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 |
|---|---|---|---|---|
537,968 | <p>Let $|\alpha|<1$ and $\psi_{\alpha}(z)=(\alpha-z)/(1-\bar\alpha z)$. I want to prove that $$\frac 1 {\pi} \int\int_{\mathbb{D}}|{\psi_{\alpha}}^{'}|dxdy = \frac{1-|\alpha|^2}{|\alpha|^2}\log\frac{1}{1-|\alpha|^2}$$</p>
<p>I calculated ${\psi_\alpha}^{'}(z)=(|\alpha|^2-1)/(1-\bar\alpha z)^2$. I substituted it and... | Ron Gordon | 53,268 | <p>Convert to polars, as mentioned by Daniel Fischer. The integral over the disk becomes</p>
<p>$$\frac{1-|\alpha|^2}{\pi} \int_0^1 dr \, r \, \int_0^{2 \pi} \frac{d\theta}{|1-\bar{\alpha} r e^{i \theta}|^2} $$</p>
<p>Now,</p>
<p>$$|1-\bar{\alpha} r e^{i \theta}|^2 = (1-\bar{\alpha} r e^{i \theta})(1-\alpha r e^{-i... |
2,329,600 | <p>I haven't studied any maths since I was at university 20 years ago. Yesterday, however, I came across a pair of equations in an online article about gaming and I couldn't understand how they'd been derived. </p>
<p>Here's the scenario. If we make a single trial of generating a number between 1 and 20, there's an ev... | Graham Kemp | 135,106 | <p>Let $X_1, X_2$ be the random variables, with the independent uniform discrete distributions on the support $\{1..20\}$ . Let $k$ be a target number within that support.</p>
<p>$$\begin{align}\mathsf P(\min\{X_1,X_2\}\geq k) &= \mathsf P(X_1\geq k, X_2\geq k) \tag 1 \\[1ex] &= \mathsf P(X_1\geq k)\mat... |
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... | Christian Blatter | 1,303 | <p>At the end the tea cup is as full as at the start. This implies that the added wine is exactly outweighed by the tea that has disappeared.</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... | Nuclear Hoagie | 240,172 | <p><strong>Argument by symmetry</strong></p>
<p>One way to approach the problem is to recognize the importance of the fact that you're expected to find a solution under the assumption that the tea-wine mixture in the teacup is <em>nonuniform</em>. In other words, it's impossible to know whether you're transferring a sp... |
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... | marshal craft | 167,793 | <p>The way I see it intuitively as a venn diagram. Two spheres represent the arbitrary amount moved around, I. This case a tea spoon amount. So when they overlap, you ask which area is greatest of the two spheres which aren't overlapping. But you see any area taken from one must be taken from the other and the area is ... |
7,223 | <p>I want to produce a <em>Mathematica</em> Computable Document in which <code>N</code> appears as a variable in my formulae. But <code>N</code> is a reserved word in the <em>Mathematica</em> language. Is there a way round this other than using a different symbol? It seems a severe limitation if you cannot use <em>Math... | DavidC | 173 | <p>You could use capital Nu, <code>\[CapitalNu]</code>, from the Greek alphabet. It is visually almost identical to capital N from the Roman alphabet. But it has no predetermined assignment.</p>
<pre><code>\[CapitalNu] = 5
2 \[CapitalNu]
</code></pre>
<p>The following shows how the input is displayed on screen.</p>
... |
3,446,663 | <blockquote>
<p>Let <span class="math-container">$\sigma\in S_{14}$</span> which is an even permutation of the order of <span class="math-container">$28$</span>.<br> Prove that exist <span class="math-container">$x\in \left\{ 1,...,14 \right\}$</span> such that <span class="math-container">$\sigma(x)=x$</span>.</p>
<... | Eric Towers | 123,905 | <p>It is not required that three elements are fixed. Consider
<span class="math-container">$$ (1\ 2\ 3\ 4\ 5\ 6\ 7)(8\ 9\ 10\ 11)(12\ 13)(14) \text{.} $$</span></p>
<p>If the order of the cycle is <span class="math-container">$28$</span>, there is at least a <span class="math-container">$7$</span>-cycle and at lea... |
3,446,663 | <blockquote>
<p>Let <span class="math-container">$\sigma\in S_{14}$</span> which is an even permutation of the order of <span class="math-container">$28$</span>.<br> Prove that exist <span class="math-container">$x\in \left\{ 1,...,14 \right\}$</span> such that <span class="math-container">$\sigma(x)=x$</span>.</p>
<... | Shaun | 104,041 | <p>You already have the prime factorisation of <span class="math-container">$28$</span>. To get an element of order <span class="math-container">$28$</span>, you need to partition <span class="math-container">$14$</span> into divisors of <span class="math-container">$28$</span> (namely, <span class="math-container">$1$... |
2,454,895 | <p>I don't know how to solve this equation:$$(1)\quad e^ {-x} = -\ln x$$</p>
<p>$x$ should be the abscissa of the point $P$ where the two functions meet on the plan and $$ P \in f(x) :y=x$$</p>
<p>so $(1)$ should be equal to $$ e^{-x}=x=-\ln x$$ </p>
<p>How do I solve this?</p>
| Donald Splutterwit | 404,247 | <p>Your equation can be rearrange to $x=e^{-e^{-x}}$. Now define $f(x)=e^{-x}$ so we are looking for solutions to $x=f(f(x))$. If we have a solution to $x=f(x)$ then this will also be a solution to $x=f(f(x))$ so to obtain a solution it suffice to solve $x=e^{-x}$. This can easily be achieved by the Lambert $W$ functio... |
631,053 | <p>I have a container of 100 yellow items.</p>
<p>I choose 2 at random and paint each of them blue.</p>
<p>I return the items to the container.</p>
<p>If I repeat this process, on average how many cycles will I make before all 100 items are painted?</p>
<p>It is obviously 50 (100/2) if there is no replacement. But ... | joriki | 6,622 | <p>By inclusion-exclusion, the probability that all items have been painted after $m$ batches of $b$ items each have been painted is</p>
<p>$$
\sum_{k=0}^{100}(-1)^k\binom{100}k\left(\frac{\binom kb}{\binom{100}b}\right)^m\;.
$$</p>
<p>Thus the expected number of cycles required is</p>
<p>\begin{align}
\sum_{m=0}^\i... |
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 g... | Timbuc | 118,527 | <p>$$\frac1{z(1-2z)}=\frac1z+\frac2{1-2z}=\frac1{\frac12+\left(z-\frac12\right)}-\frac1{z-\frac12}=\frac2{1+2\left(z-\frac12\right)}-\frac1{z-\frac12}=$$</p>
<p>$$2\left(1-2\left(z-\frac12\right)+4\left(z-\frac12\right)^2-\ldots\right)-\frac1{z-\frac12}=$$</p>
<p>$$=-\frac1{z-\frac12}+2\sum_{n=0}^\infty (-1)^n2^n\lef... |
3,077,312 | <p>The proof given in my book (and I came up with as well) is:</p>
<p><a href="https://i.stack.imgur.com/H6eqf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/H6eqf.png" alt="Proof"></a></p>
<p>However, the part that throws me off is line #3 where they do <span class="math-container">$\Sigma A_{jk}... | José Carlos Santos | 446,262 | <p>Wrong. The <span class="math-container">$ij$</span> entry of the matrix <span class="math-container">$A^tB^t$</span> is <span class="math-container">$\sum_kA_{kj}B_{ik}$</span>, not <span class="math-container">$\sum_kA_{jk}B_{ki}$</span>.</p>
|
95,965 | <p>Joyal's <a href="http://en.wikipedia.org/wiki/Combinatorial_species" rel="nofollow">combinatorial species</a>, endofunctors in the category of finite sets with bijections $\mathbf B$ have found numerous applications. One generalisation is given by so-called "tensor species" (also "tensorial species", or, "linear sp... | Bruce Westbury | 3,992 | <p>The theory of tensor species is equivalent to the theory of polynomial functors; so to this extent there is no call for a theory of tensor species as the theory of polynomial functors is well-developed. However this is, I suspect, missing the point of your question. My understanding is that the focus in combinatoria... |
167,013 | <p>I don't understand this behavior: why does <code>Limit[z/(z - a), z -> 0]</code> give zero and not a condition depending on <code>a</code>, provided it has not been defined before? is there a way to make it work properly? (By working properly I mean give the correct result, namely 1 if $a=0$ and 0 otherwise.) </p... | José Antonio Díaz Navas | 1,309 | <p>You can use <code>GeneratingConditions</code> option so MMA can look for conditionals:</p>
<pre><code>Limit[z/(z - a), z -> 0, GenerateConditions -> True]
(* ConditionalExpression[0, a != 0] *)
</code></pre>
|
1,908,820 | <p>The question is: </p>
<p>If $$\int_3^9f(x) dx = 7$$
evaluate
$$\int_3^9 2f(x)+1 dx$$</p>
<p>I know that you can factor the 2 outside of the integral. But, then I am still left with a '+ 1' inside the integral that when I take the integral of becomes $x$. So then would I proceed to stating this:
$$=2\int_3^9 f(x... | NotAName | 365,006 | <p>Your textbook is right, if you apply the 2 to the 1 in the last step you should also have applied it to the 1 at factoring it out, thus having 1/2 in the second integral. This should also lead to your textbook's answer.</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>$$\left(\frac{x}{3}\right)^{2}+\left(\frac{y}{2}\right)^{... | Community | -1 | <p>Use $\in$. And be careful with your commas: $\{-2, -1 \color{red}, \dots \color{red}, 3\}$</p>
<p>In general:</p>
<p>$x \in S$ means "$x$ is an element of $S$" or "$x$ is in $S$." (The $\LaTeX$ code to produce "$\in$" is <code>$\in$</code>.)</p>
<p>$x = s$ means "$x$ is $S$" or "$x$ is equal to $S$." In this ... |
3,608,114 | <p>Consider the example where I have a matrix <span class="math-container">$\mathbf{D}$</span> in <span class="math-container">$-1/1$</span> coding with <span class="math-container">$5$</span> columns,</p>
<p><span class="math-container">$$D = \begin{bmatrix}-1&-1&-1&1&1\\1&-1&-1&-1&1\\... | mathreadler | 213,607 | <p>There is a mutuality in linear dependence.</p>
<p>If a vector <span class="math-container">$\bf a$</span> is linearly dependent of <span class="math-container">$\bf b$</span> and <span class="math-container">$\bf c$</span>, then <span class="math-container">$\bf b$</span> is linearly dependent on <span class="math-... |
575,513 | <p>Can someone help me find the density function $f_X$ for $X$ and hence find $E(X)$ and $Var(X)$ of the following distribution function $F_X$ given by:</p>
<p>$F_X(x)=\begin{cases}
1-(1+x)e^{-x} & x>0 \\
0 & otherwise.
\end{cases}$</p>
<p>$X$ is a continuous random variable.</p>
<p>From memory, do I h... | Sam Wong | 507,382 | <p>Let me propose a more elementary example.</p>
<p>Let <span class="math-container">$R$</span> be <span class="math-container">$(\mathbb Z_6,+,\times)$</span>. Let <span class="math-container">$A=R$</span>.</p>
<p>Then <span class="math-container">$A$</span> is a <span class="math-container">$R$</span>-module.</p>
<p>... |
1,812,468 | <p>Let $x=\{a,b\}$ be a set. Then, $x\in\{a,b\}$?</p>
<p>I think: Yes. So, why?</p>
| saz | 36,150 | <p>The Brownian Motion $(W_t)_{t \geq 0}$ has (almost surely) continuous sample paths. Consequently, we have by the extrem value theorem </p>
<p>$$M(T,\omega) := \sup_{t \leq T} |W_t(\omega)|<\infty$$</p>
<p>for all $T \geq 0$ and (almost) all $\omega \in \Omega$. This implies $$X_s^2(\omega) = e^{2a W_s(\omega)^... |
3,335,892 | <p>If I have two injective functions <span class="math-container">$f : A \to B$</span> and <span class="math-container">$g : B \to A$</span>, as Schröder-Bernstein (SB) says, then there is a function <span class="math-container">$h : A \to B$</span> which is bijective.</p>
<p>As for a proof, my reasoning goes somethin... | Acccumulation | 476,070 | <p>So, your argument is that when you surround <span class="math-container">$A$</span> and <span class="math-container">$B$</span> with vertical lines, and then put a horizontal line with two diagonal lines that meet on the right above it between them, this results in a "true" statement, and thus it follows that <span ... |
2,864,585 | <p>I tried to calculate the Hessian matrix of linear least squares problem (L-2 norm), in particular:</p>
<p>$$f(x) = \|AX - B \|_2$$
where $f:{\rm I\!R}^{11\times 2}\rightarrow {\rm I\!R}$</p>
<p>Can someone help me?<br>
Thanks a lot.</p>
| lynn | 234,414 | <p>Define a new matrix $P=(AX-B)$ and write the function as
$$f=\|P\|_F^2 = P:P$$
where the colon denotes the trace/Frobenius product, i.e. $\,\,A:B={\rm tr}(A^TB)$</p>
<p>Find the differential and gradient of $f$
$$\eqalign{
df &= 2P:dP = 2P:A\,dX = 2A^TP:dX \cr
G &= \frac{\partial f}{\partial X} = 2A^TP \c... |
1,657,115 | <p>$$\int \frac{2x + 10}{(x^2 + 5x + 8) ^ 2}dx$$
we can rewrite as:
$\int \frac{2x + 5}{(x^2 + 5x + 8) ^ 2}dx$ + $\int \frac{5}{(x^2 + 5x + 8) ^ 2}dx$</p>
<p>first one is easy. What can we do with the second one?</p>
| Michael | 289,620 | <p>Coefficient of x in the equation of line PD is $-1$. Negative reciprocal of that is $1$. So, your teacher is sadly wrong :/ since this tells us that $4x-7=1$</p>
|
858,716 | <p>I'm self-studying real analysis using Abbott's text "Understanding Analysis." I'm trying to think out/prove as much on my own as I can, so I am working on proving the Nested Interval Property (Theorem 1.4.1 in the book) using "just" the Axiom of Completeness. The author does prove it in the book, but as I say, I lik... | Surb | 154,545 | <p>$$\left.\begin{array}{rcl}\frac{\partial}{\partial x}\left( \frac{x}{x-y}\right) = \frac{(x-y)-x}{(x-y)^2} &=&\frac{-y}{(x-y)^2}\\
\frac{\partial}{\partial y} \left(\frac{x}{x-y}\right) &=& \frac{x}{(x-y)^2}\end{array}\right\} \Longrightarrow \nabla \left(\frac{x}{x-y}\right) = \frac{1}{(x-y)^2}\begi... |
814,020 | <p><strong>Preamble</strong></p>
<p>The <a href="http://mathworld.wolfram.com/CassiniOvals.html" rel="nofollow">Cassinian curves</a> are the pre-images of concentric circles (centered at $1+0\,i$) under the map $z\mapsto z^2$. Using this fact and the fact that complex polynomials are conformal we can deduce that the o... | Dhruv Kohli | 97,188 | <p>Too late but I thought may be this naive approach is worth writing here.</p>
<p>Preimage of the concentric circles under the mapping $z \mapsto z^2$ form the Cassinian curves with focii $\pm 1 + 0i$ (preimages of the centre of the circles). The equation of the Cassinian curves for some positive constant $k$ will th... |
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>
| Taladris | 70,123 | <p>A cylinder of radius <span class="math-container">$R$</span> and height <span class="math-container">$h$</span> can be obtained by rotating the graph of <span class="math-container">$f(x)=R$</span>, <span class="math-container">$0\le x\le h$</span> about the <span class="math-container">$x$</span>-axis so it is a so... |
186,240 | <p>I need some notion about topology(I'm very interested in boundary points, open sets) and few examples of solved exercises about limits of functions($f:\mathbb{R}^{n}\rightarrow \mathbb{R}^m$) using $\epsilon, \delta$ and also some theory for continous functions.
Please give me some links or name of the books which ... | andreas.vitikan | 22,704 | <p>Well, most analysis books that I've seen (anything above high-school level maths) includes in the first and second chapter a discussion about $\mathbb{R}$ and the axioms, as well some notions about sets and point-set-topology, and then goes on to functions, limits, the $\epsilon - \delta$ criterion, and then in vari... |
186,240 | <p>I need some notion about topology(I'm very interested in boundary points, open sets) and few examples of solved exercises about limits of functions($f:\mathbb{R}^{n}\rightarrow \mathbb{R}^m$) using $\epsilon, \delta$ and also some theory for continous functions.
Please give me some links or name of the books which ... | Karatuğ Ozan Bircan | 12,686 | <p><a href="http://rads.stackoverflow.com/amzn/click/3540908927" rel="nofollow"><em>Topology</em> by Klaus Janich</a> is a good one as a general Topology textbook. </p>
|
2,262,011 | <p>This might be a somewhat stupid question, but I've been wondering if it is possible to define some other topology on $\mathrm{Spec} (A)$ other than Zariski topology in a way that it has some interesting properties as well.</p>
<p>First of all, I am new as this is my first encounter with anything close or related to... | Georges Elencwajg | 3,217 | <p>As far as I'm aware there is no other interesting topology on $\mathrm{Spec} (A)$. However: </p>
<p>1) There are the so called "Grothendieck topologies" on any $\mathrm{Spec} (A)$ or more generally on an arbitrary scheme.<br>
They are generalizations of the usual notion of a topology and compensate for the coarse... |
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) $$</p>
<p>For the base case: $n = 2$, $d = 2$<br>
$2\mid 2(2+1)$ which is true.<br></p>
<p>Now, the confusion begins! I assume I would need to use the second induction p... | ashley | 50,188 | <p>This is one that iterates on the relative values between n & d for the inductive steps--
not on values n, n+1, n+k, n+k+1 as the usual case. </p>
<p>The proof is by looking at the values of n and d.
Say n=dt+x whenever n>d. </p>
<p>For the base case, check whether x=0 satisfies-- which it does. </p>
<p>For ... |
253,359 | <p>I'm trying to prove by induction the following statement without success:<br>
$$\forall n \ge 2, \;\forall d \ge 2 : d \mid n(n+1)(n+2)...(n+d-1) $$</p>
<p>For the base case: $n = 2$, $d = 2$<br>
$2\mid 2(2+1)$ which is true.<br></p>
<p>Now, the confusion begins! I assume I would need to use the second induction p... | Amr | 29,267 | <p>Consider the numbers $n \mod d,(n+1)\mod d,...,(n+d-1)\mod d$ these numbers are all equal and form a subset of {$0,1,...,d-1$}. Since both sets are equal on size, therefore both sets are equal. Therefore one of the numbers $n,n+1,...,n+d$ is divisible by $d$. Thus, $d|n(n+1)...(n+d-1)$</p>
|
2,161,294 | <p>I was wondering... $1$, $\phi$ and $\frac{1}{\phi}$, they have something in common: they share the same decimal part with their inverse. And here it comes the question:</p>
<p>Are these numbers unique? How many other members are in the set if they exist? If there are more than three elements: is it finite or infin... | David K | 139,123 | <p>Note that $x$ and $\frac1x$ always have the same sign.
Also if $\lvert x\rvert > 1,$ then $0 < \left\lvert\frac1x\right\rvert < 1$
and vice versa.</p>
<p>So the pairs will always be two positive numbers as solved in the other answers, or two negative numbers which you can get from one of those solutions ju... |
1,735,910 | <p>In <a href="https://www.youtube.com/watch?v=aHU-L3BLd_w">a recent video</a> the legendary Matt Parker claimed he kept flipping a two-sided (fair) coin untill he scored a sequence of ten consecutive 'switch flips', i.e. letting $T$ denote a tail and $H$ a head, then a sequence of ten switch flips is defined to be eit... | zhoraster | 262,269 | <p>Let $a^k_{n}$ denote the number of zero-one sequences of length $n$ with longest zero run non-exceeding $k$, and $a^k_{n,m}$ denote the number of such sequences with $m$ trailing zeros, $m=0,\dots,k$. </p>
<p>Then
$$
a^{k}_{n,m} = a^{k}_{n-m,0},\ m = 1,\dots,k, n\ge m,
$$
and $a^k_{n,0} = a^k_{n-1}$, $n\ge 0$. The... |
1,252,414 | <p>In rectangle ABCD below, points F and G lie on segment AB such that AF = FG = GB and E is the midpoint of segment DC. Also, segment AC intersects segment EF at H and segment EG at J. The area of rectangle ABCD is 70. Find the area of triangle AHF.</p>
<p>(Note: This question has been slightly changed from the origi... | Alexey Burdin | 233,398 | <p>There's a "straightforward" vector solution:<br>
Let $AB=b$, $AD=d$ the basis vectors and $[x\times y]$ be the cross product, since we're given $|[AB\times AD]|= 70$.<br>
Vectors $AF=1/3AB=1/3b, AG=2/3AB=2/3b, AE=AD+1/2DE=AD+1/2AB=d+1/2b$.<br>
$X$ lies on the line $YZ$ iff $AX=t\cdot AY + (1-t)\cdot AZ$, where $t$ i... |
315,844 | <p>What is the probability P(X>Y) given that X,Y are Uniformly distributed between [0,1]?</p>
| Emanuele Paolini | 59,304 | <p>Draw the square $[0,1]\times[0,1]$ and the region $X>Y$ to <em>see</em> the answer.</p>
|
655,378 | <p>I'm new to discrete mathematics and was wondering whether the following functions are one to one:</p>
<p>$$f(x) = x - 1$$
$$f(x) = x^2 + 1$$</p>
<p>The reason I stand by this is because for the first equation:</p>
<p>$$x - 1 = y - 1\\x = y$$</p>
<p>and for the second one:</p>
<p>$$x^2 +1 = y^2 +1\\x = y$$</p>
| Claude Leibovici | 82,404 | <p>If you look at $x$ as a function of $y$, just as Artem suggested, the differential equation becomes simple. </p>
<p>The general solution is $$x =c_1 e^{\sin (y)}-2 \sin (y)-2$$ which is difficult to inverse in the most general case (the problem is only simple if $c_1=0$). </p>
<p>However, playing a little, ther... |
2,505,863 | <p>I have to find one affine transformation that maps the point P=(1,1,1) to P'=(-1,-1,-1), the point P=(-1,-1,-1)' to P=(1,1,1) and the point Q=(0,0,0) to Q'=(2,2,2).
I started with a sketch and think that it is not possible to map both points with one affine transformation, but I must somehow prove that.
So I take th... | Dr. Sonnhard Graubner | 175,066 | <p>write the term in the form
$$\frac{(\sqrt{2x+1}-3)(\sqrt{2x+1}+3)(\sqrt{x-2}+\sqrt{2})}{(\sqrt{x-2}-\sqrt{2})(\sqrt{x-2}+\sqrt{2})(\sqrt{2x+1}+3)}$$</p>
|
2,505,863 | <p>I have to find one affine transformation that maps the point P=(1,1,1) to P'=(-1,-1,-1), the point P=(-1,-1,-1)' to P=(1,1,1) and the point Q=(0,0,0) to Q'=(2,2,2).
I started with a sketch and think that it is not possible to map both points with one affine transformation, but I must somehow prove that.
So I take th... | user236182 | 236,182 | <p>$$\lim_{x\to 4} \frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}}=$$</p>
<p>$$=\lim_{x\to 4}\frac{\frac{\sqrt{2x+1}-3}{x-4}}{\frac{\sqrt{x-2}-\sqrt{2}}{x-4}}=$$</p>
<p>$$=\frac{\lim_{x\to 4}\frac{\sqrt{2x+1}-3}{x-4}}{\lim_{x\to 4}\frac{\sqrt{x-2}-\sqrt{2}}{x-4}}$$</p>
<p>If you can use derivatives, then $$=\frac{(\sqrt{2... |
3,710,018 | <p>Problem:
Find prime solutions to the equation
<strong><span class="math-container">$p^2+1=q^2+r^2$</span></strong></p>
<p>I welcome you to post your own solutions as well</p>
<p>I have found a <em>strange solution</em> which I can't understand why it works(or what's the math behind it.) Here it is through examples... | Gerry Myerson | 785,985 | <p>Let <span class="math-container">$r=193$</span>, a prime. Then <span class="math-container">$$r^2-1=9313^2-9311^2=3107^2-3101^2=323^2-259^2$$</span> are the only expressions of <span class="math-container">$r^2-1$</span> as a difference of two squares, but <span class="math-container">$9313=67\times139$</span>, and ... |
3,710,018 | <p>Problem:
Find prime solutions to the equation
<strong><span class="math-container">$p^2+1=q^2+r^2$</span></strong></p>
<p>I welcome you to post your own solutions as well</p>
<p>I have found a <em>strange solution</em> which I can't understand why it works(or what's the math behind it.) Here it is through examples... | Robert | 777,173 | <p>The below identity can be utilized to get prime solutions.</p>
<p><span class="math-container">$a^2+1=b^2+c^2$</span></p>
<p><span class="math-container">$(mp+nq)^2+(mn-pq)^2=(mn+pq)^2+(mp-nq)^2$</span> ---(1)</p>
<p>Since one of the four elements in equation (1) needs to be equal to one we take:</p>
<p><span cl... |
3,682,987 | <blockquote>
<p>Let <span class="math-container">$a$</span>, <span class="math-container">$b$</span>, and <span class="math-container">$c$</span> be positive real numbers. What is the smallest possible value of <span class="math-container">$(a+b+c)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)$</span>?</p>
</... | Zhanxiong | 192,408 | <p>AM-GM inequality is a good idea:
<span class="math-container">\begin{align}
& (a + b + c)\left(\frac{1}{a + b} + \frac{1}{a + c} + \frac{1}{b + c}\right) \\
= & \frac{1}{2}((a + b) + (a + c) + (b + c))\left(\frac{1}{a + b} + \frac{1}{a + c} + \frac{1}{b + c}\right) \\
\geq & \frac{1}{2} \times 3\sqrt[3]... |
3,684,917 | <p>Let <span class="math-container">$C_{1}$</span> and <span class="math-container">$C_{2}$</span> be polytopes in <span class="math-container">$\mathbb{R}^{n}$</span> such that
<span class="math-container">$C_{1}=conv\left( V\right) $</span> with <span class="math-container">$V$</span> being a set of vertices. If
<s... | Sil | 290,240 | <p>The number of multiples of <span class="math-container">$p$</span> in <span class="math-container">$[A,B]$</span> is easily shown to be <span class="math-container">$\Big\lfloor \frac{B}{p} \Big\rfloor-\Big\lceil \frac{A}{p} \Big\rceil+1$</span> (see for example <a href="https://math.stackexchange.com/questions/3961... |
1,419,897 | <blockquote>
<p><strong>Theorem:</strong> Let $A$ be a bounded infinite subset of $\mathbb{R}^l$. Then
it has a limit point.</p>
</blockquote>
<p>So this is the Euclidean version of the Bolzano-Weierstrass theorem, the thing is that I was trying to prove it by induction, but it doesn't help because in the case $l=... | 5xum | 112,884 | <p>You can construct a sequence of hypercubes instead of intervals.</p>
<p>For example, taking $l=2$, lets assume (without loss of generality) that $A\subseteq [0,1]^2$.</p>
<p>Then, split $[0,1]^2$ into four subsquares (a square is a $2D$ cube):</p>
<p>$$A_{11} = [0,1/2]\times [0,1/2]\\
A_{12} = [1/2, 1]\times [0,1... |
909,228 | <p>I'm trying to find a closed form for the following sum
$$\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n},$$
where $H_n=\displaystyle\sum_{k=1}^n\frac{1}{k}$ is a harmonic number.</p>
<p>Could you help me with it?</p>
| epi163sqrt | 132,007 | <p><strong>Note:</strong> Please note the top voted answer by @Tunk-Fey is regrettably <em>not correct</em>. Contrary to his claim his final expression (4) when evaluated at $x=\frac{1}{2}$ does not match @Cleo's answer but differs by $\frac{\pi^4}{120}$ from the correct identity:
\begin{align*}
\sum_{n=1}^\infty \frac... |
997,602 | <blockquote>
<p>Prove that the function <span class="math-container">$x \mapsto \dfrac 1{1+ x^2}$</span> is uniformly continuous on <span class="math-container">$\mathbb{R}$</span>.</p>
</blockquote>
<p>Attempt: By definition a function <span class="math-container">$f: E →\Bbb R$</span> is uniformly continuous iff for ... | Community | -1 | <p>$$x^2 \geq 0 \implies 1+x^2 > 1 \implies \frac{1}{1+x^2} < 1$$</p>
<p>Using the above inequality,</p>
<p>$$
\begin{align}
|f(x)-f(a)| &\leq |x - a| \frac{|x + a|}{(1 + x^2)(1 + a^2)}\\
&\leq |x - a||x+a|\\
&\leq |x - a|(|x|+|a|)
\end{align}
$$
Choose $$|x-a| \leq 1 \implies |x|\leq |a|+1$$
Now,
$... |
2,030,116 | <p>How can i prove that $\sqrt[12]{2}$ is irrational number? </p>
<p>I'm trying: </p>
<p>$$\sqrt[12]{2} = \frac{p}{q}$$ where $p$, $q$ are integers</p>
<p>it follows that :</p>
<p>$$p^{12} = 2q^{12} $$</p>
<p>What is argument of irrationality in this case?
From what we know that the right-hand side has an even n... | RGS | 329,832 | <p>You did everything fine!</p>
<p>When you get to $p^{12} = 2q^{12}$ you use the argument you mentioned. The LHS has an even number of 2s in its factorization whilst the RHS has an odd number of 2s, thus the two sides are not equal.</p>
|
2,030,116 | <p>How can i prove that $\sqrt[12]{2}$ is irrational number? </p>
<p>I'm trying: </p>
<p>$$\sqrt[12]{2} = \frac{p}{q}$$ where $p$, $q$ are integers</p>
<p>it follows that :</p>
<p>$$p^{12} = 2q^{12} $$</p>
<p>What is argument of irrationality in this case?
From what we know that the right-hand side has an even n... | Vidyanshu Mishra | 363,566 | <p>Suppose $${\sqrt[12]{2} = \frac{p}{q}}$$
where p and q are coprime
$${p^{12} = 2q^{12}}$$
$${2|p^{12}}$$
$${2|p}$$
$${2^2|p^{12}}$$
since $p^{12} = 2q^{12}$</p>
<p>$${2^2|2q^{12}}$$
$${2|q^{12}}$$
$${2|q}$$.</p>
<p>A contradiction (since we assumed that p and q are coprime)</p>
|
555,045 | <p>Let $K$ be a quadratic number field.
Let $R$ be an order of $K$, $D$ its discriminant.
By <a href="https://math.stackexchange.com/questions/546281/on-a-certain-basis-of-an-order-of-a-quadratic-number-field">this question</a>, $1, \omega = \frac{(D + \sqrt D)}{2}$ is a basis of $R$ as a $\mathbb{Z}$-module.</p>
<p>L... | Matt E | 221 | <p>Since $\omega = (D + \sqrt{D})/2$ has minimal polynomial equal to $x^2 - D x + D(D-1)/4,$ we may write $R = \mathbb Z[x]/(x^2 - Dx - D(D-1)/4),$ and so $R/2R =
\mathbb F_2[x]/(x^2 - D x - D(D-1)/4)$ (identify $\omega$ with the image of $x$).</p>
<p>If $D$ is even (and hence a multiple of $4$), the polynomial $x^2 ... |
898,683 | <p>Given a pool of 30 balls (5 of each color). When drawing 8 balls without replacement, what is the probability of getting at least one of each color?</p>
<p>Related: <a href="https://math.stackexchange.com/questions/897730/probability-of-drawing-at-least-one-red-and-at-least-one-green-ball">Probability of drawing at... | user84413 | 84,413 | <p>For the new version of the problem, we can use Inclusion-Exclusion:</p>
<p>If $E_i$ is the set of draws without at least one ball of color i, we have</p>
<p>$|E_1\cup\cdots\cup E_6|=\sum|E_i|-\sum|E_i\cap E_j|+\sum|E_i\cap E_j\cap E_k|-\cdots$</p>
<p>$=\binom{6}{1}\binom{25}{8}-\binom{6}{2}\binom{20}{8}+\binom{6}... |
973,035 | <p>I'm wondering whether there is an invertible function $f: \mathbb{R} \to \mathbb{R}$ such that $f(-1)=0$, $f(0)=1$ and $f(1)=-1$. I think it's not but I'm missing a real proof.</p>
<p>The easiest would be to show such a function cannot be injective... But I don't see how? I don't see any other way of starting this ... | Manolito Pérez | 13,293 | <p>A continuous function $f:\mathbb{R} \rightarrow \mathbb{R}$ is invertible on an interval $I$ if, and only if, it is bijective, and it is bijective if, and only if, it is monotonous. Since your function $f$ is not monotonous on the interval [-1; 1] (it increases first, and then decreases), it cannot be invertible. </... |
2,475,757 | <p>I want to determine if the following integrals converge or diverge.</p>
<ol>
<li>$\int_{0}^\infty \frac{\sqrt{x}}{\sqrt[3]{x^5+1}}dx.$</li>
<li>$\int_{0}^\infty \sin\frac{1}{x^2+1}dx$.</li>
<li>$\int_{\sqrt{2}}^2 \frac{dx}{\sqrt{x^2-2}}dx.$</li>
<li>$\int_{0}^1 \frac{\ln{x}}{x}dx.$</li>
</ol>
<hr>
<p><strong>(1):... | clark | 33,325 | <p>Let $X_1,X_2$ be your draws. Then $X_1,X_2$ are i.i.d. uniform on $\{1,2,3\}$. Now you want to compute $\mathbb{P}(Z=2)$ where $Z=\max \{X_1,X_2\}$. </p>
<p>Generally, you have $\mathbb{P}(Z\leq t) =\mathbb{P}(X_1\leq t,X_2\leq t)=\mathbb{P}(X_1\leq t)\mathbb{P}(X_2\leq t) $. Thereofore,
\begin{align}
\mathbb{P}(Z... |
2,941,106 | <p>I have tried 29.2/8.44 and tried multiplying this to get whole numbers but doesn't seem like it's working </p>
| fleablood | 280,126 | <p>Multiply both by <span class="math-container">$100$</span> to get <span class="math-container">$844:2920$</span>.</p>
<p>Then divide by the greatest common divisor. I'm too lazy to figure out what the greatest common divisor so I'll just divide both sides by <span class="math-container">$4$</span> and then keep di... |
1,722,964 | <p>Expression :$$(p\rightarrow q)\leftrightarrow(\neg q\rightarrow \neg p)$$
What does the symbol $\leftrightarrow$ mean ? Please explain by drawing the truth table for this expression and also with other examples if possible. <strong>I'm in a desperate situation so I'd really appreciate a quick response !</strong></p>... | dtldarek | 26,306 | <p>Symbol $\leftrightarrow$ or $\iff$ denote usually the equivalence, commonly known also as "NXOR", "if and only if" or "iff" for short (see also <a href="https://en.wikipedia.org/wiki/If_and_only_if" rel="noreferrer">its Wikipedia page</a>). More precisely $p \leftrightarrow q$ is equal to $$(p \to q) \land (q \to p)... |
3,347,342 | <blockquote>
<p><span class="math-container">$$\frac{2}{5}^{\frac{6-5x}{2+5x}}<\frac{25}{4}$$</span></p>
</blockquote>
<p>I can write this as
<span class="math-container">$$\frac25 ^{\frac{6-5x}{2+5x}} <\frac25 ^{-2}$$</span>
Therefore <span class="math-container">$$\frac{6-5x}{2+5x}<-2$$</span>
Solving it... | Who am I | 687,026 | <p><span class="math-container">$$\frac{6-5x}{2+5x}<-2$$</span></p>
<p>After this cross multiply</p>
<p>We get </p>
<p><span class="math-container">$$6-5x<-4-10x$$</span>
<span class="math-container">$$ 10<-5x$$</span>
<span class="math-container">$$2<-x$$</span></p>
<p>Now when we multiply or divide by... |
3,347,342 | <blockquote>
<p><span class="math-container">$$\frac{2}{5}^{\frac{6-5x}{2+5x}}<\frac{25}{4}$$</span></p>
</blockquote>
<p>I can write this as
<span class="math-container">$$\frac25 ^{\frac{6-5x}{2+5x}} <\frac25 ^{-2}$$</span>
Therefore <span class="math-container">$$\frac{6-5x}{2+5x}<-2$$</span>
Solving it... | Allawonder | 145,126 | <p>The operation you performed from step <span class="math-container">$2$</span> to step <span class="math-container">$3$</span> is taking of logarithms.</p>
<p>It is true that if <span class="math-container">$x<y,$</span> then <span class="math-container">$\log x<\log y$</span> whenever we take the base of our ... |
3,670,240 | <p>It' not a physics question, just ..coincidence ;) (i'm concerned about mathematical rightness of it)</p>
<p>Let's consider <span class="math-container">$U,T,S,P,V\in\mathbb{R_{>0}}$</span> such that
<span class="math-container">$$dU=TdS-PdV$$</span></p>
<ul>
<li>Based on this, how we can rigorously proof that <... | Giorgio Pastasciutta | 660,461 | <p>Ok, i think i finally got it</p>
<p>An important hypotesis not written is that <span class="math-container">$S, V$</span> are mutually independent</p>
<p>Let us consider
<span class="math-container">$$dU=T\,dS-P\,dV$$</span>
From this 6 cases are possible:</p>
<ol>
<li><span class="math-container">$U=U(S,V,\{X_i\... |
2,429,231 | <p>Let $X$ be a non-negative random variable with $\text{Var}(X)<\frac{1}{2}$. Show that then $P\big(-1+E(X)\le X\le 2E(X)\big)\ge \frac{1}{2}$ where $P$ is the probability measure and Var is the variance and $E(X)$ is the expectation of $X$.</p>
<p>My approach : I was thinking to compute $P(-1+E(X)\le X)$ and $P(X... | GAVD | 255,061 | <p>Some comments: In the <a href="http://mathworld.wolfram.com/ChebyshevInequality.html" rel="nofollow noreferrer">link</a>, the Chebysev's inequality is
$$P(|X-\mu|\geq k) \leq \frac{\sigma^2}{k^2}$$ or $$P(|X-\mu|< k) \geq 1 -\frac{\sigma^2}{k^2} $$</p>
<p>If $EX>1$, one has $$P(-1<X - EX \leq EX) = P(-1\l... |
2,429,231 | <p>Let $X$ be a non-negative random variable with $\text{Var}(X)<\frac{1}{2}$. Show that then $P\big(-1+E(X)\le X\le 2E(X)\big)\ge \frac{1}{2}$ where $P$ is the probability measure and Var is the variance and $E(X)$ is the expectation of $X$.</p>
<p>My approach : I was thinking to compute $P(-1+E(X)\le X)$ and $P(X... | alexjo | 103,399 | <p>Let $\mu=\Bbb E(X)$ and $\sigma^2=\mathrm{Var}(X)$. </p>
<p>From Chebyshev's Inequality we have
$$
\Bbb P(|X-k|\leq t)\geq 1-\frac{E[(X-k)^2]}{t^2}
$$
that is
$$
\Bbb P(-t+k<X<t+k)\geq 1-\frac{E[(X-k)^2]}{t^2}
$$
By setting $a=-t+k$ and $b=t+k$ we have the general Chebyshev's Inequality
$$
\Bbb P(a\leq X \... |
84,138 | <p>I recently have a paper rejected from a very good (but not the top) journal. The referee report said the result was good and certainly belong there, but he did not think I did enough to back up my claims (it was a rather long and harsh criticism at the exposition). Now I know for sure that my result is good and my p... | Brendan McKay | 9,025 | <p>A paper can be accepted (maybe subject to minor changes), rejected with a request for certain major changes, or rejected outright. If your paper fits into the middle category (which the editor will usually make clear), you should do some major surgery on it and send it again to the same journal. If it is in the last... |
3,415,624 | <p>I want to show that the sequence given in the title is convergent and find its limit. I'm not sure if I should use the monotone convergence theorem, because when I try using induction, I don't seem to get anywhere. And I also don't know how to find a suitable candidate for a limit. I do know what the definition of c... | Dr. Sonnhard Graubner | 175,066 | <p>Use that <span class="math-container">$$\sum_{i=1}^ni^2=\frac{1}{6} n (n+1) (2 n+1)$$</span></p>
|
3,415,624 | <p>I want to show that the sequence given in the title is convergent and find its limit. I'm not sure if I should use the monotone convergence theorem, because when I try using induction, I don't seem to get anywhere. And I also don't know how to find a suitable candidate for a limit. I do know what the definition of c... | marty cohen | 13,079 | <p>If <span class="math-container">$p$</span> is a polynomial of degree <span class="math-container">$d$</span>
with non-negative coefficients,
so that
<span class="math-container">$p(n)
=\sum_{k=0}^n a_kn^k
$</span>
with all <span class="math-container">$a_k \ge 0$</span>,
then
<span class="math-container">$\lim_{n \t... |
3,935,494 | <p>Usually the inverse of a square <span class="math-container">$n \times n$</span> matrix <span class="math-container">$A$</span> is defined as a matrix <span class="math-container">$A'$</span> such that:</p>
<p><span class="math-container">$A \cdot A' = A' \cdot A = E$</span></p>
<p>where <span class="math-container... | Xander Henderson | 468,350 | <p>As others have suggested, you can "just" apply Bayes' theorem. However, this problem is a relatively simple one, and so it might help to draw out a probability tree to help organize your thoughts. I will not claim that this is the "best" or most efficient way to approach the problem, but I find... |
3,085,181 | <p>I have to cope with a constraint of the form (1) in the following problem: </p>
<p><span class="math-container">$$\begin{align}\max\quad& x+y\\
\text{s.t.}\quad&
x + y \leq \max \{x,y\} &(1)\\
&0 \leq x \leq U_x&(2)\\
&0 \leq y \leq U_y&(3)\\
\end{align}$$</span></p>
<p>In the follow... | nathan.j.mcdougall | 181,447 | <p>The reformulation in the link isn't guaranteed to work. In this case, it doesn't, because the feasible region is not convex. You cannot express a non-convex feasible region with linear constraints.</p>
<p>To see that it is not convex, note that if <span class="math-container">$x\geq y$</span>, then <span class="mat... |
1,100,906 | <p>What would be the highest power of two in the given expression?</p>
<p>$32!+33!+34!+35!+...+87!+88!+89!+90!\ ?$</p>
<p>I know there are 59 terms involved. I also know the powers of two in each term. I found that $32!$ has 31 two's. If we take 32! out of every term the resulting 59 terms has 2 odd terms and 57 even... | Anurag A | 68,092 | <p>The sum
$$32!+33!+34!+35!+ \dotsb +87!+88!+89!+90!=32![1+33+33\cdot 34+33\cdot 34\cdot 35+\dotsb]$$
The expression
$$1+33+33\cdot 34+33\cdot 34\cdot 35+\dotsb \equiv 0 \pmod{2}$$
but
$$1+33+33\cdot 34+33\cdot 34\cdot 35+\dotsb \equiv 2 \pmod{4}.$$
Thus all the powers of $2$ will come from $32!$ and only one from the... |
1,100,906 | <p>What would be the highest power of two in the given expression?</p>
<p>$32!+33!+34!+35!+...+87!+88!+89!+90!\ ?$</p>
<p>I know there are 59 terms involved. I also know the powers of two in each term. I found that $32!$ has 31 two's. If we take 32! out of every term the resulting 59 terms has 2 odd terms and 57 even... | abiessu | 86,846 | <p>You can proceed this way:</p>
<p>$$32!\left(1+33+34\cdot 33+35\cdot 34\cdot 33+\dots+\frac {90!}{32!}\right)$$</p>
<p>$32!$ has factor $2^{31}$, and $1+33$ is of the form $4k+2$, and both $34\cdot 33$ and $35\cdot 34\cdot 33$ are of the form $4k+2$, and for all the terms $36!\over 32!$ and greater we have a diviso... |
1,162,161 | <p>A patient would like to take a test to determine if he has a nasty disease. Let the variable A denote that
the patient has the disease and the variable B denote a positive test. The following assumptions apply:
• The probability that the test is positive given the patient has the disease is 99%.
• The probability th... | kobe | 190,421 | <p>Since $|nx^{n-1}| \le n(9/10)^{n-1}$ and $\sum_{n = 1}^\infty n(9/10)^{n-1}$ converges (by the ratio test), by the Weierstrass $M$-test, the series $\sum_{n = 1}^\infty nx^{n-1}$ converges uniformly on $[0,9/10]$.</p>
|
1,468,208 | <p>I'm having a lot of problems with this one linear recurrence problem ... </p>
<p>First, verify that:
$x^3 − 3x − 2 = (x^2 + 2x + 1)(x − 2). $</p>
<pre><code>Then, solve the linear recurrence
f(0) = 0, f(1) = 1, f(2) = 7,
f(n) = 3f(n − 2) + 2f(n − 3).
</code></pre>
<p>I'm able to get this far but I... | D. A. | 275,736 | <p>you wrote the answer: "verify that: $x^3 − 3x − 2 = (x^2 + 2x + 1)(x − 2)$".</p>
<p>solve that polynomials separately:</p>
<p>$ x-2=0$ and $x^2 + 2x + 1=0$.</p>
|
1,468,208 | <p>I'm having a lot of problems with this one linear recurrence problem ... </p>
<p>First, verify that:
$x^3 − 3x − 2 = (x^2 + 2x + 1)(x − 2). $</p>
<pre><code>Then, solve the linear recurrence
f(0) = 0, f(1) = 1, f(2) = 7,
f(n) = 3f(n − 2) + 2f(n − 3).
</code></pre>
<p>I'm able to get this far but I... | Claude Leibovici | 82,404 | <p>For the recurrence relation $$f(n) = 3f(n − 2) + 2f(n − 3)$$ the characteristic equation is $$r^3=3r^2+2$$ which is exactly the equation you had to solve at the beginning; using D.A.'s answer, the solutions are then $r=2$, $r=-1$, $r=-1$. So, because of the double root, the solution will be $$f(n)=c_1(2)^n+c_2(-1)^n... |
3,495,852 | <p>I couldn't find an example or explanation why the following sentence is correct </p>
<p>If a Transformation is linear, and vectors <span class="math-container">$u_1$</span>,<span class="math-container">$u_2$</span>,<span class="math-container">$u_3$</span> are dependent then
<span class="math-container">$T(u_1)$</... | lonza leggiera | 632,373 | <p>You can always extend definitions of the <span class="math-container">$\ \xi_n\ $</span> from the probability space <span class="math-container">$\ (\Omega,\mathcal{F}, P)\ $</span> on which they were originally defined to (for example) the product space <span class="math-container">$\ (\Omega\times[0,1), \sigma(\ma... |
211,705 | <p>I am given a table of possible <span class="math-container">$X_1$</span> and <span class="math-container">$X_2$</span> values that can be generated in a casino. In the game, both are generated with each turn.</p>
<p><img src="https://i.stack.imgur.com/G0nLn.jpg" alt="enter image description here" /></p>
<blockquote>... | CodyBugstein | 41,803 | <p>After playing with it, I found a simpler answer:</p>
<p>Each potential payoff can simply be calculated using <span class="math-container">$E[X]$</span>. For example,</p>
<blockquote>
<p><span class="math-container">$a) \ 8X_1 = 8E[X_1]$</span></p>
<p><span class="math-container">$b) \ 4X_1 + 8(X_2)^2-\frac{51}{128} ... |
1,355,133 | <p>A while ago I asked a question about probability here <a href="https://math.stackexchange.com/questions/1353044/why-is-binomial-probability-used-here/">Why is binomial probability used here?</a></p>
<p>I get that you can find how many ways of choosing the $6$ correct out of $10$ questions.</p>
<p>But why do we <st... | Avraham | 91,378 | <p>You are correct that the simple casework will work. However, if you were to look carefully at the casework, you will notice something interesting. In each case, the actual probabilities are the same—you will have 6 correct and 4 incorrect answers. The only thing that changes is <strong>which</strong> of the 10 are c... |
2,361,516 | <p>Let $X$ be a real Hilbert space.Let $x,y \in X$ such that $\langle x,y\rangle >0$. If $\alpha \geq 1$.</p>
<p>I want to prove that $\Vert \alpha x-y \Vert \leq \Vert x-y \Vert$</p>
| John Hughes | 114,036 | <p>In $\Bbb R$, let $x = 1, y = 1, \alpha = 0.5$. The right hand side is $0$, the left hand side is $0.5$. The thing you're trying to prove is false. </p>
|
1,227,375 | <p>Show that $\sum^n_i i^4\log^2i$ = $\Theta(n^5\log^2n)$</p>
<p>I am completely lost on how to solve this problem. I understand that $\Theta$ deals with the upper and lower bounds, so do we prove both big-oh and big-omega?</p>
| Simon S | 21,495 | <p>One strategy is to rotate the coordinate system so the semi-major/minor axes are parallel to the coordinate axis.</p>
<p>To do that, write your equation as $ax^2 + 2bxy + cy^2 + dx + ey = 1$. We can rewrite the first three terms as $(x \ y)A(x \ y)^T$ for the symmetric matrix $A = \left( \begin{matrix} a & b \\... |
1,227,375 | <p>Show that $\sum^n_i i^4\log^2i$ = $\Theta(n^5\log^2n)$</p>
<p>I am completely lost on how to solve this problem. I understand that $\Theta$ deals with the upper and lower bounds, so do we prove both big-oh and big-omega?</p>
| robjohn | 13,854 | <p>In <a href="https://math.stackexchange.com/a/247534">this answer to a related question</a>, it is shown that
$$
Ax^2+Bxy+Cy^2+Dx+Ey+F=0
$$
is simply a rotated and translated version of
$$
\small\left(A{+}C-\sqrt{(A{-}C)^2+B^2}\right)x^2+\left(A{+}C+\sqrt{(A{-}C)^2+B^2}\right)y^2+2\left(F-\frac{AE^2{-}BDE{+}CD^2}{4AC... |
1,318,445 | <p>I am having difficulty with part b and c. I have got part a.</p>
<p>This is the question:</p>
<p><img src="https://i.stack.imgur.com/kT1kA.png" alt="http://i.stack.imgur.com/kT1kA.png"></p>
<p>For part a) I have $P(x=2) = (e^{-0.1} * (0.1)^2)/2!$</p>
<p>What do I do for part b?</p>
| Henry | 6,460 | <p>Hint: </p>
<ul>
<li><p>If there is a mean $0.1$ flaws per square metre, there is a mean $1$ flaw per $10$ square metres</p></li>
<li><p>If there is a mean $0.1$ flaws per square metre, there is a mean $10$ flaws per $100$ square metres</p></li>
</ul>
<p>Use the Poisson distribution again.</p>
|
1,318,445 | <p>I am having difficulty with part b and c. I have got part a.</p>
<p>This is the question:</p>
<p><img src="https://i.stack.imgur.com/kT1kA.png" alt="http://i.stack.imgur.com/kT1kA.png"></p>
<p>For part a) I have $P(x=2) = (e^{-0.1} * (0.1)^2)/2!$</p>
<p>What do I do for part b?</p>
| Stanley | 241,637 | <p>Convert the rate to $\lambda = 1$ flaw per $10m^2$ and then use the formula $\mathbb{Pr}[X = 0] = \frac{\lambda^0 e^{-\lambda}}{0!} = e^{-\lambda}$</p>
|
3,440,873 | <p>I do not how to solve this, can such equation even exist? For the root to lie on the y intercept, the line would have to pass through origin, which means one root will be 0, breaking down the whole the thing. Am I missing something here?</p>
| Mirko | 188,367 | <p>Are you sure your teacher asked for a german reference, and not for a germane justification? </p>
<p>Was this a multiple choice question? Were you expected to provide any justification for your answer, any details explaining your thinking: If the answer is "yes", why it is so, and if the answer is "no", why "no"? <... |
1,406,878 | <p>Given is following sequence:</p>
<p>$a_{n+1} = a_n - \frac{a_n - v}{s}$</p>
<p>I found out that</p>
<p>$\forall a_0, v, s \in \mathbb{R}, s>0: \lim\limits_{n \to \infty}a_n=v$</p>
<p>But I do not know why. I tried to write down $a_2$ , $a_3$, but the term becomes very long and complex, and it doesn't help me ... | Redundant Aunt | 109,899 | <p>If $s=1$ or $a_0=v$, then $a_n=v$ for all $n$ so the limit is trivial.</p>
<p>If $s≠1$, then let $a_n=b_n+v$. This yields:
$$
b_{n+1}=b_n\left(1-\frac{1}{s}\right)\implies b_n=b_0\left(1-\frac{1}{s}\right)^n
$$
So if $0.5<s$, we have $|{1-\frac{1}{s}}|<1$ and therefore $\lim_{n\to\infty}b_n=0\implies\lim_{n\t... |
1,159,860 | <p>If $$f:[a,b]\times [c,d] \to \mathbb{R}$$ is continuous and $f_{y}$ is continuous, let $$F(x,y)=\int_{a}^{x} f(t,y)dt.$$ </p>
<ol>
<li>Find $F_x$ and $F_y$.</li>
<li>If $G(x)=\int_{a}^{g(x)}f(t,x)dt$, find $G'(x)$</li>
</ol>
<p>My try: </p>
<p>For (1) $$F(x+h,y)-F(x,y)=\int_{a}^{x+h} f(t,y)dt-\int_{a}^{x}f(t,y)dt... | FAM | 216,087 | <p>Assuming X and Y to be two sets of favourable events....</p>
<p>$P(X) = a$
$P(Y)= b$</p>
<p>We are given that X and Y are independent.</p>
<p>Now,</p>
<p>X-Y defines the set of favourable events which belongs to the set X but does not belong to the set Y.</p>
<p>So, we can write $X-Y = X\cap Y^c$ </p>
<p>;whe... |
3,668,101 | <p>I know that if <span class="math-container">$n \bmod k \le k-1$</span> then this sum is converge then it has finite sum, I just guess it's <span class="math-container">$\ln(k)$</span> because when <span class="math-container">$k=1$</span> sum is <span class="math-container">$0=ln(1)$</span>. I really don't know how ... | Dr. Wolfgang Hintze | 198,592 | <p>Using the same ideas as in a recent answer (<a href="https://math.stackexchange.com/a/4024522/198592">https://math.stackexchange.com/a/4024522/198592</a>) here is a solution which is very close to that of @xpaul and @Gary but more elementary in the final step (without polygamma functions).</p>
<p>We will prove that<... |
592,912 | <p>I need to describe the minimal field extension $\mathbb Q(\sqrt[3] {2})$ of the rational numbers $\mathbb Q$ that contain $\sqrt[3] {2}$.</p>
<p>$\mathbb Q(\sqrt[3] {2}) =\{a+b\sqrt[3] {2}+c(\sqrt[3] {2})^2|a,b,c \in \mathbb{Q}\}$.</p>
<p>I tried to use the rationalization of $x^3 + y^3 + z^3 - 3xyz$ ?</p>
| gcc | 278,750 | <p>Let <span class="math-container">$\Omega$</span> be an uncountable set and <span class="math-container">$\cal A$</span> the finite-cofinite field on <span class="math-container">$\Omega$</span> (all sets which are either finite or their complement is finite)
and <span class="math-container">$\lambda:{\cal A}\to[0,+\... |
255,811 | <p>I'm recalling this question from memory, so I may be messing it up a bit.</p>
<p>Let $a/3+b/2+c=0$. Show that $ax^2+bx+c=0$ has at least one root in $[0,1]$ using the Mean Value Theorem.</p>
<p>Let $f(x)=ax^2+bc+c$. Then $f(0)=c$ and $f(1)=a+b+c$. Also $f'(x)=2ax+b$. So there exists $f(\xi)=[f(1)-f(0)]/1=a+b-c... | JavaMan | 6,491 | <p>First, if $a =0$, then we have $bx + c = 0 \implies x = - \frac{c}{b} = \frac{b/2}{b} = \frac{1}{2}$.</p>
<p>Now, suppose $a \neq 0$.
Note that $c = - \frac{a}{3} - \frac{b}{2}$, so you want to prove that the function $f(x) = ax^2 + bx - \frac{a}{3} - \frac{b}{2}$ has a root in $[0,1]$. We have $f(0) = - \frac{a}{... |
1,373,170 | <p>How can I solve $e^{k_1/x}+e^{k_2/x}+\cdots+e^{k_N/x}=1$ for $x$,</p>
<p>where $N\geq 1, k_1,\ldots,k_N \in \mathbb{R}, k_1,\ldots,k_N < 0, x\in \mathbb{R}$ and $x >0$.</p>
<p>I looked at the basic rules of exponentiation and logarithms and they do not seem to help simplify the equation in this particular ca... | Daniel | 150,142 | <p>This is equivalent to finding roots of the polynomial</p>
<p>$$y^{-k_1}+\cdots +y^{-k_N}-1=0$$</p>
<p>(by making $y:= e^{-\frac{1}{x}}$) so your problem is as difficult as finding roots of polynomials. It depends on the degree of that polynomial, in this case, it depends on the max of the $-k_i$.</p>
|
3,258,642 | <blockquote>
<p>If the roots of quadratic equation <span class="math-container">$$x^2 − 2ax + a^2 + a – 3 = 0$$</span>
are real and less than <span class="math-container">$3$</span>, find the range of <span class="math-container">$a$</span>.</p>
</blockquote>
<p>The roots are <span class="math-container">$a... | Dr. Sonnhard Graubner | 175,066 | <p>Yes , since we have <span class="math-container">$$3\geq a$$</span> you can square the inequality with <span class="math-container">$+$</span> sign and you will get <span class="math-container">$$3-a<(3-a)^2$$</span> so you will get <span class="math-container">$$0<(3-a)(3-a-1)$$</span>
<span class="math-conta... |
3,258,642 | <blockquote>
<p>If the roots of quadratic equation <span class="math-container">$$x^2 − 2ax + a^2 + a – 3 = 0$$</span>
are real and less than <span class="math-container">$3$</span>, find the range of <span class="math-container">$a$</span>.</p>
</blockquote>
<p>The roots are <span class="math-container">$a... | ArsenBerk | 505,611 | <p>Note that you will have different results while squaring both sides of </p>
<p><span class="math-container">$$ \sqrt {3 - a } \lt 3 - a$$</span></p>
<p>and</p>
<p><span class="math-container">$$- \sqrt {3 - a } \lt 3 - a$$</span></p>
<p>If you take square of both sides by taking this into consideration, it is al... |
3,258,642 | <blockquote>
<p>If the roots of quadratic equation <span class="math-container">$$x^2 − 2ax + a^2 + a – 3 = 0$$</span>
are real and less than <span class="math-container">$3$</span>, find the range of <span class="math-container">$a$</span>.</p>
</blockquote>
<p>The roots are <span class="math-container">$a... | user376343 | 376,343 | <p>The inequality <span class="math-container">$\;\underbrace{- \sqrt {3 – a } }_{{-}}\leq \underbrace{3 – a}_{+}\;$</span> holds trivially.</p>
<p>As <span class="math-container">$\;3-a\geq 0\;$</span> and <span class="math-container">$\;\sqrt {3 – a } \geq 0,\;$</span> squaring <span class="math-container">$\sqrt {... |
3,258,642 | <blockquote>
<p>If the roots of quadratic equation <span class="math-container">$$x^2 − 2ax + a^2 + a – 3 = 0$$</span>
are real and less than <span class="math-container">$3$</span>, find the range of <span class="math-container">$a$</span>.</p>
</blockquote>
<p>The roots are <span class="math-container">$a... | Tarzan | 673,567 | <p>You can avoid squareing inequality. </p>
<p>Since <span class="math-container">$2a =x_1+x_2 <6$</span> we have <span class="math-container">$a<3$</span> and <span class="math-container">$$a^2+a-3=x_1x_2 < 9$$</span> we have also <span class="math-container">$a^2+a-12 =(a+4)(a-3)<0$</span> so <span class... |
3,275,732 | <p>How can I solve it without using matrix? I tried it to solve by using systems. But I have no idea how deal with "<span class="math-container">$0$</span>"</p>
| José Carlos Santos | 446,262 | <p>Just solve the system<span class="math-container">$$\left\{\begin{array}{l}d=1\\-a+b-c+d=-2\\a+b+c+d=2\\8a+4b+2c+d=9.\end{array}\right.$$</span></p>
|
4,071,619 | <blockquote>
<p>There are two German couples, two Japanese couples and one unmarried person. If all 9 persons are two be interviewed one by one then the total number of ways of arranging their interviews such that no wife gives an interview before her husband is?</p>
</blockquote>
<p>I tried using the string method, bu... | lab bhattacharjee | 33,337 | <p>Using induction,</p>
<p>If <span class="math-container">$10^r=1+3^sk$</span> where <span class="math-container">$3\nmid k$</span></p>
<p><span class="math-container">$$10^{3r}=(10^r)^3=(1+3^sk)^3=1+3^{s+1}k+3^{2s+1}k^2+3^{3s}k^3\equiv1\pmod{3^{s+1}}$$</span></p>
<p>for <span class="math-container">$s+1\le2s+1\iff s... |
228,481 | <p>I have recently been interested in the problem of summing Combinatorials. I have been beating my brain for the past days to figure out how to find an explicit closed form of:</p>
<p>$n \choose 0 $+$ n \choose 3 $+$ n \choose 6$ + $\dots$ + $n \choose 3K$, where $3K$ is the largest multiple of $3$ less than or equal... | Raymond Manzoni | 21,783 | <p>A closed form is given by $\ \displaystyle \frac{2^n+2\cos(n\pi/3)}3$. </p>
<p>See this <a href="https://oeis.org/A024493" rel="nofollow">entry</a> of OEIS for more information.</p>
|
228,481 | <p>I have recently been interested in the problem of summing Combinatorials. I have been beating my brain for the past days to figure out how to find an explicit closed form of:</p>
<p>$n \choose 0 $+$ n \choose 3 $+$ n \choose 6$ + $\dots$ + $n \choose 3K$, where $3K$ is the largest multiple of $3$ less than or equal... | N. S. | 9,176 | <p>Let $\omega$ be a third root of 1.</p>
<p>Then</p>
<p>$$(1+1)^n = \binom{n}{0} +\binom{n}{1}+ \binom{n}{2}+ \binom{n}{3}+ ...+\binom{n}{n} \,.$$
$$ (1+\omega)^n = \binom{n}{0} + \binom{n}{1}\omega+ \binom{n}{2} \omega^2+ \binom{n}{3}+ ...+ \binom{n}{n} \omega^n \,.$$</p>
<p>$$ (1+\omega^2)^n =\binom{n}{0} + \bin... |
1,116,496 | <blockquote>
<p>Let $H$ be a Hilbert space with a countable basis $B$. Does it mean that any vector $x\in H$ can be expressed as a <strong>finite</strong> linear combination of elements from $x$, or as an <strong>infinite</strong> linear combination?</p>
</blockquote>
<p>Thanks in advance</p>
| Thomas | 128,832 | <p>In Hilbert space theory and Functional analysis the term 'basis' has a different definition than in linear algebra. For this reason one often encounters the term 'Hamel basis' for the latter and 'Hilbert space basis' in the first case. </p>
<p>In the Hilbert space case the requirement is that each vector can be rep... |
847 | <p>Apologies in advance if this is obvious.</p>
| Geoff Robinson | 14,450 | <p>This is not really an answer, but is too long for a comment. The proof given by Moonface above is given in more or less that form in the 1962 book of Curtis and Reiner. As far as I know, it is still open whether all irreducible representations of a finite group $G$ can be realized over
$\mathbb{Z}[\omega]$, where $... |
4,481,314 | <p>This is an exercise in Tristan Needham's <em>Visual Differential Geometry and Forms</em>. He uses the term <em>ultimate equality</em> to mean roughly the same thing as first order approximation, which he says is motivated by Newton's Principia. The book is dedicated to Needham's longtime personal friend Roger Penr... | Peter Plex | 1,060,638 | <p>The image of the sine function is the closed interval <span class="math-container">$[0,1]$</span>, so for a given value <span class="math-container">$L\in [0,1]$</span> there is a <span class="math-container">$b\in\mathbb{R}$</span> such that <span class="math-container">$\sin(b)=L$</span>. We may assume <span class... |
214,766 | <p>Is there an efficient way to check a number x and remove all prime factors in the number which are less than some n? For example for n = 200:</p>
<pre><code>x=88984589931961415442566827779929187431222364934742868664124547963532933
FactorInteger[x]
{{29, 2}, {31, 1}, {37, 2}, {269, 1}, {271,
1}, {3420047160553... | march | 29,734 | <p>Implement a ragged transpose after turning each sublist into a list of pairs, like so:</p>
<pre><code>list = {{1, {0}}, {2, {0}}, {3, {-2, 0, 2}}, {4, {-2, 0, 2}}, {5, {-2,0, 2}}};
Flatten[Thread /@ list, {2}]
(* {{{1, 0}, {2, 0}, {3, -2}, {4, -2}, {5, -2}},
{{3, 0}, {4, 0}, {5, 0}},
{{3, 2}, {4, 2}, {5, 2}... |
214,766 | <p>Is there an efficient way to check a number x and remove all prime factors in the number which are less than some n? For example for n = 200:</p>
<pre><code>x=88984589931961415442566827779929187431222364934742868664124547963532933
FactorInteger[x]
{{29, 2}, {31, 1}, {37, 2}, {269, 1}, {271,
1}, {3420047160553... | user1066 | 106 | <p>Adapting the <a href="https://mathematica.stackexchange.com/a/214770/106">answer</a> given by <a href="https://mathematica.stackexchange.com/users/29734/march">@march</a>, where the second argument of <code>Flatten</code> is used to transpose a ragged array (see <a href="https://mathematica.stackexchange.com/a/126/1... |
564,360 | <p>Lets take the example, if we take the expression $\frac{X!}{y_1!\cdot y_2!\cdots y_n!} $as long as summation $S=y_1+y_2+...y_n$ is less than or equals $X$, the remainder is always $0$. Thats How the permutation of $X$ things where there is $y_1$ things same , $y_2$ things same works. My question is, why does this ha... | Meow | 39,568 | <p>Define $\nu_p(n)=k$, where $k$ is the power of $p$ in the prime factorisation of $n$. Evidently $\nu_p(n!)=\sum_{k \ge 1} \left \lfloor \frac{n}{p^k} \right \rfloor$ (this essentially counts the number of multiples of $p$ that are at most $n$, then double counts for multiples of $p^2$, then double counts again for m... |
1,626,362 | <p><code>The following is a short extract from the book I am reading:</code> </p>
<blockquote>
<p>If given a Homogeneous ODE:
$$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}+5\frac{\mathrm{d} y}{\mathrm{d}x}+4y=0\tag{1}$$
Letting
$$D=\frac{\mathrm{d}}{\mathrm{d}x}$$ then $(1)$ becomes
$$D^2 y + 5Dy + 4y=(D^2+5D+4)... | sinbadh | 277,566 | <p>Note that, in general, for each $a$ we have $D-a$ is a linear transformation in the space of differentiable functions. Then, if you want to solve $(D-a)y=0$, it is the same that $Dy=ay$. That is, $a$ is an eigenvalue.</p>
<p>Now, if $a\neq b$ and $Dy_1=ay_1$ and $Dy_2=ay_2$, we have that $y_1$ and $y_2$ are eigenve... |
2,746,153 | <p>Assume $m\ \mathrm{and}\ n\ \mathrm{are\ two\ relative\ prime\ positive\ integers.}$</p>
<p>Given $x \equiv a\ \pmod m$ and $x \equiv a\ \pmod n$.</p>
<p>Prove that $x \equiv a\ \pmod {mn}\ \mathrm{by\ using\ Chinese\ Remainder\ Theorem}.$<br/></p>
<p>And I did the following:
<br>
$$ \mathrm {M_1 = }\ n\ \ and\... | fleablood | 280,126 | <p>Well every number is equivalent to itself mod any modulus. </p>
<p>So $a\equiv a \mod mn$ and $a \equiv a \mod m$ and $a \equiv a \mod n$. So $x = a \mod mn$ is <em>one</em> solution.</p>
<p>But the Chinese remainder theorem claims that the solution is <em>unique</em> $\mod mn$.</p>
<p>So $x \equiv a \mod mn$ is... |
2,332,750 | <p>At the end of chapter 5 of stein's book <a href="http://wstein.org/books/ant/ant.pdf" rel="nofollow noreferrer">A Computational Introduction to Algebraic Number Theory</a> he proves proposition 5.2.4 which states that:</p>
<p>Given a prime ideal $\mathfrak{p}$ in a Dedekind domain $R$ we have the isomorphism
$$
\fr... | Mathmo123 | 154,802 | <p>It's used, for example, to show that the ideal norm is multiplicative. This is Prop 6.3.4 in Stein's notes.</p>
<p>For a number field $K$, for each non-zero ideal $\mathfrak a\subset\mathcal O_K$, we define it's norm to be $$N(\mathfrak a)=\#(\mathcal O_K/\mathfrak a).$$
It's clear from the Chinese remainder theor... |
1,177,782 | <p>I tried to prove the following theorem and was wondering if someone could please tell me if my proof can be fixed somehow...</p>
<p>Theorem: Let $H$ be a Hilbert space and $x_n\in H$ a bounded sequence. Then $x_n$ has a weakly convergent subsequence.</p>
<p>My idea for a proof:</p>
<p>The map $\phi: H \to H^\ast$... | BigMathTimes | 220,890 | <p>Unfortunately, I do not see an easy way to salvage this proof. Your question is essentially equivalent to the <a href="http://en.wikipedia.org/wiki/Banach%E2%80%93Alaoglu_theorem" rel="nofollow">Banach-Alaoglu Theorem</a>, which states that the unit ball is weakly compact in $H$. Unfortunately, the only proof I have... |
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