 points from a bi-variate normal mixture densities.
I am trying to understand the algorithm, not just the matlab code (I have to write it, not use an existing function). I found a code on mathworks: <a href="http://www.mathworks.com/help/matlab/ref/randn.html#bufqioz-2" rel="nofollow">http://www.mathworks.com/help/matlab/ref/randn.html#bufqioz-2</a>
on the second example, but I do not understand the math behind it. If someone can help me out with that, or explain it regardless to the code, or if you have another algorithm which can be explained, I will be grateful.
Thanks a lot!</p>
| Suzu Hirose | 190,784 | <p>Let $f$ be the uniform distribution on $[0,1]$ then $P(S>x)=(1-x)$ and so according to your formula we should have $P(S>t)=\int_0^t P(S>x)dx=t-1/2t^2$. Plug in $t=1/3$ and you get $P(S>1/3)=1/3-1/18$. But clearly $P(S>1/3)=2/3$ so the equation is complete nonsense.</p>
|
226,449 | <p>Many counting formulas involving factorials can make sense for the case $n= 0$ if we define $0!=1 $; e.g., Catalan number and the number of trees with a given number of vetrices. Now here is my question:</p>
<blockquote>
<p>If $A$ is an associative and commutative ring, then we can define an
unary operation on the set of all the finite subsets of our ring,
denoted by $+ \left(A\right) $ and $\times \left(A\right)$. While it
is intuitive to define $+ \left( \emptyset \right) =0$, why should
the product of zero number of elements be $1$? Does the fact that $0! =1$ have anything to do with 1 being the multiplication unity of integers?</p>
</blockquote>
| N. S. | 9,176 | <p>I think that the definition $0!=1$ is the one which makes most of the formulas work nicely.</p>
<p>For example $(n+1)!=n! (n+1)$ also works for $n=0$, as long as $0!=1$.</p>
<p>But most importantly, $\binom{n}{k}$ also works in the case $k=0$. Note that $\binom{n}{0}$ has to be $1$, if you want to have a nice formula for $(a+b)^n$. So for convenienceence, we define $\binom{n}{0}$ to be $1$, and if you want the binomial coefficient to be given by the standard formula, then $0!$ has to be 1.</p>
|
300,253 | <p>I'm interested in invertible matrices that are built out of invertible sub-blocks. For example, four sub-blocks from $GL_n(F)$ (i.e. the group of $n \times n$ invertible matrices over a field $F$) can be assembled into a $2n \times 2n$ matrix, which may or may not be invertible.</p>
<p>Suppose that a $kn \times kn$ matrix, $M$, composed of $k^2$ invertible sub-blocks is invertible. Is it true that each sub-block of $M^{-1}$ is invertible?</p>
<p>I think that it is true (although I am happy to be shown otherwise), but I am having difficulty constructing a general proof.</p>
<p>In the case $k=1$, there is nothing to prove.</p>
<p>For the case $k=2$, we can use block-wise row reduction, as described <a href="http://en.wikipedia.org/wiki/Matrix_inversion_lemma#Blockwise_inversion" rel="nofollow">here</a>.</p>
<p>For $k>2$, I'm stuck. I've tried examining the maps $X \mapsto MX$ and $Y \mapsto M^{-1}Y$ where I suppose that some sub-block $B_{ij}$ of $M^{-1}$ is not invertible. My thought was that this might tell me something about block $B_{ji}$ of $M$, but I can't seem to make any conclusions.</p>
<p>Any suggestions?</p>
<p>Note that I am mainly interested in the case where $F = \mathrm{GF}(2)$.</p>
| Michael Hardy | 11,667 | <p>"Did" has posted a guess. Here's another. Some processes $\{X_t\}_{t\ge0}$ satisfy $\operatorname{var}(X_t)=ct$ (Poisson processes, Wiener processes, many others). Possibly it could mean $c=1$.</p>
<p>(But if something like the <a href="http://en.wikipedia.org/wiki/Ornstein%E2%80%93Uhlenbeck_process" rel="nofollow">Ornstein–Uhlenbeck process</a> is being written about, probably "Did"'s guess is right.)</p>
|
377,266 | <p>My question is very direct:</p>
<blockquote>
<p>What are the motivations for the name "jet"(subjet, superjet) in the context of viscosity solutions for second order fully nonlinear elliptic PDE?</p>
</blockquote>
<p>The definition of which can be seen in Crandall, Ishii, Lions:</p>
<p><em>Crandall, Michael G.; Ishii, Hitoshi; Lions, Pierre-Louis</em>, <a href="http://dx.doi.org/10.1090/S0273-0979-1992-00266-5" rel="nofollow noreferrer"><strong>User’s guide to viscosity solutions of second order partial differential equations</strong></a>, Bull. Am. Math. Soc., New Ser. 27, No. 1, 1-67 (1992). <a href="https://zbmath.org/?q=an:0755.35015" rel="nofollow noreferrer">ZBL0755.35015</a>.</p>
<p>see also <a href="https://arxiv.org/pdf/math/9207212.pdf" rel="nofollow noreferrer">https://arxiv.org/pdf/math/9207212.pdf</a>.</p>
| YCor | 14,094 | <p>Strictly speaking, he answer is that there is no motivation for the name "jet" in the context of viscosity solutions for second order fully nonlinear elliptic PDE, because it was initially introduced in the more basic framework of differential calculus/geometry.</p>
<p>Still one can wonder about when and where it was introduced. I suspected it was initially introduced in French and possibly by Bourbaki, and asked J-P. Serre in an email, and got the answer:</p>
<blockquote>
<p><em>Il me semble que la notion de jet, et la terminologie
correspondante, sont dus à Ehresmann (mais pas
à Bourbaki), vers 1950. Je vous envoie ci-joint un article de 1961 qui
confirme ceci.</em></p>
</blockquote>
<p>(<em>It seems to be that the notion of jet, and the corresponding terminology, are due to Ehresmann (but not Bourbaki) around 1950. Here is an article from 1961 confirming this.</em>)</p>
<p>The article is <em>Une généralisation du calcul des jets et quelques prolongements généralisés de variétés différentiables</em> (<a href="https://doi.org/10.2748/tmj/1178244307" rel="noreferrer">DOI link</a> linking to ProjectEuclid, no paywall) by M. Kawaguchi. Its first two sentences are:</p>
<blockquote>
<p><em>Le mot "calcul des jets" a fait son début dans l'article
[1] de Ch. Ehresmann. En 1951, Ch. Ehresmann s'est intéressé surtout aux
fondements de la geometrie differentielle. [1] Les prolongements d'une variété différentiable, Atti IV Congresso Unione mat.
Italiana, Taormina Ott, (1951), 1–9.</em></p>
</blockquote>
<p><em>(The phrase "jet calculus" first appeared in the article [1] by Ch. Ehresmann. In 1951, Ch. Ehresmann was especially interested in foundations of differential geometry.)</em></p>
<hr>
<p>I should add that in French "jet" has nothing to do with plane or engine.</p>
<p>"Jet" is the substantive of "jeter" = throw, launch. I can see the translations: spurt, jet, squirt, throw, spray. Also a derived phrase is "premier jet" = "first draft".</p>
|
4,035,080 | <p>Let <span class="math-container">$E$</span> be a Polish space, and let <span class="math-container">$\mu$</span> be a measure on <span class="math-container">$E$</span>. Define the following properties:</p>
<ul>
<li><span class="math-container">$E$</span> is <span class="math-container">$\sigma$</span>-<em>compact</em> if <span class="math-container">$E$</span> is the countable union of compact sets.</li>
<li><span class="math-container">$E$</span> is <em>locally compact</em> if every <span class="math-container">$x \in E$</span> has an open neighborhood <span class="math-container">$U$</span> whose closure is compact.</li>
<li><span class="math-container">$\mu$</span> is <em>locally finite</em> if for every <span class="math-container">$x \in E$</span>, there is an open set <span class="math-container">$U \subset E$</span> containing <span class="math-container">$x$</span> with <span class="math-container">$\mu(U) < \infty$</span>.</li>
<li><span class="math-container">$\mu$</span> is a <em>Borel measure</em> if for every compact <span class="math-container">$K \subset E$</span>, we have <span class="math-container">$\mu(K) < \infty$</span>.</li>
</ul>
<p>Clearly a locally finite measure is Borel. And if <span class="math-container">$E$</span> is locally compact (not even necessarily separable or complete), Borel measures are necessarily locally finite. But are there more general conditions for which Borel measures are locally finite?</p>
<p><strong>Question:</strong> If <span class="math-container">$E$</span> is a <span class="math-container">$\sigma$</span>-compact Polish space and <span class="math-container">$\mu$</span> is a Borel measure, is <span class="math-container">$\mu$</span> locally finite?</p>
<p>I can’t think of a counter example to this, but I’m having trouble proving it. My original strategy was to prove that a <span class="math-container">$\sigma$</span>-compact Polish space is locally compact. However, as the comments demonstrate, <span class="math-container">$\sigma$</span>-compact Polish spaces are not necessarily locally compact, so that strategy doesn’t work. A counter example is the subset <span class="math-container">$X \subset \ell^2$</span> given by the union of the lines <span class="math-container">$X_k = \{\lambda e_k : \lambda \in \mathbb R\}$</span>, where <span class="math-container">$\{e_k\}_{k \geq 1}$</span> is the standard orthonormal basis of <span class="math-container">$\ell^2$</span>. Then <span class="math-container">$X$</span> is <span class="math-container">$\sigma$</span>-compact, but not locally compact (at the origin specifically).</p>
<p>But I’m not sure of a measure <span class="math-container">$\mu$</span> on this space that is Borel, but not locally finite; the problem is there are compact subsets of <span class="math-container">$X$</span> containing <span class="math-container">$0$</span> and intersecting infinitely many of the <span class="math-container">$X_k$</span>. Can anyone think of a counter example?</p>
| Henno Brandsma | 4,280 | <p>I think that indeed the idea from the common thread works:</p>
<p>Let <span class="math-container">$e_n, n \in \Bbb N$</span> be the standard orthonormal base of Hilbert space <span class="math-container">$\ell^2$</span>. Let <span class="math-container">$X = \{\lambda e_n \mid n \in \Bbb N, \lambda \in \Bbb R\}$</span>. Then <span class="math-container">$X$</span> is <span class="math-container">$\sigma$</span>-compact Polish but not locally compact (at <span class="math-container">$0$</span>).</p>
<p>Let <span class="math-container">$L_k = \{\lambda e_k: \lambda \neq 0\}$</span> and <span class="math-container">$\delta_A$</span> be the Dirac measure with carrier <span class="math-container">$A$</span>: <span class="math-container">$\delta_A(B) = 1$</span> iff <span class="math-container">$A \cap B \neq \emptyset$</span> and <span class="math-container">$0$</span> otherwise, we can define <span class="math-container">$\mu = \displaystyle_{k=1}^\infty \delta_{L_k}$</span>, which is a Borel measure (not <span class="math-container">$\sigma$</span>-finite, and not locally finite at <span class="math-container">$0$</span>), but (I think) finite at compacta.</p>
<p>I haven't checked all the nitty gritty details. Others might feel inclined to add to this.</p>
|
548,902 | <p>I need to find the limit:
$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\left[ {{{(a + {1 \over n})}^2} + {{(a + {2 \over n})}^2} + ... + {{(a + {{n - 1} \over n})}^2}} \right]$ </p>
<p>any ideas here? I've tried to use "squeeze theorem" but with no luck.. </p>
| Daniel Robert-Nicoud | 60,713 | <p>Once you have shown that the function is continuous, you can certainly prove that the function is differentiable by showing that the partial derivatives agree. Indeed you have that the derivative of $f$ in direction $\mathbf{v}$ (a unit vector) at $(x,y)$ is given by
$$(\nabla f)_{(x,y)}\cdot\mathbf{v}$$
(you can try to show this: simply apply a rotation to the coordinate frame).</p>
<p>Another way is to compute directly the derivative, which gives you
$$\lim_{\|\mathbf{v}\|\rightarrow 0}\frac{f(x+v_x,y+v_y)-f(x,y)}{\|v\|}$$
Notice that in the specific case $(x,y)=(0,0)$ this gives you exactly the formula written by @Siminore.</p>
|
231,360 | <p>Let $A,B\subseteq\mathbb R^d$ with $A$ closed and $B$ open and such that $A\cap B\neq\emptyset$. Assume that there exists a sequence $(x_k)_{k\in\mathbb N}\subseteq\mathbb R^d$ conveging in Euclidean norm to some $x\in A\cap B$. Do we have $x_k\in A\cap B$ for some $k\in\mathbb N$? If $x$ is an interior point of $A$ then of course this is true, since $B$ is an open set and $A$ then contains a ball around $x$ of size $\varepsilon>0$. So the question becomes problematic if $x\in\partial A$ (boundary of $A$).</p>
| Harald Hanche-Olsen | 23,290 | <p>No. Take $A=\{0\}$ and $B=\mathbb{R}^n$, $x_k=(1/k,0,\ldots,0)$.</p>
<p>More generally, let $x$ be any non-interior point in $A$. For each $n$, pick $x_n$ with $|x_n-x|<1/n$, $x_n\notin A$, and $x_n\in B$ (the latter is automatic when $n$ is large enough).</p>
|
2,964,897 | <blockquote>
<p>Using Leibniz on <span class="math-container">$\sum_{n=1}^\infty \sin(\pi \sqrt{n^2+1})$</span></p>
</blockquote>
<p>So the question actually is how to rewrite <span class="math-container">$\sin(\pi\sqrt{n^2+1})$</span> in the form of <span class="math-container">$(-1)^n\times a_n$</span> so that I can apply Leibniz and decide the convergence or divergence?</p>
<p>I'm sorry but I'm pretty new in studying <em>series</em>.</p>
| marty cohen | 13,079 | <p>Let
<span class="math-container">$\sqrt{n^2+1}
=n+d(n)$</span>.
Then</p>
<p><span class="math-container">$\begin{array}\\
\sin(\pi\sqrt{n^2+1})
&=\sin(\pi(n+d(n))\\
&=\sin(\pi n)\cos(\pi d(n))
+\cos(\pi n)\sin(\pi d(n))\\
&=(-1)^n\sin(\pi d(n))\\
\end{array}
$</span></p>
<p>Also</p>
<p><span class="math-container">$\begin{array}\\
d(n)
&=\sqrt{n^2+1}-n\\
&=(\sqrt{n^2+1}-n)\dfrac{\sqrt{n^2+1}+n}{\sqrt{n^2+1}+n}\\
&=\dfrac{1}{\sqrt{n^2+1}+n}\\
\end{array}
$</span></p>
<p>so that
<span class="math-container">$0 < d(n)
\lt \dfrac1{2n}$</span>
and <span class="math-container">$d(n)$</span>
is decreasing
so that,
since <span class="math-container">$\sin(x)$</span> is
increasing for
<span class="math-container">$-\pi/2 < x < \pi/2$</span>,
<span class="math-container">$\sin(\pi d(n))$</span>
is also decreasing.</p>
<p>Therefore
<span class="math-container">$\sum_{n=0}^{\infty} (-1)^n \sin(\pi d(n))
$</span>
converges by the
alternating series test.</p>
|
3,910,623 | <p>There is a problem that appears in an interview<span class="math-container">$^\color{red}{\star}$</span> with <a href="https://en.wikipedia.org/wiki/Vladimir_Arnold" rel="nofollow noreferrer">Vladimir Arnol'd</a>.</p>
<blockquote>
<p>You take a spoon of wine from a barrel of wine, and you put it into your cup of tea. Then you return a spoon of the (nonuniform!) mixture of tea from your cup to the barrel. Now you have some foreign substance (wine) in the cup and some foreign substance (tea) in the barrel. Which is larger: the quantity of wine in the cup or the quantity of tea in the barrel at the end of your manipulations?</p>
</blockquote>
<p>This problem is also quoted <a href="https://math.stackexchange.com/q/895627">here</a>.</p>
<hr />
<p>Here's my solution:</p>
<p>The key is to consider the proportions of wine and tea in the second spoonful (that is, the spoonful of the nonuniform mixture that is transported from the cup to the barrel). Let <span class="math-container">$s$</span> be the volume of a spoonful and <span class="math-container">$c$</span> be the volume of a cup. The quantity of wine in this second spoonful is <span class="math-container">$\frac{s}{s+c}\cdot s$</span> and the quantity of tea in this spoonful is <span class="math-container">$\frac{c}{s+c}\cdot s$</span>. Then the quantity of wine left in the cup is <span class="math-container">$$s-\frac{s^2}{s+c}=\frac{sc}{s+c}$$</span> and the quantity of tea in the barrel now is also <span class="math-container">$\frac{cs}{s+c}.$</span> So the quantities that we are asked to compare are the same.</p>
<p>However, Arnol'd also says</p>
<blockquote>
<p>Children five to six years old like them very much and are able to solve them, but they
may be too difficult for university graduates, who are spoiled by formal mathematical training.</p>
</blockquote>
<p>Given the simple nature of the solution, I'm going to guess that there is a trick to it. How would a six year old solve this problem? My university education is interfering with my thinking.</p>
<hr />
<p><span class="math-container">$\color{red}{\star}\quad$</span> S. H. Lui, <a href="https://www.ams.org/notices/199704/arnold.pdf" rel="nofollow noreferrer">An interview with Vladimir Arnol′d</a>, Notices of the AMS, April 1997.</p>
| C.F.G | 272,127 | <ol>
<li>First We have a <span class="math-container">$B_{wine}$</span> and a <span class="math-container">$C_{tea}$</span> and a <span class="math-container">$S$</span>poon</li>
<li>Now we have <span class="math-container">$B_{wine}-S_{wine}$</span> and <span class="math-container">$C_{tea}+S_{wine}$</span></li>
<li>Then we have <span class="math-container">$B_{wine}-S_{wine}+(\frac{k}{100}S_{wine}+\frac{100-k}{100}S_{tea})$</span> and <span class="math-container">$ C_{tea}+S_{wine}-(\frac{k}{100}S_{wine}+\frac{100-k}{100}S_{tea})$</span></li>
</ol>
<p>Which shows that in cup of tea we have <span class="math-container">$\frac{100-k}{100}S_{wine}$</span> and in wine barrel we have <span class="math-container">$\frac{100-k}{100}S_{tea}$</span>. Of course <span class="math-container">$S_{tea}=S_{wine}$</span>. (Both are one spoon)!</p>
|
2,247,900 | <p>Prove the following inequality</p>
<p>$$\ln \frac{\pi + 2}{2} \cdot \frac{2}{\pi} < \int \limits_0^{\pi/2} \frac{\sin\ x}{x^2 + x} < \ln \frac{\pi + 2}{2}$$</p>
<p>I can prove that $\frac{\sin\ x}{x^2 + x} < \frac{1}{x + 1} \ \forall x \in (0, +\infty) \Rightarrow \int \limits_0^{\pi/2} \frac{\sin\ x}{x^2 + x} < \int \limits_0^{\pi/2} \frac{1}{x + 1} = \ln \frac{\pi + 2}{2}$.</p>
<p>Unfortunately, I don't know what happens when $x = 0$.</p>
<p>I can prove $\leqslant$ on LHS using Mean Value Theorem, but I have no idea how to prove that the sign is strict.</p>
| DXT | 372,201 | <p>Using $$\frac{\sin x}{x}>\frac{2}{\pi}\forall x\in \left(0,\frac{\pi}{2}\right)$$</p>
<p><a href="https://math.stackexchange.com/questions/842978/proving-frac2-pi-x-le-sin-x-le-x-for-x-in-0-frac-pi-2">Proving $\frac2\pi x \le \sin x \le x$ for $x\in [0,\frac {\pi} 2]$</a></p>
|
160,169 | <p>Consider the following implementation of the complex square root:</p>
<pre><code>f[z_]:=Sqrt[(z - I)/(z + I)]*(z + I);
</code></pre>
<p>This implementation has branch points at $\lambda=\pm i$ and a (vertical) branch cut connecting them.</p>
<p>Then</p>
<pre><code>g[z_]:=Sinc[f[z]];
</code></pre>
<p>(recalling $\mathrm{sinc}(x)=\sin(x)/x$ ) has no branch cut and it is analytic on the entire complex plane, and admits power series expansions at $\lambda=\pm i$. </p>
<p>Indeed, using Mathematica 11.0.0 (Mac OS 10.10.5) gives:</p>
<pre><code>Series[Sinc[f[z]], {z, I, 4}]
</code></pre>
<p>$1-\frac{1}{3} i (z-i)-\frac{1}{5} (z-i)^2+\frac{11}{315} i (z-i)^3+\frac{61
(z-i)^4}{5670}+O\left((z-i)^5\right)$</p>
<p>and</p>
<pre><code>SeriesCoefficient[Sinc[f[z]], {z, I, 4}]
</code></pre>
<p>gives $\frac{61}{5670}$.</p>
<p>Now, using Mathematica 11.1.1 (both on Mac OS 10.12 Sierra and Linux Ubuntu 16 LTS)</p>
<pre><code>Series[Sinc[f[z]], {z, I, 4}]
</code></pre>
<p>returns</p>
<blockquote>
<pre><code>Series[Sinc[f[z]], {z, I, 4}]
</code></pre>
</blockquote>
<p>and</p>
<pre><code>SeriesCoefficient[Sinc[f[z]], {z, I, 4}]
</code></pre>
<p>returns</p>
<blockquote>
<pre><code>SeriesCoefficient[Sinc[f[z]], {z, I, 4}].
</code></pre>
</blockquote>
<p>So neither of these stock functions work in properly in Mathematica 11.1.1. Does anyone know what is going on? Will this be fixed? They worked properly even in Mathematica 9 and also in Mathematica 11.0.0</p>
<p>Besides any information, I'd also appreciate if anyone has a workaround for this.</p>
| Carl Woll | 45,431 | <p>I think it's worth reporting this issue to support. A workaround is to use something like:</p>
<pre><code>fSeries[e_, {z_,p_,n_}] := Series[
e /. z->z+p,
{z, 0, n}
] /. Verbatim[SeriesData][x_, 0, r__] :> SeriesData[x, p, r]
</code></pre>
<p>For your example:</p>
<pre><code>fSeries[Sinc[f[z]], {z, I, 4}] //TeXForm
</code></pre>
<blockquote>
<p>$1-\frac{1}{3} i (z-i)-\frac{1}{5} (z-i)^2+\frac{11}{315} i (z-i)^3+\frac{61 (z-i)^4}{5670}+O\left((z-i)^5\right)$</p>
</blockquote>
|
2,163,494 | <p>Let $f: A\to B; \ g,h:B\to A$ and $f\circ g = I_B$ and $h \circ f = I_A$</p>
<p>I want to simply state that for any function $f$ if $f \circ h = I_A$ then it must be that $h = f^{-1}$ but that seems incomplete to me. What can I do for fixing this?</p>
| E. Joseph | 288,138 | <p><strong>Yes</strong>, this is true.</p>
<p>Since $f$ is defined on $\mathbb R^n$, and $A\subset \mathbb R^n$, let's take $F$ a close subset containing $A$ (you can take $F=\overline A$ the closure of $A$).</p>
<p>Then since $f$ is continuous on $F$, $f(F)$ is bounded because $F$ is closed.</p>
<p>So $f$ is bounded on $F$ so $f$ is bounded on $A$ because $A\subset F$.</p>
<p>So $f(A)$ is bounded.</p>
|
1,315,641 | <p>I rewrite $f(z)$ using partial fractions to get $f(z)=\frac{1}{z}+\frac{2}{1-2z}$.</p>
<p>We need powers of $\left(z-\frac{1}{2}\right)$
So how do I rewrite $\frac{1}{z}$?</p>
<p>So I rewrite it as $\frac{1}{\frac{1}{2}+\left(z-\frac{1}{2}\right)}$
And I write it as $2\left(z-\frac{1}{2}\right)$</p>
<p>How do I get the binomial expansion in powers of $\left(z-\frac{1}{2}\right)$?</p>
| Aleksandar | 240,930 | <p>The series is the following:</p>
<p>$\sum _{k=1}^\infty(-2)^{k}(z-\frac{1}{2})^{n}$.</p>
<p>That is the Laurent series for $f(z)$.</p>
<p>Hope this helped.</p>
<p>By the way:</p>
<p>Here is how to find the disk of convergence:</p>
<p>We'll find the pole of $f(z)$. To do that we must set:</p>
<p>$z(1-2z)=0$</p>
<p>This is the same as,</p>
<p>$z-2z^2=0$</p>
<p>This has two zeros at $z=\frac{1}{2}$ and $z=0$. </p>
<p>Theorem:</p>
<p>If $\beta_{1}$ and $\beta_{2}$ are two singularities of a function $f(z)$ then can be expressed as a series expansion then:</p>
<p>$|\alpha-\beta_{1}|<|z|<|\alpha-\beta_{2}|$</p>
<p>Where $\alpha$ is the point you expanded the series about. </p>
<p>In this case,</p>
<p>$0<|z|<|\frac{1}{2}|$</p>
<p>Or, $|z|<\frac{1}{2}$.</p>
<p>Hope this helps.</p>
|
3,348,780 | <p>I worked through <span class="math-container">$\int \frac{e^x}{(1-e^x)^2}dx$</span> using u-substitution, but my answer, <span class="math-container">$(1-e^x)^{-1}+C$</span> is incorrect. It should be <span class="math-container">$- \ln|1-e^x|+C$</span></p>
<blockquote>
<p><span class="math-container">$$\int \frac{e^x}{(1-e^x)^2}dx$$</span></p>
</blockquote>
<p>Let <span class="math-container">$u = 1 - e^x$</span></p>
<p>Then <span class="math-container">$du = -e^x$</span> and <span class="math-container">$dx = \frac{-du}{e^x}$</span></p>
<p>So <span class="math-container">$\int \frac{e^x}{u^2} \frac{-1}{e^x}du = -\int \frac{1}{u^2}du = -\int u^{-2}du$</span></p>
<p><span class="math-container">$$-\int u^{-2} du$$</span></p>
<p><span class="math-container">$$=-(\frac{u^{-1}}{-1})+C$$</span></p>
<p><span class="math-container">$$=u^{-1}+C$$</span></p>
<p><span class="math-container">$$=(1-e^x)^{-1}+C$$</span></p>
<p>What did I do wrong?</p>
| DonAntonio | 31,254 | <p>The result is false: in fact, </p>
<p><span class="math-container">$$\int\frac{e^x}{(1-e^x)^2}dx=\frac1{1-e^x} +C$$</span></p>
<p>This is: what you did is completely correct. There is not <span class="math-container">$\;\ln\;$</span> here.</p>
|
3,348,780 | <p>I worked through <span class="math-container">$\int \frac{e^x}{(1-e^x)^2}dx$</span> using u-substitution, but my answer, <span class="math-container">$(1-e^x)^{-1}+C$</span> is incorrect. It should be <span class="math-container">$- \ln|1-e^x|+C$</span></p>
<blockquote>
<p><span class="math-container">$$\int \frac{e^x}{(1-e^x)^2}dx$$</span></p>
</blockquote>
<p>Let <span class="math-container">$u = 1 - e^x$</span></p>
<p>Then <span class="math-container">$du = -e^x$</span> and <span class="math-container">$dx = \frac{-du}{e^x}$</span></p>
<p>So <span class="math-container">$\int \frac{e^x}{u^2} \frac{-1}{e^x}du = -\int \frac{1}{u^2}du = -\int u^{-2}du$</span></p>
<p><span class="math-container">$$-\int u^{-2} du$$</span></p>
<p><span class="math-container">$$=-(\frac{u^{-1}}{-1})+C$$</span></p>
<p><span class="math-container">$$=u^{-1}+C$$</span></p>
<p><span class="math-container">$$=(1-e^x)^{-1}+C$$</span></p>
<p>What did I do wrong?</p>
| mrtaurho | 537,079 | <p>Your answer is perfectly fine and agrees with what <a href="https://www.wolframalpha.com/input/?i=int+e%5Ex%2F%281-e%5Ex%29%5E2" rel="nofollow noreferrer">WolframAlpha</a> returns. You should give a source as the source seems to be wrong in its given answer.</p>
|
2,452,777 | <p>Let X be a topological space, $\mathcal{U} = \{U_\alpha\}_\alpha$ an open cover of $X$ and $\mathcal{F}$ a presheaf of abelian groups on $X$. Then one can define the Čech cohomology groups of $\mathcal{U}$ with values in $\mathcal{F}$:
\begin{equation}
\check{H}^k(\mathcal{U}, \mathcal{F})
\end{equation}
The Čech cohomology groups of $X$ with values in $\mathcal{F}$ are usually defined by the inductive limit over the set of open covers ordered by refinement:
\begin{equation}
\check{H}^k(X, \mathcal{F}) = \varinjlim_{\mathcal{U}} \check{H}^k(\mathcal{U}, \mathcal{F})
\end{equation}</p>
<p>But I am confused with the notion of open cover. Indeed one can consider the open cover as a set $\{U_\alpha\}_\alpha\subset \mathcal{P}(X)$ but also as a map $A\to \mathcal{P}(X)$ for some index set $A$. The open cover as a set is the range of the open cover as a map. But the distinction is real, since in the second case you can consider repeated open sets in the cover.</p>
<p>It seems to me that it is the second notion that is used to define Čech cohomology but then one cannot talk about "the set of open covers ordered by refinement" and then take the inductive limit since in the second case you don't even have a set of open overs, but rather a class.</p>
<p>So which alternative is used for open covers?</p>
| paeolo | 406,116 | <p>I found my answer in <a href="http://www.seas.upenn.edu/~jean/sheaves-cohomology.pdf" rel="nofollow noreferrer">A Gentle Introduction to Homology, Cohomology, and. Sheaf Cohomology</a> by Jean Gallier and Jocelyn Quaintance, on page 238.
As they said: "Most textboook presentations of Čech cohomology ignore this subtle point". They propose two answers, one by J.P Serre and one by Godement.</p>
<p>The idea is just to consider the second notion but to restric the index set so you get the first notion when you take the indutive limit.</p>
|
619,370 | <p>Let $C,Q$ is complex numbers Field and Rational number
Field,respectively,if $f(x),g(x)\in Q[x]$,</p>
<p>if $g(x)|f(x)$ on $C[x]$,show that
$$g(x)|f(x)$$ on $Q[x]$</p>
<p>My try: since
$g(x)|f(x)$,then we have
$$f(x)=g(x)h(x)$$
where $h(x)\in C[x]$.
Then I can't prove also have
$$g(x)|f(x)$$ on $Q[x]$.</p>
<p>Thank you</p>
| Robert Lewis | 67,071 | <p>A generalization of the stated problem from $Q$ a subfield of $C$ to arbitrary fields $F \subset E$ is resolved by means of the following </p>
<p><em><strong>Proposition:</em></strong> Let $E$ and $F$ be fields with $F$ a subfield of $E$. Let $f(x), m(x) \in F[x]$ with $f(x) = q(x)m(x)$ in $E[x]$. Then in fact $q(x) \in F[x]$.</p>
<p><em><strong>Proof:</em></strong> Let $f(x) = \sum_0^{\deg f} f_i x^i$, $m(x) = \sum_0^{\deg m} m_i x^i$, and $q(x) = \sum_0^{\deg q} q_i x^i$ with the $f_i, m_i \in F$ and the $q_i \in E$. Then since $f(x) = q(x)m(x)$, we have</p>
<p>$f_i = \sum_{j + k = i}q_j m_k \tag{7}$</p>
<p>with $j, k \ge 0$; this if you will is the definition of $q(x)m(x)$. Since $\deg f = \deg q + \deg m$, it follows that $f_{\deg f}$, the coefficient of the leading term of $f(x)$, satisfies</p>
<p>$f_{\deg f} = q_{\deg q} m_{\deg m}; \tag{8}$</p>
<p>this shows that the leading coefficient of $q(x)$, $q_{\deg q} = m_{\deg m}^{-1} f_{\deg f} \in F$. Next, we have that</p>
<p>$f_{\deg f - 1} = q_{\deg q - 1} m_{\deg m} + q_{\deg q} m_{\deg m - 1}, \tag{9}$</p>
<p>yielding </p>
<p>$q_{\deg q - 1} = m_{\deg m}^{-1}(f_{\deg f - 1} - q_{\deg q} m_{\deg m - 1}) \in F \tag{10}$</p>
<p>as well. Now (7) shows that</p>
<p>$f_{\deg f - l} = \sum_{j + k = \deg f - l}q_j m_k = m_{\deg m} q_{\deg f -l - \deg m} + \sum_{j + k = \deg f - l, k < \deg m} q_j m_k$
$= m_{\deg m} q_{\deg q -l} + \sum_{j + k = \deg f - l, k < \deg m} q_j m_k, \tag{11}$</p>
<p>for $0 \le l \le \deg q$, since $\deg q = \deg f - \deg m$. Scrutiny of (11) shows that (8) and (9) are respectively the $l = 0$ and $l = 1$ cases of this equation. Now suppose that $q_{\deg q}, q_{\deg q - 1}, . . ., q_{\deg q - r} \in F$ for some $r$ with $0 \le r < \deg q - 1$. Then by (11), </p>
<p>$f_{\deg f - r - 1} = m_{\deg m} q_{\deg q - r - 1} + \sum_{j + k = \deg f - r - 1, k < \deg m} q_j m_k, \tag{12}$</p>
<p>and since $k < \deg m$ in the summation on the right, it follows that $j > \deg f - \deg m -r - 1 = \deg q -r - 1$ for every coefficient $q_j$ appearing in this expression. If we now assume that each such $q_j \in F$, we may conclude that
$q_{\deg q - r - 1} \in F$ as well, since (12) yields</p>
<p>$q_{\deg q - r - 1} = m_{\deg m}^{-1}(f_{\deg f - r - 1} - \sum_{j + k = \deg f - r - 1, k < \deg m} q_j m_k), \tag{13}$</p>
<p>and every coefficient appearing on the right of (13) is in $F$. Thus we have completed an inductive demonstration that $q(x) \in F[x]$. <strong><em>END: Proof of Proposition.</em></strong></p>
<p>If we apply this proposition to the problem at hand, taking $E = C$, $F = Q$, $m(x) = g(x)$ and $q(x) = h(x)$, it follows immediately that $h(x) \in Q[x]$. <strong><em>QED!!!</em></strong></p>
<p>Hope this helps! Happy New Year,</p>
<p>and as always,</p>
<p><em><strong>Fiat Lux!!!</em></strong></p>
|
3,607,430 | <blockquote>
<p>Given that <span class="math-container">$a$</span>, <span class="math-container">$b$</span>, <span class="math-container">$c$</span> are the angles of a right-angled triangle, prove that:
<span class="math-container">$$\begin{align}
\sin a\sin b\sin(a-b) &+\sin b\sin c\sin(b-c)+\sin c\sin a\sin(c-a) \\ &+\sin(a-b)\sin(b-c)\sin(c-a)=0
\end{align}$$</span></p>
</blockquote>
<p>I know I'm supposed to use the properties of polynomials for this, as this question was found on the chapter on polynomials. I've tried having these values be roots of some function but don't know how to carry it out.</p>
<p>I also know that I can consider one of the angles to be <span class="math-container">$90^\circ$</span>, so the sine of that would be <span class="math-container">$1$</span>, but that dosen't really simplify it too much.</p>
| Matt | 1,042,546 | <p>So I came across a similar issue and I couldn't find any answers which were satisfactory. So I dug up as much information as I could and I found a solution. Here goes...</p>
<p>We have a known point, <span class="math-container">$(6,-1)$</span>. For any line that passes through this point, it must have an x-intercept of <span class="math-container">$(a,0)$</span> and a y-intercept of <span class="math-container">$(0,b)$</span>. We don't know what <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are as of yet, but we <em>do</em> know that their product is 3. By this definition, we also know that any line cannot be horizontal or vertical, since those lines would not have an x-intercept and y-intercept, respectively.</p>
<p>Let's take a look at our line equation: <span class="math-container">$y=mx+b$</span>. We know that <span class="math-container">$a⋅b=3$</span> so with some rearranging we have <span class="math-container">$b=\frac{3}{a}$</span>. We can sub this in for <span class="math-container">$b$</span>, such that:</p>
<p><span class="math-container">$y=mx+\frac{3}{a}$</span></p>
<p>We now need to find the slope. We can't explicitly define what the slope is yet, since that's part of what the question is asking. But we can substitute in what we <em>do</em> know. The slope is defined as:</p>
<p><span class="math-container">$m=\frac{y_1-y_0}{x_1-x_0}$</span></p>
<p>This is just the distance between our two points along the graph, in each axis. We only know one point <span class="math-container">$(6, -1)$</span> but we can substitute in our x-intercept placeholder <span class="math-container">$(a,0)$</span> so our slope becomes:</p>
<p><span class="math-container">$m=\frac{0-(-1)}{a-6}=\frac{1}{a-6}$</span></p>
<p>We can now sub this into our line equation as well, such that:</p>
<p><span class="math-container">$y=\frac{1}{a-6}x+\frac{3}{a}$</span></p>
<p>Then we myltiply by <span class="math-container">$x$</span> which is really just <span class="math-container">$\frac{x}{1}$</span> such that:</p>
<p><span class="math-container">$y=\frac{x}{a-6}+\frac{3}{a}$</span></p>
<p>We now have a fraction addition, and we do this just like any other: making the denominators the same. Multiply each fraction by the equivalent of <span class="math-container">$\frac{1}{1}$</span>, using the denominator of the other fraction:</p>
<p><span class="math-container">$y=\frac{x(a)}{(a-6)(a)}+\frac{3(a-6)}{a(a-6)}$</span></p>
<p><span class="math-container">$y=\frac{a(x) + 3(a-6)}{a(a-6)}$</span></p>
<p>After some distribution we come to:</p>
<p><span class="math-container">$y=\frac{ax+3a-18}{a^2-6a}$</span></p>
<p>Multiplying both sides by <span class="math-container">$(a^2-6a)$</span> and a bit more distribution we get:</p>
<p><span class="math-container">$y(a^2-6a)=ax+3a-18$</span></p>
<p><span class="math-container">$a^2y-6ay=ax+3a-18$</span></p>
<p>For this next step, we can sub in our known point <span class="math-container">$(6,-1)$</span> for <span class="math-container">$x$</span> and <span class="math-container">$y$</span>. Remember that in our original formula <span class="math-container">$y=mx+b$</span>, we already knew what <span class="math-container">$(x,y)$</span> was, as it was given to us as part of the question. The information we're missing are the slope <span class="math-container">$(m)$</span> and the y-intercept <span class="math-container">$(b)$</span>. In fact, we could've substituted in our known point for <span class="math-container">$(x,y)$</span> right from the very start.</p>
<p>After some more rearranging and collecting like terms, we can get to an expression in standard polynomial form:</p>
<p><span class="math-container">$a^2(-1)-6a(-1)=a(6)+3a-18$</span></p>
<p><span class="math-container">$-a^2+6a=9a-18$</span></p>
<p><span class="math-container">$-a^2-3a+18=0$</span></p>
<p>It's at this point we can use the quadratic formula to solve for a. But for clarity, let's substitute x for a:</p>
<p><span class="math-container">$-x^2-3x+18=0$</span></p>
<p>For a polynomial in standard form <span class="math-container">$(ax^2+bx+c=0)$</span> the quadratic formula is defined as:</p>
<p><span class="math-container">$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$</span></p>
<p>Looking at our equation so far, we can sub in our known values of <span class="math-container">$(a,b,c)$</span> as <span class="math-container">$(-1,-3,18)$</span> into the quadratic formula, such that:</p>
<p><span class="math-container">$x=\frac{-(-3)\pm\sqrt{(-3)^2-4(-1)(18)}}{2(-1)} \Rightarrow x=\frac{3\pm\sqrt{9+72}}{-2} \Rightarrow x=\frac{3\pm\sqrt{81}}{-2} \Rightarrow x=\frac{3\pm9}{-2}=-6,3$</span></p>
<p>At last, we have two points at which any line satisfying our conditions intersects the x-axis: <span class="math-container">$(-6,0)$</span> and <span class="math-container">$(3,0)$</span>. We now have enough information to draw our lines and find the equations. For <span class="math-container">$y=mx+b$</span> the two pieces of information we need to find are <span class="math-container">$m$</span> and <span class="math-container">$b$</span>.</p>
<p>For <span class="math-container">$m$</span>, we just use the slope formula from before:</p>
<p><span class="math-container">$m=\frac{y_1-y_0}{x_1-x_0}$</span></p>
<p>But now we can substitute all values for both our slopes. <span class="math-container">$(x_0,y_0)$</span> is our original known point, and <span class="math-container">$(x_1,y_1)$</span> are the points we just calculated. Since both points are the x-intercept, their y-coordinate will be 0:</p>
<p><span class="math-container">$m_1=\frac{0-(-1)}{-6-6} \Rightarrow m_1=-\frac{1}{12}$</span></p>
<p><span class="math-container">$m_2=\frac{0-(-1)}{3-6} \Rightarrow m_2=-\frac{1}{3}$</span></p>
<p>Finally, we just need to find <span class="math-container">$b$</span>. To do so, we just need to sub in a known <span class="math-container">$(x,y)$</span> and solve for <span class="math-container">$b$</span>. Since we know that both lines pass through <span class="math-container">$(6,-1)$</span> we can use it for both:</p>
<p><span class="math-container">$y=mx+b$</span></p>
<p><span class="math-container">$-1=-\frac{1}{12}(6)+b_1 \Rightarrow b_1=\frac{6}{12}-1 \Rightarrow b_1=-\frac{1}{2}$</span></p>
<p><span class="math-container">$-1=-\frac{1}{3}(6)+b_2 \Rightarrow b_2=\frac{6}{3}-1 \Rightarrow b_2=2-1 \Rightarrow b_2=1$</span></p>
<p>We now have all the information we need to write our equations, for all lines which pass through <span class="math-container">$(6,-1)$</span> and for which the products of their <span class="math-container">$x/y$</span> intercepts is 3:</p>
<p><span class="math-container">$y=-\frac{1}{12}x-\frac{1}{2}$</span>, and <span class="math-container">$y=-\frac{1}{3}x+1$</span></p>
<p>You can see <a href="https://www.desmos.com/calculator/q9t0yowbjl" rel="nofollow noreferrer">this Desmos graph</a> to get a visualisation.</p>
|
2,485,997 | <p><a href="https://i.stack.imgur.com/OkhoU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OkhoU.png" alt="enter image description here"></a></p>
<p>Question: What if we consider (1,4,5) and (1,5,4) as non-distinct possibilities, then what should we do?</p>
<p>$${{9}\choose{2}}-2\cdot\frac{{9}\choose{2}}{3}$$</p>
| N. F. Taussig | 173,070 | <p>Since $10$ is not a multiple of $3$, it is not possible for all three numbers to be the same.</p>
<p>However, it is possible for two numbers of the same. Since each number is positive, the repeated number must be $1$, $2$, $3$, or $4$. There are $\binom{3}{2}$ ways to choose the locations of the repeated number. Choosing the repeated number determines the third number. There are
$$4\binom{3}{2} = 12$$
such ordered positive integer solutions of the equation $a + b + c = 10$. We count each unordered solution of this type three times, depending on where the single number is located.</p>
<p>That leaves $36 - 12 = 24$ solutions of the equation $a + b + c = 10$ in which all three numbers are distinct. Since $a, b, c$ can be arranged in $3! = 6$ ways, we count each unordered solution of this type $6$ times.</p>
<p>Hence, the number of unordered positive integer solutions to the equation
$a + b + c = 10$ is
$$\frac{12}{3} + \frac{24}{6} = 4 + 4 = 8$$
As a check, we list them.
$$\{1, 2, 7\}, \{1, 3, 6\}, \{1, 4, 5\}, \{2, 3, 5\}, \{1, 1, 8\}, \{2, 2, 6\}, \{3, 3, 4\}, \{2, 4, 4\}$$</p>
|
1,722,443 | <p>I am unsure how to solve the following problem. I was able to find similar questions, but had trouble understanding them since they did not show full solutions.</p>
<p>The question:</p>
<p>Find ALL solutions (between $1$ & $40$) to the equation $25x \equiv 10 \pmod{40}$.</p>
| Peter | 82,961 | <p>It suffices to solve the equation $25x\equiv 10\ (\ mod\ 8\ )$ because modulo $5$, the equation holds no matter what $x$ is.</p>
<p>This gives $x\equiv 2\ (\ mod\ 8\ )$.</p>
<p>I think you can easily find out the solutions now.</p>
|
159,965 | <blockquote>
<p>Find limits of a function $f:\mathbb{R}^3\rightarrow \mathbb{R}$ given by formula $f(x,y,x)=x+y+z$ on set $M=\left\{ (x,y,z)\in\mathbb{R}^3:x^2+y^2\le z\le 1 \right\}$. Does $f$ reaches all its limits?</p>
</blockquote>
<p>To answer the last question I need to know if $M$ is a closed and bounded set. If it is then $f$ reaches all its limits by Bolzano-Weierstrass theorem. But I'm not sure if it is closed (bounded it is I think it just simply follows from inequalities, but probably it isn't precise proof). Previously I had only sets like $\left\{(x;y)\in\mathbb{R}^2: 4x^2+y^2\le 25\right\}$ and a reason for closedness was rather simple. I always used to state given set as an inverse image by continuous function of closed set. But in situation with $M$ I can't find similar explanation.</p>
<p>Finding limits seems to be hard too. I always used partial derivatives to find extremes in the interior of set and parameterization to explore boundary of set.
But here I'm even not sure what will be interior and what will be boundary of $M$.
Can anybody help?</p>
| Brian M. Scott | 12,042 | <p>Here’s an approach that minimizes the calculus and relies mostly on the geometry.</p>
<p>Let $R$ be the region in the $xz$-plane bounded by the curves $z=x^2$ and $z=1$; $M$ is the solid of revolution obtained by revolving $R$ about the $z$-axis. $R$ is a closed set in the plane, so $M$ is closed, and the minimum and maximum are attained.</p>
<p>The function $f(x,y,z)=x+y+z$ is constant on planes normal to the line $x=y=z$ and increases linearly along that line, so the minimum and maximum values of $f$ on $M$ will occur at points where such a plane just touches the surface of $M$; there’s no need to look for them in the interior of $M$. Since for fixed $x$ and $y$, $f(x,y,z)$ increases as $z$ increases, the maximum must occur on the top face of $M$. On that face it will occur where $x+y$ is maximal, which is evidently in the plane $x=y$, so it must occur at $\left(\frac1{\sqrt2},\frac1{\sqrt2},1\right)$. (If that’s not entirely clear, project to the $xy$-plane and observe that the maximum must occur where a line of the form $x+y=c$ is tangent to the circle $x^2+y^2=1$ in the first quadrant.) The value of $f$ at this point is $1+\sqrt2$.</p>
<p>Similarly, the minimum must also occur in the plane $x=y$, so we’re looking for $z\in[0,1]$ that minimizes $2x+z$, where $z=2x^2$. (Recall that the minimum must be on the surface of $M$, and it clearly isn’t on the top.) This is just the problem of minimizing $2x+2x^2=2x(1+x)$ for $0\le 2x^2\le 1$, or $|x|\le\frac1{\sqrt2}$, which doesn’t even require any calculus: it occurs at the vertex of the parabola, which is on the axis, $x=-\frac12$, halfway between the zeroes at $x=0$ and $x=-1$. Thus, the function $f$ takes its minimum on $M$ at $\left(-\frac12,-\frac12,\frac12\right)$. The value of $f$ at this point is $-\frac12$.</p>
|
103,970 | <p>Here's a very bizarre inconsistency I've just struggled with and I'm wondering why it exists or if I'm missing something.</p>
<p>I have some noisy data and I wish to make a framed plot of the data but allow the data to extend outside the vertical limits of the frame (for stylistic reasons). Like so:</p>
<pre><code> xs = Range[0, 0.5, 0.005];
data = Transpose[{xs, Sin[Pi xs]^2 + 0.05 RandomReal[{-1, 1}, Length[xs]]}];
opts = Sequence[PlotRange -> {{0, 0.5}, {0, 1}}, Frame -> True,
PlotRangeClipping -> False, ImagePadding -> 20];
ListLinePlot[{}, opts,
Prolog -> First@ListLinePlot[data, PlotRange -> All]
]
</code></pre>
<p><a href="https://i.stack.imgur.com/vjuqv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vjuqv.png" alt="enter image description here"></a></p>
<p>However, my dataset contains points outside my desired x limits, thus my data is more accurately given by:</p>
<pre><code> xs = Range[-0.1, 0.6, 0.005];
data = Transpose[{xs, Sin[Pi xs]^2 + 0.05 RandomReal[{-1, 1}, Length[xs]]}];
</code></pre>
<p>Which when plotted with exactly the same code gives:</p>
<p><a href="https://i.stack.imgur.com/XstNd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XstNd.png" alt="enter image description here"></a></p>
<p>which obviously extends in both the x and y directions when I only want the extension in the y. </p>
<p>My solution is to change the value of <code>PlotRange -> All</code> in the 'Prolog' 'ListLinePlot'. However, this only works in the y-direction, observe:</p>
<pre><code> Grid[{{
ListLinePlot[{}, opts,
Prolog -> First@ListLinePlot[data, PlotRange -> {All, {0, 1}}]
],
ListLinePlot[{}, opts,
Prolog -> First@ListLinePlot[data, PlotRange -> {{0, 1}, All}]
],
ListLinePlot[{}, opts,
Prolog -> First@ListLinePlot[data, PlotRange -> {{0, 1}, {0, 1}}]
]
}}]
</code></pre>
<p><a href="https://i.stack.imgur.com/YuJ95.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YuJ95.png" alt="enter image description here"></a></p>
<p>As you can see the data never obeys the PlotRange in the x direction! Looking into the content of the <code>First@ListLinePlot[data, ...]</code> we can see that the graphics items describing the data do get clipped in the y-direction:</p>
<p><a href="https://i.stack.imgur.com/Gn4wW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Gn4wW.png" alt="enter image description here"></a></p>
<p>You can see there are several instances near the beginning where the <code>Line</code>'s y-coordinate has been clipped to <code>0.</code> and many at the end where it is <code>1.</code>. </p>
<p>However if we try to restrict the graphics in the x-direction:</p>
<p><a href="https://i.stack.imgur.com/F0Zdy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/F0Zdy.png" alt="enter image description here"></a></p>
<p>We see no such clipping occurs, leading to the problems described above.</p>
<p>Why is this happening? Is there an elegant workaround? My current method to circumvent this problem is to create and <code>Interpolation</code> of may data and then use <code>Plot</code> as opposed to <code>ListLinePlot</code> in the <code>Prolog</code> but this seems like overkill for what should be a simple fix?</p>
<p>I note that merely taking a subset of my data won't work for my real data as the x-values do not lie nicely on my coordinates, ie I might not have a value at 0 so I would be left with a gap either side. </p>
<p>Many thanks. </p>
| Alexey Popkov | 280 | <p>It is another gedanken functionality in <em>basic</em> plotting functions. Using new in version 10 <code>Region*</code> functionality here is a workaround:</p>
<pre><code>xs = Range[0, 1, 0.005];
data = Transpose[{xs, Sin[Pi xs]^2 + 0.05 RandomReal[{-1, 1}, Length[xs]]}];
inf = 10;
RegionPlot[RegionIntersection[Line[data], Rectangle[{0, -inf}, {0.5, inf}]],
PlotRange -> {{0, 0.5}, {0, 1}}, PlotRangeClipping -> False, ImagePadding -> 20]
</code></pre>
<blockquote>
<p><a href="https://i.stack.imgur.com/vXs9n.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vXs9n.png" alt="enter image description here"></a></p>
</blockquote>
<p>With older <em>Mathematica</em> versions you could proceed in the following way:</p>
<pre><code>Show[Plot[Interpolation[data, InterpolationOrder -> 1][x], {x, 0, .5},
PlotRange -> {{0, 0.5}, All}], PlotRange -> {{0, 0.5}, {0, 1}},
PlotRangeClipping -> False, ImagePadding -> 20, Frame -> True, PlotRangePadding -> None]
</code></pre>
|
103,970 | <p>Here's a very bizarre inconsistency I've just struggled with and I'm wondering why it exists or if I'm missing something.</p>
<p>I have some noisy data and I wish to make a framed plot of the data but allow the data to extend outside the vertical limits of the frame (for stylistic reasons). Like so:</p>
<pre><code> xs = Range[0, 0.5, 0.005];
data = Transpose[{xs, Sin[Pi xs]^2 + 0.05 RandomReal[{-1, 1}, Length[xs]]}];
opts = Sequence[PlotRange -> {{0, 0.5}, {0, 1}}, Frame -> True,
PlotRangeClipping -> False, ImagePadding -> 20];
ListLinePlot[{}, opts,
Prolog -> First@ListLinePlot[data, PlotRange -> All]
]
</code></pre>
<p><a href="https://i.stack.imgur.com/vjuqv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vjuqv.png" alt="enter image description here"></a></p>
<p>However, my dataset contains points outside my desired x limits, thus my data is more accurately given by:</p>
<pre><code> xs = Range[-0.1, 0.6, 0.005];
data = Transpose[{xs, Sin[Pi xs]^2 + 0.05 RandomReal[{-1, 1}, Length[xs]]}];
</code></pre>
<p>Which when plotted with exactly the same code gives:</p>
<p><a href="https://i.stack.imgur.com/XstNd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XstNd.png" alt="enter image description here"></a></p>
<p>which obviously extends in both the x and y directions when I only want the extension in the y. </p>
<p>My solution is to change the value of <code>PlotRange -> All</code> in the 'Prolog' 'ListLinePlot'. However, this only works in the y-direction, observe:</p>
<pre><code> Grid[{{
ListLinePlot[{}, opts,
Prolog -> First@ListLinePlot[data, PlotRange -> {All, {0, 1}}]
],
ListLinePlot[{}, opts,
Prolog -> First@ListLinePlot[data, PlotRange -> {{0, 1}, All}]
],
ListLinePlot[{}, opts,
Prolog -> First@ListLinePlot[data, PlotRange -> {{0, 1}, {0, 1}}]
]
}}]
</code></pre>
<p><a href="https://i.stack.imgur.com/YuJ95.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YuJ95.png" alt="enter image description here"></a></p>
<p>As you can see the data never obeys the PlotRange in the x direction! Looking into the content of the <code>First@ListLinePlot[data, ...]</code> we can see that the graphics items describing the data do get clipped in the y-direction:</p>
<p><a href="https://i.stack.imgur.com/Gn4wW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Gn4wW.png" alt="enter image description here"></a></p>
<p>You can see there are several instances near the beginning where the <code>Line</code>'s y-coordinate has been clipped to <code>0.</code> and many at the end where it is <code>1.</code>. </p>
<p>However if we try to restrict the graphics in the x-direction:</p>
<p><a href="https://i.stack.imgur.com/F0Zdy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/F0Zdy.png" alt="enter image description here"></a></p>
<p>We see no such clipping occurs, leading to the problems described above.</p>
<p>Why is this happening? Is there an elegant workaround? My current method to circumvent this problem is to create and <code>Interpolation</code> of may data and then use <code>Plot</code> as opposed to <code>ListLinePlot</code> in the <code>Prolog</code> but this seems like overkill for what should be a simple fix?</p>
<p>I note that merely taking a subset of my data won't work for my real data as the x-values do not lie nicely on my coordinates, ie I might not have a value at 0 so I would be left with a gap either side. </p>
<p>Many thanks. </p>
| Edmund | 19,542 | <p>When using <code>PlotRange -> s</code> you are only specifying the range on the y-axis. Look at the <strong>Details</strong> section of the <a href="http://reference.wolfram.com/language/ref/PlotRange.html" rel="nofollow noreferrer"><code>PlotRange</code> documentation</a>. This is why only the y-axis range is being truncated.</p>
<p>You can use the format that specifies the range for both the x and y axis but this will make the frame cover those ranges; and you don't want that. You can, however, use <code>Show</code> with an empty <code>ListLinePlot</code> of your desired frame range and a second plot that takes only the points in the x-axis range. Then make use of <code>PlotRangeClipping</code> and <code>ImagePadding</code> in <code>Show</code> to get your desired result.</p>
<pre><code>xs = Range[-0.1, 0.6, 0.005];
data = Transpose[{xs, Sin[Pi xs]^2 + 0.05 RandomReal[{-1, 1}, Length[xs]]}];
Show[{
ListLinePlot[{}, PlotRange -> {{0, .5}, {0, 1}}, Frame -> True],
ListLinePlot[Select[Between[First@#, {0, .5}] &]@data, Axes -> None]},
PlotRangeClipping -> False,
ImagePadding -> {{All, All}, {Scaled[.03], Scaled[.03]}}]
</code></pre>
<p><a href="https://i.stack.imgur.com/xSxtz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xSxtz.png" alt="enter image description here"></a></p>
<p>Hope this helps.</p>
<blockquote>
<p>PS:- You can extend this by making the <code>Scaled</code> parameter a function of the data y-range above the y-axis max and the range and the y-axis. This would automate the padding so you wouldn't have to eyeball it each time. Similarly for the data y-range below the y-axis min.</p>
</blockquote>
|
133,794 | <p>Prove the existence of infinite number of infinite cardinals </p>
<p>1) $\alpha$, such that $\alpha<\alpha^\aleph$ </p>
<p>2) $\beta$, such that $\beta=\beta^\aleph$</p>
| Asaf Karagila | 622 | <p>Using $\aleph$ as a general cardinal might be ambiguous (there are places where it is used particularly for $2^{\aleph_0}$), the answer remains the same regardless to the intended use of $\aleph$.</p>
<ol>
<li><p>Recall that for every infinite $\kappa$ we have $\kappa<\kappa^{\operatorname{cf}(\kappa)}$. Simply show that there are infinitely many cardinals whose cofinality is $\aleph$. You can show that there exists a sequence $\alpha_0<\alpha_0^\aleph<\alpha_1<\ldots$ by starting the construction of $\alpha_{n+1}$ from the construction of $\alpha_n^\aleph$.</p></li>
<li><p>Cardinal exponentiation has the property $\left(\kappa^\lambda\right)^\mu=\kappa^{\lambda\cdot\mu}$. Take $\alpha$ from the previous part and take $\beta=\alpha^\aleph$, now we have $\beta^\aleph=\alpha^{\aleph\cdot\aleph}=\alpha^\aleph=\beta$.</p></li>
</ol>
<p>Note that for the second part you don't <em>really</em> need the first part, however it is easy to show that there are infinitely many $\beta$ if you use the fact that $\alpha_n^\aleph\neq\alpha_k^\aleph$ for $n\neq k$.</p>
|
2,333,857 | <p>I have this problem that I have worked out. Will someone check it for me? I feel like it is not correct. Thank you!</p>
<p>Rotate the graph of the ellipse about the $x$-axis to form an ellipsoid. Calculate the precise surface area of the ellipsoid. </p>
<p>$$\left(\frac{x}{3}\right)^{2}+\left(\frac{y}{2}\right)^{2}=1.$$</p>
<p><a href="https://i.stack.imgur.com/FtgoC.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FtgoC.jpg" alt="Surface area problem worked out"></a></p>
| orlp | 5,558 | <p>If you really want to be formal you can say that the domain of $f(x)$ is $\{x \in \mathbb{Z} \mid -2 \leq x \leq 3\}$.</p>
|
220,139 | <p>Is there a way to either make <code>FindMinimum</code> do an exact computation or <code>Minimize</code> find also the local minima? Or other ideas to find local minima exactly?</p>
<p><strong>Example:</strong> find all local minima (exact values, not approximations) of <span class="math-container">$f(x,y)=x^2 − x + 2y^2$</span> on <span class="math-container">$E=\{(x,y)\in \mathbb{R}^2 : x^2+y^2\le 1\}$</span>.</p>
| Rom38 | 10,455 | <p>The exact minimization of a function can be done using <code>Minimize</code>:</p>
<pre><code> f[x_, y_] := x^2 - x + 2 y^2
sol = Minimize[f[x, y], {x, y}]
(*{-(1/4), {x -> 1/2, y -> 0}} <= output of Minimize*)
Show[
Plot3D[f[x, y], {x, y} \[Element] Disk[{0, 0}, 1]],
Graphics3D[{Red, PointSize[Large],
Point[Flatten@{{x, y} /. sol[[2]], sol[[1]]}]}]
]
</code></pre>
<p><a href="https://i.stack.imgur.com/EoHas.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EoHas.png" alt="enter image description here"></a></p>
|
239,863 | <p>I've to study this series:</p>
<p>$$\sum_{n=1}^\infty e^{\sqrt n\,x}$$ </p>
<p>My teacher wrote that with the asymptotic comparison with this series:</p>
<p>$$\sum_{n=1}^\infty\frac{1}{n^2}$$<br>
My series converges for every </p>
<p>$$x<0$$</p>
<p>I don't understand the motivation, hoping for someone to enlighten me!</p>
<p>=) Thanks. Leonardo.</p>
| Davide Giraudo | 9,849 | <p>Let $L\in(\ell^p)'$ a linear continuous functional. We can assume that $x_n^{(k)}\to 0$ for all $k$. $L$ can be represent by an element $v$ of $\ell^q$, where $p^{-1}+q^{-1}=1$. Fix $\delta>0$ and $v'$ of finite support such that $\lVert v-v'\rVert_{\ell^q}\leqslant \delta$. Then
$$\lVert L(x_n)\rVert\leqslant\delta+|\langle v',x_n\rangle_{\ell^q,\ell^p}|,$$
giving what we want.</p>
|
3,547,989 | <p><strong>An ordinary deck of cards is dealt randomly to four players so that each player receives 13 cards.
Find the probability that each player is dealt exactly one ace.</strong></p>
<p>I was wondering if I could compare this scenario with the following:
Given 4 indistinguishable balls and 4 distinguishable boxes, if we distribute the balls randomly, what is the probability that each box has exactly one ball?
The answer to that is <span class="math-container">$\frac{1}{4+4-1\choose 4}=\frac{1}{7\choose 4}=\frac{1}{35}$</span>. So is that also the answer to the question above? If not then why?</p>
| Saaqib Mahmood | 59,734 | <p>Let us put
<span class="math-container">$$ A \colon= X \left(X^\prime X\right)^{-1} X^\prime. $$</span>
Then we note that
<span class="math-container">$$
\begin{align}
A^2 &= AA \\
&= \left[ X \left(X^\prime X\right)^{-1} X^\prime \right] \left[ X \left(X^\prime X\right)^{-1} X^\prime \right] \\
&= X \left[ \left(X^\prime X\right)^{-1} \left( X^\prime X \right) \right] \left(X^\prime X\right)^{-1} X^\prime \\
&= X I_{p\times p} \left(X^\prime X\right)^{-1} X^\prime \\
&= X \left(X^\prime X\right)^{-1} X^\prime \\
&= A.
\end{align}
$$</span></p>
<p>Now let us put
<span class="math-container">$$ B \colon= I_n-X(X^TX)^{-1}X^T. $$</span>
Then we find that
<span class="math-container">$$
\begin{align}
B^2 &= BB \\
&= \left[ I_n-X(X^TX)^{-1}X^T \right] \left[ I_n-X(X^TX)^{-1}X^T \right] \\
&= I_n \left[ I_n- X(X^TX)^{-1}X^T \right] - X(X^TX)^{-1}X^T \left[ I_n-X(X^TX)^{-1}X^T \right] \\
&= I_n I_n - I_n X(X^TX)^{-1}X^T - X(X^TX)^{-1}X^T I_n + \left[ X(X^TX)^{-1}X^T \right] \left[ X(X^TX)^{-1}X^T \right] \\
&= I_n - X(X^TX)^{-1}X^T - X(X^TX)^{-1}X^T + X\left[ (X^TX)^{-1} \left( X^T X\right) \right] \left[ (X^TX)^{-1}X^T \right] \\
&= I_n - 2 X(X^TX)^{-1}X^T + X I_n \left[ (X^TX)^{-1} X^T \right] \\
&= I_n - 2 X(X^TX)^{-1}X^T + X \left[ (X^TX)^{-1} X^T \right] \\
&= I_n - 2 X(X^TX)^{-1}X^T + X (X^TX)^{-1} X^T \\
&= I_n - X(X^TX)^{-1}X^T \\
&= B.
\end{align}
$$</span></p>
|
3,335,892 | <p>If I have two injective functions <span class="math-container">$f : A \to B$</span> and <span class="math-container">$g : B \to A$</span>, as Schröder-Bernstein (SB) says, then there is a function <span class="math-container">$h : A \to B$</span> which is bijective.</p>
<p>As for a proof, my reasoning goes something like this:</p>
<p>The injectivity of <span class="math-container">$f \implies |A| \leq |B|$</span>.
Similarly, the injectivity of <span class="math-container">$g \implies |B| \leq |A|$</span>. At this point I would say that it is perhaps obvious that <span class="math-container">$|B| = |A|$</span> in order for the prior statements to remain true.</p>
<p>With that being said, the final question is whether or not <span class="math-container">$|A| = |B| \implies $</span> that there exists a function <span class="math-container">$h : A \to B$</span> which is bijective? I am reading (perhaps somewhat naively) on wikipedia that if X and Y are finite sets then a bijection exists <span class="math-container">$ \leftrightarrow$</span> <span class="math-container">$|A| = |B|$</span>.</p>
<p>Taking the if and only if symbol as a statement of equivalence means that, at least in the finite case, considering the cardinalities of <span class="math-container">$A$</span> and <span class="math-container">$B$</span> proves the existence of <span class="math-container">$h$</span>?</p>
| BallBoy | 512,865 | <p>The conceptual order here must be a little different than you present it. If <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are <em>finite</em> sets, then we can conclude from <span class="math-container">$|A|\leq|B|$</span> and <span class="math-container">$|B|\leq|A|$</span> that <span class="math-container">$|A|=|B|$</span> (as you write, it is "perhaps obvious"). But if <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are infinite sets, it is far from obvious that we can make this jump. In fact, in the case of infinite sets, what <span class="math-container">$|A|\leq|B|$</span> <em>means</em> is that there is an injection from <span class="math-container">$A$</span> to <span class="math-container">$B$</span>, and what <span class="math-container">$|A|=|B|$</span> <em>means</em> is that there is a bijection. So in order to make the jump from <span class="math-container">$|A|\leq|B|$</span> and <span class="math-container">$|B|\leq|A|$</span> to <span class="math-container">$|A|=|B|$</span>, we first need the Schroder-Bernstein Theorem.</p>
|
2,864,585 | <p>I tried to calculate the Hessian matrix of linear least squares problem (L-2 norm), in particular:</p>
<p>$$f(x) = \|AX - B \|_2$$
where $f:{\rm I\!R}^{11\times 2}\rightarrow {\rm I\!R}$</p>
<p>Can someone help me?<br>
Thanks a lot.</p>
| Mauricio Cele Lopez Belon | 485,657 | <p>Calculate first the gradient vector: use the chain rule and calculate the partial derivatives of $f(x)$ w.r.t $x \in R^n$. You will get a function that eats a vector and produce other "vector" $g(x) \in R^n$ (well this is an abuse of notation and terminology, $g(x)$ produces a vector of functions not a vector in $R^n$ so it is really a "vector operator").</p>
<p>Then you will take the partial derivatives of $g(x)$ w.r.t $x$ again applying the chain rule. For that you can see $g(x)$ as a vector of simpler functions $g_i(x) \in R$ each of which eats a vector and produces a scalar value. </p>
<p>So for each dimension of $g(x)$ you have a function $g_i(x) \in R$. So taking the partial derivatives of $g(x)$ w.r.t $x$ amounts to taking the partial derivatives of $g_i(x) \in R$ w.r.t $x$ and put them toguether. That is the Hessian matrix.</p>
<p>In the same way that we see the derivative of $f(x)$ w.r.t $x$ is producing a vector operator, we can see the derivative of $g_i(x)$ w.r.t $x$ as producing a vector operator and hence the derivative of $g(x)$ w.r.t $x$ is producing a matrix operator named Hessian matrix.</p>
|
4,627,334 | <p>To my understanding that a primitive triple <span class="math-container">$x$</span> and <span class="math-container">$y$</span> can be written as <span class="math-container">$x = q^2 - p^2$</span> while <span class="math-container">$y=2pq$</span> for relatively prime opposite parity <span class="math-container">$q > p$</span> then the area can be calculated as: <span class="math-container">$pq(q^2 - p^2) = pq(q+p)(q-p)$</span>.
Am I missing something obvious that helps prove that a Pythagorean Triple triangle cannot have an odd area?</p>
| Dstarred | 955,900 | <p>We start off with the fact that any pythagorean triples can only consist of all even numbers <strong>or</strong> <span class="math-container">$2$</span> odd numbers and <span class="math-container">$1$</span> even number.</p>
<p>Let the numbers are denoted <span class="math-container">$x, y, z \in \mathbb{N}$</span> with <span class="math-container">$z$</span> being the hypotenuse, and <span class="math-container">$x = p^2 - q^2$</span> and <span class="math-container">$y = 2pq$</span>. Given that <span class="math-container">$y$</span> <strong>must</strong> be even, the only following cases arise:</p>
<blockquote>
<p>If <span class="math-container">$x \equiv 0 \mod 2$</span> and <span class="math-container">$y \equiv 0 \mod 2$</span></p>
</blockquote>
<p>Then, let <span class="math-container">$x = 2k, y = 2j$</span> for some <span class="math-container">$k, j$</span>.</p>
<p>It is trivial to see <span class="math-container">$\big(\frac{1}{2}xy = 2kj\big) \equiv 0 \mod 2$</span>.</p>
<blockquote>
<p>If <span class="math-container">$x \equiv 1 \mod 2$</span> and <span class="math-container">$y \equiv 0 \mod 2$</span></p>
</blockquote>
<p>Then, <span class="math-container">$x = (p + q)(p - q) \equiv 1 \mod 2 \implies \begin{cases} (p + q) \equiv 1 \mod 2 \\ (p - q) \equiv 1 \mod 2 \end{cases} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \qquad \implies \begin{cases} p \equiv 0 \mod 2, \qquad q \equiv 1 \mod 2 \\ p \equiv 1 \mod 2, \qquad q \equiv 0 \mod 2\end{cases}$</span></p>
<p>It is inevitable for <span class="math-container">$(y = 2pq) \equiv 0 \mod 4$</span> for either <span class="math-container">$p$</span> or <span class="math-container">$q$</span> even.</p>
<p>Hence, let <span class="math-container">$y = 4n$</span> for some <span class="math-container">$n$</span>.</p>
<p>Again, we see <span class="math-container">$\big(\frac{1}{2}xy = 2xn\big) \equiv 0 \mod 2 \quad \square.$</span></p>
|
4,302,855 | <p>I have tried setting up multiple systems of equations using many known volumes but I always seem to come up short. My last attempt was a hollow cylinder but that leaves you with three unknowns in only two sim. equations (for V and S.A). Can anyone help?</p>
| bubba | 31,744 | <p>You can use the basic idea described by @PC1, based on Pappus’ theorems. But use a cone instead of a torus. I think that will work.</p>
|
2,308,770 | <p>I have an equation of the form $A*i*j + B*i +C*j = N$
where I have the values of $A,B,C$ and $N$ and I want to solve for integer values of $i$ and $j$.</p>
<p>How would I approach this? I could try trial and error but the numbers I'm working with are relatively large (eg $>10^{40}$). But I'm also happy to work on understanding the process with much smaller numbers. I'm not sure if it matters, but it's also possible there will not be a solution.</p>
| Michael Burr | 86,421 | <p>Here's an idea, not a fully formed approach.</p>
<p>You wish to solve
$$
Aij+Bi+Cj=N.
$$</p>
<p>If we complete the square on the LHS, we get
$$
A\left(i+\frac{C}{A}\right)\left(j+\frac{B}{A}\right)-\frac{BC}{A}=N.
$$</p>
<p>Multiplying through by $A$, we get
$$
(Ai+C)(Aj+B)=AN+BC.
$$</p>
<p>Therefore, both $(Ai+C)$ and $(Aj+B)$ must be factors of $AN+BC$. If it is possible to write out all the factors of $AN+BC$, then for factors $x_1x_2=AN+BC$, you have $i=\frac{x_1-C}{A}$ and $j=\frac{x_2-B}{A}$, all you need to do is to check that these are integers or not.</p>
|
1,817,542 | <p><strong>Problem:</strong> Let $(X, Y)$ be uniformly distributed on the unit disk $\{ (x,y) : x^2 + y^2 \le 1\}$. Let $R = \sqrt{X^2 + Y^2}$. Find the CDF and PDF of $R$.</p>
<p><strong>Attempted Solution:</strong> First note that $r \in R = \sqrt{X^2 + Y^2}$ represents a point on $\mathbb{R}^2$ with radius $r$ about the origin. Since only points with radius $1$ had probability greater than $0$ on $(X, Y)$ we have that</p>
<p>$$
F_R(r) = \begin{cases} 0 & r < 1 \\ 1 & r \ge 1 \end{cases}
$$</p>
<p>so that</p>
<p>$$
f_R(r) = F_R'(r) = \begin{cases} 0 & r < 1 \\ \text{undefined} & r = 0 \\ 0 & r > 1 \end{cases}
$$</p>
<p>since </p>
<ol>
<li>$F'_R$ is discontinuous at $r = 0$.</li>
<li>$F_R$ is constant everywhere else (so that the derivative of a constant is $0$, and hence $F_R'$ is $0$).</li>
</ol>
<p><strong>Question:</strong> Is my reasoning correct here?</p>
| Community | -1 | <p>Your reasoning is incorrect.
First of all, by the definition of CDF,
$$
F_R(r)=P(R\leq r)=P(X^2+Y^2\leq r^2)\neq 0
$$
when $0<r<1$. </p>
<p>Second, for a probability density $f_R(r)$, one must have
$$\int_{\mathbb{R}} f_R(r)\ dr=1$$
which contradicts your calculation. </p>
|
253,359 | <p>I'm trying to prove by induction the following statement without success:<br>
$$\forall n \ge 2, \;\forall d \ge 2 : d \mid n(n+1)(n+2)...(n+d-1) $$</p>
<p>For the base case: $n = 2$, $d = 2$<br>
$2\mid 2(2+1)$ which is true.<br></p>
<p>Now, the confusion begins! I assume I would need to use the second induction principle to proof this because $P(n)$ and $P(n+1)$ are not related at all. It is also the first time I am dealing with more than one variable so it makes it harder for me.</p>
<p>I tried the following:<br>
- Trying to prove by simple induction. I did not go very far.<br>
- Trying to split my induction step in 2 parts: If $d\mid n$, I'm done. <br></p>
<p>If $d$ does not divide $n$, then I would need to do a second proof and this is where I'm blocked.</p>
<p>Anyone could tell me what's wrong in the approach I take to solve this problem?<br></p>
<p>Any help would be appreciated!</p>
| Bill Dubuque | 242 | <p><strong>Hint</strong> $\ $ Any sequence of $\rm\,d\,$ consecutive naturals has an element divisible by $\rm\,d.\,$ This has a very simple proof by induction: shifting such a sequence by one does not change its set of remainders mod $\rm\,d,\,$ since it simply replaces the old least element $\rm\:\color{#C00}n\:$ by the new greatest element $\rm\,\color{#C00}{n+d}$</p>
<p>$$\begin{array}{}\rm \:\color{#C00}n &\rm n+1 &\rm n+2 &\rm\! \cdots\! &\rm n+d-1 & \\
\to &\rm n+1 &\rm n+2\rm &\! \cdots\! &\rm n+d-1 &\rm \color{#C00}{n+d} \end{array}$$</p>
<p>Since $\rm\: \color{#C00}n\equiv \color{#C00}{n+d} \pmod{\! d},\,$ the shift does not change the remainders in the sequence. Thus the remainders are the same as the base case $\rm\ 0,1,2,\ldots,d-1\ =\: $ <em>all</em> possible remainders mod $\rm\,d.\,$ Therefore the sequence has an element with remainder $\,0,\,$ i.e. an element divisible by $\rm\,d.$</p>
|
306,011 | <p>Does anyone have a proof for $$\int_0^{\infty}\frac{\sin(x^2)}{x^2}\,dx=\sqrt{\frac{\pi}{2}}.$$
I tried to get it from contour integrating $$\frac{e^{iz^2}-1}{z^2},$$ but failed.
Thanks.</p>
| Felix Marin | 85,343 | <p>$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int_{0}^{\infty}{\sin\pars{x^{2}} \over x^{2}}\,\dd x & =
\int_{0}^{\infty}{1 \over 2}\int_{-1}^{1}\expo{\ic k x^{2}}\dd k\,\dd x =
{1 \over 2}\int_{-1}^{1}\
\overbrace{\int_{0}^{\infty}\expo{\ic k x^{2}}\dd x}
^{\ds{\begin{array}{c}
Fresnel\ Integral:
\\[1mm]
\ds{=\ {1 \over 2}\,\root{\pi \over 2}\,{\ic k + \verts{k} \over \verts{k}^{3/2}}}
\end{array}}}\,\ \dd k
\\[5mm] & =
{1 \over 2}\bracks{{1 \over 2}\root{\pi \over 2}\pars{2\int_{0}^{1}k^{-1/2}
\,\dd k}} = \bbx{\root{\pi \over 2}} \approx 1.2533
\end{align}</p>
<blockquote>
<p>See the <a href="https://en.wikipedia.org/wiki/Fresnel_integral" rel="nofollow noreferrer">Fresnel Integral</a>.</p>
</blockquote>
|
1,735,910 | <p>In <a href="https://www.youtube.com/watch?v=aHU-L3BLd_w">a recent video</a> the legendary Matt Parker claimed he kept flipping a two-sided (fair) coin untill he scored a sequence of ten consecutive 'switch flips', i.e. letting $T$ denote a tail and $H$ a head, then a sequence of ten switch flips is defined to be either $THTHTHTHTH$ or $HTHTHTHTHT$. He set up a contest and allowed each viewer to guess once at the exact amount of flips he needed to obtain such a sequence. The ten viewers with the ten closest answers would be awarded a prize.</p>
<p>The contest is over, so there is no incentive to keep a solution to the following problem to yourself. <em>What is the best number to bet?</em> Of course this somehow depends on how other viewers answer: you are more likely to win if your bet is not close to many other bets, so if a large number of viewers is mathematically inclined and bets the same number - say 1023 - then it no longer is the best (profit maximizing) bet. I've therefore simplified to the following question: let $X$ be the stochastic variable representing the number of flips needed untill a sequence of ten consecutive switch flips if obtained, then for which number $a \in \mathbb{N}$ does the expected value of the (absolute value of the) error
$$
\mathbb{E}[\vert X - a \vert]
$$
reach its minimal value? It is well-known that $a$ is the median of the (distribution of) $X$, but how can one compute it? Numerical approximations are welcome, theoretical (generalizable) results are preferred.</p>
<p>I found a way to compute the expected value of $X$ itself for general $n$ (i.e. the total number of coin flips needed to get a sequence of the form $THTHTHTH...$ or $HTHTHTHT...$ of length $n$). Let $\mathbb{E}_i$ denote the expected number of coin flips needed to get a desired sequence of length $n$, assuming we already have a sequence of length $i \in \mathbb{N}$. We immediately find
$$
\mathbb{E}_0 = 1 + \mathbb{E}_1
$$
since we are certain to have a sequence of length $1$ after one flip. Furthermore, for $1 \leq i \leq n-1$
$$
\mathbb{E}_i = \frac{1}{2}\left(\mathbb{E}_1 + 1\right) + \frac{1}{2}\left(\mathbb{E}_{i+1} + 1\right)
$$
since, given a sequence of $i$ flips which ends, say, on a tail, we have a $\frac{1}{2}$ chance to increase this to a sequence of $i+1$ flips (if we get, say, a head) and a $\frac{1}{2}$ chance to get back where we started, at $1$ flip. Using that $\mathbb{E}_n = 0$ the above gives us system of $n$ equations in the $n$ variables $\mathbb{E}_0, \ldots, \mathbb{E}_{n-1}$. One can easily check that the unique solution is given by
$$
\mathbb{E}_i = 2^{n} - 2^{i} \quad 0 \leq i \leq n
$$
Since Matt Parker started at $0$ and wanted to get $10$ flips, the expected value of the number of flips needed is $2^{10} - 1 = 1023$ and this should be a reasonable bet.</p>
<p>Does anyone know how to find the distribution of $X$ (or directly the median of $X$)? Like I said, analytical solutions are of course preferred, but any kind of method - even requiring numerical computations, but preferably not Monte Carlo simulations - would be interesting to me.</p>
<p><strong>EDIT:</strong> I found out that the problem can be reduced to a combinatorial problem. Indeed, we have that
$$
P(X \leq k) = \frac{\# \lbrace \text{sequences of length $k$ which contain a desired subsequence of length $n$}\rbrace}{2^k}
$$
where $2^k$ is the total number of sequences of length $k$, since every sequence of length $k$ is equally likely to occur. Let $S_k$ be the set of sequences of length $k$ of $0$'s and $1$'s (we identify tails with $0$ and heads with $1$). We have a map
$$
f: S_k \to S_{k-1}
$$
where, for any sequence $s \in S_k$, the $i$th element of $f(s)$ is $1$ if $s(i) \neq s(i+1)$ and $0$ is $s(i) = s(i+1)$. This map makes the desired sequences in $S_k$ correspond bijectively with the sequences in $S_{k-1}$ which contain $n-1$ zeroes in a row. Hence, it suffices to count the number of sequences of a given length $k-1$ which contain $n-1$ zeroes in a row. Any ideas on how to continue?</p>
| paw88789 | 147,810 | <p>Intuitively it seems like you should bet on this event occurring in the first 10 flips. </p>
<p>Reasoning: If you pick any particular 10 flips you have a $\frac{1}{2^9}$ probability of those particular ten flips being a sequence of interest. </p>
<p>However, we specifically want the first such sequence, which means that there will be constraints on the earlier flips (to avoid an earlier sequence of this type). Therefore your probability of some sequence of 10 being the first such sequence is less than or equal to $\frac{1}{2^9}$, with equality only for the first ten flips.</p>
|
2,814,703 | <p>I am reading <a href="https://en.wikipedia.org/wiki/Lower_limit_topology" rel="nofollow noreferrer">lower limit topology</a> on Wikipedia, which states that the lower limit topology </p>
<blockquote>
<p>[...] is the topology generated by the basis of all half-open intervals $[a,b)$, where a and b are real numbers. [...] The lower limit topology is finer (has more open sets) than the standard topology on the real numbers (which is generated by the open intervals). The reason is that every open interval can be written as a (countably infinite) union of half-open intervals.</p>
</blockquote>
<p>I cannot see how to write $(a,b)$ as a countably infinite union of half-open intervals.</p>
| nonuser | 463,553 | <p>If $M = (a,b)\cap \mathbb{Q}$ then $$(a,b) = \bigcup _{c\in M} [c,b)$$</p>
|
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