qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
2,953,837 | <p>Given <span class="math-container">$n_1$</span> number of a's, <span class="math-container">$n_2$</span> number of b's, <span class="math-container">$n_3$</span> number of c's.</p>
<p>They form a sequence using all these characters such that no two a's and no two b's are adjacent.</p>
<p>(a and b can be adjacent, ... | Ross Millikan | 1,827 | <p>I would define coupled recurrences. Let <span class="math-container">$A(x,y,z)$</span> be the number of strings <span class="math-container">$x\ a$</span>s, <span class="math-container">$y\ b$</span>s, and <span class="math-container">$z\ c$</span>s that ends in <span class="math-container">$a$</span> and similarly... |
514 | <p>I just came back from my Number Theory course, and during the lecture there was mention of the Collatz Conjecture.</p>
<p>I'm sure that everyone here is familiar with it; it describes an operation on a natural number – <span class="math-container">$n/2$</span> if it is even, <span class="math-container">$3n+1$</spa... | Patrick Da Silva | 10,704 | <p>I was so pissed after testing one of my own conjectures that I remembered this question and wanted to post it here. </p>
<p>I conjectured after numerical observations that for every prime p, and integers $k \ge 1, n \ge 1$, that
$$
p^k \, || 2^n-1 \quad \Longleftrightarrow \quad p^{k-1} \, || \, n \quad and \quad ... |
4,002 | <p>I'm trying to obtain a series of points on the unit sphere with a somewhat homogeneous distribution, by minimizing a function depending on distances (I took $\exp(-d)$). My points are represented by spherical angles $\theta$ and $\phi$, starting by choosing equidistributed random vectors:</p>
<pre><code>pts = Apply... | Jens | 245 | <p>To get arbitrarily many formal variables, you can use <code>Array</code>. But with those variables, your function definition won't work because of the <code>Apply</code> statement. So I modified your definition as follows (I reduced the point number for testing purposes):</p>
<pre><code>pts = Apply[{ArcCos[2 #2 - 1... |
4,002 | <p>I'm trying to obtain a series of points on the unit sphere with a somewhat homogeneous distribution, by minimizing a function depending on distances (I took $\exp(-d)$). My points are represented by spherical angles $\theta$ and $\phi$, starting by choosing equidistributed random vectors:</p>
<pre><code>pts = Apply... | Mark McClure | 36 | <p>Here are a few comments</p>
<p>First, I believe that you have switched the roles of $\phi$ and $\theta$ in your first definition. Thus, a slight edit of your code yields the following</p>
<pre><code>pts = Apply[{ArcCos[2 #2 - 1], 2 #1*Pi} &,
RandomReal[1, {10000, 2}], 1];
pts3D = {Sin[#[[1]]]*Cos[#[[2]]]... |
26,451 | <p>I am trying to solve the following:</p>
<p>$\begin{align*}
&X \sim N(1,1)\\
&\mathrm{cov}(X, X^3) = \text{?}
\end{align*}$</p>
<p>where $\mathrm{cov}$ is the covariance.</p>
<p>How would you do this in <em>Mathematica</em>?</p>
<p>I have tried</p>
<pre><code>X = NormalDistribution[1, 1]
cov[x_, y_] := ... | Silvia | 17 | <p><em>(This is too long for a comment.)</em></p>
<p>About your comment under 0x4A4D's answer: I think you didn't make it clear enough if your $X$ is a <strong>random variable</strong> or <strong>fixed data</strong> generated from some distribution. Usually, we interpret the notation in your question with the former m... |
3,595,622 | <p><strong>Problem: Give an example of a linear continuum which is not the real line <span class="math-container">$\mathbb{R}$</span>, nor
topologically equivalent to a subspace of <span class="math-container">$\mathbb{R}$</span>.</strong></p>
<p><strong>Definition of Linear Continuum:</strong> Let X be a linearly ord... | The_Sympathizer | 11,172 | <p>A ready example that comes to my mind from prior interests of mine and that is fairly natural, in that it is not specifically contrived for this purpose, I'd say, is the <strong>Dedekind completion of a suitably large non-Archimedean ordered field.</strong></p>
<p>Non-Archimedean ordered fields can be made of any c... |
1,401,898 | <p>I need a test for primality that I apply to $2^{255}-19$ (which is claimed to be prime) and certify to be correct with the ACL2 theorem prover. This means that I must be able to code the test in Common LISP, run it on this case in a reasonable period of time (I'd be happy if it ran in a day), and write a proof of c... | Peter | 82,961 | <p>This very small program written in PARI/GP shows the result and the time
needed for calculation. I have done this multiple times and the times the
program needed, differed, but it took never longer than $200ms$. The routine
certifies the primality using the adleman-pomerance-rumely-test (APR-test),
which is ... |
340,855 | <p>Say if there is a matrix A:</p>
<p>$$\begin{bmatrix} 1 & 2 & 0 & 2 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$</p>
<p><strong>What the column space of A?</strong> : I am confused whether to exclude NON-pivot columns.</p>
<p><strong>What is the dimension of column space?</... | Sepideh Bakhoda | 36,591 | <p>The dimension of column space of this matrix can not be 4, because dimension of column space=dimension of row space, and number of rows is 3, then the number of linearly independent rows is less than or equal to 3!</p>
|
4,167,747 | <p>I need to prove the following statement: If <span class="math-container">$1+ \alpha = \alpha$</span>, <span class="math-container">$\alpha$</span> is an infinite ordinal.</p>
<p>I am trying to use Bernstein's Theorem(CBS) to show if <span class="math-container">$1+ \alpha \leq \alpha$</span>, i.e., there is an injec... | Cooler Paradox | 654,036 | <p>In the following, we will use the von Neumann definition of ordinal numbers.</p>
<p>First suppose that <span class="math-container">$1 + \alpha = \alpha$</span>. This means that there is an order isomorphism <span class="math-container">$f: B \rightarrow \alpha$</span>, where <span class="math-container">$B$</span> ... |
897,043 | <p>I'm having issues getting my head around cartesian products and their cardinalities.</p>
<p>$A = \{0, 1, \{2, 3, 4\}\}$<br>
$B = \{1,5\}$<br>
$D = B \times N$ (where $N$ is the set of natural numbers)</p>
<p><strong>The first problem:</strong> What is the cardinality of:</p>
<p>(a) $A \times B$ (cartesian produ... | Akash Patalwanshi | 168,676 | <p>|A x B| = |A| * |B| = 6 , but |A x D| = aleph_null
i.e set A x D and set of nauturals are equipotent i.e there exists a bijection between A x D and N. note D is not a subset of N because elements of D are (1, 1) , (1,2) , (1,3),... , (5,1), (5,2),(5,3) ... i.e elements of D are in 2-tuple wheres elements of N a... |
1,848,222 | <p>Very simple and quick question. Usually distribution notation is such that you give the name of the distribution, then its mean, and finally the variance, for example for normal distribution:</p>
<p>$$N(0,1)$$</p>
<p>The 0 means that the distribution has mean zero, and the 1 tells that the variance is one. However... | parsiad | 64,601 | <p>Are you from France or somewhere in Europe? In France, you use $]0,1[$ to denote the open unit interval. Some other countries use $(0,1)$. Regardless, $U$ can be regarded as a function taking two inputs and producing a mathematical object.</p>
<p>I find $U(0,1)$ much more intuitive than $U(0.5,1/12)$. I would say i... |
4,032,767 | <blockquote>
<p>Prove that there exists no bijective function <span class="math-container">$f: \Bbb{N} \to \Bbb{N}$</span> such that <span class="math-container">$$f(mn)=f(m)+f(n)+3f(m)f(n)$$</span> for <span class="math-container">$m,n \geqslant1.$</span></p>
</blockquote>
<p>This was a problem from a Putnam practice ... | NoNames | 884,560 | <p>Let's assume <span class="math-container">$0\in\mathbb{N}$</span>, since otherwise the condition <span class="math-container">$m,n\ge1$</span> would be redundant. Then, <span class="math-container">$m=n=1$</span> implies <span class="math-container">$f(1)=0$</span>. If we define <span class="math-container">$g(n)=3\... |
307,264 | <p>A professor of mine has suggested to me to look at this theorem and to find a problem related to it to explain in a future class.
I found an understandable proof in "Linear operators" by Dunford-Schwartz and I think I studied it, so now I know how to probe Brouwer's Theorem.
Now I was thinking of some interesting re... | UwF | 59,239 | <p>How about Ky Fan's inequality? The one from his 1972 paper, "A minimax inequality and its applications" (there are several inequalities frequently called Ky Fan inequality). This Ky Fan Inequality is used to establish the existence of equilibria in various games studied in economics.</p>
<p>For the applications to ... |
1,661,986 | <p>I was out all last week sick with the flu and am trying to get caught up in my Discrete Mathematics course. One set of questions in my book goes as follows:</p>
<p>Find the least integer $n$ such that $f(x) \in O(x^n)$ for each of the functions:</p>
<p>(a). $f(x) = 8x+4$<br>
(b). $f(x) = x\sqrt(x)$<br>
(c). $f(x) ... | grand_chat | 215,011 | <p>The general concept of "Big-O" is that function $f$ is in Big-O of function $g$ if $f(x)$ grows <strong>no faster than</strong> $g(x)$ as $x\to\infty$. Generally speaking both $f$ and $g$ take positive values, so more precise ways of saying this are:</p>
<ul>
<li><p>$f\in O(g)$ if there is a constant $C>0$ such ... |
1,821,411 | <p>$f:[a,b]\rightarrow R$ that is integrable on [a,b]</p>
<p>So we need to prove:</p>
<p>$$\int_{-b}^{-a}f(-x)dx=\int_{a}^{b}f(x)dx$$</p>
<p>1.) So we'll use a property of definite integrals: (homogeny I think it's called?)</p>
<p>$$\int_{-b}^{-a}f(-x)dx=-1\int_{-b}^{-a}f(x)dx$$</p>
<p>2.) Great, now using the fun... | egreg | 62,967 | <p>Your step 1 is wrong and you can realize the error by considering $a=0$, $b=1$, $f(x)=e^x$.</p>
<p>Then
$$
\int_{0}^{1}e^x\,dx=e-1,
\qquad
-\int_{-1}^0e^x\,dx=\frac{1}{e}-1
$$
which are quite different.</p>
<p>You can prove the statement by the definition, I'll use Riemann sums. A Riemann sum for $\int_{a}^{b}f(x)... |
4,219,360 | <p>I was having some problems understanding how he found <span class="math-container">$\gamma(t)$</span> from the given <span class="math-container">$\Sigma$</span> and i was hoping someone could explain to me how if that is ok</p>
<p>So the problem goes like this:</p>
<p>Given the vector field <span class="math-contai... | RobPratt | 683,666 | <p>A correct starting point is instead:
<span class="math-container">$$\mathop{\mathbb{E}}[X^Y]
= \sum_{y=0}^\infty \mathop{\mathbb{E}}[X^Y|Y=y] \mathop{\mathbb{P}}[Y=y]
= \sum_{y=0}^\infty \mathop{\mathbb{E}}[X^y] \mathop{\mathbb{P}}[Y=y]
$$</span></p>
|
1,303,485 | <p>Evaluate the integral $$\int_{C}\frac{z^2}{z^2+9}dz$$
where C is the circle $|z|=4$</p>
<p>I know that if f is analytic in simply connected domain $D$, $C$ a simple closed positively oriented contour that lies in D and $z_o$ lies interior to $C$, then
$$\int_{C}\frac{f(z)}{z-z_o}dz=2\pi i f(z_o)$$</p>
<p>But for t... | kmitov | 84,067 | <p>$I=2\pi i (\frac{(3i)^2}{3i+3i}+\frac{(-3i)^2}{-3i-3i})=\frac{-9}{6i}+\frac{9}{6i}=0$</p>
|
4,350,450 | <p>For me, <span class="math-container">$\Bbb N$</span> includes <span class="math-container">$0$</span>. I am referencing, yet again, <a href="https://www.math.uni-leipzig.de/%7Eeisner/book-EFHN.pdf" rel="nofollow noreferrer">this</a> text, exercise <span class="math-container">$19$</span>, page <span class="math-cont... | Augusto Santos | 162,474 | <p>That <span class="math-container">$\pi(L)\subseteq K$</span> is clear.</p>
<p>Here is my take for the rest.</p>
<p>Conditioned on <span class="math-container">$\pi^{-1}(\left\{y\right\})$</span> being invariant for all <span class="math-container">$y\in K$</span>, then (and building on your derivation before identit... |
1,392,257 | <p><strong>The definition of a conjugate element</strong> </p>
<p>We say that $x$ is conjugate to $y$ in $G$ if $y = g^{-1}xg $ for some $g \in G$</p>
<p>Now for the group $G=Q_8$ , we have the group presentation $$Q_8 = \big<a,b: a^4 =1,b^2 = a^2, b^{-1}ab = a^{-1} \big>$$</p>
<p>Now the elements of $Q_8$ a... | Mark Bennet | 2,906 | <p>No. Think of an abelian group. All the conjugacy classes contain a single element. Only elements of order ($1$ or) $2$ are their own inverse.</p>
|
3,430,008 | <p>Currently stuck on the last part of 15.2.8(e) of this problem:</p>
<p><a href="https://i.stack.imgur.com/UvGJK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UvGJK.png" alt="enter image description here"></a></p>
<p>I don't know how to apply Fubini's theorem since one index relies on the other.... | Masacroso | 173,262 | <p>Note that if the terms are non-negative or the double series converges absolutely then
<span class="math-container">$$
\begin{align*}
\sum_{m\geqslant 0}d_m(x-b)^m&=\sum_{m\geqslant 0}\sum_{n\geqslant m}\binom{n}{m}(b-a)^{n-m}(x-b)^mc_n\\
&=\sum_{n\geqslant m\geqslant 0}\binom{n}{m}(b-a)^{n-m}(x-b)^mc_n\\
&a... |
482,801 | <p>What does it mean that the characteristic function <span class="math-container">$f(x)=1_{[b \le x \lt \infty]}$</span> is right continuous with left limits? Here <span class="math-container">$x ,b \in \mathbb{R}$</span>.</p>
| Stefan Hansen | 25,632 | <p>The graph of $f(x)=\mathbf{1}\{x\geq b\}$ looks like this:</p>
<p><img src="https://i.stack.imgur.com/nzR9r.png" alt="enter image description here"></p>
<p>Clearly, approaching any number from the right yields the same value of $f$ meaning that $f$ is right-continuous. That $f$ has left limits just means that the ... |
2,732,562 | <p>I'm trying to figure out how to find the number of ternary strings of length $n$ that have 3 or more consecutive 2's.
So far I've been able to establish that there is $n(2^{n-1})$ with a single 2.
And I think (but am not certain) that this can be extrapolated to the number of strings with a single group of 2's of le... | saulspatz | 235,128 | <p>Trying to count them directly will get you nowhere, because of double counting. The best approach is to come up with a recurrence relation, for the number of acceptable strings of length $n$ in terms of the number of shorter acceptable strings. Call a ternary string with at least three consecutive $2$'s "good" and... |
2,011,003 | <p>I stumbled upon this logic question in a math class recently. </p>
<p>My teacher told us that a statement that is not tested/is empty is true. For example, that if I stated that: "if the team A wins the game, I am gonna buy you a coke", and then team B goes on and wins the game, the statement would be true, indepen... | FraGrechi | 348,690 | <p>The general idea at which you are hinting is that of conditional probability, which states that for two events, $P$ and $Q$, the statement $P$ <em>implies</em> $Q$ is given by</p>
<p>$$(P \implies Q) \iff (Q \lor \lnot P),$$ </p>
<p>where "$\implies$" denotes the implication operator.</p>
<p>To better understand ... |
1,095,621 | <p>I am looking for a way to integrate $$\int \sqrt{x^2-4}\ dx $$ using trigonometric substitutions. </p>
<p>All my attempts so far lead to complicated solutions that were uncomputable.</p>
| Aaron Maroja | 143,413 | <p>Hint: Let $x = 2 \cosh \theta \Rightarrow dx = 2 \sinh \theta\ d\theta $.</p>
<p><strong>Edit:</strong> The OP said that he hasn't seen hyperbolic functions just yet. Then what about $x = 2\sec \theta \Rightarrow dx = 2\sec \theta \tan\theta\ d\theta$</p>
<p>So </p>
<p>$$2\int \sqrt{4\sec^2\theta - 4}\sec\thet... |
3,054,898 | <h3>Problem</h3>
<p>Evaluate <span class="math-container">$$\int_0^{2\pi}(t-\sin t)(1-\cos t)^2{\rm d}t.$$</span></p>
<h3>Comment</h3>
<p>It's very complicated to compute the integral applying normal method. I obtain the result resorting to the skillful formula</p>
<blockquote>
<p><span class="math-container">$$\int_0^... | egreg | 62,967 | <p>The integral
<span class="math-container">$$
\int_0^{2\pi}\sin t(1-\cos t)^2\,dt=\Bigl[\frac{(1-\cos t)^3}{3}\Bigr]_0^{2\pi}=0
$$</span>
is immediate. Thus we can concentrate on
<span class="math-container">$$
\int_0^{2\pi}t(1-\cos t)^2\,dt=[\dots t=2u \dots]=
16\int_0^{\pi}u\sin^4u\,du=\int_0^\pi u(e^{iu}-e^{-iu})^... |
2,912,152 | <p>I know there is already a question about resolving a quadrilateral from three sides and two angles, but I want to ask about a special case. Firstly, two of the sides are known to be of equal size. Secondly, I'm only interested in the area, not in the remaining angles or lengths. Can anyone suggest a simple formul... | hmakholm left over Monica | 14,366 | <p>Here's your basic mistake:</p>
<blockquote>
<p>If we don't do that, (6) is a contradiction in the case $A'$ is empty and so the whole proof is wrong. Am I the wrong one?</p>
</blockquote>
<p>There's nothing wrong with having contradictory <em>assumptions</em> at some point in a proof. On the contrary, that is a ... |
264,745 | <p>When I was learning statistics I noticed that a lot of things in the textbook I was using were phrased in vague terms of "this is a function of that" e.g. a statistic is a function of a sample from a distribution. I realized that while I know the definition of a function as a relation and I have an intuitive notion ... | Christopher A. Wong | 22,059 | <p>A function $f$ is called "a function of $x$", if, for each $x$ (in some domain $X$), there is a unique corresponding output, denoted by $f(x)$.</p>
<p>So a statistic is a function of a sample from a distribution means that, given a sample $S$, a statistic takes that sample $S$ and spits out a unique statistic value... |
2,130,807 | <p>How to Prove $G$ is connected, if $G$ is an acyclic graph on $n \ge 1$ vertices containing exactly $n − 1$ edges?</p>
| Kuifje | 273,220 | <p>If $G$ is acyclic with $n$ vertices and $n-1$ edges then $G$ is a connected tree.</p>
<p>Proof: (by induction on $n$)</p>
<p>If $n=1$ (or $n=2$), the result is trivial.</p>
<p>Suppose the property holds for a given $n$, and consider an acyclic graph with $n+1$ vertices and $n$ edges. Since $G$ is acyclic, there i... |
2,653,708 | <p>In this question I am using the euclidean metric to determine the distance between two points.</p>
<p>I want to make a function $f(x)=$ the minimum distance between $y=x$ and $y=e^x$ at each given point x, is there an efficient way of doing this?</p>
<p>Second related question if i knew $e^x$ was the shortest dist... | g.kov | 122,782 | <p>Consider the point $(x_1,g(x_1))$.
At this point
the distance to the line $y=x$
is $d(x_1)=\exp(x_1)$. </p>
<p><a href="https://i.stack.imgur.com/4rFBp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4rFBp.png" alt="enter image description here"></a></p>
<p>As we can see, for every $x$
there ar... |
116,037 | <p>I would warmly appreciate it if someone could tell me whether the following question has an affirmative answer. I am new to the field of commutative algebra, so I am simply trying to fill in some (huge) gaps. Thanks!</p>
<p>Let $ (R,{\frak{m}}) $ be a Noetherian local (commutative unital) ring. Let $ I $ be an idea... | Daniel Litt | 6,950 | <p>No. Consider $\mathfrak{m}:=(x,y,z)\subset k[x,y,z]_{(x,y,z)}=:R$. Then the kernel of the map $$R^3\to \mathfrak{m}$$ defined by the minimal generating set $x,y,z$ is minimally generated by $$k_1:=(y, -x, 0), k_2:=(z, 0, -x), k_3:=(0, z, -y).$$ But $$zk_1-yk_2-xk_3=0$$
so the submodule of $R^3$ that these generat... |
1,299,266 | <p>How many zeros are there in the number $50!$?</p>
<p>My attempt:</p>
<p>The zeros in every number come from the 10s that make up the number. The 10s are, in turn, made up of 2s and 5s.</p>
<p>So: $\frac{50}{5*2} = 5$ zeros?</p>
| 5xum | 112,884 | <p>Much more than $5$, surely. Since $50!$ is divisible by $2\cdot 5 \cdot 10\cdot 20\cdot 30\cdot 40\cdot 50$, and this number already has $6$ zeroes, you can be sure that $50!$ has at least $6$ zeroes.</p>
<p>In fact, you need to count up how many twos and how many fives appears in the factorization of $50!$. Then, ... |
1,299,266 | <p>How many zeros are there in the number $50!$?</p>
<p>My attempt:</p>
<p>The zeros in every number come from the 10s that make up the number. The 10s are, in turn, made up of 2s and 5s.</p>
<p>So: $\frac{50}{5*2} = 5$ zeros?</p>
| Community | -1 | <p>The number of 0's is equal to the powers of 5 in the expansion of 50!. This is because the prime decomposition of 50! will have more factors of 2 than factors of 5, and whenever we have a factor of 2 and 5 we can combine them and tack on a 0 at the end of the number.</p>
<p>The number of powers of 5 is $\lfloor{\fr... |
4,164,553 | <p>Can anybody enlighten me about the applications of <a href="https://en.wikipedia.org/wiki/Intuitionistic_logic" rel="noreferrer">intuitionistic logic</a>? I am familiar with this system only by <a href="https://www.maa.org/publications/maa-reviews/proof-theory" rel="noreferrer">G.Takeuti's book</a>, where it is desc... | Noah Schweber | 28,111 | <p>The <strong>internal logic of a topos</strong> is inherently intuitionistic, not classical. This means that if we want to "prove a fact in all topoi (satisfying some conditions)," we should generally look for <em>constructive</em> arguments.</p>
<p>Even if we don't care about the internal structure of a to... |
1,150,805 | <p>An unfair 3-sided die is rolled twice. The probability of rolling a 3 is $0.5$, the probability of rolling a 1 is $0.25$, and the probability of rolling a 2 is $0.25$. Let $X$ be the outcome of the first roll and $Y$ the outcome of the second.</p>
<ul>
<li><p>Find the Joint Distribution of $X$ and $Y$ in a Table.</... | Community | -1 | <p>Since the set is finite, without loss of generality, assume that $X_n = \{1,2, \ldots, n\}$. Then, note that:
$$\mathcal{P}(X_1) = \mathcal{P}(\{1\}) = \{\phi,\{1\} \} = \{\phi, X_1\}$$
$$\mathcal{P}(X_n) = \mathcal{P}(X_{n-1}) \cup \{Y \cup \{n\} : Y \in\mathcal{P}(X_{n-1})\}\; \forall \; n \ge 2$$</p>
<p>This is ... |
827,154 | <p>I need help with the definition of "within 1":</p>
<ul>
<li><p>If $x = 8$ and $y = 7$, then $x$ is "within 1" of $y$. </p></li>
<li><p>If $x = 8$ and $y = 9$, then $x$ is "within 1" of $y$.</p></li>
<li><p>If $x = 8$ and $y = 8$, is $x$ still "within 1" of $y$?</p></li>
</ul>
<p>It's my understanding that this wou... | RJ Hill | 155,968 | <p>It means that the value lies within the limits of +/− 1. </p>
<p>If you were to say 'within 1' of 20, that means that 19, 20, and 21 are all valid numbers because they're 'within 1' of 20. </p>
<p>The most common term for this is 'plus or minus 1' or whatever range you're looking in. Symbols used to denote this pa... |
2,010,069 | <p>I am looking on the solution to this problem presented in the book <em>"Fifty Challenging Problems in Probability with Solutions"</em> by Mosteller (p.18-19).</p>
<blockquote>
<p>On average, how many times must a die be thrown until one gets a 6?</p>
</blockquote>
<p>There are many ways to solve this problem as... | Somos | 438,089 | <p>I think an intuitive way to think about matrix multiplication is to regard it as a combination of coordinate extraction and scalar multiplication. First, given any basis for a finite vector space, such as $\mathbb{R}^n$, we usually express any vector $\mathbf{v}=\sum_i c_i \mathbf{e}_i=[c_1,c_2,\dots,c_n]$ as an $n$... |
2,028,703 | <p>I'm having this example for a simple <a href="https://en.wikipedia.org/wiki/Binary_symmetric_channel" rel="nofollow noreferrer">binary symmetric channel</a> (BSC) to bound the mutual information of $X$ and $Y$ as</p>
<p>\begin{align*}
I(X;Y) &= H(Y) - H(Y|X)\\
&= H(Y) - \sum p(x) H(Y \mid X = x) \\
&= H... | Stefan Falk | 70,606 | <p>I just realized that it's actually very simple to show that</p>
<p>\begin{align*}
\sum_{x \in X} \mathbb{P}[X = x]H(Y|X=x) &= H_B(p)
\end{align*}</p>
<p>We start by observing that of course</p>
<p>\begin{align*}
\sum_{x \in X} \mathbb{P}[X = x]H(Y|X=x) &= p_X(0) \cdot H(Y|X=0) + p_X(1) \cdot H(Y|X = 1) \\... |
1,476,313 | <p>I want to simplify this fraction</p>
<p>$$ \frac{\sqrt{6} + \sqrt{10} + \sqrt{15} + 2}{\sqrt{6} - \sqrt{10} + \sqrt{15} - 2} $$</p>
<p>I've tried to group up the denominator members like $ (\sqrt{6} + \sqrt{15}) - (\sqrt{10} + 2) $ and then amplify with $ (\sqrt{6} + \sqrt{15}) + (\sqrt{10} + 2) $ </p>
| Community | -1 | <p>For this <a href="https://math.stackexchange.com/a/2308483/4414">system</a> we implemented finding a split form for the denomiator:
$$a + \sqrt{p}\,b$$
Such that $\sqrt{p}$ is a new radical. For a quotient we then have:
$$\frac{c}{a + \sqrt{p}\,b} = \frac{c\,(a - \sqrt{p}\,b)}{a^2 - p\,b^2}$$
Lets give it a try:
$$\... |
1,985,552 | <p>Where $p_n \rightarrow p$. I'm trying to prove that for $E=\{ p_n : n \in \mathbb{N}$ and $lim_{n\rightarrow \infty} p_n =p \}$, then $Cl(E)=E \cup \{p \}$ and $Cl(E)$ is compact. </p>
<p>Also, I'm currently using the definition of limit points as p is a limit point if $\forall r>0, (E \cap N_r(p)) \backslash \{... | true blue anil | 22,388 | <p>It is akin to putting 30 identical balls in 4 distinct boxes</p>
<p>Put 1 in the first, 4 in the second, 6 in the third and 10 in the fourth, 9 more remain to be put.</p>
<p>Allotments in which you put $\ge 5$ in the first box and $\ge6$ in the others will be invalid.<br>
[ Note that with the given figures, you ca... |
88,156 | <p>I understand that <code>Round</code> give the nearest even integer for cases where the number is between two integers, i.e. <code>Round[2.5] = 2</code> and <code>Round[3.5] = 4</code> (see this <a href="https://mathematica.stackexchange.com/questions/29122/why-do-numberform-and-round-apparently-use-different-tie-bre... | Eric Towers | 16,237 | <p>Sjoerd C. de Vries correctly diagnoses the problem and provides a resolution. An alternative is to work in exact numbers and convert to floating point at the end.</p>
<pre><code>Floor[Range[1, 2, 1/10], 2/10]
(* Output: {1, 1, 6/5, 6/5, 7/5, 7/5, 8/5, 8/5, 9/5, 9/5, 2} *)
N[Floor[Range[1, 2, 1/10], 2/10]]
(* Out... |
2,567,607 | <p>$$\arctan 2x +\arctan 3x = \left(\frac{\pi}{4}\right)$$
$$\arctan \left(\frac{2x+3x}{1-2x*3x}\right)=\frac {\pi}{4}$$
$$\frac {5x}{1-6x^2}=\tan \frac{\pi}{4}=1$$
$$6x^2 + 5x -1 = 0$$
$$(6x-1)(x+1)=0$$
$$x=-1, \frac{1}{6}$$</p>
<p>The answer however rejects the solution $x=-1$ saying that it makes the L.H.S of the e... | Maadhav | 416,874 | <blockquote>
<p>$\arctan x + \arctan y = \arctan\left(\dfrac{x+y}{1-xy}\right), \quad\quad xy < 1$</p>
</blockquote>
<p>So</p>
<p>$x = -1$ doesn't work as $2x \times 3x = 6x^2 = 1 \nless 1$.</p>
<p>$x=\frac16$ works as $6x^2 = \frac1{6} \lt 1$.</p>
|
3,126,936 | <p>Numbers between <span class="math-container">$1 - 1000$</span> which leave no remainder when divided by <span class="math-container">$4$</span> and divided by <span class="math-container">$6$</span> but not by <span class="math-container">$21$</span>?</p>
<p>I tried <span class="math-container">$$\frac{1000}{12} = ... | Community | -1 | <p>We get that the number is a multiple of 12 but not of 21. As 12 is coprime to 21, and <span class="math-container">$12 \times 21 = 252$</span>, which has 3 multiples in 1-1000, we also know that there are 83 multiples of 12 in 1-1000. So, we get 83-3, which is equal to 3.</p>
|
422,761 | <blockquote>
<p>Prove that there is no Integer such that $x≡2 \pmod 6$ and $x≡3 \pmod 9$ are both true.</p>
</blockquote>
<p>How should I approach this question?<br>
I attempted using contra-positive proof, so $x=6p+2$ and $x=9q+3$ where $p,q$ are integers.<br>
Then $6p+2=9q+3$. </p>
| Arnab Dutta | 80,339 | <p>@TomDavies92 </p>
<p>I think the simple answer he asks for is - What does the Generating function
$ \sum\limits_{i=1}^n \ln^n3 $ means:</p>
<p>The above expression should be better written as: $ \sum\limits_{i=1}^n \ln^i3 $</p>
<p>Answer simple, Its expansion is:<br>
$ln\ 3 + ln^2\ 3 + ln^3\ 3 + ...ln^n\ 3 $
... |
4,348,969 | <p>Let <span class="math-container">$a<b\in\mathbb{R}$</span>. A sequence <span class="math-container">$P:=(p_0,\ldots,p_n)$</span> is a called a partition of <span class="math-container">$[a,b]$</span> if
<span class="math-container">$$a=p_0<\ldots<p_n=b.$$</span>
The size of <span class="math-container">$P$<... | RRL | 148,510 | <p>You gave an example of a uniform partition.</p>
<p>The dyadic partition <span class="math-container">$P_n = (x_0,x_1,\ldots,x_{2^n})$</span> with <span class="math-container">$(b-a)/2^n < \delta$</span> and</p>
<p><span class="math-container">$$x_k = a + \frac{b-a}{2^n}k, \quad k=0,1,\ldots, 2^n,$$</span></p>
<p>... |
4,348,969 | <p>Let <span class="math-container">$a<b\in\mathbb{R}$</span>. A sequence <span class="math-container">$P:=(p_0,\ldots,p_n)$</span> is a called a partition of <span class="math-container">$[a,b]$</span> if
<span class="math-container">$$a=p_0<\ldots<p_n=b.$$</span>
The size of <span class="math-container">$P$<... | B. S. Thomson | 281,004 | <p>Here is my favorite:</p>
<p><strong>LEMMA</strong>. [Cousin partitioning lemma] Let <span class="math-container">$\delta(x)$</span> be a positive function defined on some fixed interval <span class="math-container">$[a,b]$</span>. Then for any subinterval <span class="math-container">$[c,d]\subset [a,b]$</span> t... |
310,462 | <p>I am looking for an elegant proof of the fact that a countable metric space is complete iff its underlying topology is discrete.</p>
<p>It is easy to see that a discrete space is complete because its topology can be derived from the distance <span class="math-container">$d(x,y)=1$</span> iff <span class="math-conta... | Tomasz Kania | 15,129 | <p>This is not true. Take a countable successor ordinal with the order topology. This topology makes it compact hence Polish.</p>
|
484,367 | <p>I've been trying to find a tight upper bound for the series</p>
<p>$$S (x) = e^{-x} \sum_{k=0}^{\infty} \frac{x^k}{k!} \sqrt{k+1}$$</p>
<p>So far, I've managed to get a reasonable bound for small values of $x$ by using the inequality $\sqrt{k+1} \leq \sqrt{\frac{k^{2}}{4} + k + 1} = \frac{k}{2} + 1 ~\forall~k \geq... | robjohn | 13,854 | <p><strong>Upper and Lower Bounds</strong></p>
<p>Note that
$$
e^{-x}\sum_{k=0}^\infty\frac{x^k}{k!}=1\tag{1}
$$
and that
$$
e^{-x}\sum_{k=0}^\infty(k+1)\frac{x^k}{k!}=x+1\tag{2}
$$
Since $\sqrt{x}$ is concave, <a href="http://en.wikipedia.org/wiki/Jensen%27s_inequality" rel="nofollow">Jensen's Inequality</a> gives
$$... |
3,080,230 | <p>On <a href="https://en.wikipedia.org/wiki/Net_(mathematics)#Properties" rel="nofollow noreferrer">Wikipedia</a> it states that a space <span class="math-container">$X$</span> is compact if and only if every net has a convergent subnet. It then states that a net in the product topology has a limit if and only if each... | bof | 111,012 | <p>Here is an example of a sequence which has no convergent subsequence, but it has a convergent subnet <strong>assuming the axiom of choice</strong>.</p>
<p>Let <span class="math-container">$I=\{0,1\}^\mathbb N$</span>.</p>
<p>The product space <span class="math-container">$X=\{0,1\}^I$</span> is a compact space whi... |
1,822,336 | <p>My friend asked me what the roots of $y=x^3+x^2-2x-1$ was.</p>
<p>I didn't really know and when I graphed it, it had no integer solutions. So I asked him what the answer was, and he said that the $3$ roots were $2\cos\left(\frac {2\pi}{7}\right), 2\cos\left(\frac {4\pi}{7}\right)$ and $2\cos\left(\frac {8\pi}{7}\ri... | achille hui | 59,379 | <p>Let $p(x) = x^3+x^2-2x-1$, we have $$p(t + t^{-1}) = t^3 + t^2 + t + 1 + t^{-1} + t^{-2} + t^{-3} = \frac{t^7-1}{t^3(t-1)}$$</p>
<p>The RHS has roots of the form $t = e^{\pm \frac{2k\pi}{7}i}$
( coming from the $t^7 - 1$ factor in numerator )
for $k = 1,2,3$. So $p(x)$ has roots of the form $$e^{\frac{2k\pi}{7} i... |
1,659,075 | <p>In linear algebra, the Rank-Nullity theorem states that given a vector space $V$ and an $n\times n$ matrix $A$,
$$\text{rank}(A) + \text{null}(A) = n$$
or that
$$\text{dim(image}(A)) + \text{dim(ker}(A)) = \text{dim}(V).$$</p>
<hr>
<p>In abstract algebra, the Orbit-Stabilizer theorem states that given a group $G$ ... | Nick | 27,349 | <p>As was pointed out in the comments by Clement Guerin and Berci above, the Rank-Nullity Theorem is more properly seen as an immediate consequence of the First Isomorphism Theorem, which says that $\mathrm{Im}(A) \cong V / \mathrm{Ker}(A)$. Taking dimensions of these spaces gives the statement of the Rank-Nullity Theo... |
234,945 | <p>Let $a \in \mathbb R$, what values of $t$ solve the equation $at + \sin(t) = 0$?</p>
| preferred_anon | 27,150 | <p>Since we can write it as $at=-\sin(t)$, the roots lie where the line with gradient $a$ and $-\sin(t)$ intersect.<br>
If $a\le1$, the only root is $0$.<br>
For $-1<a<0$ there are finitely many solutions whose values can only be computed numerically. There will be no roots when $t>T$, where $T$ is such that $... |
234,945 | <p>Let $a \in \mathbb R$, what values of $t$ solve the equation $at + \sin(t) = 0$?</p>
| Douglas B. Staple | 65,886 | <p>The function
$$\operatorname{sinc}(x)\equiv\frac{\sin(x)}{x}$$ is called the <a href="https://en.wikipedia.org/wiki/Sinc_function" rel="nofollow">$\operatorname{sinc}$</a> function. You need the (multi-valued) inverse of this function, which is a transcendental function with no name. We could call it the "arcsinc" f... |
2,734,338 | <p>I would like to show that
$$\forall n\in\mathbb{N}^*, \quad \sqrt{\frac{n}{n+1}}\notin \mathbb{Q}$$</p>
<p>I'm interested in more ways of proofing this.</p>
<p>My method :</p>
<p>suppose that $\sqrt{\frac{n}{n+1}}\in \mathbb{Q}$ then there exist $(p,q)\in\mathbb{Z}\times \mathbb{N}^*$ such that $\sqrt{\frac{n}... | Hw Chu | 507,264 | <p>If $\displaystyle\sqrt{\frac{n}{n+1}} \in \mathbb Q$, then $\displaystyle\sqrt{n(n+1)} = (n+1)\sqrt{\frac{n}{n+1}} \in \mathbb Q$, and since $n \in \mathbb N_{>0}$, $\sqrt{n(n+1)} \in \mathbb Z$.</p>
<p>But $n^2 < n(n+1) < (n+1)^2$.</p>
<hr>
<p><em>Remark:</em> Actually the proof in your post is still in... |
1,128,623 | <p>Question: Let S be the set of sequences of $0$s and $1$s. For $x = (x_1, x_2, x_3, ...)$ and $y = (y_1, y_2, y_3, ...)$. Define</p>
<p>$d(x,y)=\sum_{i=1}^\infty \dfrac{|x_i - y_i|}{2^i}$ </p>
<p>Proof the infinite sum in the definition of $d(x,y)$ converges for all $x$ and $y$. </p>
<p>Incomplete answer: Since th... | doeo | 180,343 | <p>You have the right idea of bounding the the infinite sum. Now consider what the bounds would be if we started the series at the $n^{th}$ term. We can get that</p>
<p>$\sum_{i=n}^\infty \dfrac{|x_i-y_i|}{2^i}\leq \dfrac{1}{2^{n-1}}\to0$</p>
<p>as $n \to\infty$, which shows that the series converges.</p>
|
1,802,215 | <p>I'd like to show that $Cov(aX+b, Y+Z)=aCov(X,Y)+aCov(X,Z)$.</p>
<p>Therefore I use:</p>
<ul>
<li>$Cov(X,Y)=E(XY)-E(X)\cdot E(Y)$.</li>
</ul>
<p>So $Cov(aX+b, Y+Z)=$</p>
<p>$=E[(aX+b)(Y+Z)]-E(aX+b)E(Y+Z)$</p>
<p>$=E(aXY+aZ+bEY+bEZ)-[(aEX+b)(EY+EZ)]$</p>
<p>$=aE(XY)+aEZ+bEY+bEZ-[aEXEY+aEXEZ+bEY+bEZ]$</p>
<p>??... | Dark | 208,508 | <p>It is much faster to use the fact that $Cov$ is a bi-linear map.</p>
<p>Hence $Cov(aX+b,Y+Z) = a Cov(X,Y+Z) + Cov(b,Y+Z)$ (linearity w.r.t first variable)</p>
<p>$b$ is constant so $Cov(b,Y+Z)=0$.</p>
<p>Then $Cov(aX+b,Y+Z) = aCov(X,Y+Z) = a(Cov(X,Y)+Cov(X,Z))$. (linearity w.r.t second variable)</p>
|
1,802,215 | <p>I'd like to show that $Cov(aX+b, Y+Z)=aCov(X,Y)+aCov(X,Z)$.</p>
<p>Therefore I use:</p>
<ul>
<li>$Cov(X,Y)=E(XY)-E(X)\cdot E(Y)$.</li>
</ul>
<p>So $Cov(aX+b, Y+Z)=$</p>
<p>$=E[(aX+b)(Y+Z)]-E(aX+b)E(Y+Z)$</p>
<p>$=E(aXY+aZ+bEY+bEZ)-[(aEX+b)(EY+EZ)]$</p>
<p>$=aE(XY)+aEZ+bEY+bEZ-[aEXEY+aEXEZ+bEY+bEZ]$</p>
<p>??... | grand_chat | 215,011 | <p>Your line
$$E(aXY+aZ+bEY+bEZ)-[(aEX+b)(EY+EZ)]$$
has a mistake: it should be
$$E(aXY+a\color{red}{XZ}+bEY+bEZ)-[(aEX+b)(EY+EZ)]\tag1$$
After you fix this, you should be able to collect terms properly:
$$
\begin{align}
(1)&=aE(XY)+aEXZ+bEY+bEZ-[aEXEY+aEXEZ+bEY+bEZ]\\
&=aE(XY)-aEXEY +aEXZ-aEXEZ
\end{align}
$$... |
257,121 | <p>The question is very simple and I apologize for that, but I am not an expert of this kind of problem.
Given the polynomial
$$ P(x_1,\ldots,x_{2n})=x_1^2+\ldots+x_n^2-x_{n+1}^2-\ldots-x_{2n}^2,$$
I would like to know if there are non trivial integer roots $(y_1,\ldots, y_{2n})$ such that
$$y_1+\cdots+y_{n}=y_{n+1}+\c... | Bugs Bunny | 5,301 | <p>Yes, there are. For instance,
(1,4,6,7,2,3,5,8)</p>
<p>The general principle behind this solution is that
$$
n^2+(n+1)^2=((n-1)^2+(n+2)^2)-4.
$$
Combining two such collections on the opposite sides always gives you a solution.</p>
|
4,305,538 | <p>I am trying to compute the Fourier transform of <span class="math-container">$(x_{1}+ix_{2})^{-1}$</span> in <span class="math-container">$S'(\mathbb{R}^{2})$</span>. i.e. as a tempered distribution.</p>
<p>It might be useful to note that for <span class="math-container">$\mu \in S'(\mathbb{R}^{2})$</span> and <span... | Ninad Munshi | 698,724 | <p>To calculate this Fourier transform, we will need another well known Fourier transform pair in <span class="math-container">$1$</span>D:</p>
<p><span class="math-container">$$\int_{-\infty}^\infty e^{-|a||x|}e^{-i\lambda x}dx = \int_{-\infty}^0e^{(|a|-i\lambda)x}dx + \int_0^\infty e^{-(|a|+i\lambda)x}dx $$</span></p... |
2,508,381 | <p>Starting from the idea that $$\sum_{n=1}^\infty n = -\frac{1}{12}$$
It's fairly natural to ask about the series of odd numbers $$\sum_{n=1}^{\infty} (2n - 1)$$
I worked this out in two different ways, and get two different answers. By my first method
$$\sum_{n=1}^{\infty} (2n - 1) + 2\bigg( \sum_{n=1}^\infty n \bi... | spaceisdarkgreen | 397,125 | <p>With the usual caveat that $$ \sum_{n=1}^\infty n \ne -\frac{1}{12}$$
we can do a similar zeta function regularization for the sum of odd integers. We start with the fact that $$ \sum_{n = 1}^\infty \frac{1}{(2n-1)^s} =(1-2^{-s})\zeta(s)$$ for $\Re(s) > 1$ and then analytically continue to $s=-1$ to get $$ \sum_{... |
46,772 | <p>I have the following to set into mathematica. </p>
<p>Assume a stress(sigma) applied on a specimen which has a number of fibers within it. All fibers that have a strength less than the applied stress should fail. I set a cumulative distribution of the fiber strengths with a mean and standard deviation.</p>
<p>I wa... | george2079 | 2,079 | <p>Here is how you do this analytically: (using user2790167's <code>pull</code> in pure function form)</p>
<pre><code> nFibers = 50;
mean = 100;
stdev = 20;
fibers = Sort@RandomVariate[NormalDistribution[mean, stdev], nFibers];
Show[{
Plot[ InverseCDF[ NormalDistribution[ mean, stdev] , nweak/nFibers]
... |
983,566 | <p>Here is the problem in full:</p>
<blockquote>
<p>A heap has $x$ marbles, where $x$ is a positive integer. The following process is repeated until the heap is broken down into single marbles: choose a heap with more than 1 marble and form two non-empty heaps from it. One will contain $n$ marbles and the other $m$ ... | David Holden | 79,543 | <p>this has a computation-lite intuitive solution. think of the marbles as a squadron of $K$ tiny space-craft. suppose craft belonging to the same pile can communicate by walkie-talkie, but if any two are separated a long-range communications link is set up between them. </p>
<p>so when a pile of $m+n$ craft is split ... |
2,502,224 | <p>I don't quite understand <a href="https://math.stackexchange.com/a/1866099/185631">this example given by Mike Haskel</a>. I want to find an example about</p>
<p><span class="math-container">$$\operatorname{Hom}_R\left ( M ,\bigoplus_{i\in I} N_{i}\right )\not \cong\bigoplus_{i\in I} \operatorname{Hom}_R\left ( M ,N_... | No One | 185,631 | <p>I think they are probably isomorphic to each other. Let me know where I am wrong.</p>
<p>The dimension of LHS is $|\bigoplus_{i=1}^{\infty}\mathbb R|^{\dim(M)}=|\mathbb R|^{\dim(M)}$, while the dimension of the RHS is $|\mathbb N||\dim(M^*)|=|\mathbb N||\mathbb R ^{\dim(M)}|=|\mathbb R ^{\dim(M)}|$.</p>
<p>Referen... |
409,626 | <p>I'm studying elementary group theory, and just seeing the ways in which groups break apart into simpler groups, specifically, a group can be broken up as the sort of product of any of its normal subgroups with the quotient group of that subgroup. So I wondered how you could do the inverse of that operation:</p>
<ol... | Najib Idrissi | 10,014 | <p>Yes, the direct product $A \times B$ satisfies the property, as you've noticed. But it's not unique up to isomorphism. For example, the <a href="https://en.wikipedia.org/wiki/Dihedral_group">dihedral group</a> $D_n$ has a normal subgroup $H \simeq \mathbb Z/n \mathbb Z$, with $G/H \simeq \mathbb Z/2 \mathbb Z$, but ... |
2,003 | <p>I use some custom shortcut keys in <code>KeyEventTranslations.tr</code>. One is for the <code>Delete All Output</code> function: </p>
<pre><code>Item[KeyEvent["w", Modifiers -> {Control}],
FrontEnd`FrontEndExecute[FrontEnd`FrontEndToken["DeleteGeneratedCells"]]]
</code></pre>
<p>or simply:</p>
<pre><code>... | Szabolcs | 12 | <p>Try using this:</p>
<pre><code>FrontEndExecute[
{FrontEnd`NotebookFind[FrontEnd`SelectedNotebook[],
"Output", All, CellStyle, AutoScroll->False],
FrontEnd`FrontEndToken["Clear"]}]
</code></pre>
<p>(Untested in <code>KeyEventTranslations.tr</code>, but works as a button!)</p>
<hr... |
2,919,561 | <p>What is the following limit ? How do I solve it ?
$$\lim \limits_{x\to0}\frac{6\sin x}{x-3\tan x}$$</p>
| Matheus Andrade | 508,844 | <p>$$\begin{align*} \lim \limits_{x\to0}\frac{6\sin x}{x-3\tan x} &= \lim \limits_{x\to0} \frac{6\cos(x)}{1 - 3\sec^2(x)} \\ &=\frac{6}{1-3} \\ &= -3\end{align*}$$</p>
|
629,989 | <p>Given function sequence $\{f_n(x)\}^\infty$ defined as $f_n(x) = \frac{nx}{2 + n + x}. (0 \le x \le 1)$</p>
<p>I need to find the limit function and whether it converges uniformly or not uniformly.</p>
<p>I found that the limit is:</p>
<p>$$\lim_{n \to \infty} \frac{nx}{2 + n + x} = \lim_{n \to \infty} \frac{x}{... | Clarinetist | 81,560 | <p>Julián explains it very well in the comments, but I'll expand the discussion a bit. A sequence $\{f_{n}\}$ such that $f_{n}: E\subseteq \mathbb{R} \to \mathbb{R}$ for all $n$ converges uniformly to $f$ if for all $\epsilon > 0$ there is an $N(\epsilon) \in \mathbb{N}$ such that for all $n \geq N(\epsilon)$, $|f_{... |
3,896,314 | <p>I am trying to show that <span class="math-container">$\log 15$</span> and <span class="math-container">$\log 3$</span> + <span class="math-container">$\log 5$</span> is irrational.</p>
<p>For <span class="math-container">$\log 15$</span> I feel like I have no issues showing this is irrational.</p>
<p>By contradicti... | Community | -1 | <p>The sum of two irrational numbers can be rational. Take <span class="math-container">$a$</span> irrational, and <span class="math-container">$1-a$</span>, which is also irrational (in fact <span class="math-container">$-a$</span> is even simpler, but I guess that this would not satisfy the readers). So there is no w... |
1,694,495 | <p>Graphically, I am searching for something like this:</p>
<p><a href="https://i.stack.imgur.com/Rskpk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Rskpk.png" alt="enter image description here"></a></p>
<p>The only additional requirement would be that the elements are defined by a closed formul... | Oscar Lanzi | 248,217 | <p>This is not exactly the same thing, but consider the "Tower of Hanoi" sequence:</p>
<p>1 2 1 3 1 2 1 4 1 2 1 3 1 ...</p>
<p>$a_n = k$ when $n$ is an odd number times $2^{k-1}$</p>
<p>We need more than 1000 terms before we see any term greater than 10 (in the Tower of Hanoi puzzle that means if you have 11 layers ... |
1,311,466 | <p>My concept of real no. Is not very clear. Please also tell the logic behind the question.
The expression is true for 19, is it true for all the multiples? </p>
| gt6989b | 16,192 | <p>In $(3n)! = 1 \times 2 \times \ldots \times 3n$ there are $n$ numbers divisible by 3 directly ($3, 6, \ldots, 3n$) and additionally $n$ which are divisible by $2$, so $(3n)!$ is divisible by $(3!)^n = 3^n \cdot 2^n$.</p>
|
1,311,466 | <p>My concept of real no. Is not very clear. Please also tell the logic behind the question.
The expression is true for 19, is it true for all the multiples? </p>
| alkabary | 96,332 | <p>Here is a proof by induction </p>
<p>Base case $n=0$ , we have $$\frac{(3 \times 0)!}{(3!)^0} = \frac{0!}{6^0} = \frac{1}{1} = 1$$ because $0! = 1$ and $6^0 =1$ and so $1$ is an integral number so the base case works</p>
<p>Now assume it works for an integer $k \geq 0$ then we need to prove it for $k+1$ , here is... |
1,130,855 | <p>I've searched this website and while there are a few questions similar to mine, I couldn't find what I was looking for/a specific method for what I want to do.</p>
<p>I want to understand how one would prove that the remainder of $5^{336}$ by $23$ is $8$, or in other words, $$5^{336}\equiv 8 \pmod{23}.$$</p>
<p>Ca... | Bill Dubuque | 242 | <p><strong>Hint</strong> $\ {\rm mod}\ 23\!:\ \underbrace{\color{#c00}{5^{\large 22}}\equiv\color{#c00}1}_{\rm little\ Fermat}\Rightarrow\, 5^{\large 336}\equiv 5^{\large 6+\color{#c00}{22}(15)}\equiv \color{#0a0}{5}^{\large \color{#0a0}{2}\cdot 3}\,\color{#c00}{(5^{\large 22})}^{\large 15}\equiv \color{#0a0}2^{\large ... |
2,559,560 | <blockquote>
<p>Show that there are two distinct positive integers such that: $1394|2^a-2^b$</p>
</blockquote>
<p>I'm sure pigeon hole principle applies here,but don't recognize holes.Another problem statement is: show that there are two positive integers $a,b$ such that: $$2^a\equiv 2^b\pmod {1394}$$<br>
Of course... | Steven Alexis Gregory | 75,410 | <p>Note $1394=2 \times 17 \times 41$. </p>
<p>Since $\varphi(17 \times 41) = 16\times 40 = 640$, </p>
<p>then $2^{640} \equiv 1 \pmod{697}$.</p>
<p>Hence $2^{641} \equiv 2 \pmod{1394}$.</p>
<p>In other words, $1394 \mid 2^{641}-2^1$.</p>
<p>Note. Actually, we can do better. Checking the divisors of $640$, we find ... |
1,250,132 | <p>Below is part of a solution to a critical points question. I'm just not sure how the equation on the left becomes the equation on the right. Could someone please show me the steps in-between? Thanks.</p>
<blockquote>
<p>$$\frac{-1}{x^2}+2x=0 \implies 2x^3-1=0$$</p>
</blockquote>
| Community | -1 | <p>Just multiply both sides by $x^2$ to get $-1+2x^3=0$, then you get $2x^3-1=0$ by commutivity. </p>
|
1,014,476 | <p>I pick 6 cards from a set of 13 (ace-king). If ace = 1 and jack,queen,king = 10 what is the probability of the sum of the cards being a multiple of 6? </p>
<p><strong>Tried so far:</strong>
I split the numbers into sets with values:
6n, 6n+1, 6n+2, 6n+3
like so:</p>
<p>{6}{1,7}{2,8}{3,9}{4,10,j,q,k}{5}</p>
<p>and... | user145600 | 145,600 | <p>Here is a partial answer to your question. You want the sum of two cards to be 6, 12 or 18.</p>
<p>For the sum of 6, the possibilities are 1+5, 2+4, and 3+3</p>
<p>The probability of an ace and a 5 is 1/13 times 1/12.</p>
<p>Same with 2+4, to wit, 1/(13*12) = 1/156.</p>
<p>For two threes, we have 1/13 times 3/5... |
1,014,476 | <p>I pick 6 cards from a set of 13 (ace-king). If ace = 1 and jack,queen,king = 10 what is the probability of the sum of the cards being a multiple of 6? </p>
<p><strong>Tried so far:</strong>
I split the numbers into sets with values:
6n, 6n+1, 6n+2, 6n+3
like so:</p>
<p>{6}{1,7}{2,8}{3,9}{4,10,j,q,k}{5}</p>
<p>and... | Steve Kass | 60,500 | <p>You can “cast out” any $6$s from the values of the cards, since it will not affect whether the sum is a multiple of $6$. So you have 13 cards valued 1,2,3,4,5,0,1,2,3,4,4,4, and 4. How many ways can you make a multiple of 6 from the sum of 6 of these numbers? Consider the number of 4s used. If none, you can make a s... |
4,050,307 | <p>You have a black box function to which you can give real number inputs and from which you can receive real number outputs. <strong>How would you test whether it is likely to be a polynomial?</strong></p>
<p>One expensive idea is to use finite differences:</p>
<ol>
<li>Choose a maximum degree <em>n</em> of the "... | Petrus1904 | 808,320 | <p>I think the largest problem here is that given things like Taylor series, every function can eventually be closely approximated as a polynomial. As such, even a finite range of measured outputs of a blackbox representing a sine function will might yield a polynomial. Measuring a large range is computationally diffic... |
2,359,455 | <blockquote>
<p>Let <span class="math-container">$M$</span> be the largest subset of <span class="math-container">$\{1,\dots,n\}$</span> such that for each <span class="math-container">$x\in M$</span>, <span class="math-container">$x$</span> divides at most one other element in <span class="math-container">$M$</span>. ... | Asinomás | 33,907 | <p>Divide the $M$ integers into $\lceil\frac{n}{2}\rceil$ groups, depending on the maximum odd divisor. Notice we can take at most two from each group and some groups have size $1$, we do the analysis mod $4$.</p>
<p>$n=4k:$</p>
<p>number of groups: $2k$, number of groups of size $1: k$, upper bound: $3k=\lceil\frac{... |
2,061,363 | <p>I have the complex power series $ \sum_{k=1}^{\infty}(\frac{z^4}{4} - \frac{\pi}{7})^k$. </p>
<p>Through algebraic manipulation I obtain $ \sum_{k=1}^{\infty}(\frac{1}{4})^k(z^4 - \frac{4}{7}\pi)^k$. I now argue that this is a power series around $\frac{4}{7}\pi$ with radius of convergence R = 4, using the euler ro... | epi163sqrt | 132,007 | <blockquote>
<p><em>Hint:</em> This is a geometric series
\begin{align*}
\sum_{k=1}^\infty\left(\frac{z^4}{4}-\frac{\pi}{7}\right)^k
=\frac{1}{1-\left(\frac{z^4}{\pi}-\frac{\pi}{7}\right)}-1
\end{align*}</p>
<p>converging for $\left|\frac{z^4}{4}-\frac{\pi}{7}\right|<1$</p>
</blockquote>
|
1,809,022 | <p>I've been working on some very basic differential equations, but I came to a
problem where I need to figure out the behavior of $y(t)$ as $t \rightarrow
\infty$ Given that
$$\frac{dy}{dt} = \frac{3t}{1+2e^{y}}.$$
In this case, it was very apparent to me that I would not be able to solve for
a simple solution of $y(t... | Eric Towers | 123,905 | <p>As $t \rightarrow \infty$, the right-hand side of your (implicit) solution goes to infinity, so the left-hand side must also.</p>
<p>Note that $\dfrac{\mathrm{d}}{\mathrm{d}u} u + 2 \mathrm{e}^u = 1 + 2 \mathrm{e}^u$, which is positive for all values of $u$. Consequently, if $y(t)$ increases, the left-hand side in... |
3,547,384 | <p>I saw this equation<span class="math-container">$$S(q)=\int_a^bL(t,q(t),\dot q(t))dt$$</span>
in <a href="https://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation" rel="nofollow noreferrer">wikipedia</a>.</p>
<p>So I would think that <span class="math-container">$f(x,y)$</span> must be equal to <span class="ma... | Empy2 | 81,790 | <p>Suppose <span class="math-container">$f(x,y)$</span> is the height of a hillside above sea level, and <span class="math-container">$(x(t),y(t))$</span> the latitude and longitude of your car. Then <span class="math-container">$\frac{df}{dt}$</span> is the rate you are going up and down because the road slopes up or ... |
3,102,336 | <p>I have been looking for fixed points of <a href="https://simple.wikipedia.org/wiki/Riemann_zeta_function" rel="nofollow noreferrer">Riemann Zeta function</a> and find something very interesting, it has two fixed points in <span class="math-container">$\mathbb{C}\setminus\{1\}$</span>.</p>
<p>The first fixed point i... | Conrad | 298,272 | <p>I do not think your statement about fixed points in the plane to be true - it may be true for <span class="math-container">$Re(z)>1$</span> in the sense of being just one fixed point there, but otherwise <span class="math-container">$(s-1)\zeta(s)$</span> is an entire function of order 1 and maximal type (by the ... |
2,538,521 | <blockquote>
<p>Let $x$ be a function of $C^1(I,R)$ where $I\subset \mathbb{R}$ , such that $$x'(t)\leq a(t) x(t)+b(t),$$ where $a$ and $b$ are continuous functions on $I$ in $R$ then
$$ x(t)\leq x(t_0) \exp\left(\int_{t_0}^{t}a(s)ds\right)+\int_{t_0}^{t}\exp\left(\int_{s}^t a(\sigma)d\sigma\right)b(s)ds$$</p>
... | Robert Lewis | 67,071 | <p>I, too, first guessed this would be a problem suggesting the application of <a href="https://en.wikipedia.org/wiki/Gr%C3%B6nwall%27s_inequality" rel="nofollow noreferrer">Grownall's inequality</a>, but it seems an even more elementary solution avails itself:</p>
<p>Given that</p>
<p><span class="math-container">$x'(... |
1,662,090 | <p>First of all, hi. I am new here.</p>
<p>Let$$X_1,\dots, X_n$$
be i.i.d. exponential random variables.</p>
<p>$$
Pr({\max X_n}>{(\sum X_n-\max X_n ) }) = ?
$$</p>
<p>I think we should take integrals on exponential distribution functions over corresponding intervals but I could not make it work.</p>
<p>Thanks.<... | Intelligenti pauca | 255,730 | <p>HINT.</p>
<p>You must solve the equation $(4x^2+2)(3x^2+2)=50052$, where $50052=(75583)_9=2^2\cdot3\cdot43\cdot97$.</p>
|
24,876 | <p>As it is possible to see the last time when you or others visited M.SE, I wonder if one can see a statistics of visits of your own or of a specific user for a period of time (last year, let's say).</p>
| user642796 | 8,348 | <p>This is more of a supplement to <a href="https://math.meta.stackexchange.com/a/24877">quid's answer</a>.</p>
<p>Regular users can only see the visits information for their own account. Where in <a href="https://math.stackexchange.com/users/current?tab=profile">your own profile</a> you might see</p>
<p><a href="htt... |
4,544,787 | <p>These sums showed up in a probability problem I was working on. They're not quite the Stirling numbers of the first kind since it's possible to have e.g. <span class="math-container">$i_1 = i_2$</span>. Denoting the sum by <span class="math-container">$(k\mid n)$</span> we have the recurrence relation</p>
<p><span c... | orangeskid | 168,051 | <p>Consider the expansion</p>
<p><span class="math-container">$$X^N = \sum_{k=0}^N {N\brace k} (x)_k$$</span></p>
<p>where <span class="math-container">$(x)_k = x(x-1)\cdots (x-k+1)$</span>, and <span class="math-container">${N\brace k}$</span> is the <a href="https://en.wikipedia.org/wiki/Stirling_numbers_of_the_secon... |
1,135,045 | <p>I need to compute
\begin{align}
S = \sum_{k=-\infty}^j \sum_{m=-1}^2 w_{k,m} f_{k+m-1}
\end{align}
but I only want to access the elements of $f$ once, so I would prefer something like
\begin{align}
\sum_k f_k \sum_m ...
\end{align}
Here is what I did: substitute $l=m-1+k$ to get
\begin{align}
S &= \sum_{k=-\inf... | AlexR | 86,940 | <p>The problem is you are introducing a dependency on $k$. To get around this we define
$$s_k := \cases{1&$k\ge0$\\0&$k<0$}$$
Then we have
$$\begin{align*}
\sum_{k=-\infty}^j \sum_{m=-1}^2 w_{k,m} f_{k+m-1} & = \sum_{k=-\infty}^j w_{k,-1} f_{k-2} + w_{k,0} f_{k-1}+w_{k,1} f_k + w_{k,2} f_{k+1} \\
&= ... |
142,035 | <p>I have sets of functions of material parameters where the first input argument denotes the material. Here's a simple version:</p>
<pre><code>tf1["a", x_] = x^3;
tf1["b", x_] = x^4;
</code></pre>
<p>Now, I'd like to make a function which needs the derivative of <code>tf1</code> with respect to x. My goal is the fol... | David G. Stork | 9,735 | <pre><code>tf1[myname_String, x_] := Which[myname == "a", x^3, myname == "b", x^4];
D[tf1["a", x], x]
</code></pre>
<p>(* 3 x^2 *)</p>
<pre><code>D[tf1["b", x], x]
</code></pre>
<p>(* 4 x^3 *)</p>
<p>If you have a "default" function for all string variables not yet entered, then:</p>
<pre><code>tf1[myname_String,... |
142,035 | <p>I have sets of functions of material parameters where the first input argument denotes the material. Here's a simple version:</p>
<pre><code>tf1["a", x_] = x^3;
tf1["b", x_] = x^4;
</code></pre>
<p>Now, I'd like to make a function which needs the derivative of <code>tf1</code> with respect to x. My goal is the fol... | Pillsy | 531 | <p>I suggest doing the derivative and then substituting in a value. This can be done as follows: </p>
<pre><code>der[i_, x_] :=
D[tf1[i, \[FormalX]], \[FormalX]] /. \[FormalX] -> x;
</code></pre>
<p>Using a "formal" variable means you don't have to worry about it being defined in scope and screwing things up.</p... |
4,517,429 | <p>Let <span class="math-container">$ {\textstyle \{X_{1},\ldots ,X_{n},\ldots \}}$</span> be a sequence of independent random variables, each of those random variable follow a Gamma distribution.</p>
<p>For the summation of those random variable:</p>
<p><span class="math-container">$ {\displaystyle {\bar {X}}_{n}\equi... | heropup | 118,193 | <p>The answer depends on whether the gamma distributions have the same rate parameter.</p>
<p>if <span class="math-container">$X_i \sim \operatorname{Gamma}(a_i, b)$</span> then their sample total will also be gamma distributed: <span class="math-container">$$\sum X_i \sim \operatorname{Gamma}\left(\sum a_i, b \right)... |
3,566,603 | <p>I need a simple way to show <span class="math-container">$\mathbb R^2$</span> is not isomorphic <span class="math-container">$\mathbb{R}[x]/(x^2)$</span>. Both are not integral domains, and both are not fields, so I’m not sure how to go about it.</p>
| Geoffrey Trang | 684,071 | <p>The ring <span class="math-container">$\mathbb{R} \times \mathbb{R}$</span> has 4 ideals, namely <span class="math-container">$0$</span>, <span class="math-container">$\mathbb{R} \times 0$</span>, <span class="math-container">$0 \times \mathbb{R}$</span>, and <span class="math-container">$\mathbb{R} \times \mathbb{R... |
3,695,868 | <p>In right triangle <span class="math-container">$ABC,$</span> <span class="math-container">$\angle C = 90^\circ.$</span> Let <span class="math-container">$P$</span> and <span class="math-container">$Q$</span> be points on <span class="math-container">$\overline{AC}$</span> so that <span class="math-container">$AP = P... | Vishu | 751,311 | <p><strong>Hint:</strong></p>
<p>Suppose <span class="math-container">$AP=PQ=QC=x$</span>. Then, we know <span class="math-container">$$BC^2 = 67^2-x^2 =76^2 -(2x)^2 $$</span> You can solve for <span class="math-container">$x$</span> from here and get <span class="math-container">$AB=3x$</span>.</p>
|
277,217 | <p>I am stuck on the following problem, which I do not believe to be so difficult.</p>
<p>Let $X$ and $Y$ be Banach spaces. Let $f:X\times X\rightarrow Y$ be a function such that for any fixed $x_0$, $f(x,x_0)$ and $f(x_0,x)$ are continuous in $x$. Then is $f(x,x)$ continuous in $x$?</p>
<p>I tried taking an arbitrar... | Ittay Weiss | 30,953 | <p>Consider the function $f:\mathbb R\times \mathbb R\to \mathbb R$ given by $f(x,0)=f(0,x)=0$ for all $x\in \mathbb R$, and $f(x,y)=1$ if either $x$ or $y$ is irrational, and $f(x,y)=1/2$ otherwise. </p>
<p>Clearly, $f(x,0)$ and $f(0,x)$, being constantly $0$, are continuous. However, $f(x,x)$ is a Dirichlet function... |
1,993,217 | <p>Let $\left\{f_{n}\right\}$ be a sequence of equicontinuous functions where $f_n: [0,1] \rightarrow \mathbf{R}$. If $\{f_n(0)\}$ is bounded, why is $\left\{f_{n}\right\}$ uniformly bounded?</p>
| zhw. | 228,045 | <p>Let's do it for an equicontinuous family $\mathcal F$ of functions on $[0,1]$ such that</p>
<p>$$\sup_{f\in \mathcal F}|f(0)| =C < \infty.$$</p>
<p>Choose $m \in \mathbb N$ such that $|y-x|\le 1/m$ implies $|f(y)-f(x)| \le 1$ for all $f\in \mathcal F.$ Then for any $f\in \mathcal F,$</p>
<p>$$f(k/m) = [f(k/m) ... |
74,347 | <blockquote>
<p>Construct a function which is continuous in $[1,5]$ but not differentiable at $2, 3, 4$.</p>
</blockquote>
<p>This question is just after the definition of differentiation and the theorem that if $f$ is finitely derivable at $c$, then $f$ is also continuous at $c$. Please help, my textbook does not h... | Did | 6,179 | <p>$$\ \ \ \ \mathsf{W}\ \ \ \ $$</p>
|
521,740 | <p>$x$,$y$ are real numbers satisfying $(x-1)^{2}+4y^{2}=4$<br>
find the maximum of $xy$ and justify it without calculus.<br>
Does there exist a tricky solution using elementary inequalities (AM-GM or Cauchy-Schwarz) ?</p>
<p>I tried and got it's when $x=\dfrac{3+\sqrt{33}}{4}$</p>
| user2566092 | 87,313 | <p>If you set $y = k/x$, then you get a quartic equation in $x$ and you want to know the maximal $k$ such that there is a real solution for $x$. If you trace through all the complicated equations that define the solutions for a quartic equation, you should be able to figure it out. However I doubt this is the most effi... |
439,620 | <p>As we know, the QR-factorization <span class="math-container">$Q\cdot R=A$</span> of any real symmetric <span class="math-container">$n \times n$</span> matrix <span class="math-container">$A$</span> with full rank is <em><strong>unconditionally</strong></em> <em>numerically stable</em>. Further, when A is rank-1-up... | Federico Poloni | 1,898 | <p>[EDIT: not working in fact because the update of <span class="math-container">$Q$</span> cannot be merged efficiently, see comments] A simple eigendecomposition <span class="math-container">$A=QDQ^*$</span> should work, since <a href="https://arxiv.org/abs/1405.7537" rel="nofollow noreferrer">it can be updated</a> i... |
1,131,622 | <p>The question itself is a very easy one:<br/></p>
<blockquote>
<p>Somebody has got two kids, one of whom is a girl. Then what's the probability that he's got <strong>at least</strong> one boy?</p>
</blockquote>
<p>My answer is that, since he's already got a girl, then "he's got at least one boy" amounts to "the o... | Community | -1 | <p>A lot of great explanations about the correct answer. Your answer is wrong because it ignores that fact that people are distinguishable, but the information you got <em>doesn't</em> tell you <em>which</em> child is the boy. You are ignoring this and treating the problem as if you knew that, for example, Child 1 is a... |
394,517 | <p>How can I evaluate $\lim_{x \to \infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}\right)$?</p>
| tom | 59,101 | <p>Almost always when you have a limit of $\sqrt{A}-\sqrt{B}$ type. It is good idea to multiply it by $\frac{\sqrt{A}+\sqrt{B}}{\sqrt{A}+\sqrt{B}}$. So you have to do limit of $\frac{A-B}{\sqrt{A}+\sqrt{B}}$</p>
<p>Thus for your limit you get:
$$\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}} = \frac{ (x+\sqrt{x} ) - (x-\sqrt{x})... |
29,115 | <p>I just read a proof and, after struggling some time with a mental leap, I think that it uses tacitly the following:</p>
<p>Let $\kappa$ be a regular cardinal, $\theta > \kappa$ a regular cardinal too then:
$ S \subset \kappa$ is stationary if and only if
$\forall \mathcal{A} = (H(\theta), \in, <,..) \exists... | François G. Dorais | 2,000 | <p>Yes, the statement is true. </p>
<p>The forward direction is clear since the set
$$C_{\mathcal{A}} = \{\sup(M\cap\kappa) : M \prec \mathcal{A}, |M| < \kappa\}$$
is a club. Indeed, let $\langle M_\alpha : \alpha < \kappa \rangle$ be an elementary chain of elementary submbodels of $\mathcal{A}$ with size less t... |
3,883,164 | <p>I evaluated following limit with taylor series but for a practice I am trying to evaluate it using L'Hopital's Rule:</p>
<p><span class="math-container">$$\lim_{x\to 0}\frac{\sinh x-x\cosh x+\frac{x^3}3}{x^2\tan^3x}=\lim_{x\to0}\cfrac{f(x)}{g(x)}$$</span>
<span class="math-container">$f(x)=\sinh x-x\cosh x+\frac{x^3... | user | 505,767 | <p>To simplify the evaluation we can use that</p>
<p><span class="math-container">$$\frac{\sinh x-x\cosh x+\frac{x^3}3}{x^2\tan^3x}=\frac{x^3}{\tan^3x}\frac{\sinh x-x\cosh x+\frac{x^3}3}{x^5}$$</span></p>
<p>and since <span class="math-container">$\frac{x^3}{\tan^3x} \to 1$</span> we reduce to evaluate by l'Hospital</p... |
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