qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
1,038,060 | <p>Can anyone help me with this question: I know it before, but I have tried to solve it myself and didnt succeed.
what is the regular expression for this language: L=all words that have 00 or 11 but not both.</p>
<p>Thank you!</p>
| Community | -1 | <p>Your proof looks <strong>simple</strong> because you assumed not so <strong>simple</strong> result that $A_n $ is <strong>simple</strong> for $n\geq 5$...</p>
<p>Actually something more is true... </p>
<p>Suppose that $H\leq S_n$ of index $m $ with $m< n$ then we have homomorphism $\eta: S_n \rightarrow S_m$.</... |
1,290,316 | <p>Let $F_i$ be a family of closed sets, then we know that $\bigcup_{i=1}^nF_i$ is closed.</p>
<p>Proving that statement is equivalent to proving:</p>
<blockquote>
<p>If $p$ is a limit point of $\bigcup_{i=1}^nF_i$ then $p\in\bigcup_{i=1}^nF_i$</p>
</blockquote>
<p>It is easy to prove the contrapositive: if $p\not... | DanielWainfleet | 254,665 | <p>Use the infinite pigeon-hole principle: If aninfinite set is presented as the union of finitely many subsets,at least one subset is infinite.So if a sequence p(n) converges to p, where each p(n) belongs to some F(j), then,for at least one j, there are infinitely many n for which p(n) belongs to F(j), so p belongs t... |
4,465,504 | <p>Take an invertible formal series <span class="math-container">$f\in \mathbb{Z}_p[[T]]$</span> of inverse <span class="math-container">$g\in \mathbb{Z}_p[[T]]$</span> and let <span class="math-container">$x\in \mathbb{Z}_p$</span> such that the value of <span class="math-container">$f$</span> evaluated at <span class... | reuns | 276,986 | <p>You meant <strong>multiplicative</strong> inverse. Because the compositional inverse often exists in formal series.</p>
<p>If the series <span class="math-container">$f(x)$</span> converges then it does so to an element of <span class="math-container">$\Bbb{Z}_p$</span>, yes.</p>
<p>Then try <span class="math-contai... |
3,972,907 | <p>Reviewing Trig I come across this problem :
<span class="math-container">$\text{Solve for all real $x$ such that } 2\sqrt{2} \cos\left(\frac{x}{2}\right)=\cos(x) + 2.$</span></p>
<p>The first thing I did was use the cosine half-angle identity get this look...</p>
<p><span class="math-container">$$2\sqrt{2}\sqrt{\fra... | José Carlos Santos | 446,262 | <p>If you are trying to solve an equation of the type <span class="math-container">$f(x)=g(x)$</span>, it is perfectly fine to do <span class="math-container">$f^2(x)=g^2(x)$</span>. The solutions of the first equation will also be solutions of the second one. But there is a real possibility of creating new ones. An ex... |
3,972,907 | <p>Reviewing Trig I come across this problem :
<span class="math-container">$\text{Solve for all real $x$ such that } 2\sqrt{2} \cos\left(\frac{x}{2}\right)=\cos(x) + 2.$</span></p>
<p>The first thing I did was use the cosine half-angle identity get this look...</p>
<p><span class="math-container">$$2\sqrt{2}\sqrt{\fra... | Michael Hardy | 11,667 | <p><span class="math-container">$$2\sqrt{2} \cos\left(\frac{x}{2}\right)=\cos(x) + 2.$$</span></p>
<p>By the double-angle formula for the cosine, which says <span class="math-container">$\cos(2\theta)= 2\cos^2\theta-1,$</span> applied in the case where <span class="math-container">$\theta = x/2,$</span> we get:</p>
<p>... |
2,461,918 | <p>Often a function (real, say) is written without mentioning its domain, co-domain, just the rule $y=f(x)$ is given. In that case, how does one determine the domain, co-domain and range? For example, consider $f(x)=1/x$.</p>
| nonuser | 463,553 | <p>If you want to calculate domain it is usually problem with even root or with denominator. </p>
<p>Say you have $f:\mathbb{R}\to \mathbb{R}$ and </p>
<p>a) $f(x) = {x+3\over x-2}$, then $D_f = R\setminus \{2\}$ </p>
<p>b) $f(x) = \sqrt[4]{x-6}$, then $D_f = [6,\infty)$. </p>
<p>The hard part is range. For the fi... |
1,363,882 | <p>I am aware that the area under the curve of $\frac{1}{x}$ is infinite yet the area under the curve of $\frac{1}{x^2}$ is finite. </p>
<p>Calculus and series wise, I understand what is going on, but I can't seem to get a good geometric intuition of the problem.
Both curves can be shown to converge to $0$ (the curves... | ZenoCosini | 254,598 | <p>I do not think there is any geometric intuition behind the convergence you mentioned. As much as there is no geometric intuition which can help solving Zeno's paradox of the Tortoise and Achilles. </p>
<p>Indeed, the convergence is due to how the integral is defined - in terms of Riemann sums (here I assume you are... |
141,101 | <p>I have given a high dimensional input $x \in \mathbb{R}^m$ where $m$ is a big number. Linear regression can be applied, but in generel it is expected, that a lot of these dimensions are actually irrelevant.</p>
<p>I ought to find a method to model the function $y = f(x)$ and at the same time uncover which dimension... | shyamupa | 111,694 | <p>Have you tried using the shooting algorithm for optimizing the lasso regularized loss instead of gradient descent?</p>
<p>It involves using a coordinate wise search for the minimizer. This <a href="http://gautampendse.com/software/lasso/webpage/pendseLassoShooting.pdf" rel="nofollow">link</a> has some details and a... |
1,318,552 | <blockquote>
<p>Let $f$ be a twice differentiable function on $\left[0,1\right]$ satisfying $f\left(0\right)=f\left(1\right)=0$. Additionally $\left|f''\left(x\right)\right|\leq1$ in $\left(0,1\right)$. Prove that $$\left|f'\left(x\right)\right|\le\frac{1}{2},\quad\forall x\in\left[0,1\right]$$</p>
</blockquote>
<p>... | Ted Shifrin | 71,348 | <p>You were on the right track, but, as I suggested, the hint should have been to expand about the point $x_0$ where $|f'(x_0)|$ is a maximum.</p>
<p>Fix any $x_0\in [0,1]$ and, using Taylor's Theorem, write
$$f(x)=f(x_0) + f'(x_0)(x-x_0) + \frac12 f''(c)(x-x_0)^2\quad\text{for some $c$ between $x_0$ and $x$.}$$
Plug... |
1,905,308 | <p>In the book "A Course in Metric Geometry" (By Dmitri Burago, Yuri Burago Sergei Ivanov), there is a short proof of lower semicontinuity of length induced by a metric: (Prop 2.3.4, <a href="https://books.google.com/books?id=dRmIAwAAQBAJ&pg=PA35" rel="nofollow noreferrer">pg 35</a>)</p>
<p>Let $\gamma_j:[a,b] \to... | Asaf Shachar | 104,576 | <p>Yes, there is a mistake; See errata <a href="http://www.pdmi.ras.ru/~svivanov/papers/bbi-errata.pdf" rel="nofollow">here</a>.</p>
<p>The solution is to choose $j$ large enough, so that $d(\gamma(y_i),\gamma_j(y_i))< \frac{\epsilon}{N}$.</p>
|
159,438 | <p>Can be easily proved that the following series onverges/diverges?</p>
<p>$$\sum_{k=1}^{\infty} \frac{\tan(k)}{k}$$</p>
<p>I'd really appreciate your support on this problem. I'm looking for some easy proof here. Thanks.</p>
| Sangchul Lee | 9,340 | <p>Let $\mu$ be the <em><a href="http://mathworld.wolfram.com/IrrationalityMeasure.html">irrationality measure</a></em> of $\pi^{-1}$. Then for $s < \mu$ given, we have sequences $(p_n)$ and $(q_n)$ of integers such that $0 < q_n \uparrow \infty$ and </p>
<p>$$\left| \frac{1}{\pi} - \frac{2p_n + 1}{2q_n} \right|... |
765,738 | <p>I am trying to prove the following inequality:</p>
<p>$$(\sqrt{a} - \sqrt{b})^2 \leq \frac{1}{4}(a-b)(\ln(a)-\ln(b))$$</p>
<p>for all $a>0, b>0$.</p>
<p>Does anyone know how to prove it?</p>
<p>Thanks a lot in advance!</p>
| J. J. | 3,776 | <p>Since the inequality is homogeneous and invariant upon swapping the variables, we may assume that $b=1$ and $a \ge 1$. Then it remains to show that
$$f(a) = \frac{1}{4}(a-1)\log(a) - (\sqrt{a} - 1)^2 \ge 0.$$
Notice that $f(1) = 0$. Therefore we are done if we can show that $f$ is increasing.
Differentiating gives
$... |
870,174 | <p>If I have $6$ children and $4$ bedrooms, how many ways can I arrange the children if I want a maximum of $2$ kids per room?</p>
<p>The problem is that there are two empty slots, and these empty slots are not unique.</p>
<p>So, I assumed there are $8$ objects, $6$ kids and $2$ empties.</p>
<p>$$C_2^8 \cdot C_2^6 \... | awkward | 76,172 | <p>Here is an approach via exponential generating functions. More generally, let's say the number of ways to place $r$ children in the four rooms is $a_r$. Define $$f(x) = \sum_{r=0}^{\infty} \frac{a_r}{r!}x^r$$
In the problem where no room may remain empty, it is evident (after a little thought) that
$$f(x) = \left(... |
1,593,007 | <p>maybe this is a stupid question but I have the following expression:</p>
<p>$ 10^{-18}(e^{50,9702078⋅0,75}) = 10^{-18}(4⋅10^{16}) $</p>
<p>How would I go about simplifying the big exponent on the left to what's on the right? With the use of a calculator.</p>
<p>Thanks a lot! </p>
| Rory Daulton | 161,807 | <p><strong>SHORT ANSWER:</strong> In your first equation, $x=-1,\ y=0$ seems to be a solution. It would be, if $y'$ were also defined. However, if $y=0$ for any $x$ then $y'$ will be undefined there, so that is not actually a solution. The only restrictions on your second equation are $y=0$ and both $y$ and $y'$ are de... |
2,359,408 | <p>Question:</p>
<p>Let $\\f: \ \mathbb{R} \to \mathbb{R} \times \mathbb{R}$ via $ f(x) = (x+2, x-3)$. Is $f$ injective? Is $f$ surjective?</p>
<p>I was able to prove that $f$ is injective. However, I am not quite sure if $f$ is surjective. If it is surjective could someone please tell me how to prove that. If not, c... | Furrane | 373,901 | <p>I think you need to understand the intuition behind surjectivity :</p>
<p>A fonction is surjective if it at least fully "fills" the "end space" (if someone knows the real term feel free to comment) here $\mathbb{R} \times \mathbb{R}$. </p>
<p>But since it take $x$ values in $\mathbb{R}$, it won't be able to "fill"... |
1,910,085 | <p>For all integers $n \ge 0$, prove that the value $4^n + 1$ is not divisible by 3.</p>
<p>I need to use Proof by Induction to solve this problem. The base case is obviously 0, so I solved $4^0 + 1 = 2$. 2 is not divisible by 3.</p>
<p>I just need help proving the inductive step. I was trying to use proof by contrad... | MathIsNice1729 | 274,536 | <p>The statement is true for $n=0$. Now, let it be true for $n=k$. Also, if possible, let it be false for $n=k+1$. Then, $4^{k+1} \equiv -1 \pmod{3} \implies 4 \cdot 4^k \equiv -1 \pmod{3} \implies 4^k \equiv -4 \pmod{3} \equiv -1 \pmod{3}$ (since $4^{-1} \equiv 4 \pmod{3}$). So, $3 \mid 4^k+1$, a contradiction. Hence,... |
1,791,990 | <p>I have to prove that integral</p>
<p>$I = \int_{0}^{+\infty}\sin(t^2)dt$ is convergent. Could you tell me if it's ok?</p>
<p>Let $t^2=u$ then $dt=\frac{du}{2\sqrt{u}}$</p>
<p>Now $$I = \int_{0}^{+\infty}\frac{\sin(u)du}{2\sqrt{u}}$$</p>
<p>Which is equal to $$\int_{0}^{1}\frac{\sin(u)du}{2\sqrt{u}} + \int_{1}^{+... | Mark Viola | 218,419 | <p>To evaluate the integral, we analyze the closed-contour integral $I$ given by</p>
<p>$$I=\oint_C e^{iz^2}\,dz$$</p>
<p>where $C$ is comprised of (i) the line segment from $0$ to $R$, (ii) the circular arc from $R$ to $R(1+i)/\sqrt{2}$, and the line segment from $R(1+i)/\sqrt{2}$ to $0$. </p>
<p>Since $e^{iz^2}$ ... |
4,465,150 | <p>Let <span class="math-container">$A_1,A_2,…,A_n$</span> be events in a probability space <span class="math-container">$(\Omega,\Sigma,P)$</span>.</p>
<p>If <span class="math-container">$A_1,A_2,…,A_n$</span> are independent then <span class="math-container">$A_1^c,A_2^c,…,A_n^c$</span> are also independent, (where ... | angryavian | 43,949 | <p>Partial attempt:</p>
<p>Let <span class="math-container">$B_1 := \bigcap_{i=1}^k A_i^c$</span> and <span class="math-container">$B_2 := A_{k+1}^c$</span>.</p>
<p><span class="math-container">\begin{align}
P\left(\bigcap_{i=1}^{k+1} A_i^c\right)
&= P(B_1 \cap B_2)
\\
&= 1 - P(B_1^c \cup B_2^c)
\\
&= 1 - P... |
3,928,429 | <p>As titled, I was considering the mimimization problem where <span class="math-container">$y(x)$</span> has two endpoints fixed. That is, minimizing <span class="math-container">$$\int_a^b L(x,y(x),y'(x)) \, dx$$</span></p>
<p>where <span class="math-container">$$\ y(a)=m, y(b)=n $$</span>for all <span class="math-co... | Qmechanic | 11,127 | <ol>
<li><p>More generally, one may show that</p>
<ul>
<li>if Euler-Lagrange (EL) equations are always satisfied, and</li>
<li>if the <span class="math-container">$x$</span>- and <span class="math-container">$y$</span>-spaces are <a href="https://en.wikipedia.org/wiki/Contractible_space" rel="nofollow noreferrer">contr... |
4,173,308 | <p><a href="https://i.stack.imgur.com/G55Op.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/G55Op.png" alt="triangle of area 0.5 on a lattice grid" /></a></p>
<p>I'm trying to find the area of this triangle using the <span class="math-container">$\frac{1}{2} \times b \times h$</span> formula, but for... | DanielC | 129,267 | <p>It should be doable by a chain of substitutions, as soon as you complete the square:</p>
<p><span class="math-container">$$I=\int_0^z \sqrt{\frac{z^2}{4} -\left(x-\frac{z}{2}\right)^2} ~ dx $$</span></p>
<p>Then <span class="math-container">$ u = x-\frac{z}{2}$</span> and obviously <span class="math-container">$ dx ... |
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